# Schwarzschild extension coordinate transformation algebra

1. Mar 18, 2015

### binbagsss

So I have the metric as $ds^{2}=-(1-\frac{2m}{r})dt^{2}+(1-\frac{2m}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}$*

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 !

2. Mar 19, 2015

### binbagsss

Bump.

3. Mar 19, 2015

### Staff: Mentor

I don't see any metrics here. Did you leave them out by mistake?

4. Mar 19, 2015

### binbagsss

Apologies!
metrics are:
$ds^{2}=-(1-\frac{2M}{r})dv^{2}+2dvdr + r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}$
$ds^{2}=-(1-\frac{2M}{r})du^{2}-2dudr + r^{2}(d\theta^{2}+sin^{2}\theta d\phi^{2}$

5. Mar 19, 2015

### Staff: Mentor

Ok, those look correct.

I don't think you are. I think you are just using an opposite sign convention from Carroll for the definition of $r$ in terms of $u$ and $v$. If you look at Carroll's equation 7.74, it has $u - v$ where you have $v - u$. That's why he has an opposite sign in 7.73 from yours.

(Carroll also has a factor of 1/2 in front of 7.73, whereas you do not. That's because he also scales $u$ and $v$ differently than you are. None of these differences affect the physics; they're just different conventions for the math.)

6. Mar 20, 2015

### binbagsss

I used a different notation than Carrol $u bar=v$.
I cancelld the $1/2$ using $dudv=dvdu$

7. Mar 20, 2015

### Staff: Mentor

Hm, yes, I see. So much for that theory.

I'm not sure what's going on. I get the minus sign in the $du dv$ term the same way you do. Physically, the minus sign makes sense: a line element with $du$ and $dv$ both the same sign should lie inside one of the light cones (future for $du$ and $dv$ both positive, past for $du$ and $dv$ both negative), and so should be timelike and have a negative squared length. Unless Carroll is using a different sign convention somewhere else that I haven't spotted, the only other thing I can think of is that his equation 7.73 has a typo.

8. Apr 20, 2015

### binbagsss

Looking at the next metric, equation 7.77, it has a negative sign. (And I belive this to be corect as it agrees with http://www.damtp.cam.ac.uk/user/hsr1000/black_holes_lectures_2014.pdf , page 33, eq 2.35). Looking at how u' and v' are defined as functions of u,v, I am getting a sign change when going from the metric in u,v to u'v'. (whilst the cambridge notes, equation 2.32, defines u' with a neg sign compared to Carroll, either definition gives arise to a neg sign) and so it appears that there should be a sign change and so equation 7.73 in carroll should have a positive.

Hopefully with this information someone may find it easier to spot why I am getting a negative - been llooking for a long time.

Thanks

(Cambridge ntes and Carroll have defined their u and v, u' and v' the other way around).

Last edited: Apr 20, 2015
9. Apr 22, 2015

### binbagsss

Sorry just to add both sources are using the same metric signature - (-,+,+,+)).