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From tensor to scalar

  1. Jul 20, 2008 #1
    As I understand it, for a tensor of any rank I can produce a corresponding scalar in the following way: Create an inverted form of the tensor by lowering its upper indices and raising its lower indices, and then taking the inner product of this tensor and the original one.

    My only question is, is there a name for this result?
     
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  3. Jul 20, 2008 #2

    CompuChip

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    Invertibility?
    Metric?
    Adjointness (or whatever you call "raising/lowering indices", i.e. going from co- to contravariant and vv)
     
  4. Jul 20, 2008 #3

    robphy

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    Loosely speaking, you can call it the "square-norm" since what you are really doing is analogous to [tex] g_{ap} A^a A^p [/tex].
    (The quantity [tex] A_a= g_{ap} A^p [/tex] is called the metric-dual of [tex] A^a [/tex].)

    Given [tex]A^a{}_b{}^{cd}[/tex], you are forming the scalar using the metric [and its inverse]:
    [tex]g_{ap} g^{bq} g_{cr} g_{ds} A^a{}_b{}^{cd} A^p{}_q{}^{rs}[/tex]
     
  5. Jul 20, 2008 #4
    Thanks, robphy.

    I was wondering because in another thread someone mentioned using [tex]R^{abcd}R_{abcd}[/tex] (the "Kretschmann scalar"?) in order to show that the event horizon of a Schwarzschild black hole is not a real singularity, and it occurred to me that such a thing might be useful in other circumstances as well, so surely there must be a name for it...

    By the way, am I correct in my belief that the "square-norm" of any metric tensor is the number of dimensions of its manifold?
     
  6. Jul 20, 2008 #5

    CompuChip

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    Yes, you can show that very easily. By definition, [itex]g^{ab} g_{bc} = \delta^a_c[/itex] so if you contract a with c you get [itex]g^{ab} g_{ba} = \sum_{i = 1}^d 1 = d [/itex].
     
  7. Jul 20, 2008 #6
    Ah, yes, of course. Thanks, CompuChip.
     
  8. Jul 20, 2008 #7

    robphy

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    As seen from CompuChip's response, the metric tensor must have an inverse for that calculation. (Note: The metric of a Galilean spacetime is degenerate.)
     
  9. Jul 21, 2008 #8
    I don't know what this means. Are you referring to a Euclidean metric? Minkowskian? And whether or not their matrix representations have inverses? Don't they?
     
  10. Jul 21, 2008 #9

    robphy

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    The nondegenerate (i.e. invertible) metric of an n-dimensional Euclidean space has
    the diagonal form (+1,+1,...,+1,+1) in rectangular coordinates.

    The nondegenerate (i.e. invertible) metric of an n-dimensional Minkowskian spacetime has
    the diagonal form (-1,+1,...,+1,+1) in rectangular coordinates.

    The degenerate (i.e. non-invertible) metrics of an n-dimensional Galilean spacetime has
    the diagonal forms (0,+1,...,+1,+1) [for the spatial metric] and (+1,0,...,0,0) [for the temporal metric] in rectangular coordinates.
     
  11. Jul 21, 2008 #10
    Oh, OK. Thanks.
     
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