differentiation - Total and partial derivatives in thermodynamics and Maxwell relations

Consider the expression $$dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$

I'm trying to understand how to derive an expression for $\left( \frac{\partial S}{\partial V} \right)_P$ and how is it related to $\left( \frac{\partial S}{\partial V} \right)_T$.

I tried the following:

Method 1

i) Divide both sides by dV $$\frac{dS}{dV}=\left(\frac{\partial S}{\partial T}\right)_V\frac{dT}{dV}+\left(\frac{\partial S}{\partial V}\right)_T\frac{dV}{dV}$$

ii) and at const. P

$$\left(\frac{dS}{dV}\right)_P=\left(\frac{\partial S}{\partial T}\right)_V\left(\frac{dT}{dV}\right)_P+\left(\frac{\partial S}{\partial V}\right)_T\left(\frac{dV}{dV}\right)_P$$

$$\left(\frac{dS}{dV}\right)_P=\left(\frac{\partial S}{\partial T}\right)_V\left(\frac{dT}{dV}\right)_P+\left(\frac{\partial S}{\partial V}\right)_T$$

Question 1: how does

$$\left(\frac{dS}{dV}\right)_P=\left(\frac{\partial S}{\partial T}\right)_V\left(\frac{dT}{dV}\right)_P+\left(\frac{\partial S}{\partial V}\right)_T$$

become

$$\left(\frac{\partial S}{\partial V}\right)_P=\left(\frac{\partial S}{\partial T}\right)_V\left(\frac{\partial T}{\partial V}\right)_P+\left(\frac{\partial S}{\partial V}\right)_T???$$

Method 2:

Differentiate both side wrt V, holding P const. and use product rule

$$\frac{\partial}{\partial V}\left(dS\right)_P=\frac{\partial}{\partial V}\left(\left(\frac{\partial S}{\partial T}\right)_VdT\right)_P+\frac{\partial}{\partial V}\left(\left(\frac{\partial S}{\partial V}\right)_TdV\right)_P$$

$$\left(\frac{\partial dS}{\partial V}\right)_P=\left(\left(\frac{\partial^2 S}{\partial V \partial T}\right)_V\right)_PdT+\left(\frac{\partial S}{\partial T}\right)_V\left(\frac{\partial dT}{\partial V}\right)_P+\left(\left(\frac{\partial^2 S}{\partial V^2}\right)_T\right)_PdV+\left(\frac{\partial S}{\partial V}\right)_T\left(\frac{\partial dV}{\partial V}\right)_P$$

Question 2: I got so many extra terms, and how to deal with these $$\left(\frac{\partial \text{ d blah}_1}{\partial \text{ blah}_2}\right)_{\text{blah}_3}$$ terms?

Also, on more general grounds:

Question 3: How to partially differentiate a total differential rigorously?

Question 4: Are partial derivatives that differ in only the kept const. term identical in general?

Answer

$$\frac{dS}{dV}=\left(\frac{\partial S}{\partial T}\right)_V\frac{dT}{dV}+\left(\frac{\partial S}{\partial V}\right)_T\frac{dV}{dV}$$

This doesn't make much sense, because is not a well defined expression. The differential $$ \tag{A} dS=\left(\frac{\partial S}{\partial T}\right)_VdT+\left(\frac{\partial S}{\partial V}\right)_TdV$$ is just telling you that if you impose a slight variation on $V$ keeping $T$ constant, $S$ changes accordingly by an amount we denote with $ \left( \frac{\partial S}{\partial V}\right)_T $, so if you ask what is the partial derivative of $S$ with respect to $V$, the answer is tautologically $ \left( \frac{\partial S}{\partial V}\right)_T $ (and of course all the reasonings are the same with $V$ and $T$ exchanged).

You can safely take this (in this context) as the definition of the differential expression (A). In other words, writing (A) is exactly the same as stating that $S$ is a function of the two variables $T$ and $V$.

What is then the meaning of the expression $\left( \frac{\partial S}{\partial V} \right)_P $ ?

It means that you are now considering $T$ itself as a function of $P$ and $V$, call it $\tilde{T}(P,V)$, and effectively asking for the partial derivative of the function $\tilde{S}$ defined by $$ \tilde{S}(P,V) \equiv S(\tilde{T}(P,V),V) $$ with respect to $V$, which is the quantity: $$ \frac{\partial \tilde{S}}{\partial V} (P,V) \equiv \left( \frac{\partial \tilde{S}}{\partial V} \right)_P \equiv \lim_{\epsilon \to 0} \frac{\tilde{S}(P,V+\epsilon) - \tilde{S}(P,V)}{\epsilon}$$ using the usual chain rule you obtain $$ \frac{\partial \tilde{S}}{\partial V} (P,V) = \frac{\partial S}{\partial T}(P,V) \frac{\partial T}{\partial V}(P,V) + \frac{\partial S}{\partial V}(P,V) $$ and this is what is meant by the less rigorous, shorter expression $$ \left( \frac{\partial S}{\partial V} \right)_P = \left( \frac{\partial S}{\partial T} \right)_V \left(\frac{\partial T}{\partial V} \right)_P + \left( \frac{\partial S}{\partial V} \right)_T $$

how does $$\left(\frac{dS}{dV}\right)_P=\left(\frac{\partial S}{\partial T}\right)_V\left(\frac{dT}{dV}\right)_P+\left(\frac{\partial S}{\partial V}\right)_T$$ become $$\left(\frac{\partial S}{\partial V}\right)_P=\left(\frac{\partial S}{\partial T}\right)_V\left(\frac{\partial T}{\partial V}\right)_P+\left(\frac{\partial S}{\partial V}\right)_T???$$

They are the same thing. You have a function $S$ of the two variables $T$ and $V$. What you can do is just differentiating with respect to one or the other, obtaining: $$\frac{\partial S}{\partial T} (T,V) \qquad \text{and} \qquad \frac{\partial S}{\partial V} (T,V) $$ which means (taking the first one for example) the partial derivative of $S$ with respect to $T$, evaluated at the point $T,V$, i.e. the object defined by $$ \tag{B} \frac{\partial S}{\partial T} (T,V) \equiv \lim_{\epsilon \to 0} \frac{S(T+\epsilon,V) - S(T,V)}{\epsilon}$$ Writing this with $d$ instead of $\partial$, in this context, is just notation: $$ \frac{\partial S}{\partial T} (T,V) \equiv \left( \frac{\partial{S}}{\partial T} \right)_V \equiv \frac{d S}{d T} (T,V) $$ the notation with the $(\cdot)_V$ is useful to remark which variable/variables is/are kept constant while deriving (as you can see in (B), where $V$ is not varied).

Also related:

1. 
   

   Determine the Dependence of S (Entropy) on V and T

   
   


2. 
   

   What exactly is the difference between a derivative and a total derivative

   
   

$$\left(\frac{\partial dS}{\partial V}\right)_P$$

Please, do not ever write something like this :). A partial derivative is an operation that you can apply to (multi-variable) functions. A differential is not a (multi-variable) function, and its partial derivatives are not defined.

$dS$ means "a little variation of the variable $S$", which can be caused by a corresponding variation of the parameters on which it depends. If you ask what is the variation of $S$ while keeping some other quantity constant, you just divide $dS$ by that quantity (say $dV$) and impose the constraint you want (which is the first method you mentioned). Or for a more rigorous (and more clumsy) approach, you do the partial derivatives of the $\tilde{S}$ defined above. The result is the same.

Are partial derivatives that differ in only the kept const. term identical in general?

No they are not. Consider the following example: let $F$ be a function of the two variables $A$ and $B$, and suppose that $B$ is also a function of other variables, say $A$ and $C$: $$F = F(A,B), \qquad B = B(A,C)$$ Then we have $$ \left( \frac{\partial F}{\partial A} \right)_B \equiv \lim_{\epsilon \to 0} \frac{F(A+\epsilon,B) - F(A,B) }{\epsilon} $$ but with $\left( \frac{\partial F}{\partial A} \right)_C$ we also have to consider that the second argument $B$ of $F$ changes with $A$, then: $$ \left( \frac{\partial F}{\partial A} \right)_C = \left( \frac{\partial F}{\partial A} \right)_B + \left( \frac{\partial F}{\partial B} \right)_A \left( \frac{\partial B}{\partial A} \right)_C $$ The point is that we are now actually deriving another function, call it $\tilde{F}$, defined by $$ \tilde{F}(A,C) \equiv F(A,B(A,C)) $$ Note that the $B$ in $F(A,B)$ and the $B$ in the above definition are very different objects: the former is a number (an independent variable), the latter is a function of the two variables $A$ and $C$.

Note that in all of these reasonings you are always dealing with partial derivatives of functions of many variables (for example the $S(T,V)$ above). What makes it confusing (and I remember being confused myself by this when dealing for the first time with this subject) is the fact that you implicitly, when needed, consider some variables as functions themselves of other variables (just like $T$ that becomes $\tilde{T}(P,V)$ above). This feels pretty natural when you understand what is the exact meaning of the expressions, but can also be confusing at first.