# if $dU=TdS-PdV$ then $U=U(S,V)$ - rigorous proof?

It' not a physics question, just ..coincidence ;) (i'm concerned about mathematical rightness of it)

Let's consider $$U,T,S,P,V\in\mathbb{R_{>0}}$$ such that $$dU=TdS-PdV$$

• Based on this, how we can rigorously proof that $$U=U(S,V)$$?

Attempt 1: (probably inconclusive, see 'Attempt 2')

Let us consider $$A, X, Y \in \mathbb{R}\;\;\mid\;\; A=A(X,Y)\;\;\;\wedge\;\;\; dA=dU$$

Then $$dA=\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY$$ Requirement $$dA=dU$$ implies $$\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY=TdS-PdV$$ or $$\frac{\partial A}{\partial X}\bigg|_Y\,dX+\frac{\partial A}{\partial Y}\bigg|_X\,dY-TdS+PdV=0$$ Now, since $$dX, dY, dS$$ and $$dV$$ are arbitrary, to make the sum null, what they multiply must be zero, and since $$T,P$$ are not null by definition, only possibilities are that $$X=S\;\wedge\;Y=V \qquad\text{or}\qquad Y=S\;\wedge\;X=V$$ in either case, we obtain $$\frac{\partial A}{\partial S}\bigg|_V=T,\qquad\frac{\partial A}{\partial V}\bigg|_S=-P$$ (I've considered $$A$$ being just function of two variables $$X,Y$$, but this is not restrictive since if more than two variables were present in $$A$$ dependencies, the result woudn't change, as the additional partial derivatives appearing in $$dA$$ expansion would have been necessarily set to $$0$$, eliminating thus their dependency in $$A$$)

Also follows that

$$A=A(S, V)$$

Then, being $$dA=dU\,[..]\Rightarrow\,U=U(S,V)$$

1. How to properly carry on last step, if all was correct so far? (simply saying that $$A$$ and $$U$$ differ by a constant as a consequence to mean value theorem? but how we can say this if still we don't know $$U$$ dependencies..?)
2. Has sense to look for $$A$$ such that $$dA=dU$$ if $$A$$ initially is not function of the same variables as $$U$$?
3. Seems that to make the above reasoning work, $$X$$ and $$Y$$ have to be independent one with respect to the other, but what if we cannot require this for $$S$$ and $$V$$?

Attempt 2: (als inconclusive see 'Attempt 3')

From $$dU=TdS-PdV$$, we have $$\frac{dU}{dS}=T-P\,\frac{dV}{dS}\qquad\text{and}\qquad\frac{dU}{dV}=T\,\frac{dS}{dV}-P$$ Then $$\frac{dU}{dS}\bigg|_{V}=\Bigg(T-P\,\frac{dV}{dS}\bigg)\Bigg|_{V}=T\qquad\text{and}\qquad\frac{dU}{dV}\bigg|_{S}=\Bigg(T\,\frac{dS}{dV}-P\bigg)\Bigg|_{S}=P$$ Eventually $$dU=\frac{dU}{dS}\bigg|_{V}\,dS+\frac{dU}{dV}\bigg|_{S}\,dV$$

But here arises the problem, if i were sure that $$U$$ would just depend on $$S,\,V$$, we could have written (you can check wikipedia page on this) $$dU=\frac{\partial U}{\partial S}\,dS+\frac{\partial U}{\partial V}\,dV$$ and maybe arrive to the conclusion $$U=U(S,V)$$ in some way, but being the reasoning 'circular' we cannot do so..

So also this way seems inconclusive.. i wrote it in the hope of maybe clicking some ideas in the answerer, thanks!

• Studying thermodynamics eh? First off, if a combination of $dX,dY,dS,dV$ is $0$, you cannot say they are independently arbitrary without the coefficients all being $0$. Instead, you're assuming $dX$ and $dY$ are independent and $dS$ and $dV$ depend on them (so, 2DoF). Mathematically, this does not imply $\{X,Y\}$ is the same as $\{S,V\}$. You would need to write out $X=X(S,V)$ and $Y=Y(S,V)$ and use chain rule to expand $dS,dV$ in terms of $dX,dY$ and partial derivatives, then combine like terms. May 11, 2020 at 21:16
• For other derivatives, check out this paper. It was very helpful for me. May 11, 2020 at 21:33
• @runway44 Yea quite an old question i never understood completely.. So if i understood correcly, what you say would lead to $\frac{\partial A}{\partial X}\frac{\partial X}{\partial S}\bigg|_V+\frac{\partial A}{\partial Y}\frac{\partial Y}{\partial S}\bigg|_V=T$ and the same for $A$ wrt $V$ being equal to $-P$, then conclusion about partial derivatives $A$ is the usual May 11, 2020 at 21:52
• @runway44 Actually a problem with this requirement on $X$ and $Y$ (i.e. that $X=X(S,V)$ and $Y=Y(S,V)$ ) would be that $A$ should be defined as $A, X(S,V), Y(S,V) \in \mathbb{R}\;\;\mid\;\; A=A(X(S,V),\,Y(S,V))\;\;\;\wedge\;\;\; dA=dU$, which is a stronger requirement for $A$ and such $A$ might not exist (or should be proved to) Jun 7, 2020 at 17:45

Ok, i think i finally got it

An important hypotesis not written is that $$S, V$$ are mutually independent

Let us consider $$dU=T\,dS-P\,dV$$ From this 6 cases are possible:

1. $$U=U(S,V,\{X_i\})$$ where $$\{X_i\}=\{X_1,X_2,..,X_n\}$$ is a subset of all additional independent variables different from $$S,V,U$$ (note: if one of this additional variables had some dependencies from $$S$$ and/or $$V$$ it should not be included among U dependencies, if instead $$S$$ and/or $$V$$ had some dependencies from one or more $$X_i$$, then $$S$$ and/or $$V$$ are totally determined by a particular set of $$X_i$$s, and then $$S,V$$ should be not included in $$U$$ dependencies, but case 2 already possibly deals with this situation)
2. $$U=U(\{X_i\})$$
3. $$U=U(S,V)$$
4. $$U=U(S)$$
5. $$U=U(V)$$
6. $$U$$ has no dependencies

Case 1 - $$U=U(S,V,\{X_i\})$$

Let's calculate $$\frac{\partial U}{\partial X_i}=\frac{dU}{dX_i}\Bigg|_{S,V,\{X_{j}\}-X_i}=\bigg(T\,\frac{dS}{dX_i}-P\,\frac{dV}{dX_i}\bigg)\Bigg|_{S,V,\{X_{j}\}-X_i}=0$$ Thus we conclude that if $$U=U(S,V,\{X_i\})$$, $$U$$ cannot be function of any additional variable $$X_i$$, then Case 1 reduces to one of the remaining cases.

Case 2 - $$U=U(\{X_i\})$$

Let's calculate $$\frac{\partial U}{\partial X_i}=\frac{dU}{dX_i}\Bigg|_{\{X_{j}\}-X_i}=T\,\frac{dS}{dX_i}\Bigg|_{\{X_{j}\}-X_i}-P\,\frac{dV}{dX_i}\Bigg|_{\{X_{j}\}-X_i}$$ Now, if $$S,V$$ are not dependent from any $$X_i$$, then $$dS$$ and $$dV$$ are just arbitrary increments and then we can choose them to be null, making expression above to be zero. In this eventuality, we conclude that if $$U=U(\{X_i\})$$, $$U$$ cannot be function of any variable $$X_i$$, then this eventuality reduces to Case 6.

If instead $$S$$ is determined by a certain set $$\{X_i\}'\subset\{X_i\}$$, we cannot make the expression above to be zero, but certanly since $$\frac{d U}{d S}\bigg|_{\{X_i\}-\{X_i\}'}=T\neq 0$$ and since S is totally determined by $$\{X_i\}'$$, we could equivalently consider Case 1, Case 3 and Case 4 instead. The same goes for the situation in which $$V$$ is determined by $$\{X_i\}''\subset\{X_i\}$$, we could equivalently consider Case 1, Case 3 and Case 5.

In conclusion, considering what already concluded for Case 1, Case 2 reduces to one of the remaining cases.

Case 4 - $$U=U(S)$$

If $$U$$ is solely a function of $$S$$, then for any variable $$A\neq S$$ we should have $$\frac{dU}{dA}\Big|_S=0$$

But for $$A=V$$ $$\frac{dU}{dV}\Bigg|_S=-P\neq 0$$ Thus, we conclude that Case 4 is NOT possible.

Case 5 - $$U=U(V)$$

If $$U$$ is solely a function of $$V$$, then for any variable $$A\neq S$$ we should have $$\frac{dU}{dA}\Big|_V=0$$

But for $$A=S$$ $$\frac{dU}{dS}\Bigg|_V=T\neq 0$$ Thus, we conclude that Case 5 is NOT possible.

Case 6 - $$U$$ has no dependencies

If $$U$$ has no dependencies, then we might choose $$dU$$ arbitrarly, in particular we could choose it such that for any variable $$A$$ we have $$\frac{dU}{dA}=0$$

But for $$A=V$$ $$\frac{dU}{dV}=T\,\frac{dS}{dV}-P\neq 0$$ Having selected $$dS=0$$ since arbitrary, as not dependent on V.

Thus, we conclude that Case 6 is NOT possible.

Case 3 - $$U=U(S,V)$$

Only case left, we can finally conclude that $$U=U(S,V)$$

The OP's solution is rigorous and sound. Here I would like to point out that $$U=U(S,V)$$ by construction.

The most complete equation for the internal energy would read

$$dU = d(TS) - d(PV)$$

In this case, the internal energy is dependent on all four variables with $$d(TS)$$ being the change in heat of the system while $$d(PV)$$ is the change in pressure-volume work done by the system. The sign of $$d(PV)$$ is convention and changes between chemistry and physics uses. Now apply chain rule

$$dU = TdS + SdT - PdV - VdP$$

It just so happens, it is a heck of a lot easier to design experimental systems that are isothermal (constant temperature, $$T$$) as opposed to isentropic (constant entropy, $$S$$) and isobaric (constant pressure, $$P$$) systems are much safer to handle than isochoric (constant volume, $$V$$) systems. Hence, the most applicable equation for internal energy would be a system that is at constant temperature and pressure thereby obeying the formula

$$dU = TdS - PdV$$

You could, on the other hand, study a system that is isothermal and isobaric (like an insulated jar with the top on). That would yield a new equation

$$dU = TdS - VdP$$

In this case $$U=U(S,P)$$, making the variable dependencies by construction. OP's proof should work for every case chosen for $$U$$.

• The point is that some theory start with defining $dU=\delta Q - \delta W$, then with few passages more reach $dU=TdS-PdV$, but starting straight on with a differential, is not so intuitive for understand $U$ dependencies.. Also i studied some theory ('Physical foundations of continuum mechanics - A.I. Murdoch') where local form of energy balance (which integrated can be shown to be reconducted to first law of thermodynamcis) is obtained from purely molecular dynamics considerations, and so all quantities that appear in the final equation are function of $(\mathbf{x},t)$.. Jun 8, 2020 at 11:14
• .. so construction of $U$ might be different, but my claim is that in either cases can be shown that if $dU=TdS-PdV$ equation holds (or similar), then $U=U(S,V)$ (or related) Jun 8, 2020 at 11:17

Thinking back, i couldn't accept that such a simple question had such a long and complicated proof, so eventually i come up with:

Assume $$U=U(S,V)$$, this implies $$\frac{dU}{dA}\bigg|_{S,\,V}=0\qquad\forall\, A\;\text{not dependent on S,V,U}$$ Substituting, we have $$\frac{dU}{dA}\bigg|_{S,\,V}=\bigg(T\;\frac{dS}{dA}-P\;\frac{dV}{dA}\bigg)\bigg|_{S,\,V}=0-0=0$$ and since this holds for any A not dependent on S,V,U, assumption is proved.

Note: it does not matter if $$S$$ and/or $$V$$ depend on A, their variation must be 0 anyways.