
#37
Jan1711, 11:48 AM

Sci Advisor
HW Helper
P: 11,866

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



#38
Jan1811, 03:28 PM

P: 1,431

[latex]T_{ac;b}=(\rho+p)(u_{a;b}u_c+u_au_{c;b})[/latex] Anyway, I get a ridiculously long expression: [latex]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}[/latex] But all the terms on the RHS cancel and so [latex]\Rightarrow T_{[ac;b]}=0[/latex] But if [latex]T_{ac;b}=(\rho+p)(u_{a;b}u_c+u_au_{c;b})[/latex] then we can conclude [latex]u_{[a;b}u_{c]}+u_{[a}u_{c;b]}=0[/latex] which isn't quite what we want is it? We want to show [latex]u_{[a;b}u_{c]}=0[/latex], no? 



#39
Jan1811, 04:29 PM

Sci Advisor
HW Helper
P: 11,866

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

P: 1,431





#41
Jan1811, 05:04 PM

Sci Advisor
HW Helper
P: 11,866

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

P: 1,431

Now the Einstein equations read: [latex]8 \pi ( \rho + p) u_a u_b + 8 \pi p g_{ab} = R_{ab}  \frac{1}{2} R g_{ab}[/latex] The only constructive thing I can think to do with this is to antisymmetrise it. This tells us [latex]u_au_b=0[/latex] i.e. the product [latex]u_au_b[/latex] is symmetric but I'm not sure that really helps, does it? 



#43
Jan1911, 05:25 AM

Sci Advisor
HW Helper
P: 11,866

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

P: 1,431

How do we go from the fact that [latex]T_{ab}=(\rho + p) u_au_b + pg_{ab}[/latex] to deducing that [latex]u_a[/latex] and [latex]k_a[/latex] are parallel? Thanks and sorry for dragging this out! 



#45
Jan1911, 04:47 PM

Sci Advisor
HW Helper
P: 11,866

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

P: 1,431





#47
Jan2011, 12:09 PM

Sci Advisor
HW Helper
P: 11,866

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

P: 1,431

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



#49
Jan2111, 10:42 AM

Sci Advisor
HW Helper
P: 11,866

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




#50
Jan2111, 11:42 AM

P: 1,431

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



#51
Jan2111, 01:17 PM

Sci Advisor
HW Helper
P: 11,866

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: 1,431

However, for 2. I get: [latex]k_aR^a{}_{[b}k_{c]}=0[/latex] [latex]\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[/latex] I'm getting pretty confused as to where this is going though! 



#53
Jan2111, 03:40 PM

Sci Advisor
HW Helper
P: 11,866

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




#54
Jan2111, 05:45 PM

P: 1,431

Is it then just a case of contracting with [latex]k^c[/latex] since that gives [latex]u_b k_c k^c=u_c k^c k_b \Rightarrow u_b = \lambda k_b[/latex] since [latex]k_ck^c, u_c k^c[/latex] are both scalars, right? 


Register to reply 
Related Discussions  
No difference between covectors and functions?  Differential Geometry  4  
Analogy between vectors and covectors  Differential Geometry  9  
Vectors and Covectors  Differential Geometry  5  
Basis vectors and covectors  Differential Geometry  19 