As far as I remember, many textbooks on statistical physics introduce temperature as a condition of equilibrium of a composite thermodynamic system. E.g., if the system consists of two parts with energies $E_1,E_2$, and the total energy is fixed, $E=E_1+E_2$, then maximizing the entropy (and neglecting the small contributions due to the interactions on the systems borders) we have: $$ \frac{d}{dE_1}\left[S_1(E_1) + S_1(E-E_1)\right]=S_1'(E_1) - S_2'(E-E_1)=0\\\Rightarrow S_1'(E_1) = S_2'(E-E_1)\\\Rightarrow E_1^*, \frac{1}{T}=S_1'(E_1^*)=S_2'(E-E_1^*) $$ The temperature is thus a value of the derivative of the entropy (logarithm of the number of the microstates at thermal equilibrium, but not the derivative itself (the difference between a function and a value of this function at a specific point, contrary to what is suggested in this question, which motivated me to ask mine.)

What I miss here is the transition from this definition of temperature to the condition of thermal equilibrium as the equality of temperatures - this means that the temperature must be defined for each subsystem independently. So, is the temperature a function (of energy) or a value of a function at thermal equilibrium?

  • $\begingroup$ I don't understand. If you see $T$ as a function, i.e. $T: E\mapsto T(E)$, then it maps an energy to the (equilibrium) temperature of a system. Then $T(E)$ is the temperature of the system, while $T$ is a function which maps an energy to the associated temperature. At least in equilibrium thermodynamics, all quantities anyway are defined for equilibrium only. Maybe you can clarify to me what I am missing. $\endgroup$ Commented Jun 8, 2022 at 16:59
  • $\begingroup$ The condition on equilibrium could then be stated like: The temperature function of system 1, $T_1$, evaluated at $E_1$ must be equal to the temperature function of system 2. $T_2$ evaluated at $E_2=E-E_1$, i.e. $T_1 (E_1 ) = T_2(E_2)$. $\endgroup$ Commented Jun 8, 2022 at 17:04
  • $\begingroup$ @JasonFunderberker if it is a function, the temperature can be defined even for an isolated system (mucrocanonical ensemble). And it is not specifically related to equilibrium - it is essentially the density of states $\endgroup$
    – Roger V.
    Commented Jun 8, 2022 at 17:04
  • $\begingroup$ But if you have the function, you have all temperature values for the corresponding energies and vice versa: Define $1/T:=S^\prime$ and the equilibrium condition you derive follows. If you start from your definition (i.e. start from the value of the derivative of $S$), then you can do it for a different range of energies, thereby defining a function by setting $1/T(E) = S^\prime (E)$. BTW: In principle you should have two temperature functions, no? Because $S_1 \neq S_2$ as functions, in general and the same should hold for their corresponding derivatives. $\endgroup$ Commented Jun 8, 2022 at 17:10
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    $\begingroup$ "I am just surprized that such a widely used concept is not clearly defined." The various definitions of entropy are worse. 😉 $\endgroup$ Commented Jun 8, 2022 at 19:31

1 Answer 1


The definition of temperature is $$ T = \left(\frac{\partial S}{\partial E}\right)_{V,N} $$ and as such it is a function of $E$, $V$ and $N$: $$ T = T(E,V,N) $$ The value of $T$ at a particular state is the numerical value of this function at the energy, volume and number of moles in that state.

If you have two systems you can can calculate their temperatures. If they are equal, the systems are in thermal equilibrium.

In other words, the equilibrium condition requires the value of tyemperature to be the same in both systems, not the functions: $$ \underbrace{T(E_1,V_1,N_1)}_{T_1} = \underbrace{T(E_2,V_2,N_2)}_{T_2} $$ which we write more conventionally as $$T_1 = T_2$$

I suspect the confusion arises from the common mathematical notation, $v = v(t)$, which assigns the same symbol $v$ to both the function (e.g., velocity as a function of time) and the value of the function at some specific $t$ (as in $v=0$ at $t=0$).


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