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Parallel transport to explain motion of light near black hole?

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  1. Apr 21, 2015 #1

    bcrowell

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    I'm currently teaching a gen ed course called Relativity for Poets. This is the first semester I've taught it, and it's been a ton of fun so far. If anyone is curious, here is the class's web page with links to the syllabus and lecture notes. The required texts are Takeuchi, Stannard, and Ostriker and Mitton. We have currently gone through SR (Takeuchi) and most of GR (Stannard plus my lecture notes). Although this is a class for non-science majors, I have decided that any conceptual apparatus that doesn't require an equation is fair game, so I've taught them some pretty advanced ideas, including parallel transport and Penrose diagrams.

    In my office hours this week, one of my students asked what parallel transport was really good for, which seemed like a fair question at this point in the course. So far I've explained it using pictures on a sphere and used it as a qualitative explanation for the geodetic effect as a test of GR. One additional application I had in mind was that one can use it to answer the FAQ about where the universe gets the energy required for accelerating expansion -- but we haven't done cosmology yet, so they aren't really ready for that.

    What we are doing right now is black holes, and off the cuff, I proposed parallel transport to my student as a way of explaining how the speed of light near a black hole can appear "wrong" to a distant observer. Now that I think about it some more, I'm not so sure that this explanation makes sense. There are certainly difficulties in GR with defining how an observer describes a velocity vector that exists at a distant point in space. However, parallel transport preserves the norm of a vector, so if a ray of light at the event horizon of a black hole has a lightlike (unnormalized) velocity vector according to a local observer, then no matter where we transport the vector or what path we transport it along, it will still be lightlike (although possibly rescaled).

    Is there any way to salvage my idea of applying parallel transport to this topic?

    The only book I could find that had a relatively elementary treatment of this in any detail was Taylor and Wheeler, Exploring Black Holes. They refer to Schwarzschild coordinates as "bookkeper coordinates," and say that the bookkeeper's low speed of light is what is observed in the Shapiro time delay.

    This seems fine, but doesn't necessarily fit the pedagogy of my course, because I never even define the Schwarzschild coordinates or metric. We have discussed the fact that the event horizon is lightlike and the singularity is spacelike, and they've also seen the Penrose diagram for a black hole.

    Does this topic work at all as an application of the concept of parallel transport, or if not, is there some other application to black holes?
     
    Last edited: Apr 21, 2015
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  3. Apr 21, 2015 #2

    PeterDonis

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    I'm not sure parallel transport will help in explaining how the apparent speed of light changes, because a null vector parallel transported anywhere is still a null vector, so it will still appear to have a speed of ##c## when compared to the 4-velocity of an observer. What will change is the apparent frequency of the light. So you could explain gravitational redshift this way, but not an apparent change in the speed of light.
     
  4. Apr 21, 2015 #3

    bcrowell

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    Yeah, that was essentially what I realized after saying the wrong thing to my student.

    Hm...this might work. I've taught them energy-momentum vectors, but not frequency vectors. So I could say that the Doppler shift is what you get when you parallel-transport the p vector of a ray of light along its own path. Conceptually it might seem confusing to them that you can rescale the p vector through parallel transport, while not changing its (zero) norm. However, they already sort of know about this Doppler shift because both Stannard and I present it as an example of gravitational time dilation. (Stannard and I both present this by using elevator arguments and the e.p. to show that gravitational time dilation depends on height. I don't go into the details of how to generalize "height" to a nonuniform field.)
     
    Last edited: Apr 21, 2015
  5. Apr 21, 2015 #4

    bcrowell

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    Maybe a simple and reassuring example would be to discuss parallel-transporting the energy-momentum vector of a massive test particle, say a comet, along its geodesic world-line. Since parallel transport preserves the norm, the comet's mass is unchanged. Common sense, hurray! However, the direction of the momentum 3-vector relative to the stars does change over time. One could also discuss the point of view of an observer who coasts alongside the comet, and says nothing changes at all.

    The stranger example of parallel-transporting the p vector of a ray of light might then be easier to absorb. Its mass is zero and doesn't change. An observer can't fly alongside it, so there is no reason that the vector can't be rescaled.
     
  6. Apr 21, 2015 #5

    PAllen

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    I don't think you need any idea of rescaling the parallel transported (null) 4-momentum. The 'scale' of a null 4-momentum (i.e. the magnitude of the time component in some local frame) only has meaning in relation to local frame of some measuring apparatus. Thus, if you imagine parallel transport of both the light 4-momentum and the emitter 4-velocity, you find nothing has changed per the 'transported emitter' (parallel transport of two vectors preserves their dot product, which is the time component of the null momentum in the emitter's tetrad). What changes is really the 'spacetime direction' of the emitter 4-momentum (the direction that measures the originally emitted energy). Without curvature, you expect that if the emitter and receiver maintain constant distance from each other, then receiver would correspond to the transported emitter 4-velocity, and thus measure unchanged energy. Curvature changes this expectation: despite constant distance between these static observers, at the receiver event, it is a non-static observer that would measure the emitted energy - specifically, a receiver with motion corresponding to the transported emitter 4-velocity.
     
  7. Apr 21, 2015 #6

    bcrowell

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    Hmm...suppose I start with a null vector at event A and parallel-transport it via two different paths to event B. Then isn't it possible for the two copies of the vector, on arrival at B, to be parallel but to differ by a scaling factor? (Maybe we wouldn't expect this to happen if we were parallel-transporting energy-momentum vectors of light rays along geodesics, as in gravitational lensing. In that example, I would expect the two copies to be non-parallel, in which case the comparison of their lengths would be observer-dependent.)
     
  8. Apr 21, 2015 #7

    PAllen

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    I implicitly assumed transport along the null geodesic path from your comment "when you parallel-transport the p vector of a ray of light along its own path.". True, with gravitational lensing, you can still end up with different results for two paths, but that doesn't necessarily mean anything changed about the light along the way: each path preserves the energy of the light relative to the emitter 4-velocity transported along the same path. If you want to say there is rescaling, you have to say in relation to what. For each path, at the reception event, if the receive is moving in the same way as the transported emitter velocity, they see no change in the light [from that direction].
     
  9. Apr 21, 2015 #8

    PAllen

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    Note, I do not dispute the ability to consistently treat 'photons' as losing/gaining energy in relation to a potential for stationary spacetimes. However, that is a special case. Without a stationary spacetime, what chooses the reference observers against which energy change is measured?
     
  10. Apr 21, 2015 #9

    pervect

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    The explanation I'm fond of is to consider a naval ship sailing east near the north pole and near the equator. I liken the coordinate speed to the number of degrees of longitude the ship crosses in an hour. The main point is that the ships have the same physical speed relative to the water (which would be a static observer in the black hole ), but different coordinate speeds (degrees of lattitude per hour). The point is that physical speeds are not the same as coordiate speeds. The approach demands a small amount of math - one has to be able to use the metric to convert coordinate changes into distances. Which may be too much math for a poet :(.

    There is a somewhat subtle issue here - for the analogy to make total sense, one really has to think of one of the dimensions on the sphere representing time while the way it's presented, this point is not made explicity.

    I'm not sure if this idea will fit into your parallel transport idea or not - probably not. I'd describe the issue as being due to coordinate choices and the metric coefficients rather than parallel transport.

    Basically, because of gravitational time dilation, one proper second deep in a gravity well doesn't correspond to one coordinate second. So proper speeds and coordinate speeds are different, with d proper speeds being physically more significant.

    I personally am rather fond of descrbing parallel transport as following logically from the sides of a small quadrilateral with equal opposing sides being a parallelogram - so that tells us how to geometrically construct a parallel transport, in an approach rather similar to Schild's ladder. But I haven't seen this exact approach used. It only works if you don't have torsion, it's based on the fact that without torsion, parallel trasport yields closed parallelograms to the third order, the effects of curvature being second order, somethig I first read in Penrose "Road to Reality". Turning this idea backwards, I use the property of closure of small enough parallelograms to define a geometrical procedure for parallel transport.
     
  11. Apr 21, 2015 #10

    bcrowell

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    What I'm saying (which may be wrong) is that if the two copies of the same vector transported along different paths end up parallel to one another, then you can say whether they're rescaled in relation to *each other*. But as I think about it, I'm not so sure that this is actually possible. It might violate a symmetry of the Riemann tensor or something.
     
  12. Apr 22, 2015 #11

    bcrowell

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    This is actually a really nice analogy. The thing is, the way I've taught the whole course really de-emphasizes coordinates. As much as possible, I've tried to do things in a coordinate-free approach, as in, e.g., Bertel Laurent's SR book. So if I say to them, "Hey, coordinate velocities aren't meaningful, so don't worry about the coordinate velocity of light not being c," they would probably feel that I was responding to something that wasn't even a concern that had occurred them. What they do know is that, for example, a vertical line on a spacetime diagram represents an object at rest, so that if we represent a black hole's event horizon as a vertical line on such a diagram, it means that light rays trapped on it are at rest, which is weird. The two ways I've been addressing this have been (a) to draw tipping light cones and stress locality, and (b) to introduce Penrose diagrams.
     
  13. Apr 22, 2015 #12

    stevendaryl

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    I might be missing something, but are you considering something to "involve parallel transport" in the case of parallel transport around a loop, or does something involve parallel transport whenever connection coefficients are used? The connection coefficients are defined via parallel transport. So the fact that the geodesic equations of motion includes [itex]\Gamma^\mu_{\lambda \nu}[/itex] terms is due to parallel transport, indirectly. So the fact the coordinate speed of light varies with [itex]r[/itex] in Schwarzschild coordinates is indirectly due to parallel transport.
     
  14. Apr 22, 2015 #13

    PAllen

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    Given the discussion of transporting on null geodesics, I think it is impossible. It would mean that there are two different geodesics from a given point with a given tangent, and (at least per the parallel transport definition of geodesic), this is impossible. It would mean spacetime is acting as beam splitter (running things back from the reception event).
     
  15. Apr 22, 2015 #14

    PeterDonis

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    Yes, this is a case where something has to give: either you have to relax the definition of "at rest" so that a lightlike object can be at rest, or you have to switch to a coordinate chart where lightlike curves are never at constant spatial coordinates.
     
  16. Apr 22, 2015 #15

    bcrowell

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    Right, I agree that it can't happen if you're transporting along null geodesics. I'm thinking it probably also can't happen at all; you would then have a case where transporting a vector around a closed path results in rescaling the vector, which would be pretty weird.
     
  17. Apr 22, 2015 #16

    bcrowell

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    Right, and the latter is essentially what I've done by showing them the Penrose diagram, although there is no discussion of coordinates per se.

    There is also a third option, which is actually the one that I've pushed the hardest: simply deny that it's meaningful to talk about an observer's perception of the velocity of a distant object.
     
  18. Apr 22, 2015 #17

    bcrowell

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    Right. Of course this particular way of expressing it is much too technical for these folks. The way I presented this to them was an example in which a bug is confined to the surface of a sphere. The bug pushes a frictionless hockey puck forward, and then follows the puck as a way to navigate and avoid getting lost -- he can tell he's going along a geodesic. The puck parallel-transports its own momentum vector.

    Relating this to coordinates and coordinate velocities is something I haven't done for them, mainly because we simply aren't using coordinates much. Keep in mind that this is essentially a relativity course without equations. (We do have a few equations here and they're, but they're few and far between, and not emphasized.)
     
  19. Apr 23, 2015 #18

    PAllen

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    Thinking about this more, I am not so sure there is any reason this can't happen in GR for arbitrary paths. Obviously, transporting a vector around a closed path can change the vector - that is curvature. (There a limits in what can change - timelike remains timelike; null remains null; norm is preserved; dot product of two vectors transported together remains constant). Further, transporting a null vector, in general can rescale it (that's one way of doing gravitational red/blue shift). So, the question is whether transporting it on some arbitrary, closed, spacelike path can end up with nothing but rescaling (that is, the spatial direction has returned what it started at, but a rescaling has occurred). I'm coming up dry on any principle of differential geometry that would prohibit this. Null vectors are funny - the scale has nothing to do with the norm, and is coordinate dependent.
     
  20. Apr 23, 2015 #19

    PeterDonis

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    The only way this could happen is if the "spacetime direction" of the vector changed--i.e., it would have to be tangent to a different spacelike hypersurface passing through the same point after being transported around the closed path. I can't come up with a specific example, but, like you, I can't think of anything that would prohibit it in principle.
     
  21. Apr 23, 2015 #20

    bcrowell

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    I could be getting this wrong, but I think the following argument shows that you can't rescale a null vector under parallel transport around a closed path. The Riemann tensor has the symmetry [itex]R_{abcd}=-R_{bacd}[/itex]. So suppose you start with a null vector that points in the direction a, and you parallel-transport it along legs of an infinitesimal parallelogram that point in directions c and d, and then back along the opposite sides of the parallelogram. Then the growth or shrinkage of the vector is proportional to [itex]R_{aacd}[/itex], but this is zero by symmetry.
     
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