Chemical Potential: sign of external potential energy

Question

In statistical mechanics the total chemical potential $\mu _{\text{Total}}=\mu _{\text{Internal}}+\mu_{\text{External}}$. In the examples I have done so far $\mu_{\text{External}}$ is the external (effective) potential energy acting on the system. My question is why do you sometimes add this quantity, and sometimes subtract it?

For example, on the centrifuge problem, the centrifugal potential energy is $-mr^2\omega ^2/2$, so we have the equilibrium chemical potentials: $$\tau \ln \left[\frac{n(0)}{n_Q}\right]=\tau \ln \left[\frac{n(r)}{n_Q}\right]-\frac{mr^2\omega ^2}{2}$$

On the other-hand on the tree-sap problem, the gravitational potential is $Mgh$, and we have the equilibrium chemical potentials: $$\tau \ln \left[\frac{n(0)}{n_Q}\right]=\tau \ln \left[\frac{n(r)}{n_Q}\right]+Mgh$$

Is it something as simple as I am getting confused on directions, or what am I missing here?

Answer 1

I know this is a bit of an old question, but since I've just finished the same problem with the centrifuge, so I can give you some insight into why we use a negative side for $\mu_{ext}$.

For the rotating rod, the centripetal force acting on the particles is $F(r) = \frac{mv^2}{r}$. Furthermore, since the rotation of the rod is constant, $\omega = v/r$, so $F(r) = m\omega^2 r^2$. Lastly, we need to find the associated potential with that force. As such, we get:

$$\mu_{ext} = - \int F(r) dr = - \int m\omega^2 r^2 = -\frac{1}{2} m\omega^2 r^2.$$

Note here that I set the constant in the integration to zero for my reference frame (the centre of the rod). The moral of the the story is that the chemical potential here is negative due to the way we set up the problem and integrated. The reason the potential energy associated with gravity doesn't include a negative sign has to do with where the reference is, and how we perform the integral. So we aren't subtracting $\mu_{ext}$ here. It's just that it's a negative quantity.