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Computing Christoffel Symbols HELP!! (Relativity / Maths)

Hey everyone! Need help in understanding how these Christoffel Symbels were computed? I know they used the formula for the "connection" but still can't seem to get my head round how it works? Or does anyone have an alternative method? THANKSS!!
Could someone also explain how the got the metrics. Many thankss
Reply 2
Original post by Nazzy_HCrest
Could someone also explain how the got the metrics. Many thankss


Well, from the very little that I've just learnt through Google - one of the ways to get the 2-sphere is to imaging the 3-sphere then restrict it to constant radius.

Since, the 3-sphere has ds2=dr2+r2dθ2+r2sin2θdϕ2ds^2 = dr^2 +r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2 then restricting r=Rr=R (the unit 2-sphere):

ds2=R2dθ2+R2sin2θdϕ2ds^2 = R^2 d \theta^2 + R^2 \sin^2 \theta d\phi^2, as a matrix, you can represent this as:

gij=(R200R2sin2θ)\displaystyle g_{ij} = \begin{pmatrix} R^2 & 0 \\ 0 & R^2 \sin^2 \theta\end{pmatrix} which reduces to your matrix if we take R=1R=1 (i.e: the unit 2-sphere)

The inverse is readily obtainable as gij=(1R2001R2sin2θ)g^{ij} = \begin{pmatrix} \frac{1}{R^2} & 0 \\ 0 & \frac{1}{R^2 \sin^2 \theta} \end{pmatrix}.

Although there are more intuitive ways to get this metic, such as specifying the symemtries you require, three independent rotations and then finding the 2-sphere via that. Or you could look for 2-dim spaces with constant curvature or compute the metric for a general 2-dim geometry and then impose constant curvature that gives a set of DE's which you can then use to get this metric.
Original post by Zacken
Well, from the very little that I've just learnt through Google - one of the ways to get the 2-sphere is to imaging the 3-sphere then restrict it to constant radius.

Since, the 3-sphere has ds2=dr2+r2dθ2+r2sin2θdϕ2ds^2 = dr^2 +r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2 then restricting r=Rr=R (the unit 2-sphere):

ds2=R2dθ2+R2sin2θdϕ2ds^2 = R^2 d \theta^2 + R^2 \sin^2 \theta d\phi^2, as a matrix, you can represent this as:

gij=(R200R2sin2θ)\displaystyle g_{ij} = \begin{pmatrix} R^2 & 0 \\ 0 & R^2 \sin^2 \theta\end{pmatrix} which reduces to your matrix if we take R=1R=1 (i.e: the unit 2-sphere)

The inverse is readily obtainable as gij=(1R2001R2sin2θ)g^{ij} = \begin{pmatrix} \frac{1}{R^2} & 0 \\ 0 & \frac{1}{R^2 \sin^2 \theta} \end{pmatrix}.

Although there are more intuitive ways to get this metic, such as specifying the symemtries you require, three independent rotations and then finding the 2-sphere via that. Or you could look for 2-dim spaces with constant curvature or compute the metric for a general 2-dim geometry and then impose constant curvature that gives a set of DE's which you can then use to get this metric.


That's fantastic! Could you possibly explain how you represented it as the matrix? Thanks a million for taking the time to help!!
Reply 4
Original post by Nazzy_HCrest
That's fantastic! Could you possibly explain how you represented it as the matrix? Thanks a million for taking the time to help!!


I'm only an A-Level student, so I'm likely to be extremely useless, but I believe that the matrix comes by considering the basis vectors of your metric. We, have: gij=eiejg_{ij} = \mathbf{e}_i \cdot \mathbf{e}_{j}, does this seem familiar to you?

In general: ds2=gijdθidϕj=g11(dθ)2+2g12dθdϕ+g22(dϕ)2ds^2 = g_{ij}d\theta^i d\phi^j = g_{11}(d\theta)^2 + 2g_{12}d\theta d\phi + g_{22}(d\phi)^2 but for the 2-sphere, we have ds2=R2dθ2+R2sin2θdϕ2ds^2 = R^2 d\theta^2 +R^2 \sin^2 \theta d\phi^2, so we get: g22=R2sin2θg_{22} = R^2 \sin^2 \theta and g11=R2g_{11} = R^2 which allows us to write down our matrix right away.
(edited 8 years ago)

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