General Relativity Calculations

[latentcorpse]

So I am trying to verify a couple of things on p59 of this document
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf

How do we show eqn (3.16) implies (3.17)?

I keep getting bogged down on the maths. Could someone outline the method?

 

[kreil]

It looks like they just integrated 3.16 and defined some new quantities(3.18) to get the compact form of 3.17

 

[latentcorpse]

It looks like they just integrated 3.16 and defined some new quantities(3.18) to get the compact form of 3.17

Yeah but actually doing that integration is not going well. I'm trying to do it with partial fractions. Do you reckon that's the best approach?

 

[kreil]

At a glance that does appear to be the best approach. Equation 3.8 tells you that

$$r^2-2Mr+Q^2= (r-r_+)(r-r_-)=r^2(1-2M/r+Q^2/r^2)$$

which means we can rewrite the integrand as

$$\frac{1}{1-\frac{2M}{r}+\frac{Q^2}{r^2}}=\frac{r^2}{(r-r_+)(r-r_-)}$$

 

[dextercioby]

Well, you can do the integration in 3.16 with the variables M and Q and then reverse 3.9 to obtain the equation 3.17 with the definitions of 3.18.

 

[latentcorpse]

Well, you can do the integration in 3.16 with the variables M and Q and then reverse 3.9 to obtain the equation 3.17 with the definitions of 3.18.

Ok. I'll have a go at that. I have a couple of other physical questions though:

(i) On page 67, just before section 3.3, he asks
whether or not it is science fiction that a criminal could possibly escape justice by traveling from universe I to universe I' on a timelike path (see diagram on p66).
I'm sure he explained why this is in fact science fiction and unphysical but I can't remember and cannot figure it out for myself. I mean, everything looks OK since he is on a timelike path. What's the problem with this? Is it because we have to cross a Cauchy horizon?

(ii) Assuming it were possible to to cross the Cauchy horizon, we are now in the $r < r_{-}$ region where there is a curvature singularity. Surely now that we are in a black hole region, there is no way to cross the horizon and get into region I'?

Thanks.

Edit: I had a go at getting the integral:

$\frac{dr^*}{dr}=\frac{r^2}{(r-r_+)(r-r_-)} = \frac{A}{r-r_+} + \frac{B}{r-r_-}$
Then, by heavyside cover up method, we get
$A=\frac{r_+^2}{r_+-r_-} \quad B = \frac{r_-^2}{r_--r_+}$
So
$\frac{dr^*}{dr} = \frac{r_+^2}{r_+-r_-} \frac{1}{r-r_+} + \frac{r_-^2}{r_--r_+} \frac{1}{r-r_-}$
$\Rightarrow r^*(r) = \frac{r_+^2}{r_+-r_-} \ln{|r-r_+|} + \frac{r_-^2}{r_--r_+} \ln{|r-r_-|}$

But it's still a little bit off...

 

[fzero]

Edit: I had a go at getting the integral:

$\frac{dr^*}{dr}=\frac{r^2}{(r-r_+)(r-r_-)} = \frac{A}{r-r_+} + \frac{B}{r-r_-}$
Then, by heavyside cover up method, we get
$A=\frac{r_+^2}{r_+-r_-} \quad B = \frac{r_-^2}{r_--r_+}$

This should be

$A=\frac{r^2}{r_+-r_-} \quad B = \frac{r^2}{r_--r_+}$

He also shifts the integration constant so that the arguments of the logarithms are dimensionless.

As for the criminal, section 3.3 explains the physics.

 

[latentcorpse]

This should be

$A=\frac{r^2}{r_+-r_-} \quad B = \frac{r^2}{r_--r_+}$

He also shifts the integration constant so that the arguments of the logarithms are dimensionless.

As for the criminal, section 3.3 explains the physics.

How did you get that, we had:

$\frac{dr^*}{dr}=\frac{r^2}{(r-r_+)(r-r_-)} = \frac{A}{r-r_+} + \frac{B}{r-r_-}$
if we multiply through by $r-r_+$ we get:
$\frac{r^2}{(r-r_-)}=A + \frac{B(r-r_+)}{r-r_-}$
Then to get rid of the B term we set $r=r_+$ giving
$\frac{r_+^2}{r_+-r_-}=A$

 

[latentcorpse]

How did you get that, we had:

$\frac{dr^*}{dr}=\frac{r^2}{(r-r_+)(r-r_-)} = \frac{A}{r-r_+} + \frac{B}{r-r_-}$
if we multiply through by $r-r_+$ we get:
$\frac{r^2}{(r-r_-)}=A + \frac{B(r-r_+)}{r-r_-}$
Then to get rid of the B term we set $r=r_+$ giving
$\frac{r_+^2}{r_+-r_-}=A$

And as for the criminal explanation in section 3.3, I assume you're referring to the first paragraph of 3.3 (as the rest doesn't seem that relevant) but I don't see how this explains it?

 

[fzero]

How did you get that, we had:

$\frac{dr^*}{dr}=\frac{r^2}{(r-r_+)(r-r_-)} = \frac{A}{r-r_+} + \frac{B}{r-r_-}$
if we multiply through by $r-r_+$ we get:
$\frac{r^2}{(r-r_-)}=A + \frac{B(r-r_+)}{r-r_-}$
Then to get rid of the B term we set $r=r_+$ giving
$\frac{r_+^2}{r_+-r_-}=A$

That's true when $r=r_+$, but we need an expression that's valid for all $$r$$ in order to integrate. Multiply through by $$r-r_-$$ as well and match coefficients.

 

[latentcorpse]

That's true when $r=r_+$, but we need an expression that's valid for all $$r$$ in order to integrate. Multiply through by $$r-r_-$$ as well and match coefficients.

But that gives

$r^2=A(r-r_-)+B(r-r_+) \Rightarrow r^2=(A+B)r - A r_- - Br_+$
Doesn't that seem to imply that $A=B=0$?

 

[fzero]

But that gives

$r^2=A(r-r_-)+B(r-r_+) \Rightarrow r^2=(A+B)r - A r_- - Br_+$
Doesn't that seem to imply that $A=B=0$?

No, because $$A,B$$ can depend on $$r$$. I gave you the answer already, at least try to verify that it works.

 

[latentcorpse]

No, because $$A,B$$ can depend on $$r$$. I gave you the answer already, at least try to verify that it works.

Hi. I've just reviewed partial fractions and still don't see how you managed to keep that as an r and not an r_+?

Am I right to set $\frac{r^2}{(r-r_-)(r-r_+)}$?
If so, I don't see how we can avoid having to set $r=r_\pm$ when we solve for either A or B?

 

[fzero]

Hi. I've just reviewed partial fractions and still don't see how you managed to keep that as an r and not an r_+?

Am I right to set $\frac{r^2}{(r-r_-)(r-r_+)}$?
If so, I don't see how we can avoid having to set $r=r_\pm$ when we solve for either A or B?

It's much less confusing to expand

$\frac{1}{(r-r_-)(r-r_+)}$

and multiply by the $$r^2$$ afterwards.

 

[latentcorpse]

It's much less confusing to expand

$\frac{1}{(r-r_-)(r-r_+)}$

and multiply by the $$r^2$$ afterwards.

Ok. So (finally!) I got to

$\frac{dr^*}{dr} = \frac{1}{r_--r_+} \int \frac{r^2}{r-r_-}dr + \frac{1}{r_+-r_-} \int \frac{r^2}{r-r_+} dr$

Now I'm sure there's something else I need to do before I integrate this so that I can get the r term on its own like in the answer. Any advice?

 

[fzero]

No, you just need to do the integrals. Townsend shifts the integration constant to make the arguments of the logarithms dimensionless, but this does not change the terms proportional to $$r$$.

 

[latentcorpse]

No, you just need to do the integrals. Townsend shifts the integration constant to make the arguments of the logarithms dimensionless, but this does not change the terms proportional to $$r$$.

So to do those integrals I need to first use algebraic long division, right?

For the + term I get

$\frac{r^2}{r-r_+}=r+r_++\frac{r_+^2}{r-r_+}$
This integrates to

$\int \frac{r^2}{r-r_+}=\frac{r^2}{2} + rr_+ + r_+^2 \ln{|r-r_+|}$
I checked this on Wolfram Alpha and they have the same.

So using the prefactor we would get:

$\frac{1}{r_+-r_-} \int \frac{r^2}{r-r_+} dr = \frac{r^2}{2(r_+-r_-)}+\frac{rr_+}{r_+-r_-}+\frac{r_+^2}{r_+-r_-} \ln{|r-r_+|}$

Now, one problem is that my ln term doesn't have a factor of $\frac{1}{r_+}$?

 

[fzero]

Now, one problem is that my ln term doesn't have a factor of $\frac{1}{r_+}$?

As I told you, that comes from shifting the integration constant.

 

[latentcorpse]

As I told you, that comes from shifting the integration constant.

As in $c \rightarrow c' - \ln{r_+}$?