This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solution for Chapter 4 (compiled by Xinkai Wu, revised by Kip Thorne) A. Exercise 4.1 Pressure Measuring Device [by Kip Thorne] We could choose our pressure measuring device (pressure meter) to exchange only volume and not heat with its reservoir, but that leads to some tricky subtleties as we see in Exercise 4.8; so it is better (more straightforward) to let the meter exchange both volume and heat with its reservoir. This is analogous to the meter that Kip discussed in class on Monday, which exchanged particles and heat. Then the probability for the meter to be in a (quantum) state with vol ume Δ V and energy Δ ˜ E (probability measured either using an ensemble of meter&reservoir systems or via the fraction of the time the meter spends in a given state) is given by the meter’s Gibb’s distribution: ρ = K exp[( Δ ˜ E P Δ V ) /kT ] ; or equivalently ln ρ = ( Δ ˜ E P Δ V ) /kT + constant . Here K is the normalization constant and P and T are the reservoir’s pressure and temperature. Suppose that the reservoir plus meter has total energy ˜ E o , volume ˜ V o , and number of particles N o ; and the meter plus reservoir is a closed system. Then by energy and volume conservation, when the meter has volume Δ V and energy Δ E , the reservoir has volume V r = V o Δ V and energy E r = E o Δ E . Denote by N r ( ˜ E r ,V r ) the total number of states of the reservoir that have energy E r , volume V r and particle number N r , and similarly N m (Δ E, Δ V ) for the meter. (The meter might not be made of the same kind of particles as the reservoir; it might, for example, just be a massspring system that can be heated; the only variables we care about for it are energy and volume.) Then the probability ρ is proportional to the following product of numbers of states for the reservoir and meter: ρ ∝ N r ( E o Δ E,V o Δ V,N o ) × N m (Δ E, Δ V ) . Because the meter is tiny compared to the reservoir, when we take the logarithm of this ρ , the changes in the meter term, as energy and volume flow back and forth, are negligible compared to the changes in the reservoir term, so ln ρ = ln[ N r ( E o Δ E,V o Δ V,N o )] + constant . This logarithm of the number of reservoir states is proportional to the reservoir entropy, so ln ρ = S r ( E o Δ E,V o Δ V,N o ) /k + constant . 1 Expanding to first order in the meter energy and volume we obtain ln ρ = ∂S r ∂E r V r ,N r Δ E ∂S r ∂V r E r ,N r Δ V . Comparing with our previous expression for ln ρ we see that the pressure and temperature that appear in the meter’s Gibbs distribution are related to the reservoir’s entropy by the usual reservoir thermodynamic relations 1 T = ∂S r ∂E r V r ,N r , P = T ∂S r ∂V r ˜ E r ,N r = ∂ ˜ E r ∂S r !...
View
Full
Document
This document was uploaded on 04/17/2010.
 Spring '09
 Physics, Heat

Click to edit the document details