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**Here is the problem:**(Question 5-2 in Binney&Tremaine Galactic Dynamics)

Consider a homogeneous self-gravitating fluid of uniform density [tex]\rho_0[/tex] contained within a rotating cylinder of radius [tex]R_0[/tex]. The cylinder and the fluid rotate at angular speed [tex]\Omega[/tex] about the axis of the cylinder, which we take to be the [tex]z[/tex]-axis, so [tex]{\bf\Omega} = \Omega\hat{\bf z}[/tex].

(a) Show that the gravitational force per unit mass at distance [tex]r[/tex] from the axis is

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

directed radially toward the axis and perpendicular to it.

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

[tex]\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}[/tex]

Find the condition on [tex]\Omega[/tex] 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 [tex] \nabla P [/tex] which leaves you with

[tex]\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}[/tex]

I know you are also supposed to set [tex]v = 0 [/tex] (I know this because I have an answer key that says set [tex]v = 0 [/tex])

**but why?**

**My question:**

What part of this question implies that [tex]v = 0 [/tex]? 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.