-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathnotes-lecture-nuclear-burning.html
1484 lines (1305 loc) · 58.6 KB
/
notes-lecture-nuclear-burning.html
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
386
387
388
389
390
391
392
393
394
395
396
397
398
399
400
401
402
403
404
405
406
407
408
409
410
411
412
413
414
415
416
417
418
419
420
421
422
423
424
425
426
427
428
429
430
431
432
433
434
435
436
437
438
439
440
441
442
443
444
445
446
447
448
449
450
451
452
453
454
455
456
457
458
459
460
461
462
463
464
465
466
467
468
469
470
471
472
473
474
475
476
477
478
479
480
481
482
483
484
485
486
487
488
489
490
491
492
493
494
495
496
497
498
499
500
501
502
503
504
505
506
507
508
509
510
511
512
513
514
515
516
517
518
519
520
521
522
523
524
525
526
527
528
529
530
531
532
533
534
535
536
537
538
539
540
541
542
543
544
545
546
547
548
549
550
551
552
553
554
555
556
557
558
559
560
561
562
563
564
565
566
567
568
569
570
571
572
573
574
575
576
577
578
579
580
581
582
583
584
585
586
587
588
589
590
591
592
593
594
595
596
597
598
599
600
601
602
603
604
605
606
607
608
609
610
611
612
613
614
615
616
617
618
619
620
621
622
623
624
625
626
627
628
629
630
631
632
633
634
635
636
637
638
639
640
641
642
643
644
645
646
647
648
649
650
651
652
653
654
655
656
657
658
659
660
661
662
663
664
665
666
667
668
669
670
671
672
673
674
675
676
677
678
679
680
681
682
683
684
685
686
687
688
689
690
691
692
693
694
695
696
697
698
699
700
701
702
703
704
705
706
707
708
709
710
711
712
713
714
715
716
717
718
719
720
721
722
723
724
725
726
727
728
729
730
731
732
733
734
735
736
737
738
739
740
741
742
743
744
745
746
747
748
749
750
751
752
753
754
755
756
757
758
759
760
761
762
763
764
765
766
767
768
769
770
771
772
773
774
775
776
777
778
779
780
781
782
783
784
785
786
787
788
789
790
791
792
793
794
795
796
797
798
799
800
801
802
803
804
805
806
807
808
809
810
811
812
813
814
815
816
817
818
819
820
821
822
823
824
825
826
827
828
829
830
831
832
833
834
835
836
837
838
839
840
841
842
843
844
845
846
847
848
849
850
851
852
853
854
855
856
857
858
859
860
861
862
863
864
865
866
867
868
869
870
871
872
873
874
875
876
877
878
879
880
881
882
883
884
885
886
887
888
889
890
891
892
893
894
895
896
897
898
899
900
901
902
903
904
905
906
907
908
909
910
911
912
913
914
915
916
917
918
919
920
921
922
923
924
925
926
927
928
929
930
931
932
933
934
935
936
937
938
939
940
941
942
943
944
945
946
947
948
949
950
951
952
953
954
955
956
957
958
959
960
961
962
963
964
965
966
967
968
969
970
971
972
973
974
975
976
977
978
979
980
981
982
983
984
985
986
987
988
989
990
991
992
993
994
995
996
997
998
999
1000
<!DOCTYPE html>
<html lang="en">
<head>
<!-- 2025-02-06 Thu 09:04 -->
<meta charset="utf-8" />
<meta name="viewport" content="width=device-width, initial-scale=1" />
<title>400A - Nuclear burning</title>
<meta name="author" content="Mathieu Renzo" />
<meta name="generator" content="Org Mode" />
<link rel="stylesheet" href="./css/style.css" />
<link rel="stylesheet" href="./fontawesome-free-6.7.2-web/css/all.min.css">
<meta name="keywords" content="Mathieu, Renzo, Mathieu Renzo,
stellar evolution, 400A, University of
Arizona, Steward Observatory, stars,
theoretical astrophysics">
<script type="text/x-mathjax-config">
MathJax.Hub.Config({
displayAlign: "center",
displayIndent: "0em",
"HTML-CSS": { scale: 100,
linebreaks: { automatic: "false" },
webFont: "TeX"
},
SVG: {scale: 100,
linebreaks: { automatic: "false" },
font: "TeX"},
NativeMML: {scale: 100},
TeX: { equationNumbers: {autoNumber: "AMS"},
MultLineWidth: "85%",
TagSide: "right",
TagIndent: ".8em"
}
});
</script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/mathjax/2.7.0/MathJax.js?config=TeX-AMS_HTML"></script>
</head>
<body>
<div id="preamble" class="status">
<!-- Preamble -->
<!-- The header -->
<div class="header">
<!-- The site name -->
<div class="site-name">
<a id="top" href="./index.html">Stellar Evolution</a>
</div>
<!-- The hamburger -->
<div class="hamburger">
<div id="myLinks" class="menu">
<a href="./index.html">Home</a>
<a href="./syllabus.html">Syllabus</a>
<a href="./lectures.html">Lectures</a>
<a href="./projects.html">Projects</a>
</div>
<a href="javascript:void(0);" class="icon" onclick="HamburgerMenuFunction()">
<i class="fa fa-bars"></i>
</a>
</div>
<div class="navbar">
<a href="./syllabus.html">Syllabus</a>
<a href="./lectures.html">Lectures</a>
<a href="./projects.html">Projects</a>
</div>
</div>
<!-- scripts -->
<script>
function HamburgerMenuFunction() {
var x = document.getElementById("myLinks");
if (x.style.display === "block") {
x.style.display = "none";
} else {
x.style.display = "block";
}
}
</script>
<script>
window.onload = () => {
const toggleButton = document.getElementById("light-dark-toggle");
const body = document.body;
const html = document.documentElement;
// Check localStorage for user preference
if (localStorage.getItem("theme") === "dark") {
body.classList.add("dark-mode");
html.classList.toggle("dark-mode");
}
// Toggle theme on click
toggleButton.addEventListener("click", () => {
body.classList.toggle("dark-mode");
html.classList.toggle("dark-mode");
// Save user preference
if (body.classList.contains("dark-mode")) {
localStorage.setItem("theme", "dark");
} else {
localStorage.setItem("theme", "light");
}
});
};
</script>
<!-- end scripts-->
</div>
<div id="content" class="content">
<p>
<b>Materials:</b> Onno Pol's lecture notes Chapter 6.1 and 6.2, Kippenhahn
book Chapter 9 and 18, Cox & Giuli vol. I, Chapter 17.7, Clayton
Chapter 4, <a href="https://ui.adsabs.harvard.edu/abs/1928Natur.122..805G/abstract">Gamow 1928</a>, <a href="https://ui.adsabs.harvard.edu/abs/1999RPPh...62..395A/abstract">Arnould & Takahashi 1999</a>, <a href="https://ui.adsabs.harvard.edu/abs/2011RPPh...74i6901J/abstract">Jose & Iliadis 2011</a>
</p>
<div id="outline-container-orge274d95" class="outline-2">
<h2 id="orge274d95"><a href="#orge274d95">Microphysics of nuclear burning</a></h2>
<div class="outline-text-2" id="text-orge274d95">
</div>
<div id="outline-container-orgbae2050" class="outline-3">
<h3 id="orgbae2050"><a href="#orgbae2050">Summary of where we are</a></h3>
<div class="outline-text-3" id="text-orgbae2050">
<p>
We have four equations of the stellar <i>structure</i> assuming spherical
symmetry, LTE, and hydrostatic (∂ t ≡ 0) equilibrium:
</p>
</div>
<div id="outline-container-orgc721979" class="outline-4">
<h4 id="orgc721979"><a href="#orgc721979">Mass conservation</a></h4>
<div class="outline-text-4" id="text-orgc721979">
<div class="latex" id="org28f39e4">
\begin{equation}\label{eq:mass_cont}
\frac{dr}{dm} = \frac{1}{4\pi r^{2}\rho}\ \ .
\end{equation}
</div>
</div>
</div>
<div id="outline-container-orgd6e4d33" class="outline-4">
<h4 id="orgd6e4d33"><a href="#orgd6e4d33">Hydrostatic equilibrium</a></h4>
<div class="outline-text-4" id="text-orgd6e4d33">
<div class="latex" id="org41a1949">
\begin{equation}\label{eq:HSE}
\frac{dP}{dm} = -\frac{Gm}{4\pi r^{4}} \ \ ,
\end{equation}
</div>
<p>
which follows from the momentum conservation equation.
</p>
</div>
</div>
<div id="outline-container-org62916c2" class="outline-4">
<h4 id="org62916c2"><a href="#org62916c2">Equation of state</a></h4>
<div class="outline-text-4" id="text-org62916c2">
<div class="latex" id="orgc2cd06b">
\begin{equation}
P_\mathrm{tot} = P_\mathrm{gas} + P_\mathrm{rad} = \frac{\rho}{\mu m_{u}}k_{B}T + P_{QM} + \frac{1}{3}aT^{4} \ \ .
\end{equation}
</div>
</div>
</div>
<div id="outline-container-orga92cdb2" class="outline-4">
<h4 id="orga92cdb2"><a href="#orga92cdb2">Energy transport</a></h4>
<div class="outline-text-4" id="text-orga92cdb2">
<div class="latex" id="orgea984c4">
\begin{equation}
\frac{dT}{dm} = \frac{T}{P}\frac{dP}{dm}\nabla
\end{equation}
</div>
<p>
where ∇ =∂ log(T)/∂ log(P) is the local temperature gradient, equal to
the radiative gradient in stably stratified regions:
</p>
<div class="latex" id="orge174931">
\begin{equation}
\nabla \equiv \nabla_\mathrm{rad} = \frac{3 P}{16\pi acGm T^{4}}\kappa L
\end{equation}
</div>
<p>
with κ = (1/κ<sub>rad</sub> + 1/κ<sub>cond</sub>)<sup>-1</sup> the combination "in parallel" of the
radiative and conductive opacity (assumed to be known from atomic
physics), and ∇ ≡ ∇<sub>ad</sub> the adiabatic gradient (within ∼10<sup>-7-8</sup>
precision) for convective regions. We also have a criterion
(Schwarzschild or Ledoux) to determine which region is which.
</p>
</div>
</div>
<div id="outline-container-org3752f52" class="outline-4">
<h4 id="org3752f52"><a href="#org3752f52">Energy conservation</a></h4>
<div class="outline-text-4" id="text-org3752f52">
<div class="latex" id="org021efed">
\begin{equation}
\frac{dL}{dm} = \varepsilon_\mathrm{nuc} -\varepsilon_{\nu} + \varepsilon_\mathrm{grav} \ \ .
\end{equation}
</div>
<p>
with ε<sub>grav</sub> = T∂ s/∂ t the change in internal energy (typically ≅ 0 in
gravothermal equilibrium, that is for phases when the evolution is
much slower than τ<sub>KH</sub>), and ε<sub>ν</sub>>0.
</p>
</div>
</div>
</div>
<div id="outline-container-orgfea5221" class="outline-3">
<h3 id="orgfea5221"><a href="#orgfea5221">The next step</a></h3>
<div class="outline-text-3" id="text-orgfea5221">
<p>
With all the equations above we almost have a complete set of solvable
differential equations. What is left to discuss is the specific energy
generation per unit time ε<sub>nuc</sub> due to nuclear burning (which will
occupy us in this and <a href="./notes-lecture-nuclear-cycles.html">the next lecture</a>) and the specific energy loss
ε<sub>ν</sub> due to neutrino losses (which we will treat <a href="./notes-lecture-neutrinos.html">later</a>).
</p>
<p>
In doing so, we will introduce another time-dependent equation
(besides the implicit dependence in ε<sub>grav</sub>): as nuclear burning
proceeds, the chemical composition changes, on a timescale that we
will see is extremely long compared to the thermal and dynamical
timescales.
</p>
</div>
</div>
<div id="outline-container-org75630b9" class="outline-3">
<h3 id="org75630b9"><a href="#org75630b9">Energy generation as a self-regulating process</a></h3>
<div class="outline-text-3" id="text-org75630b9">
<p>
We have seen in the <a href="./notes-lecture-VirTheo.html">virial theorem lecture</a> that one can derive from
first principle that stars have a "negative heat capacity" and obtain
a scaling between mass, radius and average temperature:
</p>
<div class="latex" id="org5520176">
\begin{equation}
\langle T \rangle = \frac{2\mu m_{u}}{3} C \frac{GM^{}}{R} \propto \frac{\mu M}{R} \ \ ,
\end{equation}
</div>
<p>
where C was a constant depending on the details of the mass
distribution.
</p>
<p>
The star has a finite temperature, thus it radiates away thermal
energy at the surface in the form of a luminosity L. This means the
gravitational potential energy is decreasing (because of the virial
theorem!), thus it must contract leading to an increase in T. For the
Sun this would go on for ∼ 15 Myrs before it runs out of all of its
internal energy, which is way too short a timescale to explain many
Earthly observables.
</p>
<p>
We know this must stop either because the ideal gas EOS does not work
(QM effects → degeneracy pressure → white dwarfs), or because some
other energy source intervenes, balancing out the energy losses at the
surface. <i>Stars don't shine because they burn, stars burn because they
shine</i>.
</p>
<p>
Initially, people considered chemical reactions and radioactivity as a
possible energy sources, but it was shown very quickly that these are
insufficient.
</p>
<ul class="org-ul">
<li><b>Q</b>: how would you show that these sources are insufficient? (<b>Hint</b>:
consider the φ factor for the amount of mass released in the
equations below)</li>
</ul>
<p>
It took major advances in quantum mechanics (QM) and
their application to atomic and nuclear theory to work out in the late
1930s that the energy source is nuclear burning, a result obtained by
<a href="https://en.wikipedia.org/wiki/Hans_Bethe">Hans Bethe</a> and
collaborators.
</p>
<p>
In a sense, you can think of a star as an <i>inertial confinement nuclear
reactor</i> where the confinement is provided by the self-gravity. This
also implies that nuclear burning in a star is a <i>self-regulating</i>
process: because nuclear reactions are a consequence of the energy
losses, during long-lived equilibrium phases of the stellar life, they
produce just enough energy to balance the losses! In other words,
L<sub>nuc</sub> = L ( + L<sub>ν</sub>).
</p>
<p>
If a star were to not produce enough energy to verify that (i.e.,
L>L<sub>nuc</sub>), then it will lose too much energy, meaning it will contract,
and increase it's temperature because of the virial theorem, which in
turn regulates the energy generation by nuclear reactions (as we will
see in a moment), until L<sub>nuc</sub> reaches L.
Viceversa, if L<sub>nuc</sub> > L for some reason (e.g., there is a thermonuclear
explosion in the star), than the extra energy release will cause an
expansion of the star and by the virial theorem lower the mean
temperature until L<sub>nuc</sub> = L.
</p>
</div>
</div>
<div id="outline-container-org8a824df" class="outline-3">
<h3 id="org8a824df"><a href="#org8a824df">The nuclear timescale</a></h3>
<div class="outline-text-3" id="text-org8a824df">
<p>
Naturally, nuclear energy generation consumes nuclear fuel: to provide
L<sub>nuc</sub> the composition of the star slightly changes in time. This drives
the <i>evolution</i> of star.
</p>
<p>
We can estimate the timescale for this assuming the star is in
gravothermal equilibrium, so L=L<sub>nuc</sub>. The <i>nuclear timescale</i> is the time
it takes to lose the energy generated by nuclear reactions:
</p>
<div class="latex" id="orgb67426a">
\begin{equation}
\tau_\mathrm{nuc} = \varphi f_\mathrm{burn} \frac{Mc^{2}}{L_\mathrm{nuc}} \equiv \varphi f_\mathrm{burn} \frac{Mc^{2}}{L_{}} \ \ ,
\end{equation}
</div>
<p>
where φ is the fraction of rest mass of nuclei converted in energy by
nuclear burning, f<sub>burn</sub> is the fraction of the stellar mass M that is
affected by burning (we need a stellar model to estimate that).
For the Sun, f<sub>burn</sub> ≅ 0.1 from detailed models (as you can verify
with your <code>MESA-web</code> model!).
</p>
<p>
φ depends on the nuclear physics details. For hydrogen burning into
helium (the two lightest elements), the proton mass is m<sub>p</sub> = 1.0081 m<sub>u</sub>
(where the atomic mass unit m<sub>u</sub> is defined in such a way that the mass
of \(^{12}\mathrm{C}\) is exactly 12m<sub>u</sub> - this is more convenient to measure
experimentally to make a standard), and the mass of helium 4 is m(<sup>4</sup>He)
= 4.0039m<sub>u</sub>, so the fraction of rest mass of 4 protons turning into a
helium nucleus is:
</p>
<div class="latex" id="org11c32d4">
\begin{equation}\label{eq:mass_defect}
\varphi = \frac{4m_{p} - m(^{4}\mathrm{He})}{4m_{p}} = \frac{2.85 \times 10^{-2} m_{u} c^{2}}{4m_{p}} \simeq 0.007 \ \ .
\end{equation}
</div>
<p>
Note that to turn 4 protons into a helium, because of charge and
leptonic number conservation, there needs to be 2 positrons and 2
neutrinos produced! Plugging in φ and M<sub>☉</sub> and L<sub>☉</sub> we get for the nuclear
timescale:
</p>
<div class="latex" id="org9f8c48d">
\begin{equation}\label{eq:tau_nuc_scaling}
\tau_\mathrm{nuc} \simeq 10^{10} \frac{f_\mathrm{burn}}{0.1} \frac{M}{M_{☉}} \frac{L}{L_{☉}} \ \mathrm{yr} \ \ .
\end{equation}
</div>
<p>
So we now have estimates that allow us to see the complete
ordering for the main <i>global</i> timescales for stellar evolution:
</p>
<div class="latex" id="org7f78fad">
\begin{equation}\label{eq:timescale_ordering}
\tau_\mathrm{ff} \ll \tau_\mathrm{KH} \ll \tau_\mathrm{nuc} \ \ .
\end{equation}
</div>
<p>
Moreover, Eq. \ref{eq:tau_nuc_scaling} shows that when accounting for
nuclear energy generation (which we have implicitly assumed here to
give a value of φ), the nuclear timescale matches the timescale found
from geological evidence on Earth, and the ordering \ref{eq:timescale_ordering}
tells us that the evolution of the Sun is on a very slow timescale
compared to thermal and dynamical timescale, validating the
assumptions we made so far to derive the equations for stellar
<i>structure</i> and allowing for a <i>quasi-static</i> approximation to deal with
the stellar <i>evolution</i>.
</p>
<p>
Let's now dive into the details of the nuclear physics that allow for
nuclear burning.
</p>
</div>
</div>
<div id="outline-container-org46be4ee" class="outline-3">
<h3 id="org46be4ee"><a href="#org46be4ee">The energy reservoir: binding energy per nucleon</a></h3>
<div class="outline-text-3" id="text-org46be4ee">
<p>
The energy reservoir that stars tap into is the nuclear binding
energy: nuclear reactions rearrange nucleons in nuclei to create more
bound configuration and extract binding energy as heat source.
Therefore, it is useful to consider the nuclear binding energy per
nucleon of all nuclei in the periodic table:
</p>
<figure id="orgfb74904">
<img src="./images/nuclear_binding.png" alt="nuclear_binding.png" width="100%">
<figcaption><span class="figure-number">Figure 1: </span>Nuclear binding energy per nucleon as a function of atomic mass A. From <a href="https://etd.adm.unipi.it/theses/available/etd-05062015-125630/unrestricted/Thesis_colored_10052015.pdf">Renzo 2015</a>.</figcaption>
</figure>
<p>
This plot shows empirical data, and there are several notable features
coming from the internal structure of the nuclei that any model of
nuclear (strong) interactions needs to explain:
</p>
<ul class="org-ul">
<li>the binding energy per nucleon raises sharply with A among the light
nuclei;</li>
<li>the helium nucleus (α particle) has an exceptionally high binding
energy per nucleon of ∼ 7.5MeV/nucleon;</li>
<li>there are peaks for nuclei with A divisible by 4 and N=Z, that is
nuclei that can be approximately thought as bound collections of α
particles (for example \(^{12}\mathr{C}\) ∼ 3 α particles bound together,
\(^{16}\mathrm{O}\) ∼4 α particles, etc..);</li>
<li>the nuclear binding energy is roughly constant at about ∼
8MeV/nucleon for most heavy nuclei;</li>
<li>the slight drop in B/A after the maximum is caused by the
Coulomb repulsion between the protons in heavy nuclei, and this is
why those nuclei require a higher number of neutrons than protons to
hold together, N>Z.</li>
<li>there is a local <i>maximum</i> of the binding energy per nucleon at the
iron group, around $<sup>56</sup>\mathr{Fe}\(/\)<sup>62</sup>\mathrm{Ni}$.</li>
</ul>
<p>
The presence of a maximum implies that there cannot be energy release
from the <i>fusion</i> of nuclei heavier than iron: indeed those heavy nuclei
are typically fuel for nuclear fission reactors, where energy is
extracted breaking them apart. Stars, which do nuclear burning to
<i>release</i> energy and balance the losses at the surface, have no interest
in producing element heavier than iron during their stable/hydrostatic
lifetime. The question of the formation of elements heavier than iron
(which clearly exist!) is something that is actively worked on and
requires out-of-equilibrium processes that can only occur in the
presence of a neutron rich environment (e.g., AGB stellar winds,
neutron stars interacting with something else, etc.). We may have a
guest lecture on this later on, and there are projects related to
these.
</p>
<p>
Conversely, moving leftward on this chart, by fusing together light
nuclei, stars can release nuclear binding energy and sustain
themselves. This is also convenient since stars are mostly made of H
and He, so they have a lot of light elements available as energy
sources.
</p>
<p>
The change in binding energy per nucleon Δ(B/A) is what powers stars,
and we can write the binding energy as the difference between the sum
of the masses of the Z protons (m<sub>p</sub>) and N=A-Z neutrons (m<sub>n</sub>) minus the
mass of the nucleus (measured experimentally):
</p>
<div class="latex" id="org1c7e36c">
\begin{equation}
B = \left(Zm_{p}+(A-Z)m_{n} - m_\mathrm{nucleus}\right)c^{2} \simeq \left(Zm_{H}+(A-Z)m_{n} - m_\mathrm{nucleus}\right)c^{2} >0 \ \ ,
\end{equation}
</div>
<p>
which is a positive quantity precisely because the strong interaction
between protons and neutrons keeps them bound. In the second step, we
approximate the proton mass with the hydrogen atom mass, making a
mistake of the order of ∼ 10 eV by neglecting the binding energy of
the electron (which is acceptable since we are dealing with MeV =
10<sup>6</sup>eV as the relevant nuclear scale).
</p>
<p>
Because of the apparent peaks in the B/A vs. A curve, stars will tend
to produce more of the elements with the most tight nuclei, whose
production releases more energy! <i>The nuclear structure determines the
most abundant chemicals in the Universe</i>.
</p>
<p>
Moreover, from the fact that there is a sharp jump form H to He, and
then the curve rapidly flattens, we know that the rate at which nuclei
are converted to achieve the equilibrium condition L<sub>nuc</sub> = L will be
slower when H is converted into He, and then it will need to rapidly
speed up when He is converted in carbon and onwards (because the
energy release per nucleon significantly drops lowering the factor φ
in τ<sub>nuc</sub>).
</p>
<p>
In the very late evolutionary phases (post helium burning), the
required burning rate may become so high that τ<sub>nuc</sub> becomes shorter
than the <i>global</i> thermal timescale (but luckily not of the <i>local</i>
thermal timescale, so all the equations we have derived so far still
hold, since they are differential equations that describe <i>local</i>
quantities). We will return on this <a href="https://www.as.arizona.edu/~mrenzo/materials/cores_of_massive_stars.pdf">later</a>.
</p>
</div>
<div id="outline-container-org9f3f74a" class="outline-4">
<h4 id="org9f3f74a"><a href="#org9f3f74a">Q-value</a></h4>
<div class="outline-text-4" id="text-org9f3f74a">
<p>
For any nuclear reaction that may happen a+X → b+Y with a,b,X,Y
generic particles, it is useful to define the so called Q-value as the
mass difference between the reagents and the products:
</p>
<div class="latex" id="orgac0febb">
\begin{equation}
Q = \left((m_{a} + m_{X}) -(m_{b} + m_{Y}) \right)c^{2} = \Delta\left(\frac{B}{A}\right) \ \ ,
\end{equation}
</div>
<p>
which, if Q>0 is the energy <i>released</i> by the reaction which is thus
<i>exoenergetic</i> - of the kind that stars need to generate energy
and sustain themselves against gravity, or if Q<0 is the energy input
needed to get the <i>endoenergetic</i> reaction going.
</p>
<p>
<b>N.B.:</b> In exoenergetic nuclear reactions, the energy release comes from
the <i>mass defect</i>, caused by the increase in binding energy per nucleon:
the total mass of the outgoing particles is lower than the ones
incoming because their binding energy is higher or in other words
their total energy is more negative.
</p>
<p>
The energy release by nuclear reaction per unit time and unit mass is
just:
</p>
<div class="latex" id="org4d8053f">
\begin{equation}\label{eq:eps_nuc}
\varepsilon_\mathrm{nuc} = \sum_{i} \frac{Q_{i} r_{i}}{\rho} \ \ ,
\end{equation}
</div>
<p>
where the sum runs over all the possible reactions, r<sub>i</sub> is the rate per
unit time and volume of the reactions, and the division by the mass
density ρ gives the right dimensions [ε<sub>nuc</sub>] = [E]/([t][M]).
</p>
<p>
So, what is left to do is calculate the volumetric reaction rate r<sub>i</sub>
that can occur in a star.
</p>
</div>
</div>
</div>
<div id="outline-container-org44d2c35" class="outline-3">
<h3 id="org44d2c35"><a href="#org44d2c35">Variety of possible nuclear reactions</a></h3>
<div class="outline-text-3" id="text-org44d2c35">
<p>
A generic nuclear reaction X+a → Y+b is often written as X(a,b)Y to
make it easy to express chains of reactions, e.g.,
X(a,b)Y(c,d)Z(e,f)A…
</p>
<p>
Depending on the nature of the incoming particle (X and a in our
generic reaction), or in other words on the microphysics that
determines the interaction, there can be of various kinds of
reactions.
</p>
</div>
<div id="outline-container-orgd37bf92" class="outline-4">
<h4 id="orgd37bf92"><a href="#orgd37bf92">Charged-particles reactions</a></h4>
<div class="outline-text-4" id="text-orgd37bf92">
<p>
When X and a are charged nuclei, then the reaction can only occur if
something allows them to overcome the Coulomb repulsion. These can be
resonant or not (the distinction will come back later).
</p>
<p>
Example:
</p>
<div class="latex" id="org515d66c">
\begin{equation}
^{12}\mathrm{C}(\alpha,\gamma)^{16}\mathrm{O}
\end{equation}
</div>
</div>
</div>
<div id="outline-container-org7ff1417" class="outline-4">
<h4 id="org7ff1417"><a href="#org7ff1417">Reactions involving neutrons</a></h4>
<div class="outline-text-4" id="text-org7ff1417">
<p>
In this case the force involved is the strong force, and there is no
Coulomb repulsion to overcome. However, these require an environment
that is neutron rich, which is astrophysically a rare occurrence,
since the neutron is an unstable particle that decays in ∼15min to
a proton n→ p+e<sup>-</sup>+\(\bar{\nu_{e}}\). This half-life however can significantly
change for neutrons bound in nuclei as opposed to free neutrons, that
is the β<sup>-</sup> decay time of a neutron rich nucleus can be much longer than
the half-life of a free neutron.
</p>
<p>
Depending on the available flux of neutrons in the environment, we
distinguish:
</p>
<ul class="org-ul">
<li><b>r-process</b> for rapid neutron captures (i.e., the rate of neutron
captures is high w.r.t. the rate of neutron decays)</li>
<li><b>s-process</b> for slow neutron captures (i.e., each nucleus captures at
best one neutron before decaying).</li>
</ul>
<p>
These processes are involved in the formation of elements heavier
than iron, but they require particular astrophysical environment
(e.g., the merger of two neutron stars or a neutron star with the core
of another star, or the envelope of an AGB star).
</p>
</div>
</div>
<div id="outline-container-org02b0c35" class="outline-4">
<h4 id="org02b0c35"><a href="#org02b0c35">Weak reactions</a></h4>
<div class="outline-text-4" id="text-org02b0c35">
<p>
These can typically be spotted by the presence of a neutrino and/or
the conversion of a nucleon from one eigenstate of isospin to another
(in simpler words, the conversion of a proton into a neutron or
viceversa).
</p>
<p>
Example:
</p>
<div class="latex" id="orga81002e">
\begin{equation}
p+e^{-} \rightarrow n+\nu_{e} \ \mathrm{or} \ p(e^{-},\nu_{e})n
\end{equation}
</div>
</div>
</div>
<div id="outline-container-org55f4be4" class="outline-4">
<h4 id="org55f4be4"><a href="#org55f4be4">Photodisintegrations</a></h4>
<div class="outline-text-4" id="text-org55f4be4">
<p>
When one of the particles is a photon and the outgoing particles can
be seen as "fragments" of the ingoing nucleus. These can occur when
very energetic γ ray photons, because their energy needs to be
comparable to the binding energy of nuclei, of the order of ∼8MeV × A,
can encounter particles. This can occur for example at the very late
moments of massive star evolution.
</p>
<p>
Example:
</p>
<div class="latex" id="org31ea985">
\begin{equation}
^{56}\mathrm{Fe}+\gamma \rightarrow 14\alpha
\end{equation}
</div>
</div>
</div>
</div>
<div id="outline-container-org5d936da" class="outline-3">
<h3 id="org5d936da"><a href="#org5d936da">Nuclear reactions in stars</a></h3>
<div class="outline-text-3" id="text-org5d936da">
<p>
All of the types of reactions listed above (and more) can occur at
some point in the evolution (and explosion!) of stars. For example,
during hydrogen core burning (which we have used to estimate φ and
thus τ<sub>nuc</sub>) the star burns 4 protons into α particles:
</p>
<div class="latex" id="org7a0aa6e">
\begin{equation}\label{eq:overall_h_burn}
4p \rightarrow ^{4}\mathrm{He}+2e^{+} + 2\nu_{e} \ \ ,
\end{equation}
</div>
<p>
where the positrons need to be there to conserve electric charge
throughout the process, and the neutrinos need to be there for
conserving the leptonic number (+1 for the leptons electron e<sup>-</sup>, muon
μ<sup>-</sup>, tau τ<sup>-</sup> and the corresponding neutrinos ν<sub>e</sub>, ν<sub>μ</sub>, ν<sub>τ</sub> and -1
for their antiparticles positron e<sup>+</sup>, positive muon μ<sup>+</sup>, and positive
τ<sup>+</sup> and the corresponding antineutrinos).
</p>
<p>
<b>N.B.:</b> because of the phenomenon of neutrino oscillations (i.e., the
conversion of ν<sub>e</sub> ↔ ν<sub>μ</sub> or ν<sub>τ</sub>) the leptonic number
conservation is not exact in nature, or in other words, the symmetry
associated to this conservation law is not exact. While this was
discovered through the "<a href="https://www.nobelprize.org/prizes/themes/solving-the-mystery-of-the-missing-neutrinos/">missing solar neutrinos problem</a>", and it is
thus related to stellar/solar physics, it requires the propagation of
neutrinos over distances much longer than the size of nuclei,
therefore, for the purpose of treating nuclear reactions we can assume
conservation of the leptonic number.
</p>
<p>
From Eq. \ref{eq:overall_h_burn} we can see that:
</p>
<ul class="org-ul">
<li>protons need to encounter each other. Statistically, 4 protons are
unlikely to meet each other at a point in space for reaction Eq.
\ref{eq:overall_h_burn} to occur. Eq. \ref{eq:overall_h_burn} is a
compound reaction that "summarizes" the more complex burning
process of H into He that we will detail later. Nevertheless, the
process will necessarily involve charged particle reactions.</li>
<li>ν<sub>e</sub> appear ⇒ there will be weak reactions involved</li>
<li>as already seen earlier, we can calculate the φ factor (cf. Eq.
\ref{eq:mass_defect}), and thus the Q-value for the overall reaction
Q<sub>H burn</sub> ≅ 26.5 MeV. Note that the Q-value does not really depend on
the details of the burning process.</li>
</ul>
<p>
<b>N.B.:</b> we discuss Eq. \ref{eq:overall_h_burn} because H is the most
abundant element in the Universe, that most stars are made of, and
thus this is (typically) the first process occurring in stars. It is
also the one with the highest Q-value (cf. B/A vs. A plot!), thus the
one that for a given luminosity L of the star can proceed at the
slowest rate and last the longest.
</p>
</div>
</div>
<div id="outline-container-orgab7ff2f" class="outline-3">
<h3 id="orgab7ff2f"><a href="#orgab7ff2f">Charged particle reactions</a></h3>
<div class="outline-text-3" id="text-orgab7ff2f">
<p>
For the rest of this lecture we will focus mostly on charged particle
reactions: as we just saw these are involved since H core burning, and
they are the main reactions during the hydrostatic lifetime of stars.
Moreover, breakthroughs in QM (by <a href="https://en.wikipedia.org/wiki/George_Gamow">George Gamow</a>) applied
to the interaction of charged particles are what opened the way to the
discovery that the energy source in stars are nuclear reactions.
</p>
</div>
<div id="outline-container-orgc7f43a2" class="outline-4">
<h4 id="orgc7f43a2"><a href="#orgc7f43a2">Bohr's approximation</a></h4>
<div class="outline-text-4" id="text-orgc7f43a2">
<p>
To discuss them, we will implicitly use <i>Bohr's approximation</i>, which is
not a completely accurate physical picture, but it is intuitive and
allows to describe the main processes occurring in nuclei. In this
approximation, we treat the generic reaction between charged particles
X(a,b)Y as if it went through an intermediate step of forming a
compound nucleus C:
</p>
<div class="latex" id="org0145e6d">
\begin{equation}
X + a \rightarrow C^{\star} \rightarrow Y + b \ \ ,
\end{equation}
</div>
<p>
where the nucleus C is formed in an excited state C<sup>*</sup> since in
the reference frame of X the particle a arrives with its own kinetic
energy and internal binding energy that (together with X's internal
energy) are generally not exactly the total energy for the compound
nucleus C.
</p>
<p>
The second step is the decay of this fictional compound excited
particle C<sup>*</sup> in the products Y and b.
</p>
<p>
In this approximation, the second step of the decay does <i>not</i> depend on
the first step (in other words, C<sup>*</sup> loses the memory of how it
formed) as long as the half-life of C<sup>*</sup> is long compared to the
light-crossing time of C<sup>*</sup> itself. This is because we consider
charged particles by hypothesis, so electromagnetic forces mediated by
photons are what determines the interactions of the building blocks of
C<sup>*</sup>, and on a timescale long compared to the light-crossing time
they will equilibrate and lose memory of how they came to be.
</p>
<p>
The light crossing time of a nucleus can be estimated starting from
the experimental result on the size of nuclei (something that also
needs to be explained by models of the strong force):
</p>
<div class="latex" id="org6e418de">
\begin{equation}
r_\mathrm{n} = r_{0} A^{1/3} \simeq 1.4 \times 10^{-13} A^{1/3} \ \mathrm{cm} \Rightarrow \tau_\mathrm{light\ cross} = \frac{r_\mathrm{n}}{c} \simeq 10^{-23} A^{1/3} \ \mathrm{s} \ \ .
\end{equation}
</div>
<p>
Any compound nucleus C<sup>* </sup>with lifetime longer than this allows us
to use this two step approximation to treat the problem, where the two
incoming particles X and a come "into contact" (i.e., within their
r<sub>n</sub>), form an excited compound nucleus C<sup>*</sup>, which then de-excites
in the final products Y and b independently of how it formed.
</p>
<p>
<b>N.B.:</b> the nuclear radius dependence on A can be flipped around to
infer that the average density of nuclei is constant as A increases:
ρ<sub>n</sub> ≅ Am<sub>u</sub> / (4π/3 × r<sub>n</sub><sup>3</sup>) ≅ 2×10<sup>14</sup> A g cm<sup>-3</sup>.
</p>
<p>
<b>N.B.:</b> this is necessarily an oversimplified picture, since nuclei are
described by QM and don't have a "hard" boundary at r<sub>n</sub>, but rather
their constituent nucleons have decaying wave-functions that solve the
Schrodinger equation with the nuclear potential for their
interactions. In reality, each particle can be described as a wave
with De Broglie wavelength λ = h/p, and a physically more accurate
picture should treat all the particles involved accounting for their
wave nature.
</p>
<p>
To understand how two charged nuclei, both with positive charge, can "come into contact"
within r<sub>n</sub>, we need to consider the potential governing their
interaction:
</p>
<div class="latex" id="org37743eb">
\begin{equation}
V \equiv V(r) = V_\mathrm{EM} + V_\mathrm{nuc} + \frac{\ell(\ell+1)\hbar^{2}}{2m_{aX}r^{2}} \ \ ,
\end{equation}
</div>
<p>
where the last term is the centrifugal potential in the rest-frame of
the target nucleus X which depends on the reduced mass m<sub>aX</sub> =
m<sub>a</sub>m<sub>X</sub>/(m<sub>a</sub>+m<sub>X</sub>) and quantum number ℓ which determines the order of the
wave-function of the system a+X. For simplicity, we can limit ourselves to
consider ℓ=0: we already have a repulsive Coulomb term to win over,
and any extra repulsive term such a centrifugal barrier is only going
to lower the reaction rate. The most important reactions are going to
have ℓ = 0, that is head-on collisions between a and X.
</p>
</div>
</div>
<div id="outline-container-org1edd63e" class="outline-4">
<h4 id="org1edd63e"><a href="#org1edd63e">Electromagnetic potential term</a></h4>
<div class="outline-text-4" id="text-org1edd63e">
<p>
For the electromagnetic term we can write:
</p>
<div class="latex" id="orgc159e8d">
\begin{equation}
V_\mathrm{EM} = \frac{Z_{a}Z_{X}e^{2}}{r} - \{\mathrm{electron\ screening\ term}\} \ \ ,
\end{equation}
</div>
<p>
where the first part is >0 and describes the Coulomb repulsion between
the two nuclei of chage Z<sub>a</sub>e and Z<sub>X</sub>e (both positive), and the electron
screening term <i>reduces</i> the repulsion of the nuclei: in the stellar
plasma we expect each nucleus to be statistically surrounded by a
"cloud" of electrons of radius of the order of the Debye length of the
plasma:
</p>
<div class="latex" id="org3e37b1b">
\begin{equation}
r_\mathrm{Debye} = \sqrt{\frac{k_{B}T}{4\pi e N\chi}} \ \ ,
\end{equation}
</div>
<p>
where N is the total number of particles in absence of screening
(nuclei/ions+electrons), and χ = ∑<sub>i</sub> Z<sub>i</sub><sup>2</sup> (N<sub>i</sub>/N) + N<sub>e</sub>/N with N<sub>i</sub> and N<sub>e</sub>
number of ions and electrons in absence of screening.
</p>
<p>
For distances between a and X larger than r<sub>Debye</sub> the electron
screening reduces the Coulomb repulsion between the nuclei.
</p>
</div>
</div>
<div id="outline-container-org68db4ae" class="outline-4">
<h4 id="org68db4ae"><a href="#org68db4ae">Nuclear potential term</a></h4>
<div class="outline-text-4" id="text-org68db4ae">
<p>
Finally, for the nuclear potential, there isn't a well known
functional form from first principles, and it is typically derived
experimentally. This is because the interactions between nucleons
(=protons and neutrons) cannot be treated in a perturbative theory of
the strong force. For more details than necessary to understand
thermonuclear reactions, see also <a href="#org36c2a8d">this appendix</a> and nuclear physics
textbooks such as "<i>Introductory nuclear physics</i>" by K. Krane.
</p>
<p>
Besides the nuclear physics details here, the important point is that
the nuclear potential is going to be attractive at short range
(V<sub>nuc</sub>(r≤ r<sub>n</sub>)<0), but it has a repulsive core (that is there is a
certain r<sub>nuc,core</sub> below which V<sub>nuc</sub> becomes very large and positive),
otherwise the nuclei would not have a finite approximately constant
density, and goes to zero at large distances (the strong force has a
short range).
</p>
</div>
</div>
<div id="outline-container-org3e2d83d" class="outline-4">
<h4 id="org3e2d83d"><a href="#org3e2d83d">Combining electromagnetic and nuclear potential</a></h4>
<div class="outline-text-4" id="text-org3e2d83d">
<p>
Putting things together we can sketch the following graph for the
potential felt by particle a and generated by the strong and
electromagnetic force by particle X:
</p>
<ul class="org-ul">
<li>at distances r ≫ r<sub>Debye</sub> electron screening nullifies the repulsive Coulomb potential</li>
<li>for a relative energy at infinity of E, there is a distance of
classical minimum approach r<sub>c</sub></li>
<li>just outside r<sub>n</sub>≡ r<sub>n</sub> there is the maximum height of the Coulomb barrier
E<sub>C</sub> = Z<sub>a</sub> Z<sub>X</sub> e<sup>2</sup>/r<sub>n</sub>.</li>
<li>inside r<sub>n</sub> (the nuclear radius of particle X), the potential is
attractive, and allows for bound states with quantized energy levels.</li>
<li>if the two nuclei get too close to each other, there is a repulsive
core of the nuclear force that dominates over any electromagnetic
effect.</li>
<li>For 0<E<E<sub>c</sub>, there are <i>metastable</i> energy level possible (represented
in the figure by the gray bands). What makes them metastable is also
what allows nuclear burning: quantum tunnelling through the Coulomb
barrier.</li>
</ul>
<figure id="org061f019">
<img src="./images/nuc_pot.png" alt="nuc_pot.png" width="100%">
<figcaption><span class="figure-number">Figure 2: </span>Top: Interaction potential generated by the nucleus X and felt by the nucleus a considering nuclear and electromagnetic interactions. For r>r<sub>Debye</sub> the potential deviates from a r<sup>-1</sup> Coulomb potential (sketched in red) because of electron screening. The gray vertical band marks the repulsive core of the nuclear force (necessary to obtain a constant nuclear density), purple shading marks metastable states of the compound nucleus C<sup>*</sup> where a is trapped inside the nuclear potential well of X. Bottom: qualitative sketch of the wave function of particle a in the potential caused by particle X (blue). The region between r<sub>n</sub> and r<sub>c</sub> is the "classically forbidden region". The wavefunction needs to be smooth at both those radii and the solution where quantum tunnelling succeeds allow for a large amplitude of the wave function inside r<r<sub>n</sub>. Modified from Fig. 6.2 of Onno Pols' lecture notes.</figcaption>
</figure>
</div>
</div>
<div id="outline-container-org5bf4a8d" class="outline-4">
<h4 id="org5bf4a8d"><a href="#org5bf4a8d">Impossibility of nuclear reactions without QM</a></h4>
<div class="outline-text-4" id="text-org5bf4a8d">
<p>
Without QM, for a nuclear reaction to happen (assuming Bohr's
approximation), the two charged particles would need a relative energy
at infinity higher than the maximum of the Coulomb barrier, so that r<sub>c</sub>
≤ r<sub>n</sub>. Assuming the energy is just coming from the thermal energy of
the gas:
</p>
<div class="latex" id="org6db71c0">
\begin{equation}
E \simeq k_{B} T \geq E_{C} = \frac{Z_{a}Z_{X}e^{2}}{r_{n}} \geq \frac{e^{2}}{r_{0}} \Rightarrow T\geq \frac{e^{2}}{r_{0} k_{B}} \simeq 10^{10} \ \mathrm{K} \ \ ,
\end{equation}
</div>
<p>
where we assume Z<sub>a</sub> = Z<sub>x</sub> = A = 1 to minimize the Coulomb barier, so