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Suppose I have $U$ in it's natural variables that is $U(S,V)$ and lets say somehow I get the same internal energy in terms $U(V,T)$, then does it make sense to say:

$$U(V,T) = U(S,V). \tag{1}$$

And, also $(\frac{\partial U}{\partial V}_T )dV +( \frac{\partial U}{\partial T}_V )dT = (\frac{\partial U}{\partial S}_V) dS + (\frac{\partial U}{\partial V})_S dV. \tag{2}$

It sounds like mathematical gibberish but it makes physical sense to me for the following reasons: In (1) even though functional dependence is different ultimately for a given state of one substance, two variables imply all the others (eg: take ideal gas, having $P$, $V$ implies $T$) and for (2) if we change from a state $(S,V,T)$ to $(S+dS,V+dV,T+dT)$ then the change in internal energy calculated by the two expression must be equal.

However, this seems mathematically weird to do because we are equating two functions of different dependencies (albeit they are supposed to represent same thing). So, are the above equations correct and what property of the thermodynamic functions allow them to be equated as so (if it is correct)?

Qmechanic
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Clemens Bartholdy
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2 Answers2

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We can't directly write $U(V,T)=U'(S,V)$. There should be a map from $\phi:(S,V)\to (V,T)$ such that: $U'(S,V)=U(\phi(S,V))$

Consider a parametrized curve $\gamma:t\to\mathbb R^2$ s.t. $\gamma(t)=(\sin(t),\cos(t))$ where $t\in[0,\infty)$. It traces a circle starting second-quadrant from positive y-axis
Now consider another parameterized curve $\gamma ':t'\to\mathbb R^2$ s.t. $\gamma '(t)=(\cos(t'),\sin(t'))$ where $t'\in [0,\infty)$.
It traces a circle starting first quadrant from positive x-axis Now consider a reparameterization map $\phi:t'\to t\;\text{so that}\;\phi(t')=t+\pi/2$. It is translation of $t'$ to $t'+\frac{\pi}{2}$.
So, we can see that $\gamma(\phi(t'))=\gamma '(t')$.

So we can't directly equate $\gamma (t)=\gamma '(t')$. There should be a reparameterization map connecting the two.

DanielC
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Iti
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Yes. Both Eq.(1) and Eq.(2) are ok. They simply the calculus chain-rule and change variables. Nothing wrong with that. You don't need a thermodynamic property to validate these equations. It is rigorous mathematical relation. \begin{align} & f(x(\xi, \eta), y(\xi, \eta)) = f(\xi, \eta).\\ df &= \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy;\\ &=\frac{\partial f}{\partial x} \left\{\frac{\partial x}{\partial \xi} d\xi + \frac{\partial x}{\partial \eta} d\eta \right\} +\frac{\partial f}{\partial y} \left\{\frac{\partial y}{\partial \xi} d\xi + \frac{\partial y}{\partial \eta} d\eta \right\};\\ &=\left\{\frac{\partial f}{\partial x}\frac{\partial x}{\partial \xi} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \xi} \right\}d\xi +\left\{\frac{\partial f}{\partial x}\frac{\partial x}{\partial \eta} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial \eta} \right\}d\eta;\\ &= \frac{\partial f}{\partial \xi} d\xi + \frac{\partial f}{\partial \eta} d\eta \end{align}

But the natural variable had been set by the first law of thermodynamic, it favors the expression in rightt-hand-side of Eq.(2) making it fruitful:

\begin{align} dQ & = P dV + dU - \mu dN; \, \text{ and } \\ dU & = T dS - p dV + \mu dN. \tag{3} \end{align}

Therefore if you wrote the differential form in the right-hand-side of your Eq.(2), you can match the expression with Eq.(3), the first law. It allows you to further develope into various Maxwell relations:

\begin{align} T & = \, \frac{\partial U}{\partial S}\big |_{V, N};\\ P & = -\frac{\partial U}{\partial V}\big |_{S, N};\\ \mu & = \,\frac{\partial U}{\partial N}\big |_{S, V}; \end{align}

If you wrote in the form of left-hand-side, it is hard to apply the first law to develope further relations. The preference of the right-hand-side of Eq.(2) is physical reason, not mathematical.

ytlu
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