Normalization of 4-velocity in general relativity

[George Jones]

It seems that you are treating $R$ as a constant, but I still don't see how to use

$$\bar{t} = t \sqrt{1-R_s/R}$$

$$\bar{r} = \frac{r}{\sqrt{1-R_s/R}}$$

to go from

Here is the "standard" Schawzschild metric :

$$ds^2 = -\left( 1-R_s/r \right) dt^2 + \left( 1-R_s/r \right)^{-1} dr^2 + r^2 d\Omega^2$$

to

then I get the following metric :

$$ds^2 = -\frac{(1-R_s/\bar{r})}{(1-R_s/R)} d\bar{t}^2 + \frac{(1-R_s/R)}{(1-R_s/\bar{r})}d\bar{r}^2 + \bar{r}^2 (1-R_s/R) d\Omega^2$$

For example, using

$$dt = \frac{d\bar{t}}{\sqrt{1-R_s/R}}$$

$$r = \bar{r} \sqrt{1-R_s/R}$$

in the first term of the standard Schwarzschild metric gives

$$\left( 1-R_s/r \right) dt^2 = \frac{\left( 1 - \frac{R_s}{\bar{r} \sqrt{1-R_s/R}} \right)}{1-R_s/R} d\bar{t}^2 .$$

 

[Altabeh]

Here are my new variables :

$$\bar{t} = t \sqrt{1-R_s/R}$$

$$\bar{r} = \frac{r}{\sqrt{1-R_s/R}}$$

then I get the following metric :

$$ds^2 = -\frac{(1-R_s/\bar{r})}{(1-R_s/R)} d\bar{t}^2 + \frac{(1-R_s/R)}{(1-R_s/\bar{r})}d\bar{r}^2 + \bar{r}^2 (1-R_s/R) d\Omega^2$$

But, like George, I don't understand how this is obtained by the transformation proposed above. Nor do I get the meaning of $$R$$ in your equations. It is most likely that a factor $$(\sqrt{1-R_s/R}})$$ is missing in the denominator of $$R/\bar{r}$$ from the time component of the metric.

AB

 

[jmlaniel]

I have to agree with George and Altabeh... I did not see the effect of the new r in the metric factor. My solution is entirely wrong But thanks a lot for pointing it out!

I have tried for the last hour to find a way to fix this and I am unable to do so. I will have to conlude that kev metric is suspicious... Unless anybody can find a way to justify this metric, I will have to say that it is imposisble to get from the standard Schwarzschild metric with a change of variable.

I really tried to accept this metric, but I can figure it out.