# Hydrostatic equilibrium - does a partial derivative imply a dependence on some hidden variable?

while I was getting through a derivation of the hydrostatic equilibrium equation, I stumbled upon the following formulation:

$\ddot r = -\frac{Gm}{r^2}-\frac{1}{\rho}\frac{\partial P}{\partial r}$ , where $\ddot r= \frac{\partial^2r}{\partial t^2}$

The author then goes on and states:

When accelerations are negligible, the equation describes a state of hydrostatic equilibrium, with gravitational and pressure forces exactly in balance: $\frac{dP}{dr}=-\frac{\rho Gm}{r^2}$

I have three questions:

1. Why is $r$ partially differentiated? What possible other dependence, except for the temporal one, can this mass element position specifier have? I saw a different derivation of that equation and the total derivative was used. Is this a misuse of notation?

2. Why does $\frac{\partial P}{\partial r}$ transform into $\frac{dP}{dr}$ when we get rid of temporal dependencies? Does the author suggest some time-dependence of pressure?

3. Does this partial to total derivative transition, when some condition is fulfilled (or we say that some dependence no longer exists), signify the transition from the function of two variables to the function of one variable?

## 2 Answers

Well of course pressure can have an angular dependence, especially if you have a rotating sphere of fluid (e.g. any star).

When you remove acceleration that includes centripetal acceleration, which removes the angular dependence.

Pressure only depends on $r$ in cases of spherical symmetry.

As an aside, the equation of hydrostatic equilibrium in such circumstances is actually $$\frac{dP}{dr} = -\frac{Gm\rho}{r^2}.$$

If, in the transient case, P is a function of r and t, then the partial derivative of P with respect to r is taken to be constant t. Once the time dependence vanishes, P is a function only of r (and you don't need to hold anything else constant), so the partial derivative and the total derivative become identical.