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Computing Christoffel Symbols HELP!! (Relativity / Maths) Watch

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    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!!
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    Could someone also explain how the got the metrics. Many thankss
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    (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 ds^2 = dr^2 +r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2 then restricting r=R (the unit 2-sphere):

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

    \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=1 (i.e: the unit 2-sphere)

    The inverse is readily obtainable as 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.
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    (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 ds^2 = dr^2 +r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2 then restricting r=R (the unit 2-sphere):

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

    \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=1 (i.e: the unit 2-sphere)

    The inverse is readily obtainable as 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!!
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    (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: g_{ij} = \mathbf{e}_i \cdot \mathbf{e}_{j}, does this seem familiar to you?

    In general: ds^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 ds^2 = R^2 d\theta^2 +R^2 \sin^2 \theta d\phi^2, so we get: g_{22} = R^2 \sin^2 \theta and g_{11} = R^2 which allows us to write down our matrix right away.
 
 
 
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