We can identify $k \beta = \left(\frac{\partial S}{ \partial E}\right)_V$ with the thermodynamic $k \beta = \frac{1}{T}$.

This seems to be, at least to me, very natural in the context of the canonical ensemble in which our system is in contact with a heat bath.

My question is as follows: suppose we are working in the micro-canonical ensemble and we manage to obtain the entropy as a function of the energy. Is this identification still meaningful in a physical sense? We can always define $k \beta = \left(\frac{\partial S}{ \partial E}\right)_V$ but will it be nicely related to the thermodynamic temperature in the micro-canonical ensemble as well? Where we obtain the entropy $S$ according to the formula $S=k \ln[\Omega]$. $\Omega$ is the total number of microstates.

Is the temperature well defined if our system is not in contact with a heat bath? Or does thermodynamics, in a certain sense, assume that our system is in contact with a heat bath?

I would appreciate any help you can give me.

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    $\begingroup$ The Microcanonical ensemble describes an isolated system, i.e. a system not in contact with any sort of external bath. nyu.edu/classes/tuckerman/stat.mech/lectures/lecture_3/… physics.udel.edu/~glyde/PHYS813/Lectures/chapter_6.pdf $\endgroup$ – Hasan Dec 6 '16 at 11:12
  • $\begingroup$ Thanks for your reply. I know, but does that mean that it does not have a well defined temperature? If we take the derivative$\left(\frac{\partial S}{ \partial E}\right)_V$ calculated in the microcanonical ensemble is it not identifiable in any way with the thermodynamic temperature? $\endgroup$ – Ben Stokes Dec 6 '16 at 11:15
  • $\begingroup$ Temperature is defined by the zeroth law of thermodynamics in terms of the flow of heat when you place systems in contact. If you require your system to be isolated, this definition makes no sense and your system does not have a temperature. $\endgroup$ – By Symmetry Dec 6 '16 at 11:19
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    $\begingroup$ The temperature you calculate in the microcanonical ensemble is the temperature you would have to put another body to be in thermal equilibrium if you let them exchange energy $\endgroup$ – P. C. Spaniel Dec 6 '16 at 12:31

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