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Observing a cube approaching a black hole

  1. Oct 19, 2013 #1
    If a distant observer were to observe a large cube made of strong material approach a black hole, what would he see?

    ISTM that if one of the faces of the cube were to be the nearest approaching portion, he would see the four edges of the face become shorter and curved, and he would see the center of the face bulge out. ISTM that he would also see the sides of the cube lose their parallel configuration, and that they would converge in the direction of the black hole.

    Is that correct? Or would the curvature of time affect what he sees in addition to the effects of the curvature of space?
     
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  3. Oct 19, 2013 #2

    phinds

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    It would depend on the size of the BH. For a small one, spaghettification would occur far enough outside the event horizon that the remote observer would see it and your scenario would not be correct.

    For a really big one, your scenario might be correct --- I don't know.
     
  4. Oct 19, 2013 #3

    Bill_K

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    Here's a discussion of the Special Relativistic effects on the appearance of a moving cube.
     
  5. Oct 19, 2013 #4

    PeterDonis

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    I'm not sure why you think this would happen. For one thing, I'm not sure if you think these are actual effects on the cube (i.e., an observer near the cube would see them too), or if you think they are optical effects due to the way light paths are distorted on their way out to the distant observer.

    As far as actual, local effects on the cube, there aren't any unless tidal gravity is strong enough to be significant on the length scale of the cube. If we assume the hole is large enough that that isn't the case, then an observer near the cube would see it as a normal cube.

    As far as distortion of light paths, I have not worked through that in detail, but I think it would depend on where the distant observer was. If the observer is directly above the cube, I don't think there would be any optical effects (except for apparent time dilation, see below). If the observer was looking "from the side", there might be other effects, but I'm not sure they would be similar to those you describe.

    If by "curvature of time" you mean, would the observer see the cube appear to slow down as it falls, yes, he would. But I don't think that would change the apparent shape of the cube.

    For "effects of curvature of space", as above, I don't think there are any on the cube itself, but there might be on the paths of light rays.
     
  6. Oct 19, 2013 #5

    A.T.

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    Isn't the effect of falling itself and tidal forces related to gravitational time dilation ("curvature of time").

    If you build a large rigid 3D object in flat space and move it into curved space, wouldn't it break?
     
  7. Oct 19, 2013 #6

    A.T.

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  8. Oct 19, 2013 #7

    WannabeNewton

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    You can have gravitational time dilation without any tidal forces (vanishing space-time curvature); take for example Rindler space-time.
     
  9. Oct 19, 2013 #8

    A.T.

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    Never suggested otherwise. And here we talk about a black hole.
     
  10. Oct 19, 2013 #9

    WannabeNewton

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    The point is that there is no relation. What relation is there between tidal forces and gravitational time dilation if one can exist without the other? The derivation of the gravitational time dilation formula between two static observers (and their clocks) in stationary space-times makes no reference to tidal forces.
     
  11. Oct 19, 2013 #10

    A.T.

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    Can tidal forces exist without gravitational time dilation?
     
  12. Oct 19, 2013 #11

    WannabeNewton

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    No but that doesn't imply there is a relationship between them; again, gravitational time dilation can exist without tidal forces. Can you derive a formula relating the two? Physically, gravitational time dilation depends upon the inertial/non-inertial states of the two observers involved; the gravitational field of course affects what constitutes an inertial/non-inertial frame. Consider the intimately related but more "intuitive" concept of gravitational redshift: when a light ray is emitted by a static observer and received by another static observer at a different altitude in a static gravitational field, the receiving static observer measures a frequency shift but if the receiving observer is freely falling then no frequency shift will be observed. How is this related to tidal forces, which involves the relationship between the relative acceleration of infinitesimally separated worldlines in a time-like congruence and the Riemann curvature tensor?
     
  13. Oct 19, 2013 #12

    A.T.

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    If tidal forces cannot exist without gravitational time dilation, doesn't that imply some dependency?

    Does this preclude a relationship? In differential relationships a derivative might vanish in some special cases.
     
  14. Oct 19, 2013 #13

    WannabeNewton

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    The only dependency is that if tidal forces exist then the space-time is curved and in curved space-time, gravitational time dilation can be observed in an observer dependent sense. There is no implication just from this that tidal forces and gravitational time dilation have any direct relationship.

    That is true but the case here is different: we can talk about gravitational time dilation without any need for tidal forces in full generality, not just in special cases.

    I assume you have a good physical reason for why you believe tidal forces and gravitational time dilation are directly related. What is your reasoning? I'm not saying that I'm absolutely right and that you're unequivocally wrong, hence seeing your reasoning would help fine tune the discussion. Keep in mind that you can (for example) transform away the effect of gravitational redshift in a freely falling frame but you cannot transform away the Riemann curvature tensor in a freely falling frame.
     
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