So I was reading through Atkins "Physical chemistry" 10th edition and was wondering something. Suppose you have an athermal system of $k$ particles, let us define the Gibbs-Duhem equation for this system as:

$$\text{d}P=\sum_{i=1}^k n_i \text{d}\mu_i,$$

where all $n_i's$ are coupled through some relation $f(n_i,n_{i+1},n_{i+2},...)=0$. Suppose you are interested in the following quantity:

$$\left(\frac{\partial P}{\partial\mu_1}\right)_\text{T}$$

Sometimes you see this defined as "Divide both sides by $\text{d}\mu_1$" to obtain

$$\left(\frac{\partial P}{\partial\mu_i}\right)_\text{T}=\sum_{i=1}^k n_i \left(\frac{\partial \mu_i}{\partial \mu_1}\right)_\text{T}.$$

This doesn't really makes sense to me, as through the chain rule one would expect a combination of derivatives of $n_i$ with respect to $\mu_1$ and $\mu_i$ with respect to $\mu_1$. Could anyone please explain what is happening here.

  • $\begingroup$ I guess that Atkin's definition should be "Divide both sides by d$\mu_1$" to obtain $\left(\frac{\partial P}{\partial\mu_1}\right)_\text{T}=...$". Am I wrong? $\endgroup$ Commented Nov 30, 2021 at 9:52
  • $\begingroup$ Yes sorry, type-o. I've editted this in the question. $\endgroup$
    – M . M
    Commented Nov 30, 2021 at 10:22
  • $\begingroup$ It has to do with definitions of differential and derivative. See here: physics.stackexchange.com/a/565329/247642 $\endgroup$
    – Roger V.
    Commented Nov 30, 2021 at 14:14

1 Answer 1


I think your confusion comes from the fact that you might not be familiar with the concept of the total derivative. It basically is the chain rule already. Let me make an example: Imagine you have a function $f(x,y)$. Right now $x,y$ are just some coordinates. But you could consider a specific trajectory described by $x(t),y(t)$ where $t$ is time and both coordinates are some function of time. If you now want to compute the time derivative of $f(x(t),y(t))$, you use the chain rule: $$ \frac{d}{dt}f=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}$$ This can also be seen by taking the product of the row vector $\left(\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right)$ and the column vector $\left(\frac{dx}{dt},\frac{dy}{dt}\right)^T$. This is useful because it separates the object that depends on the chosen function f (which is th row vector) from the object that depends on the explicit trajectory (the column vector). In "differential geometry notation" this row vector gets written basically as $df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy$ where $dx,dy$ can be seen as something like the basis row vectors $(1,0),(0,1)$. Please don't take this as a mathematically rigorous explanation. If you are looking for one, read a book about differential geometry (or maybe just about conventional multidimensional calculus). My answer is more about providing the heuristics to understand what the notation means.

So your $dP$ can be understood as the corresponding row vector which you have to multiply with the column vector $\left(\frac{\partial\mu_1}{\partial\lambda},...,\frac{\partial\mu_k}{\partial\lambda}\right)^T$ (where there are now partial derivatives because the things depend on more than just one variable) if you want to compute $\frac{\partial P}{\partial \lambda}$. So actually we can compare the $dP$ given with the general formula to get $n_i=\left(\frac{\partial P}{\partial \mu_i}\right)_{\mu_j,j\ne i}$ (where I mean that the partial derivative is taken while every other $\mu$ is constant). So why do we even need to write down an expression here for $\left(\frac{\partial P}{\partial \mu_i}\right)_T$ if it is just $n_1$? Well, it isn't. Now $T$ is constant, not all other $\mu$. This means we have a different "trajectory" parametrized by the same parameter ($\mu_1$) like one could also have different trajectories $x'(t),y'(t)$ which are also parametrized by time.

If you plug what we just found for the $n_i$ in you can directly see that the chain rule has been applied correctly in $$ \left(\frac{\partial P}{\partial \mu_1}\right)_T=\sum_{i=1}^k n_i \left( \frac{\partial \mu_i}{\partial \mu_1}\right)_T=\sum_{i=1}^k \left(\frac{\partial P}{\partial \mu_i}\right)_{\mu_j,j\ne i} \left( \frac{\partial \mu_i}{\partial \mu_1}\right)_T$$

  • $\begingroup$ Great answer, it makes complete sense now. Especially through the fact that you separate the trajectory-dependent terms. Thank you. $\endgroup$
    – M . M
    Commented Nov 30, 2021 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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