Four-Velocity and Schwartzchild Metric

[DRose87]

Homework Statement

What is the Schwartzchild metric. Calculus the 4-velocity of a stationary observer in this spacetime (u). Show that u² = c².

Homework Equations

Schwartzchild Metric
$$d{s^2} = {c^2}\left( {1 - \frac{{2\mu }}{r}} \right)d{t^2} - {\left( {1 - \frac{{2\mu }}{r}} \right)^{ - 1}}d{r^2} - {r^2}d{\theta ^2} - {r^2}{\sin ^2}\theta d{\phi ^2}$$

The Attempt at a Solution

I already turned this in... I just want to make sure I understand this properly. I'm pretty sure that $\mathbf{u}\cdot\mathbf{u}$ is always equal to c^2 in the rest frame of a particle... So I think the point of the problem is to figure out the components of the four-velocity based on this fact. So here is my attempt at a solution... how ever I actually arrived at the solution in going from the bottom to the top.

So here's how I attempted to solve it:
From the Schwartzchild metric:
$$g_{00} = c^2\left(1 - \frac{2\mu }{r}\right)$$

The four-velocity u of a stationary observer is given by:
$$\begin{gathered}<br /> \left[ {{u^\mu }} \right] = \left( {{{\left( {1 - \frac{{2\mu }}{r}} \right)}^{ - 1/2}},0,0,0} \right) \hfill \\<br /> {c^2}\left( {1 - \frac{{2\mu }}{r}} \right){\left[ {{{\left( {1 - \frac{{2\mu }}{r}} \right)}^{ - 1/2}}} \right]^2} = {c^2} \hfill \\<br /> \hfill \\<br /> \end{gathered}$$

It follows that:
$${\mathbf{u}} \cdot {\mathbf{u}} = {g_{ab}}{u^a}{u^b} = {g_{00}}{\left( {{u^0}} \right)^2} = {c^2}\left( {1 - \frac{{2\mu }}{r}} \right){\left[ {{{\left( {1 - \frac{{2\mu }}{r}} \right)}^{ - 1/2}}} \right]^2} = {c^2}$$

 

[Orodruin]

Looks fine to me.