# Black Holes

I'm trying the exercises on p35 of the following:
http://arxiv.org/PS_cache/gr-qc/pdf/9707/9707012v1.pdf

I have done parts (i) and (ii) but was wondering if anybody can help me with (iii)?

I had $k \cdot D k^\mu |_{U'=0} = k^\nu D_\nu k^\mu |_{U'=0} = k^V' D_{V'} k^\mu$ since setting $U'=0$ means only the $\nu=V'$ componeent will contribute.

Subbing in we end up with

$\kappa V' ( k^\mu{}_{, V'} + \Gamma^\mu{}_{V' \rho} k^\rho)$
I can't really make anything useful out of this now though?

Thanks.

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fzero
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You might want to compute each component. Did you compute $$\Gamma^\mu{}_{V' \rho}$$?

You might want to compute each component. Did you compute $$\Gamma^\mu{}_{V' \rho}$$?
Is $\Gamma^\mu{}_{V' \rho} = \frac{1}{2} g^{\mu \lambda \left( g_{V' \lambda , \rho} + g_{\rho \lambda, V'} - g_{V' \rho, \lambda} \right)$

Ok but $ds^2=-dU'dV'$
which means that $g_{\mu \nu} = \begin{pmatrix} 0 & -1 \\ -1 & 0 \end{pmatrix}$

So regardless of what happens, we're always going to be taking the derivative of either -1 or 0 in the three terms in the Christoffel symbol. So am I right in saying that $\Gamma^\mu{}_{V' \rho} =0$?

That would mean that $(k \cdot D k^\mu)|_{U'=0}=k^V' k^\mu{}_{, V'} = \kappa V' k^\mu{}_{,V'}$

But $k^\mu{}_{,V'} = \kappa \frac{\partial}{\partial V'} + \kappa V' \frac{\partial^2}{\partial^2 V'}$

It may well have worked if that 2nd derivative term wasn't there....hmmm....

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fzero
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Yes it's a flat metric. However $$k^\mu$$ refers to the components of the vector field $$k$$, so we're not dealing with a derivative operator.

Yes it's a flat metric. However $$k^\mu$$ refers to the components of the vector field $$k$$, so we're not dealing with a derivative operator.
Excellent!

One other thing though. At the top of p36, he claims that $a^\mu=D_{(\lambda)} u^\mu$
I don't understand why this defines an acceleration? I though $a^\mu=\frac{d x^\mu(\lambda)}{d \lambda}$? Is there a way of showing these definitions are equivalent?

Also, how does that formula for the acceleration (2.121) then become equation (2.122)?

fzero
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Excellent!

One other thing though. At the top of p36, he claims that $a^\mu=D_{(\lambda)} u^\mu$
I don't understand why this defines an acceleration? I though $a^\mu=\frac{d x^\mu(\lambda)}{d \lambda}$? Is there a way of showing these definitions are equivalent?
$a^\mu=\frac{d x^\mu(\lambda)}{d \lambda}$

is not covariant.

Also, how does that formula for the acceleration (2.121) then become equation (2.122)?
Use (2.120).

$a^\mu=\frac{d x^\mu(\lambda)}{d \lambda}$

is not covariant.
Sure. But why do we use $a^\mu=D_{(\tau)}u^\mu = u \cdot D u^\mu$ instead of just a plain old covariant derivative $a^\mu=D u^\mu$

And on p40 there are a few things I'm not quite grasping. I can follow up to eqn (2.147).

Now in the line below he works out $T_0$ for Rindler by substituting in $T$ from Schwarzschild. How is this allowed? Isn't he mixing and matching results from two different spacetimes? I know Rindler essentially covers the region near r=2M of Schwarzschild but still......

Ok. He then presents an acceptable argument on the enxt few lines as to why if $x \rightarrow \infty$ we expect T to go to zero. Why does this mean the Hawking temp will go to zero? Or is the T in eqn (2.142) actually the Hawking temp?

So it makes sense that the temp should go to zero because RIndler is just Minkowski in different coords so it's a vacuum soln i.e. there is nothing in the spacetime to radiate any hear. I don't understand the last line though....

why does $T_{\text{local}} \rightarrow T_H$ if a black hole is present? Perhaps I should just leave that just now....he does say we shall see that later on!

fzero
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Sure. But why do we use $a^\mu=D_{(\tau)}u^\mu = u \cdot D u^\mu$ instead of just a plain old covariant derivative $a^\mu=D u^\mu$
Well we still want the proper acceleration. $a^\mu=\frac{d x^\mu(\lambda)}{d \lambda}$ would be correct in special relativity, where we neglect curvature. So it's obvious that we need to move to the corresponding covariant derivative.

And on p40 there are a few things I'm not quite grasping. I can follow up to eqn (2.147).

Now in the line below he works out $T_0$ for Rindler by substituting in $T$ from Schwarzschild. How is this allowed? Isn't he mixing and matching results from two different spacetimes? I know Rindler essentially covers the region near r=2M of Schwarzschild but still......
He's not mixing and matching results at all. He's simply using (2.146) to find $$T_0$$ and notes that it agrees with the Schwarzschild result.

Ok. He then presents an acceptable argument on the enxt few lines as to why if $x \rightarrow \infty$ we expect T to go to zero. Why does this mean the Hawking temp will go to zero? Or is the T in eqn (2.142) actually the Hawking temp?
The Hawking temperature is the temperature measured at spatial infinity, see the comment below (2.148).

So it makes sense that the temp should go to zero because RIndler is just Minkowski in different coords so it's a vacuum soln i.e. there is nothing in the spacetime to radiate any hear. I don't understand the last line though....

why does $T_{\text{local}} \rightarrow T_H$ if a black hole is present? Perhaps I should just leave that just now....he does say we shall see that later on!
$T_{\text{local}} \rightarrow T_H$ is explained by the comment below (2.142) about the asymptotic behavior of the timelike Killing vector field.

Well we still want the proper acceleration. $a^\mu=\frac{d x^\mu(\lambda)}{d \lambda}$ would be correct in special relativity, where we neglect curvature. So it's obvious that we need to move to the corresponding covariant derivative.
So I agree that the derivative wrt lambda is the special relativity result. So in GR we need something that takes into account spacetime curvature, such as the covariant derivative. Why then, is the acceleration not simply $D u^\mu$? Why do we need to dot procuvt that with u to get $u \cdot D u^\mu$?

$T_{\text{local}} \rightarrow T_H$ is explained by the comment below (2.142) about the asymptotic behavior of the timelike Killing vector field.
Hmmm.....what's $T_{\text{local}}$ in all of this? And why does this relationship tells us that the BH must be radiating?

fzero
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So I agree that the derivative wrt lambda is the special relativity result. So in GR we need something that takes into account spacetime curvature, such as the covariant derivative. Why then, is the acceleration not simply $D u^\mu$? Why do we need to dot procuvt that with u to get $u \cdot D u^\mu$?
We need an expression that reduces to the physical result in flat space. Besides $D u^\mu$ makes no sense, since you've dropped the index on the derivative without explaining why. You could write $v \cdot D u^\mu$ for some vector $$v$$ but then you must explain why you chose that $$v$$. $$v=u$$ is the obvious choice.

Hmmm.....what's $T_{\text{local}}$ in all of this? And why does this relationship tells us that the BH must be radiating?
$T_{\text{local}}$ is defined by (2.142) in terms of the temperature $$T_0$$ measured at spatial infinity and the timelike Killing vector field. $$T_0$$ is determined by considering the Euclideanized metric and demanding that it is complete, which is what was done in the discussion leading to (2.136) for the Schwarzschild BH.

Radiation follows from coupling the BH to a quantum system at equilibrium via (2.139). At the lowest approximation, you will conclude that there is a black-body spectrum of radiation due to the finite temperature.

So I agree that the derivative wrt lambda is the special relativity result. So in GR we need something that takes into account spacetime curvature, such as the covariant derivative. Why then, is the acceleration not simply $D u^\mu$? Why do we need to dot procuvt that with u to get $u \cdot D u^\mu$?
you do realize that the acceleration is just the equation of motion i.e the geodesic which is equivalent to $$\nabla_{u} u^{\mu} = u^{\nu} \nabla_{\nu} u^{\mu}$$

$T_{\text{local}}$ is defined by (2.142) in terms of the temperature $$T_0$$ measured at spatial infinity and the timelike Killing vector field. $$T_0$$ is determined by considering the Euclideanized metric and demanding that it is complete, which is what was done in the discussion leading to (2.136) for the Schwarzschild BH.

Radiation follows from coupling the BH to a quantum system at equilibrium via (2.139). At the lowest approximation, you will conclude that there is a black-body spectrum of radiation due to the finite temperature.
Thanks for that.

you do realize that the acceleration is just the equation of motion i.e the geodesic which is equivalent to $$\nabla_{u} u^{\mu} = u^{\nu} \nabla_{\nu} u^{\mu}$$
Ok. I definitely know/recognise this but can't remember how to show it.

We're talking about a massive particle so we know it must be travelling on a timelike geodesic. That's us established it is on a geodesic.
But for geodesics with tangent vector X, we know $\nabla_X X=0$ i.e. $X^\nu \nabla_\nu X^\mu=0$

In this case the tangent vector is u instead of X so surely we should have $u^\nu \nabla_\nu u^\mu=0$
But you're telling me that $a^\mu=u^\nu \nabla_\nu u^\mu$ which would mean it isn't accelerating (which i guess is true for a geodesic)

Is this correct? Or have I missed the reason why $a^\mu = u \cdot D u^\mu$ altogether?

Thanks.

well if it's a free particle then its acceleration will be zero. Anyway 0's or not that is how proper acceleration is defined

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We need an expression that reduces to the physical result in flat space. Besides $D u^\mu$ makes no sense, since you've dropped the index on the derivative without explaining why. You could write $v \cdot D u^\mu$ for some vector $$v$$ but then you must explain why you chose that $$v$$. $$v=u$$ is the obvious choice.

$T_{\text{local}}$ is defined by (2.142) in terms of the temperature $$T_0$$ measured at spatial infinity and the timelike Killing vector field. $$T_0$$ is determined by considering the Euclideanized metric and demanding that it is complete, which is what was done in the discussion leading to (2.136) for the Schwarzschild BH.

Radiation follows from coupling the BH to a quantum system at equilibrium via (2.139). At the lowest approximation, you will conclude that there is a black-body spectrum of radiation due to the finite temperature.
Hi there, could you take a look at the diagram on p45 please?
Am I right in saying that compared with teh diagram on p44 (for Minkowski), there is now this additional left hand side region because we no longer have the $r \geq 0$ i.e. $\tilde{V} \geq \tilde{U}$ constraint that Minkowski had?

Also, he indicates that the right hand rhombus represents the Rindler spacetime - how does he know this?

He then shows an orbit of $\frac{\partial}{\partial T}$ and also the line $X=0$. According to eqn (2.117), surely these are using Rindler coordinates so why aren't they a curve in the right hand Rindler rhombus?

Finally, he draws an orbit of $\partial}{\partial t}$ (where he now claims $t$ is the Rindler time coordinate and puts that in the Rindler rhombus. However, this curve will no longer be a timelike geodesic, will it? (Here I am using the definition on p273 Wald that a timelike geodesic will begin at past temporal infinity, $i^-$ and end at future temporal infinity, $i^+$).

Thanks.

fzero
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Hi there, could you take a look at the diagram on p45 please?
Am I right in saying that compared with teh diagram on p44 (for Minkowski), there is now this additional left hand side region because we no longer have the $r \geq 0$ i.e. $\tilde{V} \geq \tilde{U}$ constraint that Minkowski had?
Yes.

Also, he indicates that the right hand rhombus represents the Rindler spacetime - how does he know this?
From (2.113) the Rindler variables only cover $$U'<0, V'>0$$.

He then shows an orbit of $\frac{\partial}{\partial T}$ and also the line $X=0$. According to eqn (2.117), surely these are using Rindler coordinates so why aren't they a curve in the right hand Rindler rhombus?
$$T,X$$ are Minkowski coordinates, $$x,t$$ are the Rindler coordinates.

Finally, he draws an orbit of $\partial}{\partial t}$ (where he now claims $t$ is the Rindler time coordinate and puts that in the Rindler rhombus. However, this curve will no longer be a timelike geodesic, will it? (Here I am using the definition on p273 Wald that a timelike geodesic will begin at past temporal infinity, $i^-$ and end at future temporal infinity, $i^+$).

Thanks.
Geodesics have zero proper acceleration. It was shown that the orbits of the timelike Killing vector field have a proper acceleration proportional to $$x^{-1}$$.

From (2.113) the Rindler variables only cover $$U'<0, V'>0$$.
Hw do you know that the right rhombus corresponds to the region $U'<0,V'>0$?

$$T,X$$ are Minkowski coordinates, $$x,t$$ are the Rindler coordinates.
How do we know what X=0 looks like and what the orbit of $\frac{\partial}{\partial T}$ should look like?

Thanks!

From (2.113) the Rindler variables only cover $$U'<0, V'>0$$.
Hw do you know that the right rhombus corresponds to the region $U'<0,V'>0$?

$$T,X$$ are Minkowski coordinates, $$x,t$$ are the Rindler coordinates.
How do we know what X=0 looks like and what the orbit of $\frac{\partial}{\partial T}$ should look like?

Thanks!

fzero
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Hw do you know that the right rhombus corresponds to the region $U'<0,V'>0$?

How do we know what X=0 looks like and what the orbit of $\frac{\partial}{\partial T}$ should look like?

Thanks!
For both of these, you use the coordinate transformations (2.162) and (2.116). The $$\tilde{V}$$ coordinate runs at $$\pi/4$$ from the vertical axis of the Penrose diagram, while $$\tilde{U}$$ runs at $$-\pi/4$$.

For both of these, you use the coordinate transformations (2.162) and (2.116). The $$\tilde{V}$$ coordinate runs at $$\pi/4$$ from the vertical axis of the Penrose diagram, while $$\tilde{U}$$ runs at $$-\pi/4$$.
I don't really follow what you're saying here. By analogy with the diagram on p44 I can see that $\tilde{V}$ is at $\frac{\pi}{2}$ on future null infinity and that $\tilde{U}$ is at $\frac{\pi}{2}$ on past null infinity. Where does the $\frac{\pi}{4}$ come from?

Also, will the boundary lines on the LHS of the diagram also be past/future null infinities? Or are these only defined on the RHS?

fzero
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I don't really follow what you're saying here. By analogy with the diagram on p44 I can see that $\tilde{V}$ is at $\frac{\pi}{2}$ on future null infinity and that $\tilde{U}$ is at $\frac{\pi}{2}$ on past null infinity. Where does the $\frac{\pi}{4}$ come from?
$\frac{\pi}{4}=45^\circ$ is the angle the $\tilde{U},\tilde{V}$ axes make with the vertical axis of the diagram.

Also, will the boundary lines on the LHS of the diagram also be past/future null infinities? Or are these only defined on the RHS?
Yes, just look at the radial null geodesics on the LHS.

$\frac{\pi}{4}=45^\circ$ is the angle the $\tilde{U},\tilde{V}$ axes make with the vertical axis of the diagram.
But the two diagonal lines in the middle of the diagram correspond to the $\tilde{U}$ and $\tilde{V}$ axes, right? We also know that these corrdinates both range from $- \frac{\pi}{2}$ to $\frac{\pi}{2}$. Surely, since the dotted vertical line intersects these diagonal lines in half, we would expect $\tilde{U}=\tilde{V}=0$ at the centre of the diagram?

fzero
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But the two diagonal lines in the middle of the diagram correspond to the $\tilde{U}$ and $\tilde{V}$ axes, right?
I have been talking about the diagonal lines.

We also know that these corrdinates both range from $- \frac{\pi}{2}$ to $\frac{\pi}{2}$. Surely, since the dotted vertical line intersects these diagonal lines in half, we would expect $\tilde{U}=\tilde{V}=0$ at the centre of the diagram?
I've never said anything to the contrary.

I have been talking about the diagonal lines.
I've never said anything to the contrary.
So I'm now ok with why the right rhombus is the Rindler spacetime.

But what were you saying about $\frac{\pi}{4}$ then? I didn't follow that stuff?

And also, I still don't get how to see that X=0 is that vertical line down the middle...

Thanks.

fzero
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So I'm now ok with why the right rhombus is the Rindler spacetime.

But what were you saying about $\frac{\pi}{4}$ then? I didn't follow that stuff?
What angle do the diagonal lines make with the vertical line down the middle?

And also, I still don't get how to see that X=0 is that vertical line down the middle...

Thanks.
Can you express $$X$$ in terms of $$\tilde{U},\tilde{V}$$?

What angle do the diagonal lines make with the vertical line down the middle?

Can you express $$X$$ in terms of $$\tilde{U},\tilde{V}$$?
So I think I have it:

$U'=\tan{ \tilde{U}} , V'=\tan{ \tilde{V}}$

and $T-X=U' , \quad T+X = V'$

so $X=\frac{1}{2} ( \tan{ \tilde{V} } - \tan{ \tilde{U} } )$

so $X=0 \Rightarrow \tan{ \tilde{V}} = \tan{ \tilde{U}} \Rightarrow \tilde{V}=\tidle{U}$ and that it is true down that dotted vertical line.
Did I get it right?

So the final things to establish is why that orbit of $\frac{\partial}{\partial T}$ goes from $i^-$ to $i^+$. I guess, by defn, of curves that begin and end at these points, it is sufficient to show that $\xi=\frac{\partial}{\partial T}$ is a timelike curve.

Well $g_{\mu \nu} \xi^\mu \xi^\nu = g_{TT}$ since $\xi^\mu = (1,0)$ in coordinates $(T,X)$

Now if I had managed to get an answer of $g_{ \tilde{U} \tilde{V}}$ then that would have been awesome because $g_{ \tilde{U} \tilde{V}}=-1$ which would have meant timelike......