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2 Metric tensor question~

  1. Nov 3, 2005 #1
    Q1 If given a 2D Riemannian space, ds^2 = dx^2 + x^2dy^2, do the componets of the metric tensor are these:
    g_11 = 1, g_12 = 0
    g_21 = o, g_22 = x^2 ?

    In addition, I got a question from my lecturer:
    Q2. 2 metrics, defined in a Riemannian space, are given by ds^2 = g_ijdx^idy^j
    and ds'^2 = g'_ij dx^idy^j = e^z g_ijdx^idy^j , respectively, where z is a function of the coordinates x^i.
    Find the relation between the Chritoffel symbols corresponding to the 2 two metrics~~~
    I have no ideal how to solve it and what is e here? treat it as a function or is it represents the persudo tensor????

    Can any one help me~~

  2. jcsd
  3. Nov 3, 2005 #2
    Yes, this is correct.

    I assume you mean: [itex]ds^2 = g_{ij}dx^idx^j[/itex] and [itex]ds'^2 = g'_{ij}dx^idx^j = e^z g_{ij}dx^idx^j[/itex]. (Note that I have replaced references to [itex]y^j[/itex] with references to [itex]x^j[/itex].)

    The [itex]e^z[/itex] here is the exponential function. Note that its partial derivatives are (,i is shorthand for [itex]\partial / \partial x^i[/itex]):

    [itex]e^z{}_{,i} = z_{,i}e^z[/itex]

    The formula for the Christoffel symbol in terms of the metric tensor is:

    [itex]\Gamma^m{}_{ij} = \frac{1}{2}g^{km}(g_{ik,j} + g_{jk,i} - g_{ij,k})[/itex]

    This should be enough to get you started. If you still have trouble, post again.
    Last edited: Nov 3, 2005
  4. Nov 3, 2005 #3
    To jimmysnyder:

    Hope I can understand what you mean~~
    I work out the steps, am I correct?

    [itex]\Gamma^m{}_{ij} = \frac{1}{2}g^{km} (-g_{ij,k}+g_{jk,i}+g_{ki,j})[/itex]
    [itex]\Gamma'^m{}_{ij} = \frac{1}{2}g'^{km} (-g'_{ij,k}+g'_{jk,i}+g'_{ki,j})[/itex]
    [itex]\Gamma'^m{}_{ij} = \frac{1}{2}e^{z(i)}g^{km} (-e^{z(i)}g_{ij,k}+e^{z(i)}g_{jk,i}+e^{z(i)}g_{ki,j}[/itex]
    [itex]\Gamma'^m{}_{ij} = \frac{1}{2}e^{z(i)}g^km{}[-e^{z(i)}g_{ij,k}+z(i)_{,i}e^{z(i)}g_{jk,i}+e^{z(i)}g_{jk,i}+e^{z(i)}g_{ki,j}][/itex]
    [itex]\Gamma'^m{}_{ij} = e^{2z(i)}[\Gamma^m{}_{ij} + \frac{1}{2}g^{km}z(i)_{,i}g_{jk,i}][/itex]

    Thx so much and you are so smart that you can correct my mistake as what i want to write is dx^j not dy^j~

    Last edited: Nov 3, 2005
  5. Nov 3, 2005 #4
    So far, so good.

    The exponential factor that goes with [itex]g^{km}[/itex] should be [itex]e^{-z}[/itex], because you want [itex]g'^{np}g'_{pm} = \delta^n{}_m[/itex]. Also in order to make things more clear, I would remove the parameter from z and add parentheses as follows:
    [itex]\Gamma'^m{}_{ij} = \frac{1}{2}e^{-z}g^{km} (-(e^{z}g_{ij})_{,k} + (e^{z}g_{jk})_{,i} + (e^{z}g_{ki})_{,j})[/itex]

    The next steps are not correct, but if you start from the equation I have given you, I think you can get the rest.
    Last edited: Nov 3, 2005
  6. Nov 3, 2005 #5
    Once more question is :

    Given a fame S' which is falling along -z-axis with constant acceleration in an inertial frame S. Find a form of metric in the S' frame, assume in Newtonian approximation of the absolute time (t=t').

    I just know a definition that
    a = sqrt(g_ij dx^idx^j) but how to get the g_ij??

  7. Nov 3, 2005 #6
    I think my answer is correct~~
    as z is only a function of coordinate x^i , derative of z wrt x^j & x^k will be zero ~ right? Do I make the mistake there?
  8. Nov 3, 2005 #7
    Yes, you have made a mistake. In this case, i is an index that takes all 4 values, 0, 1, 2, and 3. In your original post you have written "z is a function of the coordinates x^i". Note coordinates, not coordinate. That means it is a function of all 4 coordinates and partials must be taken with respect to each of them. The letters k and j are just different index letters that also take on the 4 values 0, 1, 2, and 3.
    Last edited: Nov 3, 2005
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