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Dropping charges to Rindler horizon

  1. Jun 11, 2015 #1
    Let's say I keep on dropping electrons on one spot of a Rindler horizon. Does the charge of the spot increase without a limit?

    When the charge of the spot is very large, does the spot exert a Coulomb force on the electron I'm about to drop, causing the electron to start moving away from the spot, when I let go of the electron?
     
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  3. Jun 11, 2015 #2

    PeterDonis

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    You can't. The Rindler horizon is not a "place in space" that you can pick out a spot on. Draw a spacetime diagram of this scenario and you will see why.
     
  4. Jun 12, 2015 #3

    pervect

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    I'm not sure how to interpret the question, the closest I can come is to imagining a line-charge in flat Minkowskii space with some unspecified force holding it together. Perhaps the charges are implanted in a solid dielectric. Then we can ask how the line-charge looks to an accelerating observer. In Rindler time, the charge will not reach the event horizon at a finite value of the Rindler time coordinate, because the Rindler coordinates don't cover all of space-time.
     
  5. Jun 14, 2015 #4
    I'll answer myself, then people can correct.

    I'm in an accelerating rocket dropping electrons.

    A momentarily co-moving and co-located inertial observer says there are many fast moving dropped electrons nearby, with Lorentz-contracted electric fields.

    So there are many electrons nearby, but no large Coulomb-force.
     
  6. Jun 14, 2015 #5

    PeterDonis

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    No, he doesn't. He says there's one electron that's at rest relative to him (the one that was dropped at the instant he is co-moving with the accelerating rocket), another just behind him that's moving backwards a little bit relative to him, another a bit further behind him that's moving backwards a little faster, etc., etc. The Coulomb fields of these electrons are not significantly contracted.

    No; there are many electrons spread all over space behind the momentarily co-moving and co-located inertial observer, because they were all launched with different velocities relative to that observer and at different times in the past. So their Coulomb fields are at varying distances from the observer; by the time we get to electrons moving near the speed of light relative to that observer, they are also very far away, so their fields are much weaker even apart from Lorentz contraction.

    Once again, drawing a spacetime diagram of the scenario will make all this obvious.
     
  7. Jun 14, 2015 #6

    Nugatory

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    Another way of thinking about this problem:

    It is the exact same flat Minkowski spacetime whether you use Rindler coordinates or ordinary Minkowski x,y,z,t coordinates to do your calculations, just as the surface of a plane and all the shapes drawn on it are the same whether you're using Cartesian or polar coordinates. The only reason to use one coordinate system instead of another for a particular problem is that you always want to choose the coordinate system to make the problem easy.

    Now, Rindler coordinates are great for working out what the person accelerating in the spaceship observes, but not so great for figuring out how the electrons dropped from the spaceship behave So what you do is work out the behavior of the electrons using the ordinary Cartesian coordinates of an observer who is not accelerating... And then transform that result into Rindler coordinates when you're done if you want to.
     
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