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Question about the metric tensor

  1. Mar 8, 2015 #1
    Hello

    Say, the metric tensor is diagonal, ##g=\mbox{diag}(g_{11}, g_{22},...,g_{NN})##. The (null) geodesic equations are

    ##\frac{d}{ds}(2g_{ri} \frac{dx^{i}}{ds})-\frac{\partial g_{jk}}{\partial x^{r}}\frac{dx^{j}}{ds}\frac{dx^{k}}{ds} = 0##

    These are ##N## equations containing ##N## partial derivatives ##\frac{\partial g_{rr}}{\partial x^{l}}##.

    The question is - does this mean ##g_{rr}## (a total of ##N## of them) can be functions of up to one coordinate variable each?
    Say, in cyl. coordinates ##ds^2=g_{11}(r)dr^2+g_{22}(\theta)d\theta^2+g_{33}(z)dz^2+g_{44}(t)dt^2##
    What is your understanding - can say, ##g_{22}## be a function of ##t##? Or could ##g_{11}## be a function of ##z##?

    It just seems that if in the most general case ##g_{rr}=g_{rr}(x^1,x^2,...,x^N)## the geodesic equations should be at least ##N^2##, to carry the information for all possible partial derivatives...

    Any thoughts?
     
  2. jcsd
  3. Mar 8, 2015 #2

    Mentz114

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    Gold Member

    I'm not sure if I understand the question. I think you've confused the coordinate r with the index r in the equation.

    Here it is with m replacing r as the surviving index.

    ##
    \frac{d}{ds}(2g_{mi} \frac{dx^{i}}{ds})-\frac{\partial g_{jk}}{\partial x^{m}}\frac{dx^{j}}{ds}\frac{dx^{k}}{ds} = 0
    ##

    There is no reason why the metric coefficients should not be functions of all or any of the coordinates.
     
    Last edited: Mar 8, 2015
  4. Mar 8, 2015 #3
    Then we would have ##N^2## nonzero partial derivatives ##\frac{\partial g_{rr}}{\partial x^l}##, while the geodesic equations are only ##N##.
     
  5. Mar 8, 2015 #4

    stevendaryl

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    Staff Emeritus
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    The geodesic equation is for determining [itex]x^\mu(\tau)[/itex] given [itex]g_{\mu \nu}[/itex]. It's not for determining [itex]g_{\mu \nu}[/itex]. There are 4 equations and four unknowns:

    [itex]\frac{d^2 x^\mu}{d\tau^2} =...[/itex]
     
  6. Mar 8, 2015 #5
    My bad, should have used something else instead of r.

    Ah, I see. There are ##N## coefficients of the metric tensor which, through the ##N## geodesic equations output ##N## parametric equations for the geodesic. This makes sense :smile: I shouldn't have looked at the number of derivatives.
     
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