Solutions for the Homework 7

Statistical Physics: November 7, 2012

Problem 7.39: Change variables in equation 7.83 to λ = hc/ǫ, and thus derive a formula for the photon spectrum as a function of wavelength. Plot this spectrum, and ﬁnd a numerical formula for the wavelength where the spectrum peaks, in terms of hc/kT . Explain way the peak does not occur at hc/(2.82kT ). Solution: Substitute ǫ = hc/λ to the equation 7.83: ∞ U 8πǫ3/(hc)3 0 8π (hc)3 1 hc = dǫ = dλ (1) ǫ/kT 3 3 hc/kT λ 2 V 0 e 1 ∞ (hc) λ e 1 −λ Z ∞ − Z ∞ − ∞ 8πhc 1 8π(kT )4 1/x5 = dλ = dx = u(x) dx. (2) λ5 ehc/kT λ 1 (hc)3 e1/x 1 Z0 − Z0 − Z0 From the equation below, we can ﬁnd the wavelength where the spectrum peaks: du(x) = 0 x = 0.201405. (3) dx ⇒

uHxL

kT x= Λ 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 hc

Then, compare the peaks of photon energy and wavelength: kT kT λ = (0.201405) , ǫ = (0.35461) . (4) max × hc max × hc The difference between these values caused by the different measure of integral. Because of the measure of photon energy dǫ changes to the dλ with the factor hc/λ2, there relation is not linear, so the peak of energy density can be different in general.

1 Statistical Physics: November 7, 2012

Problem 7.41: Consider any two internal states, s1 and s2, of an atom. Let s2 be the higher-energy state, so that E(s ) E(s ) = ǫ for some positive constant ǫ. If the atom is currently in state 2 − 1 s2, then there is a certain probability per unit time for it to spontaneously decay down to state s1, emitting a photon with energy ǫ. This probability per unit time is called the Einstein A coeﬃcient:

A = probability of spontaneous decay per unit time. (5)

On the other hand, if the atom is currently in state s1 and we shine light on it with frequency f = ǫ/h, then there is a chance that it will absorb a photon, jumping into state s2. The probability for this to occur is proportional not only to the amount of time elapsed but also to the intensity of the light, or more precisely, the energy density of the light per unit frequency, u(f). (This is the function which, when integrated over any frequency interval, gives the energy per unit volume within that frequency interval. For our atomic transition, all that matters is the value of u(f) at f = ǫ/h.) The probability of absorbing a photon, per unit time per unit intensity, is called the Einstein B coeﬃcient: probability of absorption per unit time B = . (6) u(f)

Finally, it is also possible for the atom to make a stimulated transition from s2 down to s1, again with a probability that is proportional to the intensity of light at frequency f. (Stimulated emission is the fundamental mechanism of the laser: Light Amplification by Stimulated Emission of Radiation.) Thus we define a third coefficient, B′, that is analogous to B:

′ probability of stimulated emission per unit time B = . (7) u(f) As Einstein showed in 1917, knowing any one of these three coeﬃcients is as good as knowing them all.

(a) Imagine a collection of many of these atoms, such that N1 of them are in state s1 and N2 ′ are in state s2. Write down a formula for dN1/dt in terms of A, B, B , N1, N2, and u(f).

Solution: The number of photon at state s1 is N1 and each coefficients are add or subtract the photons. A and B′ add the photons and B subtract the photons because of it’s definition. Since the coefficients are related to the probabilities per unit time, the change of particles are these coefficients times number of particles in the state where the photon came and alto multiply the time t:

dN ′ 1 = AN + B u(f)N Bu(f)N . (8) dt 2 2 − 1

(b) Einstein’s trick is to imagine that these atoms are bathed in thermal radiation, so that u(f) is the Planck spectral function. At equilibrium, N1 and N2 should be constant in time, with their ratio given by a simple Boltzmann factor. Show, then, that the coeﬃcients must be related by

3 ′ A 8πhf B = B and = . (9) B c3

2 Statistical Physics: November 7, 2012

Solution: First, ﬁnd the function u(f) from the equation 7.83 with the substitution ǫ = hf. (We should careful because the measure of integral is also changed and it gives some factor. So the function u is not a ﬁxed, in general.) Then,

∞ ∞ U 8πǫ3/(hc)3 8πh f 3 = dǫ = df. (10) V eǫ/kT 1 c3 eǫ/kT 1 Z0 − Z0 − So, the u(f) is

8πh f 3 u(f) = . (11) c3 eǫ/kT 1 − Now, the ratio of N1 and N2 is given by the Boltzmann factor:

N e−E(s1)/kT 1 = = ehc/kT . (12) −E(s2)/kT N2 e Finally, at equilibrium,

dN ′ 1 = AN + B u(f)N Bu(f)N 0. (13) dt 2 2 − 1 ≡ From this,

′ A = (Behc/kT B )u(f). (14) − As we can see, left-hand side is independent of temperature, so right-hand side also should be. Suppose the coeﬃcients are independent of temperature. Then,

∂ ′ (Behc/kT B )u(f) (15) ∂T − h i hc/kT hc ′ e hc = Behc/kT u(f) + (Behc/kT B ) u(f) (16) − kT 2 − ehc/kT 1 kT 2 − ✘ hc/kT ✘✘ ✘✘ ′ e hc = ✘Be✘ hc/kT + B + ✘Be✘hc/kT B u(f) (17) − − ehc/kT 1 kT 2 h hc/kT i − ′ e hc = (B B ) u(f) 0. (18) − ehc/kT 1 kT 2 ≡ − Above relation leads that the coeﬃcients B and B′ are same. Using this,

3 3 ′ 8πh f 8πhf A = (Behc/kT B )u(f) = B(ehc/kT 1) = B . (19) − − c3 eǫ/kT 1 c3 − Thus,

A 8πhf 3 = (20) B c3

3 Statistical Physics: November 7, 2012

Problem 7.45: Use the formula P = (dU/dV ) to show that the pressure of a photon gas is 1/3 − S,N times the energy density (U/V ). Compute the pressure exerted by the radiation inside a kiln at 1500 K, and compare to the ordinary gas pressure exerted by the air. Then compute the pressure of the radiation at the center of the sun, where the temperature is 15 million K. Compare to the gas pressure of the ionized hydrogen, whose density is approximately 105kg/m3 . Solution: U/V is given in the equation 7.86 and to ﬁnd the pressure, substitute the temperature as a function of S which we will ﬁx and it is given in the equation 7.89. Then,

5 4 4/3 4/3 8π k 3S 3S − U = VT 4 = aT 4 = a = a 1/3. (21) 15 (hc)3 4a 4 Since ∂a/∂V = a/V ,

dU a 3S 4/3 U P = = = (22) −dV 3V 4a 3V S,N

When T = 1500K, P = 0.00127672 P a and T = 1.5 107 K, P = 1.27672 1013 P a. First one × × is lower by a factor 108 times and second one is greater by a factor 108 times with respect to the ordinary gas pressure 1 atm. For the ionized hydrogen case, let’s calculate the pressure of the ionized hydrogen:

NkT − P = = (105kg/m3)/(1.673533 10 27kg) kT = 1.23749 1016 P a (23) V × × × But there is also the same amount of electrons came from each of hydrogens, so the pressure will be twice, that is 2.47498 1016 P a. This is greater by a factor 2000 times with respect to the × T = 1.5milion K case.

Problem 7.47: In the text I claimed that the universe was ﬁlled with ionized gas until its temperature cooled to about 3000 K. To see why, assume that the universe contains only photons and hydrogen atoms, with a constant ratio of 109 photons per hydrogen atom. Calculate and plot the fraction of atoms that were ionized as a function of temperature, for temperatures between 0 and 6000 K. How does the result change if the ratio of photons to atoms is 108, or 1010? (Hint: Write everything in terms of dimensionless variables such as t = kT/I, where I is the ionization energy of hydrogen.) Solution: To solve this problem, recall the Ionization of Hydrogen in the section 5.6. Then, we can see the Saha equation which is

3/2 P kT 2πm kT − p = e e I/kT . (24) P P h H e Suppose each of gases are ideal, then the equation is modiﬁed to

N V 1 − p = e I/kT , (25) NH Ne vQ

where the vQ is deﬁned in the equation 7.18 and NP is the number of ionized hydrogen, Ne is the number of electrons which is same as Np and NH is the number of hydrogen. Now, Consider

4 Statistical Physics: November 7, 2012

the relations given in the problem. Since the total number of hydrogen and ionized hydrogen is same as the number of photon times R, the ratio of photons to hydrogen atoms deﬁned by R = (Np + NH )/Nγ and using the result of the problem 7.44,

kT 3 N + N = RN = R 8πV (2.40411). (26) p H γ × hc ×

We want to ﬁnd the fraction F = Np/(Np + NH ) = Np/RNγ as a function of temperature T , so combine these three equations. I ﬁrst eliminate Np using the fraction in terms of Nγ, then there are two equations remain:

2 2 2 V −I/kT RNγ = FRNγ + NH and F R Nγ = NH e . (27) vQ

After that, substitute NH from the one of that equations to another equation. Then,

2 2 2 V −I/kT 2 vQ I/kT F R Nγ = (1 F )RNγ e F RNγ e + F 1 = 0. (28) − vQ ⇒ V − This is quadratic equation of F and we know the other things. Solve this equation using the mathematica and change the variable t = kT/I for plotting. Then, the fraction F is a function of t and R.

fraction 1.0

0.8

- R=10 8 0.6 - R=10 9

0.4 - R=10 10

0.2

kT t = 0.005 0.010 0.015 0.020 0.025 0.030 0.035 I

From this ﬁgure, we can see that the fraction of ionized hydrogen is only 1 percent at T = 3140.36 K. And if the ratio of photons to hydrogen atoms, R, is increase, then the hydrogen prefer to unionized at more high temperature.

Problem 7.48: In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos (v) and antineutrinos (v), currently at an eﬀective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarization state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless.

5 Statistical Physics: November 7, 2012

(a) It is reasonable to assume that for each species, the concentration of neutrinos equals the concentration of antineutrinos, so that their chemical potentials are equal: µν = µν. Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction ν + ν 2γ (29) ↔ (where γ is a photon). Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that µ = 0 for both the neutrinos and the antineutrinos. Solution: By the the chemical potential of neutrino and antineutrino are same and the chemical potential of photon is zero, µν = µν = 0 from the conservation of chemical potential. (b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino-antineutrino background radiation. (Hint: There are very few differences between this ”neutrino gas” and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of 2. To account for the three species, just multiply by 3.) To evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a table or consult Appendix B. Solution: Follow the page 291 on the textbook. Then only differences are the factor 3 which came from the three types of neutrinos and statistics. (Spin state is similar to the polarization, so it is already calculated.) Then, ∞ ∞ U 8πǫ3/(hc)3 24π(kT )4 x3 = 3 dǫ = dx (30) V eǫ/kT + 1 (hc)3 ex + 1 Z0 Z0 24π(kT )4 π4 7 7π5(kT )4 = = . (31) (hc)3 15 8 5(hc)3 See the Appendix B. for the result of unfamiliar integration. (c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of 1.95 K. Solution: Similarly change the result of the Problem 7.44 and use it: ∞ kT 3 x2 N = 3 8πV dx (32) × hc ex + 1 Z0 kT 3 = 24πV (1.80309) = 3.38407 108. (33) hc × (d) It is possible that neutrinos have very small, but nonzero, masses. This wouldn’t have affected the production of neutrinos in the early universe, when mc2 would have been negligible compared to typical thermal energies. But today, the total mass of all the background neutrinos could be significant. Suppose, then, that just one of the three species of neutrinos (and the corresponding antineutrino) has a nonzero mass m. What would mc2 have to be (in eV ), in order for the total mass of neutrinos in the universe to be comparable to the total mass of ordinary matter? Solution: Present mass density of universe is 9.9 10−27kg/m3. Then, the total number of × neutrino times 1/3 will gives the total neutrino mass and so, − Nǫ/3 = 9.9 10 27kg/m3 c2 ǫ = 49.2303 eV. (34) × × ⇒

6 Statistical Physics: November 7, 2012

So, if one of the three species of neutrinos has a mass around 50eV/c2, then the total mass of neutrino will be compatible to the total mass of ordinary matter.

Problem 7.49: For a brief time in the early universe, the temperature was hot enough to produce large numbers of electron-positron pairs. These pairs then constituted a third type of “background radiation,” in addition to the photons and neutrinos (see Figure 7.21). Like neutrinos, electrons and positrons are fermions. Unlike neutrinos, electrons and positrons are known to be massive (each with the same mass), and each has two independent polarization states. During the time period of interest the densities of electrons and positrons were approximately equal, so it is a good approximation to set the chemical potentials equal to zero as in the previous problem. Recall from special relativity that the energy of a massive particle is ǫ = (pc)2 + (mc2)2. (a) Show that the energy density of electrons and positrons at temperatupre T is given by U 16π(kT )4 = u(f), (35) V (hc)3

where

∞ x2 x2 + (mc2/kT )2 u(T ) = dx. (36) √x2+(mc2/kT )2 Z0 e p + 1

Solution: Similar to the previous problem, follow the page 291 on the textbook (with the factor 4 came from the spin states and two particles):

U 4 ǫ(n) = dv (37) ǫ(n)/kT V V V e + 1 Z ∞ 4 π/2 π/2 ǫ(n) = dn dθ dφ n2 sin θ (38) ǫ(n)/kT V 0 0 0 e + 1 Z ∞ Z Z 16π n2 hcn/L + (mc2)2 = dn (39) V √hcn/L+(mc2)2/kT Z0 e p + 1 ∞ 16π(kT )4 x2 x2 + (mc2/kT )2 = dx, (40) (hc)3 √x2+(mc2/kT )2 Z0 e p + 1 where x = hcn/LkT . So the given relation is satisﬁed. (b) Show that u(T ) goes to zero when kT mc2, and explain why this is a reasonable result. ≪ Solution: Suppose the limit of temperature goes to zero. Then, we can treat the square root term as

mc2 2 mc2 1 kT x 2 x2 + 1 + (x2). (41) kT ≈ kT 2 mc2 ≈ O s " #

7 Statistical Physics: November 7, 2012

So,

∞ (x4) u(T ) dx (42) mc2/kT (1+(OkT x)2/2(mc2)2) → 0 e e + 1 Z ∞ (x4) O dx (43) ≈ emc2/kT (1 + (kT x)2/2(mc2)2) + 1 Z0 ∞ 4 −mc2/kT (x ) e O ✘2 ✘✘ dx (44) ≈ 2 ✘−✘mc /kT Z0 (x ) + e − 2 O x=∞ e mc /kT (x3) 0. when T 0 (45) ≈ O x=0 ≈ →

I drop the exponential factor because that is very small factor and each step, you should consider the temperature term also but it’s not important to the ﬁnal result. This is compatible with our think that when the temperature is very small, there is no energy to make a pair of electron and positron and so the energy density of this pair will be zero. (c) Evaluate u(T) in the limit kT mc2, and compare to the result of the previous problem for ≫ the neutrino radiation. Solution: In the limit T , the function u(T ) goes to some number. → ∞ ∞ ∞ x2 x2 + (mc2/kT )2 x3 7π4 u(T ) = dx dx = (46) √x2+(mc2/kT )2 → ex + 1 24 5 Z0 e p + 1 Z0 × Then the total energy per unit volume will same with the result of previous neutrino problem and only diﬀerence is the factor 2/3 which came from the species of particle. So,

U 2 7π5(kT )4 14π5(kT )4 = . (47) V → 3 5(hc)3 15(hc)3

(d) Use a computer to calculate and plot u(T ) at intermediate temperatures. Solution: Use a mathematica, plot the graph.

uHt L

kT t = 0.5 1.0 1.5 2.0 2.5 3.0 mc2

8 Statistical Physics: November 7, 2012

(e) Use the method of Problem 7.46, part (d), to show that the free energy density of the electron-positron radiation is F 16π(kT )4 = f(T ), (48) V − (hc)3 where ∞ − 2 2 2 f(T ) = x2 ln 1 + e √x +(mc /kT ) dx. (49) 0 Z Evaluate f(T ) in both limits, and use a computer to calculate and plot f(T ) at intermediate temperatures. Solution: This is for fermion, so the partition function and the free energy are − − Z(n) = 1 + e ǫ(n)/kT F (n) = kT ln(1 + e ǫ(n)/kT ). (50) ⇒ − Using this, similar to the ﬁnding energy, sum all modes for n. Then,

∞ π/2 π/2 − F = kT dn dθ dφ n2 sin θ ln(1 + e ǫ(n)/kT ) (51) − 0 0 0 Z 4 Z ∞ Z 16π(kT ) − 2 2 2 = x2 ln(1 + e √x +(mc /kT ) ) dx. (52) − (hc)3 Z0

fHt L 2.0

1.5

1.0

0.5

kT t = 0.5 1.0 1.5 2.0 2.5 3.0 mc2

This is the graph of f(T ). When the limit of temperature goes to zero, ∞ − 2 2 2 2 f(T ) x2 ln(1 + e mc /kT e(1+(kT x) /2(mc ) ) dx (53) → 0 Z ∞ − 2 2 2 2 e mc /kT x2e(1+(kT x) /2(mc ) ) dx (54) ≈ 0 Z ∞ − 2 e mc /kT (x4) dx (55) ≈ 0 O − 2 Z x=∞ e mc /kT (x5) 0. when T 0 (56) ≈ O x=0 ≈ →

9 Statistical Physics: November 7, 2012

I use the relation ln(1 + x) x for small x. Now, take the limit of temperature goes to infinity, ≈ ∞ ∞ 3 −x 2 −x x e f(T ) x ln(1 + e ) dx = −x dx (57) → 0 0 3 1 + e Z ∞ Z x3 1 1 π4 7 7π4 = dx = = . (58) 3 ex + 1 3 15 8 360 Z0 For the first relation, I use the integral by parts. (f) Write the entropy of the electron-positron radiation in terms of the functions u(T ) and f(T ). Evaluate the entropy explicitly in the high-T limit. Solution: From F = U TS, − U F V 16π(kT )4 S = − = (u(T ) + f(T )) . (59) T T (hc)3 From the result of (b) and (e), the entropy will zero when the temperature goes to zero. Similarly from the result of (c) and (e), when the temperature goes to infinity, the entropy will be

V 16π(kT )4 7π4 56π5 kT 3 T , S = V k . (60) → ∞ ⇒ → T (hc)3 90 45 hc Problem 7.50: The results of the previous problem can be used to explain why the current temperature of the cosmic neutrino background (Problem 7.48) is 1.95 K rather than 2.73 K. Originally the temperatures of the photons and the neutrinos would have been equal, but as the universe expanded and cooled, the interactions of neutrinos with other particles soon became negligibly weak. Shortly thereafter, the temperature dropped to the point where kT/c2 was no longer much greater than the electron mass. As the electrons and positrons disappeared during the next few minutes, they “heated” the photon radiation but not the neutrino radiation. (a) Imagine that the universe has some ﬁnite total volume V , but that V is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of V and T , using the auxiliary functions u(T ) and f(T ) introduced in the previous problem. Argue that this total entropy would have been conserved in the early universe, assuming that no other species of particles interacted with these. Solution: The entropy of electron and positron is obtained at the previous problem 7.49 (e) and the entropy of photon is given by the equation 7.89. Then, the total entropy is V 16π(kT )4 32π5 kT 3 S = S + S + + S = (u(T ) + f(T )) + V k (61) e e γ T (hc)3 45 hc kT 3 2π4 = 16πV k u(T ) + f(T ) + constant. (62) hc 45 ≡ (b) The entropy of the neutrino radiation would have been separately conserved during this time period, because the neutrinos were unable to interact with anything. Use this fact to show that the neutrino temperature Tν and the photon temperature T are related by T 3 2π4 + u(T ) + f(T ) = constant (63) T 45 ν

10 Statistical Physics: November 7, 2012

as the universe expands and cools. Evaluate the constant by assuming that T = Tν when the temperatures are very high. Solution: As we can see in the all entropy values, neutrino entropy also expected proportional to some factor,

kT 3 S 16πV k ν constant, (64) ν ∼ hc ≡ and it is constant because of the neutrinos are unable to interact with anything. Then, we can easily see that

T 3 2π4 + u(T ) + f(T ) = constant. (65) T 45 ν Now, set the temperatures are same with high temperature limit. Then we can use the result of problem 7.49 (f) and so the constant is

2π4 2π4 7π5 11π4 + u(T ) + f(T ) + = . (66) 45 → 45 90 90

(c) Calculate the ratio T/Tν in the limit of low temperature, to conﬁrm that the present neutrino temperature should be 1.95 K. Solution: We know that the function u(T ) and f(T ) goes to zero when the low temperature limit. So, the relation is

T 3 2π4 11π4 T 3 11 4T = = T = = 1.95 K, (67) T 45 90 ⇒ T 4 ⇒ ν 11 ν ν where T = 2.73 K, the temperature of universe. 2 (d) Use a computer to plot the ratio T/Tν as a function of T , for kT/mc ranging from 0 to 3. Solution: Use a mathematica, plot the graph.

TTΝ 1.4

1.3

1.2

1.1

1.0

kT t = 0.5 1.0 1.5 2.0 2.5 3.0 mc2

11 Problem 7.39

In[1]:= maximum = FindMaximum@ 1 Hx^5 HE^H1 xL - 1LL, 8x, 0.1

PlotA 1 Hx^5 HE^H1 xL - 1LL, 8x, 0, 1.5<, GridLines ® 88 8maximum, 8Dashed, Red<<<, 8<<, kT PlotRange ® 80, 22<, AxesLabel ® 9"x= Λ", "uHxL"=, PlotStyle ® 8Black

Out[2]= 10

kT x= Λ 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 hc Problem 7.47

In[3]:= h = 6.62607 * 10^H-34L; k = 1.38065 * 10^H-23L; c = 299 792 458; me = 9.10938 * 10^H-31L; eV = 1.6021766 * 10^H-19L; ionE = 13.6 eV; V = 1;

vq@t_D = Hh Sqrt@2 Pi me ionE tDL^3; Ngamma@t_D = 8 Pi V HHionE t Hh cLL^3 L * 2.40411; tmax = k 6000 ionE; 2 solution_7.nb

In[13]:= fraction@R_, t_D = Solve@F^2 R Ngamma@tD Hvq@tD VL Exp@1 tD + F - 1 0, F, 8R, t

Needs@"PlotLegends`"D PlotA8fraction@10^H-8L, tD, fraction@10^H-9L, tD, fraction@10^H-10L, tD<, -8 -9 -10 8t, 0, tmax<, PlotLegend ® 9"R=10 ", "R=10 ", "R=10 "=, LegendShadow ® None, LegendSize ® 0.5, PlotStyle ® 88Red, Dashed<, Black, 8Blue, Dashed<<, kT LegendPosition ® 8-0.7, -0.25<, AxesLabel ® 9"t= ", "fraction"=E I

t1 = FindRoot@fraction@10^H-9L, tD 0.01, 8t, 0.02

0.8

- R=10 8 0.6 - Out[15]= R=10 9

0.4 - R=10 10

0.2

kT t= 0.005 0.010 0.015 0.020 0.025 0.030 0.035 I

Out[16]= 3140.36 Problem 7.49

In[17]:= u@T_D := NIntegrate@x^2 Sqrt@x^2 + 1 T^2D HE^H Sqrt@x^2 + 1 T^2DL + 1L, 8x, 0, Infinity<, AccuracyGoal ® 100D;

infu = Integrate@ x^2 Sqrt@x^2 + 1 T^2D HE^H Sqrt@x^2 + 1 T^2DL + 1L . T ® Infinity, 8x, 0, Infinity

uHtL

Out[19]= 3

kT t= 0.5 1.0 1.5 2.0 2.5 3.0 mc2 solution_7.nb 3

In[20]:= f@T_D := NIntegrate@x^2 Log@1 + E^H-Sqrt@x^2 + 1 T^2DLD, 8x, 0, Infinity<, AccuracyGoal ® 100D;

inff = Integrate@x^2 Log@1 + E^H-Sqrt@x^2 + 1 T^2DLD . T ® Infinity, 8x, 0, Infinity

fHtL 2.0

1.5

Out[22]= 1.0

0.5

kT t= 0.5 1.0 1.5 2.0 2.5 3.0 mc2 Problem 7.50

In[23]:= tempratio@T_D := HH11 Pi^4 90L H2 Pi^4 45 + u@TD + f@TDLL^H1 3L;

PlotAtempratio@TD, 8T, 0, 3<, kT PlotRange ® 8Full, 80.9, 1.45<<, AxesLabel ® 9"t= ", "TTΝ"=, mc2 PlotStyle ® 8Black<, GridLines ® 88<, 881, 8Red, Dashed<<<

TTΝ

1.4

1.3

1.2 Out[24]=

1.1

1.0

kT t= 0.5 1.0 1.5 2.0 2.5 3.0 mc2