3
$\begingroup$

Blau, in his GR book, says that a stationary and spherically symmetric metric is automatically static. He says this easily follows from the fact that for a stationary metric, and in spherical symmetry, in coordinates $(t,r,\theta,\phi)$ suitable for expressing both these facts, the only allowed off-diagonal $g_{tk}$-term of the metric is $C(r)=g_{tr}(r)$, so that the $(t,r)$-part of the metric takes the form $$ds^2=-A(r)\,dt^2+B(r)\,dr^2+2C(r)\,dtdr.$$

He adds then $C(r)$ can be eliminated by a coordinate transformation $T(t,r)=t+\psi(r)$, and $\partial_t=\partial_T$ is thus orthogonal to the surfaces of constant $T$.

My question how can one know that $\partial_t=\partial_T$ is thus orthogonal to the surfaces of constant $T$?

$\endgroup$
2
  • $\begingroup$ Are you supposed to pick a $\Psi (r)$ to make it work or did you forget to tell us what $\Psi (r)$ is? $\endgroup$ – Timaeus Jun 9 '15 at 22:45
  • 2
    $\begingroup$ Beyond: I can't find any book by Blau online. There are lecture notes though. I strongly encourage you to precisely cite and link any mentioned work, so other users can easily find it. $\endgroup$ – magma Jun 10 '15 at 9:51
0
$\begingroup$

That's just a very basic concept of how you choose (curved) coordinates.

Maybe think first of the example of an Euclidean coordinate system. There your coordinate basis vector $\partial_x$ is also orthogonal to surfaces of constant $x$.

The same holds also true for curved coordinates.

If you like you can also parametrize your surface $T=const.$ as a function of $S_T=S_T(r,\theta,\phi)$, build the tangential vectors $\partial_r S_T, \partial_\theta S_T, \partial_\phi S_T$ and check that these are orthogonal to $\partial_T$ (use that the metric is diagonal).

Another easy exercise is just to build the normal vector to the surface: Characterize the surface as a scalar function of the coordinates $f(T,r,\theta,\phi)$ and the requirement $ f=0$. Calculate the normal vector which is defined by: $$ n^\nu=g^{\nu\mu}\partial_\mu f$$

(Of course both 'calculations' are equivalent.)

$\endgroup$
0
$\begingroup$

As I said in my comment above, I could not find any book by Blau. In his lecture notes at page 481 he makes the transformation you mention. In detail he starts with the metric in $(t,r)$ in eq 23.1 which has diagonal terms, makes the transformation 23.3 which introduces $T$, then he finds a suitable $\psi$, and finally he rewites the metric in (T and r) without the diagonal terms , but...- without saying this explicitly - instead of using the new $T$ he renames it again $t$ in eq 23.5. It is just a renaming operation using an old variable, nothing fancy. Not nice, but unfortunately common. These are notes for advanced students who are supposed to see this in a blink. He does this again a few lines later when he introduces $R(r)$ only to rename it again $r$ in the next equation.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.