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Annoying Differential Geometry/tensor question in GR

  • #1
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i'm working through appendix A of the paper "Vacuum Spacetimes with Two-Parameter Spacelike Isometry Groups and Compact Invariant Hypersurfaces:Topologies and Boundary Conditions" by Robert H. Gowdy

anyway i. using a orthonormal basis method to get the curvature tensors and hence the einstein eqns

i have [itex]w^0=Le^adt, w^1=Le^a d \theta, w= \left( \begin{array}{c} w^2 \\ w^3 \end{array} \right)=L \left( \begin{array}{cc} M_{22} & M_{23} \\ M_{32} & M_{33} \end{array} \right)[/itex]

the structure eqn is [itex]dw^{\mu}=w^\mu_\nu \wedge w^\nu[/itex] and is rewritten in terms of the one-form matrices [itex]\Omega_0=\left( \begin{array}{c} w^2_0 \\ w^3_0 \end{array} \right), \Omega_1 \left( \begin{array}{c} w^2_1 \\ w^3_1 \end{array} \right), \Omega=w^2_3 \epsilon[/itex] where [itex]\epsilon= \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)[/itex]

the structure eqns become

[itex]dw^0=w^0_1 \wedge w^1 + \hat{\Omega_0} \wedge w[/itex] (1)
[itex]dw^1=w^0_1 \wedge w^0 - \hat{\Omega_1} \wdge w[/itex] (2)
[itex]dw=\Omega_0 \wedge w^0 + \Omega_1 \wedge w^1 + \Omega \wedge w[/itex]
where the antisymmetry [itex]w_{\mu \nu}=-w_{\nu \mu}[/itex] has been used.

the antisymmetric derivatives are copmuted to be

[itex]dw^0=L^{-1} a' e^{-a} w^1 \wedge w^0[/itex]
[itex]dw^1=L^{-1} \dot{a} e^{-a} w^0 \wedge w^1[/itex]
[itex]dw=L^{-1} e^{-a} (Pw^0+uw^1) \wedge w[/itex]
where [itex]P=\dot{M}M^{-1},U=M'M^{-1}[/itex]

i should mention that L is a constant and a is a function of [itex]\theta,t[/itex]
so i'm having a few problems with what should be straightforward differential geometry:

(i) in (1) and (2) why is there a hat on the Omega matrices? what does this signify? some sort of antisymmetric version of Omega?

(ii) in (2), if you plug the numbers into the structure eqns, the first term should just be [itex]w^1_0 \wedge w^1[/itex] but they somehow rearrange to [itex]w^0_1 \wedge w^1[/itex]. How does this work?

(iii) finally, and probably most importantly, how do they get [itex]dw^0,dw^1[/itex] and [itex]dw[/itex]?
 

Answers and Replies

  • #2
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ok. i've actually managed (iii) myself. i still don't get (i) or (ii) though.

i have a bit of difficulty with another bit though:
we then solve the structure eqns for the connection forms giving
[itex]w^0{}_1=-L^{-1}e^{-a}(a'w^0+ \dot{a} w^1)[/itex] i can get this one

i can't get these next three though:
[itex]\Omega_0=-L^{-1}e^{-a}P_{sym}w, \Omega_1=-L^{-1}e^{-a}U_{sym}w, \Omega=L^{-1}e^{-a}(Pw^0+Uw^1)_{asym}[/itex]
in particular why we take the symmetric and asymmetric components of those matrices.
can anybody help here?

thanks.
 
Last edited:

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