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Covectors physics help

by latentcorpse
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dextercioby
#37
Jan17-11, 11:48 AM
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Quote Quote by latentcorpse View Post

So we have

[itex]T_{ac;b}=(\rho+p) u_{a;b}u_c+u_au_{c;b}[/itex] (1)

since

[itex]g_{ac;b}=0[/itex]
If [itex] p [/itex] and [itex] \rho [/itex] 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
#38
Jan18-11, 03:28 PM
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Quote Quote by bigubau View Post
If [itex] p [/itex] and [itex] \rho [/itex] 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:

[itex]T_{ac;b}=(\rho+p)(u_{a;b}u_c+u_au_{c;b})[/itex]

Anyway, I get a ridiculously long expression:

[itex]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}[/itex]
But all the terms on the RHS cancel and so
[itex]\Rightarrow T_{[ac;b]}=0[/itex]

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

then we can conclude

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

which isn't quite what we want is it? We want to show [itex]u_{[a;b}u_{c]}=0[/itex], no?
dextercioby
#39
Jan18-11, 04:29 PM
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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
#40
Jan18-11, 04:58 PM
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Quote Quote by bigubau View Post
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
#41
Jan18-11, 05:04 PM
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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
#42
Jan19-11, 03:57 AM
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Quote Quote by bigubau View Post
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 [itex]u_a[/itex] is parallel to [itex]k_a[/itex] then it will also be hypersurface orthogonal and hence it will also satisfy [itex]u_{[a;b}u_{c]}=0[/itex]. Am I correct that this is what we are ultimately trying to prove?

Now the Einstein equations read:

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

The only constructive thing I can think to do with this is to antisymmetrise it. This tells us [itex]u_au_b=0[/itex] i.e. the product [itex]u_au_b[/itex] is symmetric but I'm not sure that really helps, does it?
dextercioby
#43
Jan19-11, 05:25 AM
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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

[tex] k_a = p u_a [/tex]

This means that everything you have proven for [itex]k_a[/itex] applies for [itex]u_a [/itex].
latentcorpse
#44
Jan19-11, 09:37 AM
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Quote Quote by bigubau View Post
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

[tex] k_a = p u_a [/tex]

This means that everything you have proven for [itex]k_a[/itex] applies for [itex]u_a [/itex].
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 [itex]T_{ab}=(\rho + p) u_au_b + pg_{ab}[/itex] to deducing that [itex]u_a[/itex] and [itex]k_a[/itex] are parallel?

Thanks and sorry for dragging this out!
dextercioby
#45
Jan19-11, 04:47 PM
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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 paralel 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
#46
Jan20-11, 11:11 AM
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Quote Quote by bigubau View Post
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 paralel 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
#47
Jan20-11, 12:09 PM
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You need to use the following things

1. [tex] E_{ab} = kT_{ab} [/tex]

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

3. [tex] \mbox{a ext b} = 0 \Leftrightarrow \mbox{a} = \lambda \mbox{b} [/tex]

3. contains the 1-forms a and b, lambda is a 0-form.
latentcorpse
#48
Jan21-11, 10:29 AM
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Quote Quote by bigubau View Post
You need to use the following things

1. [tex] E_{ab} = kT_{ab} [/tex]

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

3. [tex] \mbox{a ext b} = 0 \Leftrightarrow \mbox{a} = \lambda \mbox{b} [/tex]

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

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

Thanks.
dextercioby
#49
Jan21-11, 10:42 AM
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To use 2. you have to enter the Ricci curvature in terms of the 4-velocity. Can you do that ?
latentcorpse
#50
Jan21-11, 11:42 AM
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Quote Quote by bigubau View Post
To use 2. you have to enter the Ricci curvature in terms of the 4-velocity. Can you do that ?
Is it as follows?

[itex]R_{ab}=8 \pi ( \rho + p) u_au_b + ( 8 \pi p + \frac{1}{2} R) g_{ab}[/itex]
[itex]R^a{}_b=8 \pi ( \rho + p) u^a u_b + (8 \pi p + \frac{1}{2} R) \delta^a{}_b[/itex]
dextercioby
#51
Jan21-11, 01:17 PM
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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
#52
Jan21-11, 03:33 PM
P: 1,441
Quote Quote by bigubau View Post
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:

[itex]k_aR^a{}_{[b}k_{c]}=0[/itex]
[itex]\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[/itex]

I'm getting pretty confused as to where this is going though!
dextercioby
#53
Jan21-11, 03:40 PM
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What can you derive of what you've written ? The second term of the sum vanishes, right ? So ?
latentcorpse
#54
Jan21-11, 05:45 PM
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Quote Quote by bigubau View Post
What can you derive of what you've written ? The second term of the sum vanishes, right ? So ?
[itex]u_b k_c=u_ck_b[/itex]?

Is it then just a case of contracting with [itex]k^c[/itex] since that gives

[itex]u_b k_c k^c=u_c k^c k_b \Rightarrow u_b = \lambda k_b[/itex] since [itex]k_ck^c, u_c k^c[/itex] are both scalars, right?


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