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Minkowski metric

  1. Nov 9, 2012 #1
    hello

    Whic one of these to metric are Minkowski metric
    ds^2 =-(cdt)^2+(dX)^2

    ds^2 =(cdt)^2-(dX)^2

    and what about timelike (ds^2<0) and spacelike (ds^2>0) for each metric?

    With my appreciation to those who answer
     
  2. jcsd
  3. Nov 9, 2012 #2

    bcrowell

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    Welcome to PF!

    These are both ways of writing the Minkowski metric. People refer to this as (-+++) and (+---) signatures. It doesn't matter which one you use as long as you're consistent. It can be confusing reading the literature, because different people use different signatures.
     
  4. Nov 9, 2012 #3

    Dale

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    Both are considered the same metric, just with a different sign convention. My personal preference is the first one, but both are well accepted.

    When I want to use a metric with positive timelike intervals squared I tend to use [itex]c^2d\tau^2=c^2dt^2-dX^2[/itex]. It is just a convention, but that is my preference.
     
    Last edited: Nov 9, 2012
  5. Nov 9, 2012 #4

    Fredrik

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    The first convention seems to be more popular in texts about classical SR and GR. The second convention seems to be more popular in books on quantum field theory.
     
  6. Nov 9, 2012 #5

    robphy

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  7. Nov 10, 2012 #6
    bcrowell, DalesPam, robphy Thank you very much

    my queastion is

    if I use first one

    ds^2 =-(cdt)^2+(dX)^2

    spacelike: (ds^2>0) then -(cdt)^2+(dX)^2>0 then (dX/dt)>c



    if I use second one

    ds^2 =(cdt)^2-(dX)^2

    spacelike: (ds^2>0) then (cdt)^2-(dX)^2>0 then (dX/dt)<c


    we know in light cone spacelike out of cone that mean (dX/dt)>c but why second one (dX/dt)<c
     
  8. Nov 10, 2012 #7

    George Jones

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    No, for this convention for the metric, spacelike means ds^2 < 0.
     
  9. Nov 10, 2012 #8

    robphy

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    spacelike: outside the lightcone... so "faster than light"
    (dx/dt)^2 > c^2 (i.e. either (dx/dt) > c or (dx/dt) < -c).

    Thus, spacelike means dx^2 > c^2 dt^2
    ("larger square-of-the-magnitude of the spatial-part than that of c-times-the-temporal-part")

    So, spacelike is dx^2 - c^2 dt^2 > 0.

    Now on to the conventions...
    If ds^2 = -c^2 dt^2 + dx^2 (-+++), then spacelike is ds^2 > 0 (in -+++... that is, "+ for space").

    If ds^2 = c^2 dt^2 - dx^2 (+---), then spacelike is ds^2 < 0 (in +---... that is "- for space").
     
  10. Nov 10, 2012 #9
    thanks
     
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