Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

2 Questions

  1. Jul 24, 2006 #1
    I have two questions...let's suppose we have a metric in the form:

    [tex] ds^2 =f(t)dt^2 +g(x)dx^2 +H(y)dy^2 [/tex]

    So every element of the metric only depend on a variable..my question is..does this mean that the Einstein Equations (vaccuum) are of the form:

    [tex] R_ii =0 [/tex] i=t,x,y ?..

    -And the second question is i know that [tex] det(g_ab )=f(t)g(x)H(y) [/tex] but ..what's the form of the Lagrangian?..i guess:

    [tex] L= \int_ V dVf(t)g(x)H(y)(f(t)R_00 +g(x)R_11+ H(y)R_22 ) [/tex]:rolleyes: :cool: :frown:
  2. jcsd
  3. Jul 24, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Yep. I threw this line element (with a k(z) dz^2 term added) into GrTensor II, the Riemann was zero, as well as the Ricci and the Einstein.
  4. Jul 24, 2006 #3
    And a "Mixed" one?.. [tex] ds^2 = f(t)dt^2 + g_ij dx^i dx^j [/tex] Where Einstein summation is assumed....
  5. Jul 24, 2006 #4
    This line element is just a simple diffeomorphism of normal Minkowski space. Consider changing the coordinates to T, X, and Y, such that:

    [tex] dT = \sqrt{-f(t)} dt[/tex]
    [tex]dX = \sqrt{g(x)} dx[/tex]
    [tex]dY = \sqrt{H(y)} dy[/tex]

    In these coordinates, the line element becomes:
    [tex] ds^2 = - dT^2 + dX^2 + dY^2[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook