I have a question about the definition temperature, given by $\frac{\partial S}{\partial E}(E,V,N) = \frac{1}{T}$

Is this valid only for isolated systems (and not applicable, for instance, to a (small) closed system in contact with a heat-bath)?

From the above formula (if it's applicable to any thermodynamic system in equilibrium), it would seem like temperature is a deterministic function of $(E,V,N)$. --- which would contradict the Maxwell-Boltzman distribution of energies, of a system with fixed (T,V,N).

  • $\begingroup$ What you think of as a contradiction is just a misunderstanding on your part. When you are applying Maxwell-Boltzmann distribution, you have connected it to a constant $T$ heat bath of infinite energy. That is a totally different situation from when you have a fixed $E,V,N$ isolated system. Mathematically you cannot say that they are the same, and conceptually they are very different. Physically, however, we establish rigorously that the actual system will behave the same way as if both definitions give the same results. $\endgroup$ Commented Feb 25 at 4:47
  • $\begingroup$ Thanks; I am trying to precisely nail down my misunderstanding. Could you clarify your last statement please? Do you mean that partial-derivative definition can also be applied to the canonical ensemble (non-isolated, temperature controlled system)? Perhaps I am miss-interpreting ... (I understand that the infinite heat-bath may be regarded as "approximately isolated", and hence partial-derivative definition applies to it). $\endgroup$ Commented Feb 26 at 0:02
  • $\begingroup$ For large enough systems that thermodynamic arguments can be safely applied, the partial derivative definition in the microcanonical ensemble will get the (temperature) numbers in agreement with the canonical ensemble's, if you just so happen to have those number-data you want to compare. $\endgroup$ Commented Feb 26 at 0:06
  • $\begingroup$ No, you are not trying to do the partial derivative on the infinite heat bath. You are taking partial derivative of the system that we are studying and controlling. $\endgroup$ Commented Feb 26 at 0:06

1 Answer 1


I think you are mixing Thermodynamics and Statistical Mechanics concepts.

From the point of view of Thermodynamics, there is a set of definitions and relations between functions of the state that hold for all systems at equilibrium, independently of the boundary conditions. Entropy and other state functions are well-defined for an isolated system (fixed energy $E$) and for a system in contact with a thermostat (fixed temperature $T$). Also, a relation like $\frac{1}{T}=\left.\frac{\partial S}{\partial E}\right|_{V,N}$ is universally valid although its meaning will be slightly different. For an isolated system, the temperature is provided by the knowledge of the fundamental equation $S(E,V,N)$. For a thermostatted system, it gives the value of the derivative of entropy wrt energy from an external parameter (temperature).

Things are slightly different in statistical mechanics, mainly if applied to finite systems. In such a case, relative fluctuations do not vanish, and some quantities may not be well-defined for small systems. However, thermodynamics is not intended to be applied to small systems in its standard form. And it is precisely the case of a finite, small system, where the statistical mechanics formulas in different ensembles are not equivalent. However, this is not the condition of the validity of thermodynamics.

  • $\begingroup$ Your answers are very helpful. Would you mind elaborating on how you would interpret $\frac{1}{𝑇}=\left.\frac{∂𝑆}{∂𝐸}\right\vert∣_{𝑉,𝑁}$ for a non-isolated system in contact with a thermostat? Which value of E would you use in this equation? (Can we use E corresponding to any micro-state in the ensemble? I guess not...). Thanks in advance... $\endgroup$ Commented Feb 25 at 23:58
  • $\begingroup$ @VaridhiShayana Within Equilibrium Thermodynamics, there are only macrostates. $E$ is the internal energy, which is a function of the state. Whether the state is obtained with a thermostat or other ways doesn't matter. It is only at the level of Statistical Mechanics of finite systems that one introduces microstates and fluctuations. Since Statistical Mechanics provides Thermodynamics only at the thermodynamic limit, if you insist on Statistical Mechanics, you may think $E$ as the most probable energy or the average energy among the energies of all the available microstates if $N$ is large.. $\endgroup$ Commented Feb 26 at 7:47
  • $\begingroup$ I think I understand, but just to confirm, let me ask a modified version of my original question. Consider a small statistical system, that is NOT a thermodynamic-limit system. Suppose the system is isolated, with fixed $(E,V,N)$. Let $\Omega(E,V,N)$ denote the number (/volume/measure) of micro-states corresponding to the macro-state $(E,V,N)$. Define $S(E,V,N) = k_B \ln \Omega(E,V,N)$. ...continued below ... $\endgroup$ Commented Feb 26 at 23:05
  • $\begingroup$ continued. .... Now consider the partial derivative $\left\{\frac{\partial S}{\partial E}(E,V,N)\right\}^{-1}$: Does this represent the (ensemble-)average of system-temperature, or the temperature of EVERY micro-state corresponding to $(E,V,N)$? Thanks again ... $\endgroup$ Commented Feb 26 at 23:06
  • $\begingroup$ @VaridhiShayana It is possible to show that in the microcanonical ensemble $\frac{1}{\frac{\partial S}{\partial E}}=\langle K \rangle$, where $K$ is the kinetic energy of each configuration. There is no temperature for a microstate; microstates have kinetic energy, and different microstates may have different kinetic energy, $\endgroup$ Commented Feb 26 at 23:29

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