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Homework Help: Special relativity: components of a metric tensor

  1. Jun 25, 2010 #1

    Uku

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    1. The problem statement, all variables and given/known data

    An interval in Minkovski space is given in spheric coordinates as;

    [tex]ds^{2}=c^{2}dt^{2}-dr^{2}-r^{2}d\theta^{2}-r^{2}sin^{2}\theta d\phi^{2}[/tex]

    Now I have to find the covariant and contravariant components of the metric tensor.

    2. Relevant equations

    General expression of a metric tensor is:
    [tex]G=g_{\mu\nu}dx^{\mu}dx^{\nu}[/tex] which also equals [tex]ds^{2}[/tex]

    3. The attempt at a solution
    I have some messy answer written down from the lecture, but cant get it clear.
    Seems that I have written down the covariant components of the metric as
    [tex]g_{00}=1[/tex]
    [tex]g_{11}=-1[/tex]
    [tex]g_{22}=-r^{2}[/tex]
    [tex]g_{33}=-r^{2}sin^{2}\theta[/tex]
    Are these the covariant components that the problem asks me to find? How are they found?

    I see a correlation in the Schwarzchild metric

    G=[PLAIN]http://rqgravity.net/images/gravitation/Gravitation-92.gif [Broken]

    After this it seems that the lecturer has written down that

    [tex]g_{\mu\sigma} g^{\mu\rho}=\delta^{\rho}_{\sigma}[/tex]

    Is this correct?

    So I get a matrix on which all the components on the main diagonal are 1. From there I can derive that the contra-variant components of the metric are simply inverses of the above covariant metric.
     
    Last edited by a moderator: May 4, 2017
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  3. Jun 25, 2010 #2

    vela

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    If you expand the implied summations in [itex]ds^2 = g_{\mu\nu} dx^\mu dx^\nu[/tex], you get

    [tex]ds^2 = g_{00} dx^0 dx^0 + g_{01}dx^0dx^1 + \cdots + g_{32}dx^3dx^2 + g_{33}dx^3dx^3[/tex]

    Keeping in mind that [itex]x^0 = t[/itex], [itex]x^1 = r[/itex], [itex]x^2 = \theta[/itex], and [itex]x^3 = \phi[/itex], compare that expression to

    [tex]ds^2 = dt^2 - dr^2 - r^2 d\theta^2 - r^2\sin^2\theta\,d\phi^2[/tex]

    You can just read off the components of [itex]g_{\mu\nu}[/itex].

    Your notes are correct in that [itex]g_{\mu\sigma}g^{\mu\rho}=\delta^\rho_\sigma[/itex]. Consider

    [tex]x_\rho = g_{\rho\sigma} x^\sigma=g_{\rho\sigma}g^{\sigma\mu}x_\mu[/tex]

    You can see that all [itex]g_{\rho\sigma}g^{\sigma\mu}[/itex] does is change the label of the index on [itex]x_\mu[/itex]. In other words, it's just the identity.
     
    Last edited: Jun 25, 2010
  4. Jun 25, 2010 #3

    Uku

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    Thanks!

    A well, actually I have a question still.
    The bit:

    [tex]ds^{2}=g_{\mu\nu}dx^{\mu} dx^{\nu}[/tex]

    indicates summation over 16 (4x4) values, yes? With only the main diagonal being significant on the g.

    EDIT: yes, that is what happens.
     
    Last edited: Jun 25, 2010
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