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

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?

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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).

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Thanks for the answer. In my books including Carolls notes, the four vector for the static killing vector is given by $K_{\mu}=(-f,0,0,0)$. So for e.g. for the Schwarchild metric it is $K_{\mu}=(-(1-\frac{2GM}{r}),0,0,0)$ Why is this? How do I get the f factor? –  ramanujan_dirac Jul 1 '13 at 21:15
In the same section Wald says $h^{-\frac{1}{2}}(dr)_a=h^{\frac{1}{2}}(\frac{\partial}{\partial r})_a$. How do I calculate this? –  ramanujan_dirac Jul 1 '13 at 21:28
To find the functions $f$ and $h$, you have to solve the field equations. The solution will depend on the equation of state, and it may not be possible to find it in closed form. –  Ben Crowell Jul 1 '13 at 21:39
I understand. My question is why is $(\frac{\partial}{\partial r})_a=h (dr)_a$? Sorry I have put the constants the other way round in the above comment. I don't want to calculate f, and h. –  ramanujan_dirac Jul 1 '13 at 21:44
Are those both lower indices? It would make more sense to me if one was upper and one was lower. Then you'd be using the metric to raise or lower an index. –  Ben Crowell Jul 1 '13 at 22:50
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