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Solution Exercise 7 StatisticalThermodynamics SolutionExercise7 HS2015 Solution Exercise 7 Problem 1: Photon Gas in a Box a) The question is: How many states exist between ω and ω + dω. We can see that to each ω there is a surface of a sphere associated in m-space because the equation ω = cπ 2 2 2 cπ L mx + my + mz = L r is allied to a sphere. Since mx,my,mz can only assume posi- tivep values we are dealing with the positive octant. The number of states in the volume spanned by ω and ω +dω must be equivalent to the number of states in the volume spanned by r and r +dr. The approach becomes useful if the involved m are so large as to neglect the fact that the individual m are discrete. In this case the numbers of states for a given 2 2 2 ω are well described by the surface area in m-space. We find for mx + my + mz ≫ 1 or cπ ω ≫ L p 1 1 Lω 2 L 1 L3 f(ω)dω = A (r)dr = 4πr2dr = 4π dω = ω2dω. (1.1) surf 8 8 cπ cπ 2 π2c3 1 If we consider that there are two modes (TE, TM) per angular frequency, the factor of 2 in above equation is eliminated, leaving us with L3 f(ω)dω = ω2dω. (1.2) π2c3 In order to relate the condition of large ω to an experimental condition it is helpful to rewrite c ω =2π λ in terms of the wavelength λ c L λ =2π = . (1.3) mx,my,mz 2 2 2 ωmx,my,mz mx + my + mz p From above expression it is intuitive to see that as long as the electromagnetic wavelengths under investigation are much smaller than the relevant dimensions of the enclosure, equation 1.2 should hold. b) In case of Bose-Einstein statistics we have learned that the population per state is given as Ai Ni(ǫi)= , (1.4) B exp( ǫi ) − 1 kB T where B = exp(−α). The parameter α is the Lagrange multiplier which introduces the particle conservation constraint. It can be shown that α = µ with µ being the chemical kB T potential. We can therefore rewrite equation 1.4 to Ai Ai Ni(ǫi)= − = , (1.5) exp( ǫi µ ) − 1 exp( ǫi ) − 1 kB T kB T where we have set in the second step µ = 0. We have therefore for our modes ~ Ai f( ω) ~ Eω = Niǫi = ǫω = ~ ω. (1.6) exp( Eω ) − 1 exp( ω ) − 1 kB T kB T page 1 of 4 StatisticalThermodynamics SolutionExercise7 HS2015 In the last step of above equation we have implicitly changed from Ai to its continuous form f(E). The density of states f(~ω) is identical to f(ω) calculated in a). c) In order to calculate the spectral energy density we have to combine the results of a) and b) to find an expression for the total energy and modify this expression such that the integration variable is changed towards the frequency ν. In a first step we insert equation 1.2 into equation 1.6 and substitute V = L3 the resulting expression is ∞ V ~ ω3 u(T )= dω. (1.7) Z π2c3 exp( ~ω ) − 1 0 kB T It follows for the energy density u ∞ ~ ω3 uV (T )= = dω. (1.8) V Z π2c3 exp( ~ω ) − 1 0 kB T We can now substitute ω =2πν to obtain ∞ ~ 8π3ν3 ∞ 8πh ν3 uV (T )= 2πdν = dν, (1.9) Z π2c3 exp( hν ) − 1 Z c3 exp( hν ) − 1 0 kB T 0 kB T 8πh ν3 where we can identify u (ν,T )= 3 . V c exp( hν )−1 kB T d) With the same steps taken in c) and substituting the frequency ν with ν = c/λ we find 8πhc 1 uV (λ,T )= . (1.10) λ5 exp( hc ) − 1 λkB T If we plot equation 1.10 for different temperatures we find that the peak value of uV (λ,T ) depends strongly on the temperature. The peak spectral emission changes by 4 orders of magnitude between ambient temperature (300 K) and the melting point of platinum (2041 K). This explains why the historic experiments to determine the black body radiation were performed around the melting point of platinum. 4 300 K 3 1000 K )] 4 2041 K 2 1 ) [log10(J/m λ 0 −1 log10(u −2 −3 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1 wavelength [mm] Figure 1-1: Spectral energy density uV (λ) for different temperatures. From the scientific article provided on the exercise sheet we can deduce that the original design consisted of a 40 cm x 4 cm cylinder made from platinum. It is therefore clear that page 2 of 4 StatisticalThermodynamics SolutionExercise7 HS2015 the assumption of large ω done in a) is justified, since the majority of the radiation is emitted at wavelengths below 0.1 mm meaning that the physical dimensions of the black body are by a factor of 500 larger than the relevant wavelengths. Problem 2: Thermodynamical State Functions of the Photon Gas a) The heat capacity at constant volume is given as the derivative of the internal energy towards the temperature: ∂u 32 π5k4 c = =4bV T 3 = B VT 3. (2.1) v ∂T 15 h3c3 We want to compare now the energy required to heat the platinum enclosure and the energy required to heat the photon gas from 300 to 301 K. For the photon gas one finds T2 5 4 32 π kB 3 −14 ∆Ephotons = cvdT ≈ cv(300K)∆T = 3 3 VinsideT300K ∆T =4.17 · 10 J (2.2) ZT1 15 h c and equivalently for the heating of the enclosure it can be calculated 25◦C 25◦C 25◦C ∆Eenclosure = mpcp ∆T = VPtρPtcp ∆T =(Vcube − Vinside)ρPtcp ∆T =1.39J, (2.3) the quantities Vcube and Vinside correspond to the total volume of the cube and the volume of the cavity inside respectively. We find that the change in energy of the system due to the presence of the photon gas is negligible compared to the energy required to heat up the enclosure. b) In order to calculate the entropy we use the definition of the Clausius entropy T T cv ′ ′2 ′ 4 3 4u s = ′ dT = 4bV T dT = bV T = . (2.4) Z0 T Z0 3 3T c) The free energy can be calculated as 4u 1 f = u − Ts = u − T = − u. (2.5) 3T 3 d) Taking the derivative of the free energy towards the volume yields the associated pressure, the radiation pressure: 5 4 ∂f 1 4 8π kB 4 p = − = bT = 3 3 T . (2.6) ∂V T 3 45h c The condition to make a person with two black bodies strapped to their feet levitate is 1 mg p(T )= F = , (2.7) g 2A 2A where A is the effective emission area per foot. It follows immediately page 3 of 4 StatisticalThermodynamics SolutionExercise7 HS2015 1 1 45h3c3 4 Tfloat = mg 5 4 . (2.8) 2A 8π kB According to Wikipedia the average European weighs about 70 kg. This yields for the 5 necessary temperature Tfloat = 3 · 10 K which might be a minor technical obstacle given that the temperature on the surface of our sun is about 6 · 103 K. e) Lets discuss how many particles are hitting one of the walls being located in the y-z plane of our coordinate system. To do so we freeze our black body in time and look at all the photons knowing their velocities and positions. How many photons nhit will hit the x-z wall after a time interval ∆t after unfreezing the system? The answer is: Every photon that fulfills vx∆t ≥ s with s being the distance from the wall. Since we are dealing with a statistical ensemble where we only know the absolute velocity |v| = c we can take the root mean square velocity hvxiRMS but have to take into account that only half of the photons in the associated volume will hit, since the other half will travel in the opposite direction. It follows that the number of hits on the wall after the time interval ∆t is half the photons in the volume spanned by the area of the wall and hvxiRMS∆t. 1 1 n (ω)= ρ(ω)L2hv i ∆t = ρ (ω)L2c∆t, (2.9) hit 2 x RMS 6 Photon 1 1 where we have used that hvxiRMS = 3 hviRMS = 3 c. We find therefore for the hit rate n (ω) 1 n˙ (ω) ≈ hit = ρ (ω)L2c, (2.10) hit ∆t 6 Photon and conversely for the pressure per angular frequency ~ F (ω) n˙ hit(ω) 2 ω 1 ~ Pwall(ω)= = 2 = ρPhoton(ω) ω. (2.11) Awall L c 3 We can find the number density starting from the photon numbers as derived in 1b) f(~ω) N(ω,T )= . (2.12) exp( ~ω ) − 1 kB T As we assume this mode occupation to be related to ballistic traveling photons being isotrop- ically distributed we can calculate the number density N(ω,T ) f(~ω) ρPhoton(ω)= = . (2.13) V V (exp( ~ω ) − 1) kB T Inserting above equation into equation 2.11, substituting f(~ω) and integrating over all omega we obtain 1 u P = (2.14) total 3 V being the result we anticipated. page 4 of 4.
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