Solve Physics Q1 from 2007 Paper61. Cambridge Maths Postgrad

[dextercioby]

A Killing vector is divergence-free, as it can be seen by contracting the 2 indices in their defining equation.

The 6th term is minus the 4th:

$$k_{b;c} k_{a}^{~;c} = k_{a}^{~;c} k_{b;c} = k_{a;c} k_{b}^{~;c} = - k_{c;a} k_{b}^{~;c}$$

Ok ?

I leave it to you to figure out why the first term is 0.

 

[latentcorpse]

A Killing vector is divergence-free, as it can be seen by contracting the 2 indices in their defining equation.

The 6th term is minus the 4th:

$$k_{b;c} k_{a}^{~;c} = k_{a}^{~;c} k_{b;c} = k_{a;c} k_{b}^{~;c} = - k_{c;a} k_{b}^{~;c}$$

Ok ?

I leave it to you to figure out why the first term is 0.

Yeah I made a stupid mistake when trying to get the 4th to cancel the 6th...

As for Killing vectors beign divergence free:
we contract the defining equation to get $\nabla^a k_a + \nabla_a k^a=0 \Rightarrow \nabla^a k_a + \nabla^a k_a = 0 \Rightarrow \nabla^a k_a =0$
i.e. $k_c{}^{;c}=0$
and hence the second term, $k_{a;b} k_c{}^{;c}$, will vanish.

The first term remains a mystery. Am I right in thinking that we want to show it's equal to negative itself and hence is zero?
The natural way to approach that would be to say
$k_{a;b}{}^{;c}k_c=-k_{b;a}{}^{;c}k_c$
Now the best I can come up with is arguing that whilst the two c indices are symmetric under exchange, the a and b indices are antisymmetric under exchange and therefore we have a symmetric term multiplying an antisymmetric term and so the whole thing will be zero. How's that? I have a feeling it's wrong because the c index terms aren't next to each other!

 

[dextercioby]

I will give you a hint, not post the solution this time. You need to use one of the 2 points already proven in this problem.

 

[latentcorpse]

I will give you a hint, not post the solution this time. You need to use one of the 2 points already proven in this problem.

Well, using the last thing we proved, we have that

$k_{a;b}{}^{;c}k_c=k_{a;bc}k^c=R_{abc}{}^dk^ck_d$

But the Riemann tensor is antisymmetric under exchange of indices $c \leftrightarrow d$ but $k^ck_d$ is symmetric and hence the whole thing vanishes.

Is that correct?

 

[dextercioby]

Yes. Finally :)

 

[latentcorpse]

Yes. Finally :)

Now for the final part, it is asking us to show that $u_a$ is parallel to $k_a$. But we have a static spacetime and so $k_a$ will be hypersurface orthogonal. This means we are asked to show that $u_a$ is also hypersurface orthogonal.

So the question is asking us to show that $u_{[a;b}u_{c]}=0$, correct?

Well we know that this spacetime will have energy momentum tensor $T_{ab}=(\rho+p)u_au_b+pg_{ab}$

So we have $T_{ac;b}=(\rho+p) u_{a;b}u_c+u_au_{c;b}$ since $g_{ac;b}=0$

Then I tried writing $u_{[a;b}u_{c]}$ in terms of $T$'s but it didn't get me anywhere. Can you give me a hint please?

 

[dextercioby]

So we have

$T_{ac;b}=(\rho+p) u_{a;b}u_c+u_au_{c;b}$ (1)

since

$g_{ac;b}=0$

If $p$ and $\rho$ are totally unrelated, then (1) is more that enough to prove your statement.

The hint is to consider a full antisymmetrization on all 3 terms of (1).

 

[latentcorpse]

If $p$ and $\rho$ are totally unrelated, then (1) is more that enough to prove your statement.

The hint is to consider a full antisymmetrization on all 3 terms of (1).

I misstyped - it should be:

$T_{ac;b}=(\rho+p)(u_{a;b}u_c+u_au_{c;b})$

Anyway, I get a ridiculously long expression:

$T_{ac;b}+T_{cb;a}+T_{ba;c}-T_{ca;b}-T_{ab;c}-T_{bc;a}=(\rho+p)(u_{a;b}u_c+u_{b;c}u_a+u_{c;a}u_b-u_{b;a}u_c-u_{a;c}u_b-U_{c;b}u_a+u_au_{c;b}+u_cu_{b;a}+u_bu_{a;c}-u_cu_{a;b}-u_au_{b;c}-u_bu_{c;a}$
But all the terms on the RHS cancel and so
$\Rightarrow T_{[ac;b]}=0$

But if $T_{ac;b}=(\rho+p)(u_{a;b}u_c+u_au_{c;b})$

then we can conclude

$u_{[a;b}u_{c]}+u_{[a}u_{c;b]}=0$

which isn't quite what we want is it? We want to show $u_{[a;b}u_{c]}=0$, no?

 

[dextercioby]

The energy momentum tensor is the RHS of the Einstein field equations, so it must be a symmetric second order tensor. Antisymmetrizing over the 2 free indices it would give 0. It's no need to give a proof for it. It's a trivial thing.

 

[latentcorpse]

The energy momentum tensor is the RHS of the Einstein field equations, so it must be a symmetric second order tensor. Antisymmetrizing over the 2 free indices it would give 0. It's no need to give a proof for it. It's a trivial thing.

Ok. Yes, I should have spotted that. Althoughw e still arrive at the wrong conclusion do we not?

 

[dextercioby]

The last point of this problem is really nice. I'll give you a hint: What does it mean for the 4-velocity u_a to be parallel to a covariant Killing vector k_a ? After that, I tell you that the solution of the problem involves the Einstein equations.

 

[latentcorpse]

The last point of this problem is really nice. I'll give you a hint: What does it mean for the 4-velocity u_a to be parallel to a covariant Killing vector k_a ? After that, I tell you that the solution of the problem involves the Einstein equations.

Well if $u_a$ is parallel to $k_a$ then it will also be hypersurface orthogonal and hence it will also satisfy $u_{[a;b}u_{c]}=0$. Am I correct that this is what we are ultimately trying to prove?

Now the Einstein equations read:

$8 \pi ( \rho + p) u_a u_b + 8 \pi p g_{ab} = R_{ab} - \frac{1}{2} R g_{ab}$

The only constructive thing I can think to do with this is to antisymmetrise it. This tells us $u_au_b=0$ i.e. the product $u_au_b$ is symmetric but I'm not sure that really helps, does it?

 

[dextercioby]

If they're parallel, then they must have the same direction, which mean that they (up to the modulus) should coincide when subject to a parallel transport.

This means that there exists a constant 'p' such that

$$k_a = p u_a$$

This means that everything you have proven for $k_a$ applies for $u_a$.

 

[latentcorpse]

If they're parallel, then they must have the same direction, which mean that they (up to the modulus) should coincide when subject to a parallel transport.

This means that there exists a constant 'p' such that

$$k_a = p u_a$$

This means that everything you have proven for $k_a$ applies for $u_a$.

I understand that argument but I'm not sure that we have actually answered what the question is asking us, have we?

How do we go from the fact that $T_{ab}=(\rho + p) u_au_b + pg_{ab}$ to deducing that $u_a$ and $k_a$ are parallel?

Thanks and sorry for dragging this out!

 

[dextercioby]

The hint I'm giving you is to use Einstein's equation, one of the points already proved in the problem and one more fact: you know that in R^3 euclidean geometry, 2 vectors are parallel iff their cross product is 0.

This last fact can be extended to the space-time geometry determined by the existence of that particular T_ab.

 

[latentcorpse]

The hint I'm giving you is to use Einstein's equation, one of the points already proved in the problem and one more fact: you know that in R^3 euclidean geometry, 2 vectors are parallel iff their cross product is 0.

This last fact can be extended to the space-time geometry determined by the existence of that particular T_ab.

Sorry. I'm just not seeing it. Please don't give me the answer, but could you offer a bit of extra help?

 

[dextercioby]

You need to use the following things

1. $$E_{ab} = kT_{ab}$$

2. $$k_{a} R^{a}_{~[b} k_{c]} = 0$$

3. $$\mbox{a ext b} = 0 \Leftrightarrow \mbox{a} = \lambda \mbox{b}$$

3. contains the 1-forms a and b, lambda is a 0-form.

 

[latentcorpse]

You need to use the following things

1. $$E_{ab} = kT_{ab}$$

2. $$k_{a} R^{a}_{~[b} k_{c]} = 0$$

3. $$\mbox{a ext b} = 0 \Leftrightarrow \mbox{a} = \lambda \mbox{b}$$

3. contains the 1-forms a and b, lambda is a 0-form.

Sorry again!

Starting from $R_{ab}-\frac{1}{2}Rg_{ab}=8 \pi \rho u_au_b + 8 \pi p u_au_b + 8 \pi pg_{ab}$, what should I do next?
Antisymmetrise? Or Raise an index with a metric and then contract with $k_a$? Or something else entirely?

Thanks.

 

[dextercioby]

To use 2. you have to enter the Ricci curvature in terms of the 4-velocity. Can you do that ?

 

[latentcorpse]

To use 2. you have to enter the Ricci curvature in terms of the 4-velocity. Can you do that ?

Is it as follows?

$R_{ab}=8 \pi ( \rho + p) u_au_b + ( 8 \pi p + \frac{1}{2} R) g_{ab}$
$R^a{}_b=8 \pi ( \rho + p) u^a u_b + (8 \pi p + \frac{1}{2} R) \delta^a{}_b$

 

[dextercioby]

The separation is not fully done (there's some explicit and some implicit dependence of the Ricci tensor of u), however the second equation will lead you to a successful use of 1. and 2. in what I wrote above.

So post your final equation.

 

[latentcorpse]

The separation is not fully done (there's some explicit and some implicit dependence of the Ricci tensor of u), however the second equation will lead you to a successful use of 1. and 2. in what I wrote above.

So post your final equation.

Well, I don't knkow how to use that in 1.

However, for 2. I get:

$k_aR^a{}_{[b}k_{c]}=0$
$\Rightarrow 4 \pi ( \rho + p) k_a u^a ( u_b k_c - u_c k_b) + ( 4 \pi p + \frac{1}{4} R)(k_bk_c-k_ck_b)=0$

I'm getting pretty confused as to where this is going though!

 

[dextercioby]

What can you derive of what you've written ? The second term of the sum vanishes, right ? So ?

 

[latentcorpse]

What can you derive of what you've written ? The second term of the sum vanishes, right ? So ?

$u_b k_c=u_ck_b$?

Is it then just a case of contracting with $k^c$ since that gives

$u_b k_c k^c=u_c k^c k_b \Rightarrow u_b = \lambda k_b$ since $k_ck^c, u_c k^c$ are both scalars, right?

 

[dextercioby]

Yes, finally. :)

Did you solve the rest of the exercises, in case you needed it ?