Schwarzchild metric - rescaled coordinates

radiogaga35
Messages
34
Reaction score
0
Schwarzschild metric - rescaled coordinates

Hi,

I've been working through a problem (no. 14 in ch. 9) of Alan Lightman's book of GR problems. I can't understand one of the results that are stated without proof. Basically it amounts to a rescaling of coordinates.

I know that to first order in {\textstyle{M \over r}} \ll 1 (weak gravitational field), the standard Schwarzschild metric can be written

d{s^2} = - (1 - {\textstyle{{2M} \over r}})d{t^2} + (1 + {\textstyle{{2M} \over r}})d{r^2} + {r^2}d{\Omega ^2}

in units where G = c = 1, and where d{\Omega ^2} = d{\theta ^2} + {\sin ^2}\theta d{\phi ^2}.

Lightman states that, in an "appropriate" coordinate system, the Schwarzschild metric (again to lowest order in {\textstyle{M \over r}} \ll 1) can be written

d{s^2} = - (1 - {\textstyle{{2M} \over r}})d{t^2} + (1 + {\textstyle{{2M} \over r}})(d{x^2} + d{y^2} + d{z^2}),

where {r^2} = {x^2} + {y^2} + {z^2}.

I've been pulling my hair out trying to derive Lightman's form from the standard form I gave above. Apparently it's just a simple rescaling of the radial coordinate but I've had no luck. Any help would be appreciated!

Thanks in advance.
 
Last edited:
Physics news on Phys.org
Indeed, for this problem it's convenient to introduce a re-scaling ##r \rightarrow \tilde{r}## which satisfies ##r = \tilde{r} (1+M/2\tilde{r})^2##, which can also be re-arranged for ##1-2M/r = (1-M/2\tilde{r})^2/(1+M/2\tilde{r})^2##. Take the derivation,\begin{align*}
dr = (1+M/2\tilde{r})(1-M/2\tilde{r}) d\tilde{r}
\end{align*}Consider the classic Schwarzschild metric\begin{align*}
g &= -(1-2M/r) dt^2 + (1-2M/r)^{-1} dr^2 + r^2 d\Omega^2 \\ \\
&= - \dfrac{(1-M/2\tilde{r})^2}{(1+M/2\tilde{r})^2} dt^2 + \dfrac{(1+M/2\tilde{r})^2}{(1-M/2\tilde{r})^2} dr^2 + r^2 d\Omega^2 \\ \\
&= - \dfrac{(1-M/2\tilde{r})^2}{(1+M/2\tilde{r})^2} dt^2 + (1+M/2\tilde{r})^4 d\tilde{r}^2 + \tilde{r}^2 (1+M/2\tilde{r})^4 d\Omega^2 \\ \\
&= - \dfrac{(1-M/2\tilde{r})^2}{(1+M/2\tilde{r})^2} dt^2 + (1+M/2\tilde{r})^4 \left\{ d\tilde{r}^2 + \tilde{r}^2 d\Omega^2 \right\}
\end{align*}To first order in ##M/\tilde{r}##, one has simply\begin{align*}
g = (1-2M/\tilde{r}) dt^2 + (1+2M/\tilde{r})\{ d\tilde{r}^2 + \tilde{r}^2 d\Omega^2 \}
\end{align*}Defining new Cartesian-ish coordinates ##(x,y,z) = (\tilde{r} \sin{\theta} \cos{\phi}, \tilde{r} \sin{\theta} \sin{\phi}, \tilde{r}\cos{\theta})## puts the metric into this familiar form\begin{align*}
g = (1-2M/\tilde{r}) dt^2 + (1+2M/\tilde{r})\{ dx^2 + dy^2 + dz^2 \}
\end{align*}
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top