Statistical Physics: November 7, 2012 Solutions for the Homework 7 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 20 15 10 5 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).