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Need to find the riemann curvature for the following metric

  1. May 11, 2012 #1
    1. The problem statement, all variables and given/known data

    Calculate the Riemann curvature for the metric:
    ds2 = -(1+gx)2dt2+dx2+dy2+dz2 showing spacetime is flat

    2. Relevant equations

    Riemann curvature eqn:

    [itex]\Gamma[/itex]αβγδ=(∂[itex]\Gamma[/itex]αβδ)/∂xγ)-(∂[itex]\Gamma[/itex]αβγ)/∂xδ)+([itex]\Gamma[/itex]αγε)(Rεβδ)-([itex]\Gamma[/itex]αδε)([itex]\Gamma[/itex]εβγ)

    3. The attempt at a solution

    I know that the non-vanishing Christoffel components are as follows:

    [itex]\Gamma[/itex]=sinθcosθ
    [itex]\Gamma[/itex]θ=[itex]\Gamma[/itex]θ=cotθ


    My guess is that the middle terms disappear creating:
    -cos2θ+sin2θ-(-sinθcosθ)(cosθ/sinθ)
    The sinθ's cos2θ's cancel each other out making the answer sin2θ

    Is this answer correct? My confusion is that I received this answer for the curvature for a different metric (namely, ds2=R22+R2sin2θd[itex]\vartheta[/itex]). Will I always receive the answer sin2θ? I am not understanding fully what the Riemann curvature is...

    Any help would be greatly appreciated!!
     
  2. jcsd
  3. May 12, 2012 #2
    How did you find those Christoffel symbols? Since the metric is almost flat already, maybe you should try to find a coordinate transformation which transforms the metric to ds2=-dT2+dx2+dy2+dz2. Then it's manifestly flat and you know that the Riemann tensor vanishes.
     
  4. May 14, 2012 #3
    Thanks for your help! I found the Christoffel symbols from an appendix in the back of my textbook. Yes, this sounds like it would work because the second part of the problem asks us to transform the metric into Minkowski form (I have figured this part out). However, I think the questioning is asking to explicitly show that the Riemann curvature makes the metric flat. I've spent hours trying to understand this Riemann equation and how to work with it, but I just can't seem to fully understand.

    So working with the equation ds2=-dt2+dx2+dy2+dz2, would the Riemann curvature equalling 0 be equivalent to the tensor vanishing? If so, how would I show this?

    I guess you could say that I am completely confused about the basic mathematics of applying the Riemann curvature on a metric.
     
  5. May 14, 2012 #4
    So remember that the curvature tensor itself is independent of coordinates. You can write it as
    [tex] R = R_{\alpha \beta \gamma \delta} dx^{\alpha} dx^{\beta} dx^{\gamma} dx^{\delta}. [/tex]
    Now, if for some coordinate system the componenents [itex] R_{\alpha \beta \gamma \delta} [/itex] vanish identically, that means that the entire tensor is 0 as well, and the components must vanish in all coordinate systems.

    The connection coefficients on the other hand do not form a tensor, so you may initially have nonvanishing connection, but then find a coordinate transformation which causes all [itex] \Gamma^{\rho}_{\mu \nu} = 0 [/itex], and as Riemann tensor is proportional to the connection, the calculation becomes trivial.
     
  6. May 14, 2012 #5
    This helps a lot - I really appreciate your help!!
     
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