Static Spherical symmetric solution of Einstein's equations with a perfect fluid

Question

I am reading Wald for the interior solutions of a static spherical metric. Assume it to be of the form $$ds^2 = -f(r)dt^2 + h(r)dr^2 + r^2 ( d{\theta^2} \sin^2{\theta}d{\phi^2})$$

Wald states: For a perfect fluid tensor $T_{ab}= \rho u_a u_b + P ( g_{ab}+ u_{ab})$

In order to be compatible with the static symmetry of space time, the four velocity of the fluid should point in the direction of the static killing vector $\xi^a$

i.e. $u^a=-(e_0)^a=-f^{\frac{1}{2}}(dt)^a$

EDIT: It also seems $(e_0)_a=f^{\frac{1}{2}}(dt)_a=f^{-\frac{1}{2}}(\frac{\partial}{\partial t})_a$. Please could someone tell, why this is so?

1. First, why is the the static killing vector $\frac{\partial}{\partial t}$ equal to $-f^{\frac{1}{2}} dt$?

2. Second, why is the velocity, along the killing time vector? What would happen if there is a component perpendicular to it? Does this mean, the fluid doesn't move through space?

Answer 1

1. First, why is the the static killing vector $\frac{\partial}{\partial t}$ equal to $-f^{\frac{1}{2}} dt$?

The vectors $\partial/\partial t$ and $-f^{\frac{1}{2}} dt$ are not equal to each other. They're parallel to each other, and the factor of $-f^{1/2}$ is just so that the four-velocity is properly normalized. If you plug $u$ into the metric, you have to get 1.

1. Second, why is the velocity, along the killing time vector? What would happen if there is a component perpendicular to it? Does this mean, the fluid doesn't move through space?

The Killing vector supplies a preferred frame of reference, one in which the observables (e.g., curvature scalars) stay constant. A perfect fluid is one for which there exists a frame such that the stress-energy tensor is diagonal; in this frame, there is no spatial flux of energy-momentum. So basically this means that in this frame, the fluid doesn't move through space (in the sense that there's no flux).