For a fluid in equilibrium, the level surfaces of $\rho$ and $\Psi$ must coincide?

Question

I am reading through An Introduction to Fluid Dynamics by G.K. Batchelor.

Question background:

Batchelor states that if a fluid is in equilibrium, then everywhere in the fluid we have $$ \rho\mathbf{F} = \nabla p$$ where $\rho = \rho(\mathbf{x},t)$, $\mathbf{F}=\mathbf{F}(\mathbf{x},t)$, and $p=p(\mathbf{x},t)$. In the case of a conservative field $\mathbf{F}=-\nabla\Psi$, the condition for equilibrium is $$-\rho\nabla\Psi=\nabla p$$ and taking the curl of both sides yields $$\nabla \rho \times \nabla \Psi = 0 $$

Question:

Batchelor then concludes that the level-surfaces of $\rho$ and $\Psi$ must coincide, and that these same level surfaces are the level surfaces of $p$ (everything makes sense up to this point) and we can write $$ dp/d\Psi = -\rho(\Psi)$$

What does it mean to take the derivative of $p$ with respect to $\Psi$? The pressure is a function of space and time, so how can it be written in terms of $\Psi$ for taking a derivative? Can someone help me understand the mathematics and, if possible, the physical meaning behind this relationship?

Answer 1

To make this clear, one should distinguish between the pressure field $p: (\mathbf{x},t) \mapsto p(\mathbf{x},t)$ and a function $\hat{p}: \mathbb{R} \to \mathbb{R}$ such that

$$\hat{p}(\Psi(\mathbf{x},t)) = p(\mathbf{x},t)$$

Such a function is well-defined (single-valued) since level surfaces of $p$ and $\Psi$ (for fixed $t$) coincide. This follows because given $\mathbf{x} \neq\mathbf{x}'$ on the same level surface where $ p(\mathbf{x},t) = p(\mathbf{x}',t)$, we have $ \Psi(\mathbf{x},t) = \Psi(\mathbf{x}',t)$ and $\hat{p}(\Psi(\mathbf{x},t)) = \hat{p}(\Psi(\mathbf{x}',t)) = p(\mathbf{x},t)$. In other words, we cannot have $\Psi_1 = \Psi_2$ and $\hat{p}(\Psi_1) \neq \hat{p}(\Psi_2)$.

Also the level surfaces of $\rho$ coincide with those of $p$ and $\Psi$, so we define $\hat{\rho}$ the same way as $\hat{p}$. By the chain rule for multi-variate functions it follows that

$$\nabla p(\mathbf{x},t) = \hat{p}'(\Psi(\mathbf{x},t)) \nabla \Psi(\mathbf{x},t) = -\hat{\rho}(\Psi(\mathbf{x},t))\nabla \Psi(\mathbf{x},t) ,$$

and this implies wherever $\nabla \Psi \neq 0$,

$$\frac{d \hat{p}}{d \Psi} = \hat{p}'(\Psi) = - \hat{\rho}(\Psi)$$