So I have the metric as ##ds^{2}=-(1-\frac{2m}{r})dt^{2}+(1-\frac{2m}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}##*(adsbygoogle = window.adsbygoogle || []).push({});

I have transformed to coordinate system ##u,r,\phi, \theta ##, where ##u=t-r*##(2),

where ##r*=r+2m In(\frac{r}{2m}-1)##

and to the coordinate system ##v,r,\phi, \theta ##,

where ##v=t+r*##,(1)

From (1) and (2) I see that ##dt=dv-\frac{dr}{(1-\frac{2m}{r})}## and ##dt=du+\frac{dr}{(1-\frac{2m}{r})}##

(On a side note, what is the proper name of these types of derivative expressions?)

Substituting these into * in turn it is easy enough to get the metrics:

(which I believe are correct?).

Question:

I am now want to get the metric using both \(v\) and \(u\) in favour of \(r\) and \(t\).

To do this I make use of:

##\frac{1}{2}(v-u)=r+2M In(\frac{r}{2M}-1) ##

therefore ##\frac{1}{2}(dv-du)(1-\frac{2m}{r})=dr##

and I sub this into either (1) or (2),

say (1) , I then get:

##ds^{2}= - (1-\frac{2M}{r}) dudv+r^{2}d\Omega^{2}##

And the first term is a minus sign out.

(in accord to source sean m carroll lecture notes on general relativity eq.7.73.)

I have no idea why I am a sign out,

Thanks,your assistance is greatly appreciated !

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# Schwarzschild extension coordinate transformation algebra

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