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Local detectability of frame-dragging

  1. Nov 14, 2009 #1

    bcrowell

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    I'm puzzling over a certain aspect of the interpretation of frame-dragging.

    Frame-dragging says that the angular momentum of a body makes itself felt in a certain way in the curvature of the spacetime surrounding it. In GR, you typically can't point to a certain spacelike surface with a boundary around it, and say, "The amount of angular momentum inside the boundary is x." This only becomes possible under certain assumptions, e.g., asymptotic flatness. MTW have a nice argument to this effect, which is that in a closed universe, the boundary has two sides to it, so the flux of some quantity passing through the boundary cannot unambiguously be attributed to either of the two regions on the two sides.

    In an experiment to detect frame-dragging, it therefore seems to me that you must carry out some measuring operations that depend on asymptotic flatness. For example, you could build two gyroscopes A and B out in the flat region, then carry A in close to the rotating body, loop it around once in the equatorial plane, and then transport it back out to the flat region and compare it with B. Call this experiment #1.

    On the other hand, suppose you have two satellites, C and D, one in a prograde equatorial orbit and one in a retrograde orbit. Frame dragging causes them to have different orbital periods, and I think this is *locally* measurable. E.g., you can have the satellites depart from a certain starting point in opposite directions, then reunite on the other side, and I think you would see a different amount of proper time on their clocks. Call this experiment #2.

    In practical terms, Gravity Probe B has verified frame-dragging to 15%. It used a distant star as a reference point, so it certainly wasn't carried out entirely locally.

    The interpretation that I'm thinking is correct is that although experiment #2 is purely local, in a universe without asymptotic flatness the results can't be interpreted unambiguously as a measurement of the angular momentum contained *inside* the orbit. Is this correct?

    TIA! -Ben
     
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  3. Nov 14, 2009 #2

    pervect

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    Let's think about the analogous situation with mass. Finding the mass of a body requires certain assumptions, depending on the sort of mass, just as finding the angular momentum does.

    But if we have a small region of space, we know that if it contains matter the Ricci will be nonzero, and if it doesn't it will be zero, a purely local measurement.


    This sounds similar to the situation with mass - one reason I chose it as an analogy is that I'm more familiar with it, but I think you are on the right track.
     
  4. Nov 14, 2009 #3

    bcrowell

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    Thanks for the reply, Pervect, that's good food for thought! I agree that there's nothing special about angular momentum. You could just as well talk about electric charge, and I think all the issues would be the same. But I'm not convinced that using mass is a simplification. The local value of the Einstein tensor lets you infer the local value of Tab, which isn't the same as measuring the local density of mass. For one thing, Tab is frame-dependent, and if you don't have an asymptotically flat background, you can't even decree a standard Lorentz frame in which to measure Tab.

    The other issue that occurs to me is that if you want to determine a body's total mass, charge, angular momentum, ... you can't necessarily do it by internal measurements. In the case of the earth it's impractical, and in the case of a black hole it's not even theoretically possible. But if you measure the Ricci or Einstein tensor on the exterior, you get zero, which doesn't tell you anything.
     
  5. Nov 14, 2009 #4

    pervect

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    Probably it was a mistake to use mass for the example - perhaps energy would be a better choice. T_00 measures the local energy density - but to determine the total energy of the system, you still need to look at a special space time (asymptotically flat, or perhaps a stationary one).

    The issue is that you can't get a sensible answer for "total energy" just by multiplying T_00 by the volume and integrating - the energies are all defined in different tangent spaces and don't add in that manner.
     
  6. Nov 14, 2009 #5

    bcrowell

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    Okay, it sounds like we've converged on an interpretation we agree on. Thanks!
     
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