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Question. Why internal energy $U(S, V, N)$ is a homogeneous function of $S$, $V$, $N$? How to prove that? That is why does $\lambda\cdot U(S,V,n)=U(\lambda\cdot S, \lambda\cdot V, \lambda\cdot N)$?

Actuality. Homogeniety of internal energy is used to prove $U=TS-PV+\sum_i \mu_i n_i$ like here: Why does $U = T S - P V + \sum_i \mu_i N_i$?.

I know that function homogeneity is related to function extensionality.

P.S.: I am chemist, so things that are obvious to physicist might be not obvious to me. So I prefer proofs. Proof is sequence of formulas where each of them is axiom or hypothesis or derived from previous steps by inference rules. I prefer Fitch notation.

Update: From Roger Vadim answer I can deduce the following:

  1. $U_1=U(S_1;V_1;N_1)$

  2. $U_1=U(S(N_1);V(N_1);N_1)=U(N_1)$ because $S_1$, $V_1$ are functions of $N_1$

  3. $U_1=U(S(N_2);V(N_2);N_2)=U(N_2)$ similarly

  4. $N_2=N_1 k$ property of numbers

  5. $U_2=U(N_2)=U(N_1 k)=kU_1$ because $U$ is extensive variable

  6. $S_2=S(N_2)=S(N_1 k)=S(N_1) k$ because $S$ is extensive variable

  7. $V_2=V(N_2)=V(N_1 k)=V(N_1) k$ because $V$ is extensive variable

  8. $k U(S(N_1);V(N_1);N_1)=U(S(N_2);V(N_2);N_2)$ from 2,3,5 using simple algebra

  9. $k U(S(N_1);V(N_1);N_1)=U(S(k N_1);V(k N_1);k N_1)$ from 4,8 using simple algebra

  10. $k U(S(N_1);V(N_1);N_1)=U(k S(N_1);k V(N_1);k N_1)$ from 6,7,9 using simple algebra

  11. $k U(S_1;V_1;N_1)=U(k S_1;k V_1;k N_1)$ from 10 similarly as in 2 step

Summary: If function $F(x_1, x_2...)$ is extensive and $x_1, x_2...$ are extensive than $F(x_1, x_2...)$ is homogenous function of order one of variables $x_1,x_2...$.

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1 Answer 1

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In physical terms the homogenuity of the internal energy, entropy, and some other function is due to it being an extensive variable, i.e., variable whose value increases proportionally to the size of the system. This is opposed to intensive variables, such as temperature or pressure.

In other words, if we take a system consisting of two parts, its total energy is the sum of the energy of the parts. It is obviously not true, but this is where the beauty of the thermodynamical/statistical reasoning comes in: while the energy of the parts scales proportionally to their volume (i.e., proportional to the number of particles, $V\propto N$), their interaction energy scales proportionally to their surface area, $S\propto N^{2/3}$, and can be neglected in thermodynamic limit.

All texts on statistical physics/thermodynamics address this point, although not necessarily in terms of homogeneous functions, i.e., not necessarily in terms of equations - as it is often the case in this field, where many results are obtained via logical reasoning, rather than equation manipulation.

Remarks
In practice, in many problems thermodynamics/stat phsyics is reduced to non-interacting Hamiltonians, for which the additivity of energy is exact.

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  • $\begingroup$ Ok. So how to prove that if U is extensive variable then U(S,T,N) is homogeneous function of S, T, N? It is not self evident for me, I can't deduce it from your post. I used to study logic by myself, for me equation manipulation and logical reasoning it is the same thing. $\endgroup$
    – Alex Alex
    Commented Nov 18, 2021 at 15:01
  • $\begingroup$ There is actually a typo in the first sentence in your question: $U(S, V, N)$ is a function of $S, V,N$ (rather than $S, T, N$), all of which are extensive variables. E.g., if we take two identical systems, they will have twice as many particles, twice as big a volume and twice the entropy of any one of them. But they have the same teperature $\endgroup$
    – Roger V.
    Commented Nov 18, 2021 at 15:35
  • $\begingroup$ Thank you. I corrected typo. $\endgroup$
    – Alex Alex
    Commented Nov 18, 2021 at 15:37
  • $\begingroup$ Is it clear now? $\endgroup$
    – Roger V.
    Commented Nov 18, 2021 at 15:38
  • $\begingroup$ Does that mean that there is a criterion that defines what can be a state function other than the fact that its value cannot depend on the path taken (i.e. it must either be extensive or intensive and never in between) at the thermodynamic limit given that the system can reduce to one with non-interacting Hamiltonians at the limit? (I'm asking this since I'm never convinced that the commonly-used state functions must be extensive or intensive) $\endgroup$ Commented Mar 5, 2022 at 5:40

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