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Experiment with light

  1. Oct 25, 2005 #1
    Hello excuse my bad english

    u take two large masses for example two black holes.
    Now we imagine we could bring the two masses very near together.
    And now we send an laserstream(lightstream) exactly in the middle of the distance between the 2 masses.
    The question is what happens to the stream? On left and right side is a very huge gravity-potential. What does the light here? Bcause if the photon is an particle, such an huge gravity must have any effects on this particle isnt it?
    Or if light is in this case wave what happens then? Einstein says the light follows the room-crookedness (raumkrümmung) this would mean that nothing happens? Is someone there how knows mor about this?
  2. jcsd
  3. Oct 25, 2005 #2


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    Welcome to these Forums silici!!

    You have certainly started with an interesting question.
    The solution to which is not trivial as it is an example of the two body problem and in GR that in general can only be solved numerically.

    We assume the two masses are not in orbit with each other or individually rotating - that would introduce further complications. In a gedanken, or thought experiment, you could just release the two masses so they are momentarily at rest as the light beam is shone exactly down the middle of the two.

    At the midpoint, and along the axis of symmetry of the system, the gravitational fields do not cancel out. There is no space curvature at the mid-point although there would be (for close black holes) considerable space-time curvature - the components of the Riemannian tensor are not all zero.

    The effect would be that the light beam is not deflected in space but it is delayed in its passage between the point of emission A & and point of reception B. The spatial distance between the two points A & B is increased by the curvature of space-time compared to the distance between the same points without the masses.

    I hope this helps.

    Last edited: Oct 25, 2005
  4. Oct 26, 2005 #3
    Hello Garth

    thank you very much for your fast answer!
    You re the first one i know who has an answer for this question.
    It is very plausible how you told it.
    If a have any probs like t5his, now i know where i ve to go.

  5. Nov 23, 2005 #4


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    It should be obvious by symmetry that the photon must travel undeflected through the strait between objects. I think that the only effects observable by a distant receiver would be the time delay.

    I suspect that if the objects were Solar Mass objects (less intense than black holes), the time delay would be essentially (a "wee bit" more than) double the Shapiro portion of the time delay for a single (offset) object.
    (I THINK the only other portion of the time delay, in standard GR, is due to geometric deflection).

    I googled "shapiro time delay" and came upon this recent article by Clifford Will (a reputable source): http://arxiv.org/PS_cache/astro-ph/pdf/0301/0301145.pdf [Broken]
    C.Will calculates the delay of light that grazes the Sun, and gives the Shapiro Time Delay as 70 microseconds.
    It seems that this should be pretty easy to verify, since there are pulsars within ¼ degree of the ecliptic, but I don't know whether it has been done yet.
    C.Will writes the Shapiro time delay as:
    \Delta = - \frac{2 G m}{c^3} ln [ |r_s| - \hat{k} \cdot x_r ] [/tex] . With a solar mass as 2E30 kg , the constant term is 9.88 microseconds.
    The log term is perplexing, with units that don't cancel! Maybe these are all supposed to be unit vectors; but then, why write "|r_s|" instead of "1"?

    Now, if these are supposed to be unit vectors, the argument of the log is just (1 - cos theta), with theta = d/r being the anglular radius of the Sun as seen from Earth (about ¼ degree). Putting that in radians, I get ln [ 10.9E-6 ] = -11.4 . So I compute a time delay of -113 microseconds ; quite a bit different from C.Will's .
    C.Will derived (and probably used) an approximate form for this time delay:
    \Delta = - \frac{2 G m}{c^3} ln [ \frac{2 r_s}{d^2}] [/tex] .
    In small-angle approximation, [tex] (1 - cos \theta) \approx \frac{1}{2} {\theta}^2 [/tex] ; it seems that he reciprocated the argument of the log to get a negative.
    But HIS argument is still missing a "r", so it STILL has units!
    And 5 mrad is small enough that the approximation should be okay.

    Anybody know what he's doing with the logarithm?
    Or why I'm getting such a different answer than he does?
    Last edited by a moderator: May 2, 2017
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