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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 $x_1,x_2...$.

Roger V.
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Alex Alex
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1 Answers1

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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.

Roger V.
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