In The Feynman Lectures on Physics, Vol. II, page 515, the author derived the equation describing a fluid in a hydrostatic situation:

$$-\nabla p -\rho \nabla \phi =0,$$

where $p$ is the pressure, $\phi$ is a potential from which the force counterbalancing the pressure force derive (for example, the gravitational potential) , and $\rho$ is the density of the fluid.

He then said, about this equation,

In general, it has no solution. If the density varies in space in an arbitrary way, there is no way for the forces to be in balance, and the fluid cannot be in static equilibrium. Convection currents will start up. We can see this from the equation since the pressure term is a pure gradient, whereas for variable $\rho$ the other term is not. Only when $\rho$ is a constant is the potential term a pure gradient.

Why can't there be a $\phi$ such that $-\rho \nabla \phi$ counterbalances $-\nabla p$ at each point in the fluid?

Why $\rho \nabla \phi$ cannot be a pure gradient? I don't really understand what he means by pure when referring to these gradients.


Suppose that $\rho = \rho(\mathbf{x})$ and that there exists some potential $\Phi$ such that $$ \rho(\mathbf{x}) \nabla \phi = \nabla \Phi.$$

It is well known that the curl of a gradient is zero:

$$\nabla \times \nabla \Phi = 0,$$ so let's calculate this curl explicitly: \begin{align} \nabla \times \nabla \Phi &= \nabla \times [\rho(\mathbf{x}) \nabla \phi] \\ &= \rho(\mathbf{x}) (\nabla \times \nabla \phi) + \nabla( \rho(\mathbf{x})) \times \nabla\phi \\ &= 0 + \nabla( \rho(\mathbf{x})) \times \nabla\phi. \end{align} For an arbitrary $\rho(\mathbf{x})$ we cannot grantee that is has no gradient. Further, if $\rho(\mathbf{x})$ has no gradient then it is (at least locally) a constant. Therefore, a general potential can only exist if $\rho(\mathbf{x})$ is constant, as Feynman states.

  • $\begingroup$ A potential can also exist if $\rho$ is a function of $p$. $\endgroup$
    – Hilbert
    Oct 23 '19 at 7:47

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