Annoying Differential Geometry/tensor question in GR

  • Thread starter Thread starter latentcorpse
  • Start date Start date
  • Tags Tags
    Differential Gr
latentcorpse
Messages
1,411
Reaction score
0
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 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)

the structure eqn is dw^{\mu}=w^\mu_\nu \wedge w^\nu and is rewritten in terms of the one-form matrices \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 where \epsilon= \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right)

the structure eqns become

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

the antisymmetric derivatives are copmuted to be

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

i should mention that L is a constant and a is a function of \theta,t
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 w^1_0 \wedge w^1 but they somehow rearrange to w^0_1 \wedge w^1. How does this work?

(iii) finally, and probably most importantly, how do they get dw^0,dw^1 and dw?
 
Physics news on Phys.org
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
w^0{}_1=-L^{-1}e^{-a}(a'w^0+ \dot{a} w^1) i can get this one

i can't get these next three though:
\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}
in particular why we take the symmetric and asymmetric components of those matrices.
can anybody help here?

thanks.
 
Last edited:
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top