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I don't understnad how to get the 4-volume element for a Minkowksi space

  1. Nov 1, 2011 #1
    I see that it is [itex]\sqrt{|-g|}[/itex][itex]d^{4}[/itex]x but I'm not sure why it needs to be multiplied by the square root of the determinant of -g. It must be the Jacobian of [itex]g_{μ\nu}[/itex] or something right? So I guess I'm asking how do you calculate the Jacobian of the metric.
     
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  3. Nov 1, 2011 #2
    In Minkowski space the coordiates are orthonormal. The diagonal of the metric is (-1,1,1,1) with all other entries zero.

    The expression you gave should be [itex]\sqrt{-g}[/itex], where [itex]g[/itex] is the determinate of the metric. It means you take the square root of the negative of the determinate of the metric. The determinate is -1, so the volume element in Minkowski space is just multipled by one.

    [tex]d^4 x' = \sqrt{-g}d^4 x[/tex]

    What we are doing with [itex]\sqrt{-g}[/itex] is translating between a volume element in x' coordinates to orthonormal minkowski volume elements in the unprimed frame.
     
    Last edited: Nov 1, 2011
  4. Nov 1, 2011 #3
    Yes but why do we use the square root of the determinate of -g to transform between volume elements in different frames? I'd like to see the math of it.
     
  5. Nov 1, 2011 #4

    Matterwave

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    The volume element in the Minkowski metric is invariant under Lorentz transformations. You can show this by explicitly computing the Jacobian for a Lorentz Transformation.

    That factor is redundant in Minkowski space-time.
     
  6. Nov 1, 2011 #5

    pervect

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    Consider the transformation x' = ax

    Consider how the metric transforms under this transformation (from t,x,y,z to t,x',y,z), and how the volume element tranforms under this transformation.

    Hint:

    A unit volume 4- cube has dt, dy, and dz the same, but dx' = a dx.

    We can use the tensor transformation law to transform the metric, or just algebra. To use just algebra, we write the line element in terms of dx,dy,dz and we use again use the formula dx' = a dx.

    So a line element of -dt^2 + dx^2 + dy^2 + dz^2 transforms into -dt^2 + dx'^2/a^2 + dy^2 + dz^2

    This isn't a completely general derivation , but should (hopefully) give you a good intiutive feeling for questions like "where did that square root come from"
     
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