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Help with Diffeomorphisms

by latentcorpse
Tags: diffeomorphisms
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fzero
#19
Apr13-11, 12:40 PM
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Quote Quote by latentcorpse View Post
Ok. I tried taking [itex]f=y^\alpha[/itex] but then on the LHS, this cancelled the d/dy i had and similarly on the RHS it gets pulled back to an [itex]x^\mu[/itex] which cancels the d/dx we have?
Leave [tex]f[/tex] arbitrary, but note that you have [tex](\partial/\partial y^\alpha )f(y^\alpha)[/tex] on the LHS and [tex](\partial/\partial x^\mu) f(y^\alpha)[/tex] 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 [itex]X=\frac{\partial}{\partial r} \Rightarrow X^\mu(0,1,0,0)[/itex] in Schwarzschild coordinates?
Yes.

If so, when I calculate [itex]g_{rr}X^rX^r[/itex] 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 [tex]r>2M[/tex] and timelike for [tex]r<2M[/tex].

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
#20
Apr13-11, 01:24 PM
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Leave [tex]f[/tex] arbitrary, but note that you have [tex](\partial/\partial y^\alpha )f(y^\alpha)[/tex] on the LHS and [tex](\partial/\partial x^\mu) f(y^\alpha)[/tex] on the RHS. You want to use the chain rule on the RHS and equate coefficients.
[/QUOTE]
What happened to your [itex]\phi^*(f)[/itex] on the RHS - you have just made it [itex]f[/itex], no?


Quote Quote by fzero View Post
You might want to show your work. You should find that it's spacelike for [tex]r>2M[/tex] and timelike for [tex]r<2M[/tex].
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 [itex]g_{\mu \nu} X^\mu X^\nu[/itex] with [itex]X^\mu=(0,1,0,0)[/itex]
so we would have [itex]g_{\mu \nu} X^\mu X^\nu=g_{rr}X^rX^r=g_{rr}=0[/itex]?

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


Quote Quote by fzero View Post
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
#21
Apr13-11, 02:04 PM
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Quote Quote by latentcorpse View Post
What happened to your [itex]\phi^*(f)[/itex] on the RHS - you have just made it [itex]f[/itex], no?
It's been evaluated. [itex]\phi^*(f)[/itex] takes [tex]x^\mu\in M[/tex] to [tex]f(y^\alpha)[/tex].

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 [tex]r>2M[/tex] region does not connect to the [tex]r<2M[/tex] 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 [itex]g_{\mu \nu} X^\mu X^\nu[/itex] with [itex]X^\mu=(0,1,0,0)[/itex]
so we would have [itex]g_{\mu \nu} X^\mu X^\nu=g_{rr}X^rX^r=g_{rr}=0[/itex]?

I'm sure I've done this before. I can't understand what I am doing wrong!
The radial vector is [itex]X^\mu=(0,1,0,0)[/itex] in Schwarzschild coordinates, but not in EF.

The simplest way I'd work this out is by using the Schwarzschild coordinates for the region [tex]r<2M[/tex]. 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 [tex]r=2M[/tex], but I don't see anything wrong with using EF to extend through [tex]r=2M[/tex] 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
#22
Apr13-11, 02:56 PM
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Quote Quote by fzero View Post
It's been evaluated. [itex]\phi^*(f)[/itex] takes [tex]x^\mu\in M[/tex] to [tex]f(y^\alpha)[/tex].
So did you apply the LHS to [itex]y^\alpha[/itex] seeing as this is something in [itex]N[/itex] and the RHS to [itex]x^\mu[/itex] since these are the coordinates on [itex]M[/itex]? How is that allowed, surely we have to act with the same thing on both sides?

Quote Quote by fzero View Post
Everything you say above is correct. The Schwarzschild coordinates are not complete because the [tex]r>2M[/tex] region does not connect to the [tex]r<2M[/tex] interior. However, as you say we can use ingoing EF coordinates to connect the two regions and transform back.



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

The simplest way I'd work this out is by using the Schwarzschild coordinates for the region [tex]r<2M[/tex]. 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 [tex]r=2M[/tex], but I don't see anything wrong with using EF to extend through [tex]r=2M[/tex] 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 [itex]X^\mu=(0,1,0,0)[/itex]. But it doesn't make sense to use the EF metric with this (since the EF metric involves [itex]v[/itex] 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
#23
Apr13-11, 04:26 PM
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Quote Quote by latentcorpse View Post
So did you apply the LHS to [itex]y^\alpha[/itex] seeing as this is something in [itex]N[/itex] and the RHS to [itex]x^\mu[/itex] since these are the coordinates on [itex]M[/itex]? 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 [tex]A: M \rightarrow T[/tex] and a function [tex]B:N\rightarrow T[/tex], it's perfectly acceptable to say that [tex]A(x) = B(y)[/tex] as an equality on [tex]T[/tex], even though these act on different spaces. A map [tex]\phi:M\rightarrow N[/tex] 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 [itex]X^\mu=(0,1,0,0)[/itex]. But it doesn't make sense to use the EF metric with this (since the EF metric involves [itex]v[/itex] 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 [itex]X^\mu=(0,1,0,0)[/itex] 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 [itex]X^\mu X_\mu[/itex] has the same value in both regions. Just try to compute it and see for yourself.
latentcorpse
#24
Apr13-11, 04:50 PM
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Quote Quote by fzero View Post
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 [tex]A: M \rightarrow T[/tex] and a function [tex]B:N\rightarrow T[/tex], it's perfectly acceptable to say that [tex]A(x) = B(y)[/tex] as an equality on [tex]T[/tex], even though these act on different spaces. A map [tex]\phi:M\rightarrow N[/tex] can be used to explore such an equality and that is the whole point of the discussion.
Think I have it:
[itex]( \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)[/itex]
[itex]( \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)[/itex]
using [itex] (\phi^*f)(x^\nu)=f(\phi(x^\nu))=f(y^\beta)[/itex]
so [itex]( \phi_*(X))^\alpha = X^\mu \left( \frac{\partial y^\alpha}{\partial x^\mu} \right)_p [/itex]

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?

Quote Quote by fzero View Post
I'm saying first that you would have to work out the transformation of [itex]X^\mu=(0,1,0,0)[/itex] 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 [itex]v=t+r_*=t+r+2M \log{ | \frac{r}{2M}-1 |}[/itex]
Taking t=0 and r=1 I find
[itex]v=1+2M \log{ | \frac{1-2M}{2M} | }[/itex]
So [itex]X^\mu=(1+2M \log{ | \frac{1-2M}{2M} | },1,0,0)[/itex]
Surely those components are wrong?

Quote Quote by fzero View Post
Secondly, that the metric is the "same" does not mean that [itex]X^\mu X_\mu[/itex] 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 [itex]X^\mu X_\mu[/itex] has the same value whether we compute wrt the EF metric (with [itex]X^\mu[/itex] expressed in EF coords) or if we compute wrt the Schwarzschild metric (with [itex]X^\mu[/itex] expressed in Schwarzschild coords). But clearly evaluating in the Schwarzschild gives it as being spacelike?
fzero
#25
Apr13-11, 05:27 PM
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Quote Quote by latentcorpse View Post
Think I have it:
[itex]( \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)[/itex]
[itex]( \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)[/itex]
using [itex] (\phi^*f)(x^\nu)=f(\phi(x^\nu))=f(y^\beta)[/itex]
so [itex]( \phi_*(X))^\alpha = X^\mu \left( \frac{\partial y^\alpha}{\partial x^\mu} \right)_p [/itex]

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

[tex] \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)} [/tex]

Ok. So EF coords require us to find [itex]v=t+r_*=t+r+2M \log{ | \frac{r}{2M}-1 |}[/itex]
Taking t=0 and r=1 I find
[itex]v=1+2M \log{ | \frac{1-2M}{2M} | }[/itex]
So [itex]X^\mu=(1+2M \log{ | \frac{1-2M}{2M} | },1,0,0)[/itex]
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 [itex]X^\mu X_\mu[/itex] has the same value whether we compute wrt the EF metric (with [itex]X^\mu[/itex] expressed in EF coords) or if we compute wrt the Schwarzschild metric (with [itex]X^\mu[/itex] 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
#26
Apr13-11, 06:48 PM
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Quote Quote by fzero View Post
That's not how the components of a vector transform under a change of coordinates.
[itex]X'^\mu = A^\mu{}_\nu X^\nu[/itex]
where [itex]A^\mu{}_\nu = \frac{\partial x'^\mu}{\partial x^\nu}[/itex]

So what is this [itex]A[/itex] matrix?

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

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

Quote Quote by fzero View Post
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
#27
Apr13-11, 08:16 PM
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Quote Quote by latentcorpse View Post
[itex]X'^\mu = A^\mu{}_\nu X^\nu[/itex]
where [itex]A^\mu{}_\nu = \frac{\partial x'^\mu}{\partial x^\nu}[/itex]

So what is this [itex]A[/itex] matrix?

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

but then we just have a identity matrix which doesn't seem very useful?
[tex]\partial v/\partial r[/tex] 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
#28
Apr14-11, 04:43 AM
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Quote Quote by fzero View Post
[tex]\partial v/\partial r[/tex] is non-vanishing too.



I expect so.
I'm finding [itex]X'^\mu= \begin{pmatrix} 2 + \frac{2M}{r-2M} \\ 1 \\ 1 \\ 1 \end{pmatrix}[/itex]

But things don't really simplify much when I work out [itex]g_{\mu \nu} X'^\mu X'^\nu[/itex]?


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 [itex]\frac{dr}{d \tau}=1[/itex] for these radial null geodesics and so wouldn't they have increasing r i.e. be able to escape from the black hole?
fzero
#29
Apr14-11, 11:40 AM
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Quote Quote by latentcorpse View Post
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 [itex]\frac{dr}{d \tau}=1[/itex] 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 [tex]r<2M[/tex] 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
#30
Apr14-11, 11:54 AM
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Quote Quote by fzero View Post
If you just blindly apply (259), then you'll find that for [tex]r<2M[/tex] 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 [itex]\frac{dt}{d \tau}[/itex] 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 [itex]\frac{dr}{d \tau}=1[/itex] 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
#31
Apr14-11, 12:11 PM
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Quote Quote by latentcorpse View Post
Yeah we have the plot which does make it clear. I'm just trying to convince myself with the maths. So, the expression for [itex]\frac{dt}{d \tau}[/itex] 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 [tex]\tau[/tex].

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 [tex]U^\mu[/tex] is wrong, since I don't think the angular components change. I think you chose the wrong sign in the log too.
latentcorpse
#32
Apr14-11, 02:05 PM
P: 1,442
Quote Quote by fzero View Post
Change the sign of [tex]\tau[/tex].
Why do we do that????
Surely [itex]\tau[/itex], being the affine parameter, can only increase?



Quote Quote by fzero View Post
I expect that your expression for [tex]U^\mu[/tex] 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 [itex]X'^\mu = A^\mu_\nu X^\nu[/itex]
where [itex]A^\mu _\nu = \frac{\partial x^\mu}{\partial x^\nu}[/itex]

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

So [itex]X'^1=X'^2=X'^3=1[/itex] (I'm hoping my [itex]X'^1[/itex] 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 [itex]\frac{\partial v}{\partial x^\nu}[/itex]
but [itex]v=t+r_*=t+r+2M \log{| \frac{r}{2M}-1|}[/itex]
so [itex]\frac{\partial v}{\partial t}=1[/itex] and [itex]\frac{\partial v}{\partial r} = 1 +\frac{2M}{\frac{r}{2M}-1} \frac{1}{2M}=1+\frac{2M}{r-2M}[/itex]

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


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
#33
Apr14-11, 04:27 PM
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Quote Quote by latentcorpse View Post
Why do we do that????
Surely [itex]\tau[/itex], 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 [itex]X'^\mu = A^\mu_\nu X^\nu[/itex]
where [itex]A^\mu _\nu = \frac{\partial x^\mu}{\partial x^\nu}[/itex]

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

So [itex]X'^1=X'^2=X'^3=1[/itex] (I'm hoping my [itex]X'^1[/itex] 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 [tex]X^\mu = (0,1,0,0)[/tex].

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

so presumably these add to give [itex]\frac{\partial v}{\partial x^\nu}=2 + \frac{2M}{r-2M}[/itex]?
[tex]\log{\left| \frac{r}{2M}-1\right|} = \log\left(1-\frac{r}{2M}\right)[/tex]

when [tex]r<2M[/tex].

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
#34
Apr14-11, 06:49 PM
P: 1,442
Quote Quote by fzero View Post
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 [itex]\tau[/itex] increasing in which case we still have the result [itex]\frac{dr}{d \tau}=+1[/itex]. No?



Quote Quote by fzero View Post
You were originally transforming a vector [tex]X^\mu = (0,1,0,0)[/tex].
Ok. So I think I finally have it.

I find [itex]X'^\mu=(1-\frac{2M}{2M-r},1,0,0)[/itex]

and then [itex]g(X,X)= \frac{4M^2}{(2M-r)r} - \frac{6M^3}{(2M-r)r^2} + 1 - \frac{2M}{2M-r}[/itex]
I find this can be rewritten as
[itex]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)[/itex]

Now everything in the brackets is positive and [itex]\frac{2M}{2M-r}>1[/itex] for [itex]r<2M[/itex] so we get [itex]g(X,X)<0[/itex]
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 [itex]r[/itex] to be spacelike. I'm curious as to how they agree!


Quote Quote by fzero View Post
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 [itex]\delta[/itex] not hit the metric in front of the brackets?

Cheers.
fzero
#35
Apr14-11, 09:10 PM
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Quote Quote by latentcorpse View Post
This just seems like we're cheating almost. Surely we're also free to take [itex]\tau[/itex] increasing in which case we still have the result [itex]\frac{dr}{d \tau}=+1[/itex]. No?
If you get the sign of [tex]\tau[/tex] 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 [itex]X'^\mu=(1-\frac{2M}{2M-r},1,0,0)[/itex]

and then [itex]g(X,X)= \frac{4M^2}{(2M-r)r} - \frac{6M^3}{(2M-r)r^2} + 1 - \frac{2M}{2M-r}[/itex]
I find this can be rewritten as
[itex]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)[/itex]

Now everything in the brackets is positive and [itex]\frac{2M}{2M-r}>1[/itex] for [itex]r<2M[/itex] so we get [itex]g(X,X)<0[/itex]
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 [itex]r[/itex] to be spacelike. I'm curious as to how they agree!
In Schwarzschild coordinates you find

[tex] - \frac{2M}{2M-r}[/tex]

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

Sorry! I meant (355). The first line of that. Why does the [itex]\delta[/itex] 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

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

but these all vanish because he takes [tex] g_{\sigma\nu,\rho} =0[/tex]. If you used general coordinates, the effect of the terms (*) would be to generate the covariant derivatives in the 2nd line of (355).
latentcorpse
#36
Apr15-11, 03:44 AM
P: 1,442
Quote Quote by fzero View Post
If you get the sign of [tex]\tau[/tex] 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...

Quote Quote by fzero View Post
In Schwarzschild coordinates you find

[tex] - \frac{2M}{2M-r}[/tex]

so there's probably something wrong with your derivation above.
Here's what I did:

[itex]v=t+r+2M \log{1-\frac{r}{2M}}[/itex]
[itex]\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}[/itex]
so [itex]X'^\mu=(1 - \frac{2M}{2M-r},1,0,0)[/itex]

so [itex]g(X,X)=\begin{pmatrix} 1-\frac{2M}{2M-r} & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} - (1-\frac{2M}{r}) & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1 - \frac{2M}{2M-r} \\ 1 \\ 0 \\ 0 \end{pmatrix}[/itex]
[itex]=\begin{pmatrix} 1-\frac{2M}{2M-r} & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} - ( 1 - \frac{2M}{r} )( 1 - \frac{2M}{2M-r} ) + 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}[/itex]
[itex]=\begin{pmatrix} 1-\frac{2M}{2M-r} & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} \frac{2M}{2M-r}+\frac{2M}{r} - \frac{4M^2}{(2M-r)r} \\ 1 \\ 0 \\ 0 \end{pmatrix}[/itex]
[itex] = (1-\frac{2M}{2M-r} ) ( \frac{2M}{2M-r}+\frac{2M}{r} - \frac{4M^2}{(2M-r)r})+1[/itex]
[itex]=\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[/itex]

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

Also, I tried to show (399):

[itex]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[/itex]
[itex]=-\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[/itex]
[itex]=-\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[/itex]

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?


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