Schwarzschild Black Hole

  • #1
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Show that in region 2 of the Kruskal manifold, one may regard r as a time coordinate and introduce a new set of spatial coordinates x such that

[itex]ds^2=-\frac{dr^2}{(\frac{2M}{r}-1)} + ( \frac{2M}{r}-1) dx^2 + r^2 d \Omega^2[/itex]

Hence show that every timelike curve in region 2 intersects the singularity at r=0 within a proper time no grater than [itex]\pi M[/itex]

So I did the first bit by observing that using the standard Schwarzschild metric, with r>2M, [itex]g_{tt}u^tu^t<0,g_{rr}u^ru^r>0[/itex] which tells us that t behaves as a timelike coordinate here and r as a spacelike coordinate as expected. However, for r<2M, [itex]g_{tt}u^tu^t>0,g_{rr}u^ru^r<0[/itex] and so t behaves in a spacelike manner and r in a timelike manner. Hence we can treat r as a time coordinate.

I have some doubts about this though. Firstly, I have "used" the Schwarzschild metric in the region r<2M where it doesn't apply. Surely, I should want to use ingoing Eddington Finkelstein coordinates or something - these have the advantage of being applicable for r<2M but if I use them then I get [itex]g_{rr}u^ru^r=0[/itex] since there is no rr component!!!
What have I done wrong here? If anything?

And does anyone have any advice on how to do the next two bits?

Thanks very much!
 

Answers and Replies

  • #2
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Not really sure about all the "region 2" stuff but I know how to show a timelike particle will reach r=0 within a proper time M*pi (take a look at q9 of this example sheet http://www.damtp.cam.ac.uk/user/examples/D22c.pdf, I have a solution if you need).
 
  • #3
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Not really sure about all the "region 2" stuff but I know how to show a timelike particle will reach r=0 within a proper time M*pi (take a look at q9 of this example sheet http://www.damtp.cam.ac.uk/user/examples/D22c.pdf, I have a solution if you need).

Hey. Thanks for this. So far I have:

for radial case: [itex]1=F \dot{t}^2 - \frac{\dot{r}^2}{F}[/itex]

Now [itex]F=1-\frac{r_s}{r} \leq 0 [/itex] since [itex]r \leq r_s[/itex] as we're in or on horizon of BH.
And since [itex]\dot{t}^2 \geq 0[/itex] obviously, this gives [itex]F \dot{t}^2 \leq 0 \Rightarrow -\frac{\dot{r}^2}{F} \geq 1 \Rightarrow \dot{r}^2 \geq - F[/itex]

Now consider non radial motion. Then [itex]\dot{\theta}^2,\dot{\phi}^2 \geq 0[/itex] and obviously [itex]r^2 , \sin^2{\theta} \geq 0[/itex]
so we have [itex]-r^2 ( \dot{\theta}^2 + \sin^2{\theta} \dot{\phi}^2 ) \leq 0[/itex]
which means [itex]F \dot{t}^2 - r^2 ( \dot{\theta}^2 + \sin^2{\theta} \dot{\phi}^2 ) \geq 0[/itex]

and so [itex]-\frac{r^2}{F} \geq 1 \Rightarrow \dot{r}^2 \geq -F[/itex] again.

However I don't know how to treat an accelerating observer? What do I do for that?
And are there any other cases I should consider before I conclude that [itex]\dot{r}^2 \geq -F[/itex] always?

And then I'm fairly sure I'm messing up the last part.....

if [itex]\dot{r}^2 \geq -F[/itex]
then [itex]\dot{r}^2 \geq \frac{r_s}{r}-1[/itex]
then I did some other stuff that was definitely wrong (as I ended up with completely the wrong answer!)
What's the next step?

Cheers!
 
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  • #4
148
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Hey. Thanks for this. So far I have:

for radial case: [itex]1=F \dot{t}^2 - \frac{\dot{r}^2}{F}[/itex]

Now [itex]F=1-\frac{r_s}{r} \leq 0 [/itex] since [itex]r \leq r_s[/itex] as we're in or on horizon of BH.
And since [itex]\dot{t}^2 \geq 0[/itex] obviously, this gives [itex]F \dot{t}^2 \leq 0 \Rightarrow -\frac{\dot{r}^2}{F} \geq 1 \Rightarrow \dot{r}^2 \geq - F[/itex]

Now consider non radial motion. Then [itex]\dot{\theta}^2,\dot{\phi}^2 \geq 0[/itex] and obviously [itex]r^2 , \sin^2{\theta} \geq 0[/itex]
so we have [itex]-r^2 ( \dot{\theta}^2 + \sin^2{\theta} \dot{\phi}^2 ) \leq 0[/itex]
which means [itex]F \dot{t}^2 - r^2 ( \dot{\theta}^2 + \sin^2{\theta} \dot{\phi}^2 ) \geq 0[/itex]

and so [itex]-\frac{r^2}{F} \geq 1 \Rightarrow \dot{r}^2 \geq -F[/itex] again.

However I don't know how to treat an accelerating observer? What do I do for that?
And are there any other cases I should consider before I conclude that [itex]\dot{r}^2 \geq -F[/itex] always?

And then I'm fairly sure I'm messing up the last part.....

if [itex]\dot{r}^2 \geq -F[/itex]
then \dot{r}^2 \geq \frac{r_s}{r}-1[/itex]
then I did some other stuff that was definitely wrong (as I ended up with completely the wrong answer!)
What's the next step?

Cheers!

Basically you're almost done, you've showed [tex]\dot{r}^2\geq -F[/tex], you don't need to consider accelerating, non-radiating etc because the question says "show that whatever happens...this is true" and you've got the inequality without assuming anything with what happens inside the horizon, so all that's left to do is solve the equation.

So you have [tex]\dot{r}^2 \geq \frac{r_s}{r}-1[/tex], square root both sides and integrate, eventually you'll get r=r(tau) so then you just set r=0 and solve for tau.
 
  • #5
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Basically you're almost done, you've showed [tex]\dot{r}^2\geq -F[/tex], you don't need to consider accelerating, non-radiating etc because the question says "show that whatever happens...this is true" and you've got the inequality without assuming anything with what happens inside the horizon, so all that's left to do is solve the equation.

So you have [tex]\dot{r}^2 \geq \frac{r_s}{r}-1[/tex], square root both sides and integrate, eventually you'll get r=r(tau) so then you just set r=0 and solve for tau.

But we can't do

[itex]\int dr = \int \sqrt{ \frac{r_s}{r}-1} d \tau[/itex] since we haven't moved all the r's to the one side. This is what I was having problems with last time!

Also, can you clarify, does the geodesic equation work for both accelerating and non accelerating particles?
 
  • #6
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But we can't do

[itex]\int dr = \int \sqrt{ \frac{r_s}{r}-1} d \tau[/itex] since we haven't moved all the r's to the one side. This is what I was having problems with last time!

Also, can you clarify, does the geodesic equation work for both accelerating and non accelerating particles?

You're gonna kick yourself, you have to divide both sides by [tex]\sqrt{ \frac{r_s}{r}-1} [/tex] and integrate with respect to r! It's not the easiest integral to do but it comes out in about 4 or 5 lines with the correct trig substitution.

I think so yes, the Sch. metric is the unique spherically symmetric vacuum metric and it holds regardless of what the particle is doing, accelerating or non-accelerating it does not matter.
 
  • #7
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You're gonna kick yourself, you have to divide both sides by [tex]\sqrt{ \frac{r_s}{r}-1} [/tex] and integrate with respect to r! It's not the easiest integral to do but it comes out in about 4 or 5 lines with the correct trig substitution.

I think so yes, the Sch. metric is the unique spherically symmetric vacuum metric and it holds regardless of what the particle is doing, accelerating or non-accelerating it does not matter.

Oh yeah.... what an idiot.

However, doing the integral, we have:

[itex]\int_{r_s}^0 \frac{dr}{ \sqrt{ \frac{r}{r_s} - 1 }} = \int_0^\tau d \tau[/itex]
[itex]\tau = \int_{r_s}^0 \frac{dr}{ \sqrt{ \frac{r}{r_s} - 1 }} [/itex]

Then I substituted [itex]r=r_s \sinh^2{\theta}[/itex]
[itex]\frac{dr}{d \theta} = 2 r_s \cosh{\theta} \sinh{\theta}[/itex]
[itex]r=0 \Rightarrow \sinh^2{\theta}=0 \Rightarrow \theta=0[/itex]
[itex]r=r_s \Rightarrow \sinh^2{\theta}=1 \sinh{\theta}=1 \Rightarrow \theta = \sinh^{-1}{\theta}[/itex] However I have assumed here that [itex]\sqrt{1}=1[/itex] and ignored the negative case - possibly a mistake.

And so we sub it all back in we get

[itex]\tau = \int_{\sinh^{-1}{\theta}}^0 \frac{2 r_s \cosh{\theta} \sinh{\theta} d \theta}{\sqrt{\sinh^2{\theta}-1}}= \int_{\sinh^{-1}{\theta}}^0 \frac{2 r_s \cosh{\theta} \sinh{\theta} d \theta}{\cosh{\theta}}= \int_{\sinh^{-1}{\theta}}^0 2r_s \sinh{\theta} d \theta [/itex]

But integrating sinh is really complicated so I assumed I have made a mistake at this point?
 
  • #8
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Oh yeah.... what an idiot.

However, doing the integral, we have:

[itex]\int_{r_s}^0 \frac{dr}{ \sqrt{ \frac{r}{r_s} - 1 }} = \int_0^\tau d \tau[/itex]
[itex]\tau = \int_{r_s}^0 \frac{dr}{ \sqrt{ \frac{r}{r_s} - 1 }} [/itex]

Then I substituted [itex]r=r_s \sinh^2{\theta}[/itex]
[itex]\frac{dr}{d \theta} = 2 r_s \cosh{\theta} \sinh{\theta}[/itex]
[itex]r=0 \Rightarrow \sinh^2{\theta}=0 \Rightarrow \theta=0[/itex]
[itex]r=r_s \Rightarrow \sinh^2{\theta}=1 \sinh{\theta}=1 \Rightarrow \theta = \sinh^{-1}{\theta}[/itex] However I have assumed here that [itex]\sqrt{1}=1[/itex] and ignored the negative case - possibly a mistake.

And so we sub it all back in we get

[itex]\tau = \int_{\sinh^{-1}{\theta}}^0 \frac{2 r_s \cosh{\theta} \sinh{\theta} d \theta}{\sqrt{\sinh^2{\theta}-1}}= \int_{\sinh^{-1}{\theta}}^0 \frac{2 r_s \cosh{\theta} \sinh{\theta} d \theta}{\cosh{\theta}}= \int_{\sinh^{-1}{\theta}}^0 2r_s \sinh{\theta} d \theta [/itex]

But integrating sinh is really complicated so I assumed I have made a mistake at this point?

Your first line is wrong, first of all you should have no limits as you're just going for the anti-derivatives and then you add in the +c at the end. Second it is r_s/r NOT r/r_s so the LHS should be:

[tex]\int \frac{dr}{ \sqrt{ \frac{r_s}{r} - 1 }}=\int\sqrt{\frac{r}{r_s-r}}dr[/tex] you then use the substitution [tex]r=r_{s}sin^2\theta[/tex]
 
  • #9
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Your first line is wrong, first of all you should have no limits as you're just going for the anti-derivatives and then you add in the +c at the end. Second it is r_s/r NOT r/r_s so the LHS should be:

[tex]\int \frac{dr}{ \sqrt{ \frac{r_s}{r} - 1 }}=\int\sqrt{\frac{r}{r_s-r}}dr[/tex] you then use the substitution [tex]r=r_{s}sin^2\theta[/tex]

so [itex]dr=2r_s \sin{\theta} \cos{\theta} d \theta[/itex]

so [itex]\tau = \int \sqrt{ \frac{r}{r_s-1}} dr = \int \sqrt{ \frac{r_s \sin^2{\theta}}{r_s \cos^2{\theta}} 2r_s \sin{\theta} \cos{\theta} d \theta = 2 r_s \int \sin^2{\theta} = 2 r_s \int \frac{1}{2}( 1 + \cos{2 \theta} )d \theta[/itex]
[itex]\tau = r_s ( \theta + \frac{1}{2} \sin{2 \theta} ) + c = r_s \theta + \frac{r_s}{2} \sin{2 \theta} + c[/itex]

Now we know that [itex]\tau(r=0)=\tau(\theta = \frac{\pi}{2})=0[/itex]

so [itex]r_s \frac{\pi}{2} + \frac{r_s}{2} \sin{\pi} + c = 0 \Rightarrow c = -\frac{\pi r_s}{2}[/itex]

But then [itex]\tau(\theta)=r_s \theta + \frac{r_s}{2} \sin{2 \theta} - \frac{\pi r_s}{2}[/itex]

so [itex]\tau(r=0)=\tau(\theta=0)=-\frac{\pi r_s}{2}[/itex]

So clearly I have missed something again but I cannot see it at all!

My way to "force" the right answer would be to use [itex]\theta= \pi[/itex] corresponding to [itex]r=0[/itex]. This way [itex]\tau(r=0)=\tau(\theta=\pi)=\pi r_s - \frac{\pi r_s}{2}=\frac{\pi r_s}{2}[/itex] as required. How do we justify the use of [itex]\theta=\pi[/itex] instead of [itex]\theta=0[/itex] though? And surely, by extension of this process, we could take [itex]\theta[/itex] even larger and thus extend the proper time as much as we want? Or is this forbidden by the constraint [itex]\theta \in [0,\pi][/itex] - I'm not sure if this is the right constraint on [itex]\theta[/itex] here though - can you confirm?

Thanks.
 
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  • #10
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so [itex]dr=2r_s \sin{\theta} \cos{\theta} d \theta[/itex]

so [itex]\tau = \int \sqrt{ \frac{r}{r_s-1}} dr = \int \sqrt{ \frac{r_s \sin^2{\theta}}{r_s \cos^2{\theta}} 2r_s \sin{\theta} \cos{\theta} d \theta = 2 r_s \int \sin^2{\theta} = 2 r_s \int \frac{1}{2}( 1 + \cos{2 \theta} )d \theta[/itex]
[itex]\tau = r_s ( \theta + \frac{1}{2} \sin{2 \theta} ) + c = r_s \theta + \frac{r_s}{2} \sin{2 \theta} + c[/itex]

Now we know that [itex]\tau(r=0)=\tau(\theta = \frac{\pi}{2})=0[/itex]

so [itex]r_s \frac{\pi}{2} + \frac{r_s}{2} \sin{\pi} + c = 0 \Rightarrow c = -\frac{\pi r_s}{2}[/itex]

But then [itex]\tau(\theta)=r_s \theta + \frac{r_s}{2} \sin{2 \theta} - \frac{\pi r_s}{2}[/itex]

so [itex]\tau(r=0)=\tau(\theta=0)=-\frac{\pi r_s}{2}[/itex]

So clearly I have missed something again but I cannot see it at all!

[tex]sin^2\theta=\frac{1}{2}( 1 - \cos{2 \theta} )[/tex]. Note the minus sign, you have a + sign in your 2nd line.
 
  • #11
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[tex]sin^2\theta=\frac{1}{2}( 1 - \cos{2 \theta} )[/tex]. Note the minus sign, you have a + sign in your 2nd line.

Ooops! But I don't think that helps...

we still end up with [itex]\tau = 2 r_s \int \sin^2{\theta} d \theta = r_s \int (1-\cos{2 \theta}) = r_s ( \theta - \frac{1}{2} \sin{2 \theta}) + c[/itex]

So at [itex]r=r_s, \theta=\frac{\pi}{2}[/itex] and [itex]\tau=0[/itex] so

[itex]\frac{r_s \pi}{2} + \frac{r_s}{2} \sin{\pi} + c =0[/itex]
which gives [itex]c=-\frac{\pi r_s}{2}[/itex] as before
 
  • #12
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Ooops! But I don't think that helps...

we still end up with [itex]\tau = 2 r_s \int \sin^2{\theta} d \theta = r_s \int (1-\cos{2 \theta}) = r_s ( \theta - \frac{1}{2} \sin{2 \theta}) + c[/itex]

So at [itex]r=r_s, \theta=\frac{\pi}{2}[/itex] and [itex]\tau=0[/itex] so

[itex]\frac{r_s \pi}{2} + \frac{r_s}{2} \sin{\pi} + c =0[/itex]
which gives [itex]c=-\frac{\pi r_s}{2}[/itex] as before

Yes that's fine, the key then is to note that when r=0, theta=pi.
 
  • #13
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Yes that's fine, the key then is to note that when r=0, theta=pi.

Yes but why do we chose [itex]\theta=\pi[/itex] and not [itex]\theta=0[/itex] or [itex]\theta=2 \pi[/itex]? Isn't [itex]r=0[/itex] part of the manifold where [itex]\theta[/itex] is undefined?
 
  • #14
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Yes but why do we chose [itex]\theta=\pi[/itex] and not [itex]\theta=0[/itex] or [itex]\theta=2 \pi[/itex]? Isn't [itex]r=0[/itex] part of the manifold where [itex]\theta[/itex] is undefined?

Yeah that's the fudge, can't really say with any certainty why theta has to be pi but I'm pretty sure what we've done is the right method to go about the problem.
 

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