Ph 136: Solution 4 for Chapter 4 and 5

Size: px

Start display at page:

Download "Ph 136: Solution 4 for Chapter 4 and 5"

Jemima Hoover
5 years ago
Views:

3 Ph 136: olution 4 for Chapter 4 and Grand Canonical Ensemble for Ideal Relativistic Gas [by Alexei Dvoretskii, edited by Geoffrey Lovelace (a) uppose there are N identical particles in the volume. ince the gas is in a classical regime, the average occupation number for each state is low ( η 1), so it is highly unlikely that a state will contain more than one quanta. A given configuration thus has N particles in N states, which can be done in N! different ways. Therefore M N! (b) Each particle can move in three dimensions; since the gas is monatomic, there are no rotational or vibrational degrees of freedom. Therefore, the total number of degrees of freedom is W 3N. We know that N 1, so, using Mh 3N the result from part (a), N 1. The particles travel freely for most of the N!h 3N time, so all of the energy is kinetic: E N N A1 p A + m (c) o far, we ve been treating N as a constant. However, in reality the system may have any number of particles, from to. For each system with N particles, we must sum over the possible states, weighting each term by the onential given in Eq. (3.4a). The sum over all states of a system with N particles confined to a volume each confined to a volume can be ressed as dnstates. Therefore, using Eq. (3.4a), we have Z n ( ) E + µr N n N dn states,n e E state +µ R N. Consider a single particle in the gas (say, particle A). It is confined to the volume that the gas occupies, so dn states,n,a d3 p A h 4πp 3 A dp A. The number of states available to the entire gas of N particles, then, is N A1 dn states,n 4πp A dp A N h 3N. N! Note that we have divided by N!, since the particles are identical; interchanging any two particles does not give a new state. We already found E E state in part (b).

4 Inserting these ression into the equation for Z, we have Z N A1 4πp A dp A N N N N e µ R N e µ R N N N!h 3N A1 e «PNA1 p A +m +µ R N h 3N N! N [ e p A +m 4πp Adp A N N!h 3N [ e N p +m 4πp dp In the last line, we note that dp A is just a dummy variable. (d) For the non-relativistic limit, write p + m m + p m, which is the non-relativistic energy. Then we have e p +m 4πp dp Let µ µ R m, so that Z N e m p m 4πp dp e m/ (πm ) 3/. N ( ) µn e N!h3N (πm ) 3N/ h 3 e µ (πm ) 3/. From Z, it is straightforward to get the grand potential Ω: Z e Ω/ log Z Ω e µ h 3 (πm )3/. This is Eq. 3.47a. For the ultrarelativistic limit, let m, so that the particles travel near the speed of light. Then e p +m 4πp dp e p/ 4πp dp 8π( ) 3. It is straightforward to evaluate this integral. The rest of the problem proceeds analogously to the nonrelativistic case, except you keep the relativistic chemical potential: log Z Ω 8π h 3 ( ) 4 e µ R/ (e) The quantities you are asked to calculate in part (e) are all partial derivatives of the potential you calculated in part (d). Eqs. (3.47c) tells you how to get N,, and P : Take minus the partial derivative of Ω with respect to

5 µ R, T. and, respectively. The results are then 8π ( )3 N P To calculate Ē, use Eq. (3.43): Then it immediately follows that h 3 e µ R/ 8π( )4 h 3 e µ R/ k N(4 µ R ). Ē Ω + T + µ R N 3 N. Ē/ N 3 and Ē/ 3P 5. Energy Representation for a Nonrelativistic Monatomic Gas [by Dan Grin (a) We begin with an ression for the fundamental potential E of a nonrelativistic gas in the energy representation, (see Eq. 4.9c in the text) ( ) ( ) 3h /3 ( ) E(,, N) N 4πm N N 5/3. (1) To derive the desired relations, we need only substitute Eqn. (1) into the following ressions for the intensive variables in terms of variables of the fundamental potential (Eqns. 4.1a in the text): T ( E ),N, µ ( E N ),, P ( ) E. (),N In the case of temperature, this derivative is trivial, as E only depends on through the onent, and yields ) /3 [ T ( h N πk B m N 5/3. In the case of pressure, the derivative is trivial, as E only depends on through the pre-factor in front, and so, P ( E ),N ( h N ) 5/3 [ πm N 5/3. The chemical potential is a little trickier, but we can simplify our lives a little by re-writing E in the following form ( ) 5/3 ( ) [ N 3h E 4πm N 5/3 (3)

6 Then, calling on the product (Leibniz) rule, we see that µ ( ) E N N µ 5, 3 ( ) 5/3 N ( ) /3 N 3h [ 4πm [ N 5/3 ( h N ) /3 ( ) 4πm 5 k B N [ N 5/3 N 5/3. (4) (b) We verify that the Maxwell relations are satisfied by taking the appropriate derivatives of the Eqns. derived in part a). ( ) P ( ) µ N, N, ( ( h h ) ( N πnm ) ( N 6πmN ) 5/3 [ ) 5/3 ( 5 ) k B N N 5/3 ( ) T,N [ N 5/3 ( ) P N (5) (6)., ( ) T N, { h ( ) 1/3 N 3πmk B [ N 5/3 h 3πmkB N ( ) µ ( ) } /3 N. (7) N, (c) To derive the ideal gas gas equation, we solve the temperature equation for the onential ression to obtain [ N 5/3 πk ( ) /3 BmT h. (8) N Plugging this into the pressure equation, oodles of factors cancel out to yield the desired ideal gas law: P Nk B T/. 4.1 Primordial Element Formation [by Alexei Dvoretskii, edited by Geoffrey Lovelace, Dan Grin, Nate Bode In the early universe, the protons and neutrons travelled at non-relativistic speeds, so they can be described as a monatomic, non-relativistic gas. The photons travel at the speed of light and make up an ultrarelativistic gas. We solve the problem by considering the gas in two different epochs and suppose that the transition is rapid. Initially we have a gas made up of solely neutrons

7 and protons. The entropy per protons and neutrons is (ignoring the small mass difference between the proton and neutron) ( [ ( ) ) 3/ 5 σ init 4 + ln m p πmp f ρ h [ ( ) 3/ m p πmp f ln ρ h In the final state the entropy will be given by the α-particle entropy per α- particle added to the photon entropy per photon. Note that the 7 Me is the binding energy per nucleon. Therefore, the total binding energy of an α-particle is 8 Me (so the problem incorrectly gives 7 Me). Therefore [ σ final 5 ( ) 3/ + ln 8m p 8πmp f ρ h + σ γ, where σ γ is the photon entropy per photon and can be roughly found from the thermodynamic equation σ γ γ /k U/T 8 Me/ ergs/. (9) Now we must calculate ρ at the transition point. As nucleosynthesis occurs after inflation the universe ands adiabatically except when particle species annihilate. At the Me temperatures under consideration, the last annihilation event (e + e γγ) has occurred, so we may treat the ansion of the universe as adiabatic. This means that the density of baryons is given by ρ baryon,init ρ baryon,now ( Tinit T now ) 3. (1) Currently the total matter density is ρ total,now g/cm 3, and the mass fraction of baryons is about %. Therefore to we need to equate σ init and σ final, plugging in Eqns. 1 and 9. olving using your favorite numerical solver gives T crit K. The time of nucleosynthesis can be found by plugging this temperature into T (t) T ( ) init Tinit t crit t init (11) t/tinit Therefore, t crit 1 s. T crit In Fig. 1, we can see that at early times (high temperatures), the higher entropy (preferred) state is p + n, while at late times (low temperatures T < T crit ), the higher entropy (preferred) state is α + γ, indicating that helium production does not occur until T T crit.

8 Σ particle 1 k Figure 1: Particle Entropy per Particle in units of k 1 as a function of temperature in Kelvin. Red curve shows entropy per particle for collection of protons and neutrons only (initial state), while the green curve shows entropy per particle for α-particles and photons in the final state.

( ) ( )( k B ( ) ( ) ( ) ( ) ( ) ( k T B ) 2 = ε F. ( ) π 2. ( ) 1+ π 2. ( k T B ) 2 = 2 3 Nε 1+ π 2. ( ) = ε /( ε 0 ).

( ) ( )( k B ( ) ( ) ( ) ( ) ( ) ( k T B ) 2 = ε F. ( ) π 2. ( ) 1+ π 2. ( k T B ) 2 = 2 3 Nε 1+ π 2. ( ) = ε /( ε 0 ). PHYS47-Statistical Mechanics and hermal Physics all 7 Assignment #5 Due on November, 7 Problem ) Problem 4 Chapter 9 points) his problem consider a system with density of state D / ) A) o find the ermi