# Explain this equation mathematically

$$\Bigl( \frac{\partial S}{\partial T} \Bigr)_H = \Bigl( \frac{\partial S}{\partial T} \Bigr)_M + \Bigl( \frac{\partial S}{\partial M} \Bigr)_T \Bigl( \frac{\partial M}{\partial T} \Bigr)_H$$

How can one go from LHS to RHS. I understand chain rule and partial derivatives but unable to figure out how can this be written. Explanations are welcome!

• Please do not use images/ screenshots but Mathjax. Commented Apr 6, 2022 at 6:47
• Consider $S(T,M(T,H))$ Commented Apr 6, 2022 at 7:03

Dealing with this kind of equations, I like to think of differential forms.

There are essentially two rules that solve most of the problems:

1. Algebraically, you can treat differential $$dH$$ forms as vectors, i.e. you can add them, multiply by scalars and decompose one form as a linear combination of others.
2. The differential of a function $$f$$ is a sum of terms: partial derivative $$\left(\frac{\partial f}{\partial x_i}\right)_{x_1,\ldots,x_n}$$ times the differential of a variable $$dx_i$$. For example, suppose you want to write a differential of $$S(T,H)$$. You have: $$$$dS = \left(\frac{\partial S}{\partial T}\right)_H dT + \left(\frac{\partial S}{\partial H}\right)_T dH$$$$

Now, you want to find $$\left(\frac{\partial S}{\partial T}\right)_H$$. This is simple - you need to decompose $$dS$$ into a linear combination of $$dT$$ and $$dH$$ and whatever stands next to $$dT$$ is your derivative.

On the RHS you have terms where $$S$$ is a function of $$T$$ and $$M$$, so a good starting point would be $$$$dS = \left(\frac{\partial S}{\partial T}\right)_M dT + \left(\frac{\partial S}{\partial M}\right)_T dM.$$$$ And now we only need to get rid of $$dM$$, so we express it as: $$$$dM = \left(\frac{\partial M}{\partial T}\right)_H dT + \left(\frac{\partial M}{\partial H}\right)_T dH.$$$$ I will let you solve the rest of the problem.

This is probably not as fast as a simple chain rule but gives you a better control of what's happening. I think it allows for easier backtracking of your steps and avoiding silly mistakes.

You can deduce the involved functions by looking at terms like $$\frac{\partial S}{\partial T}|_M$$. This tells you that there is a function $$S$$ of variables $$(T,M)$$. Likewise $$\frac{\partial M}{\partial T}|_H$$. This tells us that there is a function $$M(T,H)$$.

The equation is then obtained by using the the total differentials of the functions, $$S(T,H), \ S(T,M), \ M(T,H)$$ Expand $$dS(T,H)=dS(T,M(T,H))$$ and compare the coefficient in front of $$dT$$. This yields your equation. Alternatively you can take the partial derivative and use the chainrule $$\frac{\partial }{\partial T}S(T,H)=\frac{\partial }{\partial T}S(T, M(T,H))$$.