Can someone explain why p terms are not canceling in this Diffeomorphisms proof?

[latentcorpse]

So I'm trying to prove eqn (223) in the notes attached in this thread:
https://www.physicsforums.com/showthread.php?t=457123

I took the equation ( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X))
and expanded in a coordinate basis as follows

( \phi^* ( \eta) )_\mu dx^\mu X = \eta_\alpha dy^\alpha \phi_*(X)

So to get the result it seems like it should be a simple case of cross multiplying but unfortunately the X doesn't cancel the \phi_*(X)

Thanks for any help.

 

[fzero]

So I'm trying to prove eqn (223) in the notes attached in this thread:
https://www.physicsforums.com/showthread.php?t=457123

I took the equation ( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X))
and expanded in a coordinate basis as follows

( \phi^* ( \eta) )_\mu dx^\mu X = \eta_\alpha dy^\alpha \phi_*(X)

So to get the result it seems like it should be a simple case of cross multiplying but unfortunately the X doesn't cancel the \phi_*(X)

Thanks for any help.

The notation ( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X)) always means that indices are contracted in the coordinate basis, so

( \phi^* ( \eta) ) ( X) = ( \phi^* ( \eta) )_\mu X^\mu,

\eta ( \phi_* (X)) = \eta_\alpha ( \phi_* (X))^\alpha.

Just think of the usual expression for a 1-form acting on a vector: \omega(X) = \omega_\mu X^\mu.

 

[latentcorpse]

The notation ( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X)) always means that indices are contracted in the coordinate basis, so

( \phi^* ( \eta) ) ( X) = ( \phi^* ( \eta) )_\mu X^\mu,

\eta ( \phi_* (X)) = \eta_\alpha ( \phi_* (X))^\alpha.

Just think of the usual expression for a 1-form acting on a vector: \omega(X) = \omega_\mu X^\mu.

Ok. So then we get the equality:
( \phi^*( \eta ) )_\mu X^\mu = \eta_\alpha ( \phi_* ( X ))^\alpha
I don't see how we can get this into the form of (223)?Additionally, I am trying to prove (187). Given the hint, I am able to multiply by those vectors and get

\nabla_X \nabla_Y Z^a- \nabla_Y \nabla_X Z^a = R^a{}_{bcd}Z^bX^cY^d = (R(X,Y)Z)^a. But according to teh defn of the Riemann tensor (see (178)), I am missing a \nabla_{[X,Y]} piece. Any ideas?

 

[fzero]

Ok. So then we get the equality:
( \phi^*( \eta ) )_\mu X^\mu = \eta_\alpha ( \phi_* ( X ))^\alpha
I don't see how we can get this into the form of (223)?

You can use (221).

Additionally, I am trying to prove (187). Given the hint, I am able to multiply by those vectors and get

\nabla_X \nabla_Y Z^a- \nabla_Y \nabla_X Z^a = R^a{}_{bcd}Z^bX^cY^d = (R(X,Y)Z)^a. But according to teh defn of the Riemann tensor (see (178)), I am missing a \nabla_{[X,Y]} piece. Any ideas?

You have to use the product rule correctly on the terms \nabla_b\nabla_c(Z^a X^bY^c) and (135).

 

[latentcorpse]

You can use (221).

Ok. That works. But how do I prove (221)? There is no identity to help me there like there was for (223).

You have to use the product rule correctly on the terms \nabla_b\nabla_c(Z^a X^bY^c) and (135).

So would it go something like this:

\nabla_c ( Z^a{}_{;d} X^c Y^d + Z^a X^c{}_{;d} Y^d + \nabla_Y Z^a X^c) - \nabla_d ( Z^a{}_{;c} X^c Y^d + \nabla_X Z^a Y^d + Z^a X^c Y^d{}_{;c} )?

If I expand the next derivative, I'll get terms like (Z^a{}_{;dc}-Z^a{}_{;cd})X^cY^d but I can't cancel them as the second covariant derivatives don't commute for tensor fields, do they? Maybe I'm still doing something wrong...

 

[fzero]

You can use (221).

Ok. That works. But how do I prove (221)? There is no identity to help me there like there was for (223).

(221) follows in one or two lines from (220).

So would it go something like this:

\nabla_c ( Z^a{}_{;d} X^c Y^d + Z^a X^c{}_{;d} Y^d + \nabla_Y Z^a X^c) - \nabla_d ( Z^a{}_{;c} X^c Y^d + \nabla_X Z^a Y^d + Z^a X^c Y^d{}_{;c} )?

If I expand the next derivative, I'll get terms like (Z^a{}_{;dc}-Z^a{}_{;cd})X^cY^d but I can't cancel them as the second covariant derivatives don't commute for tensor fields, do they? Maybe I'm still doing something wrong...

OK, it actually works out much faster if you multiply from the left. Acting with the derivatives the way I originally suggested probably does give extra terms. From the left, you end up with

Y^d \nabla_X \nabla_d Z^a - X^c \nabla_Y \nabla_c Z^a

that you have to manipulate with the product rule.

 

[latentcorpse]

(221) follows in one or two lines from (220).

I get
( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)
Again, I'm having trouble getting rid of the \phi^* on the right

OK, it actually works out much faster if you multiply from the left. Acting with the derivatives the way I originally suggested probably does give extra terms. From the left, you end up with

Y^d \nabla_X \nabla_d Z^a - X^c \nabla_Y \nabla_c Z^a

that you have to manipulate with the product rule.

Ok. You appear to have multiplied from the left by Y^dX^c on the first term but X^cY^d on the second term. Now we are using abstract index notation so these both represent vectors rather than the components of vectors. How do you know those vectors commute?

Thanks.

 

[fzero]

I get
( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)
Again, I'm having trouble getting rid of the \phi^* on the right

You need to use that \phi is the map x^\mu\rightarrow y^\alpha and the chain rule.

Ok. You appear to have multiplied from the left by Y^dX^c on the first term but X^cY^d on the second term. Now we are using abstract index notation so these both represent vectors rather than the components of vectors. How do you know those vectors commute?

Thanks.

They represent the components of vectors and commute. Look at the derivation of (130), the 2nd line follows from the first by commuting the components of X and Y.

 

[latentcorpse]

You need to use that \phi is the map x^\mu\rightarrow y^\alpha and the chain rule.

well \phi has components y^\alpha(x^\mu)
I can see we want to introduce something like dy/dx d/dy on the RHS but I just don't see how?

They represent the components of vectors and commute. Look at the derivation of (130), the 2nd line follows from the first by commuting the components of X and Y.

Right. But there he is using basis indices (see chapter 7).
Although he does say in Chapter 7 that we are free to change abstract indices to basis indices to get an equation that is true in an arbitrary basis so I guess everything is ok providing we work this out in greek indices?

Anyway I tried working out the following using the product rule:
\nabla_Y (X^\mu \nabla_\mu Z^\nu) = Y^\rho \nabla_\rho (X^\mu \nabla_\mu Z^\nu) = Y^\rho X^\mu{}_{; \rho} \nabla_\mu Z^\nu + Y^\rho X^\mu \nabla_\rho \nabla_\mu Z^\nu
in the hope that one of those terms we get would be useful but I don't see it?

 

[fzero]

well \phi has components y^\alpha(x^\mu)
I can see we want to introduce something like dy/dx d/dy on the RHS but I just don't see how?

Use \phi^*(f) = f\circ \phi applied to the coordinates x^\mu.

Right. But there he is using basis indices (see chapter 7).
Although he does say in Chapter 7 that we are free to change abstract indices to basis indices to get an equation that is true in an arbitrary basis so I guess everything is ok providing we work this out in greek indices?

Anyway I tried working out the following using the product rule:
\nabla_Y (X^\mu \nabla_\mu Z^\nu) = Y^\rho \nabla_\rho (X^\mu \nabla_\mu Z^\nu) = Y^\rho X^\mu{}_{; \rho} \nabla_\mu Z^\nu + Y^\rho X^\mu \nabla_\rho \nabla_\mu Z^\nu
in the hope that one of those terms we get would be useful but I don't see it?

I'm not sure why you chose that expression. We had

<br /> Y^d \nabla_X \nabla_d Z^a - X^c \nabla_Y \nabla_c Z^a,<br />

so you want to re-express

Y^d \nabla_X \nabla_d Z^a = \nabla_X(Y^d \nabla_d Z^a) - ( \nabla_X Y^d )(\nabla_d Z^a)

and similarly the other term.

 

[latentcorpse]

Use \phi^*(f) = f\circ \phi applied to the coordinates x^\mu.

Sorry I still don't get. The \phi's are charts so they have components x^\mu so how would this work?

 

[latentcorpse]

Use \phi^*(f) = f\circ \phi applied to the coordinates x^\mu.



I'm not sure why you chose that expression. We had

<br /> Y^d \nabla_X \nabla_d Z^a - X^c \nabla_Y \nabla_c Z^a,<br />

so you want to re-express

Y^d \nabla_X \nabla_d Z^a = \nabla_X(Y^d \nabla_d Z^a) - ( \nabla_X Y^d )(\nabla_d Z^a)

and similarly the other term.

Sorry, I have been working on some other stuff for the last few days and didn't have time to look at it again until now. I still do not get how either of these work? The second one I keep getting bogged down with what to do with algebra. The first one I don't really follow what you want me to do.

Thanks again.

 

[fzero]

There really isn't any algebra to get bogged down with,

<br /> \nabla_X(Y^d \nabla_d Z^a) = \nabla_X \nabla_Y Z^a,<br />

while ( \nabla_X Y^d ) is one term in the expression (130) for [X,Y]^d.

As for the other part, if \phi(x^\mu) = y^\alpha, then what is (f\circ \phi)(x^\mu)?

 

[latentcorpse]

There really isn't any algebra to get bogged down with,

<br /> \nabla_X(Y^d \nabla_d Z^a) = \nabla_X \nabla_Y Z^a,<br />

while ( \nabla_X Y^d ) is one term in the expression (130) for [X,Y]^d.

As for the other part, if \phi(x^\mu) = y^\alpha, then what is (f\circ \phi)(x^\mu)?

Ok. I get us down to

\nabla_X \nabla_Y Z^a - \nabla_Y \nabla_X Z^a - ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a)

Now from our Riemann tensor we want those last terms to give us a -\nabla_{[X,Y]}Z^a

So I tried working backwards:

\nabla_{[X,Y]}Z^a=\nabla_{XY-YX}Z^a = \nabla_{XY}Z^a-\nabla_{YX}Z^a

Now I don't know how to get to the next line there? Could you explain please?

As for the other bit I would find that (f \circ \phi)(x^\mu) = f(y^\alpha)

Thanks.

 

[fzero]

Ok. I get us down to

\nabla_X \nabla_Y Z^a - \nabla_Y \nabla_X Z^a - ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a)

Now from our Riemann tensor we want those last terms to give us a -\nabla_{[X,Y]}Z^a

Just write

- ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a) = - ( \nabla_X Y^d-\nabla_Y X^d)(\nabla_d Z^a)

and use (130).

So I tried working backwards:

\nabla_{[X,Y]}Z^a=\nabla_{XY-YX}Z^a = \nabla_{XY}Z^a-\nabla_{YX}Z^a

The last part makes no sense at all, since XY and YX are not tangent vectors.

As for the other bit I would find that (f \circ \phi)(x^\mu) = f(y^\alpha)

So you need to go back a few posts and figure out what that's good for.

 

[latentcorpse]

Just write

- ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a) = - ( \nabla_X Y^d-\nabla_Y X^d)(\nabla_d Z^a)

and use (130).
The last part makes no sense at all, since XY and YX are not tangent vectors.
So you need to go back a few posts and figure out what that's good for.

Ok. So I have got the RIcci identity to work out.

As for the transformation, I assume I am meant to go back to( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)

and get rid of the \phi^*(f) on the RHS.
Should I act both sides on the coordinates x^\mu? That's the only way I can see of being able to use the f(x^\mu)=y^\alpha property.Also, on p90, he says that we should think of the singularity as a time rather than a place inside the black hole because r plays the role of the time coordinate. How does this work?

Thanks.

 

[fzero]

Ok. So I have got the RIcci identity to work out.

As for the transformation, I assume I am meant to go back to


( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)

and get rid of the \phi^*(f) on the RHS.
Should I act both sides on the coordinates x^\mu?

Then try doing that. It's much better to try to work it out before asking the question.

That's the only way I can see of being able to use the f(x^\mu)=y^\alpha property.

Hopefully you mean \phi(x^\mu)=y^\alpha

Also, on p90, he says that we should think of the singularity as a time rather than a place inside the black hole because r plays the role of the time coordinate. How does this work?

Thanks.

You should be able to use the metric to show that the radial vector is timelike for r&lt;2M.

 

[latentcorpse]

Then try doing that. It's much better to try to work it out before asking the question.



Hopefully you mean \phi(x^\mu)=y^\alpha



You should be able to use the metric to show that the radial vector is timelike for r&lt;2M.

Ok. I tried taking f=y^\alpha but then on the LHS, this canceled the d/dy i had and similarly on the RHS it gets pulled back to an x^\mu which cancels the d/dx we have?

This may be a stupid question but what is the form of the radial vector? Would it just be X=\frac{\partial}{\partial r} \Rightarrow X^\mu(0,1,0,0) in Schwarzschild coordinates?
If so, when I calculate g_{rr}X^rX^r I find it to be spacelike so I'm guessing that's wrong.
Moreover, how does showing a vector is timelike tell us it behaves like a time coordinate/that t doesn't behave like a time coordinate?

 

[fzero]

Ok. I tried taking f=y^\alpha but then on the LHS, this canceled the d/dy i had and similarly on the RHS it gets pulled back to an x^\mu which cancels the d/dx we have?

Leave f arbitrary, but note that you have (\partial/\partial y^\alpha )f(y^\alpha) on the LHS and (\partial/\partial x^\mu) f(y^\alpha) on the RHS. You want to use the chain rule on the RHS and equate coefficients.

This may be a stupid question but what is the form of the radial vector? Would it just be X=\frac{\partial}{\partial r} \Rightarrow X^\mu(0,1,0,0) in Schwarzschild coordinates?

Yes.

If so, when I calculate g_{rr}X^rX^r I find it to be spacelike so I'm guessing that's wrong.

You might want to show your work. You should find that it's spacelike for r&gt;2M and timelike for r&lt;2M.

Moreover, how does showing a vector is timelike tell us it behaves like a time coordinate/that t doesn't behave like a time coordinate?

How else would you decide that something is or isn't a time-coordinate?

 

[latentcorpse]

Leave f arbitrary, but note that you have (\partial/\partial y^\alpha )f(y^\alpha) on the LHS and (\partial/\partial x^\mu) f(y^\alpha) on the RHS. You want to use the chain rule on the RHS and equate coefficients.
[/QUOTE]
What happened to your \phi^*(f) on the RHS - you have just made it f, no?

You might want to show your work. You should find that it's spacelike for r&gt;2M and timelike for r&lt;2M.

Well in the notes it says to work with the Schwarzschild coordinates so I used the Schwarzschild metric. Now I know that the metric that covers the r<2M is the one defined using ingoing Eddington Finkelstein coordinates but, we know that if we're in the region r<2M and we change back from EF to Schwarz coordinates we should get the Schwarz metric back again (but now applied to the r<2M region)> Isn't this guaranteed by Birkhoff's theorem which tells us that Schwarzschild is unique? So I guess my question is, why does it not work if you use the Schwarzschild metric?

Anyway, let's say I try with the metric in EF coords...

we have g_{\mu \nu} X^\mu X^\nu with X^\mu=(0,1,0,0)
so we would have g_{\mu \nu} X^\mu X^\nu=g_{rr}X^rX^r=g_{rr}=0?

I'm sure I've done this before. I can't understand what I am doing wrong!

How else would you decide that something is or isn't a time-coordinate?

[/QUOTE]
I assume that checking it's timelike is the only way? Presumably then t would be spacelike in this region? How would we prove this since there is no t coordinate in the EF coord metric?

Thanks again.

 

[fzero]

What happened to your \phi^*(f) on the RHS - you have just made it f, no?

It's been evaluated. \phi^*(f) takes x^\mu\in M to f(y^\alpha).

Well in the notes it says to work with the Schwarzschild coordinates so I used the Schwarzschild metric. Now I know that the metric that covers the r<2M is the one defined using ingoing Eddington Finkelstein coordinates but, we know that if we're in the region r<2M and we change back from EF to Schwarz coordinates we should get the Schwarz metric back again (but now applied to the r<2M region)> Isn't this guaranteed by Birkhoff's theorem which tells us that Schwarzschild is unique?

Everything you say above is correct. The Schwarzschild coordinates are not complete because the r&gt;2M region does not connect to the r&lt;2M interior. However, as you say we can use ingoing EF coordinates to connect the two regions and transform back.

So I guess my question is, why does it not work if you use the Schwarzschild metric?

Anyway, let's say I try with the metric in EF coords...

we have g_{\mu \nu} X^\mu X^\nu with X^\mu=(0,1,0,0)
so we would have g_{\mu \nu} X^\mu X^\nu=g_{rr}X^rX^r=g_{rr}=0?

I'm sure I've done this before. I can't understand what I am doing wrong!

The radial vector is X^\mu=(0,1,0,0) in Schwarzschild coordinates, but not in EF.

The simplest way I'd work this out is by using the Schwarzschild coordinates for the region r&lt;2M. Then it's almost trivial to show that the radial and time coordinates get switched. One can complain that the coordinates are not valid past r=2M, but I don't see anything wrong with using EF to extend through r=2M and then transforming back.

One can obviously show the same thing in EF coordinates, it just seems like it's going to be a lot harder. One can see how it works qualitatively by looking at the light cones in the Finklestein diagram on pg 18 of Townsend's BH notes. It's actually a little bit complicated to work those lightcones out, but is sort of a useful exercise if your morals prevent you from believing my suggestion above.

I assume that checking it's timelike is the only way? Presumably then t would be spacelike in this region? How would we prove this since there is no t coordinate in the EF coord metric?

Thanks again.

 

[latentcorpse]

It's been evaluated. \phi^*(f) takes x^\mu\in M to f(y^\alpha).

So did you apply the LHS to y^\alpha seeing as this is something in N and the RHS to x^\mu since these are the coordinates on M? How is that allowed, surely we have to act with the same thing on both sides?

Everything you say above is correct. The Schwarzschild coordinates are not complete because the r&gt;2M region does not connect to the r&lt;2M interior. However, as you say we can use ingoing EF coordinates to connect the two regions and transform back.



The radial vector is X^\mu=(0,1,0,0) in Schwarzschild coordinates, but not in EF.

The simplest way I'd work this out is by using the Schwarzschild coordinates for the region r&lt;2M. Then it's almost trivial to show that the radial and time coordinates get switched. One can complain that the coordinates are not valid past r=2M, but I don't see anything wrong with using EF to extend through r=2M and then transforming back.

One can obviously show the same thing in EF coordinates, it just seems like it's going to be a lot harder. One can see how it works qualitatively by looking at the light cones in the Finklestein diagram on pg 18 of Townsend's BH notes. It's actually a little bit complicated to work those lightcones out, but is sort of a useful exercise if your morals prevent you from believing my suggestion above.

So you're telling me it's easier to work it out in Schwarzschild coords where X^\mu=(0,1,0,0). But it doesn't make sense to use the EF metric with this (since the EF metric involves v which isn't a Schwarzschild cooridnate) so I would have to use the Schwarzschild metric but then I have the same problem as in my last post? The Schwarzschild metric is going to be the same in the region r<2M and the region r>2M, isn't it? We only introduce the EF coordinates for the purposes of showing we can smoothly extend through the r=2M coordinate singularity, don't we?

 

[fzero]

So did you apply the LHS to y^\alpha seeing as this is something in N and the RHS to x^\mu since these are the coordinates on M? How is that allowed, surely we have to act with the same thing on both sides?

I'd suggest you reread pg 67 in detail again, because I can't explain it any better without reproducing all of that again or drawing pictures. If we have a function A: M \rightarrow T and a function B:N\rightarrow T, it's perfectly acceptable to say that A(x) = B(y) as an equality on T, even though these act on different spaces. A map \phi:M\rightarrow N can be used to explore such an equality and that is the whole point of the discussion.

So you're telling me it's easier to work it out in Schwarzschild coords where X^\mu=(0,1,0,0). But it doesn't make sense to use the EF metric with this (since the EF metric involves v which isn't a Schwarzschild cooridnate) so I would have to use the Schwarzschild metric but then I have the same problem as in my last post? The Schwarzschild metric is going to be the same in the region r<2M and the region r>2M, isn't it? We only introduce the EF coordinates for the purposes of showing we can smoothly extend through the r=2M coordinate singularity, don't we?

I'm saying first that you would have to work out the transformation of X^\mu=(0,1,0,0) under the EF coordinate change. If you did this, you could presumably use the EF coordinates to compute the inner product.

Secondly, that the metric is the "same" does not mean that X^\mu X_\mu has the same value in both regions. Just try to compute it and see for yourself.

 

[latentcorpse]

I'd suggest you reread pg 67 in detail again, because I can't explain it any better without reproducing all of that again or drawing pictures. If we have a function A: M \rightarrow T and a function B:N\rightarrow T, it's perfectly acceptable to say that A(x) = B(y) as an equality on T, even though these act on different spaces. A map \phi:M\rightarrow N can be used to explore such an equality and that is the whole point of the discussion.

Think I have it:
( \phi_*(X))^\alpha \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)} f(y^\beta) = X^\mu \left( \frac{\partial}{\partial x^\mu} \right)_p (\phi^* f)(x^\nu)
( \phi_*(X))^\alpha \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)} f(y^\beta) = X^\mu \left( \frac{\partial}{\partial x^\mu} \right)_p f(y^\beta)
using (\phi^*f)(x^\nu)=f(\phi(x^\nu))=f(y^\beta)
so ( \phi_*(X))^\alpha = X^\mu \left( \frac{\partial y^\alpha}{\partial x^\mu} \right)_p

One little problem though is that in the 2nd last line I had the d/dy evaluated at phi(p) but then when I cross multiplied I now miraculously have it evaluated at p. How does that work?

I'm saying first that you would have to work out the transformation of X^\mu=(0,1,0,0) under the EF coordinate change. If you did this, you could presumably use the EF coordinates to compute the inner product.

Ok. So EF coords require us to find v=t+r_*=t+r+2M \log{ | \frac{r}{2M}-1 |}
Taking t=0 and r=1 I find
v=1+2M \log{ | \frac{1-2M}{2M} | }
So X^\mu=(1+2M \log{ | \frac{1-2M}{2M} | },1,0,0)
Surely those components are wrong?

Secondly, that the metric is the "same" does not mean that X^\mu X_\mu has the same value in both regions. Just try to compute it and see for yourself.

But what about in the same region? If we are in r<2M then surely the X^\mu X_\mu has the same value whether we compute wrt the EF metric (with X^\mu expressed in EF coords) or if we compute wrt the Schwarzschild metric (with X^\mu expressed in Schwarzschild coords). But clearly evaluating in the Schwarzschild gives it as being spacelike?

 

[fzero]

Think I have it:
( \phi_*(X))^\alpha \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)} f(y^\beta) = X^\mu \left( \frac{\partial}{\partial x^\mu} \right)_p (\phi^* f)(x^\nu)
( \phi_*(X))^\alpha \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)} f(y^\beta) = X^\mu \left( \frac{\partial}{\partial x^\mu} \right)_p f(y^\beta)
using (\phi^*f)(x^\nu)=f(\phi(x^\nu))=f(y^\beta)
so ( \phi_*(X))^\alpha = X^\mu \left( \frac{\partial y^\alpha}{\partial x^\mu} \right)_p

One little problem though is that in the 2nd last line I had the d/dy evaluated at phi(p) but then when I cross multiplied I now miraculously have it evaluated at p. How does that work?

I'd say that

\left( \frac{\partial}{\partial x^\mu} \right)_p = \left( \frac{\partial y^\alpha}{\partial x^\mu} \right)_p \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)}

Ok. So EF coords require us to find v=t+r_*=t+r+2M \log{ | \frac{r}{2M}-1 |}
Taking t=0 and r=1 I find
v=1+2M \log{ | \frac{1-2M}{2M} | }
So X^\mu=(1+2M \log{ | \frac{1-2M}{2M} | },1,0,0)
Surely those components are wrong?

That's not how the components of a vector transform under a change of coordinates.

But what about in the same region? If we are in r<2M then surely the X^\mu X_\mu has the same value whether we compute wrt the EF metric (with X^\mu expressed in EF coords) or if we compute wrt the Schwarzschild metric (with X^\mu expressed in Schwarzschild coords). But clearly evaluating in the Schwarzschild gives it as being spacelike?

It is not spacelike. This would be immediately apparent if you would actually do the computation.

 

[latentcorpse]

That's not how the components of a vector transform under a change of coordinates.

X&#039;^\mu = A^\mu{}_\nu X^\nu
where A^\mu{}_\nu = \frac{\partial x&#039;^\mu}{\partial x^\nu}

So what is this A matrix?

I find it has non zero cpts A=\text{diag} ( \frac{\partial v}{\partial t} , 1 , 1 , 1)
but then v=t+r+2M \log{ | \frac{r}{2M}-1 |} \Rightarrow \frac{\partial v}{ \partial t}=1

but then we just have a identity matrix which doesn't seem very useful?

It is not spacelike. This would be immediately apparent if you would actually do the computation.

So when I get the calculation to work, I should find they should give identical results (hopefully)?

 

[fzero]

X&#039;^\mu = A^\mu{}_\nu X^\nu
where A^\mu{}_\nu = \frac{\partial x&#039;^\mu}{\partial x^\nu}

So what is this A matrix?

I find it has non zero cpts A=\text{diag} ( \frac{\partial v}{\partial t} , 1 , 1 , 1)
but then v=t+r+2M \log{ | \frac{r}{2M}-1 |} \Rightarrow \frac{\partial v}{ \partial t}=1

but then we just have a identity matrix which doesn't seem very useful?

\partial v/\partial r is non-vanishing too.

So when I get the calculation to work, I should find they should give identical results (hopefully)?

I expect so.

 

[latentcorpse]

\partial v/\partial r is non-vanishing too.
I expect so.

I'm finding X&#039;^\mu= \begin{pmatrix} 2 + \frac{2M}{r-2M} \\ 1 \\ 1 \\ 1 \end{pmatrix}

But things don't really simplify much when I work out g_{\mu \nu} X&#039;^\mu X&#039;^\nu?Also, in Chapter 34, he says that outgoing radial null geodesics inside the region r<2M will have decreasing r and therefore reach the r=0 singularity in finite affine parameter. This makes sense to me physically but I don't see how the maths backs it up. We have shown that \frac{dr}{d \tau}=1 for these radial null geodesics and so wouldn't they have increasing r i.e. be able to escape from the black hole?

 

[fzero]

Also, in Chapter 34, he says that outgoing radial null geodesics inside the region r<2M will have decreasing r and therefore reach the r=0 singularity in finite affine parameter. This makes sense to me physically but I don't see how the maths backs it up. We have shown that \frac{dr}{d \tau}=1 for these radial null geodesics and so wouldn't they have increasing r i.e. be able to escape from the black hole?

If you just blindly apply (259), then you'll find that for r&lt;2M coordinate time is decreasing along the affine parameter. When you fix this, you'll find that the radius is now decreasing. Alternatively, you can simply plot the geodesics (268) which is apparently done for you in the text (the version I downloaded doesn't display any figures).

 

[latentcorpse]

If you just blindly apply (259), then you'll find that for r&lt;2M coordinate time is decreasing along the affine parameter. When you fix this, you'll find that the radius is now decreasing. Alternatively, you can simply plot the geodesics (268) which is apparently done for you in the text (the version I downloaded doesn't display any figures).

Yeah we have the plot which does make it clear. I'm just trying to convince myself with the maths. So, the expression for \frac{dt}{d \tau} makes it clear that for r<2M, we have the coordinate time decreasing along the affine parameter.

What do you mean by fix it?

The only way I can think of to see what the radial coordinate is doing along the geodesic is to find its rate of change wrt the affine parameter and that is given as \frac{dr}{d \tau}=1 so why does that not mean the geodesic will have increasing r?

And do you have any advice about my attempt to do that calculation in the last post? Am I going about it on the right lines?

Thanks again!

 

[fzero]

Yeah we have the plot which does make it clear. I'm just trying to convince myself with the maths. So, the expression for \frac{dt}{d \tau} makes it clear that for r<2M, we have the coordinate time decreasing along the affine parameter.

What do you mean by fix it?

Change the sign of \tau.

And do you have any advice about my attempt to do that calculation in the last post? Am I going about it on the right lines?

I expect that your expression for U^\mu is wrong, since I don't think the angular components change. I think you chose the wrong sign in the log too.

 

[latentcorpse]

Change the sign of \tau.

Why do we do that?
Surely \tau, being the affine parameter, can only increase?

I expect that your expression for U^\mu is wrong, since I don't think the angular components change. I think you chose the wrong sign in the log too.

Ok. so we had X&#039;^\mu = A^\mu_\nu X^\nu
where A^\mu _\nu = \frac{\partial x^\mu}{\partial x^\nu}

Now our sets of coords are (v,r,theta,phi) and (t,r,theta,phi)

So X&#039;^1=X&#039;^2=X&#039;^3=1 (I'm hoping my X&#039;^1 component is ok there. I don't think r in EF varies with t, theta or phi so I think it's ok.

So we need to work out \frac{\partial v}{\partial x^\nu}
but v=t+r_*=t+r+2M \log{| \frac{r}{2M}-1|}
so \frac{\partial v}{\partial t}=1 and \frac{\partial v}{\partial r} = 1 +\frac{2M}{\frac{r}{2M}-1} \frac{1}{2M}=1+\frac{2M}{r-2M}

so presumably these add to give \frac{\partial v}{\partial x^\nu}=2 + \frac{2M}{r-2M}?


One small problem I encountered whilst reading is why in the 1st line of (35) do we only vary the derivative terms and not the metric outside the brackets?

Cheers.

 

[fzero]

Why do we do that?
Surely \tau, being the affine parameter, can only increase?

You're free to choose the affine parameter however you want. If you're choosing it as a proper time, you'd generally want the direction to go the same way as the coordinate time, if there's no reason why it shouldn't.

Ok. so we had X&#039;^\mu = A^\mu_\nu X^\nu
where A^\mu _\nu = \frac{\partial x^\mu}{\partial x^\nu}

Now our sets of coords are (v,r,theta,phi) and (t,r,theta,phi)

So X&#039;^1=X&#039;^2=X&#039;^3=1 (I'm hoping my X&#039;^1 component is ok there. I don't think r in EF varies with t, theta or phi so I think it's ok.

You were originally transforming a vector X^\mu = (0,1,0,0).

So we need to work out \frac{\partial v}{\partial x^\nu}
but v=t+r_*=t+r+2M \log{| \frac{r}{2M}-1|}
so \frac{\partial v}{\partial t}=1 and \frac{\partial v}{\partial r} = 1 +\frac{2M}{\frac{r}{2M}-1} \frac{1}{2M}=1+\frac{2M}{r-2M}

so presumably these add to give \frac{\partial v}{\partial x^\nu}=2 + \frac{2M}{r-2M}?

\log{\left| \frac{r}{2M}-1\right|} = \log\left(1-\frac{r}{2M}\right)

when r&lt;2M.

One small problem I encountered whilst reading is why in the 1st line of (35) do we only vary the derivative terms and not the metric outside the brackets?

Cheers.

I've no idea what equation (35) you're referring to. In my copy of Reall's notes, (35) is a product between a covector and a vector.

 

[latentcorpse]

You're free to choose the affine parameter however you want. If you're choosing it as a proper time, you'd generally want the direction to go the same way as the coordinate time, if there's no reason why it shouldn't.

This just seems like we're cheating almost. Surely we're also free to take \tau increasing in which case we still have the result \frac{dr}{d \tau}=+1. No?

You were originally transforming a vector X^\mu = (0,1,0,0).

Ok. So I think I finally have it.

I find X&#039;^\mu=(1-\frac{2M}{2M-r},1,0,0)

and then g(X,X)= \frac{4M^2}{(2M-r)r} - \frac{6M^3}{(2M-r)r^2} + 1 - \frac{2M}{2M-r}
I find this can be rewritten as
g(X,X)=1-\frac{2M}{2M-r} \left( \left( \sqrt{3} \frac{M}{r} - \frac{1}{\sqrt{3}} \right)^2 + \frac{8}{9} \right)

Now everything in the brackets is positive and \frac{2M}{2M-r}&gt;1 for r&lt;2M so we get g(X,X)&lt;0
Does this look ok?

How about doing the same thing in Schwarzschild coordinates? What did I do wrong the 1st time I tried this and ended up finding r to be spacelike. I'm curious as to how they agree!

I've no idea what equation (35) you're referring to. In my copy of Reall's notes, (35) is a product between a covector and a vector.

Sorry! I meant (355). The first line of that. Why does the \delta not hit the metric in front of the brackets?

Cheers.

 

[fzero]

This just seems like we're cheating almost. Surely we're also free to take \tau increasing in which case we still have the result \frac{dr}{d \tau}=+1. No?

If you get the sign of \tau wrong, you'll find that r increases as you travel backwards in coordinate time on the outgoing geodesics.

Ok. So I think I finally have it.

I find X&#039;^\mu=(1-\frac{2M}{2M-r},1,0,0)

and then g(X,X)= \frac{4M^2}{(2M-r)r} - \frac{6M^3}{(2M-r)r^2} + 1 - \frac{2M}{2M-r}
I find this can be rewritten as
g(X,X)=1-\frac{2M}{2M-r} \left( \left( \sqrt{3} \frac{M}{r} - \frac{1}{\sqrt{3}} \right)^2 + \frac{8}{9} \right)

Now everything in the brackets is positive and \frac{2M}{2M-r}&gt;1 for r&lt;2M so we get g(X,X)&lt;0
Does this look ok?

How about doing the same thing in Schwarzschild coordinates? What did I do wrong the 1st time I tried this and ended up finding r to be spacelike. I'm curious as to how they agree!

In Schwarzschild coordinates you find

- \frac{2M}{2M-r}

so there's probably something wrong with your derivation above.

Sorry! I meant (355). The first line of that. Why does the \delta not hit the metric in front of the brackets?

Cheers.

You should reread the remarks above (355) about the normal coordinates he's using. When you act on the inverse metric, you find terms like

\delta g^{\mu\sigma} \left( g_{\sigma\nu,\rho} +\cdots \right), ~~(*)

but these all vanish because he takes g_{\sigma\nu,\rho} =0. If you used general coordinates, the effect of the terms (*) would be to generate the covariant derivatives in the 2nd line of (355).

 

[latentcorpse]

If you get the sign of \tau wrong, you'll find that r increases as you travel backwards in coordinate time on the outgoing geodesics.

I'm still confused as to what's going on here...

In Schwarzschild coordinates you find

- \frac{2M}{2M-r}

so there's probably something wrong with your derivation above.

Here's what I did:

v=t+r+2M \log{1-\frac{r}{2M}}
\frac{\partial v}{\partial r} = 1 + \frac{2M}{1-\frac{r}{2M}} ( - \frac{1}{2M}) = 1 - \frac{1}{1-\frac{r}{2M}} = 1 - \frac{2M}{2M-r}
so X&#039;^\mu=(1 - \frac{2M}{2M-r},1,0,0)

so g(X,X)=\begin{pmatrix} 1-\frac{2M}{2M-r} &amp; 1 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} - (1-\frac{2M}{r}) &amp; 1 &amp; 0 &amp; 0 \\ 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} 1 - \frac{2M}{2M-r} \\ 1 \\ 0 \\ 0 \end{pmatrix}
=\begin{pmatrix} 1-\frac{2M}{2M-r} &amp; 1 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} - ( 1 - \frac{2M}{r} )( 1 - \frac{2M}{2M-r} ) + 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}
=\begin{pmatrix} 1-\frac{2M}{2M-r} &amp; 1 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} \frac{2M}{2M-r}+\frac{2M}{r} - \frac{4M^2}{(2M-r)r} \\ 1 \\ 0 \\ 0 \end{pmatrix}
= (1-\frac{2M}{2M-r} ) ( \frac{2M}{2M-r}+\frac{2M}{r} - \frac{4M^2}{(2M-r)r})+1
=\frac{2M}{2M-r} + \frac{2M}{r} - \frac{4M^2}{(2M-r)r} - \frac{4M^2}{(2M-r)^2} - \frac{4M^2}{(2M-r)r} + \frac{8M^3}{(2M-r)^2r}+1

This is slightly different from what I got last time (I realized I made a mistake) but it still isn't right...

Also, I tried to show (399):

g_{ab}u^au^b = - \left( \frac{dt}{d \tau} \right)^2 + L^2 \cosh^2{\frac{t}{L}} \left( \frac{\partial \chi}{\partial t} \right)^2
=-\left( \frac{\partial t}{\partial \tau} \right)^2 + L^2 \cosh^2{\frac{t}{L}} \left( \frac{1}{1-e^{2t}} e^t \left( \frac{dt}{d \tau} \right)^2
=-\left( \frac{dt}{d \tau} \right)^2 + \frac{L^2}{2} \left( e^{t/L} + e^{-t/L} \right) \frac{2 e^{-t}e^t}{e^{-t}-e^t} \left( \frac{dt}{d \tau} \right)^2

but that second term gives a cosh/sinh and also the arguments are different: one is t/L and the other is just t so I don't see how I'm going to get them to cancel?

 

[fzero]

I had a small mistake, the Schwarzschild coordinate result is

\frac{1}{1-\frac{2M}{r}} = -\frac{r}{2M-r} .

There are some simplifications that make your calculation simpler:

X&#039;^\mu=(- \frac{r}{2M-r},1,0,0)

g(X,X)=\begin{pmatrix} -\frac{r}{2M-r} &amp; 1 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} \frac{2M-r}{r} &amp; 1 &amp; 0 &amp; 0 \\ 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} - \frac{r}{2M-r} \\ 1 \\ 0 \\ 0 \end{pmatrix}

If you expand the above out, you should find

g(X,X)=- \frac{r}{2M-r} .

 

[latentcorpse]

I had a small mistake, the Schwarzschild coordinate result is

\frac{1}{1-\frac{2M}{r}} = -\frac{r}{2M-r} .

There are some simplifications that make your calculation simpler:

X&#039;^\mu=(- \frac{r}{2M-r},1,0,0)

g(X,X)=\begin{pmatrix} -\frac{r}{2M-r} &amp; 1 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} \frac{2M-r}{r} &amp; 1 &amp; 0 &amp; 0 \\ 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 \end{pmatrix} \begin{pmatrix} - \frac{r}{2M-r} \\ 1 \\ 0 \\ 0 \end{pmatrix}

If you expand the above out, you should find

g(X,X)=- \frac{r}{2M-r} .

Great! So if you were asked to show this result (that r behaves as the time coordiante), you would just say that all we need to do is demonstrate d/dr is timelike. It's then easiest to show this in Schwarzschild cooridnates isn't it as we just take the components of X as (0,1,0,0) and use the Schwarzschild metric to get your quoted result.

I think I have figured everything else out except the changing the tau thing on the outgoing geodesic in the r<2M region.

Thans.

 

[fzero]

Great! So if you were asked to show this result (that r behaves as the time coordiante), you would just say that all we need to do is demonstrate d/dr is timelike. It's then easiest to show this in Schwarzschild cooridnates isn't it as we just take the components of X as (0,1,0,0) and use the Schwarzschild metric to get your quoted result.

I was saying that the Schwarzschild calculation was easiest all along. I just canceled off the wrong bit when I wrote the answer down.

I think I have figured everything else out except the changing the tau thing on the outgoing geodesic in the r<2M region.

Thans.

The way to think of it is the following. We know that geodesics satisfy

\frac{dt}{d\tau} = \frac{r}{r-2M} , ~~ \frac{dr}{d\tau} = \pm 1,

but it's up to us to figure out which of \pm is ingoing or outgoing.

For r&gt;2M, dt/d\tau &gt;0, so increasing \tau means advancing into the future coordinate time. Then dr/d\tau &lt;0 is ingoing.

For r&lt;2M, dt/d\tau &lt;0, so decreasing \tau means advancing into the future coordinate time and dr/d\tau &gt;0 is ingoing. It's better to send \tau \rightarrow -\tau in this region so that we have the same inequalities as for r&gt;2M, but it's not necessary if you study the behavior along geodesics. It also makes sense to compare the conclusions with the actual solutions for geodesics.

 

[latentcorpse]

I was saying that the Schwarzschild calculation was easiest all along. I just canceled off the wrong bit when I wrote the answer down.



The way to think of it is the following. We know that geodesics satisfy

\frac{dt}{d\tau} = \frac{r}{r-2M} , ~~ \frac{dr}{d\tau} = \pm 1,

but it's up to us to figure out which of \pm is ingoing or outgoing.

For r&gt;2M, dt/d\tau &gt;0, so increasing \tau means advancing into the future coordinate time. Then dr/d\tau &lt;0 is ingoing.

For r&lt;2M, dt/d\tau &lt;0, so decreasing \tau means advancing into the future coordinate time and dr/d\tau &gt;0 is ingoing. It's better to send \tau \rightarrow -\tau in this region so that we have the same inequalities as for r&gt;2M, but it's not necessary if you study the behavior along geodesics. It also makes sense to compare the conclusions with the actual solutions for geodesics.

Thanks. At the bottom of p119, he says that de Sitter spacetime has constant curvature 1/L^2. How do we work this out? Do we have to go to (382) and work out a component of the riemann tensor. that seems quite a lot of work.

 

[fzero]

Thanks. At the bottom of p119, he says that de Sitter spacetime has constant curvature 1/L^2. How do we work this out? Do we have to go to (382) and work out a component of the riemann tensor. that seems quite a lot of work.

It's enough in this case to show that the Ricci tensor satisfies R_{\mu\nu} = c g_{\mu\nu} for some constant c.

 

[latentcorpse]

It's enough in this case to show that the Ricci tensor satisfies R_{\mu\nu} = c g_{\mu\nu} for some constant c.

but to work out the ricci tensor won't i have to work out the riemann tensor anyway?

also if R_{abcd}=Kg_{a[c}g_{b]d} then R_{ab}=3Kg_{ab} so I'd have to divide that value by 3.

 

[fzero]

but to work out the ricci tensor won't i have to work out the riemann tensor anyway?

also if R_{abcd}=Kg_{a[c}g_{b]d} then R_{ab}=3Kg_{ab} so I'd have to divide that value by 3.

I forgot that the Riemann tensor has that form, which makes it easier. There is a lot that ends up requiring computing the Riemann tensor and there's not much that you can do about it other than learn how to use Mathematica, Maple or something else.

 

[latentcorpse]

I forgot that the Riemann tensor has that form, which makes it easier. There is a lot that ends up requiring computing the Riemann tensor and there's not much that you can do about it other than learn how to use Mathematica, Maple or something else.

I'm not sure what you mean there. You say this form makes it easier but then also that it's a hard enough calculation to merit maple or the likes! is this dooable by picking a particular component and trying and working it out or is it something that would take so long that they would never be able to ask it in an exam?

 

[fzero]

I'm not sure what you mean there. You say this form makes it easier but then also that it's a hard enough calculation to merit maple or the likes! is this dooable by picking a particular component and trying and working it out or is it something that would take so long that they would never be able to ask it in an exam?

If you wanted to prove that the Riemann tensor was proportional to that combination of metric components, you'd need to compute all independent components. When I said easier, I mean recognizing that saves having to compute the trace that gives the Ricci tensor. It doesn't sound like a useful exercise to ask on an exam, since it only tests grunt work and not concepts.

 

[latentcorpse]

If you wanted to prove that the Riemann tensor was proportional to that combination of metric components, you'd need to compute all independent components. When I said easier, I mean recognizing that saves having to compute the trace that gives the Ricci tensor. It doesn't sound like a useful exercise to ask on an exam, since it only tests grunt work and not concepts.

Hmm. But how could working out the Ricci tensor be easier than getting the riemann tensor? surely we'd have to get riemann tensor first and then contract? I might be missing what you're saying here.

Then on p114, he has an example where he says the metric ds^2=dx^2+d \theta^2 is flat and hence constant curvature. Well it's clearly a cylinder which we know to be flat but how do you actually show this? Would you pull back the euclidean metric?

 

[fzero]

Hmm. But how could working out the Ricci tensor be easier than getting the riemann tensor? surely we'd have to get riemann tensor first and then contract? I might be missing what you're saying here.

I'm saying that if you compute the Riemann tensor and noticed that symmetry, then you wouldn't have to compute the Ricci tensor. Remember that constant curvature means that the curvature scalar is a constant, so in general we need to compute two traces of the Riemann tensor.

Then on p114, he has an example where he says the metric ds^2=dx^2+d \theta^2 is flat and hence constant curvature. Well it's clearly a cylinder which we know to be flat but how do you actually show this? Would you pull back the euclidean metric?

It's easiest just to note that since the metric is constant in these coordinates, the Christoffel symbols vanish and so does the Riemann tensor. Hence the scalar curvature vanishes in these coordinates and, since it is an invariant, must vanish in any coordinates.

 

[latentcorpse]

I'm saying that if you compute the Riemann tensor and noticed that symmetry, then you wouldn't have to compute the Ricci tensor. Remember that constant curvature means that the curvature scalar is a constant, so in general we need to compute two traces of the Riemann tensor.



It's easiest just to note that since the metric is constant in these coordinates, the Christoffel symbols vanish and so does the Riemann tensor. Hence the scalar curvature vanishes in these coordinates and, since it is an invariant, must vanish in any coordinates.

Great. I am trying this question (question 2) though:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2009/Paper54.pdf

So the first two bits are easy enough. Now for the angular momentum bit.

I didn't really get the whole point of taking two separate surfaces - why do we do that?

Anyway I find (taking F^{ab}=\xi^{a;b}

J_S=-\frac{1}{8 \pi} \int_\Omega \xi^{a;b}{}_{;b} d^3 \Sigma_a
But from the previous part we have \xi^{a;b}{}_{;b}=-R^{ab}\xi_b
Now it says we have to use the Einstein equations.
Seeing as we are in an asymptotically flat region of spacetime, I took this to mean that T_{ab}=0 as we are far away from matter.
So the einstein eqns give \xi^{a;b}{}_{;b}=-\frac{1}{2} R g^{ab} \xi_b
So we get
J_S = \frac{1}{16 \pi} \int_\Omega R \xi^a d^3 \Sigma_a
So if we use the other identity given with X^a=R \xi^a we get
J_S=\frac{1}{16 \pi} \int_V (R \xi^a)_{;a} d^4 \Sigma

Is this going right? Presumably I now need to use the fact that \xi=\frac{\partial}{\partial \phi}?

 

[fzero]

Great. I am trying this question (question 2) though:
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2009/Paper54.pdf

So the first two bits are easy enough. Now for the angular momentum bit.

I didn't really get the whole point of taking two separate surfaces - why do we do that?

I don't think that you can guarantee that the boundary of a generic 3-manifold \Omega is a single spacelike hypersurface. Just consider the case where some of \Omega is in the casual past of another part.

Anyway I find (taking F^{ab}=\xi^{a;b}

J_S=-\frac{1}{8 \pi} \int_\Omega \xi^{a;b}{}_{;b} d^3 \Sigma_a
But from the previous part we have \xi^{a;b}{}_{;b}=-R^{ab}\xi_b
Now it says we have to use the Einstein equations.
Seeing as we are in an asymptotically flat region of spacetime, I took this to mean that T_{ab}=0 as we are far away from matter.
So the einstein eqns give \xi^{a;b}{}_{;b}=-\frac{1}{2} R g^{ab} \xi_b
So we get
J_S = \frac{1}{16 \pi} \int_\Omega R \xi^a d^3 \Sigma_a
So if we use the other identity given with X^a=R \xi^a we get
J_S=\frac{1}{16 \pi} \int_V (R \xi^a)_{;a} d^4 \Sigma

Is this going right? Presumably I now need to use the fact that \xi=\frac{\partial}{\partial \phi}?

If T_{ab}=0, you can use the trace of the Einstein equation to show that R=0.

 

[latentcorpse]

I don't think that you can guarantee that the boundary of a generic 3-manifold \Omega is a single spacelike hypersurface. Just consider the case where some of \Omega is in the casual past of another part.

So will this mean that all the rest of my work is wrong?

If T_{ab}=0, you can use the trace of the Einstein equation to show that R=0.

I thought about this but surely it can't be that easy. We haven't made use of the fact that \xi=\partial_\phi which seemed to be central to the question!Also, in the notes, could you give me some advice on how to derive 415?

And on p128 he says that the universe will expand forever even though the rate of expansion is decreasing (\ddot{a}&lt;0). But surely if \ddot{a}&lt;0 then at some point \dot{a} will change from being positive and become negative and then we will have a contracting universe?

And is there a difference between "the past light cone at p" and the "particle horizon at p"?

Thanks.