In Yung Kuo Lim's book of exercises in thermodynamics and Stat. Physics I have found more than once the following notation for partial derivatives (ex. 1081 page 79):

$$ \left(\frac{\partial T}{\partial L}\right)_S = \frac{\partial(T,S)}{\partial(L,S)} $$

I don't think that this is only a matter of notation, because the authors use it as a way to solve the exercises, not just as a different way to write something.

What does the writing on the RHS of the equation stand for?

EDIT: The author uses both notations, that's why I wondered if the second one has something more to say than the first one. Here's another example of how he uses it:

$$ \left( \frac{\partial T}{\partial H} \right)_S = \frac{\partial(T,S)}{\partial(H,S)} = \frac{\partial(H,T)}{\partial (H,S)} \cdot \frac{\partial(H,M)}{\partial(H,T)} \cdot \frac{\partial(T,S)}{\partial(H,M)} $$

with T the temperature, S the entropy, H the magnetic field and M the magnetization. If it is only a matter of notation, how is it the second one more useful than the first one?

  • 1
    $\begingroup$ @PPeg: In most areas that use partial derivatives, the number of variables is the same as the number of dimensions, and $\delta z/\delta x$ is unambiguous. Not for thermodynamics. How would you write it if you didn't use this notation? If you just write $\delta T/\delta L$, how do I know which variable to hold constant? If $T, L, S$ and $P$ are four different variables in a 3-dimensional space, then $\left(\delta T/\delta L\right)_S \neq \left(\delta T/\delta L \right)_P$. $\endgroup$ – Peter Shor Aug 16 '16 at 18:56
  • $\begingroup$ @PeterShor That's fine, but doesn't the first notation say that already? Why the need for the second? $\endgroup$ – garyp Aug 16 '16 at 20:14
  • $\begingroup$ Question for the OP: does the author use both notations, or only one of them, and you are filling in the meaning. If he uses only one, I can understand. But if he uses both notations at the same time, I'm wondering if he's trying to make some distinction between the two. $\endgroup$ – garyp Aug 16 '16 at 20:17
  • $\begingroup$ @garyp: sometimes it's useful to actually have a notation for infinitesimals and not just for partial derivatives. Do you complain about unnecessary duplication of notation when people use both $y'$ and $dy/dt$? $\endgroup$ – Peter Shor Aug 16 '16 at 21:07
  • $\begingroup$ @garyp I edited my question. $\endgroup$ – PPeg Aug 17 '16 at 7:59

I think this is the answer: $\frac{\partial(y,x)}{\partial (z,w)}\equiv -\frac{\partial y}{\partial x}\Bigr |_{all~other~ variables~ constant}\Bigr/\frac{\partial z}{\partial w}\Bigr |_{all~other~ variables~ constant}$. I will explain below why minus sign is required. Whether or not my answer is right, a book with such crappy notation that it requires considerable effort to understand what the notation means is not worth spending your time over. I would recommend Thermodynamics by Callen. In particular read the chapter on Maxwell relations.

I make use of the identity: $\frac{\partial y}{\partial x}\Bigr |_z=-\frac{\frac{\partial y}{\partial z}\Bigr |_x}{\frac{\partial x}{\partial z}\Bigr |_y}$. This is where the minus sign comes in.

If my definition is correct then:

$\frac{\partial T}{\partial H}\Bigr |_{S,M}=-\frac{\frac{\partial T}{\partial S}\Bigr |_{H,M}}{\frac{\partial H}{\partial S}\Bigr |_{T,M}}=\frac{\partial (T,S)}{\partial (H,S)}$

Taking the expression on the extreme right:

$\frac{\partial (H,T)}{\partial (H,S)}\frac{\partial (H,M)}{\partial (H,T)}\frac{\partial (T,S)}{\partial (H,M)}=-\frac{\frac{\partial H}{\partial T}}{\frac{\partial H}{\partial S}} \frac{\frac{\partial H}{\partial M}}{\frac{\partial H}{\partial T}} \frac{\frac{\partial T}{\partial S}}{\frac{\partial H}{\partial M}}=-\frac{\frac{\partial T}{\partial S}}{\frac{\partial H}{\partial S}}=\frac{\partial T}{\partial H}\Bigr |_{S,M}$

where in the intermediate steps I have suppressed names of variables which are held constant during partial differentiation to avoid clumsiness of appearance.

  • $\begingroup$ This definition along with the identity you used seems to work. The book in which I have found this notation is this, and it is a book of solved exercises. $\endgroup$ – PPeg Aug 17 '16 at 12:30
  • $\begingroup$ I'm sitting without access to a pencil, so I'm following along in my head. Looks good, and explains the value of the notation, while reminding me why I did poorly in physical chemistry (5.61, for those who recognize that notation). $\endgroup$ – garyp Aug 17 '16 at 12:30

I have found the answer to my own question in Callen, Thermodynamics and Thermo-statistics:

\begin{equation} \frac{\partial(u,v,\ldots,w)}{\partial(x,y,\ldots,z)} \equiv \det \left(\matrix{\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} &\ldots &\frac{\partial u}{\partial z}\\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} &\ldots &\frac{\partial v}{\partial z}\\ \vdots &\vdots &\ddots &\vdots\\ \frac{\partial w}{\partial x} & \frac{\partial w}{\partial y} &\ldots &\frac{\partial w}{\partial z}\\}\right), \end{equation}

which is just a Jacobian matrix. The following identities hold:

\begin{align} \frac{\partial(u,v,\ldots,w)}{\partial(x,y,\ldots,z)} &= -~\frac{\partial(v,u,\ldots,w)}{\partial(x,y,\ldots,z)}\\ \frac{\partial(u,v,\ldots,w)}{\partial(x,y,\ldots,z)} &=\frac{\partial(u,v,\ldots,w)}{\partial(r,s,\ldots,t)} \frac{\partial(r,s,\ldots,t)}{\partial(x,y,\ldots,z)}\\ \frac{\partial(u,v,\ldots,w)}{\partial(x,y,\ldots,z)} &=1\left/\frac{\partial(x,y,\ldots,z)}{\partial(u,v,\ldots,w)}\right.. \end{align}

The link with thermodynamic relations comes from the identity:

\begin{equation} \left(\frac{\partial u}{\partial x}\right)_{y,\ldots,z} = \frac{\partial(u,y,\ldots,z)}{\partial(x,y,\ldots,z)}, \end{equation}

hence the relation of the example in the question.


The notation stands for the following: $$ \partial \left( \left[\text{variable to differentiate} \right], \left[\text{variable to hold constant} \right] \right) $$

As @PeterShor pointed out, a multi-variate system requires one to worry about which variables to hold constant during partial differentiation.

You could equivalently write this as: $$ \left( \frac{ \partial \left[\text{variable to differentiate} \right] }{ \partial \left[ \text{variable of differentiation} \right] } \right)_{\left[\text{variable to hold constant} \right]} $$ as there are multiple forms of notation.

Note that partial derivatives require you to specify with respect to which variables to hold constant but total derivatives differentiate every variable of which the differentiated variable is a function.

Side Note: There is a specific name for the variable in the denomenator, which I am not recalling off hand. Perhaps a more mathematically inclined individual could help?

  • $\begingroup$ I edited the question to clarify what I would like to understand. $\endgroup$ – PPeg Aug 17 '16 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.