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#37
Jan1711, 11:48 AM

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


#38
Jan1811, 03:28 PM

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[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_bu_{b;a}u_cu_{a;c}u_bU_{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? 


#39
Jan1811, 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.



#40
Jan1811, 04:58 PM

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#41
Jan1811, 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 4velocity 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.



#42
Jan1911, 03:57 AM

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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? 


#43
Jan1911, 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]. 


#44
Jan1911, 09:37 AM

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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! 


#45
Jan1911, 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 spacetime geometry determined by the existence of that particular T_ab. 


#46
Jan2011, 11:11 AM

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#47
Jan2011, 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 1forms a and b, lambda is a 0form. 


#48
Jan2111, 10:29 AM

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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. 


#49
Jan2111, 10:42 AM

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



#50
Jan2111, 11:42 AM

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[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] 


#51
Jan2111, 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. 


#52
Jan2111, 03:33 PM

P: 1,437

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_ck_ck_b)=0[/itex] I'm getting pretty confused as to where this is going though! 


#53
Jan2111, 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 ?



#54
Jan2111, 05:45 PM

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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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