# Fluids in a rotating frame

Here is the problem: (Question 5-2 in Binney&Tremaine Galactic Dynamics)
Consider a homogeneous self-gravitating fluid of uniform density $$\rho_0$$ contained within a rotating cylinder of radius $$R_0$$. The cylinder and the fluid rotate at angular speed $$\Omega$$ about the axis of the cylinder, which we take to be the $$z$$-axis, so $${\bf\Omega} = \Omega\hat{\bf z}$$.

(a) Show that the gravitational force per unit mass at distance $$r$$ from the axis is

$$\begin{displaymath} -\nabla\phi = -2\pi G\rho_0 (x\hat{\bf x} + y\hat{\bf y}), \end{displaymath}$$

directed radially toward the axis and perpendicular to it.

(b) Euler's equation for the fluid in the rotating frame is

$$\begin{displaymath} \frac{\partial{\bf v}}{\partial t} + ({\bf v}\cdot\nabla) \bf v = - \frac{\nabla P} {\rho}- \nabla \phi - 2 \Omega \times{\bf v} + \Omega^2 (x\hat{\bf x} + y\hat{\bf y})\,. \end{displaymath}$$

Find the condition on $$\Omega$$ such that the fluid is in equilibrium with no pressure gradients (i.e. no Jeans swindle).

My problem:
So I understand part a but I am having difficulty with part b.

There are no pressure gradients so you set $$\nabla P$$ which leaves you with
$$\begin{displaymath} \frac{\partial{\bf v}}{\partial t} + ({\bf v}\cdot\nabla) \bf v = - \nabla \phi - 2 \Omega \times{\bf v} + \Omega^2 (x\hat{\bf x} + y\hat{\bf y})\,. \end{displaymath}$$

I know you are also supposed to set $$v = 0$$ (I know this because I have an answer key that says set $$v = 0$$) but why?

My question:
What part of this question implies that $$v = 0$$? Is it zero because it is in equilibrium? What does it mean for the fluid to be in equilibrium? I would like a little help with this concept please.