Graduate Kruskal Coordinates in Schwartzchild metric

[mad mathematician]

I am reading the book by Birrell and Davies on QFT on curved spacetime.
On page 40 they write the following metric:

and on the following page they provide the Kruskal Coordiantes and the changed metric:

Now, I get something a little bit different than theirs.
According to my computations, if we change $2M/r$ to $(1-2M/r)$ and $e^{-r/2M}$ to $e^{-r^*/2m}$, then I get that by Kruskal transformation that (3.20) is equivalent to (3.18).
Am I right?

 

[haushofer]

If I understand you correctly: wouldn't that mean that your version of the Kruskal metric still has a coordinate singularity at r=2M?

So no, that doesn't seem right.

 

[JimWhoKnew]

Now, I get something a little bit different than theirs.

If you'll post your calculations, we'll be able to comment (and if you do so, please use LaTeX (which enables quoting)).

 

[mad mathematician]

OK, here are my calculations. Let me know where did I get it wrong.
So we have the following: $d\bar{u}=e^{-u/4M}du$ and $d\bar{v}=e^{v/4M}dv$.
Now, we should multiply $d\bar{u}d\bar{v}=e^{(v-u)/4M}dudv$.
Now, $du=dt-dr^* , dv=dt+dr^*$, so multiply and get:
$d\bar{u}d\bar{v}=e^{(v-u)/4M}(dt^2-(dr^*)^2)$, notice that: $v-u=2r^*$, and $dr^*=dr+dr/(r/2M-1)=dr/(1-2M/r)$, plug the last two equations and get: $d\bar{u}d\bar{v}=e^{r^*/2M}(dt^2-dr^2/(1-2M/r)^2)$.
So as we can see, I hope, we get back the Schwarzchild metric (eq. 3.18), if we multiply $d\bar{u}d\bar{v}$ by $e^{-r^*/2M}(1-2M/r)$.

Your comments are well appreciated, thanks!

Edit: changed what @JimWhoKnew rightly so remarked my mistake.

 

[JimWhoKnew]

Shouldn't it be
$dr^*=dr+dr/(r/2M-1)=dr/(1-2M/r)$
?

I didn't read beyond that. I'll do, if necessary, after your reply.

 

[mad mathematician]

Shouldn't it be
$dr^*=dr+dr/(r/2M-1)=dr/(1-2M/r)$
?

I didn't read beyond that. I'll do, if necessary, after your reply.

good catch, I'll change it. you are right.

 

[JimWhoKnew]

good catch, I'll change it. you are right.

I lost track.
Can you see that after the correction, you have$$\frac{2M}r e^{-r/2M}d\bar{u}d\bar{v}=\frac{2M}r e^{-r/2M}e^{r^*/2M}(dt^2-dr^2/(1-2M/r)^2)$$which correctly yields the corresponding terms in Schwarzschild metric?
(when $~r>2M~$)

 

[mad mathematician]

I lost track.
Can you see that after the correction, you have$$\frac{2M}r e^{-r/2M}d\bar{u}d\bar{v}=\frac{2M}r e^{-r/2M}e^{r^*/2M}(dt^2-dr^2/(1-2M/r)^2)$$which correctly yields the corresponding terms in Schwarzschild metric?
(when $~r>2M~$)

There should be a difference between $r$ and $r^*$, they aren't the same.

 

[JimWhoKnew]

There should be a difference between $r$ and $r^*$, they aren't the same.

Of course they are not. That's how you can get the factors right.

Hint: use $~e^{\ln{x}}=x~$ in the equation in #7.

 

[mad mathematician]

Thanks that cleared my perplexion.