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Is it possible to define a notion of temperature in a microcanonical ensemble?

+ 1 like - 0 dislike
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I am thinking of a mircrocanonical ensemble as a finite system for which the number of particles, volume and the total energy is fixed. Is there a more refined view of this?

Can I think of temperature of this system as the average kinetic energy for all the particles?


To put the thing in a larger context - what is the implicit choice of ensemble when people define the entropy of a black-hole either as (1) horizon area or (2) via microstate counting or (3) macroscopically via quantum entropy function ?

Naively it feels that for a black-hole entropy one must think of a microcanonical ensemble because its not clear to me as to with what "bath" will it be able to exchange anything to maintain any chemical potentials for the conserved charges, volume or energy...may be I am being too naive...

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user user6818
asked Apr 14, 2014 in Theoretical Physics by user6818 (955 points) [ no revision ]
What about the standard definition of temperature for a microcanonical ensemble (appearing in every textbook on statistical mechanics) $\frac{1}{T} = \frac{\partial S}{\partial E}$? Where $S$ is the entropy of the system and $E$ its energy.

This post imported from StackExchange Physics at 2014-04-14 17:03 (UCT), posted by SE-user V. Moretti

1 Answer

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In a microcanonical system, one defines the entropy $S$ as $k$ times the logarithm of the microcanonical partition function (depending on energy $E$, particle number $N$, and volume $V$) and the temperature $T$ as $[(dS/dE)_{N,V}]^{-1}$. See, e.g., 

http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.73.2875

However, in order that thermodynamical variables satisfy the usual thermodynamic relations (and hence have the usual meaning!), one must take the thermodynamic limit of infinitely many particles. For finite systems, these relations are only approximately valid. 

answered Apr 14, 2014 by Arnold Neumaier (12,355 points) [ revision history ]

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