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I Why ##ds^2=0## in the Metric

  1. Dec 31, 2017 #1
    In FLWR metric or in Minkowski metric or in any general metric can we say that ##ds^2=0## cause speed of light should be constant to all observers ?

    Or theres another reason ?
     
  2. jcsd
  3. Dec 31, 2017 #2

    Orodruin

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    What do you mean? What ##ds^2## is depends on the world-line you are considering. ##ds^2 = 0## corresponds to a light-like curve. This is a coordinate independent statement and it will be true for the light-like curve regardless of the coordinate system you use.
     
  4. Dec 31, 2017 #3
    In FLRW metric when we measure the redshift we assume ##ds^2=0##. Like also in minkowski metric ##ds^2=0## cause only in that case we can get c=dx/dt.

    Probably I should add to the question why for a light ##ds^2=0##...
     
  5. Dec 31, 2017 #4

    Orodruin

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    This is not a measurement of redshift. It is a computation of the redshift based on the FLRW universe. Studying light, it is quite clear that we must use a light-like geodesics.

    Light is massless and moves along null geodesics.
     
  6. Dec 31, 2017 #5
    Hmm I see, I ll do more research on light-like geodesic case.

    I havent seen it yet so I was confused about the reason. I guess I can understand why we use ##ds^2=0##.
     
  7. Dec 31, 2017 #6

    Dale

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    The redshift you are talking about is for light. Light travels on null geodesics. Therefore ##ds^2=0##. It is not a general statement about the metric, it is a specific statement about light.

    Same thing here. For light ##ds^2=0## for the reason you gave. But for massive objects ##ds^2<0## and for hypothetical tachyons ##ds^2>0##.

    If you write down the metric, set ##ds^2=0##, then what are you left with? The equation of a sphere of radius ##ct##. This is something traveling at c in all directions, which is the second postulate. Therefore, ##ds^2=0## for light is the mathematical statement of the second postulate.
     
  8. Dec 31, 2017 #7

    Orodruin

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    I think it should be qualified that whether ##ds^2 > 0## or ##ds^2 < 0## for time-like world lines depends on the sign convention for the metric. Mathematicians and GR people generally prefer ##ds^2 < 0## while particle physicists prefer ##ds^2 > 0##. Always check which convention is being used in the particular text. Of course, this does not affect ##ds^2 = 0## for null world lines.
     
  9. Dec 31, 2017 #8

    Dale

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    Yes, good point. My preferred convention is to write ##ds^2## when I am using the (-+++) convention and to write ##d\tau^2## when I am using the (+---) convention
     
  10. Dec 31, 2017 #9
    Yes I tried to mean that, as I understood from your post for light ##ds^2=0##. But Is this true for all of the metrics ?
    In example For Minkowski metric we are working on light rays then ##ds^2=0##, in FLRW metric , we are working on light then ##ds^2=0## etc. ?

    Or there could be a metric where we cant set ##ds^2=0## for light ?

    I see, thanks
    Also,
    I understand it now
     
  11. Dec 31, 2017 #10

    Dale

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    Yes, it is true for all metrics and all spacetimes
     
  12. Dec 31, 2017 #11
    Thanks a lot
     
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