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Hovering in gravity makes curved space flat?

  1. Sep 10, 2004 #1
    Have I got it right this time?

    Red-shifting is determined by the difference between the space-time curvature at the light source and the observer's space-time curvature.

    For the light source, all that matters is the distance from the center of gravity -- I think the motion of the source is irrelevant.

    For the observer, if he is free-falling, then his curvature varies with distance from the center of gravity, increasing from flat at infinity to infinite at the singularity.

    But if the observer is hovering at rest relative to the center of mass that means that his rocket thrust is unbending the curvature of gravity so that his space-time is made flat -- the same as the space-time of the free-faller at infinity. The distance of the hoverer doesn't matter. Whether he hovers just above the event horizon or at infinity or anywhere in between, his space-time will be flat, because his rockets are exactly cancelling out the gravity-caused curvature wherever he is.

    Flat space-time means that there is no net acceleration relative to the center of mass, however it feels like acceleration to the hovering observer -- his reference frame is not inertial. The free-faller at infinity has no net acceleration relative to the center of mass and he feels no acceleration -- his reference frame is inertial.

    The difference in curvature between the source at the event horizon and an observer in flat space-time is just enough so that the observer sees time stopped at the event horizon.
     
  2. jcsd
  3. Sep 11, 2004 #2
    In the previous post I neglected to mention that I was talking about red-shifting due to a black hole.

    When I said that I think the motion of the source doesn't matter, I should have said that I think the acceleration of the source doesn't matter. Very vast velocity of the source relative to the observer would produce a special-relativistic red-shift. But in this thought experiment, let's say that the source, the free-falling observers, and the hovering observers are all initially at rest with respect to the center of gravity.
     
  4. Sep 11, 2004 #3

    pervect

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    This doesn't work as written, but something roughly similar can be said.
    *IF* and only if, the metric components are not a function of time,
    red shifting is determined by the ratio (not the difference) of the metric tensor coefficient g00 at the source and destination.

    Specifically, [tex]\frac{\lambda_{local}}{\sqrt{g_{00}}}[/tex] is a constant when the metric coefficients (all of them) are not functions of time, but only functions of space.

    So the ratio between the wavelength measured at point 1 and the wavelength measured at point 2, measured locally in both cases, can be given as the ratio of the square roots of g00 at point 2 and point 1.

    g00 should not be described as 'curvature' though. For a static metric, like that around a Schwarzschild black hole, you can think of the redshift between a local observer and an observer at infinity to be a measure of 1/g00^2. You might say informally that g00 is "gravitational redshift", or "gravitational time dilation". Calling g00 a curvature is not right and will cause major confusion.

    This is correct.

    This is correct. The curvature here is a different entity than the g00 that determines redshift (which I mentioned should not be called a curvature). This curvature is known as the Riemann tensor. One specific component of the Riemann can be measured (for instance) by measuring the tension on a 1 meter rod connecting two 1kg masses when the rod is pointed radially. There are a total of 20 components in the complete Riemann tensor. Six of them can be measured by measuring the stretching, compression, and torques on a rod. All 20 of them can be deduced by studying the geodesic paths of nearby observers moving at different velocities.

    No. Technically, the Riemann tensor would be measured by "cutting the jets" for a very brief time period to measure the force on the rod. One could also look for the effect of the Riemann without actually cutting out the jets simply by subtracting the known forces due to the acceleration, and looking for any remaining forces which are unaccounted for by the local acceleration.

    The Riemann tensor in general can depend on the velocity of the observer. For the special case of a Schwarzschild metric, the radial component of the velocity does not affect the Riemann tensor, so it doesn't matter whether or not the observer is falling in, or momentarily at rest. Velocity components in the other (non-radial) directions do affect the Riemann tensor if they are high enough to be relativistic, even for a black hole.

    If I understand this corretly, it's right. I'd say this more succienctly as Flat space-time means no tidal forces.
     
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