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I am new in thermal physics and I started studying recently the canonical ensemble.

Until now, I understand that, in thermal equilibrium, the temperature and energy content of two systems placed in thermal contact with each other but thermally isolated from the surroundings will no longer be changing with time (in the thermodynamic limit, because there will still be somo fluctuations, but they are negligible).

We also define temperature in a statistical manner, and we relate it to the number of microstates the system can be with the energy available, more precisely:

$$ \frac{1}{k_bT} = \frac{\text d(ln(\Omega))}{\text dE} $$

So we relate the temperature with a microscopic property... And my question is:

We relate the temperature with a “microscopic configuration of the system”, does that mean the temperature is a microscopic property of the system?

Thank you for your time!

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    $\begingroup$ What do you mean by a "microscopic property"? $\endgroup$ – Aaron Stevens Dec 10 '19 at 22:43
  • $\begingroup$ In the same way that we can define the system by the microstate and / or the macrostate $\endgroup$ – Leibniz Rib Dec 10 '19 at 22:45
  • $\begingroup$ I suppose that a system have some macroscopics properties such as the pressure or the volume that we can actually measure, and then we have the microscopic properties such as the description of the motion of each individual molecule $\endgroup$ – Leibniz Rib Dec 10 '19 at 22:47
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    $\begingroup$ No, temp. is a macrostate. $\endgroup$ – aditya_stack Dec 11 '19 at 7:01
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Entropy isn't a property of a microstate (after there is only one of it so its entropy would be zero), but of the ensemble of states available under some set of conditions.

By the "some set of conditions" you consider in counting the states are generally defined in terms of macroscopic variables.

That is: the defining condition for the entropy is a macrostate.

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