Solutions for the Homework 7

Transcription

1 Statistical Physics: November 7, 212 Solutions for the Homework 7 Problem 7.39: Change variables in equation 7.83 to λ = /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 /kt. Explain way the peak does not occur at /(2.82kT). Solution: Substitute ǫ = /λ to the equation 7.83: U 8π () 3 ( = () 3 λ 3 = 8πǫ 3 /() 3 e ǫ/kt 1 dǫ = 8π λ 5 1 e /ktλ 1 dλ = 1 e /ktλ 1 )( 8π(kT) 4 () 3 1/x 5 e 1/x 1 dx = ) λ 2 dλ From the equation below, we can find the wavelength where the spectrum peaks: du(x) dx (1) u(x) dx. (2) = x = (3) u x kt x Λ Then, compare the peaks of photon energy and wavelength: λ max = (.2145) kt, ǫ max = (.35461) kt. (4) 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 /λ 2, there relation is not linear, so the peak of energy density can be different in general. 1

2 Statistical Physics: November 7, 212 Problem 7.41: Consider any two internal states, s 1 and s 2, of an atom. Let s 2 be the higher-energy state, so that E(s 2 ) E(s 1 ) = ǫ for some positive constant ǫ. If the atom is currently in state s 2, then there is a certain probability per unit time for it to spontaneously decay down to state s 1, 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 s 1 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 s 2. 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: B = probability of absorption per unit time. (6) u(f) Finally, it is also possible for the atom to make a stimulated transition from s 2 down to s 1, 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: B = probability of stimulated emission per unit time. (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 N 1 of them are in state s 1 and N 2 are in state s 2. Write down a formula for dn 1 /dt in terms of A, B, B, N 1, N 2, and u(f). Solution: The number of photon at state s 1 is N 1 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 dt = AN 2 +B u(f)n 2 Bu(f)N 1. (8) (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, N 1 and N 2 should be constant in time, with their ratio given by a simple Boltzmann factor. Show, then, that the coefficients must be related by B = B and A B = 8πhf3 c 3. (9) 2

3 Statistical Physics: November 7, 212 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 = 8πǫ 3 /() 3 e ǫ/kt 1 dǫ = 8πh c 3 f 3 e ǫ/kt 1 df. (1) So, the u(f) is Now, the ratio of N 1 and N 2 is given by the Boltzmann factor: Finally, at equilibrium, u(f) = 8πh c 3 f 3 e ǫ/kt 1. (11) N 1 N 2 = e E(s1)/kT e E(s 2)/kT = e/kt. (12) From this, dn 1 dt = AN 2 +B u(f)n 2 Bu(f)N 1. (13) A = (Be /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, T [ ] (Be /kt B )u(f) ( ) = Be /kt kt 2 u(f)+(be /kt B e /kt ( ) e /kt 1 kt 2 [ = Be /kt +B + Be /kt B ] e /kt ( e /kt 1 kt 2 e /kt = (B B ) e /kt 1 (15) ) u(f) (16) ) u(f) (17) ( ) kt 2 u(f). (18) Above relation leads that the coefficients B and B are same. Using this, Thus, A = (Be /kt B )u(f) = B(e /kt 1) 8πh c 3 f 3 e ǫ/kt 1 = B ( 8πhf 3 c 3 ). (19) A B = 8πhf3 c 3 (2) 3

4 Statistical Physics: November 7, 212 Problem 7.45: Use the formula P = (du/d) S,N to show that the pressure of a photon gas is 1/3 times the energy density (U/). Compute the pressure exerted by the radiation inside a kiln at 15 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 1 5 kg/m 3. Solution: U/ 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 Then, U = 8π5 15 k 4 () 3T4 = at 4 = a ( ) 3S 4/3 = 4a ( ) 3S 4/3 a 1/3. (21) 4 Since a/ = a/, P = du d = a S,N 3 ( ) 3S 4/3 = U 4a 3 (22) When T = 15K, P = Pa and T = K, P = Pa. First one is lower by a factor 1 8 times and second one is greater by a factor 1 8 times with respect to the ordinary gas pressure 1 atm. For the ionized hydrogen case, let s calculate the pressure of the ionized hydrogen: P = NkT = (1 5 kg/m 3 )/( kg) kt = Pa (23) But there is also the same amount of electrons came from each of hydrogens, so the pressure will be twice, that is Pa. This is greater by a factor 2 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 cooledtoabout3k.toseewhy,assumethattheuniversecontainsonlyphotonsandhydrogen atoms, with a constant ratio of 1 9 photons per hydrogen atom. Calculate and plot the fraction of atoms that were ionized as a function of temperature, for temperatures between and 6 K. How does the result change if the ratio of photons to atoms is 1 8, or 1 1? (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 P p P H = kt P e ( ) 2πme kt 3/2 e I/kT. (24) h Suppose each of gases are ideal, then the equation is modified to N p N H = N e 1 v Q e I/kT, (25) where the v Q is defined in the equation 7.18 and N P is the number of ionized hydrogen, N e is the number of electrons which is same as N p and N H is the number of hydrogen. Now, Consider 4

5 Statistical Physics: November 7, 212 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 = (N p +N H )/N γ and using the result of the problem 7.44, N p +N H = RN γ = R 8π ( ) kt 3 (2.4411). (26) We want to find the fraction F = N p /(N p +N H ) = N p /RN γ as a function of temperature T, so combine these three equations. I first eliminate N p using the fraction in terms of N γ, then there are two equations remain: RN γ = FRN γ +N H and F 2 R 2 N 2 γ = N H v Q e I/kT. (27) After that, substitute N H from the one of that equations to another equation. Then, F 2 R 2 N 2 γ = (1 F)RN γ v Q e I/kT F 2 RN γ v Q ei/kt +F 1 =. (28) 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 R 1 8 R 1 9 R kt t I From this figure, we can see that the fraction of ionized hydrogen is only 1 percent at T = 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

6 Statistical Physics: November 7, 212 (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: µ ν = µ ν. 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 = 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, µ ν = µ ν = 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 = 3 = 24π(kT)4 () 3 8πǫ 3 /() 3 24π(kT)4 e ǫ/kt dǫ = +1 () 3 ( ) π See the Appendix B. for the result of unfamiliar integration. x 3 e x +1 dx (3) = 7π5 (kt) 4 5() 3. (31) (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 x 2 N = 3 8π e x dx (32) +1 ( ) kt 3 = 24π (1.839) = (33) (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 mc 2 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 mc 2 have to be (in e), 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 kg/m 3. Then, the total number of neutrino times 1/3 will gives the total neutrino mass and so, Nǫ/3 = kg/m 3 c 2 ǫ = e. (34) 6

7 Statistical Physics: November 7, 212 So, if one of the three species of neutrinos has a mass around 5e/c 2, 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 +(mc 2 ) 2. (a) Show that the energy density of electrons and positrons at temperature T is given by U = 16π(kT)4 () 3 u(f), (35) where u(t) = x 2 x 2 +(mc 2 /kt) 2 e x 2 +(mc 2 /kt) 2 +1 dx. (36) 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 = 4 = 16π ǫ(n) e ǫ(n)/kt dv (37) +1 dn = 16π(kT)4 () 3 π/2 dθ π/2 where x = n/lkt. So the given relation is satisfied. dφ n 2 ǫ(n) sinθ e ǫ(n)/kt +1 (38) n 2 n/l+(mc 2 ) e 2 dn (39) n/l+(mc 2 ) 2 /kt +1 x 2 x 2 +(mc 2 /kt) 2 dx, (4) e x 2 +(mc 2 /kt) 2 +1 (b) Show that u(t) goes to zero when kt mc 2, 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 ( ) [ mc x mc ( ) ] ktx 2 kt kt 2 mc 2 O(x 2 ). (41) 7

8 Statistical Physics: November 7, 212 So, u(t) e mc2 /kt O(x 4 ) e mc2 /kt e (1+(kTx)2 /2(mc 2 ) 2) dx +1 (42) O(x 4 ) e mc2 /kt (1+(kTx) 2 /2(mc 2 ) 2 dx )+1 (43) O(x 4 ) dx (44) O(x 2 )+ e mc2 /kt e mc2 /kt O(x 3 ) x= x=. when T (45) 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 mc 2, 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. u(t) = x 2 x 2 +(mc 2 /kt) 2 e x 2 +(mc 2 /kt) 2 +1 dx x 3 e x +1 dx = 7π 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, (46) U 2 7π 5 (kt) 4 3 5() 3 = 14π5 (kt) 4 15() 3. (47) (d) Use a computer to calculate and plot u(t) at intermediate temperatures. Solution: Use a mathematica, plot the graph. u t t kt mc 2 8

9 Statistical Physics: November 7, 212 (e) Use the method of Problem 7.46, part (d), to show that the free energy density of the electron-positron radiation is where f(t) = F = 16π(kT)4 () 3 f(t), (48) ( x 2 ln 1+e x 2 +(mc 2 /kt) 2) dx. (49) 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 ). (5) Using this, similar to the finding energy, sum all modes for n. Then, F = kt dn = 16π(kT)4 () 3 π/2 dθ π/2 dφ n 2 sinθln(1+e ǫ(n)/kt ) (51) x 2 ln(1+e x 2 +(mc 2 /kt) 2 ) dx. (52) f t t kt This is the graph of f(t). When the limit of temperature goes to zero, mc 2 f(t) e mc2 /kt e mc2 /kt x 2 ln(1+e mc2 /kt e (1+(kTx)2 /2(mc 2 ) 2) dx (53) x 2 e (1+(kTx)2 /2(mc 2 ) 2) dx (54) O(x 4 ) dx (55) e mc2 /kt O(x 5 ) x= x=. when T (56) 9

10 Statistical Physics: November 7, 212 I use the relation ln(1+x) x for small x. Now, take the limit of temperature goes to infinity, f(t) = x 2 ln(1+e x ) dx = For the first relation, I use the integral by parts. x e x +1 dx = 1 π x 3 3 e x dx (57) 1+e x 7 8 = 7π4 36. (58) (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, S = U F T = T 16π(kT) 4 () 3 (u(t)+f(t)). (59) 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 T, S T 16π(kT) 4 () 3 7π 4 9 = 56π5 45 k ( kt ) 3. (6) Problem 7.5: 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/c 2 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, but that is increasing with time. Write down a formula for the total entropy of the electrons, positrons, and photons as a function of 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 Then, the total entropy is S = S e +S e + +S γ = 16π(kT) 4 ( ) T () 3 (u(t)+f(t))+ 32π5 kt 3 45 k (61) ( ) kt 3 ] = 16πk [u(t)+f(t)+ 2π4 constant. (62) 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 T ν and the photon temperature T are related by ( ) T 3 [ ] 2π u(t)+f(t) = constant (63) T ν 1

11 Statistical Physics: November 7, 212 as the universe expands and cools. Evaluate the constant by assuming that T = T ν when the temperatures are very high. Solution: As we can see in the all entropy values, neutrino entropy also expected proportional to some factor, ( ) 3 ktν S ν 16πk constant, (64) and it is constant because of the neutrinos are unable to interact with anything. Then, we can easily see that ( ) T 3 [ ] 2π u(t)+f(t) = constant. (65) T 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 2π u(t)+f(t) 2π π5 9 = 11π4 9. (66) (c) Calculate the ratio T/T ν 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 2π 4 ( ) 45 = 11π4 T 3 = 11 T ν = 4T = 1.95 K, (67) T ν where T = 2.73 K, the temperature of universe. T ν (d) Use a computer to plot the ratio T/T ν as a function of T, for kt/mc 2 ranging from to 3. Solution: Use a mathematica, plot the graph. T T Ν t kt mc 2 11

12 In[1]:= Problem 7.39 maximum FindMaximum 1 x^5 E^ 1 x 1, x,.1 2, 1, 2 ; Plot 1 x^5 E^ 1 x 1, x,, 1.5, GridLines maximum, Dashed, Red,, PlotRange, 22, AxesLabel "x kt Λ", "u x ", PlotStyle Black u x 2 15 Out[2]= 1 5 In[3]:= Problem 7.47 h ^ 34 ; k ^ 23 ; c ; me ^ 31 ; e ^ 19 ; ione 13.6 e; 1; kt x Λ vq t_ h Sqrt 2 Pi me ione t ^3; Ngamma t_ 8 Pi ione t h c ^ ; tmax k 6 ione;

13 2 solution_7.nb In[13]:= fraction R_, t_ Solve F^2 R Ngamma t vq t Exp 1 t F 1, F, R, t 2, 1, 2 ; Needs "PlotLegends`" Plot fraction 1^ 8, t, fraction 1^ 9, t, fraction 1^ 1, t, t,, tmax, PlotLegend "R 1 8 ", "R 1 9 ", "R 1 1 ", LegendShadow None, LegendSize.5, PlotStyle Red, Dashed, Black, Blue, Dashed, LegendPosition.7,.25, AxesLabel "t kt ", "fraction" I t1 FindRoot fraction 1^ 9, t.1, t,.2 1, 2 ione k fraction 1..8 Out[15]=.6.4 R 1 8 R 1 9 R kt t I Out[16]= In[17]:= Problem 7.49 u T_ : NIntegrate x^2 Sqrt x^2 1 T^2 E^ Sqrt x^2 1 T^2 1, x,, Infinity, AccuracyGoal 1 ; infu Integrate x^2 Sqrt x^2 1 T^2 E^ Sqrt x^2 1 T^2 1. T Infinity, x,, Infinity ; Plot u T, T,, 3, PlotRange Full, Full, AxesLabel "t kt ", "u t ", mc2 GridLines, infu, Red, Dashed, PlotStyle Black u t 5 4 Out[19]= t kt mc 2

14 solution_7.nb 3 In[2]:= f T_ : NIntegrate x^2 Log 1 E^ Sqrt x^2 1 T^2, x,, Infinity, AccuracyGoal 1 ; inff Integrate x^2 Log 1 E^ Sqrt x^2 1 T^2. T Infinity, x,, Infinity ; Plot f T, T,, 3, PlotRange Full,, 2, AxesLabel "t kt ", "f t ", mc2 GridLines, inff, Red, Dashed, PlotStyle Black f t Out[22]= t kt Problem 7.5 mc 2 In[23]:= tempratio T_ : 11 Pi^4 9 2 Pi^4 45 u T f T ^ 1 3 ; Plot tempratio T, T,, 3, PlotRange Full,.9, 1.45, AxesLabel "t kt mc ", "T T Ν", 2 PlotStyle Black, GridLines, 1, Red, Dashed T T Ν Out[24]= t kt mc 2