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If Geodesic are light cone, how can be test particle trajectiories

  1. Jan 2, 2012 #1
    I had this doubt studing GR, but lets consider SR for semplicity,
    where [itex]g_{\mu\nu}=\eta_{\mu\nu}[/itex]the geodesics are
    [itex]0=ds^2=dt^2-dr^2[/itex]
    we obtain the constraint we obtain the constraint r=(+/-)t

    So it is a well known light cone, but in SR we have that a (test?)particle can always move in a line described by r=vt with v<c

    Where this come from? seems to me that geodesics are just light cone and not trajectories allowed by common test particles.

    In particular, given the metric tensor and the initial conditions of a particle(speed and position) how can be derived the line he follows?

    I know there questions are trivial, but for a new player as me, GR seems very subtle and elusive.
     
  2. jcsd
  3. Jan 2, 2012 #2

    pervect

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    In general, geodesics are not necessarily light cones.

    Consider the geodesic equation http://en.wikipedia.org/w/index.php?title=Solving_the_geodesic_equations&oldid=425732007

    In flat space-time the equations are particularly easy: If we have [itex]x^0,x^1,x^2,x^3[/itex] as our coordinates then the a geodesic is defined by

    [tex]\frac{d^2 x^0} {ds^2} = \frac{d^2 x^1}{ds^2} = \frac{d^2 x^2}{ds^2} = \frac{d^2 x^3} {ds^2} = 0[/tex]

    The general form of the solution is that all the [itex]x^i[/itex] are linear functions of s. Note that this is a parametric curve, and s is a parameter - if you select a value of s, you select one particular point on the geodesic curve.

    You can break down the complete set of geodesic solutions into several subcategories:

    1) timelike geodeics. These are timelike paths that a particle might follow In non-parametric form, these curves are just objects moving with a constant velocity

    2) null geodesics - the previously mentioned light cone

    3) space-like geodesics. An example of a space-like geodesic would be lines of constant [itex]x^i[/itex]
     
  4. Jan 5, 2012 #3
    Your assumption is wrong: geodesic and light cones are not the same, but different concepts.
     
  5. Jan 6, 2012 #4
    Huh, what should be the different,

    as I just understood, with the geodesics equation I have a first order differential equation that should give me a "line" in the manifold given an initial point and an initial vector (velocity).

    As i understood light cones are given by imposing [itex]0=ds^2=g_{\mu\nu}dx^\mu dx^\nu[/itex] that means the length of the curve is
    [itex]\int{ds^2}=0[/itex] and so,
    what is the meaning of the integral??

    that, as I notice just now, it depends only from the metric tensor!
    how this integral is related to light cones (if it is..but i'm quite sure it is related)?
    what is the meaning of the integral, if ds does not vanish?
     
  6. Jan 6, 2012 #5

    pervect

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    I'm not sure what your question is, spocchio.

    Netheril's remark is correct. To oversimplify slightly, all light cones are geodsics (this isn't quite right, actually, but it's very close *), but not all geodesics are light cones.

    * A light cone is actually a surface traced out by geodiesics, not a geodesic itself.

    I don't really have the time to repeat myself, trying different words - you can reread my post, or maybe someone else can explain it more clearly.
     
  7. Jan 6, 2012 #6

    Mentz114

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    As Pervect has stated, it's necessary to distinguish between null geodesics, time-like geodesics and the space-like kind.

    As we all know, null geodesics are massless and travelling at c, time-like geodesics are the paths of massive bodies in free-fall and their proper length is the time on the observers clocks.

    [I started this post without reading Pervects post properly - but if a point is worth making, it's worth making twice].
     
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