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

[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 $X'^\mu = A^\mu_\nu X^\nu$
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 $X'^1=X'^2=X'^3=1$ (I'm hoping my $X'^1$ 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 $X'^\mu = A^\mu_\nu X^\nu$
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 $X'^1=X'^2=X'^3=1$ (I'm hoping my $X'^1$ 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 $$r<2M$$.

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 $X'^\mu=(1-\frac{2M}{2M-r},1,0,0)$

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 $\frac{2M}{2M-r}>1$ for $r<2M$ so we get $g(X,X)<0$
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 $X'^\mu=(1-\frac{2M}{2M-r},1,0,0)$

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 $\frac{2M}{2M-r}>1$ for $r<2M$ so we get $g(X,X)<0$
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 $X'^\mu=(1 - \frac{2M}{2M-r},1,0,0)$

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

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:

$$X'^\mu=(- \frac{r}{2M-r},1,0,0)$$

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

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:

$$X'^\mu=(- \frac{r}{2M-r},1,0,0)$$

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

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 $$r>2M$$, $$dt/d\tau >0$$, so increasing $$\tau$$ means advancing into the future coordinate time. Then $$dr/d\tau <0$$ is ingoing.

For $$r<2M$$, $$dt/d\tau <0$$, so decreasing $$\tau$$ means advancing into the future coordinate time and $$dr/d\tau >0$$ is ingoing. It's better to send $$\tau \rightarrow -\tau$$ in this region so that we have the same inequalities as for $$r>2M$$, 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 $$r>2M$$, $$dt/d\tau >0$$, so increasing $$\tau$$ means advancing into the future coordinate time. Then $$dr/d\tau <0$$ is ingoing.

For $$r<2M$$, $$dt/d\tau <0$$, so decreasing $$\tau$$ means advancing into the future coordinate time and $$dr/d\tau >0$$ is ingoing. It's better to send $$\tau \rightarrow -\tau$$ in this region so that we have the same inequalities as for $$r>2M$$, 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 ($\ddot{a}<0$). But surely if $\ddot{a}<0$ 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.

 

[fzero]

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

Not necessarily, though you should be a bit more careful about setting things up. For example $$\xi^{a;b}$$ is not itself an antisymmetric 2-form.

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!

Well you do use the fact that it's a Killing vector. The fact that it's spacelike should also be tied to the fact that you're using a spacelike hypersurface.

Also, in the notes, could you give me some advice on how to derive 415?

The stress tensor for a perfect fluid is (175).

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

The solutions for $$a(\eta)$$ are (427). What you claim clearly doesn't happen for $$\eta>0$$ in the flat and open universes.

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

Thanks.

Yes, the horizon is composed of the boundary of the present positions of particles which were in casual contact with us at some time in the past. Again, I don't have the figures, but the one on p. 130 is probably relevant.

 

[latentcorpse]

Not necessarily, though you should be a bit more careful about setting things up. For example $$\xi^{a;b}$$ is not itself an antisymmetric 2-form.
Well you do use the fact that it's a Killing vector. The fact that it's spacelike should also be tied to the fact that you're using a spacelike hypersurface.

Surely from Killing's equation, $\nabla_a \xi_b = - \nabla_b \xi_a$ will be antisymmetric? Do I need to do anything about the two surfaces then?

The stress tensor for a perfect fluid is (175).

I have been trying to take the $\nu=0$ component of $T^{\mu \nu}{}_{; \mu}=0$ but it simply will not work! Is this the way I should go about it?

The solutions for $$a(\eta)$$ are (427). What you claim clearly doesn't happen for $$\eta>0$$ in the flat and open universes.

But just from a plot of sinh, it looks like a smiley face (i.e. positive second derivatie), no?

Yes, the horizon is composed of the boundary of the present positions of particles which were in casual contact with us at some time in the past. Again, I don't have the figures, but the one on p. 130 is probably relevant.

So the particle horizon is the boundary of the past light cone?

 

[fzero]

Surely from Killing's equation, $\nabla_a \xi_b = - \nabla_b \xi_a$ will be antisymmetric? Do I need to do anything about the two surfaces then?

OK that'll do. As for the two surfaces, you need to figure out what your surface $$\Omega$$ is. Is $$\partial \Omega$$ necessarily connected?

I have been trying to take the $\nu=0$ component of $T^{\mu \nu}{}_{; \mu}=0$ but it simply will not work! Is this the way I should go about it?

It looks like you need to compute a trace of one of the Christoffel symbols.

But just from a plot of sinh, it looks like a smiley face (i.e. positive second derivatie), no?

$$\ddot{a}$$ is the 2nd derivative w.r.t. coordinate time, not $$\eta$$. I can't see the figures, so I don't know if they should be helpful to you.

So the particle horizon is the boundary of the past light cone?

The figure is probably clearer than I would put into words. The horizon is a boundary on spacetime at our present expansion parameter. To determine the points that make up the horizon, you need to follow the worldlines of the particles that crossed our past light cone at some earlier time.

 

[latentcorpse]

OK that'll do. As for the two surfaces, you need to figure out what your surface $$\Omega$$ is. Is $$\partial \Omega$$ necessarily connected?

I thought S is the surface given in the formula for J_S. Then $\partial \Omega=S$ and $\Omega$ would just be the enclosed 3-sphere?

The figure is probably clearer than I would put into words. The horizon is a boundary on spacetime at our present expansion parameter. To determine the points that make up the horizon, you need to follow the worldlines of the particles that crossed our past light cone at some earlier time.

Well from the figure, it appears to me that the particle horizon actually is the boundary of the past light cone. But if this is so, why don't they just define it as this? This is why I'm having doubts about my interpretation. But I can't think of an example where they wouldn't be equal.

Also, for the $\nabla_\mu T^{\mu \nu}$ question, can I take $u^\mu=(1,0,0,0)$ as we are assuming a comoving observer. And apparently they have $u^\alpha=\delta^\alpha{}_0$.
However, even though our notes say we can do this for a comoving observer, I find that
$\nabla_\mu T^{\mu \nu}=0$
$\nabla_\mu ( \rho u^\mu u^\nu ) + \nabla_\mu ( p u^\mu u^\nu) - \nabla_\mu (pg^{\mu \nu})=0$
Taking $\nu=0$ we get
$\nabla_0 ( \rho u^0) + \nabla_i ( \rho u^i) + \nabla_0 ( pu^0) + \nabla_i ( p u^i) - \nabla_0p$
$\frac{\partial \rho}{\partial t}=0$
which is clearly incomplete so something isn't right...

Thanks.

 

[fzero]

Also, for the $\nabla_\mu T^{\mu \nu}$ question, can I take $u^\mu=(1,0,0,0)$ as we are assuming a comoving observer. And apparently they have $u^\alpha=\delta^\alpha{}_0$.
However, even though our notes say we can do this for a comoving observer, I find that
$\nabla_\mu T^{\mu \nu}=0$
$\nabla_\mu ( \rho u^\mu u^\nu ) + \nabla_\mu ( p u^\mu u^\nu) - \nabla_\mu (pg^{\mu \nu})=0$
Taking $\nu=0$ we get
$\nabla_0 ( \rho u^0) + \nabla_i ( \rho u^i) + \nabla_0 ( pu^0) + \nabla_i ( p u^i) - \nabla_0p$
$\frac{\partial \rho}{\partial t}=0$
which is clearly incomplete so something isn't right...

Thanks.

If you take the calculation another line you should find something like $$\dot{\rho} + (\rho +p){\Gamma^0}_{00}=0$$.

 

[latentcorpse]

If you take the calculation another line you should find something like $$\dot{\rho} + (\rho +p){\Gamma^0}_{00}=0$$.

How did you manage to get any $p$ terms surviving? All of mine cancel.