├── figures ├── fx.pdf ├── fx2.pdf ├── gx.pdf ├── h2p.pdf ├── sp.pdf ├── psig.pdf ├── psiu.pdf ├── sync1.pdf ├── sync5.pdf ├── sync9.pdf ├── CO_Car.png ├── bondijsh.pdf ├── channel.pdf ├── fortrat.pdf ├── fortrat.png ├── sync01.pdf ├── sync05.pdf ├── sync09.pdf ├── sync095.pdf ├── sync99.pdf ├── syncct.pdf ├── syncct_1.pdf ├── syncct_5.pdf ├── syncct_9.pdf ├── syncfig.pdf ├── crab_icon.png ├── crab_scale.jpg ├── isothermal.pdf ├── syncct_0.1.pdf ├── syncct_95.pdf ├── syncct_99.pdf ├── week9cross.pdf ├── 0052_xray_lg.jpg ├── Crab_Nebula.jpg ├── Tumbler_Snapper_rope_tricks.jpg └── sync.tex ├── astrophysical_processes_notes.pdf ├── README.md ├── sync ├── psync_0.5 ├── psync_0.9 ├── psync_0.1 ├── psync_0.99 ├── lsyncg.dat └── lsyncf.dat ├── bremms ├── bremmst1.dat ├── bremmst5.dat ├── bremms1.dat ├── bremms0.dat └── bremms.dat ├── book_defs.tex ├── chapters ├── chap15.tex ├── app_math.tex ├── chap14.tex └── chap10.tex ├── morse ├── xegx.dat ├── xeux.dat └── xmorsex.dat ├── sedov ├── sedovc2.dat ├── sedovp.dat ├── sedovv.dat ├── sedovrho.dat ├── sedov.dat └── sedov_nd.dat ├── solutions ├── chap5sol.tex ├── chap6sol.tex ├── chap14sol.tex ├── chap9sol.tex ├── chap8sol.tex ├── chap7sol.tex ├── chap10sol.tex ├── chap3sol.tex ├── chap1sol.tex ├── chap12sol.tex ├── chap11sol.tex ├── chap13sol.tex ├── chap4sol.tex └── chap2sol.tex ├── astrophysical_processes_notes.tex ├── channel_dat ├── chan.res ├── chan_t.res ├── chan_m.res └── chan_y.res └── astrobookshelf.cls /figures/fx.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/fx.pdf -------------------------------------------------------------------------------- /figures/fx2.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/fx2.pdf -------------------------------------------------------------------------------- /figures/gx.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/gx.pdf -------------------------------------------------------------------------------- /figures/h2p.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/h2p.pdf -------------------------------------------------------------------------------- /figures/sp.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sp.pdf -------------------------------------------------------------------------------- /figures/psig.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/psig.pdf -------------------------------------------------------------------------------- /figures/psiu.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/psiu.pdf -------------------------------------------------------------------------------- /figures/sync1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync1.pdf -------------------------------------------------------------------------------- /figures/sync5.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync5.pdf -------------------------------------------------------------------------------- /figures/sync9.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync9.pdf -------------------------------------------------------------------------------- /figures/CO_Car.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/CO_Car.png -------------------------------------------------------------------------------- /figures/bondijsh.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/bondijsh.pdf -------------------------------------------------------------------------------- /figures/channel.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/channel.pdf -------------------------------------------------------------------------------- /figures/fortrat.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/fortrat.pdf -------------------------------------------------------------------------------- /figures/fortrat.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/fortrat.png -------------------------------------------------------------------------------- /figures/sync01.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync01.pdf -------------------------------------------------------------------------------- /figures/sync05.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync05.pdf -------------------------------------------------------------------------------- /figures/sync09.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync09.pdf -------------------------------------------------------------------------------- /figures/sync095.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync095.pdf -------------------------------------------------------------------------------- /figures/sync99.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/sync99.pdf -------------------------------------------------------------------------------- /figures/syncct.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncct.pdf -------------------------------------------------------------------------------- /figures/syncct_1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncct_1.pdf -------------------------------------------------------------------------------- /figures/syncct_5.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncct_5.pdf -------------------------------------------------------------------------------- /figures/syncct_9.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncct_9.pdf -------------------------------------------------------------------------------- /figures/syncfig.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncfig.pdf -------------------------------------------------------------------------------- /figures/crab_icon.png: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/crab_icon.png -------------------------------------------------------------------------------- /figures/crab_scale.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/crab_scale.jpg -------------------------------------------------------------------------------- /figures/isothermal.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/isothermal.pdf -------------------------------------------------------------------------------- /figures/syncct_0.1.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncct_0.1.pdf -------------------------------------------------------------------------------- /figures/syncct_95.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncct_95.pdf -------------------------------------------------------------------------------- /figures/syncct_99.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/syncct_99.pdf -------------------------------------------------------------------------------- /figures/week9cross.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/week9cross.pdf -------------------------------------------------------------------------------- /figures/0052_xray_lg.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/0052_xray_lg.jpg -------------------------------------------------------------------------------- /figures/Crab_Nebula.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/Crab_Nebula.jpg -------------------------------------------------------------------------------- /astrophysical_processes_notes.pdf: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/astrophysical_processes_notes.pdf -------------------------------------------------------------------------------- /figures/Tumbler_Snapper_rope_tricks.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/Open-Astrophysics-Bookshelf/astrophysical_processes_notes/HEAD/figures/Tumbler_Snapper_rope_tricks.jpg -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | # Astrophysical Processes Notes 2 | 3 | These are lecture notes developed for a course on high-energy astrophysics at UBC. You can access the PDF file directly at this [link](astrophysical_processes_notes.pdf). 4 | -------------------------------------------------------------------------------- /sync/psync_0.5: -------------------------------------------------------------------------------- 1 | 0 0.008276 2 | 1 0.06603 3 | 2 0.03184 4 | 3 0.01019 5 | 4 0.002621 6 | 5 0.0005752 7 | 6 0.0001104 8 | 7 1.862e-05 9 | 8 2.73e-06 10 | 9 3.373e-07 11 | 10 3.317e-08 12 | 11 2.196e-09 13 | 12 1.387e-10 14 | 13 7.358e-11 15 | 14 4.614e-11 16 | 15 2.492e-11 17 | 16 7.659e-12 18 | 17 9.55e-13 19 | 18 8.071e-13 20 | 19 9.062e-13 21 | 20 3.269e-13 22 | -------------------------------------------------------------------------------- /sync/psync_0.9: -------------------------------------------------------------------------------- 1 | 0 0.002663 2 | 1 0.007069 3 | 2 0.007516 4 | 3 0.006841 5 | 4 0.00564 6 | 5 0.004321 7 | 6 0.003114 8 | 7 0.002124 9 | 8 0.001376 10 | 9 0.0008487 11 | 10 0.0004993 12 | 11 0.0002809 13 | 12 0.0001518 14 | 13 7.931e-05 15 | 14 4.047e-05 16 | 15 2.046e-05 17 | 16 1.04e-05 18 | 17 5.371e-06 19 | 18 2.816e-06 20 | 19 1.477e-06 21 | 20 7.552e-07 22 | -------------------------------------------------------------------------------- /sync/psync_0.1: -------------------------------------------------------------------------------- 1 | 0 0.001415 2 | 1 0.2874 3 | 2 0.007139 4 | 3 0.0001001 5 | 4 1.047e-06 6 | 5 1.185e-08 7 | 6 7.376e-11 8 | 7 4.09e-11 9 | 8 3.062e-11 10 | 9 2.063e-11 11 | 10 1.501e-11 12 | 11 1.132e-11 13 | 12 8.885e-12 14 | 13 7.103e-12 15 | 14 5.833e-12 16 | 15 4.867e-12 17 | 16 4.083e-12 18 | 17 3.522e-12 19 | 18 3.035e-12 20 | 19 2.671e-12 21 | 20 2.355e-12 22 | -------------------------------------------------------------------------------- /sync/psync_0.99: -------------------------------------------------------------------------------- 1 | 0 0.0002807 2 | 1 0.0005771 3 | 2 0.0006027 4 | 3 0.0006299 5 | 4 0.0006544 6 | 5 0.0006736 7 | 6 0.0006857 8 | 7 0.0006896 9 | 8 0.0006849 10 | 9 0.0006717 11 | 10 0.0006507 12 | 11 0.000623 13 | 12 0.0005898 14 | 13 0.0005526 15 | 14 0.0005128 16 | 15 0.0004715 17 | 16 0.0004301 18 | 17 0.0003892 19 | 18 0.0003497 20 | 19 0.0003121 21 | 20 0.0002768 22 | -------------------------------------------------------------------------------- /bremms/bremmst1.dat: -------------------------------------------------------------------------------- 1 | -1.9 .5024721646e-1 2 | -1.8 .5961172061e-1 3 | -1.7 .7045924242e-1 4 | -1.6 .8293197857e-1 5 | -1.5 .9714811209e-1 6 | -1.4 .1131813611 7 | -1.3 .1310328052 8 | -1.2 .1505933679 9 | -1.1 .1715957462 10 | -1.0 .1935558180 11 | -.9 .2157062137 12 | -.8 .2369297247 13 | -.7 .2557085293 14 | -.6 .2701175698 15 | -.5 .2779058694 16 | -.4 .2767230961 17 | -.3 .2645473230 18 | -.2 .2403298167 19 | -.1 .2047642657 20 | 0. .1609022853 21 | .1 .1141445094 22 | .2 .7114397420e-1 23 | .3 .3764008894e-1 24 | .4 .1618293957e-1 25 | .5 .5351057332e-2 26 | .6 .1269491026e-2 27 | .7 .1979798908e-3 28 | .8 .1817883835e-4 29 | .9 .8555612174e-6 30 | 1.0 .1733456786e-7 -------------------------------------------------------------------------------- /bremms/bremmst5.dat: -------------------------------------------------------------------------------- 1 | -1.9 .5494936175e-1 2 | -1.8 .6552086064e-1 3 | -1.7 .7787958967e-1 4 | -1.6 .9224015163e-1 5 | -1.5 .1088072168 6 | -1.4 .1277554971 7 | -1.3 .1492000875 8 | -1.2 .1731539870 9 | -1.1 .1994692601 10 | -1.0 .2277586654 11 | -.9 .2572965503 12 | -.8 .2869029760 13 | -.7 .3148256530 14 | -.6 .3386530733 15 | -.5 .3553205748 16 | -.4 .3613049692 17 | -.3 .3531255340 18 | -.2 .3282400137 19 | -.1 .2862824376 20 | 0. .2302888114 21 | .1 .1671684213 22 | .2 .1065258424 23 | .3 .5754999153e-1 24 | .4 .2522632380e-1 25 | .5 .8489207572e-2 26 | .6 .2045811266e-2 27 | .7 .3234685348e-3 28 | .8 .3005679661e-4 29 | .9 .1429005913e-5 30 | 1.0 .2920186852e-7 31 | -------------------------------------------------------------------------------- /book_defs.tex: -------------------------------------------------------------------------------- 1 | \def\dddot#1{{\mathop{#1}\limits^{\vbox{ \hbox{\rm ...}}}}} 2 | \def\gtrsim{\mathop{\sim}\limits^{>}} 3 | \def\lesssim{{\mathop{\sim}\limits^{<}}} 4 | \newcommand{\be}{\begin{equation}} 5 | \newcommand{\ee}{\end{equation}} 6 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 7 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 8 | \newcommand{\comment}[1]{\relax} 9 | \newcommand{\dd}[2]{\frac{d {#1}}{d {#2}}} 10 | \newcommand{\pp}[2]{\frac{\partial {#1}}{\partial {#2}}} 11 | \newcommand{\betabold}{\mbox{\boldmath $\beta$}} 12 | \newcommand{\epbold}{\mbox{\boldmath $\epsilon$}} 13 | \newcommand{\lambdabare}{\lambda_e\!\!\!\!\!\!^-~} 14 | \newcommand{\ie}{{\em i.e.}} 15 | -------------------------------------------------------------------------------- /chapters/chap15.tex: -------------------------------------------------------------------------------- 1 | \chapter{Cosmic Rays} 2 | 3 | Up to this point the book has focussed either on the propagation and 4 | production of light that we ultimately detect, or the properties of 5 | matter and its motion that produce the light. The reader might get 6 | the mistaken impression that light is the only messenger from the 7 | matter in the cosmos. In fact the matter itself is often a messenger 8 | in the form of meteorites, interplanetary and interstellar dust. 9 | However, here the focus will be on the elementary particles and nuclei 10 | that travel to Earth at nearly the speed of light. 11 | 12 | \section{Cosmic Ray Production} 13 | 14 | \section{Cosmic Ray Propagation} 15 | 16 | 17 | %%% Local Variables: 18 | %%% TeX-master: "book" 19 | %%% End: -------------------------------------------------------------------------------- /morse/xegx.dat: -------------------------------------------------------------------------------- 1 | 1 0.2116 2 | 1.1 0.1473 3 | 1.2 0.09759 4 | 1.3 0.05889 5 | 1.4 0.02865 6 | 1.5 0.004986 7 | 1.6 -0.01351 8 | 1.7 -0.0279 9 | 1.8 -0.03901 10 | 1.9 -0.04746 11 | 2 -0.05377 12 | 2.1 -0.05834 13 | 2.2 -0.06149 14 | 2.3 -0.06349 15 | 2.4 -0.06454 16 | 2.5 -0.06483 17 | 2.6 -0.0645 18 | 2.7 -0.06366 19 | 2.8 -0.06243 20 | 2.9 -0.06088 21 | 3 -0.05908 22 | 3.1 -0.0571 23 | 3.2 -0.05498 24 | 3.3 -0.05276 25 | 3.4 -0.05048 26 | 3.5 -0.04817 27 | 3.6 -0.04586 28 | 3.7 -0.04355 29 | 3.8 -0.04128 30 | 3.9 -0.03904 31 | 4 -0.03687 32 | 4.1 -0.03475 33 | 4.2 -0.0327 34 | 4.3 -0.03073 35 | 4.4 -0.02884 36 | 4.5 -0.02703 37 | 4.6 -0.0253 38 | 4.7 -0.02365 39 | 4.8 -0.02209 40 | 4.9 -0.02061 41 | 5 -0.0192 42 | 5.1 -0.01788 43 | 5.2 -0.01663 44 | 5.3 -0.01546 45 | 5.4 -0.01435 46 | 5.5 -0.01332 47 | 5.6 -0.01235 48 | 5.7 -0.01144 49 | 5.8 -0.0106 50 | 5.9 -0.009805 51 | 6 -0.009068 52 | -------------------------------------------------------------------------------- /morse/xeux.dat: -------------------------------------------------------------------------------- 1 | 1 1.045 2 | 1.1 0.9179 3 | 1.2 0.8106 4 | 1.3 0.7191 5 | 1.4 0.6406 6 | 1.5 0.5726 7 | 1.6 0.5133 8 | 1.7 0.4613 9 | 1.8 0.4155 10 | 1.9 0.3751 11 | 2 0.3391 12 | 2.1 0.3072 13 | 2.2 0.2786 14 | 2.3 0.253 15 | 2.4 0.23 16 | 2.5 0.2094 17 | 2.6 0.1907 18 | 2.7 0.1739 19 | 2.8 0.1587 20 | 2.9 0.1449 21 | 3 0.1324 22 | 3.1 0.1211 23 | 3.2 0.1108 24 | 3.3 0.1014 25 | 3.4 0.09285 26 | 3.5 0.08505 27 | 3.6 0.07794 28 | 3.7 0.07144 29 | 3.8 0.06551 30 | 3.9 0.06008 31 | 4 0.05511 32 | 4.1 0.05057 33 | 4.2 0.04641 34 | 4.3 0.04259 35 | 4.4 0.0391 36 | 4.5 0.03589 37 | 4.6 0.03295 38 | 4.7 0.03026 39 | 4.8 0.02778 40 | 4.9 0.02551 41 | 5 0.02343 42 | 5.1 0.02152 43 | 5.2 0.01976 44 | 5.3 0.01814 45 | 5.4 0.01666 46 | 5.5 0.0153 47 | 5.6 0.01405 48 | 5.7 0.0129 49 | 5.8 0.01184 50 | 5.9 0.01087 51 | 6 0.009979 52 | -------------------------------------------------------------------------------- /morse/xmorsex.dat: -------------------------------------------------------------------------------- 1 | 0.9 0.2335 2 | 1 0.1766 3 | 1.1 0.1289 4 | 1.2 0.08922 5 | 1.3 0.0563 6 | 1.4 0.02915 7 | 1.5 0.006939 8 | 1.6 -0.01109 9 | 1.7 -0.02555 10 | 1.8 -0.03701 11 | 1.9 -0.04593 12 | 2 -0.05271 13 | 2.1 -0.05769 14 | 2.2 -0.06118 15 | 2.3 -0.06342 16 | 2.4 -0.06463 17 | 2.5 -0.065 18 | 2.6 -0.06468 19 | 2.7 -0.06381 20 | 2.8 -0.06249 21 | 2.9 -0.06083 22 | 3 -0.0589 23 | 3.1 -0.05677 24 | 3.2 -0.0545 25 | 3.3 -0.05213 26 | 3.4 -0.04971 27 | 3.5 -0.04726 28 | 3.6 -0.04482 29 | 3.7 -0.04239 30 | 3.8 -0.04001 31 | 3.9 -0.03768 32 | 4 -0.03542 33 | 4.1 -0.03323 34 | 4.2 -0.03111 35 | 4.3 -0.02908 36 | 4.4 -0.02713 37 | 4.5 -0.02527 38 | 4.6 -0.02349 39 | 4.7 -0.0218 40 | 4.8 -0.02019 41 | 4.9 -0.01866 42 | 5 -0.01721 43 | 5.1 -0.01585 44 | 5.2 -0.01455 45 | 5.3 -0.01333 46 | 5.4 -0.01218 47 | 5.5 -0.01109 48 | 5.6 -0.01007 49 | 5.7 -0.009111 50 | 5.8 -0.008207 51 | 5.9 -0.007357 52 | 6 -0.00656 53 | -------------------------------------------------------------------------------- /bremms/bremms1.dat: -------------------------------------------------------------------------------- 1 | -4.9 .9998470652 2 | -4.8 .9998165909 3 | -4.7 .9997801194 4 | -4.6 .9997364889 5 | -4.5 .9996843164 6 | -4.4 .9996219584 7 | -4.3 .9995474622 8 | -4.2 .9994585095 9 | -4.1 .9993523506 10 | -4.0 .9992257260 11 | -3.9 .9990747771 12 | -3.8 .9988949394 13 | -3.7 .9986808191 14 | -3.6 .9984260498 15 | -3.5 .9981231254 16 | -3.4 .9977632081 17 | -3.3 .9973359063 18 | -3.2 .9968290178 19 | -3.1 .9962282378 20 | -3.0 .9955168242 21 | -2.9 .9946752182 22 | -2.8 .9936806147 23 | -2.7 .9925064781 24 | -2.6 .9911220082 25 | -2.5 .9894915367 26 | -2.4 .9875738756 27 | -2.3 .9853216067 28 | -2.2 .9826803126 29 | -2.1 .9795877751 30 | -2.0 .9759731394 31 | -1.9 .9717560751 32 | -1.8 .9668459610 33 | -1.7 .9611411363 34 | -1.6 .9545282726 35 | -1.5 .9468819346 36 | -1.4 .9380644206 37 | -1.3 .9279259866 38 | -1.2 .9163055834 39 | -1.1 .9030322540 40 | -1.0 .8879273606 41 | -.9 .8708078243 42 | -.8 .8514905606 43 | -.7 .8297982964 44 | -.6 .8055669184 45 | -.5 .7786544523 46 | -.4 .7489516756 47 | -.3 .7163942439 48 | -.2 .6809760126 49 | -.1 .6427630104 50 | 0. .6019072302 51 | .1 .5586590911 52 | .2 .5133771010 53 | .3 .4665329809 54 | .4 .4187103518 55 | .5 .3705951228 56 | .6 .3229560594 57 | .7 .2766147103 58 | .8 .2324050146 59 | .9 .1911244190 60 | 1.0 .1534801287 61 | 1.1 .1200358597 62 | 1.2 .9116580859e-1 63 | 1.3 .6702297985e-1 64 | 1.4 .4752812193e-1 65 | 1.5 .3238305925e-1 66 | 1.6 .2110836262e-1 67 | 1.7 .1310072660e-1 68 | 1.8 .7701232468e-2 69 | 1.9 .4263158389e-2 70 | 2.0 .2208152121e-2 71 | -------------------------------------------------------------------------------- /bremms/bremms0.dat: -------------------------------------------------------------------------------- 1 | -4.9 .3735217174e-1 2 | -4.8 .4045769726e-1 3 | -4.7 .4380332562e-1 4 | -4.6 .4740522029e-1 5 | -4.5 .5128029205e-1 6 | -4.4 .5544617815e-1 7 | -4.3 .5992120972e-1 8 | -4.2 .6472436401e-1 9 | -4.1 .6987520003e-1 10 | -4.0 .7539377385e-1 11 | -3.9 .8130053073e-1 12 | -3.8 .8761617009e-1 13 | -3.7 .9436147959e-1 14 | -3.6 .1015571333 15 | -3.5 .1092234490 16 | -3.4 .1173800995 17 | -3.3 .1260457712 18 | -3.2 .1352377634 19 | -3.1 .1449715230 20 | -3.0 .1552601040 21 | -2.9 .1661135484 22 | -2.8 .1775381767 23 | -2.7 .1895357849 24 | -2.6 .2021027376 25 | -2.5 .2152289534 26 | -2.4 .2288967795 27 | -2.3 .2430797518 28 | -2.2 .2577412446 29 | -2.1 .2728330158 30 | -2.0 .2882936629 31 | -1.9 .3040470087 32 | -1.8 .3200004519 33 | -1.7 .3360433267 34 | -1.6 .3520453332 35 | -1.5 .3678551204 36 | -1.4 .3832991225 37 | -1.3 .3981807791 38 | -1.2 .4122802937 39 | -1.1 .4253551131 40 | -1.0 .4371413456 41 | -.9 .4473563462 42 | -.8 .4557027292 43 | -.7 .4618740556 44 | -.6 .4655624413 45 | -.5 .4664682659 46 | -.4 .4643120922 47 | -.3 .4588487575 48 | -.2 .4498834102 49 | -.1 .4372890030 50 | 0. .4210244382 51 | .1 .4011521867 52 | .2 .3778538190 53 | .3 .3514415247 54 | .4 .3223634421 55 | .5 .2912005497 56 | .6 .2586531178 57 | .7 .2255153478 58 | .8 .1926379348 59 | .9 .1608798542 60 | 1.0 .1310525785 61 | 1.1 .1038619163 62 | 1.2 .7985430649e-1 63 | 1.3 .5937520636e-1 64 | 1.4 .4254667151e-1 65 | 1.5 .2926905892e-1 66 | 1.6 .1924806270e-1 67 | 1.7 .1204363415e-1 68 | 1.8 .7132828439e-2 69 | 1.9 .3975600376e-2 70 | 2.0 .2072144148e-2 71 | -------------------------------------------------------------------------------- /sedov/sedovc2.dat: -------------------------------------------------------------------------------- 1 | 0.0114572 28405.9 2 | 0.0229144 5971.61 3 | 0.0343716 2398.21 4 | 0.0458287 1255.38 5 | 0.0572859 759.848 6 | 0.0687431 504.16 7 | 0.0802003 356.401 8 | 0.0916575 263.911 9 | 0.103115 202.471 10 | 0.114572 159.738 11 | 0.126029 128.907 12 | 0.137486 105.987 13 | 0.148943 88.5191 14 | 0.160401 74.924 15 | 0.171858 64.1511 16 | 0.183315 55.4803 17 | 0.194772 48.406 18 | 0.206229 42.5644 19 | 0.217687 37.6889 20 | 0.229144 33.5809 21 | 0.240601 30.0896 22 | 0.252058 27.0993 23 | 0.263515 24.5201 24 | 0.274972 22.2809 25 | 0.28643 20.3256 26 | 0.297887 18.6088 27 | 0.309344 17.0939 28 | 0.320801 15.7509 29 | 0.332258 14.5551 30 | 0.343716 13.4861 31 | 0.355173 12.527 32 | 0.36663 11.6633 33 | 0.378087 10.8831 34 | 0.389544 10.1761 35 | 0.401001 9.53359 36 | 0.412459 8.94805 37 | 0.423916 8.41503 38 | 0.435373 7.92528 39 | 0.44683 7.47575 40 | 0.458287 7.06221 41 | 0.469745 6.68096 42 | 0.481202 6.32886 43 | 0.492659 6.00303 44 | 0.504116 5.70097 45 | 0.515573 5.42047 46 | 0.527031 5.1596 47 | 0.538488 4.91657 48 | 0.549945 4.68986 49 | 0.561402 4.47806 50 | 0.572859 4.27992 51 | 0.584316 4.09434 52 | 0.595774 3.92029 53 | 0.607231 3.75688 54 | 0.618688 3.60328 55 | 0.630145 3.45874 56 | 0.641602 3.32259 57 | 0.65306 3.19423 58 | 0.664517 3.07309 59 | 0.675974 2.95866 60 | 0.687431 2.85049 61 | 0.698888 2.74814 62 | 0.710345 2.65123 63 | 0.721803 2.5594 64 | 0.73326 2.47233 65 | 0.744717 2.38972 66 | 0.756174 2.31129 67 | 0.767631 2.2368 68 | 0.779089 2.16598 69 | 0.790546 2.09864 70 | 0.802003 2.03457 71 | 0.81346 1.97358 72 | 0.824917 1.91551 73 | 0.836374 1.86017 74 | 0.847832 1.80743 75 | 0.859289 1.75713 76 | 0.870746 1.70914 77 | 0.882203 1.66334 78 | 0.89366 1.61958 79 | 0.905118 1.57776 80 | 0.916575 1.53776 81 | 0.928032 1.49946 82 | 0.939489 1.46275 83 | 0.950946 1.42753 84 | 0.962404 1.39369 85 | 0.973861 1.36111 86 | 0.985318 1.3297 87 | 0.996775 1.29936 88 | 89 | -------------------------------------------------------------------------------- /sedov/sedovp.dat: -------------------------------------------------------------------------------- 1 | 0.0114572 0.306195 2 | 0.0229144 0.306195 3 | 0.0343716 0.306195 4 | 0.0458287 0.306195 5 | 0.0572859 0.306195 6 | 0.0687431 0.306195 7 | 0.0802003 0.306195 8 | 0.0916575 0.306195 9 | 0.103115 0.306195 10 | 0.114572 0.306195 11 | 0.126029 0.306195 12 | 0.137486 0.306195 13 | 0.148943 0.306195 14 | 0.160401 0.306195 15 | 0.171858 0.306195 16 | 0.183315 0.306195 17 | 0.194772 0.306195 18 | 0.206229 0.306195 19 | 0.217687 0.306195 20 | 0.229144 0.306195 21 | 0.240601 0.306195 22 | 0.252058 0.306195 23 | 0.263515 0.306195 24 | 0.274972 0.306195 25 | 0.28643 0.306195 26 | 0.297887 0.306195 27 | 0.309344 0.306195 28 | 0.320801 0.306195 29 | 0.332258 0.306195 30 | 0.343716 0.306195 31 | 0.355173 0.306195 32 | 0.36663 0.306195 33 | 0.378087 0.306195 34 | 0.389544 0.306195 35 | 0.401001 0.306195 36 | 0.412459 0.306195 37 | 0.423916 0.307411 38 | 0.435373 0.307641 39 | 0.44683 0.307909 40 | 0.458287 0.308217 41 | 0.469745 0.30857 42 | 0.481202 0.308976 43 | 0.492659 0.309439 44 | 0.504116 0.309966 45 | 0.515573 0.310565 46 | 0.527031 0.311244 47 | 0.538488 0.312012 48 | 0.549945 0.312879 49 | 0.561402 0.313854 50 | 0.572859 0.314951 51 | 0.584316 0.316182 52 | 0.595774 0.31756 53 | 0.607231 0.319103 54 | 0.618688 0.320827 55 | 0.630145 0.32275 56 | 0.641602 0.324895 57 | 0.65306 0.327285 58 | 0.664517 0.329944 59 | 0.675974 0.332903 60 | 0.687431 0.336193 61 | 0.698888 0.33985 62 | 0.710345 0.343915 63 | 0.721803 0.348431 64 | 0.73326 0.353451 65 | 0.744717 0.359032 66 | 0.756174 0.365238 67 | 0.767631 0.372143 68 | 0.779089 0.37983 69 | 0.790546 0.388396 70 | 0.802003 0.39795 71 | 0.81346 0.408618 72 | 0.824917 0.420544 73 | 0.836374 0.433895 74 | 0.847832 0.448864 75 | 0.859289 0.465675 76 | 0.870746 0.484589 77 | 0.882203 0.50591 78 | 0.89366 0.529995 79 | 0.905118 0.557265 80 | 0.916575 0.588212 81 | 0.928032 0.623423 82 | 0.939489 0.663594 83 | 0.950946 0.709558 84 | 0.962404 0.762314 85 | 0.973861 0.823068 86 | 0.985318 0.893284 87 | 0.996775 0.97475 88 | -------------------------------------------------------------------------------- /sedov/sedovv.dat: -------------------------------------------------------------------------------- 1 | 0.0114572 0.00916575 2 | 0.0229144 0.0183315 3 | 0.0343716 0.0274972 4 | 0.0458287 0.036663 5 | 0.0572859 0.0458287 6 | 0.0687431 0.0549945 7 | 0.0802003 0.0641602 8 | 0.0916575 0.073326 9 | 0.103115 0.0824917 10 | 0.114572 0.0916575 11 | 0.126029 0.100823 12 | 0.137486 0.109989 13 | 0.148943 0.119155 14 | 0.160401 0.12832 15 | 0.171858 0.137486 16 | 0.183315 0.146652 17 | 0.194772 0.155818 18 | 0.206229 0.164983 19 | 0.217687 0.174149 20 | 0.229144 0.183315 21 | 0.240601 0.192481 22 | 0.252058 0.201646 23 | 0.263515 0.210812 24 | 0.274972 0.219978 25 | 0.28643 0.229144 26 | 0.297887 0.238309 27 | 0.309344 0.247475 28 | 0.320801 0.256641 29 | 0.332258 0.265807 30 | 0.343716 0.274972 31 | 0.355173 0.284138 32 | 0.36663 0.293304 33 | 0.378087 0.30247 34 | 0.389544 0.311635 35 | 0.401001 0.320801 36 | 0.412459 0.329967 37 | 0.423916 0.3395 38 | 0.435373 0.348747 39 | 0.44683 0.358009 40 | 0.458287 0.367289 41 | 0.469745 0.376589 42 | 0.481202 0.385912 43 | 0.492659 0.395261 44 | 0.504116 0.404641 45 | 0.515573 0.414054 46 | 0.527031 0.423505 47 | 0.538488 0.433001 48 | 0.549945 0.442544 49 | 0.561402 0.452143 50 | 0.572859 0.461803 51 | 0.584316 0.471532 52 | 0.595774 0.481338 53 | 0.607231 0.491228 54 | 0.618688 0.501214 55 | 0.630145 0.511304 56 | 0.641602 0.521509 57 | 0.65306 0.531843 58 | 0.664517 0.542317 59 | 0.675974 0.552946 60 | 0.687431 0.563745 61 | 0.698888 0.57473 62 | 0.710345 0.585918 63 | 0.721803 0.597327 64 | 0.73326 0.608978 65 | 0.744717 0.62089 66 | 0.756174 0.633085 67 | 0.767631 0.645587 68 | 0.779089 0.658419 69 | 0.790546 0.671605 70 | 0.802003 0.685171 71 | 0.81346 0.699142 72 | 0.824917 0.713545 73 | 0.836374 0.728403 74 | 0.847832 0.743744 75 | 0.859289 0.759589 76 | 0.870746 0.775961 77 | 0.882203 0.792879 78 | 0.89366 0.810359 79 | 0.905118 0.828416 80 | 0.916575 0.847057 81 | 0.928032 0.866286 82 | 0.939489 0.886103 83 | 0.950946 0.9065 84 | 0.962404 0.927465 85 | 0.973861 0.948981 86 | 0.985318 0.971025 87 | 0.996775 0.993568 88 | -------------------------------------------------------------------------------- /bremms/bremms.dat: -------------------------------------------------------------------------------- 1 | -2.9 .1661135484 .9946752182 2 | -2.8 .1775381767 .9936806147 3 | -2.7 .1895357849 .9925064781 4 | -2.6 .2021027376 .9911220082 5 | -2.5 .2152289534 .9894915367 6 | -2.4 .2288967795 .9875738756 7 | -2.3 .2430797518 .9853216067 8 | -2.2 .2577412446 .9826803126 9 | -2.1 .2728330158 .9795877751 10 | -2.0 .2882936629 .9759731394 11 | -1.9 .3040470087 .9717560751 12 | -1.8 .3200004519 .9668459610 13 | -1.7 .3360433267 .9611411363 14 | -1.6 .3520453332 .9545282726 15 | -1.5 .3678551204 .9468819346 16 | -1.4 .3832991225 .9380644206 17 | -1.3 .3981807791 .9279259866 18 | -1.2 .4122802937 .9163055834 19 | -1.1 .4253551131 .9030322540 20 | -1.0 .4371413456 .8879273606 21 | -.9 .4473563462 .8708078243 22 | -.8 .4557027292 .8514905606 23 | -.7 .4618740556 .8297982964 24 | -.6 .4655624413 .8055669184 25 | -.5 .4664682659 .7786544523 26 | -.4 .4643120922 .7489516756 27 | -.3 .4588487575 .7163942439 28 | -.2 .4498834102 .6809760126 29 | -.1 .4372890030 .6427630104 30 | 0. .4210244382 .6019072302 31 | .1 .4011521867 .5586590911 32 | .2 .3778538190 .5133771010 33 | .3 .3514415247 .4665329809 34 | .4 .3223634421 .4187103518 35 | .5 .2912005497 .3705951228 36 | .6 .2586531178 .3229560594 37 | .7 .2255153478 .2766147103 38 | .8 .1926379348 .2324050146 39 | .9 .1608798542 .1911244190 40 | 1.0 .1310525785 .1534801287 41 | 1.1 .1038619163 .1200358597 42 | 1.2 .7985430649e-1 .9116580859e-1 43 | 1.3 .5937520636e-1 .6702297985e-1 44 | 1.4 .4254667151e-1 .4752812193e-1 45 | 1.5 .2926905892e-1 .3238305925e-1 46 | 1.6 .1924806270e-1 .2110836262e-1 47 | 1.7 .1204363415e-1 .1310072660e-1 48 | 1.8 .7132828439e-2 .7701232468e-2 49 | 1.9 .3975600376e-2 .4263158389e-2 50 | 2.0 .2072144148e-2 .2208152121e-2 51 | -------------------------------------------------------------------------------- /sedov/sedovrho.dat: -------------------------------------------------------------------------------- 1 | 0.0114572 6.32453e-10 2 | 0.0229144 1.43108e-08 3 | 0.0343716 8.87307e-08 4 | 0.0458287 3.23816e-07 5 | 0.0572859 8.8388e-07 6 | 0.0687431 2.00775e-06 7 | 0.0802003 4.01762e-06 8 | 0.0916575 7.32712e-06 9 | 0.103115 1.24486e-05 10 | 0.114572 1.99999e-05 11 | 0.126029 3.07111e-05 12 | 0.137486 4.54301e-05 13 | 0.148943 6.51288e-05 14 | 0.160401 9.09085e-05 15 | 0.171858 0.000124005 16 | 0.183315 0.000165794 17 | 0.194772 0.000217795 18 | 0.206229 0.000281679 19 | 0.217687 0.000359269 20 | 0.229144 0.000452546 21 | 0.240601 0.000563657 22 | 0.252058 0.000694913 23 | 0.263515 0.000848796 24 | 0.274972 0.00102797 25 | 0.28643 0.00123526 26 | 0.297887 0.0014737 27 | 0.309344 0.00174649 28 | 0.320801 0.00205702 29 | 0.332258 0.0024089 30 | 0.343716 0.00280591 31 | 0.355173 0.00325204 32 | 0.36663 0.00375148 33 | 0.378087 0.00430865 34 | 0.389544 0.00492814 35 | 0.401001 0.0056148 36 | 0.412459 0.00637367 37 | 0.423916 0.00723531 38 | 0.435373 0.00816327 39 | 0.44683 0.00918251 40 | 0.458287 0.0102997 41 | 0.469745 0.0115219 42 | 0.481202 0.0128565 43 | 0.492659 0.0143114 44 | 0.504116 0.0158952 45 | 0.515573 0.0176168 46 | 0.527031 0.0194858 47 | 0.538488 0.0215127 48 | 0.549945 0.0237086 49 | 0.561402 0.0260854 50 | 0.572859 0.0286563 51 | 0.584316 0.0314353 52 | 0.595774 0.034438 53 | 0.607231 0.0376812 54 | 0.618688 0.0411836 55 | 0.630145 0.0449655 56 | 0.641602 0.0490498 57 | 0.65306 0.0534616 58 | 0.664517 0.058229 59 | 0.675974 0.0633835 60 | 0.687431 0.0689604 61 | 0.698888 0.0749997 62 | 0.710345 0.0815465 63 | 0.721803 0.088652 64 | 0.73326 0.0963748 65 | 0.744717 0.104782 66 | 0.756174 0.11395 67 | 0.767631 0.123967 68 | 0.779089 0.134936 69 | 0.790546 0.146977 70 | 0.802003 0.160226 71 | 0.81346 0.174846 72 | 0.824917 0.191026 73 | 0.836374 0.208991 74 | 0.847832 0.229003 75 | 0.859289 0.251376 76 | 0.870746 0.276481 77 | 0.882203 0.304763 78 | 0.89366 0.336757 79 | 0.905118 0.373103 80 | 0.916575 0.414579 81 | 0.928032 0.462129 82 | 0.939489 0.516904 83 | 0.950946 0.580318 84 | 0.962404 0.654113 85 | 0.973861 0.740451 86 | 0.985318 0.842031 87 | 0.996775 0.962242 88 | -------------------------------------------------------------------------------- /solutions/chap5sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \usepackage{graphicx} 4 | \newcommand{\be}{\begin{equation}} 5 | \newcommand{\ee}{\end{equation}} 6 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 7 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 8 | \newcommand{\comment}[1]{\relax} 9 | \begin{document} 10 | \fi 11 | \section{Chapter 5} 12 | \begin{enumerate} 13 | \item{\bf Bremsstrahlung:} 14 | 15 | Consider a sphere of ionized hydrogen plamsa that is undergoing 16 | spherical gravitational collapse. The sphere is held at uniform 17 | temperature, $T_0$, uniform density and constant mass $M_0$ during the 18 | collapse and has decreasing radius $R_0$. The sphere cools by 19 | emission of bremsstrahlung radiation in its interior. At $t=t_0$ the 20 | sphere is optically thin. 21 | \begin{enumerate} 22 | \item What is the total luminosity of the sphere as a function of 23 | $M_0, R(t)$ and $T_0$ while the sphere is optically thin? 24 | \item 25 | What is the luminosity of the sphere as a function of time after it 26 | becomes optically thick in terms of $M_0, R(t)$ and $T_0$? 27 | \item 28 | Give an implicit relation in terms of $R(t)$ for the time $t_1$ when 29 | the sphere becomes optically thick. 30 | \item 31 | Draw a curve of the luminosity as a function of time. 32 | \end{enumerate} 33 | 34 | {\bf Answer:} 35 | 36 | \begin{enumerate} 37 | \item 38 | $L = \epsilon^{ff} \frac{4}{3} \pi R^3 = \frac{2^5 \pi 39 | e^6}{3 h m c^3} \left ( \frac{2\pi k T}{3m} \right )^{1/2} 40 | \left (\frac{M}{m_p \frac{4}{3} \pi R^3} \right)^2 {\bar g}_{B} \frac{4}{3} \pi R^3 41 | $, so $L \propto R^{-3}$. 42 | \item 43 | $L=\sigma_{SB} T^4 4 \pi R^2$ 44 | \item 45 | $ 46 | \sigma_{SB} T^4 4 \pi R^2 = \frac{2^5 \pi 47 | e^6}{3 h m c^3} \left ( \frac{2\pi k T}{3m} \right )^{1/2} 48 | \left (\frac{M}{m_p}\right)^2 \frac{3}{4 \pi R^3} {\bar g}_{B} 49 | $ 50 | \item 51 | Draw your graph with luminosity increasing with time as $R(t)^{-3}$ 52 | and then decreasing after a certain time as $R(t)^2$. 53 | \end{enumerate} 54 | \end{enumerate} 55 | \ifx\bookloaded\undefined 56 | \end{document} 57 | \fi 58 | %%% Local Variables: 59 | %%% TeX-master: "book" 60 | %%% End: 61 | -------------------------------------------------------------------------------- /solutions/chap6sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \usepackage{graphicx} 4 | \input book_defs 5 | \begin{document} 6 | \fi 7 | \section{Chapter 6} 8 | 9 | \begin{enumerate} 10 | \item{\bf Synchrotron Radiation:} 11 | 12 | An ultrarelativistic electron emits synchrotron radiation. Show that 13 | its energy decreases with time according to 14 | \begin{equation} 15 | \gamma = \gamma_0 \left ( 1 + A \gamma_0 t \right )^{-1}, A=\frac{2e^4 16 | B_\perp^2}{3m^3 c^5}. 17 | \label{eq:395} 18 | \end{equation} 19 | Here $\gamma_0$ is the initial value of $\gamma$ and $B_\perp = B 20 | \sin\alpha$. Show that the time for the electron to lose half its 21 | energy is 22 | \begin{equation} 23 | t_{1/2} = \left (A\gamma_0\right)^{-1} 24 | \label{eq:396} 25 | \end{equation} 26 | How do you reconcile the decrease of $\gamma$ with the result of 27 | constant $\gamma$ for motion in a magnetic field? 28 | 29 | {\bf Answer:} 30 | 31 | $$ 32 | P = -\dd{E}{t} = -m_e c^2 \dd{\gamma}{t} = \frac{2}{3} r_0^2 c \beta^2 33 | \gamma^2 B_\perp^2 34 | $$ 35 | so 36 | $$ 37 | \dd{\gamma}{t} = -\frac{2}{3} \frac{e^4}{m_e^3 c^5} B_\perp^2 \gamma^2 38 | $$ 39 | where we have taken $\beta\approx 1$. If the Lorentz factor 40 | $\gamma=\gamma_0$ at $t=0$, integrating this yields 41 | $$ 42 | \frac{1}{\gamma} - \frac{1}{\gamma_0} = \frac{2}{3} \frac{e^4}{m_e^3 43 | c^5} B_\perp^2 t, 44 | $$ 45 | and rearranging yields the answers above. 46 | 47 | \item{\bf Synchrotron Cooling More Precisely:} 48 | 49 | Derive the evolution of the energy of the electron (or $\gamma$) 50 | evolves in time without making the ultrarelativistic approximation. 51 | 52 | {\bf Answer:} 53 | 54 | Let's start with 55 | $$ 56 | \dd{\gamma}{t} = -\frac{2}{3} \frac{e^4}{m_e^3 c^5} B_\perp^2 \beta^2 57 | \gamma^2 = -A (\gamma^2 -1 ) 58 | $$ 59 | so 60 | $$ 61 | -A dt = \frac{1}{2} \left [ \frac{d\gamma}{\gamma-1} - 62 | \frac{d\gamma}{\gamma+1} \right ] 63 | $$ 64 | and the answer upon integrating is 65 | $$ 66 | \gamma = \coth \left ( \coth^{-1} \gamma_0 + A t \right ) . 67 | $$ 68 | 69 | \comment{\item{\bf Power-Law Distribution More Precisely:} 70 | 71 | Calculate the photon spectrum for a power-law distribution of electron 72 | energies as in \S~\ref{sec:spectral-index-power} including the 73 | normalization and polarization. } 74 | \end{enumerate} 75 | 76 | \ifx\bookloaded\undefined 77 | \end{document} 78 | \fi 79 | %%% Local Variables: 80 | %%% TeX-master: "book" 81 | %%% End: 82 | -------------------------------------------------------------------------------- /astrophysical_processes_notes.tex: -------------------------------------------------------------------------------- 1 | \documentclass{astrobookshelf} 2 | \makeindex 3 | \def\bookloaded{\true} 4 | \usepackage{graphicx} 5 | \graphicspath{{figures/}} 6 | %\usepackage[paperwidth=7in,paperheight=10in,bottom=2cm,top=2.5cm]{geometry} 7 | \usepackage{makeidx} 8 | \usepackage{bm} 9 | %\usepackage[colorlinks=true]{hyperref} 10 | %\usepackage{times} %charter font 11 | \usepackage[OT1, T1]{fontenc} %use TeX encoding then Type 1 12 | \usepackage{tikz} 13 | \usetikzlibrary{snakes} 14 | \usetikzlibrary{arrows} 15 | \usetikzlibrary{shapes} 16 | \usetikzlibrary{backgrounds} 17 | 18 | % \ifx\pdftexversion\undefined 19 | % \usepackage[dvips]{graphicx} 20 | % \else 21 | % \usepackage[pdftex]{graphicx} 22 | % \fi 23 | \input book_defs 24 | %\includeonly{solutions/chap2sol} 25 | %\includeonly{chap11} 26 | %\includeonly{chap11,chap12,chap12sol} 27 | %\includeonly{chap4,chap4sol} 28 | %\includeonly{chap7} 29 | %\includeonly{chap8} 30 | %\includeonly{chap11,chap11sol} 31 | %\includeonly{preface,intro,chap1,chap2,chap3,chap4,chap5,chap6,chap7,chap8,chap9,chap10,chap11,chap12,chap13,chap14,app_math} 32 | \pagenumbering{roman} 33 | \title{Astrophysical\\\noindent Processes} 34 | \author{Jeremy Heyl} 35 | \begin{document} 36 | 37 | \maketitle 38 | \vspace*{3in} 39 | \begin{center} 40 | Copyright \copyright\ 2018 \\ 41 | Jeremy Heyl 42 | \end{center} 43 | \tableofcontents 44 | 45 | %\listoffigures 46 | 47 | %\listoftables 48 | \newpage 49 | \include{preface} 50 | \include{intro} 51 | \pagenumbering{arabic} 52 | \include{chapters/chap1} 53 | \include{chapters/chap2} 54 | \include{chapters/chap3} 55 | \include{chapters/chap4} 56 | \include{chapters/chap5} 57 | \include{chapters/chap6} 58 | \include{chapters/chap7} 59 | \include{chapters/chap8} 60 | \include{chapters/chap9} 61 | \include{chapters/chap10} 62 | \include{chapters/chap11} 63 | \include{chapters/chap12} 64 | \include{chapters/chap13} 65 | \include{chapters/chap14} 66 | %\include{chapters/chap15} 67 | 68 | \appendix 69 | 70 | \include{chapters/app_math} 71 | \chapter{Selected Solutions} 72 | 73 | \include{solutions/chap1sol} 74 | \include{solutions/chap2sol} 75 | \include{solutions/chap3sol} 76 | \include{solutions/chap4sol} 77 | \include{solutions/chap5sol} 78 | \include{solutions/chap6sol} 79 | \include{solutions/chap7sol} 80 | \include{solutions/chap8sol} 81 | \include{solutions/chap9sol} 82 | \include{solutions/chap10sol} 83 | \include{solutions/chap11sol} 84 | \include{solutions/chap12sol} 85 | \include{solutions/chap13sol} 86 | \include{solutions/chap14sol} 87 | 88 | \printindex 89 | 90 | \end{document} 91 | \end 92 | 93 | %%% Local Variables: 94 | %%% mode: latex 95 | %%% TeX-master: book 96 | %%% End: 97 | -------------------------------------------------------------------------------- /figures/sync.tex: -------------------------------------------------------------------------------- 1 | % Sketch output, version 0.3 (build 2d, Fri Jun 3 09:55:31 2011) 2 | % Output language: PGF/TikZ,LaTeX 3 | \begin{tikzpicture}[line join=round] 4 | \tikzstyle{vector}=[-latex',very thick] 5 | \tikzstyle{angle}=[->,shorten >=1pt]\filldraw[draw=gray,fill=white](5.333,-1.778)--(5.496,-1.72)--(5.647,-1.659)--(5.786,-1.595)--(5.913,-1.529)--(6.027,-1.46)--(6.128,-1.388)--(6.215,-1.315)--(6.288,-1.24)--(6.347,-1.163)--(6.391,-1.086)--(6.421,-1.007)--(6.436,-.928)--(6.436,-.849)--(6.421,-.771)--(6.391,-.692)--(6.347,-.615)--(6.288,-.538)--(6.215,-.463)--(6.128,-.39)--(6.027,-.318)--(5.913,-.249)--(5.786,-.182)--(5.647,-.118)--(5.496,-.058)--(5.333,0)--(5.161,.054)--(4.978,.104)--(4.786,.151)--(4.586,.193)--(4.379,.231)--(4.164,.265)--(3.944,.294)--(3.719,.318)--(3.489,.338)--(3.257,.353)--(3.022,.363)--(2.785,.368)--(2.548,.368)--(2.312,.363)--(2.077,.353)--(1.844,.338)--(1.615,.318)--(1.389,.294)--(1.169,.265)--(.955,.231)--(.747,.193)--(.547,.151)--(.355,.104)--(.173,.054)--(0,0)--(-.162,-.058)--(-.313,-.118)--(-.452,-.182)--(-.579,-.249)--(-.694,-.318)--(-.794,-.39)--(-.882,-.463)--(-.955,-.538)--(-1.014,-.615)--(-1.058,-.692)--(-1.088,-.771)--(-1.103,-.849)--(-1.103,-.928)--(-1.088,-1.007)--(-1.058,-1.086)--(-1.014,-1.163)--(-.955,-1.24)--(-.882,-1.315)--(-.794,-1.388)--(-.694,-1.46)--(-.579,-1.529)--(-.452,-1.595)--(-.313,-1.659)--(-.162,-1.72)--(0,-1.778)--(.173,-1.832)--(.355,-1.882)--(.547,-1.929)--(.747,-1.971)--(.955,-2.009)--(1.169,-2.043)--(1.389,-2.072)--(1.615,-2.096)--(1.844,-2.116)--(2.077,-2.13)--(2.312,-2.14)--(2.548,-2.145)--(2.785,-2.145)--(3.022,-2.14)--(3.257,-2.13)--(3.489,-2.116)--(3.719,-2.096)--(3.944,-2.072)--(4.164,-2.043)--(4.379,-2.009)--(4.586,-1.971)--(4.786,-1.929)--(4.978,-1.882)--(5.161,-1.832)--cycle; 6 | \draw[vector](0,0)--(1.071,2.059); 7 | \draw(-.082,1.225)--(.429,.824); 8 | \draw[vector](0,0)--(-3.333,-1.111); 9 | \draw[vector](0,0)--(1.667,-.556); 10 | \draw[vector](0,0)--(4,-1.333); 11 | \draw[vector](0,0)--(-2.043,1.605); 12 | \draw[vector](0,0)--(0,4.444); 13 | \draw(-.511,.401)--(-.082,1.225); 14 | \draw(2.667,-.889)--(-1.047,-1.107); 15 | \draw[vector](-1.047,-1.107)--(-.556,-2.036); 16 | \draw[angle] (-2,-.667)..controls (-1.879,.286).. node[midway,left] {$\theta$}(-1.532,1.203);\draw[angle] (1.333,-.444)..controls (.982,-.607).. node[midway,left] {$vt'/a$}(.81,-.998);\path (-3.333,-1.111) node[below] {$\bf{x}$} 17 | (4,-1.333) node[below] {$\bf{y}$} 18 | (0,4.444) node[above] {$\bf{z}$} 19 | (-2.043,1.605) node[above] {$\bf{n}$} 20 | (1.667,-.556) node[above] {$\epbold_\perp$} 21 | (1.071,2.059) node[above] {$\epbold_\parallel$} 22 | (-.556,-2.036) node[below] {$\bf{v}$} 23 | (.845,-1.052) node[below] {$a$};\end{tikzpicture}% End sketch output 24 | -------------------------------------------------------------------------------- /channel_dat/chan.res: -------------------------------------------------------------------------------- 1 | -0.99 1.00023 2 | -0.98 1.00023 3 | -0.97 1.00023 4 | -0.96 1.00023 5 | -0.95 1.00023 6 | -0.94 1.00023 7 | -0.93 1.00023 8 | -0.92 1.00023 9 | -0.91 1.00023 10 | -0.9 1.00023 11 | -0.89 1.00022 12 | -0.88 1.00022 13 | -0.87 1.00022 14 | -0.86 1.00022 15 | -0.85 1.00022 16 | -0.84 1.00022 17 | -0.83 1.00021 18 | -0.82 1.00021 19 | -0.81 1.00021 20 | -0.8 1.0002 21 | -0.79 1.0002 22 | -0.78 1.00019 23 | -0.77 1.00019 24 | -0.76 1.00018 25 | -0.75 1.00017 26 | -0.74 1.00016 27 | -0.73 1.00015 28 | -0.72 1.00014 29 | -0.71 1.00013 30 | -0.7 1.00011 31 | -0.69 1.00009 32 | -0.68 1.00007 33 | -0.67 1.00005 34 | -0.66 1.00002 35 | -0.65 0.999989 36 | -0.64 0.999955 37 | -0.63 0.999917 38 | -0.62 0.999875 39 | -0.61 0.999827 40 | -0.6 0.999774 41 | -0.59 0.999715 42 | -0.58 0.99965 43 | -0.57 0.999577 44 | -0.56 0.999497 45 | -0.55 0.999408 46 | -0.54 0.999311 47 | -0.53 0.999204 48 | -0.52 0.999086 49 | -0.51 0.998458 50 | -0.5 0.998317 51 | -0.49 0.998165 52 | -0.48 0.997999 53 | -0.47 0.997819 54 | -0.46 0.997624 55 | -0.45 0.997413 56 | -0.44 0.997186 57 | -0.43 0.996942 58 | -0.42 0.99668 59 | -0.41 0.996399 60 | -0.4 0.9961 61 | -0.39 0.995781 62 | -0.38 0.995441 63 | -0.37 0.99508 64 | -0.36 0.994697 65 | -0.35 0.994294 66 | -0.34 0.993867 67 | -0.33 0.993421 68 | -0.32 0.99295 69 | -0.31 0.992458 70 | -0.3 0.991946 71 | -0.29 0.991411 72 | -0.28 0.990857 73 | -0.27 0.990282 74 | -0.26 0.989689 75 | -0.25 0.98908 76 | -0.24 0.988452 77 | -0.23 0.987813 78 | -0.22 0.987154 79 | -0.21 0.986472 80 | -0.2 0.98578 81 | -0.19 0.985084 82 | -0.18 0.984384 83 | -0.17 0.983684 84 | -0.16 0.982991 85 | -0.15 0.982304 86 | -0.14 0.981628 87 | -0.13 0.980969 88 | -0.12 0.980329 89 | -0.11 0.979713 90 | -0.1 0.979128 91 | -0.09 0.978575 92 | -0.08 0.978058 93 | -0.07 0.977583 94 | -0.06 0.977154 95 | -0.05 0.976775 96 | -0.04 0.976447 97 | -0.03 0.976175 98 | -0.02 0.975961 99 | -0.01 0.975805 100 | 0 0.975712 101 | 0.01 0.97568 102 | 0.02 0.975712 103 | 0.03 0.975805 104 | 0.04 0.975961 105 | 0.05 0.976175 106 | 0.06 0.976447 107 | 0.07 0.976775 108 | 0.08 0.977154 109 | 0.09 0.977583 110 | 0.1 0.978058 111 | 0.11 0.978575 112 | 0.12 0.979128 113 | 0.13 0.979713 114 | 0.14 0.980329 115 | 0.15 0.980969 116 | 0.16 0.981628 117 | 0.17 0.982304 118 | 0.18 0.982991 119 | 0.19 0.983684 120 | 0.2 0.984384 121 | 0.21 0.985084 122 | 0.22 0.98578 123 | 0.23 0.986472 124 | 0.24 0.987154 125 | 0.25 0.987813 126 | 0.26 0.988452 127 | 0.27 0.98908 128 | 0.28 0.989689 129 | 0.29 0.990282 130 | 0.3 0.990857 131 | 0.31 0.991411 132 | 0.32 0.991946 133 | 0.33 0.992458 134 | 0.34 0.99295 135 | 0.35 0.993421 136 | 0.36 0.993867 137 | 0.37 0.994294 138 | 0.38 0.994697 139 | 0.39 0.99508 140 | 0.4 0.995441 141 | 0.41 0.995781 142 | 0.42 0.9961 143 | 0.43 0.996399 144 | 0.44 0.99668 145 | 0.45 0.996942 146 | 0.46 0.997186 147 | 0.47 0.997413 148 | 0.48 0.997624 149 | 0.49 0.997819 150 | 0.5 0.997999 151 | 0.51 0.998165 152 | 0.52 0.998317 153 | 0.53 0.998458 154 | 0.54 0.999086 155 | 0.55 0.999204 156 | 0.56 0.999311 157 | 0.57 0.999408 158 | 0.58 0.999497 159 | 0.59 0.999577 160 | 0.6 0.99965 161 | 0.61 0.999715 162 | 0.62 0.999774 163 | 0.63 0.999827 164 | 0.64 0.999875 165 | 0.65 0.999917 166 | 0.66 0.999955 167 | 0.67 0.999989 168 | 0.68 1.00002 169 | 0.69 1.00005 170 | 0.7 1.00007 171 | 0.71 1.00009 172 | 0.72 1.00011 173 | 0.73 1.00013 174 | 0.74 1.00014 175 | 0.75 1.00015 176 | 0.76 1.00016 177 | 0.77 1.00017 178 | 0.78 1.00018 179 | 0.79 1.00019 180 | 0.8 1.00019 181 | 0.81 1.0002 182 | 0.82 1.0002 183 | 0.83 1.00021 184 | 0.84 1.00021 185 | 0.85 1.00021 186 | 0.86 1.00022 187 | 0.87 1.00022 188 | 0.88 1.00022 189 | 0.89 1.00022 190 | 0.9 1.00022 191 | 0.91 1.00022 192 | -------------------------------------------------------------------------------- /channel_dat/chan_t.res: -------------------------------------------------------------------------------- 1 | -0.99 0.999997 2 | -0.98 0.999996 3 | -0.97 0.999996 4 | -0.96 0.999995 5 | -0.95 0.999994 6 | -0.94 0.999992 7 | -0.93 0.999991 8 | -0.92 0.999989 9 | -0.91 0.999987 10 | -0.9 0.999984 11 | -0.89 0.999981 12 | -0.88 0.999977 13 | -0.87 0.999973 14 | -0.86 0.999967 15 | -0.85 0.999961 16 | -0.84 0.999954 17 | -0.83 0.999945 18 | -0.82 0.999935 19 | -0.81 0.999924 20 | -0.8 0.99991 21 | -0.79 0.999895 22 | -0.78 0.999876 23 | -0.77 0.999855 24 | -0.76 0.999831 25 | -0.75 0.999803 26 | -0.74 0.999771 27 | -0.73 0.999734 28 | -0.72 0.999692 29 | -0.71 0.999644 30 | -0.7 0.999589 31 | -0.69 0.999527 32 | -0.68 0.999456 33 | -0.67 0.999376 34 | -0.66 0.999285 35 | -0.65 0.999183 36 | -0.64 0.999067 37 | -0.63 0.998938 38 | -0.62 0.998794 39 | -0.61 0.998631 40 | -0.6 0.998451 41 | -0.59 0.998249 42 | -0.58 0.998025 43 | -0.57 0.997776 44 | -0.56 0.997502 45 | -0.55 0.997197 46 | -0.54 0.996861 47 | -0.53 0.996492 48 | -0.52 0.996085 49 | -0.51 0.99564 50 | -0.5 0.995152 51 | -0.49 0.994619 52 | -0.48 0.994038 53 | -0.47 0.993404 54 | -0.46 0.992717 55 | -0.45 0.99197 56 | -0.44 0.991159 57 | -0.43 0.990283 58 | -0.42 0.989339 59 | -0.41 0.988317 60 | -0.4 0.987222 61 | -0.39 0.986044 62 | -0.38 0.984779 63 | -0.37 0.983423 64 | -0.36 0.981972 65 | -0.35 0.980429 66 | -0.34 0.978772 67 | -0.33 0.977019 68 | -0.32 0.97514 69 | -0.31 0.973152 70 | -0.3 0.971048 71 | -0.29 0.96881 72 | -0.28 0.966446 73 | -0.27 0.963948 74 | -0.26 0.961313 75 | -0.25 0.958541 76 | -0.24 0.955607 77 | -0.23 0.952542 78 | -0.22 0.949308 79 | -0.21 0.945926 80 | -0.2 0.942378 81 | -0.19 0.938673 82 | -0.18 0.934792 83 | -0.17 0.930744 84 | -0.16 0.926541 85 | -0.15 0.922162 86 | -0.14 0.917599 87 | -0.13 0.912866 88 | -0.12 0.907959 89 | -0.11 0.902874 90 | -0.1 0.897633 91 | -0.09 0.892211 92 | -0.08 0.886603 93 | -0.07 0.880846 94 | -0.06 0.874911 95 | -0.05 0.868857 96 | -0.04 0.862604 97 | -0.03 0.856232 98 | -0.02 0.849723 99 | -0.01 0.843025 100 | 0 0.836219 101 | 0.01 0.829493 102 | 0.02 0.822268 103 | 0.03 0.815129 104 | 0.04 0.807875 105 | 0.05 0.8006 106 | 0.06 0.793251 107 | 0.07 0.785808 108 | 0.08 0.778377 109 | 0.09 0.770865 110 | 0.1 0.763354 111 | 0.11 0.755814 112 | 0.12 0.748304 113 | 0.13 0.740832 114 | 0.14 0.733372 115 | 0.15 0.725969 116 | 0.16 0.718629 117 | 0.17 0.71136 118 | 0.18 0.704197 119 | 0.19 0.697144 120 | 0.2 0.690181 121 | 0.21 0.683341 122 | 0.22 0.676655 123 | 0.23 0.670105 124 | 0.24 0.66372 125 | 0.25 0.657486 126 | 0.26 0.651453 127 | 0.27 0.645564 128 | 0.28 0.639892 129 | 0.29 0.634403 130 | 0.3 0.629106 131 | 0.31 0.624011 132 | 0.32 0.619107 133 | 0.33 0.614427 134 | 0.34 0.609941 135 | 0.35 0.605641 136 | 0.36 0.601579 137 | 0.37 0.59769 138 | 0.38 0.594025 139 | 0.39 0.590542 140 | 0.4 0.587253 141 | 0.41 0.584155 142 | 0.42 0.581243 143 | 0.43 0.578515 144 | 0.44 0.575947 145 | 0.45 0.573557 146 | 0.46 0.571323 147 | 0.47 0.569243 148 | 0.48 0.567316 149 | 0.49 0.565531 150 | 0.5 0.56388 151 | 0.51 0.562356 152 | 0.52 0.560952 153 | 0.53 0.559664 154 | 0.54 0.558483 155 | 0.55 0.557401 156 | 0.56 0.556415 157 | 0.57 0.555517 158 | 0.58 0.554699 159 | 0.59 0.55396 160 | 0.6 0.55329 161 | 0.61 0.552684 162 | 0.62 0.55214 163 | 0.63 0.55165 164 | 0.64 0.55121 165 | 0.65 0.550817 166 | 0.66 0.550467 167 | 0.67 0.550153 168 | 0.68 0.549874 169 | 0.69 0.549627 170 | 0.7 0.549409 171 | 0.71 0.549216 172 | 0.72 0.549045 173 | 0.73 0.548895 174 | 0.74 0.548764 175 | 0.75 0.548648 176 | 0.76 0.548548 177 | 0.77 0.54846 178 | 0.78 0.548383 179 | 0.79 0.548317 180 | 0.8 0.548259 181 | 0.81 0.54821 182 | 0.82 0.548167 183 | 0.83 0.54813 184 | 0.84 0.548098 185 | 0.85 0.548071 186 | 0.86 0.548047 187 | 0.87 0.548028 188 | 0.88 0.548011 189 | 0.89 0.547996 190 | 0.9 0.547984 191 | 0.91 0.547974 192 | -------------------------------------------------------------------------------- /channel_dat/chan_m.res: -------------------------------------------------------------------------------- 1 | -0.99 0.999997 2 | -0.98 0.999996 3 | -0.97 0.999996 4 | -0.96 0.999995 5 | -0.95 0.999994 6 | -0.94 0.999992 7 | -0.93 0.999991 8 | -0.92 0.999989 9 | -0.91 0.999987 10 | -0.9 0.999984 11 | -0.89 0.999981 12 | -0.88 0.999977 13 | -0.87 0.999973 14 | -0.86 0.999967 15 | -0.85 0.999961 16 | -0.84 0.999954 17 | -0.83 0.999945 18 | -0.82 0.999935 19 | -0.81 0.999924 20 | -0.8 0.99991 21 | -0.79 0.999895 22 | -0.78 0.999876 23 | -0.77 0.999855 24 | -0.76 0.999831 25 | -0.75 0.999803 26 | -0.74 0.999771 27 | -0.73 0.999734 28 | -0.72 0.999692 29 | -0.71 0.999644 30 | -0.7 0.999589 31 | -0.69 0.999527 32 | -0.68 0.999456 33 | -0.67 0.999376 34 | -0.66 0.999285 35 | -0.65 0.999183 36 | -0.64 0.999067 37 | -0.63 0.998938 38 | -0.62 0.998794 39 | -0.61 0.998631 40 | -0.6 0.998451 41 | -0.59 0.998249 42 | -0.58 0.998025 43 | -0.57 0.997776 44 | -0.56 0.997502 45 | -0.55 0.997197 46 | -0.54 0.996861 47 | -0.53 0.996492 48 | -0.52 0.996085 49 | -0.51 0.99564 50 | -0.5 0.995152 51 | -0.49 0.994619 52 | -0.48 0.994038 53 | -0.47 0.993404 54 | -0.46 0.992717 55 | -0.45 0.99197 56 | -0.44 0.991159 57 | -0.43 0.990283 58 | -0.42 0.989339 59 | -0.41 0.988317 60 | -0.4 0.987222 61 | -0.39 0.986044 62 | -0.38 0.984779 63 | -0.37 0.983423 64 | -0.36 0.981972 65 | -0.35 0.980429 66 | -0.34 0.978772 67 | -0.33 0.977019 68 | -0.32 0.97514 69 | -0.31 0.973152 70 | -0.3 0.971048 71 | -0.29 0.96881 72 | -0.28 0.966446 73 | -0.27 0.963948 74 | -0.26 0.961313 75 | -0.25 0.958541 76 | -0.24 0.955607 77 | -0.23 0.952542 78 | -0.22 0.949308 79 | -0.21 0.945926 80 | -0.2 0.942378 81 | -0.19 0.938673 82 | -0.18 0.934792 83 | -0.17 0.930744 84 | -0.16 0.926541 85 | -0.15 0.922162 86 | -0.14 0.917599 87 | -0.13 0.912866 88 | -0.12 0.907959 89 | -0.11 0.902874 90 | -0.1 0.897633 91 | -0.09 0.892211 92 | -0.08 0.886603 93 | -0.07 0.880846 94 | -0.06 0.874911 95 | -0.05 0.868857 96 | -0.04 0.862604 97 | -0.03 0.856232 98 | -0.02 0.849723 99 | -0.01 0.843025 100 | 0 0.836219 101 | 0.01 0.829493 102 | 0.02 0.836219 103 | 0.03 0.843025 104 | 0.04 0.849723 105 | 0.05 0.856232 106 | 0.06 0.862604 107 | 0.07 0.868857 108 | 0.08 0.874911 109 | 0.09 0.880846 110 | 0.1 0.886603 111 | 0.11 0.892211 112 | 0.12 0.897633 113 | 0.13 0.902874 114 | 0.14 0.907959 115 | 0.15 0.912866 116 | 0.16 0.917599 117 | 0.17 0.922162 118 | 0.18 0.926541 119 | 0.19 0.930744 120 | 0.2 0.934792 121 | 0.21 0.938673 122 | 0.22 0.942378 123 | 0.23 0.945926 124 | 0.24 0.949308 125 | 0.25 0.952542 126 | 0.26 0.955607 127 | 0.27 0.958541 128 | 0.28 0.961313 129 | 0.29 0.963948 130 | 0.3 0.966446 131 | 0.31 0.96881 132 | 0.32 0.971048 133 | 0.33 0.973152 134 | 0.34 0.97514 135 | 0.35 0.977019 136 | 0.36 0.978772 137 | 0.37 0.980429 138 | 0.38 0.981972 139 | 0.39 0.983423 140 | 0.4 0.984779 141 | 0.41 0.986044 142 | 0.42 0.987222 143 | 0.43 0.988317 144 | 0.44 0.989339 145 | 0.45 0.990283 146 | 0.46 0.991159 147 | 0.47 0.99197 148 | 0.48 0.992717 149 | 0.49 0.993404 150 | 0.5 0.994038 151 | 0.51 0.994619 152 | 0.52 0.995152 153 | 0.53 0.99564 154 | 0.54 0.996085 155 | 0.55 0.996492 156 | 0.56 0.996861 157 | 0.57 0.997197 158 | 0.58 0.997502 159 | 0.59 0.997776 160 | 0.6 0.998025 161 | 0.61 0.998249 162 | 0.62 0.998451 163 | 0.63 0.998631 164 | 0.64 0.998794 165 | 0.65 0.998938 166 | 0.66 0.999067 167 | 0.67 0.999183 168 | 0.68 0.999285 169 | 0.69 0.999376 170 | 0.7 0.999456 171 | 0.71 0.999527 172 | 0.72 0.999589 173 | 0.73 0.999644 174 | 0.74 0.999692 175 | 0.75 0.999734 176 | 0.76 0.999771 177 | 0.77 0.999803 178 | 0.78 0.999831 179 | 0.79 0.999855 180 | 0.8 0.999876 181 | 0.81 0.999895 182 | 0.82 0.99991 183 | 0.83 0.999924 184 | 0.84 0.999935 185 | 0.85 0.999945 186 | 0.86 0.999954 187 | 0.87 0.999961 188 | 0.88 0.999967 189 | 0.89 0.999973 190 | 0.9 0.999977 191 | 0.91 0.999981 192 | -------------------------------------------------------------------------------- /sync/lsyncg.dat: -------------------------------------------------------------------------------- 1 | -3 -3.06463 2 | -2.98 -3.04929 3 | -2.96 -3.03395 4 | -2.94 -3.01861 5 | -2.92 -3.00326 6 | -2.9 -2.98792 7 | -2.88 -2.97258 8 | -2.86 -2.95724 9 | -2.84 -2.9419 10 | -2.82 -2.92656 11 | -2.8 -2.91123 12 | -2.78 -2.89589 13 | -2.76 -2.88055 14 | -2.74 -2.86522 15 | -2.72 -2.84988 16 | -2.7 -2.83455 17 | -2.68 -2.81922 18 | -2.66 -2.80389 19 | -2.64 -2.78856 20 | -2.62 -2.77323 21 | -2.6 -2.7579 22 | -2.58 -2.74258 23 | -2.56 -2.72725 24 | -2.54 -2.71193 25 | -2.52 -2.69661 26 | -2.5 -2.68129 27 | -2.48 -2.66597 28 | -2.46 -2.65066 29 | -2.44 -2.63535 30 | -2.42 -2.62004 31 | -2.4 -2.60473 32 | -2.38 -2.58943 33 | -2.36 -2.57412 34 | -2.34 -2.55883 35 | -2.32 -2.54353 36 | -2.3 -2.52824 37 | -2.28 -2.51295 38 | -2.26 -2.49767 39 | -2.24 -2.48239 40 | -2.22 -2.46711 41 | -2.2 -2.45184 42 | -2.18 -2.43657 43 | -2.16 -2.42131 44 | -2.14 -2.40605 45 | -2.12 -2.39081 46 | -2.1 -2.37556 47 | -2.08 -2.36032 48 | -2.06 -2.3451 49 | -2.04 -2.32987 50 | -2.02 -2.31466 51 | -2 -2.29945 52 | -1.98 -2.28426 53 | -1.96 -2.26907 54 | -1.94 -2.25389 55 | -1.92 -2.23873 56 | -1.9 -2.22357 57 | -1.88 -2.20843 58 | -1.86 -2.1933 59 | -1.84 -2.17818 60 | -1.82 -2.16308 61 | -1.8 -2.14799 62 | -1.78 -2.13292 63 | -1.76 -2.11787 64 | -1.74 -2.10283 65 | -1.72 -2.08782 66 | -1.7 -2.07282 67 | -1.68 -2.05785 68 | -1.66 -2.0429 69 | -1.64 -2.02797 70 | -1.62 -2.01307 71 | -1.6 -1.9982 72 | -1.58 -1.98336 73 | -1.56 -1.96855 74 | -1.54 -1.95377 75 | -1.52 -1.93902 76 | -1.5 -1.92431 77 | -1.48 -1.90964 78 | -1.46 -1.89502 79 | -1.44 -1.88043 80 | -1.42 -1.86589 81 | -1.4 -1.85141 82 | -1.38 -1.83697 83 | -1.36 -1.82259 84 | -1.34 -1.80826 85 | -1.32 -1.794 86 | -1.3 -1.7798 87 | -1.28 -1.76568 88 | -1.26 -1.75162 89 | -1.24 -1.73764 90 | -1.22 -1.72375 91 | -1.2 -1.70994 92 | -1.18 -1.69622 93 | -1.16 -1.6826 94 | -1.14 -1.66908 95 | -1.12 -1.65567 96 | -1.1 -1.64237 97 | -1.08 -1.6292 98 | -1.06 -1.61615 99 | -1.04 -1.60324 100 | -1.02 -1.59047 101 | -1 -1.57785 102 | -0.98 -1.56539 103 | -0.96 -1.5531 104 | -0.94 -1.54098 105 | -0.92 -1.52906 106 | -0.9 -1.51733 107 | -0.88 -1.50581 108 | -0.86 -1.49451 109 | -0.84 -1.48344 110 | -0.82 -1.47262 111 | -0.8 -1.46206 112 | -0.78 -1.45178 113 | -0.76 -1.44178 114 | -0.74 -1.43208 115 | -0.72 -1.4227 116 | -0.7 -1.41367 117 | -0.68 -1.40499 118 | -0.66 -1.39668 119 | -0.64 -1.38877 120 | -0.62 -1.38128 121 | -0.6 -1.37422 122 | -0.58 -1.36763 123 | -0.56 -1.36152 124 | -0.54 -1.35593 125 | -0.52 -1.35088 126 | -0.5 -1.34639 127 | -0.48 -1.3425 128 | -0.46 -1.33925 129 | -0.44 -1.33665 130 | -0.42 -1.33475 131 | -0.4 -1.33358 132 | -0.38 -1.33318 133 | -0.36 -1.33359 134 | -0.34 -1.33486 135 | -0.32 -1.33701 136 | -0.3 -1.34011 137 | -0.28 -1.3442 138 | -0.26 -1.34932 139 | -0.24 -1.35554 140 | -0.22 -1.36291 141 | -0.2 -1.37148 142 | -0.18 -1.38131 143 | -0.16 -1.39247 144 | -0.14 -1.40503 145 | -0.12 -1.41905 146 | -0.1 -1.43462 147 | -0.08 -1.45179 148 | -0.06 -1.47066 149 | -0.04 -1.49131 150 | -0.02 -1.51382 151 | 2.39392e-15 -1.53829 152 | 0.02 -1.56482 153 | 0.04 -1.5935 154 | 0.06 -1.62444 155 | 0.08 -1.65775 156 | 0.1 -1.69355 157 | 0.12 -1.73195 158 | 0.14 -1.77309 159 | 0.16 -1.8171 160 | 0.18 -1.86411 161 | 0.2 -1.91428 162 | 0.22 -1.96775 163 | 0.24 -2.02469 164 | 0.26 -2.08525 165 | 0.28 -2.14962 166 | 0.3 -2.21798 167 | 0.32 -2.29053 168 | 0.34 -2.36745 169 | 0.36 -2.44896 170 | 0.38 -2.53529 171 | 0.4 -2.62665 172 | 0.42 -2.7233 173 | 0.44 -2.82548 174 | 0.46 -2.93346 175 | 0.48 -3.04752 176 | 0.5 -3.16794 177 | 0.52 -3.29503 178 | 0.54 -3.4291 179 | 0.56 -3.57049 180 | 0.58 -3.71954 181 | 0.6 -3.87662 182 | 0.62 -4.04212 183 | 0.64 -4.21642 184 | 0.66 -4.39995 185 | 0.68 -4.59314 186 | 0.7 -4.79645 187 | 0.72 -5.01037 188 | 0.74 -5.23538 189 | 0.76 -5.47203 190 | 0.78 -5.72086 191 | 0.8 -5.98244 192 | 0.82 -6.25738 193 | 0.84 -6.54631 194 | 0.86 -6.84989 195 | -------------------------------------------------------------------------------- /sync/lsyncf.dat: -------------------------------------------------------------------------------- 1 | -3 -2.37984 2 | -2.98 -2.36476 3 | -2.96 -2.34968 4 | -2.94 -2.33461 5 | -2.92 -2.31955 6 | -2.9 -2.3045 7 | -2.88 -2.28946 8 | -2.86 -2.27443 9 | -2.84 -2.25941 10 | -2.82 -2.2444 11 | -2.8 -2.2294 12 | -2.78 -2.21441 13 | -2.76 -2.19943 14 | -2.74 -2.18447 15 | -2.72 -2.16951 16 | -2.7 -2.15457 17 | -2.68 -2.13965 18 | -2.66 -2.12473 19 | -2.64 -2.10983 20 | -2.62 -2.09494 21 | -2.6 -2.08007 22 | -2.58 -2.06522 23 | -2.56 -2.05037 24 | -2.54 -2.03555 25 | -2.52 -2.02074 26 | -2.5 -2.00595 27 | -2.48 -1.99118 28 | -2.46 -1.97642 29 | -2.44 -1.96169 30 | -2.42 -1.94697 31 | -2.4 -1.93227 32 | -2.38 -1.9176 33 | -2.36 -1.90294 34 | -2.34 -1.88831 35 | -2.32 -1.8737 36 | -2.3 -1.85912 37 | -2.28 -1.84456 38 | -2.26 -1.83002 39 | -2.24 -1.81551 40 | -2.22 -1.80103 41 | -2.2 -1.78657 42 | -2.18 -1.77215 43 | -2.16 -1.75775 44 | -2.14 -1.74339 45 | -2.12 -1.72905 46 | -2.1 -1.71475 47 | -2.08 -1.70048 48 | -2.06 -1.68625 49 | -2.04 -1.67205 50 | -2.02 -1.6579 51 | -2 -1.64378 52 | -1.98 -1.62969 53 | -1.96 -1.61565 54 | -1.94 -1.60166 55 | -1.92 -1.5877 56 | -1.9 -1.5738 57 | -1.88 -1.55994 58 | -1.86 -1.54612 59 | -1.84 -1.53236 60 | -1.82 -1.51865 61 | -1.8 -1.505 62 | -1.78 -1.4914 63 | -1.76 -1.47785 64 | -1.74 -1.46437 65 | -1.72 -1.45095 66 | -1.7 -1.43759 67 | -1.68 -1.42429 68 | -1.66 -1.41107 69 | -1.64 -1.39792 70 | -1.62 -1.38483 71 | -1.6 -1.37183 72 | -1.58 -1.3589 73 | -1.56 -1.34605 74 | -1.54 -1.33329 75 | -1.52 -1.32061 76 | -1.5 -1.30803 77 | -1.48 -1.29553 78 | -1.46 -1.28314 79 | -1.44 -1.27084 80 | -1.42 -1.25865 81 | -1.4 -1.24656 82 | -1.38 -1.23458 83 | -1.36 -1.22272 84 | -1.34 -1.21098 85 | -1.32 -1.19936 86 | -1.3 -1.18788 87 | -1.28 -1.17652 88 | -1.26 -1.1653 89 | -1.24 -1.15423 90 | -1.22 -1.1433 91 | -1.2 -1.13253 92 | -1.18 -1.12192 93 | -1.16 -1.11147 94 | -1.14 -1.1012 95 | -1.12 -1.0911 96 | -1.1 -1.08119 97 | -1.08 -1.07148 98 | -1.06 -1.06196 99 | -1.04 -1.05265 100 | -1.02 -1.04356 101 | -1 -1.0347 102 | -0.98 -1.02607 103 | -0.96 -1.01768 104 | -0.94 -1.00954 105 | -0.92 -1.00167 106 | -0.9 -0.994072 107 | -0.88 -0.986758 108 | -0.86 -0.97974 109 | -0.84 -0.973031 110 | -0.82 -0.966643 111 | -0.8 -0.960589 112 | -0.78 -0.954883 113 | -0.76 -0.94954 114 | -0.74 -0.944574 115 | -0.72 -0.940002 116 | -0.7 -0.935841 117 | -0.68 -0.932107 118 | -0.66 -0.928819 119 | -0.64 -0.925996 120 | -0.62 -0.923657 121 | -0.6 -0.921825 122 | -0.58 -0.920521 123 | -0.56 -0.919767 124 | -0.54 -0.919588 125 | -0.52 -0.92001 126 | -0.5 -0.921058 127 | -0.48 -0.922761 128 | -0.46 -0.925147 129 | -0.44 -0.928247 130 | -0.42 -0.932094 131 | -0.4 -0.936721 132 | -0.38 -0.942162 133 | -0.36 -0.948456 134 | -0.34 -0.95564 135 | -0.32 -0.963756 136 | -0.3 -0.972846 137 | -0.28 -0.982955 138 | -0.26 -0.994131 139 | -0.24 -1.00642 140 | -0.22 -1.01988 141 | -0.2 -1.03456 142 | -0.18 -1.05052 143 | -0.16 -1.06782 144 | -0.14 -1.08652 145 | -0.12 -1.10668 146 | -0.1 -1.12839 147 | -0.08 -1.1517 148 | -0.06 -1.1767 149 | -0.04 -1.20346 150 | -0.02 -1.23208 151 | 2.39392e-15 -1.26263 152 | 0.02 -1.29521 153 | 0.04 -1.32992 154 | 0.06 -1.36686 155 | 0.08 -1.40613 156 | 0.1 -1.44786 157 | 0.12 -1.49214 158 | 0.14 -1.53912 159 | 0.16 -1.58892 160 | 0.18 -1.64167 161 | 0.2 -1.69751 162 | 0.22 -1.7566 163 | 0.24 -1.81909 164 | 0.26 -1.88515 165 | 0.28 -1.95495 166 | 0.3 -2.02866 167 | 0.32 -2.10648 168 | 0.34 -2.18861 169 | 0.36 -2.27525 170 | 0.38 -2.36662 171 | 0.4 -2.46295 172 | 0.42 -2.56447 173 | 0.44 -2.67145 174 | 0.46 -2.78414 175 | 0.48 -2.90281 176 | 0.5 -3.02776 177 | 0.52 -3.15928 178 | 0.54 -3.2977 179 | 0.56 -3.44334 180 | 0.58 -3.59655 181 | 0.6 -3.75769 182 | 0.62 -3.92716 183 | 0.64 -4.10533 184 | 0.66 -4.29264 185 | 0.68 -4.48952 186 | 0.7 -4.69643 187 | 0.72 -4.91384 188 | 0.74 -5.14227 189 | 0.76 -5.38223 190 | 0.78 -5.63428 191 | 0.8 -5.89899 192 | 0.82 -6.17697 193 | 0.84 -6.46886 194 | 0.86 -6.77531 195 | -------------------------------------------------------------------------------- /channel_dat/chan_y.res: -------------------------------------------------------------------------------- 1 | -1 4.53999e-06 2 | -0.99 5.53962e-06 3 | -0.98 6.74584e-06 4 | -0.97 8.19829e-06 5 | -0.96 9.94356e-06 6 | -0.95 1.20363e-05 7 | -0.94 1.45403e-05 8 | -0.93 1.75302e-05 9 | -0.92 2.10927e-05 10 | -0.91 2.53284e-05 11 | -0.9 3.03539e-05 12 | -0.89 3.63039e-05 13 | -0.88 4.33335e-05 14 | -0.87 5.16208e-05 15 | -0.86 6.13703e-05 16 | -0.85 7.28153e-05 17 | -0.84 8.6222e-05 18 | -0.83 0.000101893 19 | -0.82 0.000120172 20 | -0.81 0.000141447 21 | -0.8 0.000166156 22 | -0.79 0.000194791 23 | -0.78 0.000227904 24 | -0.77 0.000266114 25 | -0.76 0.00031011 26 | -0.75 0.000360656 27 | -0.74 0.000418604 28 | -0.73 0.000484892 29 | -0.72 0.000560554 30 | -0.71 0.000646728 31 | -0.7 0.000744658 32 | -0.69 0.000855705 33 | -0.68 0.000981346 34 | -0.67 0.00112319 35 | -0.66 0.00128296 36 | -0.65 0.00146253 37 | -0.64 0.00166391 38 | -0.63 0.00188923 39 | -0.62 0.00214078 40 | -0.61 0.00242097 41 | -0.6 0.00273237 42 | -0.59 0.00307766 43 | -0.58 0.00345966 44 | -0.57 0.0038813 45 | -0.56 0.00434563 46 | -0.55 0.00485578 47 | -0.54 0.00541499 48 | -0.53 0.00602652 49 | -0.52 0.00669372 50 | -0.51 0.00741993 51 | -0.5 0.0082085 52 | -0.49 0.00906273 53 | -0.48 0.00998586 54 | -0.47 0.010981 55 | -0.46 0.0120513 56 | -0.45 0.0131994 57 | -0.44 0.014428 58 | -0.43 0.0157394 59 | -0.42 0.0171358 60 | -0.41 0.0186188 61 | -0.4 0.0201897 62 | -0.39 0.0218493 63 | -0.38 0.0235982 64 | -0.37 0.0254361 65 | -0.36 0.0273624 66 | -0.35 0.0293758 67 | -0.34 0.0314743 68 | -0.33 0.0336553 69 | -0.32 0.0359155 70 | -0.31 0.038251 71 | -0.3 0.040657 72 | -0.29 0.0431279 73 | -0.28 0.0456576 74 | -0.27 0.0482391 75 | -0.26 0.0508648 76 | -0.25 0.0535261 77 | -0.24 0.0562142 78 | -0.23 0.0589194 79 | -0.22 0.0616313 80 | -0.21 0.0643393 81 | -0.2 0.067032 82 | -0.19 0.0696979 83 | -0.18 0.072325 84 | -0.17 0.0749012 85 | -0.16 0.0774142 86 | -0.15 0.0798516 87 | -0.14 0.0822012 88 | -0.13 0.0844509 89 | -0.12 0.0865888 90 | -0.11 0.0886034 91 | -0.1 0.0904837 92 | -0.09 0.0922194 93 | -0.08 0.0938005 94 | -0.07 0.0952181 95 | -0.06 0.096464 96 | -0.05 0.097531 97 | -0.04 0.0984127 98 | -0.03 0.099104 99 | -0.02 0.0996008 100 | -0.01 0.0999 101 | 7.5287e-16 0.1 102 | 0.01 0.0999 103 | 0.02 0.0996008 104 | 0.03 0.099104 105 | 0.04 0.0984127 106 | 0.05 0.097531 107 | 0.06 0.096464 108 | 0.07 0.0952181 109 | 0.08 0.0938005 110 | 0.09 0.0922194 111 | 0.1 0.0904837 112 | 0.11 0.0886034 113 | 0.12 0.0865888 114 | 0.13 0.0844509 115 | 0.14 0.0822012 116 | 0.15 0.0798516 117 | 0.16 0.0774142 118 | 0.17 0.0749012 119 | 0.18 0.072325 120 | 0.19 0.0696979 121 | 0.2 0.067032 122 | 0.21 0.0643393 123 | 0.22 0.0616313 124 | 0.23 0.0589194 125 | 0.24 0.0562142 126 | 0.25 0.0535261 127 | 0.26 0.0508648 128 | 0.27 0.0482391 129 | 0.28 0.0456576 130 | 0.29 0.0431279 131 | 0.3 0.040657 132 | 0.31 0.038251 133 | 0.32 0.0359155 134 | 0.33 0.0336553 135 | 0.34 0.0314743 136 | 0.35 0.0293758 137 | 0.36 0.0273624 138 | 0.37 0.0254361 139 | 0.38 0.0235982 140 | 0.39 0.0218493 141 | 0.4 0.0201897 142 | 0.41 0.0186188 143 | 0.42 0.0171358 144 | 0.43 0.0157394 145 | 0.44 0.014428 146 | 0.45 0.0131994 147 | 0.46 0.0120513 148 | 0.47 0.010981 149 | 0.48 0.00998586 150 | 0.49 0.00906273 151 | 0.5 0.0082085 152 | 0.51 0.00741993 153 | 0.52 0.00669372 154 | 0.53 0.00602652 155 | 0.54 0.00541499 156 | 0.55 0.00485578 157 | 0.56 0.00434563 158 | 0.57 0.0038813 159 | 0.58 0.00345966 160 | 0.59 0.00307766 161 | 0.6 0.00273237 162 | 0.61 0.00242097 163 | 0.62 0.00214078 164 | 0.63 0.00188923 165 | 0.64 0.00166391 166 | 0.65 0.00146253 167 | 0.66 0.00128296 168 | 0.67 0.00112319 169 | 0.68 0.000981346 170 | 0.69 0.000855705 171 | 0.7 0.000744658 172 | 0.71 0.000646728 173 | 0.72 0.000560554 174 | 0.73 0.000484892 175 | 0.74 0.000418604 176 | 0.75 0.000360656 177 | 0.76 0.00031011 178 | 0.77 0.000266114 179 | 0.78 0.000227904 180 | 0.79 0.000194791 181 | 0.8 0.000166156 182 | 0.81 0.000141447 183 | 0.82 0.000120172 184 | 0.83 0.000101893 185 | 0.84 8.6222e-05 186 | 0.85 7.28153e-05 187 | 0.86 6.13703e-05 188 | 0.87 5.16208e-05 189 | 0.88 4.33335e-05 190 | 0.89 3.63039e-05 191 | 0.9 3.03539e-05 192 | 0.91 2.53284e-05 193 | -------------------------------------------------------------------------------- /solutions/chap14sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \ifx\pdftexversion\undefined 4 | \usepackage[dvips]{graphicx} 5 | \else 6 | \usepackage[pdftex]{graphicx} 7 | \fi 8 | \newcommand{\be}{\begin{equation}} 9 | \newcommand{\ee}{\end{equation}} 10 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 11 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 12 | \newcommand{\comment}[1]{\relax} 13 | \newcommand{\dd}[2]{\frac{d {#1}}{d {#2}}} 14 | \newcommand{\pp}[2]{\frac{\partial {#1}}{\partial {#2}}} 15 | \begin{document} 16 | \fi 17 | 18 | \section{Chapter 14} 19 | \begin{enumerate} 20 | \item{\bf X-ray Bursts:} 21 | 22 | We will try to model Type-I X-ray bursts using a simple model for the instability. We will calculate how much material will accumulate on a neutron star before it bursts. 23 | \begin{enumerate} 24 | \item Let us assume that the star accretes pure helium, that the 25 | temperature of the degenerate layer is constant down to the core 26 | ($T_c$), how much luminosity emerges from the surface of the star? 27 | 28 | \item Let us assume that the helium layer has a mass, $dM$, and that the enregy generation rate for helium burning is given by 29 | $$ 30 | \epsilon_{3\alpha} = 3.5 \times 10^{20} T_9^{-3} \exp(-4.32/T_9) \mathrm{erg s}^{-1} \mathrm{g}^{-1} 31 | $$ 32 | where $T_9=T/10^9 \mathrm{K}$. The energy generation rate is a 33 | function of density too, but let's forget about that to keep things 34 | simple. How much power does the helium layer generate as a function of 35 | $dM$? 36 | 37 | \item Equate your answer to (a) to the answer to (b) and solve for 38 | $dM$. This is the thickness of a layer in thermal equilibrium. 39 | 40 | \item Let's assume that the potential burst starts by the temperature 41 | in the accreted layer jiggling up by a wee bit. If the surface 42 | luminosity increases faster with temperature than the helium burning 43 | rate, then the layer is stable. Calculate $dL_\mathrm{surface}/dT$ and 44 | $dP_\mathrm{helium}/dT$. 45 | 46 | \item Calculate the value of $dM$ for which $dP_\mathrm{helium}/dT$ 47 | exceeds $dL_\mathrm{surface}/dT$ and the layer bursts. 48 | 49 | \item Equate your value of $dM$ in (c) and (e) and solve for $T$. What 50 | is $dM$? How long will it take for such a layer to accumulate if the 51 | star is accreting at one-tenth of the Eddington accretion rate? 52 | 53 | \end{enumerate} 54 | {\bf Answer:} 55 | \begin{enumerate} 56 | \item If you assume free-free opacity you get using results from 57 | Chapter 1 58 | $$ 59 | L_{\gamma,ff} = 2.35 \times 10^{8} \textrm{erg/s} \left ( 60 | \frac{T}{1\textrm{K}}\right )^{7/2} 61 | $$ 62 | or if you used the black-body formula you get 63 | $$ 64 | L_{\gamma,BB} = 7 \times 10^{8} \textrm{erg/s} \left ( 65 | \frac{T}{1\textrm{K}}\right )^4 66 | $$ 67 | \item 68 | $$ 69 | P_\textrm{\scriptsize He} = \epsilon_{3\alpha} dM = 70 | 3.5 \times 10^{20} T_9^{-3} \exp(-4.32/T_9) \mathrm{erg s}^{-1} 71 | \mathrm{g}^{-1} dM 72 | $$ 73 | \item 74 | $$ 75 | dM_{ff} = 2.12 \times 10^{19} T_9^{13/2} \exp (4.32/T_9) \textrm{g} 76 | $$ 77 | and 78 | $$ 79 | dM_{BB} = 2 \times 10^{24} T_9^{7} \exp (4.32/T_9) \textrm{g} 80 | $$ 81 | \item 82 | $$ 83 | \frac{dL_{\gamma,ff}}{dT} = 8.2 \times 10^8 \textrm{erg/s/K} \left ( 84 | \frac{T}{1\textrm{K}} \right)^{5/2} 85 | $$ 86 | and 87 | $$ 88 | \frac{dL_{\gamma,BB}}{dT} = 2.8 \times 10^9 \textrm{erg/s/K} \left ( 89 | \frac{T}{1\textrm{K}} \right)^3. 90 | $$ 91 | For the helium burning we get 92 | $$ 93 | \frac{dP_\textrm{\scriptsize He}}{dT} = 4.2 \times 10^{10} 94 | \textrm{erg/s/g/K} T_9^{-5} \exp (4.32/T_9) (36 - 25 T_9) dM. 95 | $$ 96 | \item 97 | Let's solve for $dM$ again where the various derivatives are equal 98 | $$ 99 | dM_{ff} = 6.19 \times 10^{20} T_9^{15/2} \exp(4.32/T_9) (36 - 25 100 | T_9)^{-1} \textrm{g} 101 | $$ 102 | and 103 | $$ 104 | dM_{ff} = 6.67 \times 10^{25} T_9^{8} \exp(4.32/T_9) (36 - 25 105 | T_9)^{-1} \textrm{g}. 106 | $$ 107 | \item 108 | We find that $T_9 = 0.664$ for the free-free opacity and $T_9=0.617$ 109 | for the BB-case (no insulation). The layer thicknesses are 110 | $$ 111 | dM_{ff} = 10^{21}~\textrm{g} 112 | $$ 113 | and 114 | $$ 115 | dM_{BB} = 7 \times 10^{25}~\textrm{g}, 116 | $$ 117 | yielding accretion times of 2.8 hours and 24 years, respectively. The insulation of the envelope makes a big difference. Type-I bursts typically recur on a timescale of hours at one-tenth of the Eddington accretion rate. 118 | \end{enumerate} 119 | \end{enumerate} 120 | \ifx\bookloaded\undefined 121 | \end{document} 122 | \end 123 | \fi 124 | %%% Local Variables: 125 | %%% TeX-master: "book" 126 | %%% End: -------------------------------------------------------------------------------- /solutions/chap9sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \ifx\pdftexversion\undefined 4 | \usepackage[dvips]{graphicx} 5 | \else 6 | \usepackage[pdftex]{graphicx} 7 | \fi 8 | \newcommand{\be}{\begin{equation}} 9 | \newcommand{\ee}{\end{equation}} 10 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 11 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 12 | \newcommand{\comment}[1]{\relax} 13 | \begin{document} 14 | \fi 15 | 16 | \section{Chapter 9} 17 | \begin{enumerate} 18 | \item{\bf Lifetime} 19 | 20 | Derive the lifetime of the $n=2, l=1, m=0$ state of hydrogen to emit a 21 | photon and end up in the $n=1, l=0, m=0$ state. 22 | 23 | {\bf Answer:} 24 | 25 | The Einstein $A$-coefficient gives the rate of spontaneous emission 26 | for a state 27 | \begin{equation} 28 | A_{21} = \frac{2h\nu^3}{c^2} B_{21} = \frac{32 \pi^3 \nu^3}{3 \hbar 29 | c^3} |{\bf d}_{if}|^2 = \frac{32 \pi^3 \nu^3}{3 \hbar 30 | c^3} e^2 a_0^2 \left |\frac{{\bf r}_{if}}{a_0} \right |^2 31 | \end{equation} 32 | Let's calculate everything expect the matrix element to be sure of the 33 | units. We know that 34 | \begin{equation} 35 | h\nu = 2\pi \hbar \nu = \frac{e^2}{2 a_0} \left ( 1 - \frac{1}{2^2} \right ) = \frac{3 36 | e^2}{8 a_0} 37 | \end{equation} 38 | so we get 39 | \begin{equation} 40 | A_{21} = \frac{9 e^8}{128 \hbar^4 c^3 a_0} \left |\frac{{\bf 41 | r}_{if}}{a_0} \right |^2 = \frac{9 \alpha^4}{128} \frac{c}{a_0} \left |\frac{{\bf 42 | r}_{if}}{a_0} \right |^2 = 1.13 \times 10^9 \rmmat{s}^{-1} \left |\frac{{\bf 43 | r}_{if}}{a_0} \right |^2 44 | \end{equation} 45 | where we used $\alpha=e^2/(\hbar c)$, so the units are clearly right! 46 | 47 | The last step is to calculate the matrix element. We will choose the 48 | electron to initially be in the $m=0$ state so the $x$ and $y$ 49 | components of the dipole matrix element will be zero, so we are left with 50 | \begin{eqnarray} 51 | \frac{{\bf r}_{if}}{a_0} &=& 52 | \frac{2\pi}{a_0^5} \int_0^\infty r^2 dr \int_{-1}^1 d\mu \frac{1}{\sqrt{\pi}} e^{-r} (r\mu) 53 | \frac{1}{4\sqrt{2\pi}} r e^{-r/2} \mu \\ 54 | &=& \frac{1}{2 \sqrt{2} a_0^5} \int_0^\infty dr r^4 e^{-3r/2} 55 | \int_{-1}^1 d\mu \mu^2 = \frac{2^7\sqrt{2}}{3^5} = 0.745\\ 56 | \end{eqnarray} 57 | The lifetime is 58 | \begin{equation} 59 | \frac{1}{A_{21}} = \left ( \frac{3}{2} \right )^8 \frac{a_0}{c} 60 | \frac{1}{\alpha^4} = \left ( \frac{3}{2} \right )^8 \frac{\hbar}{m_e 61 | c^2} \frac{1}{\alpha^5} = 1.58~\rmmat{ns} 62 | \end{equation} 63 | \item{\bf Hydrogen-Like Absorption} 64 | 65 | How much energy does a photon need to ionize the following atoms by 66 | removing a K-shell electron? 67 | 68 | Hydrogen, Helium, Carbon, Oxygen, Iron 69 | 70 | Using the formula that I derived in class, draw an energy diagram that 71 | shows the total cross section for one gram of gas as a function of 72 | energy between 10eV and 10keV. It would be great if you used the 73 | initial expression in Eq. (72) for the dipole matrix element rather 74 | than the final answer given by Eq. (73). 75 | 76 | Consider that the mass fraction of the different atoms are hydrogen 77 | (0.7), helium (0.27), carbon (0.008), oxygen (0.016) and iron (0.004). 78 | 79 | {\bf Answer: } 80 | 81 | Let's first get the units right like in the previous quesiton. Using 82 | equations (72) and (61) we get 83 | \begin{equation} 84 | \sigma_{bf} = \frac{2 p V m \omega}{3 c \hbar^3} 85 | \frac{256\pi}{V} \left ( \frac{Z}{a_0^2} \right )^5 86 | \left ( \frac{Z}{a_0^2} + q^2 \right )^{-6} e^2 q^2 87 | \end{equation} 88 | Let's relate $p=\hbar q$ to the energy of the photon, we have 89 | \begin{equation} 90 | E = \hbar \omega = \frac{p^2}{2 m} + E_I = \frac{p^2}{2 m} + 91 | \frac{Z^2 \alpha^2}{2} 92 | m c^2 93 | \end{equation} 94 | Let's define $x=E/E_I$ to get 95 | \begin{equation} 96 | \sigma_{bf} = \frac{256 \pi}{3 Z^2} \frac{1}{\alpha} \left ( 97 | \frac{\hbar}{m c} \right )^2 ( x - 1 )^{3/2} x^{-5} = \frac{32}{Z^2 98 | \alpha^3} \sigma_T ( x - 1 )^{3/2} x^{-5} 99 | \end{equation} 100 | I know that $\sigma_T/m_p = 0.4~\rmmat{cm}^2 \rmmat{g}^{-1}$ so the 101 | total cross section per gram of matertial is 102 | \begin{equation} 103 | \sigma_{bf,\rmscr{Total}} = \sum_i X_i \frac{\sigma_T}{m_p} 104 | \frac{32}{A_i Z_i^2 105 | \alpha^3} ( x_i - 1 )^{3/2} x_i^{-5} 106 | \end{equation} 107 | where $X_i$ is the mass fraction of the species, $A_i$ is its atomic 108 | weight, $Z_i$ is its atomic number and $x_i=E/(Z^2 13.6~\rmmat{eV})$. 109 | 110 | { 111 | \centering 112 | \includegraphics[width=0.9\textwidth]{week9cross} } 113 | \end{enumerate} 114 | 115 | \ifx\bookloaded\undefined 116 | \end{document} 117 | \end 118 | \fi 119 | -------------------------------------------------------------------------------- /astrobookshelf.cls: -------------------------------------------------------------------------------- 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 2 | % 3 | % astro-bookshelf.cls 4 | % 5 | % Formating class for notes posted on Open-Astrophysics-Bookshelf. 6 | % 7 | % Edward Brown 8 | % Michigan State University 9 | % 10 | \def\fileversion{1.0} 11 | \def\filedate{2015/5/27} 12 | % 13 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 14 | 15 | \NeedsTeXFormat{LaTeX2e} 16 | \ProvidesClass{astro-bookshelf}[\filedate, \fileversion] 17 | \RequirePackage{ifxetex} 18 | \RequirePackage{xifthen} 19 | 20 | \typeout{% 21 | Document Class: `astro-bookshelf' v\fileversion \space <\filedate>} 22 | 23 | % Flag to indicate we are making a handout 24 | \newboolean{@handout} 25 | \setboolean{@handout}{false} 26 | \DeclareOption{handout}{\setboolean{@handout}{true}} 27 | 28 | % Define fonts. The option profonts specifies the use of Chapparal Pro, Raleway Medium, 29 | % and Source Code Pro for the serif, san serif, and fixed width fonts. 30 | \newboolean{@profonts} 31 | \setboolean{@profonts}{false} 32 | \DeclareOption{profonts}{\setboolean{@profonts}{true}} 33 | \newboolean{@stix} 34 | \setboolean{@stix}{false} 35 | \DeclareOption{stix}{\setboolean{@stix}{true}} 36 | 37 | \ProcessOptions 38 | \ifthenelse{\boolean{@handout}} 39 | { 40 | \LoadClass{tufte-handout} 41 | % \DeclareOption*{\PassOptionsToClass{\CurrentOption}{tufte-handout}} 42 | } 43 | { 44 | \LoadClass{tufte-book} 45 | % \DeclareOption*{\PassOptionsToClass{\CurrentOption}{tufte-book}} 46 | } 47 | \RequirePackage{natbib} 48 | \RequirePackage{amssymb} 49 | \ifxetex 50 | \RequirePackage{mathspec} 51 | \fi 52 | 53 | \RequirePackage{graphicx} 54 | \RequirePackage{wallpaper} 55 | \RequirePackage{ccicons} 56 | \RequirePackage{fancyvrb} 57 | 58 | \ifxetex 59 | \defaultfontfeatures{Scale=MatchLowercase} 60 | \renewcommand{\allcapsspacing}[1]{{\addfontfeature{LetterSpace=20.0}#1}} 61 | \renewcommand{\smallcapsspacing}[1]{{\addfontfeature{LetterSpace=5.0}#1}} 62 | \renewcommand{\textsc}[1]{\smallcapsspacing{\textsmallcaps{#1}}} 63 | \renewcommand{\smallcaps}[1]{\smallcapsspacing{\scshape\MakeTextLowercase{#1}}} 64 | 65 | \ifthenelse{\boolean{@profonts}} 66 | { 67 | \typeout{loading Chaparral Pro, Source Code Pro, and Raleway Medium} 68 | \setmainfont[Mapping=tex-text]{Chaparral Pro} 69 | \setmonofont[Mapping=tex-text]{Source Code Pro} 70 | \setsansfont[Mapping=tex-text]{Raleway Medium} 71 | \setmathsfont(Digits,Latin)[Numbers={Lining,Proportional}, Scale=MatchLowercase, Mapping=tex-text]{Chaparral Pro} 72 | \setmathrm{Chaparral Pro} 73 | }{ 74 | \typeout{loading Palatino, Courier, and Helvetica} 75 | \setmainfont[Mapping=tex-text]{TeX Gyre Pagella} 76 | \setmonofont[Mapping=tex-text]{TeX Gyre Cursor} 77 | \setsansfont[Mapping=tex-text]{TeX Gyre Heros} 78 | \setmathsfont(Digits,Latin)[Numbers={Lining,Proportional}, Scale=MatchLowercase, Mapping=tex-text]{TeX Gyre Pagella} 79 | \setmathrm{TeX Gyre Pagella} 80 | } 81 | \ifthenelse{\boolean{@stix}} 82 | { 83 | \setmathsfont(Greek)[Scale=MatchLowercase,Mapping=tex-text]{STIXGeneral} 84 | % for bold Greeks 85 | \fontspec{STIXGeneral} 86 | \SetSymbolFont{Greek:m:n}{bold}{TU}{\zf@family}{bx}{n} 87 | \SetSymbolFont{Greek:m:it}{bold}{TU}{\zf@family}{bx}{it} 88 | }{} 89 | \fi 90 | 91 | 92 | % for having inline author citations 93 | \newcommand{\citetalt}[1]{\citeauthor{#1}\cite{#1}} 94 | 95 | % for code listings 96 | \fvset{numbers=left} 97 | 98 | % Offset box for inline exercises 99 | \ifthenelse{\boolean{@handout}} 100 | { 101 | \newcounter{exercise} 102 | } 103 | { 104 | \newcounter{exercise}[chapter] 105 | \renewcommand{\theexercise}{\thechapter.\arabic{exercise}} 106 | } 107 | 108 | \newcommand{\listexercisename}{List of Exercises} 109 | \newcommand\listofexercises{% 110 | \ifthenelse{\equal{\@tufte@class}{book}}% 111 | {\chapter*{\listexercisename}}% 112 | {\section*{\listexercisename}}% 113 | \@starttoc{loe}% 114 | } 115 | % \newlistof[chapter]{exercise}{loe}{\listofexercises} 116 | 117 | \newcommand{\exercisedescription}[1]{% 118 | \addcontentsline{loe}{exercise}{\protect\numberline{\theexercise}{\ignorespaces #1}} 119 | } 120 | 121 | \newenvironment{exercisebox}[1][]{% 122 | \vspace{\baselineskip} 123 | \noindent\begin{minipage}{\linewidth}\small\raggedright 124 | \refstepcounter{exercise} 125 | \ifnum\value{exercise} = 1% 126 | \addtocontents{loe}{\protect\addvspace{10\p@}}% 127 | \fi 128 | \noindent\rule[0.6ex]{\linewidth}{0.5pt} 129 | \allcaps{exercise \theexercise---} 130 | \exercisedescription{#1}} 131 | {\noindent\rule[0.6ex]{\linewidth}{0.5pt} 132 | \end{minipage} 133 | } 134 | 135 | \let\l@exercise\l@figure 136 | -------------------------------------------------------------------------------- /solutions/chap8sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \usepackage{graphicx} 4 | \newcommand{\be}{\begin{equation}} 5 | \newcommand{\ee}{\end{equation}} 6 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 7 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 8 | \newcommand{\comment}[1]{\relax} 9 | \begin{document} 10 | \fi 11 | 12 | \section{Chapter 8} 13 | \begin{enumerate} 14 | \item{\bf Particles in a Box} 15 | 16 | A reasonable model for the neutrons and protons in a nucleus is that 17 | they are confined to a small region. Let's take a one-dimensional 18 | model of this. The potential is $V(x)$ is zero everywhere for $0l$. 24 | What are the energy levels of this system? 25 | 26 | 27 | {\bf Answer:} 28 | The harmonic functions the sine and cosine have the property that the 29 | second derivative is proportional to the function itself. We have 30 | $\psi=0$ at $x=0$ and at $x=l$ so 31 | \begin{equation} 32 | \psi_n = N \sin \left ( \frac{\pi n x}{l} \right ) 33 | \end{equation} 34 | where $n=1,2,3,\ldots$. Let's calculate, 35 | \begin{equation} 36 | -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = \frac{\hbar^2}{2m} 37 | \frac{\pi^2 n^2}{l^2} N \sin \left ( \frac{\pi n x}{l} \right ) = \frac{\hbar^2}{2m} 38 | \frac{\pi^2}{l^2} n^2 \psi 39 | \end{equation} 40 | so 41 | \begin{equation} 42 | E_n = \frac{\hbar^2}{2m} 43 | \frac{\pi^2}{l^2} n^2 44 | \end{equation} 45 | 46 | \item{\bf Hyperfine Transition} 47 | 48 | Calculate the energy and wavelength of the hyperfine transition of the 49 | hyodrgen atom. You may use the following formula for the energy of 50 | two magnets near to each other 51 | \begin{equation} 52 | E = -\frac{{\bf \mu}_1 \cdot {\bf \mu}_2}{r^3} 53 | \end{equation} 54 | We are looking for an order of magnitude estimate of the wavelength. 55 | I got 151~cm which is in the ballpark. 56 | 57 | {\bf Answer:} 58 | First let's write the values of the magnetic moments, 59 | \begin{equation} 60 | \mu_1 = \mu_p = g_p \frac{e}{2 M c} \frac{\hbar}{2} 61 | \end{equation} 62 | and 63 | \begin{equation} 64 | \mu_2 = \mu_e = g_e \frac{e}{2 m c} \frac{\hbar}{2} 65 | \end{equation} 66 | The spins can be aligned or antialigned so the energy difference 67 | is $2 \mu_1 \mu_2 / r^3$ so we get 68 | \begin{equation} 69 | \Delta E \sim \frac{g_p g_e}{8} \frac{e^2}{m c^2} \frac{\hbar^2}{M r^3} 70 | \end{equation} 71 | Let's take $r=a_0=\hbar^2/(me^2)$ to get 72 | \begin{equation} 73 | \Delta E \sim \frac{g_p g_e}{8} \frac{e^2}{m c^2} \frac{m^3 e^6}{\hbar^4 M} = 74 | \frac{g_p g_e}{8} \frac{\alpha \hbar c}{m c^2} \frac{m^3 \alpha^3 75 | \hbar^3 c^3}{\hbar^4 M} 76 | = \frac{g_p g_e}{8} \alpha^4 \frac{m}{M} m c^2 = 10^{-6}~\rmmat{eV} 77 | \end{equation} 78 | so $\lambda=123$~cm. 79 | 80 | {\bf A Better Answer:} 81 | 82 | First let's write the values of the magnetic moments, 83 | \begin{equation} 84 | \mu_1 = \mu_p = g_p \frac{e}{2 M c} \hbar {\bf s}_1 85 | \end{equation} 86 | and 87 | \begin{equation} 88 | \mu_2 = \mu_e = g_e \frac{e}{2 m c} \hbar {\bf s}_2 89 | \end{equation} 90 | so we get 91 | \begin{equation} 92 | E = \frac{g_p g_e}{4} \frac{e^2}{m c^2} \frac{\hbar^2}{M r^3} {\bf 93 | s}_1 \cdot {\bf s}_2 94 | \end{equation} 95 | Let's take $r=a_0=\hbar^2/(me^2)$ to get 96 | \begin{equation} 97 | E = \frac{g_p g_e}{4} \frac{e^2}{m c^2} \frac{m^3 e^6}{\hbar^4 M} = 98 | \frac{g_p g_e}{4} \frac{\alpha \hbar c}{m c^2} \frac{m^3 \alpha^3 99 | \hbar^3 c^3}{\hbar^4 M} 100 | = \frac{g_p g_e}{4} \alpha^4 \frac{m}{M} m c^2 \left ({\bf 101 | s}_1 \cdot {\bf s}_2 \right ). 102 | \end{equation} 103 | Let's calculate ${\bf F} = {\bf s}_1 + {\bf s}_2$ and square it 104 | \begin{eqnarray} 105 | {\bf F} \cdot {\bf F} &=& 106 | \left ( {\bf s}_1 + {\bf s}_2 \right )^2 = 107 | {\bf s}_1^2 + {\bf s}_2^2 + 2 {\bf s}_1 \cdot {\bf s}_2 \\ 108 | F (F+1) &=& S_1 (S_1 + 1) + S_2 (S_2 + 1) + 2 {\bf s}_1 \cdot {\bf 109 | s}_2 \\ 110 | F(F+1) &=& \frac{3}{4} + \frac{3}{4} + 2 {\bf s}_1 \cdot {\bf 111 | s}_2 112 | \end{eqnarray} 113 | so 114 | \begin{equation} 115 | {\bf s}_1 \cdot {\bf s}_2 = \frac{1}{2} F(F+1) - \frac{3}{4} 116 | = -\frac{3}{4}, \frac{1}{4} 117 | \end{equation} 118 | so 119 | \begin{equation} 120 | \Delta E_{F=0,F=1} = 121 | \frac{g_p g_e}{4} \alpha^4 \frac{m}{M} m c^2 = 2 \times 10^{-6} \rmmat{eV} 122 | \end{equation} 123 | and $\lambda = 60$~cm. 124 | 125 | \item{\bf Density and Ionization} 126 | 127 | Calculate the ionized fraction of pure hydrogen as a function of the 128 | density for a fixed temperature. You may take $U(T)=g_0=2$ and 129 | $U^+(T)=g_0^+=2$. 130 | 131 | {\bf Answer:} 132 | 133 | Let's take the Saha equation, 134 | \begin{equation} 135 | \frac{N^+ N_e}{N} = \left ( \frac{2\pi m_e kT}{h^2} \right)^{3/2} 136 | \frac{2 U^+(T)}{U(T)} e^{-E_I/kT}. 137 | \end{equation} 138 | Let $\xi$ be the ionized fraction, 139 | \begin{equation} 140 | \xi = \frac{N^+}{N+N^+} = \frac{N^+}{N_\rmscr{tot}} 141 | \end{equation} 142 | so using the values of $U(T)$ and $U^+(T)$ given in the problem 143 | \begin{equation} 144 | \frac{\xi^2 N_\rmscr{tot}^2}{(1-\xi) N_\rmscr{tot}} = 2 \left ( \frac{2\pi m_e kT}{h^2} \right)^{3/2} e^{-E_I/kT}. 145 | \end{equation} 146 | Rearranging 147 | \begin{equation} 148 | \frac{\xi^2}{1-\xi} = 2 \frac{2}{N_\rmscr{tot}} \left ( \frac{2\pi m_e 149 | kT}{h^2} \right)^{3/2} e^{-E_I/kT} = 2 y 150 | \end{equation} 151 | so 152 | \begin{equation} 153 | \xi = \sqrt{y^2+2y} - y \approx \sqrt{2y} \propto N_\rmscr{tot}^{-1/2} 154 | \end{equation} 155 | \end{enumerate} 156 | \ifx\bookloaded\undefined 157 | \end{document} 158 | \end 159 | \fi 160 | -------------------------------------------------------------------------------- /sedov/sedov.dat: -------------------------------------------------------------------------------- 1 | # t=1 rs=1.23195 s=0.49278 alpha=0.3524 2 | # r V-V_min rho p u 3 | 0.01 6.99415e-11 1.27089e-07 0.0445071 0.00210526 4 | 0.02 6.99415e-11 1.28098e-06 0.0445071 0.00421053 5 | 0.03 1.04912e-10 4.94894e-06 0.0445071 0.00631579 6 | 0.04 9.09239e-10 1.29114e-05 0.0445071 0.00842105 7 | 0.05 3.05994e-09 2.71649e-05 0.0445071 0.0105263 8 | 0.06 8.07824e-09 4.98822e-05 0.0445071 0.0126316 9 | 0.07 1.82984e-08 8.33876e-05 0.0445071 0.0147368 10 | 0.08 3.72788e-08 0.000130139 0.0445071 0.0168421 11 | 0.09 6.97928e-08 0.000192715 0.0445071 0.0189474 12 | 0.1 1.2231e-07 0.000273805 0.0445071 0.0210526 13 | 0.11 2.03399e-07 0.000376198 0.0445071 0.0231579 14 | 0.12 3.23444e-07 0.000502781 0.0445071 0.0252632 15 | 0.13 4.95819e-07 0.000656527 0.0445071 0.0273684 16 | 0.14 7.3616e-07 0.000840494 0.0445071 0.0294737 17 | 0.15 1.06361e-06 0.00105782 0.0445071 0.0315789 18 | 0.16 1.50059e-06 0.00131172 0.0445071 0.0336842 19 | 0.17 2.0734e-06 0.00160548 0.0445071 0.0357895 20 | 0.18 2.81241e-06 0.00194245 0.0445071 0.0378947 21 | 0.19 3.75253e-06 0.00232606 0.0445071 0.04 22 | 0.2 4.93323e-06 0.00275978 0.0445071 0.0421053 23 | 0.21 6.39934e-06 0.00324718 0.0445071 0.0442105 24 | 0.22 8.20143e-06 0.00379184 0.0445071 0.0463158 25 | 0.23 1.03956e-05 0.00439745 0.0445071 0.0484211 26 | 0.24 1.30445e-05 0.00506772 0.0445071 0.0505263 27 | 0.25 1.62174e-05 0.00580642 0.0445071 0.0526316 28 | 0.26 1.99906e-05 0.00661738 0.0445071 0.0547368 29 | 0.27 2.44479e-05 0.00750448 0.0445071 0.0568421 30 | 0.28 2.96809e-05 0.00847165 0.0445071 0.0589474 31 | 0.29 3.57896e-05 0.00952287 0.0445071 0.0610526 32 | 0.3 4.28827e-05 0.0106622 0.0445071 0.0631579 33 | 0.31 5.10776e-05 0.0118936 0.0445071 0.0652632 34 | 0.32 6.05018e-05 0.0132213 0.0445071 0.0673684 35 | 0.33 7.12924e-05 0.0146494 0.0445071 0.0694737 36 | 0.34 8.3597e-05 0.0161822 0.0445071 0.0715789 37 | 0.35 9.75738e-05 0.0178239 0.0445071 0.0736842 38 | 0.36 0.000113392 0.0195787 0.0445071 0.0757895 39 | 0.37 0.000131234 0.021451 0.0445071 0.0778947 40 | 0.38 0.000151293 0.0234452 0.0445071 0.08 41 | 0.39 0.000173774 0.0255656 0.0445071 0.0821053 42 | 0.4 0.000198897 0.0278169 0.0445071 0.0842105 43 | 0.41 0.000226894 0.0302826 0.0446657 0.0864088 44 | 0.42 0.000258011 0.0328272 0.0446875 0.0885294 45 | 0.43 0.000292511 0.0355198 0.0447117 0.0906521 46 | 0.44 0.000330668 0.0383657 0.0447385 0.0927771 47 | 0.45 0.000372775 0.0413702 0.044768 0.0949046 48 | 0.46 0.000419137 0.0445387 0.0448006 0.0970349 49 | 0.47 0.000470079 0.0478771 0.0448365 0.0991683 50 | 0.48 0.00052594 0.0513909 0.0448759 0.101305 51 | 0.49 0.00058708 0.0550862 0.044919 0.103446 52 | 0.5 0.000653871 0.0589692 0.0449662 0.10559 53 | 0.51 0.00072671 0.0630463 0.0450177 0.107739 54 | 0.52 0.000806007 0.0673239 0.0450739 0.109893 55 | 0.53 0.000892195 0.0718089 0.0451351 0.112052 56 | 0.54 0.000985724 0.0765084 0.0452016 0.114217 57 | 0.55 0.00108707 0.0814298 0.0452738 0.116387 58 | 0.56 0.00119672 0.0865806 0.0453522 0.118565 59 | 0.57 0.00131518 0.091969 0.045437 0.12075 60 | 0.58 0.00144301 0.0976032 0.0455288 0.122942 61 | 0.59 0.00158074 0.103492 0.0456279 0.125143 62 | 0.6 0.00172896 0.109645 0.0457349 0.127353 63 | 0.61 0.00188828 0.116071 0.0458504 0.129573 64 | 0.62 0.00205932 0.12278 0.0459747 0.131803 65 | 0.63 0.00224272 0.129784 0.0461085 0.134044 66 | 0.64 0.00243917 0.137094 0.0462525 0.136298 67 | 0.65 0.00264936 0.144721 0.0464071 0.138564 68 | 0.66 0.00287401 0.152678 0.0465732 0.140844 69 | 0.67 0.00311386 0.160979 0.0467514 0.143139 70 | 0.68 0.0033697 0.169637 0.0469426 0.145449 71 | 0.69 0.00364232 0.178668 0.0471474 0.147776 72 | 0.7 0.00393254 0.188089 0.0473667 0.150121 73 | 0.71 0.00424121 0.197916 0.0476015 0.152485 74 | 0.72 0.00456919 0.208167 0.0478528 0.154869 75 | 0.73 0.0049174 0.218863 0.0481214 0.157274 76 | 0.74 0.00528673 0.230024 0.0484087 0.159702 77 | 0.75 0.00567815 0.241674 0.0487156 0.162153 78 | 0.76 0.00609261 0.253835 0.0490435 0.16463 79 | 0.77 0.0065311 0.266535 0.0493937 0.167134 80 | 0.78 0.00699462 0.2798 0.0497677 0.169666 81 | 0.79 0.00748421 0.293662 0.0501668 0.172228 82 | 0.8 0.0080009 0.308153 0.0505928 0.174822 83 | 0.81 0.00854575 0.323307 0.0510474 0.177448 84 | 0.82 0.00911982 0.339162 0.0515324 0.18011 85 | 0.83 0.00972419 0.35576 0.0520499 0.182808 86 | 0.84 0.0103599 0.373143 0.052602 0.185544 87 | 0.85 0.0110282 0.391361 0.053191 0.188321 88 | 0.86 0.0117299 0.410464 0.0538194 0.19114 89 | 0.87 0.0124663 0.430509 0.0544899 0.194004 90 | 0.88 0.0132383 0.451557 0.0552053 0.196913 91 | 0.89 0.014047 0.473673 0.0559688 0.19987 92 | 0.9 0.0148934 0.496929 0.0567836 0.202878 93 | 0.91 0.0157785 0.521404 0.0576535 0.205937 94 | 0.92 0.0167032 0.547181 0.0585822 0.209051 95 | 0.93 0.0176683 0.574354 0.059574 0.212221 96 | 0.94 0.0186747 0.603024 0.0606335 0.215449 97 | 0.95 0.0197232 0.633299 0.0617656 0.218737 98 | 0.96 0.0208143 0.6653 0.0629755 0.222087 99 | 0.97 0.0219486 0.699157 0.0642692 0.225501 100 | 0.98 0.0231266 0.735013 0.0656527 0.22898 101 | 0.99 0.0243485 0.773024 0.0671329 0.232526 102 | 1 0.0256147 0.81336 0.0687171 0.236141 103 | 1.01 0.0269252 0.856208 0.0704133 0.239826 104 | 1.02 0.0282798 0.901771 0.0722301 0.243582 105 | 1.03 0.0296784 0.950274 0.074177 0.247411 106 | 1.04 0.0311205 1.00196 0.076264 0.251313 107 | 1.05 0.0326054 1.0571 0.0785023 0.255288 108 | 1.06 0.0341325 1.11599 0.080904 0.259338 109 | 1.07 0.0357008 1.17894 0.0834821 0.263463 110 | 1.08 0.0373091 1.24632 0.0862509 0.267662 111 | 1.09 0.038956 1.31852 0.089226 0.271936 112 | 1.1 0.04064 1.39596 0.0924243 0.276283 113 | 1.11 0.0423595 1.47911 0.0958644 0.280703 114 | 1.12 0.0441126 1.56849 0.0995664 0.285196 115 | 1.13 0.0458972 1.66468 0.103552 0.289759 116 | 1.14 0.0477111 1.76831 0.107847 0.294391 117 | 1.15 0.0495522 1.88006 0.112476 0.29909 118 | 1.16 0.0514178 2.00072 0.117468 0.303855 119 | 1.17 0.0533057 2.13112 0.122856 0.308683 120 | -------------------------------------------------------------------------------- /sedov/sedov_nd.dat: -------------------------------------------------------------------------------- 1 | # t=1 rs=1.23195 s=0.49278 alpha=0.3524 2 | # r V rho p u 3 | 0.00811722 1.07049e-09 3.94414e-08 0.265761 0.00619472 4 | 0.0162344 1.07049e-09 3.97545e-07 0.265761 0.0123894 5 | 0.0243517 1.60574e-09 1.53588e-06 0.265761 0.0185842 6 | 0.0324689 1.39164e-08 4.007e-06 0.265761 0.0247789 7 | 0.0405861 4.68341e-08 8.43049e-06 0.265761 0.0309736 8 | 0.0487033 1.23642e-07 1.54807e-05 0.265761 0.0371683 9 | 0.0568205 2.80068e-07 2.58789e-05 0.265761 0.043363 10 | 0.0649377 5.70573e-07 4.0388e-05 0.265761 0.0495577 11 | 0.073055 1.06822e-06 5.98082e-05 0.265761 0.0557525 12 | 0.0811722 1.87202e-06 8.4974e-05 0.265761 0.0619472 13 | 0.0892894 3.11313e-06 0.000116751 0.265761 0.0681419 14 | 0.0974066 4.9505e-06 0.000156035 0.265761 0.0743366 15 | 0.105524 7.58879e-06 0.00020375 0.265761 0.0805313 16 | 0.113641 1.12673e-05 0.000260843 0.265761 0.0867261 17 | 0.121758 1.62792e-05 0.000328289 0.265761 0.0929208 18 | 0.129875 2.29674e-05 0.000407086 0.265761 0.0991155 19 | 0.137993 3.17346e-05 0.000498252 0.265761 0.10531 20 | 0.14611 4.30455e-05 0.000602829 0.265761 0.111505 21 | 0.154227 5.74346e-05 0.000721879 0.265761 0.1177 22 | 0.162344 7.55059e-05 0.000856484 0.265761 0.123894 23 | 0.170462 9.79455e-05 0.00100774 0.265761 0.130089 24 | 0.178579 0.000125527 0.00117678 0.265761 0.136284 25 | 0.186696 0.000159111 0.00136473 0.265761 0.142479 26 | 0.194813 0.000199653 0.00157274 0.265761 0.148673 27 | 0.20293 0.000248216 0.00180199 0.265761 0.154868 28 | 0.211048 0.000305967 0.00205367 0.265761 0.161063 29 | 0.219165 0.000374189 0.00232898 0.265761 0.167257 30 | 0.227282 0.000454282 0.00262913 0.265761 0.173452 31 | 0.235399 0.00054778 0.00295537 0.265761 0.179647 32 | 0.243517 0.000656343 0.00330895 0.265761 0.185842 33 | 0.251634 0.000781771 0.00369112 0.265761 0.192036 34 | 0.259751 0.000926014 0.00410317 0.265761 0.198231 35 | 0.267868 0.00109117 0.00454638 0.265761 0.204426 36 | 0.275985 0.0012795 0.00502206 0.265761 0.21062 37 | 0.284103 0.00149342 0.00553154 0.265761 0.216815 38 | 0.29222 0.00173553 0.00607614 0.265761 0.22301 39 | 0.300337 0.00200861 0.0066572 0.265761 0.229205 40 | 0.308454 0.00231562 0.00727609 0.265761 0.235399 41 | 0.316571 0.00265971 0.00793417 0.265761 0.241594 42 | 0.324689 0.00304422 0.00863282 0.265761 0.247789 43 | 0.332806 0.00347273 0.00939803 0.266708 0.254257 44 | 0.340923 0.00394901 0.0101877 0.266838 0.260497 45 | 0.34904 0.00447704 0.0110234 0.266983 0.266743 46 | 0.357158 0.00506106 0.0119066 0.267143 0.272996 47 | 0.365275 0.00570552 0.012839 0.267319 0.279256 48 | 0.373392 0.00641512 0.0138224 0.267514 0.285524 49 | 0.381509 0.00719482 0.0148584 0.267728 0.291802 50 | 0.389626 0.00804981 0.0159489 0.267963 0.298089 51 | 0.397744 0.00898558 0.0170957 0.268221 0.304388 52 | 0.405861 0.0100079 0.0183008 0.268503 0.310698 53 | 0.413978 0.0111227 0.0195661 0.26881 0.317021 54 | 0.422095 0.0123364 0.0208936 0.269146 0.323359 55 | 0.430213 0.0136555 0.0222855 0.269511 0.329711 56 | 0.43833 0.0150871 0.023744 0.269908 0.336081 57 | 0.446447 0.0166382 0.0252713 0.27034 0.342469 58 | 0.454564 0.0183164 0.0268699 0.270807 0.348876 59 | 0.462681 0.0201296 0.0285421 0.271314 0.355305 60 | 0.470799 0.022086 0.0302907 0.271862 0.361756 61 | 0.478916 0.0241941 0.0321182 0.272454 0.368233 62 | 0.487033 0.0264627 0.0340277 0.273093 0.374736 63 | 0.49515 0.0289012 0.036022 0.273782 0.381267 64 | 0.503267 0.031519 0.0381043 0.274525 0.387829 65 | 0.511385 0.0343261 0.0402779 0.275324 0.394425 66 | 0.519502 0.0373328 0.0425463 0.276183 0.401055 67 | 0.527619 0.0405499 0.0449133 0.277107 0.407724 68 | 0.535736 0.0439882 0.0473828 0.278099 0.414433 69 | 0.543854 0.0476594 0.0499589 0.279163 0.421185 70 | 0.551971 0.0515752 0.052646 0.280304 0.427983 71 | 0.560088 0.0557478 0.0554488 0.281527 0.434831 72 | 0.568205 0.0601897 0.0583724 0.282837 0.44173 73 | 0.576322 0.064914 0.0614221 0.284239 0.448686 74 | 0.58444 0.0699341 0.0646036 0.285739 0.4557 75 | 0.592557 0.0752635 0.0679231 0.287343 0.462777 76 | 0.600674 0.0809164 0.0713869 0.289058 0.469921 77 | 0.608791 0.0869072 0.0750022 0.290891 0.477135 78 | 0.616909 0.0932508 0.0787764 0.292849 0.484423 79 | 0.625026 0.0999621 0.0827177 0.29494 0.491791 80 | 0.633143 0.107057 0.0868346 0.297173 0.499242 81 | 0.64126 0.11455 0.0911366 0.299557 0.50678 82 | 0.649377 0.122458 0.0956336 0.3021 0.514412 83 | 0.657495 0.130797 0.100337 0.304815 0.52214 84 | 0.665612 0.139584 0.105257 0.307711 0.529972 85 | 0.673729 0.148834 0.110408 0.310801 0.537911 86 | 0.681846 0.158565 0.115803 0.314098 0.545963 87 | 0.689963 0.168792 0.121457 0.317615 0.554134 88 | 0.698081 0.179533 0.127386 0.321367 0.562429 89 | 0.706198 0.190803 0.133606 0.325371 0.570854 90 | 0.714315 0.202619 0.140138 0.329642 0.579414 91 | 0.722432 0.214997 0.147002 0.334201 0.588117 92 | 0.73055 0.227952 0.154219 0.339067 0.596966 93 | 0.738667 0.241499 0.161815 0.344261 0.605969 94 | 0.746784 0.255651 0.169815 0.349807 0.615131 95 | 0.754901 0.270424 0.178248 0.355729 0.624459 96 | 0.763018 0.285827 0.187145 0.362056 0.633957 97 | 0.771136 0.301874 0.196541 0.368815 0.643632 98 | 0.779253 0.318574 0.206472 0.37604 0.653489 99 | 0.78737 0.335935 0.21698 0.383765 0.663534 100 | 0.795487 0.353965 0.228107 0.392026 0.673771 101 | 0.803605 0.372668 0.239904 0.400865 0.684206 102 | 0.811722 0.392048 0.252422 0.410324 0.694843 103 | 0.819839 0.412105 0.26572 0.420453 0.705686 104 | 0.827956 0.432838 0.27986 0.431301 0.716739 105 | 0.836073 0.454244 0.294913 0.442926 0.728004 106 | 0.844191 0.476316 0.310954 0.455388 0.739485 107 | 0.852308 0.499044 0.328066 0.468754 0.751184 108 | 0.860425 0.522417 0.346341 0.483095 0.763101 109 | 0.868542 0.54642 0.365879 0.498489 0.775238 110 | 0.87666 0.571036 0.38679 0.515022 0.787594 111 | 0.884777 0.596243 0.409196 0.532787 0.800168 112 | 0.892894 0.622018 0.433228 0.551885 0.81296 113 | 0.901011 0.648336 0.459034 0.572427 0.825967 114 | 0.909128 0.675168 0.486774 0.594532 0.839186 115 | 0.917246 0.702482 0.516626 0.618334 0.852612 116 | 0.925363 0.730246 0.548786 0.643975 0.866242 117 | 0.93348 0.758423 0.583468 0.671615 0.88007 118 | 0.941597 0.786979 0.620913 0.701426 0.894091 119 | 0.949714 0.815873 0.661383 0.733597 0.908298 120 | -------------------------------------------------------------------------------- /chapters/app_math.tex: -------------------------------------------------------------------------------- 1 | \chapter{Mathematical Appendix} 2 | \label{cha:math-append} 3 | 4 | \section{The Integral of $x^3/(e^x-1)$} 5 | \label{sec:integral-b_nut} 6 | 7 | The integral can be evaluated using a Taylor series 8 | \begin{equation} 9 | \int_0^\infty \frac{x^3}{e^x - 1} d x 10 | = \int_0^\infty \frac{x^3 e^{-x}}{1-e^{-x}} d x = 11 | \int_0^\infty x^3 \sum_{n=1}^{\infty} e^{-nx} d x 12 | \label{eq:796} 13 | \end{equation} 14 | Let's look at each term in the sum (we can do this because each 15 | term in the sum is a convergent integral) 16 | \begin{equation} 17 | \int_0^\infty x^3 e^{-nx} dx = \frac{1}{n^4} \int_0^\infty u^3 e^{-u} 18 | du. 19 | \label{eq:801} 20 | \end{equation} 21 | The integral 22 | \begin{equation} 23 | \int_0^\infty u^3 e^{-u} du = \left . -u^3 e^{-u} \right |_0^\infty 24 | + 3 \int_0^\infty u^2 e^{-u} du 25 | \end{equation} 26 | and 27 | \begin{equation} 28 | \int_0^\infty u^3 e^{-u} du = 3 \left ( \left . -u^2 e^{-u} 29 | \right|_0^\infty + 2 \int_0^\infty u e^{-u} du \right ) 30 | \end{equation} 31 | and 32 | \begin{equation} 33 | \int_0^\infty u^3 e^{-u} du = 3 \times 2 \times \left ( \left . -u e^{-u} 34 | \right|_0^\infty + \int_0^\infty e^{-u} du \right ) = 6 35 | \end{equation} 36 | to yield 37 | \begin{equation} 38 | \int_0^\infty \frac{x^3}{e^x - 1} d x = \sum_{n=1}^\infty \frac{6}{n^4}. 39 | \label{eq:800} 40 | \end{equation} 41 | This result can be generalized to yield 42 | \begin{equation} 43 | \int_0^\infty \frac{x^\alpha}{e^x - 1} d x = \sum_{n=1}^\infty 44 | \frac{\Gamma(\alpha+1)}{n^\alpha} = \Gamma(\alpha+1) \zeta(\alpha+1). 45 | \label{eq:828} 46 | \end{equation} 47 | For odd positive values of $\alpha$ the summation can be solved with contour 48 | integration. Let's start with 49 | $$ 50 | \sum_{n=1}^\infty \frac{1}{n^4} 51 | $$ 52 | to evaluate. We will use a basic result from complex analysis that 53 | the integral of an analytic function around a closed contour vanishes 54 | if the contour contains no poles. Let's examine the function 55 | \begin{equation} 56 | f(z) = \frac{\pi \cot (\pi z)}{z^4} dz 57 | \label{eq:795} 58 | \end{equation} 59 | that has poles at $z= \ldots, -2, -1, 0, 1, 2 \ldots$ and for large 60 | values of $z$ $f(z)$ quickly approaches zero, so the integral 61 | \begin{figure} 62 | \begin{center} 63 | \begin{tikzpicture} 64 | \foreach \x in {1,...,4} { 65 | \fill (\x,0) circle (0.05) ; 66 | \fill (-\x,0) circle (0.05) ; 67 | } 68 | \draw (0,0) circle (0.05); 69 | \foreach \x in {-4,...,4} { 70 | \draw [->] (\x,0.2) arc (90:360:0.2) arc (0:90:0.2); 71 | } 72 | \draw [->] (-5,0) -- (5,0); 73 | \draw [->] (0,-2) -- (0,2); 74 | \end{tikzpicture} 75 | \end{center} 76 | \caption{The poles of $f(z)$ in the complex plane} 77 | \label{fig:polesfz} 78 | \end{figure} 79 | \begin{equation} 80 | \lim_{R\rightarrow \infty }\oint_{C_R} f(z) dz = 0 81 | \label{eq:797} 82 | \end{equation} 83 | where $C_R$ is a circle of radius $R$. The sum of the integrals about 84 | all of the poles must vanish. Fig.~\ref{fig:polesfz} shows all of the 85 | poles. At the poles (solid points in the figure) other than at the 86 | origin, the function is given by 87 | \begin{equation} 88 | f(z) \approx \frac{1}{n^4} \frac{1}{z-n} 89 | \label{eq:798} 90 | \end{equation} 91 | that we can integrate along the loops in the figure by substituting 92 | $z=n+R e^{i\theta}$ so $dz = i R e^{i\theta} d\theta$ and 93 | \begin{equation} 94 | \lim_{R\rightarrow 0} \oint_{C_R} f(z) dz = 95 | \lim_{R\rightarrow 0} \int_0^{2\pi} \frac{1}{n^4} \frac{1}{R e^{i\theta}} i R 96 | e^{i\theta} d\theta = \frac{i}{n^4} \int_0^{2\pi} d\theta = 2\pi i \frac{1}{n^4} 97 | \label{eq:829} 98 | \end{equation} 99 | where $C_R$ is a circle of radius $R$ centered on the pole. Let's 100 | combine this result with the integral around the large loop 101 | (Eq.~\ref{eq:797}) to give 102 | \begin{equation} 103 | 0 = 4\pi i \sum_{n=1}^\infty \frac{1}{n^4} + \lim_{R\rightarrow 0} 104 | \oint_{C_R} f(z) dz 105 | \label{eq:799} 106 | \end{equation} 107 | where the first term is the sum we seek and the second term is an 108 | integral is over a circle surrounding the origin. The leading term in 109 | the integral about the origin is proportional to $z^{-5}$ and $\oint 110 | z^{-n} dz=0$ if $n\neq 1$, so we have to look at higher order terms, 111 | specifically 112 | \begin{equation} 113 | f(z) = \frac{1}{z^5} - \frac{\pi^2}{3 z^3} - \frac{\pi^4}{45 z} + 114 | \cdots 115 | \label{eq:803} 116 | \end{equation} 117 | so we have 118 | \begin{equation} 119 | 0 = 4\pi i \sum_{n=1}^\infty \frac{1}{n^4} + 2\pi i \left ( - 120 | \frac{\pi^4}{45} \right ) 121 | \label{eq:802} 122 | \end{equation} 123 | and 124 | \begin{equation} 125 | \sum_{n=1}^\infty \frac{6}{n^4} = 6 \frac{\pi^4}{45\times 2} = \frac{\pi^4}{15}. 126 | \end{equation} 127 | 128 | \section{Parseval's Theorem} 129 | \label{sec:an-math-asid} 130 | \index{Fourier transform!Parseval's theorem} 131 | 132 | We have stated a rather useful result, 133 | \begin{equation} 134 | \int_{-\infty}^{\infty} |E(t)|^2 dt = 2\pi \int_{-\infty}^{\infty} 135 | |{\hat E}(\omega)|^2 d \omega. 136 | \label{eq:159} 137 | \end{equation} 138 | We now have the tools to prove it quickly, 139 | \begin{eqnarray} 140 | \int_{-\infty}^{\infty} |E(t)|^2 dt &=& \int_{-\infty}^{\infty} d t 141 | \int_{-\infty}^{\infty} {\hat E}(\omega') 142 | e^{-i\omega' t} d \omega'. 143 | \int_{-\infty}^{\infty} {\hat E}^*(\omega) 144 | e^{i\omega t} d \omega \\ 145 | &=& \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} 146 | \int_{-\infty}^{\infty} d t d\omega' d \omega 147 | {\hat E}(\omega') {\hat E}^*(\omega) 148 | e^{-i\omega' t} e^{i\omega t} 149 | \label{eq:160} 150 | \end{eqnarray} 151 | The integral over time is simply Fourier transform of $2\pi 152 | e^{-i\omega' t}$ which we know, 153 | \begin{eqnarray} 154 | \int_{-\infty}^{\infty} |E(t)|^2 dt &=& 2\pi 155 | \int_{-\infty}^{\infty} 156 | \int_{-\infty}^{\infty} d\omega' d \omega 157 | {\hat E}(\omega') {\hat E}^*(\omega) \delta (\omega -\omega') \\ 158 | &=& 2 \pi \int_{-\infty}^{\infty} d \omega 159 | {\hat E}(\omega) {\hat E}^*(\omega) =2 \pi \int_{-\infty}^{\infty} 160 | |{\hat E}(\omega)|^2 d \omega 161 | \label{eq:161} 162 | \end{eqnarray} 163 | 164 | 165 | %%% Local Variables: 166 | %%% TeX-master: "book" 167 | %%% End: -------------------------------------------------------------------------------- /solutions/chap7sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \usepackage{graphicx} 4 | \newcommand{\be}{\begin{equation}} 5 | \newcommand{\ee}{\end{equation}} 6 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 7 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 8 | \newcommand{\comment}[1]{\relax} 9 | \begin{document} 10 | \fi 11 | \section{Chapter 7} 12 | 13 | \begin{enumerate} 14 | \item{ \bf The Sunyaev-Zeldovich Effect} 15 | \begin{enumerate} 16 | \item Let's say that you have a blackbody spectrum of temperature $T$ 17 | of photons passing through a region of hot plasma ($T_e$). You can assume 18 | that $T \ll T_e \ll m c^2/k$ 19 | 20 | What is the brightness temperature of the photons in the 21 | Rayleigh-Jeans limits after passing through the plasma in terms of 22 | the Compton $y-$parameter? 23 | 24 | {\bf Answer:} 25 | \begin{equation} 26 | T_{b,\rmscr{initial}} = \frac{c^2}{2\nu^2 k } I_\nu 27 | \end{equation} 28 | In Compton scattering, $I = I_\nu/(h\nu)$ is constant but $\nu_f = \nu_i e^y$ so 29 | we have 30 | \begin{equation} 31 | T_{b,\rmscr{final}} = \frac{c^2}{2\nu^2 e^{2y} k } e^y I_\nu = e^{-y} T_{b,\rmscr{initial}} 32 | \end{equation} 33 | 34 | \item Let's suppose that the gas has a uniform density $\rho$ and 35 | consists of hydrogen with mass-fraction $X$ and helium with 36 | mass-fraction $Y$ and other stuff $Z$. You can assume that 37 | $Z/A=1/2$ is for the other stuff. What is the number density of 38 | electrons in the gas? 39 | 40 | {\bf Answer:} 41 | One gram of the gas has $X$ grams of hydrogen which provide $X/m_p$ 42 | electrons. It has $Y$ grams of helium which provides $(Z/A) Y/m_p= 43 | 2/4 Y/m_p$ electrons and $Z$ grams of other stuff which provides 44 | $1/2 Y/m_p$ electrons. Adding it up gives 45 | \begin{equation} 46 | n_e = \frac{\rho}{m_p} \left ( X + \frac{1}{2} Y + \frac{1}{2} Z 47 | \right ) = \frac{\rho}{2 m_p} \left ( 2 X + 1 - X \right ) = 48 | \frac{\rho}{2 m_p} (1 + X ) 49 | \end{equation} 50 | \item 51 | If you assume that the gas is spherical with radius $R$, what is the 52 | value of the Compton $y-$parameter as a function of $b$, the 53 | distance between the line of sight and the center of the cluster? 54 | You can assume that the optical depth is much less than one. 55 | 56 | {\bf Answer:} 57 | The distance through the cluster is given by 58 | \begin{equation} 59 | l = 2 \sqrt{ R^2 - b^2} 60 | \end{equation} 61 | so the optical depth is 62 | \begin{equation} 63 | \tau_{es} = l n_e \sigma_T = 2\sqrt{R^2 - b^2} \frac{\rho}{2m_p} (1 + 64 | X) \sigma_T 65 | \end{equation} 66 | so 67 | \begin{equation} 68 | y_{NR} = \frac{4kT}{mc^2} \tau_{es} = \frac{4kT}{mc^2} \sqrt{R^2 - b^2} \frac{\rho}{m_p} (1 + 69 | X) \sigma_T 70 | \end{equation} 71 | \item 72 | Let's assume that the sphere contains $10^{12}$~M$_\odot$ of gas and 73 | that the radius of the sphere is 10 Mpc, $X=0.7, Y=0.27$ and 74 | $Z=0.03$ what is the value of the $y-$parameter? 75 | 76 | {\bf Answer:} 77 | The density of the cluster gas is 78 | \begin{equation} 79 | \rho = \frac{10^{12}\rmmat{M}_\odot (2 \times 80 | 10^{33}\rmmat{g/M}_\odot)}{\frac{4}{3} \pi 81 | (10\rmmat{Mpc}(3.08 \times 10^{24}\rmmat{cm/Mpc}))^3} = 0.16 \times 82 | 10^{-31} \rmmat{g/cm}^3 83 | \end{equation} 84 | This is actually really low. A realistic cluster is more massive that 85 | this. Let's plug these values in the formula for $y_{NR}$ and pick a 86 | reasonable value for $kT = 10$~keV so we get 87 | \begin{equation} 88 | y_{NR} = 2 \times 10^{-8} M_{12} R_{10}^{-2} T_{10} 89 | \end{equation} 90 | We can estimate the temperature of the cluster gas using the virial 91 | theorem 92 | \begin{equation} 93 | 2 \frac{M}{m_p} k T \approx \frac{3}{5} \frac{G M^2}{R} 94 | \end{equation} 95 | so 96 | \begin{equation} 97 | k T \approx \frac{3}{10} \frac{G M m_p}{R} \approx 1.3 \rmmat{eV} 98 | M_{12} R_{10}^{-1} 99 | \end{equation} 100 | \item 101 | Let's suppose that the blackbody photons are from the cosmic 102 | microwave background. What is the difference in the brightness 103 | temperature of the photons that pass through the cluster and those 104 | that don't (including the sign)? How does this difference compare 105 | with the primordial fluctuations in the CMB? How can you tell this 106 | change in the spectrum due to the cluster from the primordial 107 | fluctuations? 108 | 109 | {\bf Answer:} 110 | 111 | The photons that pass through the cluster have a brightness 112 | temperature that is lower by $2y T_\rmscr{CMB}$. 113 | The fluctuations of the CMB are around $10^{-5} T_\rmscr{CMB}$, so for 114 | such a puny cluster the S-Z would be hard to see. However, clusters 115 | are generally much more massive so the S-Z dominates over the 116 | fluctuations. Furthermore, the S-Z shifts photons to higher energies 117 | which is different than CMB fluctuations which change the temperature, 118 | so observations at energies in the Rayleigh-Jeans and Wein tail of the 119 | CMB spectrum can distinguish between the S-Z effect and primordial 120 | fluctuations. 121 | \end{enumerate} 122 | 123 | \item{\bf Synchrotron Self-Compton Emission Blazars} 124 | 125 | \begin{enumerate} 126 | \item 127 | What is the synchrotron emission from a single electron passing 128 | through a magnetic field in terms of the energy density of the 129 | magnetic field and the Lorentz factor of the electron? 130 | 131 | {\bf Answer:} 132 | \begin{equation} 133 | P_B = \frac{4}{3} \gamma^2 c \beta^2 \sigma_T U_B 134 | \end{equation} 135 | \item 136 | The number density of the electrons is $n_e$ and they fill a 137 | spherical region of radius $R$. What is the energy density of photons 138 | within the sphere, assuming that it is optically thin? 139 | 140 | {\bf Answer:} 141 | $P n_e$ gives the power per unit volume. To get the energy per unit 142 | volume we have to multiply by the typical time for photons to escape the 143 | spherical region typically $R/c$ because it is optically thin so we have 144 | \begin{equation} 145 | U_\rmscr{photon} = \frac{4}{3} \gamma^2 \sigma_T c \beta^2 U_B n_e \frac{R}{c} 146 | \end{equation} 147 | \item 148 | What is the inverse Compton emission from a single electron passing 149 | through a gas of photons field in terms of the energy density of the 150 | photons and the Lorentz factor of the electron? 151 | 152 | {\bf Answer:} 153 | \begin{equation} 154 | P_\rmscr{IC} = \frac{4}{3} \gamma^2 c \beta^2 \sigma_T U_\rmscr{photon} 155 | \end{equation} 156 | \item 157 | What is the total inverse Compton emission from the region if you 158 | assume that the synchrotron emission provides the seed photons for the 159 | inverse Compton emission? 160 | 161 | {\bf Answer:} 162 | \begin{equation} 163 | P_\rmscr{IC} = \frac{4}{3} \gamma^2 c \beta^2 \sigma_T \left ( 164 | \frac{4}{3} \gamma^2 c \beta^2 \sigma_T U_B n_e \frac{R}{c} \right 165 | ) n_e V 166 | \end{equation} 167 | so 168 | \begin{equation} 169 | P_\rmscr{IC} = \frac{64}{27} \gamma^4 \beta^4 c \sigma_T^2 170 | U_B n_e^2 R^4 171 | \end{equation} 172 | 173 | \end{enumerate} 174 | \end{enumerate} 175 | \ifx\bookloaded\undefined 176 | \end{document} 177 | \end 178 | \fi 179 | -------------------------------------------------------------------------------- /solutions/chap10sol.tex: -------------------------------------------------------------------------------- 1 | 2 | \section{Chapter 10} 3 | \begin{enumerate} 4 | \item{\bf The Number of Levels} 5 | 6 | I fit a Morse function to the potential of H$_2^+$. The parameters 7 | were 8 | \begin{equation} 9 | E_{n,0} = -0.065 \frac{e^2}{a_0}, B_n = 0.07 \frac{e^2}{a_0}, \beta_n 10 | = 0.7 a_0^{-1}, R_0 = 2.5 a_0 11 | \end{equation} 12 | How many vibrational levels does H$_2^+$ have? How many rotational 13 | levels does each vibrational level typically have? 14 | 15 | {\bf Answer:} 16 | 17 | Let's first to the rotational levels. When we increase the value 18 | of the angular momentum from $L$ to $L+1$ and the energy of the 19 | molecule decreases, we have reached the maximum value of $L$. From 20 | Eq.~(14) we have 21 | \begin{equation} 22 | 4 \hbar^2 \frac{(L+1)^2}{k \mu_{AB} R_0^4} = 1 23 | \end{equation} 24 | so 25 | \begin{equation} 26 | L_\rmscr{max} = \left ( \frac{k \mu_{AB} R_0^4}{4 \hbar^2} \right )^{1/2} - 1 27 | \end{equation} 28 | We need to determine $k$. This is related to the parameters that I 29 | gave in the question, we know that $\omega^2 = k/\mu_{AB}$ and in the Morse 30 | potential $\omega^2 = 2 \beta_n^2 B_n$ so 31 | \begin{equation} 32 | k = 2\beta_n^2 B_n 33 | \end{equation} 34 | and 35 | \begin{equation} 36 | L_\rmscr{max} = \left ( \frac{2 \beta_n^2 B_n \mu_{AB} R_0^4}{4 \hbar^2} \right )^{1/2} - 1 37 | \end{equation} 38 | I'm going to substitute the units for the various quantities into the 39 | expression above 40 | \begin{equation} 41 | L_\rmscr{max} = \left ( \frac{\beta_n B_n e^2 m_p a_0 R_0^4}{4 42 | \hbar^2} \right )^{1/2} - 1 = \left ( \frac{\beta_n B_n m_p R_0^4}{4 m_e} \right )^{1/2} - 1 43 | \end{equation} 44 | where I used $\mu_{AB}=m_p/2$ and $e^2 a_0=\hbar^2/m_e$. The 45 | expression in the parenthesis is dimensionless! I get 46 | \begin{equation} 47 | L_\rmscr{max} = 23.79. 48 | \end{equation} 49 | Because $L$ ranges from zero to $L_\rmscr{max}$, I have 24 or 25 50 | levels. 51 | 52 | A formula for the number of vibrational levels is given explicitly in 53 | Eq. (19). The number of levels is 54 | \begin{equation} 55 | \frac{(2 \mu_{AB} B_n)^{1/2}}{\beta_n \hbar} + \frac{1}{2} = 56 | \frac{B_n^{1/2}}{\beta_n} \left ( m_p \frac{e^2}{a_0} a_0^2 \hbar^{-2} 57 | \right )^{1/2} + \frac{1}{2} = 58 | \frac{B_n^{1/2}}{\beta_n} \left ( \frac{m_p}{m_e} \right )^{1/2} + 59 | \frac{1}{2} = 16.7 60 | \end{equation} 61 | 62 | \item{\bf Nuclear Overlap} 63 | 64 | Consider two deuterons bound by a single electron as in question (1). 65 | What is the probability that the two deuterons lie on top of each 66 | other, {\ie} that $R<4$~fermi, the diameter of the deuteron? What is the 67 | probability if the two deuterons are bound by a single muon, $m_\mu 68 | \approx 207 m_e$? You can find the eigenfunctions of the Morse 69 | potential on Wikipedia. 70 | 71 | If you assume that whenever the deuterons overlap they fuse and that 72 | you get to ``roll the dice'' once each oscillation period, calculate 73 | the fusion rate in both cases. 74 | 75 | {\bf Answer:} 76 | \index{muon-catalysed fusion} 77 | 78 | The probability of overlap is simply the squared modulus of the 79 | nuclear wavefunction evaluated at $r=0$ integrated over the volume 80 | $4/3 \pi (4~\mathrm{fermi})^3$. The nuclear wavefunction is given by 81 | \begin{equation} 82 | \Psi_n(z) = N_n z^{\lambda - n - \frac{1}{2}} e^{-z/2} L_n^{2\lambda-2n-1}(z) 83 | \end{equation} 84 | where $\lambda=\sqrt{2 M B_n}/(\beta_n \hbar)$ and the normalization 85 | \begin{equation} 86 | N_n=n! \left [ \frac{\beta_n (2\lambda - 2 n - 1)}{\Gamma(n+1) 87 | \Gamma(2\lambda - n)} \right]^{1/2} 88 | \end{equation} 89 | and $L^{\alpha}_n$ is a Laguerre polynomial and $z=2\lambda 90 | e^{-(x-x_e)}$ and $x=\beta_n r$. This wavefunction is in terms of 91 | $r$ as a one-dimensional coordinate; it is analogous to the function 92 | $R(r)$ in the expansion of the atomic wavefunction in spherical 93 | symmetry. The complete wavefunction is 94 | \begin{equation} 95 | \psi(r,\theta,\phi) = \frac{1}{\sqrt{4\pi}} r^{-1} \Psi_0(z) , 96 | \end{equation} 97 | so the probability of the two nuclei being within 4~fermi of each 98 | other is given by 99 | \begin{equation} 100 | P = \int_0^{4~\mathrm{fermi}} d r \left | \Psi \left (2 \lambda \exp 101 | \left [ \beta_n R_0 \right ] \right ) \right |^2 = \frac{4~\mathrm{fermi}}{a_0} \left | \Psi\left(2 \lambda \exp 102 | \left [ \beta_n R_0 \right ]\right) \right |^2 103 | \end{equation} 104 | 105 | Since we are interested in the ground state, $n=0$ so 106 | \begin{equation} 107 | L_n^{2\lambda-2n-1} = 1 ~\mathrm{and}~ N_n = \left [ \frac{\beta_n ( 2 108 | \lambda -1)}{\Gamma(2\lambda)} \right ]^{1/2} 109 | \end{equation} 110 | which simplifies matters. What remains is to calculate determine how 111 | the value of $\lambda$ depends on the mass of the binding particle 112 | muon or electron. We have 113 | \begin{equation} 114 | \lambda = \frac{\sqrt{2 M A e^2/a_0}}{B a_0^{-1} \hbar} = 115 | \frac{\sqrt{2 A}}{B} \sqrt{\frac{M}{m}} 116 | \end{equation} 117 | where $M$ is the reduced mass of the pair of deuterons and $m$ is the 118 | mass of the muon or electron. The constants $A$ and $B$ are simply 119 | the numerical constants $0.07$ and $0.7$ that define the parameters of 120 | the Morse potential in dimensionless units. For the electronically bound 121 | system $\lambda=22.9$ and for the muonically bound system 122 | $\lambda=1.59$. 123 | 124 | What remains is to evaluate the wavefunctions in both cases, for the 125 | electron we have 126 | \begin{equation} 127 | \Psi(R=0) = 7.5 \times 10^{-30} 128 | \end{equation} 129 | and for the muon we have 130 | \begin{equation} 131 | \Psi(R=0) = 2.0 \times 10^{-3}. 132 | \end{equation} 133 | Converting these to probabilities yields 134 | \begin{equation} 135 | P_\mathrm{electron} = 5 \times 10^{-63}, P_\mathrm{muon} = 6 \times 10^{-8}. 136 | \end{equation} 137 | To get a fusion rate we should multiply these by the typical frequency 138 | of the systems say $\omega=2\beta_n^2 B_n/M=2 A B^2 e^2/(a_0^3 m_D/2)$ 139 | or $1.2 \times 10^{16}$~Hz for the electron and $3.6 \times 140 | 10^{19}$~Hz for the muon. Therefore, we get a rate of three deuterium 141 | fusions over the age of the universe in one ton of deuterium for 142 | electronically bound molecules or $2 \times 10^{12}$~Hz for the 143 | muonically bound molecule or about 4 million times over the 2.2$\mu$s 144 | lifetime of the muon. 145 | 146 | It turns out that the rate-limiting step in muonic fusion is the 147 | formation of muonic molecules which takes about one thousand times 148 | longer than the fusion, but even this is not the killer. It is the 149 | fact that about one percent of the time the muon stays stuck to the 150 | fusion product so cannot catalyse another reaction. The first person 151 | to consider muon-catalysed fusion was John David Jackson, and Eugene 152 | Wigner suggested that ``alpha sticking'' could be a problem. This 153 | process was the original ``cold fusion,'' and it almost breaks even 154 | (within a factor of a few). 155 | \index{cold fusion} 156 | \end{enumerate} 157 | 158 | %%% Local Variables: 159 | %%% mode: latex 160 | %%% TeX-master: "book" 161 | -------------------------------------------------------------------------------- /solutions/chap3sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \usepackage{graphicx} 4 | \usepackage{tikz} 5 | \usetikzlibrary{snakes} 6 | \usetikzlibrary{arrows} 7 | \usetikzlibrary{shapes} 8 | \usetikzlibrary{backgrounds} 9 | \input book_defs 10 | \begin{document} 11 | \fi 12 | \section{Chapter 3} 13 | \begin{enumerate} 14 | \item{\bf Constant Velocity Charge} 15 | 16 | Show that if charge is not accelerating, the electric field vector 17 | points to the current (not the retarded) position of the charge. 18 | 19 | {\bf Answer:} 20 | 21 | \begin{center} 22 | \begin{tikzpicture} 23 | \begin{scope}[rotate=-30] 24 | \draw [->] (0,0)--(2,3); 25 | \draw [->] (0,0)--(3,2); 26 | \draw [->] (3,2)--(4,2) ; 27 | \draw [->] (3,2)--(2,3) ; 28 | \draw [->,dashed] (4,2)--(3,3) ; 29 | \draw [->,dashed] (3,3)--(2,3) ; 30 | \draw [->,thick] (4,2)--(2,3) ; 31 | \draw (3,2) node [below] {${\bf r}_0$} 32 | (4,2) node [right] {$\betabold c t=\betabold R$} 33 | (2,3) node [above] {${\bf r}$} 34 | (2.5,2.5) node [left] {${\bf R}$} 35 | (4,3) node [right] {${\bf E} \| R\left ({\bf n}-\betabold\right)$}; 36 | \draw [->,gray] (4,3)--(3.1,2.5); 37 | \end{scope} 38 | \end{tikzpicture} 39 | \end{center} 40 | 41 | \setcounter{enumi}{3} 42 | 43 | \item{\bf Synchrotron Cooling:} 44 | 45 | A particle of mass $m$, charge $q$, moves in a plane perpendicular to 46 | a uniform, static, magnetic field $B$. 47 | \begin{enumerate} 48 | \item 49 | Calculate the total energy radiated per unit time, expressing it in 50 | terms of the constants already defined and the ratio 51 | $\gamma=1/\sqrt{1-\beta^2}$ of the particle's total energy to its rest 52 | energy. You can assume that the particle is ultrarelativistic. 53 | \item 54 | If at time $t=0$ the particle has a total energy $E_0=\gamma_0 m c^2$, 55 | show that it will have energy $E=\gamma m c^2 < E_0$ at a time $t$, 56 | where 57 | \[ 58 | t \approx \frac{3 m^3 c^5}{2 q^4 B^2} \left ( \frac{1}{\gamma} - 59 | \frac{1}{\gamma_0} \right ). 60 | \] 61 | \end{enumerate} 62 | 63 | {\bf Answer:} 64 | 65 | The synchrotron power is given by the power emitted by a particle 66 | performing circular motion 67 | $$ 68 | P_\perp = \frac{2}{3} \frac{q^2}{m^2 c^3} \gamma^2 \left ( \frac{d 69 | {\bf p}}{dt} \right )^2 70 | $$ 71 | where for an ultrarelativistic charged particle in a magnetic field we 72 | have 73 | $$ 74 | \left | \frac{d {\bf p}}{dt} \right | = q B 75 | $$ 76 | so 77 | $$ 78 | P_\perp = \frac{2}{3} \frac{q^5}{m^2 c^3} \gamma^2 B^2 = -\frac{d 79 | E}{dt} = -m c^2 \frac{d \gamma}{dt} 80 | $$ 81 | and 82 | $$ 83 | \frac{d \gamma}{dt} = -\frac{2}{3} \frac{q^5}{m^3 c^5} B^2 \gamma^2. 84 | $$ 85 | Separating the variables and integrating yields the required answer. 86 | 87 | \item{\bf Classical HI:} 88 | A particle of mass $m$ and charge $q$ moves in a circle due to a force 89 | ${\bf F} = -\hat{\bf r} \frac{q^2}{r^2}$. 90 | You may assume that the particle always moves non-relativistically. 91 | \begin{enumerate} 92 | \item What is the acceleration of the particle as a function of $r$? 93 | \item What is the total energy of the particle as a function of $r$? 94 | The potential energy is given by $-q^2/r$. 95 | \item What is the power radiated as a function of $r$? 96 | \item Using the fact the $P=-dE/dt$ and the answer to (2), find 97 | $dr/dt$. 98 | \item Assuming that the particle starts with $r=r_i$ at $t=0$, find 99 | the value of $t$ where $r=0$. 100 | \item Let's assume that $q=e$, the charge of the electron, and 101 | $m=m_e$, the mass of the electron. Write your answer in (4) in 102 | terms of $r_i$, $r_0$ (the classical electron radius) and $c$. 103 | \item What is the time if $r_i=0.5$\AA (for an hydrogen)? 104 | \item Compare this to the lifetime of a hydrogen atom. 105 | \end{enumerate} 106 | 107 | {\bf Answer:} 108 | 109 | \begin{enumerate} 110 | \item 111 | \[ \dot{\bf u} = -{\hat r} \frac{q^2}{r^2 m} \] 112 | \item 113 | \[ E = -\frac{q^2}{r} + \frac{1}{2} m v^2 = -\frac{q^2}{r} + \frac{1}{2} \left ( \frac{q^2}{r} \right ) = -\frac{1}{2} \frac{q^2}{r} \] 114 | where I used 115 | \[ \frac{m v^2}{r} = \frac{q^2}{r^2} \] 116 | for circular motion. 117 | \item 118 | \[ 119 | P = \frac{2 q^2 \dot{u}^2}{3 c^3} = \frac{2 q^2 }{3 c^3} \left ( \frac{q^2}{r^2 m} \right )^2 120 | \] 121 | \item 122 | \[ 123 | \frac{d E}{d t} = \frac{d}{dt} \left ( -\frac{1}{2} \frac{q^2}{r} \right ) = \frac{1}{2} \frac{q^2}{r^2} \frac{dr}{dt} 124 | \] 125 | \[ 126 | \frac{d E}{d t} = -P = -\frac{2 q^6 }{3 m^2 c^3} \frac{1}{r^4} = \frac{1}{2} \frac{q^2}{r^2} \frac{dr}{dt} 127 | \] 128 | \[ 129 | \frac{dr}{dt} = -\frac{4 q^4 }{3 m^2 c^3} \frac{1}{r^2} 130 | \] 131 | \item 132 | \[ 133 | t = \int_{r_i}^0 \frac{dt}{dr} dr = -\frac{3 m^2 c^3}{4 q^4 }\int_{r_i}^0 r^2 d r 134 | = \frac{ r_i^3 m^2 c^3}{4 q^4 } 135 | \] 136 | \item 137 | \[ 138 | t = \frac{ r_i^3 m^2 c^3}{4 e^4 } = \frac{1}{4 c} r_i \left ( \frac{r_i}{r_0} \right )^2 139 | \] 140 | \item 141 | \[ 142 | r_0 = 2.82 \times 10^{-13}~\rmmat{cm}, 1\rmmat{\AA} = 10^{-8}~\rmmat{cm} 143 | \] 144 | \[ 145 | t = \frac{1}{12 \times 10^{10} \rmmat{cm/s}} 0.5 \times 10^{-8}~\rmmat{cm} ( 17000 )^2 = 1.2 \times 10^{-11}~\rmmat{s} 146 | \] 147 | \item 148 | It is much smaller than the lifetime of a hydrogen atom. 149 | \end{enumerate} 150 | 151 | \item{\bf The Eddington Luminosity:} 152 | 153 | There is a natural limit to the luminosity a gravitationally bound 154 | object can emit. At this limit the inward gravitational force on a 155 | piece of material is balanced by the outgoing radiation 156 | pressure. Although this limiting luminosity, the Eddington 157 | luminosity, can be evaded in various ways, it can provide a useful 158 | (if not truly firm) estimate of the minimum mass of a particular 159 | source of radiation. 160 | 161 | \begin{enumerate} 162 | \item Consider ionized hydrogen gas. Each electron-proton pair has a 163 | mass more or less equal to the mass of the proton ($m_p$) and a cross 164 | section to radiation equal to the Thompson cross-section ($\sigma_T$). 165 | \item The radiation pressure is given by outgoing radiation flux over the speed of light. 166 | \item Equate the outgoing force due to radiation on the pair with the inward force of gravity on the pair. 167 | \item Solve for the luminosity as a function of mass. 168 | \end{enumerate} 169 | The mass of the sun is $2 \times 10^{33}$g. What is the Eddington 170 | luminosity of the sun? 171 | 172 | {\bf Answer:} 173 | \begin{enumerate} 174 | \item OK 175 | \item $P=\frac{F}{c} = \frac{L}{4\pi r^2 c}$ 176 | \item $F_\mathrm{out} = P \sigma_T = \frac{L \sigma_T}{4\pi r^2 c}$, 177 | $F_\mathrm{in} = \frac{G M m_p}{r^2}$ 178 | \item 179 | $F_\mathrm{in}=F_\mathrm{out}$ for $L=L_\mathrm{Edd}$ so 180 | $L_\mathrm{Edd} = \frac{4\pi c G M m_p}{\sigma_T}$ 181 | \item 182 | $L_\mathrm{Edd} = 1.26 \times 10^{38} \textrm{erg/s} \left (\frac{M}{M_\odot} 183 | \right ) = 3.2 \times 10^4 \left (\frac{M}{M_\odot} 184 | \right ) L_\odot.$ 185 | \end{enumerate} 186 | \end{enumerate} 187 | 188 | 189 | \ifx\bookloaded\undefined 190 | \end{document} 191 | \end 192 | \fi 193 | %%% Local Variables: 194 | %%% TeX-master: "book" 195 | %%% End: -------------------------------------------------------------------------------- /solutions/chap1sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \usepackage{graphicx} 4 | \newcommand{\be}{\begin{equation}} 5 | \newcommand{\ee}{\end{equation}} 6 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 7 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 8 | \newcommand{\comment}[1]{\relax} 9 | \begin{document} 10 | \fi 11 | \section{Chapter 1} 12 | 13 | \begin{enumerate} 14 | \item{\bf Hot Cloud} 15 | 16 | X-ray photons are produced in a cloud of radius $R$ at the uniform rate 17 | (photons per unit volume per unit times). the cloud is a distance $d$ 18 | away. Assume that the cloud is optically thin. A detector at Earth has 19 | an angular acceptance beam of half-angle $\Delta \theta$ and an effective area $A$. 20 | \begin{enumerate} 21 | \item If the cloud is fully resolved by the detector, what is the observed 22 | intensity of the radiation as a function of position? 23 | \item If the cloud is fully unresolved, what is the average intensity when 24 | the source is in the detector? 25 | \end{enumerate} 26 | {\bf Answer:} 27 | \begin{enumerate} 28 | \item If the source is resolved, we can discern different parts of the cloud, so the observed intensity is the integral of the emission coefficient through the cloud, 29 | \begin{equation} 30 | I = \int j ds = \int_{-\sqrt{R^2-b^2}}^{\sqrt{R^2-b^2}} \frac{\Gamma}{4\pi} ds = \frac{\Gamma}{2\pi} \sqrt{R^2-b^2} = \frac{\Gamma}{2\pi} R \sqrt{1-\frac{b^2}{R^2}} 31 | \end{equation} 32 | where $b/R$ is the relative distance between our line of sight and the centre of the cloud. 33 | \item If the source is not resolved, the observed intensity is given by the flux from the source divided by the solid angle of acceptance of the detector. 34 | \begin{equation} 35 | I = \frac{F}{\pi \Delta \theta^2} = \frac{\frac{4}{3} \pi R^3 \Gamma }{4\pi d^2} \frac{1}{\pi \Delta \theta^2} = \frac{\Gamma R^3}{3 \pi d^2 \Delta \theta^2} 36 | \end{equation} 37 | 38 | Clearly, this is the minimum value of the actual intensity of the 39 | source because it may actually subtend a smaller region of the sky 40 | that $\Delta \theta$ but we have no way of know because our detector 41 | cannot resolve below this scale. 42 | \end{enumerate} 43 | \setcounter{enumi}{2} 44 | \item{\bf Blackbody} 45 | 46 | Only one or no neutrinos can occupy a single state. Calculate the spectrum of the neutrino field in thermal equilibrium (neglect the mass of the neutrino). Neutrinos like photons have two polarization states. What is the ratio of the Stefan-Boltzmann constant for neutrinos to that of photons? 47 | 48 | {\bf Answer:} 49 | 50 | The main difference between the neutrinos and the photons is the partition function. The mean energy of the neutrinos with a certain value of $\nu$ is 51 | \begin{equation} 52 | {\bar E} = \frac{\sum_{i=0}^1 n h \nu e^{-n h\nu/kT}}{\sum_{i=0}^1 e^{-n h\nu/kT}}. 53 | \end{equation} 54 | For photons the sum is from 0 to infinity. So we have 55 | \begin{equation} 56 | B_\nu (T) = \frac{2 h}{c^2} \frac{\nu^3}{\exp ( h \nu / k T) + 1}. 57 | \end{equation} 58 | for neutrinos. The ratio of the Stefan-Boltzmann constants is 59 | \begin{equation} 60 | R = \frac{\int_0^\infty x^3 ( e^x + 1 )^{-1}}{\int_0^\infty x^3 (e^x - 1)^{-1}} = \frac{7 \pi^4/120}{\pi^4/15} = \frac{7}{8} 61 | \end{equation} 62 | \setcounter{enumi}{3} 63 | \item {\bf Surface Emission from the Crab Pulsar:} The neutron star 64 | that powers the Crab Pulsar can be assumed to have a mass of 65 | $1.4\mathrm{M}_\odot$ and a radius of 10~km with constant internal 66 | density and an effective temperature of $10^6$~K. The frequency of 67 | the Crab Pulsar is 30~Hz and its period increases by 38 ns each 68 | day. Compare the power from the surface emission to the power 69 | lost as the neutron star spins down. The total power of the Crab 70 | Nebulae is about 75,000 times that of the Sun. What is the likely 71 | source of this power? 72 | 73 | {\bf Answer:} 74 | 75 | The blackbody flux from the surface of the star is given by 76 | \begin{equation} 77 | F = 4 \pi R^2 \sigma T^4 = 7 \times 10^{32}~\mathrm{erg/s} = 7 \times 10^{25}~\mathrm{W} = 0.17 \mathrm{L}_\odot. 78 | \end{equation} 79 | 80 | As the neutron star spins down it loses kinetic energy at a rate 81 | \begin{equation} 82 | \frac{dE}{dt} = - I \Omega \dot \Omega = - 4\pi^2 \nu^3 I {\dot P} = 83 | -5 \times 10^{38}~\mathrm{erg/s} = -5 \times 10^{31}~\mathrm{W} = 10^{5}\mathrm{L}_\odot 84 | \end{equation} 85 | where $I \approx \frac{2}{5} M R^2 \approx 10^{45} \mathrm{g cm}^2$. 86 | The spin-down power is approximately the power needed to power the 87 | nebula so it is a possible source of energy. 88 | 89 | \item {\bf Power-Law Atmosphere} 90 | 91 | Assume the following 92 | \begin{itemize} 93 | \item The Rosseland mean opacity is related to the density and 94 | temperature of the gas through a power-law relationship, 95 | \begin{equation} 96 | \kappa_R = \kappa_0 \rho^\alpha T^\beta; 97 | \end{equation} 98 | \item The pressure of the gas is given by the ideal gas law; 99 | \item The gas is in hydrostatic equilibrium so $p=g\Sigma$ where 100 | $g$ is the surface gravity; and 101 | \item The gas is in radiative equilibrium with the radiation 102 | field so the flux is constant with respect to $z$ or $\Sigma$. 103 | \end{itemize} 104 | Calculate the temperature of the gas as a function of $\Sigma$. 105 | 106 | {\bf Answer:} 107 | 108 | First we take the equation of radiative transfer 109 | 110 | \begin{equation} 111 | F(z) = -\frac{16 \sigma T^3}{3\kappa_R} \frac{\partial T}{\partial \Sigma} = 112 | \frac{16 \sigma T^3}{3\kappa_0 \rho^\alpha T^\beta} \frac{\partial T}{\partial \Sigma} 113 | \end{equation} 114 | We eliminate the variable $\rho$ using the ideal gas law and the equation of 115 | hydrostatic equilibrium, 116 | \begin{equation} 117 | g_s \Sigma = \frac{1}{\mu m_p} \rho k T 118 | \end{equation} 119 | so we have 120 | \begin{equation} 121 | \frac{\partial T}{\partial \Sigma} = \frac{3\kappa_0 }{16 \sigma F} 122 | \Sigma^\alpha T^{\beta-\alpha-3} \left ( \frac{\mu m_p}{g_s k} \right )^\alpha 123 | \end{equation} 124 | which can be integrated by the separation of variables to yield 125 | \begin{equation} 126 | \frac{T^{4+\alpha-\beta}}{4+\alpha-\beta} = \frac{3\kappa_0 }{16 \sigma F} 127 | \frac{\Sigma^{\alpha+1}}{\alpha+1} \left ( \frac{\mu m_p}{g_s k} \right )^\alpha 128 | \end{equation} 129 | \setcounter{enumi}{6} 130 | \item{\bf Goggles} 131 | 132 | Calculate from thermodynamic principles how much objects are magnified or demagnified while viewed through goggles underwater. N.B. The wavenumber of a photon of a given frequency is proportional to the index of refraction. 133 | 134 | {\bf Answer:} 135 | 136 | If we have a blackbody underwater and a blackbody in air at equal temperatures, the underwater blackbody will emit 137 | \begin{equation} 138 | F_{water} = n^2 F_{air} 139 | \end{equation} 140 | energy per unit area per unit time. You can see this from the definition of the density of states 141 | \begin{equation} 142 | \rho_s = 4\pi k^2 d k = 4\pi \left ( \frac{n \nu}{c} \right )^2 d \left (\frac{n \nu}{c} \right ) 143 | \end{equation} 144 | which is larger by a factor of $n^3$, so the energy density within the water of the blackbody radiation is larger by a factor of $n^3$ than in air. However, flux is related to the intensity which is energy density times the velocity so the flux is only larger by a factor of $n^2$. 145 | 146 | For the underwater blackbody to absorb as much as radiation from the 147 | blackbody in air as the blackbody in air receives from it, the solid 148 | angle subtended by the underwater BB must be larger by $n^2$ 149 | so it is magnified linearly by a factor of$n\approx 1.33$. 150 | \end{enumerate} 151 | 152 | \ifx\bookloaded\undefined 153 | \end{document} 154 | \end 155 | \fi 156 | 157 | %%% Local Variables: 158 | %%% TeX-master: "book" 159 | %%% End: -------------------------------------------------------------------------------- /solutions/chap12sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \ifx\pdftexversion\undefined 4 | \usepackage[dvips]{graphicx} 5 | \else 6 | \usepackage[pdftex]{graphicx} 7 | \fi 8 | \newcommand{\be}{\begin{equation}} 9 | \newcommand{\ee}{\end{equation}} 10 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 11 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 12 | \newcommand{\comment}[1]{\relax} 13 | \newcommand{\dd}[2]{\frac{d {#1}}{d {#2}}} 14 | \newcommand{\pp}[2]{\frac{\partial {#1}}{\partial {#2}}} 15 | \begin{document} 16 | \fi 17 | 18 | \section{Chapter 12} 19 | \begin{enumerate} 20 | \item{\bf Shock Entropy} 21 | Show that the entropy of the fluid increases as it passes 22 | through a shock. Hint: the equation of state of an isentropic fluid 23 | is $P = K\rho^\gamma$ where the value of $K$ increases with 24 | increasing entropy. 25 | 26 | {\bf Answer:} 27 | 28 | The simplest way to solve this is to look at the shock adiabat and see 29 | that the entropy increases along it, but let's be a bit more 30 | rigorous. The value of $K$ is a function of entropy alone, so let's 31 | look at how $K$ changes across the shock. Specifically what is 32 | $P/\rho^\gamma$ on each side of the shock? We have 33 | \begin{equation} 34 | \frac{\rho_2}{\rho_1} = \frac{(\gamma+1) 35 | M_1^2}{2 + M_1^2 (\gamma-1)}, 36 | \frac{P_2}{P_1} = \frac{1-\gamma + 2 M_1^2 \gamma }{(\gamma+1)} 37 | \end{equation} 38 | so 39 | \begin{equation} 40 | \frac{K_2}{K_1} = \frac{1-\gamma + 2 M_1^2 \gamma }{(\gamma+1)} 41 | \left [ \frac{(\gamma+1) 42 | M_1^2}{2 + M_1^2 (\gamma-1)} \right ]^{-\gamma} 43 | \end{equation} 44 | Let's expand this ratio for $M_1^2 \approx 1$ to understand the change 45 | in entropy for a weak shock, 46 | \begin{equation} 47 | \frac{K_2}{K_1} = 1 + \frac{2\gamma(\gamma-1)}{3(\gamma+1)^2} \left (M_1^2 - 1\right 48 | )^3 + {\cal O} \left (M_1^2 - 1\right)^4. 49 | \end{equation} 50 | The value of $K$ increases across the shock for $\gamma>1$, therefore 51 | the entropy increases. To make this precise we know that for an ideal 52 | gas, $s = c_V \ln K + s_0$, so 53 | \begin{equation} 54 | \Delta s = c_v \frac{\Delta K}{K} = 55 | \frac{2\gamma(\gamma-1)}{3(\gamma+1)^2} \left (M_1^2 - 1\right )^3 c_V 56 | + + {\cal O} \left (M_1^2 - 1\right)^4. 57 | \end{equation} 58 | 59 | \item{\bf Bomb Yield} 60 | 61 | Fig.~\ref{fig:tumbler} shows shocked air heated to incandescence 62 | about two milliseconds after the detonation of a nuclear 63 | bomb. The height of the device was 90~meters. What 64 | was the approximate yield of the device? 65 | 66 | {\bf Answer: } 67 | 68 | Eq.~\ref{eq:710}. 69 | 70 | \item{\bf Relativistic Shock} 71 | 72 | Find the incoming and outgoing velocity of a relativistic shock in 73 | terms of the energy density and pressure on either side of the 74 | shock. 75 | 76 | {\bf Answer:} 77 | 78 | Start with Eq.~\ref{eq:855} and~\ref{eq:856}, 79 | \begin{equation} 80 | w_1 U_1 \gamma_1= w_2 U_2 \gamma_2, w_1 U_1^2 + p_1 = w_2 U_2^2 + p_2. 81 | \end{equation} 82 | Let's use the second equation to solve for $U_2^2$ 83 | \begin{equation} 84 | U_2^2 = \frac{w_1 U_1^2 + p_1 - p_2}{w_2} 85 | \end{equation} 86 | and rewrite $\gamma_2^2$ in terms of $U_2$ 87 | \begin{equation} 88 | \gamma_2^2 = \frac{1}{1-\beta^2} = \frac{\beta^2 + 1 - 89 | \beta^2}{1-\beta^2} = U_2^2 + 1. 90 | \end{equation} 91 | The square of the first equation yields 92 | \begin{eqnarray} 93 | w_1^2 U_1^2 \left ( U_1^2 + 1 \right ) &=& w_2^2 U_2^2 \left ( U_2^2 + 94 | 1 \right ) \\ 95 | w_1^2 U_1^2 \left ( U_1^2 + 1 \right ) &=& w_2^2 \frac{w_1 U_1^2 + p_1 - 96 | p_2}{w_2} \left ( \frac{w_1 U_1^2 + p_1 - p_2}{w_2} + 1 \right ) \\ 97 | U_1^2 &=& \frac{(p_1-p_2)^2 - p_1 w_2 - p_2 w_2}{w_1 \left [ w_2 - w_1 98 | + 2 ( p_1 - p_2 ) \right ]} \\ 99 | \left ( \frac{v_1}{c} \right )^2 &=& 100 | \frac{(p_1-p_2)(p_1-p_2+w_2)}{(p_1-p_2-w_1)(p_1-p_2+w_2-w_1)} \\ 101 | \left ( \frac{v_1}{c} \right )^2 &=& 102 | \frac{(p_2-p_1)(\epsilon_2+p_1)}{(\epsilon_2-\epsilon_1)(\epsilon_1+p_2)} 103 | \end{eqnarray} 104 | and we obtain $v_2$ by swapping the one and two indicies in the 105 | previous equaiton, yielding 106 | \begin{eqnarray} 107 | \frac{v_1}{c} &=& 108 | \sqrt{\frac{(p_2-p_1)(\epsilon_2+p_1)}{(\epsilon_2-\epsilon_1)(\epsilon_1+p_2)}} \\ 109 | \frac{v_2}{c} &=& 110 | \sqrt{\frac{(p_2-p_1)(\epsilon_1+p_2)}{(\epsilon_2-\epsilon_1)(\epsilon_2+p_1)}} 111 | \end{eqnarray} 112 | 113 | \item{\bf Relativistic Bernoulli} 114 | 115 | Find the relativistic generalisation of Bernoulli's equation for a 116 | streamline (you can neglect gravitiy). 117 | 118 | {\bf Answer:} 119 | 120 | For the Bernoulli equaion we must assume that all time derivatives 121 | vanish and look at the properties of the fluid along a flow line. We 122 | can use the shock jump conditions as a starting point, 123 | (e.g. Eq.~\ref{eq:854} and~\ref{eq:855}), because they must hold along 124 | a streamline as well as across a discontinuity. We have 125 | \begin{equation} 126 | U n = \textrm{constant}, w U \gamma = \textrm{constant}. 127 | \end{equation} 128 | Using the first equation to eliminate $U$ from the second yields 129 | \begin{equation} 130 | \frac{\gamma w}{n} = \textrm{constant}. 131 | \end{equation} 132 | This doesn't look much like the non-relativistic Bernoulli equation. 133 | Let's make some substitutions. We have 134 | \begin{equation} 135 | \frac{1}{n} \left ( 1 + \frac{v^2}{2 c^2}\right ) \left ( \rho c^2 + w_{NR,V} 136 | \right ) + \textrm{Higher order in velocity} = \textrm{constant}. 137 | \end{equation} 138 | Now let's divide both sides by the rest mass of the particles 139 | \begin{equation} 140 | \frac{1}{\rho} \left ( 1 + \frac{v^2}{2 c^2}\right ) \left ( \rho c^2 + w_{NR,V} 141 | \right ) + \textrm{Higher order in velocity} = \textrm{constant} 142 | \end{equation} 143 | and expand, dropping higher-order terms 144 | \begin{equation} 145 | c^2 + \frac{v^2}{2} + \frac{w_{NR,V}}{\rho} = \textrm{constant} 146 | \end{equation} 147 | and 148 | \begin{equation} 149 | \frac{v^2}{2} + w = \textrm{constant} 150 | \end{equation} 151 | where $w$ is the enthalpy per unit mass. This is the non-relativistic 152 | Bernoulli equation. For the classical result with an incompressible fluid 153 | we have $w=P/\rho$. 154 | 155 | \item{\bf Bathtub Physics} 156 | 157 | When water flows into a bathtub, a circular hydraulic jump forms 158 | around the incoming stream of water. If you assume that the flow 159 | rate is constant and the flow is initially vertical, calculate the 160 | height of the water downstream of the jump as a function of the 161 | radius of the jump and the flow rate. You may neglect friction. If 162 | the bathtub is large compared to the radius of the jump and the 163 | walls are vertical, how does the radius of the jump change with 164 | time? 165 | 166 | {\bf Answer:} 167 | 168 | Here the flow rate and downstream height are given. We have 169 | \begin{equation} 170 | j = \frac{Q}{2 \pi r} 171 | \end{equation} 172 | where $Q$ is the volumetric flow rate and $r$ is the radius of the 173 | jump. What is the height of the upsteam flow $h_1$ in terms of the 174 | downstream height $h_2$? We have 175 | \begin{equation} 176 | v_1^2 h_1 + \frac{1}{2} g h_1^2 = v_2^2 h_2 + \frac{1}{2} g h_2^2 177 | \end{equation} 178 | so 179 | \begin{equation} 180 | \frac{j^2}{h_1} + \frac{1}{2} g h_1^2 = \frac{j^2}{h_2} + \frac{1}{2} g h_2^2 181 | \end{equation} 182 | and with rearranging 183 | \begin{equation} 184 | j^2 \left ( h_1 - h_2 \right ) + \frac{1}{2} g \left ( h_2^3 h_1 - 185 | h_1^3 h_2 \right ) = 0. 186 | \end{equation} 187 | We can factor this to give 188 | \begin{equation} 189 | \frac{1}{2} \left( h_1 - h_2 \right ) \left ( g h_2 h_1^2 + g h_2^2 h_1 - 2 190 | j_2 \right ) = 0 191 | \end{equation} 192 | so we have the positive solutions 193 | \begin{equation} 194 | h_1 = h_2, h_1 = \frac{h_2}{2} \left ( \sqrt{ 1 + \frac{8 j^2}{g 195 | h_2^3}} - 1 \right ). 196 | \end{equation} 197 | Now we use that $v_1$ is constant to eliminate $h_1=j/v_1=Q/(2\pi r 198 | v_1)$. Furthermore, $h_2=Q t/A$ where $A$ is the cross-sectional area 199 | of the jump. Putting all of these into the equation above yields 200 | \begin{equation} 201 | \frac{Q}{2\pi r v_1} = \frac{Q t}{2 A} \left ( \sqrt{ 1 + 202 | \frac{8}{g} \left (\frac{Q}{2\pi r} \right )^2 \left ( \frac{A}{Q t} \right)^3 203 | } - 1 \right ). 204 | \end{equation} 205 | and solving for the radius $r$ yields 206 | \begin{equation} 207 | r = \frac{A \left ( 2 A v_1^2 - t Q g \right )}{2 \pi t^2 Q g v_1} 208 | = \frac{A^2 v_1}{Q g \pi} \frac{1}{t^2} - \frac{A}{2\pi v_1} \frac{1}{t}. 209 | \end{equation} 210 | \end{enumerate} 211 | 212 | \ifx\bookloaded\undefined 213 | \end{document} 214 | \end 215 | \fi 216 | %%% Local Variables: 217 | %%% TeX-master: "book" 218 | %%% End: -------------------------------------------------------------------------------- /solutions/chap11sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \ifx\pdftexversion\undefined 4 | \usepackage[dvips]{graphicx} 5 | \else 6 | \usepackage[pdftex]{graphicx} 7 | \fi 8 | \newcommand{\be}{\begin{equation}} 9 | \newcommand{\ee}{\end{equation}} 10 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 11 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 12 | \newcommand{\comment}[1]{\relax} 13 | \newcommand{\dd}[2]{\frac{d {#1}}{d {#2}}} 14 | \newcommand{\pp}[2]{\frac{\partial {#1}}{\partial {#2}}} 15 | \begin{document} 16 | \fi 17 | 18 | \section{Chapter 11} 19 | \begin{enumerate} 20 | 21 | \item{\bf Maximum Flux} 22 | 23 | Calculate from the Euler equation and the continuity equation, at what 24 | velocity does the flux ($\rho V$) reach its maximum for fluid flowing 25 | through a tube of variable cross-sectional area? 26 | At which velocities does the flux vanish? You can consider the flow 27 | to be adiabatic. 28 | 29 | {\bf Answer:} 30 | 31 | From Euler's equation we have 32 | \begin{equation} 33 | ({\bf v} \cdot \nabla ){\bf v}=-\frac{\nabla P}{\rho} 34 | \end{equation} 35 | so 36 | \begin{equation} 37 | v \dd{v}{x} = -\frac{1}{\rho} \dd{P}{x} 38 | \end{equation} 39 | and we have 40 | \begin{equation} 41 | \dd{P}{\rho} = c_s^2 42 | \end{equation} 43 | Combining these two gives 44 | \begin{equation} 45 | \dd{\rho}{v} = -\rho \frac{v}{c_s^2} 46 | \end{equation} 47 | We have 48 | \begin{equation} 49 | \dd{(\rho v)}{v} = \rho + v \dd{\rho}{v} = \rho \left ( 1 - 50 | \frac{v^2}{c_s^2} \right ) 51 | \end{equation} 52 | This function reaches an extremum at $v=c_s$. Because the flux is 53 | zero for $v=0$ and increases with $v$ for $v \ll c_s$, this must be a 54 | maximum for $jv$. 55 | 56 | If we assume that the sound speed is constant (isothermal gas), 57 | this integrates to give 58 | \begin{equation} 59 | \rho v = \rho_0 v e^{-v^2/(2c_s^2)} 60 | \end{equation} 61 | where $\rho_0$ is the density at zero velocity. This has a maximum of 62 | $\rho_0 c e^{-1/2}$ at 63 | $v=c_s$. We can let the gas expand and accelerate to arbitrarily high 64 | velocities. In a more realistic situation, the sound speed is a 65 | function of density 66 | \begin{equation} 67 | c_s^2 = c_{s,0}^2 \left ( \frac{\rho}{\rho_0} \right )^{\gamma-1}, 68 | \end{equation} 69 | and we have 70 | \begin{equation} 71 | \dd{j}{v} = \rho \left [ 1 - 72 | \frac{v^2}{c_{s,0}^2} \left ( \frac{\rho}{\rho_0} \right )^{1-\gamma} \right ] 73 | = \frac{j}{v} \left [ 1 - 74 | \frac{v^2}{c_{s,0}^2} (\rho_0 v)^{\gamma-1} j^{1-\gamma} \right ] 75 | \end{equation} 76 | with $j=\rho v$. This differential equation has the following 77 | solution 78 | \begin{equation} 79 | j(v) = \rho_0 v \left [ 1 + \left ( 1 - \gamma \right ) 80 | \frac{v^2}{2 c_{s,0}^2} \right ]^{1/(\gamma-1)} 81 | \end{equation} 82 | If we take $\gamma\rightarrow 1$ we get the solution above for the 83 | isothermal case. The flux reaches a maximum of 84 | \begin{equation} 85 | j_\mathrm{max} = \rho_0 c_{s,0} \left ( \frac{2^\gamma}{\gamma+1} 86 | \right )^{1/(\gamma-1)} \frac{1}{\sqrt{2\gamma+2}} 87 | \end{equation} 88 | at a velocity of 89 | \begin{equation} 90 | v = \sqrt{\frac{2}{\gamma+1}} c_{s,0}. 91 | \end{equation} 92 | Unlike the isothermal case, the flux vanishes at 93 | $v=0$ and $v=c_{s,0}\sqrt{2/(\gamma-1)}$. How can we understand this 94 | second velocity when the flux vanishes? Along a streamline of the 95 | gas we have 96 | \begin{equation} 97 | \frac{v^2}{2} + w = w_0 = \frac{P + \epsilon}{\rho} = \frac{P + 98 | (\gamma-1)^{-1} P}{\rho} = \frac{\gamma}{\gamma-1} \gamma^{-1} c_{s,0}^2 99 | \end{equation} 100 | The maximum velocity that the gas can attain is 101 | \begin{equation} 102 | v = \sqrt{2 w_0} = \sqrt{2 c_{s,0}^2/(\gamma-1)} = c_{s,0} \sqrt{2/(\gamma-1)} 103 | \end{equation} 104 | % Since we have a solution for $j(v)$, it would be nice to find the 105 | % velocity at which the momentum flux is maximized for a given mass 106 | % flux. The momentum flux is 107 | % \begin{equation} 108 | % f(v) = P + \rho v^2 = P + j(v) v = \frac{c_s^2 \rho}{\gamma} + j(v) v 109 | % = \frac{c_s^2 j(v)}{v \gamma} + j(v) v 110 | % \end{equation} 111 | % Let's take 112 | % \begin{equation} 113 | % \left . \dd{f}{v}\right|_{j(v)} = j(v) \left [ \frac{1}{\gamma} \left 114 | % ( \left(\gamma-1\right ) \frac{c_s^2}{\rho v} \left (-\rho 115 | % \frac{v}{c_s^2}\right ) - \frac{c_s^2}{v^2} \right ) + 1 \right ] 116 | % \end{equation} 117 | % \begin{equation} 118 | % \left . \dd{f}{v}\right|_{j(v)} = j(v) \left [ 1 - \frac{\gamma-1 + 119 | % \frac{c_s^2}{v^2}}{\gamma} \right ] 120 | % \end{equation} 121 | % so 122 | % \begin{equation} 123 | % v = c_s. 124 | % \end{equation} 125 | 126 | % The maximum momentum flux as a function of velocity (not for fixed mass 127 | % flux) occurs at a different velocity. We have 128 | % \begin{equation} 129 | % f(v) = \frac{c_s^2 \rho}{\gamma} - \rho v^2 130 | % \end{equation} 131 | % so 132 | % \begin{equation} 133 | % \dd{f}{v} = \left ( \frac{\gamma-1}{\gamma} c_s^2 + \frac{1}{\gamma} 134 | % c_s^2 \right - v^2) \left ( -\rho \frac{v}{c_s^2} \right ) - 2 v \rho= \rho v \left 135 | % (3-\frac{v^2}{c_s^2}\right ) 136 | % \end{equation} 137 | % and the maximum occurs at $v=\sqrt{3} c_s$. 138 | 139 | \item{\bf Stream Bed} 140 | 141 | Fig.~\ref{fig:channel} shows how the level of the surface changes 142 | for a flow passing over an obstacle. For an initial depth of 143 | $z_0=1$ and $g=10$ and a bump height of $y(x)=0.1 e^{-x^2}$, find the 144 | solutions to Bernoulli's equation (Eq.~\ref{eq:bern}) for $z$ as a 145 | function of $x$ and the initial velocity $v_0$. You may find 146 | several solutions for a given $x$. Also you should only worry 147 | about the positive real solutions for $z$. What are the values of 148 | the critical velocities $v_0$? 149 | 150 | {\bf Answer:} 151 | 152 | The solution follows from Eq.~\ref{eq:675} by plugging in the values 153 | of $y(x)$, $z_0=1$, $g=10$ and $v_0$ which you are going to vary to look 154 | at the different solutions. This yields 155 | \begin{equation} 156 | A = \frac{v_0^2}{2}, B=\frac{v_0^2}{2} + 10 \left [ 1 - 0.1 e^{-x^2} 157 | \right ], C = 10. 158 | \end{equation} 159 | Next we use Eq.~\ref{eq:cubic_sol} to find the value of $\cos 3 t$. 160 | This equation will yield several values of $3t$ because the cosine 161 | function is symmetric and periodic. They are 162 | \begin{equation} 163 | 3 t = 3 t_1, 3 \left (-t_1 \right ), 164 | 3 \left ( t_1 + \frac{2}{3} \pi \right ), 165 | 3 \left ( \frac{2}{3} \pi - t_1 \right ), 166 | 3 \left ( t_1 + \frac{4}{3} \pi \right ), 167 | 3 \left ( \frac{4}{3} \pi - t_1 \right ). 168 | \end{equation} 169 | Because we are interested in the value of $\cos t$, the first two 170 | results yield the same value. Let's draw a picture with the various 171 | possibilities numbered: 172 | 173 | \newcommand{\circlepic}[2]{ 174 | \begin{tikzpicture} 175 | \draw (0,0) circle (#2) 176 | (0,0) -- ++(#1:#2) node {1} 177 | (0,0) -- ++(-#1:#2) node {2} 178 | (0,0) -- ++(120+#1:#2) node {3} 179 | (0,0) -- ++(120-#1:#2) node {4} 180 | (0,0) -- ++(240+#1:#2) node {5} 181 | (0,0) -- ++(240-#1:#2) node {6} ; 182 | \end{tikzpicture} 183 | } 184 | \circlepic{25}{1.5} \circlepic{35}{1.5} \circlepic{60}{1.5} 185 | 186 | Because we are only interested in the $x-$coordinate (the cosine of 187 | the angle), we see that solutions 1 and 2, 3 and 6 and 4 and 5 are 188 | equivalent, so we only need to keep the solutions with $t$ between 189 | zero and $\pi$. In the left diagram we used $t=25^\circ$ so we only have 190 | a single value with $\cos t$ greater than zero. We can discard 191 | negative values of $\cos t$ because that would yield that the surface 192 | of the water lies underneath the surface of the bottom. 193 | 194 | For $t_1>30^\circ$ there are two positive solutions. The centre 195 | diagram has $t_1=35^\circ$. These solutions coincide for 196 | $t_1=60^\circ$ (right diagram). Where this condition holds the flow 197 | is travelling at the critical velocity. The value of $v_0$ that 198 | causes the flow to travel at the critical velocity over the peak of 199 | the bump is the critical value of $v_0$. In general, we only have to 200 | be concerned with solutions (1) and (4), the rest are repeats or 201 | negative. 202 | 203 | \item{\bf Sound Velocity} 204 | 205 | Show that for a linear sound wave {\em i.e.} one in which $\delta 206 | \rho \ll \rho$ that the velocity $v$ of fluid motion is much less 207 | than $c_s$. Estimate the maximum longitudinal fluid velocity in the 208 | case of a sound wave in air at STP in the case of a disturbance 209 | which sets up pressure fluctuations of order 0.1\%. 210 | 211 | {\bf Answer:} 212 | 213 | Starting with Eq.~\ref{eq:660} we can relate the velocity of the fluid 214 | in the wave to the pressure disturbance, 215 | \begin{equation} 216 | {\bf v}' = \frac{p'}{\rho_0} \frac{{\bf k} }{\omega}, v' = 217 | \frac{p'}{\rho_0} \frac{1}{c_s} = c_s \frac{\rho'}{\rho_0} = 218 | \frac{p'}{p_0} \frac{c_s}{\gamma} 219 | \end{equation} 220 | where $p' = c_s^2 \rho'$ because $c_s^2=\partial p/\partial \rho$. 221 | Furthermore, the adiabatic exponent is given by $\gamma=\partial \ln 222 | p/\partial \ln \rho=(\rho/p) c_s^2$. 223 | 224 | \end{enumerate} 225 | 226 | \ifx\bookloaded\undefined 227 | \end{document} 228 | \end 229 | \fi 230 | %%% Local Variables: 231 | %%% TeX-master: "book" 232 | %%% End: -------------------------------------------------------------------------------- /solutions/chap13sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \ifx\pdftexversion\undefined 4 | \usepackage[dvips]{graphicx} 5 | \else 6 | \usepackage[pdftex]{graphicx} 7 | \fi 8 | \newcommand{\be}{\begin{equation}} 9 | \newcommand{\ee}{\end{equation}} 10 | \newcommand{\rmmat}[1]{\hbox{\rm #1}} 11 | \newcommand{\rmscr}[1]{{\hbox{\rm \scriptsize #1}}} 12 | \newcommand{\comment}[1]{\relax} 13 | \newcommand{\dd}[2]{\frac{d {#1}}{d {#2}}} 14 | \newcommand{\pp}[2]{\frac{\partial {#1}}{\partial {#2}}} 15 | \begin{document} 16 | \fi 17 | 18 | \section{Chapter 13} 19 | \begin{enumerate} 20 | \item{\bf Exact Solutions} 21 | 22 | For which values of $\gamma$ can the Bernoulli equation 23 | (Eq.~\ref{eq:725}) be solved using elementary methods (linear, 24 | quadratic and cubic equations of the form in Eq.~\ref{eq:676}). There 25 | are many, however only a few have $1< \gamma < 5/3$. 26 | 27 | {\bf Answer: } 28 | 29 | Let's start with the Bernoulli equation, 30 | \begin{equation} 31 | \frac{v^2}{2} + \frac{c_s^2 - c_s^2(\infty)}{\gamma - 1} - \frac{G M}{r} = 0. 32 | \end{equation} 33 | First let's divide both sides by $c_s^2$ to get 34 | \begin{equation} 35 | \frac{1}{2} \frac{v^2}{c_s^2} + \frac{1 - c_s^2(\infty)/c_s^2}{\gamma - 36 | 1} - \frac{G M}{r c_s^2} = 37 | \frac{1}{2} \frac{v^2}{c_s^2} + \frac{1 - c_s^2(\infty)/c_s^2}{\gamma - 38 | 1} - \frac{r_c}{r} \frac{4}{5-3\gamma} \frac{c_s^2(\infty)}{c_s^2} = 0. 39 | \end{equation} 40 | We need to express the sound speed in terms of $r$ and $v$. We have 41 | \begin{equation} 42 | P = K \rho^\gamma 43 | \end{equation} 44 | so 45 | \begin{equation} 46 | c_s^2 = \gamma K \rho^{\gamma-1} = c_s^2(\infty) \left ( \frac 47 | {\rho}{\rho(\infty)} \right )^{\gamma - 1} 48 | \end{equation} 49 | We can relate $v$, $r$ and $\rho$ through ${\dot M}=4\pi r^2 v \rho$ 50 | We are interested in a plot of $y=v/c_s$ versus $x=r/r_c$, so let's 51 | substitute for $x$ and $y$ to get 52 | \begin{equation} 53 | \frac{y^2}{2} + \frac{1}{\gamma-1} - \left ( \frac{1}{\gamma-1} - \frac{1}{x} 54 | \frac{4}{5-3\gamma} \right ) \frac{c_s^2(\infty)}{c_s^2} = 0. 55 | \end{equation} 56 | Now we need to find $c_s^2(\infty)/c_s^2$ in terms of $x$ and $y$. We 57 | can determine $\rho$ at any point through ${\dot M}=4\pi \rho v$ and 58 | the formula above. 59 | The key is to write ${\dot M}=\alpha {\dot M}_\rmscr{crit}$. First, 60 | we have 61 | \begin{equation} 62 | \dot M = 4 \pi r^2 v \rho = 4 \pi \alpha r_c^2 c_s(r_c) \rho(r_c) 63 | \end{equation} 64 | so 65 | \begin{equation} 66 | \frac{\rho}{\rho(r_c)} = \frac{\alpha}{x^2 y} \frac{c_s(r_c)}{c_s} 67 | = \frac{\alpha}{x^2 y} \left ( 68 | \frac{\rho(r_c)}{\rho} \right )^{(\gamma-1)/2} = \left ( \frac{\alpha}{x^2 y} \right )^{2/(\gamma+1)}\kern-0.5cm. 69 | \end{equation} 70 | and 71 | \begin{equation} 72 | \frac{\rho}{\rho(\infty)} = \frac{\rho}{\rho(r_c)} 73 | \frac{\rho(r_c)}{\rho(\infty)} 74 | = \left ( \frac{\alpha}{x^2 y} \right )^{2/(\gamma+1)} \left ( \frac{2}{5-3\gamma} \right )^{1/(\gamma-1)} 75 | \end{equation} 76 | using Eq.~\ref{eq:729} and giving an expression for 77 | \begin{equation} 78 | \frac{c_s^2}{c_s^2(\infty)} = \left ( \frac{\alpha}{x^2 y} \right )^{2(\gamma-1)/(\gamma+1)} \frac{2}{5-3\gamma} 79 | \end{equation} 80 | Let's substitute this into the Bernoulli equation to yield 81 | \begin{eqnarray} 82 | y^{2(1-\gamma)/(\gamma+1)} \left ( \frac{y^2}{2} + \frac{1}{\gamma-1} 83 | \right ) \frac{2 \alpha^{2(\gamma-1)/(\gamma+1)}}{5 -3\gamma} - 84 | \nonumber \\ 85 | x^{4(\gamma-1)/(\gamma+1)}\left ( \frac{1}{\gamma-1} 86 | + \frac{4}{x(5 - 3\gamma)} \right ) 87 | =0 88 | \label{eq:789} 89 | \end{eqnarray} 90 | There could be many possibilities: we could solve for $x$ or $y$ in 91 | terms of the other, and the resulting equation could be a cubic, 92 | quadratic or linear in the three terms. Let's begin with solving for 93 | $x$. 94 | 95 | {\bf Solution for $x$:} 96 | 97 | On the second line there are two terms in $x$ that differ by a 98 | single power of $x$. We can make substitutions of the form $x=u^{\pm 2}$, 99 | $x=w^{\pm 3}$ that will transform the equation into a quadratic or cubic 100 | with the correct choice of $\gamma$, or we could solve for $x$ directly 101 | which would require that $4(\gamma-1)/(\gamma+1)=2,1,0,-1,-2$, 102 | yielding quadratic, linear, linear, quadratic and cubic equations, 103 | respectives. A value of $4(\gamma-1)/(\gamma-1)=3$ would also yield a 104 | cubic equation but not of the simply solvable form 105 | (Eq.~\ref{eq:676}). There are also simply solvable quartics, but this 106 | is beyond the scope of the question. 107 | \begin{table}[h] 108 | \centering 109 | \begin{tabular}{ccccc} 110 | $\frac{4(\gamma-1)}{\gamma-1}$ & $\gamma$ & Type & Substitution & Comment \\ \hline 111 | 2 & 3 & quadratic & $x$ & Non-Ideal \\ 112 | 1 & $5/3$ & linear & $x$ & Divergent \\ 113 | 0 & 1 & linear & $x$ & Divergent \\ 114 | $-1$ & $3/5$ & quadratic & $x$ & Unphysical\\ 115 | $-2$ & $1/3$ & cubic & $x$ & Unphysical \\ 116 | $1/2$ & $9/7$ & quadratic & $x=u^2$ & Good \\ 117 | $-1/2$ & $7/9$ & cubic & $x=u^{-2}$ & Unphysical \\ 118 | $2/3$ & $7/5$ & cubic & $x=w^3$ & Good \\ 119 | $1/3$ & $13/11$ & cubic & $x=w^{-3}$ & Good \\ 120 | \end{tabular} 121 | \end{table} 122 | 123 | {\bf Solution for $y$:} 124 | 125 | We can also solve for $y$ in terms of $x$. Here the two terms in $y$ 126 | differ by two powers of $y$; this naturally leads to a quadratic 127 | without substitution. Some of the various possibilities are listed in 128 | the following table. 129 | \begin{table}[h] 130 | \centering 131 | \begin{tabular}{ccccc} 132 | $\frac{2(1-\gamma)}{\gamma+1}$ & $\gamma$ & Type & Substitution & 133 | Comment \\ \hline 134 | 2 & 0 & quadratic & $w=y^2$ & Unphysical \\ 135 | 1 & $1/3$ & cubic & $y$ & Unphysical \\ 136 | 0 & 1 & linear & $w=y^2$ & Divergent \\ 137 | $-2/3$ & 2 & cubic & $w^3=y^2$ & Non-Ideal \\ 138 | -1 & 3 & quadratic & $y$ & Non-Ideal \\ 139 | $-4/3$ & 5 & cubic & $w^{-3}=y^2$ & Non-Ideal \\ 140 | \end{tabular} 141 | \end{table} 142 | 143 | \setcounter{enumi}{3} 144 | \item{\bf Bondi Solution} 145 | 146 | Generate a picture like the figure in the lecture notes for the Bondi 147 | solution to spherical accretion. Use $\gamma=9/7$. 148 | 149 | {\bf Answer: } 150 | 151 | The answer starts as the first question up to the Bernoulli equation 152 | (Eq.~\ref{eq:789}) where we substitute $\gamma=9/7$ to yield 153 | 154 | \begin{equation} 155 | \frac{7}{2} x^{1/2} - \frac{7}{2} x^{-1/2} - \frac{7}{8} \alpha^{1/4} 156 | y^{7/4} - \frac{49}{8} \frac{\alpha^{1/4}}{y^{7/4}} = 0 157 | \end{equation} 158 | Let $u=\sqrt{x}$ and multiply everything by $u$ to give a quadratic 159 | equation for $u$ 160 | \begin{equation} 161 | \frac{7}{2} u^2 - \left [ \frac{7}{8} \frac{\alpha^{1/4}}{y^{1/4}} 162 | (y^2 + 7 ) \right ]u - 163 | \frac{7}{2} = 0 164 | \end{equation} 165 | 166 | \item{\bf Bondi Solution --- Harder} 167 | 168 | Generate a picture like the figure in the lecture notes for the Bondi 169 | solution to spherical accretion. Use $\gamma=7/5$. 170 | 171 | {\bf Answer: } 172 | 173 | The answer starts as the first question up to the Bernoulli equation 174 | (Eq.~\ref{eq:789}) where we substitute $\gamma=7/5$ to yield 175 | \begin{equation} 176 | \frac{5}{2} x^{2/3} - 5 x^{-1/3} - \frac{5}{4} \alpha^{1/3} y^{5/3} - 177 | \frac{25}{4} \frac{\alpha^{1/3}}{y^{1/3}} = 0. 178 | \end{equation} 179 | Let $w^3=x$ and multiply everything by $u$ to give a cubic 180 | equation for $w$ 181 | \begin{equation} 182 | \frac{5}{2} w^3 - \frac{5}{4} \frac{\alpha^{1/3}}{y^{1/3}} \left ( y^2 183 | - 5 \right ) w - 5= 0. 184 | \end{equation} 185 | of the form Eq.~\ref{eq:676}. This equation can be solved analytically 186 | 187 | \item{\bf Accretion Energetics} 188 | 189 | \begin{enumerate} 190 | \item $T=\frac{G M m}{R}$ 191 | \item $T=\frac{G M m}{2 R}$ 192 | \item $T_\textrm{\scriptsize NS}/m=2 \times 10^{20}$ erg/g, 193 | $T_\textrm{\scriptsize WD}/m=8 194 | \times 10^{16}$ erg/g. The accretion energy for a neutron star 195 | greatly exceeds the nuclear energy. The opposite holds for a white 196 | dwarf. 197 | \item The total energy per gram is essentially the value given in part 198 | c). The Eddington luminosity is $1.8\times 10^{38}$~erg/s for a 199 | neutron star (see problem 3.6). This yields an Eddington accretion 200 | rate of approximately $10^{18}$~g/s. 201 | \end{enumerate} 202 | 203 | \item{\bf A simplified accretion disk} 204 | 205 | \begin{enumerate} 206 | \item $dE = -\frac{G M dm}{2 r}$ 207 | \item $\frac{d}{dr} dE = \frac{G M dm}{2 r^2}$, $\frac{dE}{dr dt}= 208 | \frac{GM}{2 r^2} \frac{dm}{dt}$ 209 | \item $\frac{dE}{dA dt}= \frac{GM}{4 \pi r^3} \frac{dm}{dt}$ 210 | \item $\sigma T^4 = \frac{GM}{4 \pi r^3} \frac{dm}{dt}$ 211 | \item $\frac{dE}{dt} = \int_{r_0}^{r_A} \frac{GM}{2 r^2} 212 | \frac{dm}{dt} dr = \frac{GM}{2} \frac{dm}{dt} \left ( r_0^{-1} - 213 | r_A^{-1} \right )$, the peak temperature is at $r_0$ and the 214 | minimum temperature is at $r_A$. 215 | \item To sketch the spectrum we will assume that the BB emission 216 | emerges exclusively at the peak of the BB, so we need to translate 217 | $dE/(dr dt)$ to $dE/(dT dt)$. 218 | $$ 219 | \frac{dP}{dT} = \frac{dP}{dr} \frac{dr}{dT} = 220 | \frac{GM}{2 r^2} \frac{dm}{dt} \left ( \frac{\sigma T^3 4 \pi r^4}{3 221 | G M {\dot m}} \right ) = \frac{4 \pi r^2 \sigma T^3}{3} 222 | $$ 223 | and substituting yields 224 | $$ 225 | \frac{dP}{dT} = \frac{4\pi \sigma}{3} \left ( \frac{G M}{4\pi \sigma} 226 | {\dot m} \right )^{2/3} T^{1/3} 227 | $$ 228 | or $F_\nu \propto \nu^{1/3}$. 229 | \item If the accretion rate exceeds the Eddington rate, some matter 230 | must be expelled. 231 | \item Viscosity 232 | \end{enumerate} 233 | \end{enumerate} 234 | 235 | \ifx\bookloaded\undefined 236 | \end{document} 237 | \end 238 | \fi 239 | 240 | %%% Local Variables: 241 | %%% TeX-master: "book" 242 | %%% End: -------------------------------------------------------------------------------- /solutions/chap4sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass[pdftex,10pt]{article} 3 | \begin{document} 4 | \fi 5 | \section{Chapter 4} 6 | \begin{enumerate} 7 | \item{\bf The Ladder and the Barn: A Spacetime Diagram:} 8 | 9 | This problem will work best if you have a sheet of graph paper. 10 | In a spacetime diagram one draws a particular coordinate (in our case 11 | $x$) along the horizontal direction and the time coordinate 12 | vertically. People also generally draw the path of a light ray at 13 | 45$^\circ$. This sets the relative units of the two axes. 14 | \begin{enumerate} 15 | \item Draw a spacetime diagram and label the axes $x$ and 16 | $t$. The $t$-axis is the path of Emma through the spacetime. 17 | 18 | \item Draw the world line of someone travelling at 19 | $\frac{3}{5}$ of the speed of light. The world line should 20 | intersect with the origin of the spacetime diagram. Label this 21 | world line $t'$. The $t'$-axis is the path of Kara through the 22 | spacetime. 23 | 24 | \item Draw the $x'$ axis on the graph. Here's a hint about 25 | where it should go. The light ray bisects the angle between the 26 | $x$ and $t$ axes. Kara who is travelling along $t'$ will find that 27 | the speed of light is the same for her, so the light ray must also 28 | bisect the angle between $x'$ and $t'$. 29 | 30 | \item Parallel to Emma's time axis draw the walls of the barn 31 | in pencil. The barn is 4.5 meters wide in Emma's frame. 32 | 33 | \item Draw Kara's ladder along Kara's $x$-axis. The ladder 34 | is 5 meters long in Kara's frame. How long is it in Emma's frame. 35 | 36 | \item Draw the world lines of the ends of Kara's ladder. 37 | These lines are parallel to Kara's time axis. 38 | 39 | \item Erase a portion of the barn walls to allow Kara's ladder 40 | to fit through. 41 | 42 | \item Using the diagram, explain how Kara and Emma can 43 | understand how the too-long ladder fits in the too-small barn. 44 | \end{enumerate} 45 | \medskip 46 | 47 | \begin{picture}(200,200)(0,0) 48 | \put(-10,205){$t$ Emma} 49 | \put(0,0){\line(0,1){200}} 50 | \put(0,0){\line(1,0){200}} 51 | \put(210,0){$x$} 52 | \put(0,0){\line(3,5){120}} 53 | \put(110,205){$t'$ Kara} 54 | \put(0,0){\line(5,3){175}} 55 | \put(175,105){ $x'$} 56 | \put(50,-10){\line(0,1){210}} 57 | \put(95,-10){\line(0,1){210}} 58 | \put(55,200){Barn} 59 | \put(53,40){\line(5,3){40}} 60 | \put(53,40){\line(3,5){100}} 61 | \put(93,64){\line(3,5){90}} 62 | \put(53,40){\line(-3,-5){24}} 63 | \put(93,64){\line(-3,-5){45}} 64 | \put(30,36){\vector(1,0){20}} 65 | \put(30,-8){\vector(1,0){20}} 66 | \put(115,67){\vector(-1,0){20}} 67 | \put(115,110){\vector(-1,0){20}} 68 | \end{picture} 69 | 70 | \medskip 71 | 72 | Erase the sections between the arrows. Emma sees the ladder inside 73 | the barn with the two doors closed at the same time. Kara sees the 74 | forward door open before the back door has shut. 75 | 76 | \item{\bf The Fermi Process:} 77 | 78 | One model to understand how cosmic rays are accelerated is through 79 | shocks. The main idea is that a charge particle can cross a shock and 80 | turned around by the tangled mangetic field and recross the shock. 81 | Each time the charge does this it gains energy. 82 | 83 | To understand this let's use a simplified model in which two mirrors 84 | are travelling toward each other at some velocity $v$. When a 85 | particle hits the mirror, its energy in the frame of the mirror 86 | remains unchanged but its velocity and therefore the spacelike 87 | components of the four-momentum change sign. 88 | \begin{enumerate} 89 | \item Draw a diagram with the two mirrors. 90 | \bigskip 91 | 92 | \begin{picture}(80,80)(0,0) 93 | \put(60,40){\line(0,1){50}} 94 | \put(60,65){\vector(1,0){20}} 95 | \put(80,65){$v= \beta c$} 96 | \put(240,40){\line(0,1){50}} 97 | \put(240,65){\vector(-1,0){20}} 98 | \put(180,65){$v=-\beta c$} 99 | \end{picture} 100 | 101 | \item For argument's 102 | sake, let's first focus on the mirror on the left and consider that 103 | the mirror on the right is moving. What is the four-velocity in this 104 | frame of the mirror on the left ($U_{l}^\mu$)? What is the four-velocity in this 105 | frame of the mirror on the right ($U_{r}^\mu$)? 106 | 107 | \begin{center} 108 | $ 109 | U_{l}^\mu = \left [ \begin{array}{c} c \\ 0 \\ 0 \\ 0 \end{array} 110 | \right ] 111 | $ 112 | \hspace{1in} 113 | $ 114 | U_{r}^\mu = \left [ \begin{array}{c} \gamma c \\ -\gamma v \\ 0 \\ 0 \end{array} 115 | \right ] 116 | $ 117 | \end{center} 118 | 119 | \item Now let's focus on the mirror on the right and consider that 120 | the mirror on the left is moving. What is the four-velocity in this 121 | frame of the mirror on the left ($U'^\mu_l$)? What is the 122 | four-velocity in this frame of the mirror on the right ($U'^\mu_r$)? 123 | 124 | \begin{center} 125 | $ 126 | U_{l}^\mu = \left [ \begin{array}{c} \gamma c \\ \gamma v \\ 0 \\ 0 \end{array} \right ] 127 | $ 128 | \hspace{1in} 129 | $ 130 | U_{r}^\mu = \left [ \begin{array}{c} c \\ 0 \\ 0 \\ 0 \end{array} 131 | \right ] 132 | $ 133 | \end{center} 134 | 135 | \item To start let's assume that the particle of mass $m$ 136 | approaches the mirror on the left at the velocity of the mirror on 137 | the right. What is the four-momemtum of the particle ($p^\mu$) in 138 | the frame of the mirror on the left? 139 | 140 | $$ 141 | p^\mu = \left [ \begin{array}{c} m\gamma c \\ -m\gamma v \\ 0 \\ 0 \end{array} 142 | \right ] 143 | $$ 144 | 145 | \item The particle bounces off of the mirror. What is its 146 | four-momentum now? 147 | 148 | $$ 149 | p^\mu = \left [ \begin{array}{c} m\gamma c \\ m\gamma v \\ 0 \\ 0 \end{array} 150 | \right ] 151 | $$ 152 | 153 | \item Now the particle is approaching the mirror on the 154 | right. What is the zeroth component of the four-momentum of the 155 | particle in the frame of the right-hand mirror? One could do a 156 | Lorentz transformation but it is easier to use $U^\mu_r p_\mu$ to 157 | determine the energy of the particle in the primed frame. 158 | 159 | $$ 160 | U^\mu_r p_\mu = 161 | \left [ \begin{array}{c} \gamma c \\ -\gamma v \\ 0 \\ 0 \end{array} 162 | \right ] 163 | \left [ \begin{array}{cccc} m\gamma c & -m\gamma v & 0 & 0 \end{array} 164 | \right ] = m \gamma^2 \left ( c^2 + v^2 \right ) 165 | $$ 166 | 167 | \item Using the answer to 6, construct the four-momentum of 168 | the particle in the frame of the right-hand mirror ($p'_\mu$). 169 | 170 | $$ 171 | p^\mu = \left [ \begin{array}{c} m c \frac{1+\beta^2}{1-\beta^2} \\ m 172 | \frac{2v}{1-\beta^2} \\ 0 \\ 0 \end{array} 173 | \right ] 174 | $$ 175 | 176 | \item The particle bounces off of the mirror. What is its 177 | four-momentum now? 178 | 179 | $$ 180 | p^\mu = \left [ \begin{array}{c} m c \frac{1+\beta^2}{1-\beta^2} \\ -mc 181 | \frac{2\beta}{1-\beta^2} \\ 0 \\ 0 \end{array} 182 | \right ] 183 | $$ 184 | 185 | \item Now the particle is approaching the mirror on the 186 | left. What is the zeroth component of the four-momentum of the 187 | particle in the frame of the left-hand mirror? Again one could 188 | do a Lorentz transformation but it is easier to use $U'^\mu_l p'_\mu$ to 189 | determine the energy of the particle in the unprimed frame. 190 | 191 | $$ 192 | U^\mu_l p_\mu = 193 | \left [ \begin{array}{c} \gamma c \\ \gamma v \\ 0 \\ 0 \end{array} 194 | \right ] 195 | \left [ \begin{array}{cccc} m c \gamma^2 (1+\beta^2) & 196 | 2 \beta \gamma^2 mc & 0 & 0 \end{array} 197 | \right ] = m c^2 \gamma^3 \left ( 1 + 3 \beta^2 \right) 198 | $$ 199 | 200 | \item Compare the energy of the particle in step (d) to the 201 | energy of the particle in step (i). Has the energy of the particle 202 | increased? Let's let the relative velocity of the mirrors approach 203 | the speed of light. 204 | $$ 205 | \beta \approx 1 - \frac{1}{2\gamma^2} 206 | $$ 207 | By what factor does the energy of the particle increase each time it 208 | goes back and forth. 209 | 210 | The energy has increased by a factor of 211 | $$ 212 | \gamma^2 \left ( 1 + 3 \beta^2 \right ) \approx 4 \gamma^2 213 | $$ 214 | \item The final element is the fact that only a tiny 215 | fraction of the particles bounce back and forth. Let's take that 216 | fraction to be $10^{-5}$ and $\gamma=100$. What can you say about 217 | the final distribution of particle energies? 218 | 219 | The final distribution will be a power-law with slope given by 220 | $$ 221 | s = \ln 10^{-5} / \ln ( 4 \gamma^2) \approx -1.1 222 | $$ 223 | \end{enumerate} 224 | \item{\bf Boosting} 225 | We are going to figure out how times and energies measured by someone in motion differ from what we might measure. 226 | \begin{enumerate} 227 | 228 | \item Use special relativity (the Minkowski metric) to figure this 229 | out. I measure a photon to have an energy $E$. What is the 230 | four-momentum of the photon? 231 | 232 | \item My pal is travelling toward me in the opposite direction of the 233 | photon at a velocity $\beta c$. What is his four-velocity? Use the 234 | definition $\gamma = \left ( 1- \beta^2\right)^{-1/2}$ to simplify 235 | the expression. What energy would he measure for the photon? What 236 | does the expression look like as $\gamma$ gets much larger than one? 237 | 238 | \item If my pal observes the photon to have an energy of 100~MeV while 239 | I say its energy is less than 500~keV, what is the minimal value of 240 | $\gamma$ for my pal (take $\beta \approx 1$ to make life easier)? 241 | 242 | \item My pal is still coming toward me at a velocity $\beta c$. When 243 | he is a distance $r$ away from me (at a time $t_0$) he emits a photon 244 | toward me. How long does it take this photon to reach me? 245 | 246 | \item From his point of view a short time $\Delta t$ later he emits 247 | another photon toward me. How long is $\Delta t$ in my frame and 248 | when do I receive the second photon? What is the difference in time 249 | between when I receive the first and second photons? What does the 250 | expression look like as $\gamma$ gets much larger than one? Compare 251 | it with you answer to (b). 252 | \end{enumerate} 253 | 254 | {\bf Answer:} 255 | \begin{enumerate} 256 | \item 257 | \begin{equation} 258 | p^\mu = \frac{E}{c} \left [ \begin{array}{c} 259 | 1 \\ 260 | {\bf n} 261 | \end{array} 262 | \right ] 263 | ~\mathrm{Take}~ 264 | p^\mu = \left [ \begin{array}{c} 265 | \frac{E}{c} \\ 266 | \frac{E}{c} \\ 267 | 0 \\ 268 | 0 \\ 269 | \end{array} 270 | \right ] 271 | \end{equation} 272 | \item 273 | \begin{equation} 274 | u^\mu = \left [ \begin{array}{c} 275 | \gamma c \\ 276 | -\beta \gamma c \\ 277 | 0 \\ 278 | 0 \\ 279 | \end{array} 280 | \right ] 281 | ~\mathrm{and}~E'=-u_\mu p^\mu = \gamma E + \beta \gamma E \approx 2 282 | \gamma E 283 | \end{equation} 284 | \item 285 | $E=500$~keV and $E'=100~\mathrm{Mev}=2\gamma (500~\mathrm{keV})$ so 286 | $\gamma_\mathrm{min} =100$. 287 | \item 288 | $t_\mathrm{Arrival} = t_0 + \frac{r}{c}$ 289 | \item 290 | \begin{eqnarray} 291 | \Delta t_\mathrm{me} &=& \gamma \Delta t_\mathrm{him} \\ 292 | t_\mathrm{Arrival,2} &=& t_0 + \gamma \Delta t_\mathrm{him} + 293 | \frac{1}{c} \left ( r - \beta c \gamma \Delta t_\mathrm{him} 294 | \right ) \\ 295 | &=& t_0 + \frac{r}{c} + \gamma \Delta t_\mathrm{him} \left ( 1 - 296 | \beta \right ) \\ 297 | \Delta t_\mathrm{Arrival} &=& \Delta t_\mathrm{him} \gamma \left (1 - \beta 298 | \right ) \\ 299 | \Delta t_\mathrm{Arrival} &=& \Delta t_\mathrm{him} \frac{1}{\gamma \left (1 + \beta 300 | \right )} \approx \Delta t_\mathrm{him} \frac{1}{2\gamma}. 301 | \end{eqnarray} 302 | where to get the penultimate result, one uses the identity 303 | $(1-\beta)(1+\beta) = \gamma^{-2}$ and in general we have 304 | \begin{equation} 305 | \frac{\Delta t_\mathrm{Arrival}}{\Delta t_\mathrm{him}} = \frac{E}{E'}. 306 | \end{equation} 307 | \end{enumerate} 308 | \end{enumerate} 309 | \ifx\bookloaded\undefined 310 | \end{document} 311 | \end 312 | \fi 313 | 314 | %%% Local Variables: 315 | %%% TeX-master: "book" 316 | %%% End: -------------------------------------------------------------------------------- /solutions/chap2sol.tex: -------------------------------------------------------------------------------- 1 | \ifx\bookloaded\undefined 2 | \documentclass{article} 3 | \usepackage{graphicx} 4 | \input book_defs 5 | \begin{document} 6 | \fi 7 | \section{Chapter 2} 8 | \begin{enumerate} 9 | \item{\bf Coulomb's Law} 10 | 11 | Derive Coulomb's law from Maxwell's Equations 12 | 13 | {\bf Answer:} 14 | 15 | The first of Maxwell's equations is 16 | \begin{equation} 17 | \nabla \cdot {\vec E} = 4 \pi \rho 18 | \end{equation} 19 | Let's assume that there is a single charge q located at r=0 and integrate over a spherical region centered on the origin we get 20 | \begin{equation} 21 | \int_V dV \nabla \cdot {\vec E} = \int dV 4 \pi \rho = 4 \pi q 22 | \end{equation} 23 | However the integral of the left-hand side is a integral of a divergence over a volume so we have 24 | \begin{equation} 25 | \int_{\partial V} dV \nabla \cdot {\vec E} = \int {\vec E} \cdot d A = |{\vec E}| 4 \pi R^2 = 4 \pi q 26 | \end{equation} 27 | so 28 | \begin{equation} 29 | {\vec E} = \frac{q}{R^2} {\hat r} 30 | \end{equation} 31 | 32 | \item{\bf Ohm's Law} 33 | 34 | In certain cases the process of aborption of radiation can be treated 35 | by means of the macroscopic Maxwell equations. For example, suppose we 36 | have a conducting medium so that the current density j is related to 37 | the electric field E by Ohm's law: 38 | ${\vec j} = \sigma {\vec E}$ where 39 | $\sigma$ is the conductivity (cgs unit = sec$^{-1}$). 40 | Investigate the propagation of electromagnetic waves in such a medium 41 | and show that: 42 | \begin{enumerate} 43 | \item 44 | The wave vector ${\vec k}$ is complex 45 | ${\vec k}^2 = \frac{\omega^2 m^2}{c^2}$ 46 | where $m$ is the complex index of refraction with 47 | \begin{equation} 48 | m^2 = \mu \epsilon \left ( 1 + \frac{4 \pi i \sigma}{\omega \epsilon} 49 | \right ) 50 | \end{equation} 51 | \item The waves are attenuated as they propagate, corresponding to an 52 | absorption coefficient. 53 | \begin{equation} 54 | \alpha = \frac{2\omega}{c} \Im (m) 55 | \end{equation} 56 | 57 | {\bf Answer:} 58 | 59 | Let's take the third and fourth of Maxwell's equations 60 | \begin{equation} 61 | {\nabla} \times {\vec E} = -\frac{1}{c} \frac{\partial \vec B}{\partial t} 62 | \end{equation} 63 | and 64 | \begin{equation} 65 | {\nabla} \times {\vec H} = \frac{4\pi}{c} {\vec J} +\frac{1}{c} 66 | \frac{\partial \vec D}{\partial t} 67 | \end{equation} 68 | Let's substitute 69 | \begin{equation} 70 | \mu {\vec H}={\vec B}, {\vec D}=\epsilon {\vec E} 71 | \end{equation} 72 | and 73 | \begin{equation} 74 | {\vec J} = \sigma {\vec E} 75 | \end{equation} 76 | to get 77 | \begin{equation} 78 | {\nabla} \times {\vec B} = \mu \sigma \frac{4\pi}{c} {\vec E} +\frac{1}{c} 79 | \mu \epsilon \frac{\partial \vec E}{\partial t} 80 | \end{equation} 81 | Let's take the curl of this equation to get 82 | \begin{equation} 83 | -{\nabla}^2 {\vec B} = \mu \sigma \frac{4\pi}{c} \nabla \times {\vec E} +\frac{1}{c} 84 | \mu \epsilon \frac{\partial \nabla \times \vec E}{\partial t} 85 | \end{equation} 86 | and substitute in the other Maxwell's equation to get 87 | \begin{equation} 88 | -{\nabla}^2 {\vec B} = -\mu \sigma \frac{4\pi}{c^2} \frac{\partial \vec B}{\partial t} - \frac{1}{c^2} 89 | \mu \epsilon \frac{\partial^2 \vec B}{\partial t^2} 90 | \end{equation} 91 | Let's substitute 92 | \begin{equation}{\vec B} = {\vec B}_0 \exp \left [ i \left ({\vec k}\cdot {\vec x} - \omega t \right ) \right ]\end{equation} 93 | to get 94 | \begin{equation} 95 | k^2 = i \mu \sigma \frac{4\pi}{c^2} \omega + \frac{1}{c^2} 96 | \mu \epsilon \omega^2 = \frac{\omega^2 m^2}{c^2} 97 | \end{equation} 98 | with 99 | \begin{equation} 100 | m^2 = \mu \epsilon \left ( 1 + i \frac{4\pi\sigma}{\omega \epsilon} \right ) 101 | \end{equation} 102 | If we substitute this into the formula for the wave we find 103 | \begin{equation} 104 | {\vec B} = {\vec B}_0 \exp \left [ i \left ({\vec k}\cdot {\vec x} - \omega t \right ) \right ] 105 | = {\vec B}_0 \exp \left [ i \left (\Re {\vec k}\cdot {\vec x} - \omega t \right ) \right ] 106 | \exp \left [ -\Im {\vec k} \cdot {\vec x} \right ] 107 | \end{equation} 108 | so the magnetic field decreases with a mean-free path $\frac{c}{\omega 109 | \Im m}$. The energy is proportional to $B^2$ so the absorption coefficient is $\frac{2\omega}{c}\Im 110 | m$ 111 | \end{enumerate} 112 | \item {\bf The Edge of the Crab} 113 | 114 | Fig.~\ref{fig:crab-edge} shows the x-ray emission of the Crab pulsar 115 | wind nebula at a distance of 2~kpc. The x-ray emitting gas is 116 | contained by magnetic fields causing the x-ray emission regions to 117 | end sharply. We can relate the frequency of the emission to the energy 118 | of the electrons and the strength of the magnetic field by 119 | \begin{equation} 120 | \omega = \left ( \frac{E}{m_e c^2} 121 | \right )^2 \frac{e B}{m_e c} 122 | \end{equation} 123 | and assume that the electrons are relativistic so their inertial mass 124 | is $E/c^2$. Use the sharpness of the emission 125 | regions to determine the energy of the electrons and the strength of 126 | the magnetic field. 127 | 128 | {\bf Answer:} 129 | 130 | Let's assume that the particles are doing cyclotron motion so 131 | \begin{equation} 132 | F = \frac{m v^2}{r} = \frac{ e v B}{c}, v\approx c, m\approx 133 | \frac{E}{c^2} 134 | \end{equation} 135 | so 136 | $$ 137 | \frac{E}{r} = e B, E = e B r 138 | $$ 139 | where $r < 2~\mathrm{kpc} \times 1~\mathrm{arcsecond} = 2000~\mathrm{AU}$ 140 | and we also know that $\omega$ lies in the X-rays, so $\hbar \omega \sim 141 | 1$~keV and 142 | $$ 143 | \omega = E^2 B \frac{e}{m_e^3 c^5} = E^2 \frac{E}{er} \frac{e}{m_e^3 c^5} 144 | $$ 145 | so 146 | $$ 147 | E^3 = m_e^3 c^5 \omega r, B^3 = \frac{m_e^3 c^5 \omega}{e^3 r^2}. 148 | $$ 149 | The fact the the edge is unresolved yields an upper limit on the energy 150 | and a lower limit on the magnetic field strength. If we use $\hbar 151 | \omega=1~\mathrm{keV}$, we obtain 152 | $$ 153 | E < 6 \times 10^{13}~\mathrm{eV}, B > 7 \times 10^{-11}~\mathrm{G} 154 | $$ 155 | 156 | 157 | \item{\bf Momenta} 158 | This problem is meant to deduce the momentum and angular momentum 159 | properties of radiation and does not recesarily represent any real 160 | physical system of interest. Consider a charge $Q$ in a viscous medium 161 | where the viscous force is proportional to velocity: 162 | \begin{equation} 163 | F_{visc} = -\beta v 164 | \end{equation} 165 | Suppose a circular polarized wave passes through the medium. The 166 | equation of motion of the charge is 167 | \begin{equation} 168 | m \frac{dv}{dt} = F_{visc} + F_{Lorentz} 169 | \end{equation} 170 | We assume that the terms on the right dominate the inertial term on the 171 | left, so that approximately 172 | \begin{equation} 173 | 0 = F_{visc} + F_{Lorentz} 174 | \end{equation} 175 | Let the frequency of the wave be $\omega$ and the strength of the 176 | electric field be E. 177 | \begin{enumerate} 178 | \item Show that to lowest order (neglecting the magnetic force) the charge 179 | moves on a circle in a plane normal to the direction of propagation of 180 | the wave with speed $QE/\beta$ and with radius $QE/(\beta \omega)$. 181 | \item Show that the power transmitted to the fliud by the wave is $Q^2 E^2/\beta$ 182 | \item. By considering the small magnetic force acting on the particle show 183 | that the momentum per unit time (force) given to the fluid by the wave 184 | is in the direction of propagation and has the magnitude 185 | $Q^2 E^2/(\beta c)$. 186 | \item Show that the angular momentum per unit time (torque) given to the 187 | fluid by the wave is in the direction of propagation and has magnitude 188 | $\pm Q^2 E^2/(\beta \omega)$ where the $+$ is for left and $-$ is for 189 | right circular polarization. 190 | \item Show that the absorption cross section of the charge is 191 | $4\pi Q^2/(\beta c)$. 192 | \item If we regard the radiation to be composed of circular polarized 193 | photons of energy $E_\gamma= h \nu$, show that these results imply that the 194 | photon has momemtum $p=h/\lambda=E_\gamma/c$ and has angular momemtum 195 | $J=\pm \hbar$ along the direction of propagation. 196 | \item Repeat this problem for a linearly polarized wave 197 | \end{enumerate} 198 | 199 | {\bf Answer:} 200 | 201 | \begin{enumerate} 202 | \item 203 | We have ${\vec v} = \frac{Q}{\beta} {\vec E}$. The electric field traces a circle so the 204 | particle traces a circle with a speed $\frac{QE}{\beta}$. The angular velocity of the particle is $\omega$ of the wave, so $\omega r=\frac{QE}{\beta}$ so $r=\frac{QE}/{\beta\omega}$. 205 | 206 | \item Power is $Q{\vec v}\cdot {\vec E} = \frac{Q^2 E^2}{\beta}$. 207 | 208 | \item The magnetic force is in the direction ${\vec v}\times {\vec B}$ but the velocity points in the 209 | direction of the electric field so the force is in the direction ${\vec E}\times {\vec B}$, the direction of propagation. The magnitude of magnetic field equals that of the electric field so we have 210 | $F = \frac{Q^2 E^2}{\beta c}$ 211 | 212 | \item Torque is ${\vec r} \times {\vec F} = \frac{Q^2 E^2}{\beta \omega}$ 213 | 214 | \item The cross section is power absorbed divided by the Poynting vector 215 | \begin{equation} 216 | \sigma = \frac{Q^2 E^2}{\beta} \left [ \frac{c}{4\pi} E^2 \right ]^{-1} = \frac{4 \pi Q^2}{\beta c} 217 | \end{equation} 218 | 219 | \item If the wave comes in energy units of $h\nu$. The ratio of the momentum unit to the energy unit must 220 | be the ratio of the force (momentum per unit time) to the power (energy per unit time), so we get 221 | \begin{equation} 222 | h\nu \frac{Q^2 E^2}{\beta c} \frac{\beta}{Q^2 E^2} = \frac{h\nu}{c} 223 | \end{equation} 224 | The ratio of the angular momentum unit to the energy unit must be the 225 | ratio of the torque (angular momentum per unit time) to the power 226 | (energy per unit time), so we get 227 | \begin{equation} 228 | h\nu \frac{Q^2 E^2}{\beta \omega} \frac{\beta}{Q^2 E^2} = \frac{h\nu}{\omega} = \hbar 229 | \end{equation} 230 | \item For the linearly polarized wave, the particle moves up and down sinusoidally. The size of the up and down path is twice the value of R above (the circle is squished along one axis to be a line). The velocity varies sinusoidally, the power magnetic force vary as $\sin^2 \omega t$. The torque vanishes. The cross section is the same as is the momentum of a photon. The angular momentum vanishes (because the torque vanishes). 231 | \end{enumerate} 232 | 233 | 234 | \item{\bf Maxwell before Maxwell} 235 | 236 | Show that Maxwell's equations before Maxwell, that is, without the 237 | ``displacement current'' term, $c^{-1} \frac{\partial D}{\partial t}$, unacceptably constrained the 238 | sources of the field and also did not permit the existence of waves. 239 | 240 | {\bf Answer:} 241 | 242 | Let's take the divergence of the Maxwell's equation 243 | \begin{equation} 244 | {\nabla} \times {\vec H} = \frac{4\pi}{c} {\vec J} +\frac{1}{c} 245 | \frac{\partial \vec D}{\partial t} 246 | \end{equation} 247 | to get 248 | \begin{equation} 249 | 0 = \frac{4\pi}{c} \nabla \cdot {\vec J} 250 | \end{equation} 251 | where we have left the displacement current out. This states that the 252 | divergence of the current must vanish, which means that either charge 253 | is not conserved or that the charge density is constant (neither is 254 | good). 255 | 256 | Let's take the curl of the Maxwell's equation 257 | \begin{equation} 258 | {\nabla}^2 {\vec B} = 0 259 | \end{equation} 260 | and we would get the same thing for the electric field. This is not a 261 | wave equation. 262 | 263 | \item {\bf Coulomb gauge} 264 | Derive the equations describing the dynamics of the electric and 265 | vector potentials in the Coulomb gauge 266 | $$ 267 | \nabla \cdot {\bf A} = 0 268 | $$ 269 | Look at the equation for the electric potential. What is the solution 270 | to the electric potential given the charge density $\rho$? Why is this 271 | called the Coulomb gauge? 272 | 273 | How does the expression for the scalar potential in the Coulomb gauge 274 | differ from that in the Lorenz gauge? What is strange about it? Is it 275 | physical? 276 | 277 | Now look at the equation for the vector potential. Show that the LHS 278 | can be arranged to be the same as in the Lorenz gauge but the RHS is 279 | not just the current but the current plus something else. 280 | 281 | Show that the RHS can be expressed as 282 | $$ 283 | \frac{4\pi}{c} \left ( {\bf J} - {\bf J}_\rmscr{long} \right ) 284 | $$ 285 | where 286 | $$ 287 | {\bf J}_\rmscr{long} = -\frac{1}{4\pi} \nabla \int \frac{\nabla' \cdot 288 | {\bf J}}{|{\bf x}-{\bf x}'|} d^3 x 289 | $$ 290 | 291 | {\bf Answer:} 292 | In the Coulomb gauge the scalar potential follows Coulomb's law 293 | $$ 294 | \nabla^2 \phi = -4\pi \rho. 295 | $$ 296 | That is why it is called the Coulomb gauge. The potential everywhere 297 | right now depends on the charge here right now, so it is acausal 298 | (strange); however, because we cannot actually measure the scalar 299 | potential the acausality has no physical consequence. 300 | 301 | Now for the vector potential 302 | $$ 303 | \nabla^2 {\bf A} - \frac{1}{c^2} \pp{^2 {\bf A}}{t^2} - \nabla \left 304 | ( \frac{1}{c} \pp{\phi}{t} \right ) = -\frac{4\pi}{c} {\bf J} 305 | $$ 306 | $$ 307 | \nabla^2 {\bf A} - \frac{1}{c^2} \pp{^2 {\bf A}}{t^2} = 308 | -\frac{4\pi}{c} \left ( {\bf J} - {\bf J}_\mathrm{long} \right ) 309 | $$ 310 | where 311 | \begin{eqnarray*} 312 | {\bf J}_\mathrm{long} &=& \frac{1}{4\pi} \nabla \left 313 | ( \pp{\phi}{t} \right ) = \frac{1}{4\pi} \nabla \left 314 | ( \int \pp{\rho}{t} \frac{1}{\left |{\bf x} - {\bf x}'\right|} d^3 x 315 | \right ) \\ 316 | &=& -\frac{1}{4\pi} \nabla \left 317 | ( \int \frac{\nabla \cdot {\bf J}}{\left |{\bf x} - {\bf x}'\right|} d^3 x 318 | \right ) = -\frac{1}{4\pi} \nabla \left 319 | ( \int \frac{\nabla' \cdot {\bf J}}{\left |{\bf x} - {\bf x}'\right|} d^3 x 320 | \right ) 321 | \end{eqnarray*} 322 | What remains of the current after subtracting the longitudinal current 323 | is the transverse current which is given by the expression 324 | $$ 325 | {\bf J}_\mathrm{trans} = \frac{1}{4\pi} \nabla \times \nabla \left 326 | ( \int \frac{{\bf J}}{\left |{\bf x} - {\bf x}'\right|} d^3 x 327 | \right ) 328 | $$ 329 | so the source for the wave equation for ${\bf A}$ is given by the 330 | transverse current alone. 331 | \end{enumerate} 332 | 333 | \ifx\bookloaded\undefined 334 | \end{document} 335 | \fi 336 | %%% Local Variables: 337 | %%% TeX-master: "book" 338 | %%% End: 339 | 340 | 341 | 342 | -------------------------------------------------------------------------------- /chapters/chap14.tex: -------------------------------------------------------------------------------- 1 | \chapter{Fluid Instabities} 2 | \label{cha:fluid-instabities} 3 | \label{sec:instabilities} 4 | \index{fluid mechanics!instabilities} 5 | \section{Gravity Waves and Rayleigh-Taylor Instability} 6 | \label{sec:gravitywaves} 7 | \index{fluid mechanics!gravity waves} 8 | \index{gravity waves} 9 | \index{Rayleigh-Taylor instability} 10 | \index{fluid mechanics!instabilities!Rayleigh-Taylor} 11 | 12 | \begin{figure} 13 | \begin{center} 14 | \begin{tikzpicture}[scale=0.95] 15 | \draw [thick] (0,2) -- (11,2) node [right] {$z=h$} (0,-2) -- (11, -2) 16 | node [right] {$z=-h'$} ; 17 | \draw plot [domain=0:11,samples=100] ( \x, {sin(32*\x+170)}) node [right] {$z=\zeta(x,t)$} ; 18 | \draw [<-] (5,1.5) -- (4,1.5) node [left] {$\rho, U$} ; 19 | \draw [<-] (5,-1.5) -- (4,-1.5) node [left] {$\rho', U'$} ; 20 | \end{tikzpicture} 21 | \end{center} 22 | \caption{Shearing flow in a stratified fluid} 23 | \end{figure} 24 | Let's imagine a different type of wave on a fluid. Let's imagine we 25 | have two fluids in a gravitational field. The lower fluid has density 26 | $\rho'$, velocity $U'$ and thickness $h'$ and the upper fluid has 27 | density $\rho$, velocity $U$ and 28 | thickness $h$. Let's have $\zeta(x,t)$ denote the displacement of the 29 | interface in the $z-$direction. Let's assume that both fluids are 30 | incompressible and the flow is irrotational, so we can define 31 | \begin{equation} 32 | {\bf v} = -\nabla \Phi ~\mathrm{and}~ {\bf v}' = -\nabla \Phi' 33 | \end{equation} 34 | where 35 | \begin{equation} 36 | \Phi = -U x + \phi~\mathrm{and}~ \Phi' = -U' x + \phi'. 37 | \end{equation} 38 | To make further progress let us assume that the displacement of the 39 | interface has the form 40 | \begin{equation} 41 | \zeta(x,t) = A \cos \left ( k x - \omega t \right ) 42 | \end{equation} 43 | and furthermore let velocity potentials also have a similar dependence 44 | \begin{equation} 45 | \phi = C \sin \left (k x - \omega t\right ) f(z)~\mathrm{and}~ 46 | \phi' = C' \sin \left (k x - \omega t\right ) f'(z). 47 | \end{equation} 48 | Because the fluids are assumed to be incompressible, we have $\nabla^2 49 | \phi=0$ and the boundary condition gives $\partial \phi/\partial z=0$ 50 | at $z=h$ and similarly for lower fluid, so we have 51 | \begin{eqnarray} 52 | \phi &=& C \sin \left (k x - \omega t\right ) \cosh \left [ k(z-h) 53 | \right ] \\ 54 | \phi' &=& C' \sin \left (k x - \omega t\right ) \cosh \left [ k(z+h') 55 | \right ]. 56 | \end{eqnarray} 57 | The Lagrangian derivative of the displacement of the interface 58 | $D \zeta(x,t)/Dt$ gives the vertical velocity of the fluid at the 59 | interface, so 60 | \begin{equation} 61 | -\pp{\phi}{z} = \pp{\zeta}{t} + U \pp{\zeta}{x}~\mathrm{and}~ 62 | -\pp{\phi'}{z} = \pp{\zeta}{t} + U' \pp{\zeta}{x}. 63 | \end{equation} 64 | This yields a relationship between the constants $A$, $C$ and $C'$, 65 | the wavenumber and frequency, 66 | \begin{eqnarray} 67 | A \left ( k U - \omega \right ) &=& - k C \sinh k h, 68 | \label{eq:868} 69 | \\ 70 | A \left ( k U' - \omega \right ) &=& k C' \sinh k h'. 71 | \label{eq:869} 72 | \end{eqnarray} 73 | We seek a relationship between $\omega$ and $k$, so we require an 74 | additional equation to eliminate the unknowns $A$, $C$ and $C'$. 75 | Specifically, the pressure on each side of the interface must be 76 | equal. To examine the pressure let's look at Euler's equation 77 | for the ideal fluid (Eq.~\ref{eq:631}) and substitute ${\bf V}=-\nabla 78 | \Phi$ to yield 79 | \begin{equation} 80 | -\pp{\nabla \Phi}{t} + \frac{1}{2} \nabla V^2 = -\frac{\nabla P}{\rho} 81 | - g {\hat z}. 82 | \end{equation} 83 | Since the fluid is incompressible we can write 84 | \begin{equation} 85 | \nabla \left [ -\pp{\Phi}{t} + \frac{V^2}{2} + \frac{P}{\rho} + g h 86 | \right ] = 0, 87 | \end{equation} 88 | so for each fluid we have 89 | \begin{eqnarray} 90 | -\rho \pp{\Phi}{t} + \rho \frac{v^2}{2} + g z \rho + p &=& B(t), \\ 91 | -\rho \pp{\Phi'}{t} + \rho' \frac{v'^2}{2} + g \rho' z + p &=& B'(t). 92 | \end{eqnarray} 93 | At the upper and lower surface the velocity of the perturbation 94 | vanishes and $p(z=-h')-p(z=h)$ must equal $g \rho h + g \rho' h'$, so 95 | \begin{equation} 96 | B(t) - B'(t) = \frac{\rho U^2}{2} - \frac{\rho' U'^2}{2}. 97 | \end{equation} 98 | Let's take the difference of the two Bernoulli equations and evaluate 99 | it at $z=\zeta(x,t)$ to yield 100 | \begin{equation} 101 | \rho \left ( -\pp{\phi}{t} - U \pp{\phi}{x} + g \zeta \right )= 102 | \rho' \left ( -\pp{\phi'}{t} - U' \pp{\phi'}{x} + g \zeta \right ). 103 | \end{equation} 104 | to first order in the small quantities $\phi$ and $\phi'$. 105 | Substituting the expressions for $\phi$, $\phi'$ and $\zeta$ yields 106 | \begin{equation} 107 | \rho \left [ C \cosh kh \left ( \omega - k U \right ) + g A \right ] = 108 | \rho' \left [ C' \cosh kh' \left ( \omega - k U' \right ) + g A \right ] . 109 | \end{equation} 110 | Combining this result with Eq.~\ref{eq:868} and~\ref{eq:869} yields 111 | the equation, 112 | \begin{equation} 113 | \rho \left ( \omega - k U \right )^2 \coth kh + 114 | \rho' \left ( \omega - k U' \right )^2 \coth kh' = k g \left (\rho' - 115 | \rho \right ), 116 | \end{equation} 117 | and the dispersion relation, 118 | \begin{eqnarray} 119 | \frac{\omega}{k} &=& \frac{\rho U \coth kh + \rho' U'\coth k h'}{\rho 120 | \coth kh + \rho' \coth k h'} \pm \nonumber \\ 121 | & &~~~ \left [ \frac{g}{k} \frac{\rho' - \rho}{\rho 122 | \coth kh + \rho' \coth k h'} - \frac{\rho \rho' \coth kh \coth kh' \left ( U - U' \right )^2}{\left (\rho 123 | \coth kh + \rho' \coth k h' \right)^2 } \right ]^{1/2}. 124 | \label{eq:870} 125 | \end{eqnarray} 126 | The first interesting limit is where $U=U'=0$ which yields the simpler expression 127 | \begin{equation} 128 | \frac{\omega^2}{k^2} = \frac{g}{k} \frac{\rho' - \rho}{\rho \coth k h + 129 | \rho' \coth k h'} 130 | \label{eq:769} 131 | \end{equation} 132 | If $\rho' > \rho$, then $\omega^2>0$ and we have a stable wave. There 133 | are several interesting limits to this result. 134 | \begin{itemize} 135 | \item If $\rho=0$, then $\omega^2 = g k \tanh kh$. 136 | \item If $\rho=0$ and $kh \gg 1$, then $\omega^2 = g k$ (deep-water waves). 137 | \item If $\rho=0$ and $kh \ll 1$, then $\omega^2 = g h k^2$ (shallow-water 138 | waves). 139 | \item If $\rho\neq 0$, $kh' \gg 1$ and $kh \gg 1$, then (both liquids 140 | very deep) 141 | \begin{equation} 142 | \omega^2 = \frac{kg (\rho' - \rho)}{\rho+\rho'} 143 | \label{eq:770} 144 | \end{equation} 145 | \item If $\rho\neq 0$, $kh' \ll 1$ and $kh \ll 1$, then (long waves) 146 | \begin{equation} 147 | \omega^2 = k^2 \frac{g (\rho' - \rho) h h'}{\rho h'+\rho' h} 148 | \label{eq:771} 149 | \end{equation} 150 | \end{itemize} 151 | 152 | On the other hand if $\rho' < \rho$, then $\omega^2 < 0$ and the 153 | perturbation simply grows (it does not oscillate). This is the 154 | Rayleigh-Taylor instability. This instability occurs whenever a low 155 | density gas underlies a higher density gas, for example in a supernova 156 | explosion. The gravitational acceleration $g$ can be due to gravity 157 | (as in a supernova) or due to a deceleration of the fluid, if a 158 | low-density fluid plows into a high-density fluid. According to 159 | Eq.~\ref{eq:771} the smallest scales have the highest growth rates. 160 | This is countered by viscosity and surface tension, so a 161 | particular scale dominates the growth at least initially. 162 | 163 | \section{Kelvin-Helmholtz or Shearing Instability} 164 | \label{sec:kelvin-helmholtz-or} 165 | \index{Kelvin-Helmholtz instability} 166 | \index{shearing instability} 167 | \index{fluid mechanics!instabilities!shearing} 168 | 169 | If we look at the term in the brackets in Eq.~\ref{eq:870} for $U \neq 170 | U'$ we see that if $g=0$ waves with all values of $k$ are unstable and if $g\neq 171 | 0$ for sufficiently large values of $k$ (small wavelengths), waves are 172 | unstable even if $\rho'>\rho$. The critical value of $k$ is 173 | \begin{equation} 174 | k_\mathrm{crit} = \frac{g}{\left (U - U'\right)^2} \frac{\rho'-\rho}{\rho \rho' } \frac{\rho 175 | \coth kh + \rho' \coth k h'}{\coth kh \coth k h'} 176 | \end{equation} 177 | and for $k>k_\mathrm{crit}$ the growth rate increases monotonically. 178 | In reality for really small wavelengths other effects come into play, 179 | such as surface tension and viscosity; therefore, unless the velocity 180 | difference is sufficiently large, waves will not grow, and furthermore 181 | a particular wavelength grow the fastest. 182 | 183 | We have also assumed that the velocity change is abrupt. It turns out 184 | that even if the velocity changes gradually with position, the flow is 185 | unstable, so we would like to get a heuristic understanding of the 186 | Kelvin-Helmholtz instability. We have two fluids moving in opposite 187 | directions along their shared interface which may be thick. We do not 188 | include gravity. 189 | 190 | Let's assume that the flow is initially steady and 191 | irrotational that so we have Euler's equation 192 | \begin{equation} 193 | ({\bf V} \cdot \nabla) {\bf V} + \frac{\nabla P}{\rho} = 0 194 | \label{eq:772} 195 | \end{equation} 196 | We know that 197 | \begin{equation} 198 | \frac{1}{2} \nabla v^2 = {\bf V} \times (\nabla \times {\bf v}) + ({\bf 199 | v} \cdot \nabla ) {\bf v} 200 | \label{eq:773} 201 | \end{equation} 202 | which yields 203 | \begin{equation} 204 | \frac{1}{2} \nabla V^2 + \frac{\nabla P}{\rho} = 0 205 | \label{eq:774} 206 | \end{equation} 207 | and 208 | \begin{equation} 209 | \frac{1}{2} V^2 + \frac{P}{\rho} = \rmmat{constant} 210 | \label{eq:775} 211 | \end{equation} 212 | for the flow. Therefore, regions where $V^2$ is large have lower 213 | pressure. In the figure we have chosen a reference frame where the fluids 214 | are moving with equal and opposite velocities. We will also assume 215 | that the depths of both fluids are really large and the densities are 216 | equal. Therefore, the picture of what goes on in one fluid is 217 | mirrored in the other. If we focus on the wrinkle in the interface 218 | on the right hand side, the upper fluid must travel a bit farther to 219 | get around the wrinkle than the lower fluid, so it must travel faster 220 | and according to Eq.~\ref{eq:775}, its pressure must drop more than 221 | the fluid below the interface. The pressure on the inside of the 222 | curve is greater than on the outside. These pressure gradient causes 223 | the wrinkle to grow. We could even imagine a rubber sheet or less 224 | dramatically a layer of fluid moving at intermediate velocities lying 225 | along the interface and the forces would still be the same, and the 226 | instability remains. 227 | 228 | \begin{figure} 229 | \begin{center} 230 | % \includegraphics[width=\columnwidth]{shear} 231 | \begin{tikzpicture} 232 | \draw [dashed] (0,0) -- (12,0) ; 233 | \draw plot [domain=0:12,samples=100] ( \x, {sin(32*\x+170)}) ; 234 | \draw [solid] (3.125,-0.85) circle (0.075) (3.125, -1.15) circle (0.075) 235 | (8.75,0.85) circle (0.075) (8.75, 1.15) circle (0.075) ; 236 | \draw [->] (3.325,-1.4) -- (3.325,-0.6) ; 237 | \draw (3.725,-1.3) node {${\bf \nabla} P$} ; 238 | \draw [->] (8.55,1.4) -- (8.55,0.6) ; 239 | \draw (8.15,1.3) node {${\bf \nabla} P$} ; 240 | \draw [->] (5.5,2) -- (6.5,2) ; 241 | \draw (7,2) node {${\bf v}/2$} ; 242 | \draw [<-] (5.5,-2) -- (6.5,-2) ; 243 | \draw (5,-2) node {$-{\bf v}/2$} ; 244 | \end{tikzpicture} 245 | \end{center} 246 | \caption{Illustration of shearing flow with pressure gradients} 247 | \end{figure} 248 | 249 | \section{Gravitational Instability} 250 | \label{sec:grav-inst} 251 | \index{fluid mechanics!instabilities!gravitational} 252 | \index{gravitational instability} 253 | 254 | Let's revisit our small sound waves but this time we will include the 255 | effects of self-gravity. 256 | have 257 | \begin{equation} 258 | \pp{\rho}{t} + \nabla \cdot (\rho {\bf V}) = \pp{\rho'}{t} + \rho_0 259 | \nabla \cdot {\bf V}' = 0 260 | \label{eq:776} 261 | \end{equation} 262 | and 263 | \begin{equation} 264 | \pp{{\bf V}}{t} + \left ( {\bf V} \cdot \nabla \right ) {\bf V} + 265 | \frac{\nabla P}{\rho} = 266 | \dd{\bf V}{t} + \frac{\nabla P}{\rho} = 267 | \pp{{\bf V}'}{t} + \frac{\nabla P'}{\rho_0} = -\nabla \phi' 268 | \label{eq:777} 269 | \end{equation} 270 | We can write $P' = (\partial P/\partial \rho)_s \rho'$ and rewrite the 271 | continuity equation to get 272 | \begin{equation} 273 | \pp{P'}{t} + \rho_0 \left ( \pp{P}{\rho} \right )_s \nabla \cdot {\bf 274 | V'} = 0 275 | \label{eq:778} 276 | \end{equation} 277 | Let's take the divergence of the Euler equation to get 278 | \begin{equation} 279 | \pp{{\bf \nabla \cdot V}'}{t} + \frac{\nabla^2 P'}{\rho_0} = -\nabla^2 \phi' 280 | \label{eq:779} 281 | \end{equation} 282 | and the time derivative of the continuity equation to get 283 | \begin{equation} 284 | \pp{^2 P'}{t^2} + \rho_0 \left ( \pp{P}{\rho} \right )_s \nabla \cdot \pp{\bf 285 | V'}{t} = 0. 286 | \label{eq:780} 287 | \end{equation} 288 | Finally we put the two together to get 289 | \begin{equation} 290 | \pp{^2 P'}{t^2} - \left ( \pp{P}{\rho} \right )_s \left ( \nabla^2 P' 291 | + \rho_0 \nabla^2 \phi' \right ) = 0. 292 | \label{eq:781} 293 | \end{equation} 294 | This is a wave equation with a sound speed of $c_s^2 = (\partial 295 | P/\partial \rho)_s$ but there is an extra term. 296 | \begin{equation} 297 | \nabla^2 \phi' = 4\pi G \rho' 298 | \label{eq:782} 299 | \end{equation} 300 | We can write $P' = c_s^2 \rho'$. This eliminates density from the 301 | equation to get 302 | \begin{equation} 303 | \pp{^2 P'}{t^2} - c_s^2 \nabla^2 P' - 4\pi G \rho_0 P' = 0. 304 | \label{eq:783} 305 | \end{equation} 306 | Let's try a trial plane wave to find a solution to this equation 307 | \begin{equation} 308 | P' = p' \exp[i({\bf k} \cdot {\bf r} - \omega t)] 309 | \label{eq:784} 310 | \end{equation} 311 | We get 312 | \begin{equation} 313 | -\omega^2 + c_s^2 k^2 - 4\pi G \rho_0 = 0 314 | \label{eq:785} 315 | \end{equation} 316 | so 317 | \begin{equation} 318 | \omega^2 = c_s^2 k^2 - 4\pi G \rho_0. 319 | \label{eq:786} 320 | \end{equation} 321 | If $k^2 > 4\pi G \rho_0/c_s^2$, then $\omega^2>0$ and the wave is 322 | stable. On the other hand, if $k^2 < 4\pi G \rho_0/c_s^2$, the 323 | perturbation will grow. We can define a Jeans length 324 | \begin{equation} 325 | l_\rmscr{Jeans} = \frac{2\pi}{k_\rmscr{crit}} = 326 | \sqrt{\frac{\pi}{G\rho_0}} c_s 327 | \label{eq:787} 328 | \end{equation} 329 | and a Jeans mass 330 | \begin{eqnarray} 331 | M_\rmscr{Jeans} &=& \frac{4}{3} \pi l_\rmscr{Jeans}^3 \rho_0 = 332 | \frac{4}{3} \sqrt{\frac{\pi^5}{G^3 \rho_0}} c_s^3 \\ 333 | &=& 1.4 \times 10^{42}\rmmat{g} 334 | \left [ \frac{\rho_0}{10^{-24} \rmmat{g cm}^{-3}} \right ]^{-1/2} \left [ 335 | \frac{c_s}{10 \rmmat{km s}^{-1}} \right ]^3 336 | \label{eq:788} 337 | \end{eqnarray} 338 | 339 | \section{Thermal Instability} 340 | \index{fluid mechanics!instabilities!thermal} 341 | \index{thermal instability} 342 | 343 | So far we have examined instabilities where energy does not leave or 344 | enter the fluid. In general hot gas emits and absorbs radiation; it 345 | may release energy through nuclear or chamical reactions as well. If 346 | the power absorbed and generated within the gas equals the power 347 | emitted by the gas, the temperature of the gas will remain constant 348 | and equilibrium is achieved. The question remains whether this 349 | equilibrium is stable. Heuristically we can see that if the cooling 350 | rate increases faster with temperature than the heating rate, then a 351 | slight increase in temperature will result in the gas cooling faster 352 | and the temperture returning to its equilibruum value. On the other 353 | hand, if the heating rate increases with temperature faster than the 354 | cooling rate, the slight temperature increase will be compounded with 355 | a further increase in temperature. 356 | 357 | 358 | \section{Problems} 359 | \begin{enumerate} 360 | \item{\bf X-ray Bursts:} 361 | 362 | We will try to model Type-I X-ray bursts using a simple model for the instability. We will calculate how much material will accumulate on a neutron star before it bursts. 363 | \begin{enumerate} 364 | \item Let us assume that the star accretes pure helium, that the 365 | temperature of the degenerate layer is constant down to the core 366 | ($T_c$), how much luminosity emerges from the surface of the star? 367 | 368 | \item Let us assume that the helium layer has a mass, $dM$, and that the enregy generation rate for helium burning is given by 369 | $$ 370 | \epsilon_{3\alpha} = 3.5 \times 10^{20} T_9^{-3} \exp(-4.32/T_9) \mathrm{erg s}^{-1} \mathrm{g}^{-1} 371 | $$ 372 | where $T_9=T/10^9 \mathrm{K}$. The energy generation rate is a 373 | function of density too, but let's forget about that to keep things 374 | simple. How much power does the helium layer generate as a function of 375 | $dM$? 376 | 377 | \item Equate your answer to (a) to the answer to (b) and solve for 378 | $dM$. This is the thickness of a layer in thermal equilibrium. 379 | 380 | \item Let's assume that the potential burst starts by the temperature 381 | in the accreted layer jiggling up by a wee bit. If the surface 382 | luminosity increases faster with temperature than the helium burning 383 | rate, then the layer is stable. Calculate $dL_\mathrm{surface}/dT$ and 384 | $dP_\mathrm{helium}/dT$. 385 | 386 | \item Calculate the value of $dM$ for which $dP_\mathrm{helium}/dT$ 387 | exceeds $dL_\mathrm{surface}/dT$ and the layer bursts. 388 | 389 | \item Equate your value of $dM$ in (c) and (e) and solve for $T$. What 390 | is $dM$? How long will it take for such a layer to accumulate if the 391 | star is accreting at one-tenth of the Eddington accretion rate? 392 | 393 | \end{enumerate} 394 | \end{enumerate} 395 | %%% Local Variables: 396 | %%% TeX-master: "book" 397 | %%% End: -------------------------------------------------------------------------------- /chapters/chap10.tex: -------------------------------------------------------------------------------- 1 | \chapter{Molecular Structure} 2 | \label{cha:molecular-structure} 3 | \index{molecular structure} 4 | We are to deal with the structure and energetics of molecules in a 5 | very heuristic fashion, deriving the energies and importance 6 | of various transitions. 7 | 8 | The simplest molecules are the diatomic molecules such as H$_2$, CO, 9 | {\em etc.}. They are also among the most abundant in the universe, so 10 | we are going to restrict our attention to these two-atom molecules. 11 | 12 | \section{The Born-Oppenheimer Approximation} 13 | \label{sec:born-oppenh-appr} 14 | \index{molecular structure!Born-Oppenheimer approximation} 15 | \index{Born-Oppenheimer approximation} 16 | 17 | The problem of understanding the structure of molecules initially 18 | appears formidable. At a basic level the equations are no longer 19 | spherically symmetric. This is a real difficulty. For diatomic 20 | molecules there is still a rotational symmetry about the line 21 | connecting the two nuclei. The key simplification is that the 22 | electrons whip around a lot faster than the nuclei, so one can 23 | approximate the sitution by assuming that the electrons sit in a 24 | particular eigenstate of the potential with the two ions fixed. 25 | The ions on the other hand experience an effective potential as a 26 | function of their separation that includes the effects of the 27 | electrons (whose state we have already calculated). This in the {\em 28 | Born-Oppenheimer} approximation. 29 | 30 | By looking a moleculle in terms of the electrons and the nuclei 31 | separately, we can estimate the energies of the various transitions of 32 | the molecule. Let's assume that the ions are separated by a distance 33 | $a \sim a_0$. By the uncertainty principle the momentum of the 34 | electrons will be on the order of $\hbar / a$, and the typical energy 35 | of electronic transitions will be 36 | \begin{equation} 37 | E_\textrm{\scriptsize elec} \sim \frac{p^2}{m} \sim 38 | \frac{\hbar^2}{m a^2} \sim \alpha^2 m c^2 \sim 1~\textrm{eV} 39 | \label{eq:10000} 40 | \end{equation} 41 | or the visibile, near-infrared and ultraviolet. 42 | 43 | The nuclei are separated by a distance of order $a$ as well and the 44 | typical energy change from moving nuclei over a distance $a$ is the 45 | electronic energy (Eq.~\ref{eq:10000}), so we can define a spring 46 | constant for the nuclei 47 | \begin{equation} 48 | k \sim \frac{E_\textrm{\scriptsize elec}}{a^2} \sim \frac{\hbar^2}{m a^4} 49 | \label{eq:10001} 50 | \end{equation} 51 | yielding a vibrational energy corresponding to changes in the distance 52 | between the nuclei of 53 | \begin{equation} 54 | E_\textrm{\scriptsize vib} \sim \hbar \omega = \hbar \left 55 | (\frac{k}{M} \right )^{1/2} 56 | = \hbar \left ( \frac{\hbar^2}{m M a^4} \right )^{1/2} 57 | = \left ( \frac{m}{M} \right )^{1/2} \frac{\hbar^2}{m a^2} 58 | \sim 0.01-0.1~\textrm{eV}. 59 | \label{eq:10002} 60 | \end{equation} 61 | These energies fall in the infrared. Finally the molecule can change 62 | its rotational state. The angular momentum of the molecule is 63 | quanized in units of $\hbar$ so we would expect transitions with 64 | energies of the order of 65 | \begin{equation} 66 | E_\textrm{\scriptsize rot} \sim \frac{\hbar^2 L \left (L+1\right) }{2 67 | I} = \frac{\hbar^2 L \left (L+1\right )}{M 68 | a^2} = \frac{m}{M} \frac{\hbar^2}{m a^2} L \left (L+1\right )\sim 10^{-3}~\textrm{eV}. 69 | \label{eq:10003} 70 | \end{equation} 71 | Because the typical energies of the various transitions are well 72 | separated we can to a good approximate consider each of them 73 | separately, justifying the Born-Oppenheimer approximation. 74 | 75 | \section{The H$_2^+$ Molecular Ion} 76 | 77 | An example which illustrates much of the physics of diatomic molecules 78 | is the hydrogen molecular ion H$_2^+$. The Schrodinger equation for 79 | this system is 80 | \begin{equation} 81 | \left [ -\frac{\hbar^2}{2 \mu_{AB}} \nabla^2_{\bf R} 82 | -\frac{\hbar^2}{2 \mu_e} \nabla^2_{\bf r} - \frac{e^2}{r_A} - 83 | \frac{e^2}{r_B} + \frac{e^2}{R} - E \right ] \psi({\bf r},{\bf R}) = 0 84 | \label{eq:573} 85 | \end{equation} 86 | $\mu_{AB}$ is the reduced mass of the two protons, $M/2$ and $\mu_e$ 87 | is the reduced mass of the electron relative to the two protons 88 | $\approx m_e$. $r_A$ and $r_B$ are the distances between the 89 | electron and the two protons and $R$ is the distance between the two 90 | protons. The key to the Born-Oppenheimer approximation is first to 91 | hold ${\bf R}$ fixed and neglect the first kinetic energy term and 92 | solve for the electronic wavefunction $\chi_j({\bf r};{\bf R})$. 93 | \begin{equation} 94 | \left [ 95 | -\frac{\hbar^2}{2 \mu_e} \nabla^2_{\bf r} - \frac{e^2}{r_A} - 96 | \frac{e^2}{r_B} + \frac{e^2}{R} - E_j({\bf R}) \right ] \chi_j({\bf r};{\bf R}) = 0 97 | \label{eq:574} 98 | \end{equation} 99 | where the semicolon in the $\chi_j$ function encourages us to think of 100 | ${\bf R}$ as a parameter. We try various values of ${\bf R}$ and 101 | solve for $\chi_j({\bf r};{\bf R})$ each time. The solutions to this 102 | equation are called {\em molecular orbitals}. 103 | 104 | After the electronic wavefuntion is calculated as a function of $R$, 105 | we can determine the proton wavefunction. The proton wavefunction 106 | satisfies the one-body Schrodinger equation 107 | \begin{equation} 108 | \left [ 109 | -\frac{\hbar^2}{2 \mu_{AB}} \nabla^2_{\bf R} + E_j({\bf R}) - E \right 110 | ] F_j({\bf R}) = 0 111 | \label{eq:575} 112 | \end{equation} 113 | 114 | Eq.~\ref{eq:574} is generally to difficult to solve directly, so one generally 115 | picks a trial wavefunction and calculates the value of the energy for 116 | this function. One can prove that the ground state eigenvalue $E$ of the 117 | Hamiltonian $H$ 118 | \begin{equation} 119 | E \leq \langle \psi | H | \psi \rangle 120 | \label{eq:576} 121 | \end{equation} 122 | where $\psi$ is any normalized wavefunction. This is the basis of the 123 | {\em Rayleigh-Ritz variational method}. 124 | 125 | For the case in point, we will guess that the wavefunction of the 126 | electron is a {\bf l}inear {\bf c}ombination of {\bf a}tomic {\bf 127 | o}rbitals (LCAO), specifically the $1s$ state of hydrogen. 128 | \begin{equation} 129 | \chi_g = \frac{1}{\sqrt{2}} \left ( \psi_{1s} (r_A) + \psi_{1s}(r_B) 130 | \right ) 131 | \label{eq:577} 132 | \end{equation} 133 | and 134 | \begin{equation} 135 | \chi_u = \frac{1}{\sqrt{2}} \left ( \psi_{1s} (r_A) - \psi_{1s}(r_B) 136 | \right ) 137 | \label{eq:578} 138 | \end{equation} 139 | The $g$ and the $u$ refer to {\em gerade} (even-parity) and {\em 140 | ungerade} (odd-parity) wavefunctions. 141 | 142 | We can substitute these trial wavefunctions into the Hamiltonian in 143 | Eq.~(\ref{eq:574}) to find an upper limit on the value of $E_j({\bf R})$. We 144 | obtain 145 | \begin{equation} 146 | E_{gu} (R) = E_{1s} + \frac{e^2}{R} \frac{(1 + 147 | R/a_0)e^{-2R/a_0}\pm[1-(2/3)(R/a_0)^2] e^{-R/a_0}}{1\pm[1+R/a_0+(R/a_0)^2/3]e^{-R/a_0}} 148 | \label{eq:579} 149 | \end{equation} 150 | where the upper (positive) sign corresponds to the {\em gerade} configuration. 151 | Fig.~\ref{fig:h2p} depicts the energy of the electronic configuration 152 | and Fig.~\ref{fig:psiug} shows the electron density for the two 153 | orbitals. We see that the only the gerade state binds in the case. 154 | From the picture of the electron probability density we can see why 155 | this is the case. In the gerade case, the electron lies in between 156 | the two ions so it can shield the charge of one ion from the other. 157 | In the ungerade state because it has odd parity, the electron cannot 158 | lie on the midplane between the ions so the shielding is much less 159 | effective. 160 | { 161 | \begin{figure} 162 | \begin{center} 163 | \begin{tikzpicture} 164 | \begin{scope}[yscale=20,smooth] 165 | \clip (0,-0.1) rectangle (6,0.2); 166 | \draw[color=red] plot file{morse/xegx.dat} ; 167 | \draw[color=blue] plot file{morse/xeux.dat} ; 168 | \draw[dashed] plot file{morse/xmorsex.dat} ; 169 | \end{scope} 170 | \begin{scope}[yscale=20] 171 | \draw[->] (-0.2,-0.1) -- (6.2,-0.1) node[right] {$\frac{R}{a_0}$}; 172 | \draw[->] (0,-0.12) -- (0,0.22) node[above] {$\frac{E a_0}{e^2}$}; 173 | \foreach \x in {1,...,6} 174 | \draw (\x,-0.1) node [below] {$\x$}; 175 | \foreach \x in {-0.1,0,0.1,0.2} 176 | \draw (-0.1,\x) node [left] {$\x$}; 177 | \end{scope} 178 | \end{tikzpicture} 179 | \end{center} 180 | %\centering 181 | %\includegraphics[width=0.45\textwidth]{h2p} 182 | \caption{$E_{u}$ (upper curve) and $E_g$ (lower curve) for 183 | H$_2^+$. The dashed curve is a well-fit Morse potential for $E_g(R)$.} 184 | \label{fig:h2p} 185 | \end{figure} 186 | } 187 | { 188 | \begin{figure} 189 | \centering 190 | \includegraphics[width=0.45\textwidth]{psig} 191 | \includegraphics[width=0.45\textwidth]{psiu} 192 | \caption{$|\psi_g|^2$ (left) and $|\psi_u|^2$ (right) for H$_2^+$ with 193 | $R=2 a_0$.} 194 | \label{fig:psiug} 195 | \end{figure} 196 | } 197 | 198 | For molecules with more than one electron, we find that the Hund's rule 199 | for the total spin of a system is reversed for molecules. The states 200 | with even parity (gerade) tend to bond. Because the spatial wavefunction is 201 | even with respect interchanging the electrons their spins must be 202 | antiparallel. 203 | 204 | \section{Molecular Excitations} 205 | \label{sec:molec-excit} 206 | \index{molecular structure!excitations} 207 | 208 | The energy states of molecule may be excited in three ways: {\em 209 | electronic}, {\em vibrational}, and {\em rotational}. Let's start 210 | with the least energetic of these. 211 | 212 | We can get a first-order understanding of the rotational states of a 213 | molecule simply looking at the Schrodinger equation for the ions 214 | \be 215 | \left [ 216 | -\frac{\hbar^2}{2 \mu_{AB}} \left ( \frac{d^2}{dR^2} - 217 | \frac{L(L+1)}{R^2} \right ) + E_j({\bf R}) - E \right 218 | ] F_j({\bf R}) = 0 219 | \label{sec:molec-excit-1}\end{equation} 220 | where we have solved the angular wavefunction in terms of spherical 221 | harmonics like we did for hydrogen. In Fig.~\ref{fig:h2p} we saw that 222 | the function $E_j({\bf R})$ varies over atomic distances $\sim a_0$. 223 | On the other hand because the mass of the ions is much larger than 224 | that of the electrons we expect the wavefunction of the ions to be 225 | localized in a region $\sim a_0 m/M \ll a_0$. Over this small region 226 | we can expand the function $E_j(R)$ about its minimum $R_0$ 227 | \begin{equation} 228 | E_j(R) = E_j(R_0) + (R-R_0) \left [ \dd{E_j(R)}{R} \right ]_{R=R_0} + 229 | \frac{1}{2} (R-R_0)^2 \left [ \frac{d^2 E_j(R)}{d R^2} \right 230 | ]_{R=R_0} + \cdots 231 | \label{eq:580} 232 | \end{equation} 233 | where the second term vanishes because $R_0$ is the minimum so we have 234 | \be 235 | \left [ 236 | -\frac{\hbar^2}{2 \mu_{AB}} \frac{d^2}{dR^2} - 237 | \frac{\hbar^2}{2\mu_{AB}} \frac{L(L+1)}{R^2} + E_j(R_0) + \frac{1}{2} 238 | k (R-R_0)^2 - E \right 239 | ] F_j({\bf R}) = 0 240 | \label{sec:molec-excit-2}\end{equation} 241 | so we have 242 | \begin{equation} 243 | E = E_j(R_0) + \hbar \omega_0 \left ( v + \frac{1}{2} \right ) + \frac{\hbar^2}{2\mu_{AB}} \frac{L(L+1)}{R_0^2} 244 | \label{eq:581} 245 | \end{equation} 246 | where $\omega_0 = (k/\mu_{AB})^{1/2}$ and 247 | we have treated the rotational motion of the molecule 248 | perturbatively. We have effectively ignored the possible centrifugal 249 | stretching of the molecule. If we were to include the stretching of 250 | the molecule we would have 251 | \begin{equation} 252 | E_\rmscr{rot} = \frac{\hbar^2}{2\mu_{AB}} \frac{L(L+1)}{R_0^2} \left [ 253 | 1 - \frac{2\hbar^2 L(L+1)}{k \mu_{AB} R_0^4} \right ] 254 | \label{eq:582} 255 | \end{equation} 256 | 257 | We can get dipole transitions between the different rotational states 258 | if 259 | \begin{equation} 260 | |d| = Z_1 e r_1 + Z_2 e r_2 + |d_e| \neq 0 261 | \label{eq:583} 262 | \end{equation} 263 | and $\Delta L = \pm 1$. We see that homonuclear diatomic molecules 264 | cannot emit dipole radiation due to changes in their rotational state. 265 | The energy of the radiation is given by 266 | \begin{equation} 267 | E_{L+1,L} = \frac{\hbar^2 (L+1)}{\mu_{AB} R_0^2} \left [ 1 - 4 268 | \frac{\hbar^2 (L+1)^2}{k \mu_{AB} R_0^4} \right ] 269 | \label{eq:584} 270 | \end{equation} 271 | This energy is $\sim e^2/a_0 (m/\mu_{AB})$ or $\sim 10^{-3}$~eV. 272 | 273 | The transitions between vibrational states has a typical energy of 274 | $\sim e^2/a_0 (m/\mu_{AB})^{1/2}$ or $\sim 10^{-1}$~eV. If one 275 | includes the centrifugal effects one finds that 276 | \begin{equation} 277 | E_v = \hbar \omega_L \left ( v + \frac{1}{2} \right ) 278 | \label{eq:585} 279 | \end{equation} 280 | where 281 | \index{Morse function} 282 | \index{molecular structure!Morse function} 283 | \begin{equation} 284 | \omega_L \approx \mu_{AB}^{-1/2} \left [ k + \frac{3 \hbar^2 285 | L(L+1)}{\mu_{AB} R_0^4} \right ]^{1/2} 286 | \label{eq:586} 287 | \end{equation} 288 | Morse found that the internuclear potential can often be well 289 | approximated by a function of the form 290 | \begin{equation} 291 | E_n(R) = E_{n,0} + B_n \left \{ 1 - \exp \left[ \beta_n \left ( R - 292 | R_0 \right )\right ] \right \}^2 293 | \label{eq:587} 294 | \end{equation} 295 | The energy eigenvalues of this potential are 296 | \begin{equation} 297 | E_{nv} = \hbar \omega_{n0} \left ( v + \frac{1}{2} \right ) - 298 | \frac{\hbar^2 \omega_0^2}{4 B_n} \left ( v + \frac{1}{2} \right )^2. 299 | \label{eq:588} 300 | \end{equation} 301 | The vibrational levels get closer together as $v$ increases and there 302 | are a finite number of vibrational levels 303 | \begin{equation} 304 | 0 \leq v \leq \frac{(2\mu_{AB} B_n)^{1/2}}{\beta_n \hbar} - \frac{1}{2} 305 | \label{eq:589} 306 | \end{equation} 307 | 308 | The selection rules for vibrational transtions are again $|d|\neq 0$ but 309 | also $d|d|/dR \neq 0$. We can change the vibrational level by $\Delta 310 | v=\pm 1$ and we must also have $\Delta L=L_\textrm{\scriptsize lower}-L_\textrm{\scriptsize upper}=+1$ ($P$ branch) or $\Delta 311 | L=-1$ ($R$ branch) or if there is an component of electronic orbital 312 | or spin angular momentum along the internuclear axis $\Delta L=0$ ($Q$ branch). 313 | 314 | Fig.~\ref{fig:molecule_trans} shows the three branches for 315 | roto-vibrational transitions, if we neglect the stretching of the 316 | molecule. We see that 317 | for the $R$ branch the transition energy decreases with increasing $L$. 318 | For the $Q$ branch it is constant, and for the $P$ branch the 319 | transition energy increases with increase $L$. The centrifugal stretching reduces the spacing of the 320 | angular momentum energy levels for large values of $L$ (Eq.~\ref{eq:584}), but it 321 | stiffens the spring constant of the vibrational statres (Eq.~\ref{eq:586}). The latter 322 | dominates, so the stretching effect 323 | tends to make the transition energies increase with increasing $L$ for 324 | large values of $L$. Fig.~\ref{fig:COCar} depicts the 325 | roto-vibrational spectrum of CO (the most commonly observed molecule 326 | in astrophysics --- it isn't the most common, what is the most common 327 | molecule and why isn't it commonly observed?) from samples of car 328 | exhaust. The CO molecule does not exhibit a $Q$-branch which would 329 | appear at about 2140~cm$^{-1}$. 330 | 331 | The Fortrat diagram (Fig.~\ref{fig:fortrat}) 332 | depicts the transition energies for various roto-vibrational 333 | transitions as a function of the rotational quantum number $L$. The 334 | figure shows the expected behaviour, especially for the ``22'' 335 | transitions for which all three branches are depicted. For small 336 | values of $L$ the $Q$-branch has constant energy with $L$ and then 337 | begins to increase with increasing $L$ The energy of the $P$-branch 338 | transitions decreases with increasing $L$, and the opposite occurs 339 | for the $R$ branch. 340 | 341 | \begin{figure} 342 | \begin{center} 343 | \begin{tikzpicture}[scale=0.95] 344 | \foreach \nqn in {0,1} 345 | \foreach \lqn in {0,1,2,3} 346 | \draw (0,\nqn*4+0.15*\lqn*\lqn+0.15*\lqn) -- (10,\nqn*4+0.15*\lqn*\lqn+0.15*\lqn) node [right] {$L=\lqn$}; 347 | \draw (0,5) node [left] {$\nu=1$} (0,1) node [left] {$\nu=0$}; 348 | \draw [->] (1,0.3) -- (1,4); 349 | \draw [->] (1.5,0.9) -- (1.5,4.3); 350 | \draw [->] (2,1.8) -- (2,4.9); 351 | \draw (1.5,0) node [below] {$\Delta L=+1$ $P$ branch}; 352 | \draw [->] (4.5,0) -- (4.5,4); 353 | \draw [->] (5,0.3) -- (5,4.3); 354 | \draw [->] (5.5,0.9) -- (5.5,4.9); 355 | \draw (5,0) node [below] {$\Delta L=0$ $Q$ branch}; 356 | \draw [->] (8,0) -- (8,4.3); 357 | \draw [->] (8.5,0.3) -- (8.5,4.9); 358 | \draw [->] (9,0.9) -- (9,5.8); 359 | \draw (8.5,0) node [below] {$\Delta L=-1$ $R$ branch}; 360 | \end{tikzpicture} 361 | \end{center} 362 | \caption{Roto-Vibrational Transitions Neglecting Stretching} 363 | \label{fig:molecule_trans} 364 | \end{figure} 365 | 366 | {\begin{figure} 367 | \centering 368 | \begin{tikzpicture}[scale=0.9] 369 | \draw (0,0) node { 370 | \includegraphics[width=0.9\textwidth,clip,trim=2.8cm 1.68cm 2.1cm 3.93cm]{CO_Car} 371 | }; 372 | \draw [->] (-6.2,-3.25) -- (-6.2,3.05); 373 | \draw [->] (-6.3,-3.15) -- (6.5,-3.15); 374 | \draw (0,-3.6) node [below] {Wavenumber [cm$^{-1}$]}; 375 | \foreach \lam in {2040,2060,...,2220} 376 | \draw (0.05893*\lam-125.5,-3.2) node [below] {\lam}; 377 | \foreach \abs in {0,0.1,0.2,0.3} 378 | \draw (-6.2,18.6*\abs-3.15) node [left] {\abs}; 379 | \end{tikzpicture} 380 | \caption{Absorption vs.\ Wavenumber for FTIR Spectroscopy of CO in car 381 | exhaust from {\tt 382 | http://home.swipnet.se/$\sim$w-74877/ftir/ftir.htm}. Green is a 383 | cold engine, blue is a warm engine, red is a calibration reading and 384 | cyan is a 100ppm calibration.} 385 | \label{fig:COCar} 386 | \end{figure}} 387 | {\begin{figure} 388 | \centering 389 | \begin{tikzpicture}[scale=0.9] 390 | \draw (0,0) node { 391 | \includegraphics[width=0.9\textwidth,clip,trim=1cm 4.6cm 0 6.7cm]{fortrat.png} 392 | }; 393 | \draw (-6.5,0.3) node [rotate=90] {Rotational Quantum Number}; 394 | \end{tikzpicture} 395 | \caption{A Fortrat diagram for $^{24}$Mg$^{35}$Cl and 396 | $^{24}$Mg$^{37}$Cl (isotope) from Gutterres et al. (2003), 397 | Braz. J. Phys. vol. 33} 398 | \label{fig:fortrat} 399 | \end{figure}} 400 | 401 | In general each vibration transition includes a rotational transition 402 | as well so one gets group of transitions. The final wrinkle is that 403 | electronic transitions in molecules whose energy $\sim 1$~eV 404 | necessarily include changes in the rotational and vibrational state of 405 | the molecule. The general electronical-vibrational-rotational 406 | spectrum takes the form of bands which can be resolved into individual 407 | lines if the broadening is weak. 408 | 409 | \section{Problems} 410 | \begin{enumerate} 411 | 412 | \item{\bf The Number of Levels} 413 | 414 | I fit a Morse function to the potential of H$_2^+$. The parameters 415 | were 416 | \begin{equation} 417 | E_{n,0} = -0.065 \frac{e^2}{a_0}, B_n = 0.07 \frac{e^2}{a_0}, \beta_n 418 | = 0.7 a_0^{-1}, R_0 = 2.5 a_0 419 | \label{eq:590} 420 | \end{equation} 421 | How many vibrational levels does H$_2^+$ have? How many rotational 422 | levels does each vibrational level typically have? 423 | 424 | \item{\bf Nuclear Overlap} 425 | 426 | Consider two deuterons bound by a single electron as in question (1). 427 | What is the probability that the two deuterons lie on top of each 428 | other, {\ie} that $R<4$~fermi, the diameter of the deuteron? What is the 429 | probability if the two deuterons are bound by a single muon, $m_\mu 430 | \approx 207 m_e$? You can find the eigenfunctions of the Morse 431 | potential on Wikipedia. 432 | 433 | If you assume that whenever the deuterons overlap they fuse and that 434 | you get to ``roll the dice'' once each oscillation period, calculate 435 | the fusion rate in both cases. 436 | 437 | \item{\bf Stretching} 438 | 439 | Calculate the value of $L$ for which the energy of the $P$ branch 440 | transitions begins to increase. 441 | 442 | \item{\bf Temperature} 443 | 444 | Using the results depicted in Fig.~\ref{fig:COCar}, estimate the 445 | temperature of the hot and cold car exhaust and the relative 446 | concentration of CO in the two cases. 447 | \end{enumerate} 448 | 449 | %%% Local Variables: 450 | %%% mode: latex 451 | %%% TeX-master: "book" 452 | --------------------------------------------------------------------------------