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B Metric Tensor and The Minkowski metric

  1. Dec 17, 2017 #1

    I have seen the general form for the metric tensor in general relativity, but I don't understand how that math would create a Minkowski metric with the diagonal matrix {-1 +1 +1 +1}. I assume that using the kronecker delta to create the metric would produce a matrix that has all positive 1s in the diagonal. Are these numbers arbitrarily picked for certain "special" spaces?
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  3. Dec 17, 2017 #2


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    In relativity you use an indefinite metric (also called pseudo-metric) which has signature +--- or -+++, depending on convention. The Kronecker delta is positive definite and therefore you cannot find any coordinate system where the metric used in GR or SR is the Kronecker delta (or, more accurately, where the metric is diagonal with ones on the diagonal - the Kronecker delta really is a (1,1) tensor, not a (0,2) tensor like the metric - however, you usually encounter it as the metric in Cartesian coordinates on a Euclidean space as it does not matter whether indices are covariant and contravariant in such coordinates).
  4. Dec 17, 2017 #3


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    Different numbers (well, functions) give different spaces - or can do. In principle you slot a stress-energy tensor into the Einstein field equation and solve the resulting system of differential equations to find the metric tensor. In practice that's quite difficult, and there is quite a lot of gauge freedom which means two different metric tensors can describe the same spacetime, just in different coordinates. You can always pick coordinates so that at one chosen point the metric is Minkowski. So everywhere the signature is -+++ (or +---). It will never be anything else.

    The reason we picked the Minkowski metric is that Minkowski noticed that it is invariant under Lorentz transforms. So what the Lorentz transforms are telling you is that you live in a universe where the basic geometry is Minkowski, not Euclidean.
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