Partial differentiation notation in thermodynamics In my thermodynamics notes, the definition of isothermal heat capacity at constant volume is
$$C_V := \frac T N \left(\frac{\partial S}{\partial T} \right)_V $$
and at constant pressure is
$$C_P := \frac T N \left(\frac{\partial S}{\partial T} \right)_P $$
where $T$ is temperature, $N$ is the number of molecules in the system and $S$ is entropy.
I don't fully understand the derivation symbols. To me, the partial derivative of entropy with respect to temperature would correspond to the "slope in the temperature direction", meaning that all other variables are considered to be "frozen". This would mean that the two partial derivatives above, and consequently $C_V$ and $C_P$, are one and the same object. But it is clearly not so.
Why is there a need to specify that we are keeping volume / pressure constant, when this is already implied in the definition of partial differentiation? How would one make this notation mathematically rigorous, i.e., what is the actual math behind this abbreviated notation? (Possibly taking differential forms into account?)
 A: Whenever we take a partial derivative, we need to specify what we are keeping constant.  We commonly just use the partial derivative notation (say $\frac{\partial }{\partial x_1}$ in $\mathbb{R}^n$) to mean "keeping the other $n-1$ coordinates constant" and if that's what is meant, we usually don't need to say anything more.  But that is not the only meaning of the partial notation.  You can keep other things constant.  For example, in $\mathbb{R}^2$, you can ask, "what is $\frac{\partial f}{\partial x_1}$ keeping $(x_1+x_2)$ constant?".
In the definitions you have given, $C_v$ is the partial derivative w.r.t. T and indeed P is allowed to vary while V is held constant.  And in the case of $C_p$, V is allowed to vary as P is held constant.  
A: The problem is that $S$ in physics can refer to several mathematical functions.
For example, for the mono-atomic ideal gas, the ones relevant here are:
$$\begin{align*}
S_1(T,V) &= k_\text{B} N \left( \frac{3}{2} \ln T  + \ln \frac{V}{N} + c \right),\\
S_2(T,P) &= k_\text{B} N \left( \frac{3}{2} \ln T + \ln \frac{k_\text{B} T}{P} + c \right)
.\end{align*}$$
Mind that the indices of $S$ are only for the purpose of this answer and are omitted in physics.
You can see how the two are connected via the thermic equation of state ($pV = N k_\text{B} T$).
Obviously, $S_1(X,Y) ≠ S_2(X,Y)$ for most arguments and thus $S_1≠S_2$.
Yet in physics, we just use $S$ to refer to both of them.
In other words, we are overloading the symbol $S$ in a way that we would never do in mathematics.
The reason why this (usually) works and you need not burn your physics textbooks is:

*

*The outcome of the function (if used as intended) is always the same physical quantity, namely the entropy.

*We usually know whether we want to calculate the entropy in dependence of $T$ and $V$ or of $T$ and $P$. This is also why physicists tend to refer to these functions as $S(T,V)$ or $S(T,P)$ instead of just $S$.

*If we use the wrong arguments, we usually get totally bogus results where even the units do not match. (Mind that in the above example, you would first need to combine all the logarithms and $c$ to get a logarithm of a unit-less quantity.)

As a result of this overloading, just writing $\frac{∂S}{∂T}$ is ambiguous: Do we mean $\frac{∂S_1}{∂T}$ or $\frac{∂S_2}{∂T}$? To resolve this ambiguity, we use the partial-derivative notation:

*

*$\left(\dfrac{∂S}{∂T}\right)_V$ refers to $\dfrac{∂S_1}{∂T}$ or $\dfrac{∂S(T,V)}{∂T}$.

*$\left(\dfrac{∂S}{∂T}\right)_P$ refers to $\dfrac{∂S_2}{∂T}$ or $\dfrac{∂S(T,P)}{∂T}$.

