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Bending of light and the travel distance

  1. Jan 1, 2010 #1
    a question has been puzzling me for a while is : "is the distance traveled by a beam of light from point A to Point B in a flat space-time differs from the distance traveled by the same beam of light from the same point A and The Same point B but in a curved space-time ?"
    In other words :
    Suppose that there is a flat space-time and there is a beam of light travels between two points [tex]A[/tex] and [tex]B[/tex] and the distance between A and B is [tex]d_0[/tex] .
    and then we put a large mass in this flat space time and make a curved one , we would find the beam of light has been bent toward the large mass while it is traveling between the same two points [tex]A[/tex] and [tex]B[/tex] which the distance between them is [tex]d[/tex] .
    my question is :
    "is the distance between the points in the flat space-time [tex]d_0[/tex] differs form the distance between the same two points in the curved-space time [tex]d[/tex] ???"

    I appreciate your comments .....
     
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  3. Jan 1, 2010 #2

    bcrowell

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  4. Jan 1, 2010 #3
    thanks for your comment ...
    but i think the delay that happens to light (The Shapiro Effect) is due to the gravitational time dilation near the sun not because of the difference between the distances in the flat version and the curved version of the space time ....
     
  5. Jan 1, 2010 #4

    Nabeshin

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    Yes. In the case of flat space time, we calculate the distance between two points A and B using the minkowski metric, namely:

    [tex]D=\int_{A}^{B}\sqrt{ds^2}=\int_{A}^{B}\sqrt{-dt^2+dr^2+r^2\left(d\theta^2 + sin^2 \theta d\phi ^2\right)}[/tex]

    But in a curved space, say, schwarzschild geometry, this becomes:

    [tex]D=\int_{A}^{B}\sqrt{-\left(1-\frac{2M}{r}\right) dt^2 + \left(1-\frac{2M}{r}\right)^{-1} dr^2 + r^2\left(d\theta^2 + sin^2 \theta d\phi ^2\right)}[/tex]

    (Note: -,+,+,+ convention and c=G=1)

    The two are clearly not the same.

    Edit: Note: the distance between any two points for a beam of light is always zero anyways.
     
    Last edited: Jan 1, 2010
  6. Jan 1, 2010 #5
    Let's assume that A and B are far from the mass M. Even simpler we would find that we can not point the light directly at B we must aim off enough so that after the effect of the mass M the photons hit B. Case 1 no mass is a straight line, Case 2 with mass has two sides of a triangle where the third side is the path from case 1. One side of a triangle is always shorter than the other two sides added together. Or a straight line is always the shortest path.
     
  7. Jan 1, 2010 #6

    bcrowell

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    I don't think there's any useful way to distinguish between the two interpretations.
     
  8. Jan 2, 2010 #7
    thanks for your comments ....
    what you both said makes perfect sense ...but ..
    let's think about the spatial distance only ( forget about the time component of the metric [tex]dt=0[/tex]) , i know that the metrics of the two versions will stay different but think about it this way :
    suppose that we have a flat piece of paper on which we marked two points A and B and we pick the shortest path between the two points (which is a straight line) and draw it and measure the length of this path ... then I curve the paper with me hands so that the straight line we draw would be curved .
    obviously the length of path when it's curved is the same as the length of it before i curve the paper , only the way it looks will differ (because of the curvature) .
    why don't we say the same about the space in GR ???

    thanks in advance ...
     
  9. Jan 2, 2010 #8

    Nabeshin

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    My preliminary response is precisely because we cannot simply "ignore" the time component of the metric. GR deals with warping of spacetime, not simply space. If you take out the time portion, you're just discussing differential geometry.
     
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