1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
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

    pervect

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: 2 Questions
  1. 2 questions (Replies: 4)

  2. 2 Questions (Replies: 2)

Loading...