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Question about equivalence principle

  1. Aug 5, 2011 #1
    Let me restate my problem in a more clear way i think.
    Situation A: a accelerated rocket in flat spacetime. the observer in the rocket could think he is in a gravitation field equivalently.All experiments he do are like in a gravitation field downward.

    SituationB: a apple falls down under the gravitation force from the tree, or "in GR way", caused by the curved spacetime which is disturbed by the earth.
    Can I say that the gravity in A is different from in B? or the gravity in B is real , in the sense that it is caused by another object. After all what we called gravity is a interaction of two bodies , no matter you intepreted as curved spacetime or force, which both demand a source (the earth in B).Obviously in case A there is no such a source.
  2. jcsd
  3. Aug 5, 2011 #2
    The idea is as you as an observer in a closed capsule cannot differentiate between Situation-A and Situation-B. This is called the principle of equivalence.

    After the formulation of space-time continuum in Special theory of relativity, the immediate afterthought was "What happens to space-time during acceleration?". The mathematics showed curvature of space-time during acceleration.

    Because of the principle of equivalence, instead working with the unknowns in gravity, we can as well work with a simple system of acceleration. Equations will be applicable in both cases.

    Einstein did this and got GTR.

    Makes sense?

  4. Aug 5, 2011 #3


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    The rindler metric has accelerated observes but the intrinsic geometry of rindler space is flat. The presence of intrinsic curvature is seen with the non -vanishing of the Riemann Curvature tensor and is related to mass - energy distributions / fluxes as per the EFEs.
  5. Aug 5, 2011 #4
    I didn't get a word you said. Yet, you must be right. The 'observable' space-time must be a 'Riemann Curvature tensor' thingy. Yes?
  6. Aug 5, 2011 #5
    acceleration do not generate curvature, it can be treat just in special relativity.And do you answer my question?
  7. Aug 5, 2011 #6


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    There are flat space - times that describe accelerated observers so acceleration does not directly generate curvature. Curvature is generated by mass - energy as is quantified by the field equations of GR. The tensor in question quantifies curvature and if it vanishes then you can say you are dealing with flat space - time, a lack of acceleration does not imply the same.
  8. Aug 6, 2011 #7
    hey everybody, please answer my question.....
  9. Aug 6, 2011 #8


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    I'm not quite sure what your question is, here.

    The first part seems to be a pretty good summary of the equivalence principle. (But not phrased as a question).

    In the second part seems to be that you're looking for GR to be just like Newtonian theory, with some sort of "force between bodies" defined.

    So I guess my best shot at an answer at this point would be that you're on the right track in part 1 as far as understanding what is usually meant by the equivalence principle in basic textbooks, and getting off on a speculative track in part 2, where you start talking about "force between bodies".

    You could conceivably be trying to talk about Mach's principle in part 2, I suppose.
  10. Aug 6, 2011 #9


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    In GR, real gravity that cannot be mimicked by acceleration is called "tidal gravity". One has to do "non-local" experiments to see tidal gravity, whereas the equivalence principle applies only to "local" experiments. In realistic cases, tidal gravity indicates the presence of matter somewhere in the universe. However, GR does contain solutions in which tidal gravity is present even without matter.

    Last edited by a moderator: Apr 26, 2017
  11. Aug 6, 2011 #10
    my question is that is the gravity the guy in rocket feel is real? in the 2 art I give what i mean by real.
  12. Aug 6, 2011 #11
    then what the gravity in case A is? if I am a man in the rocket, i want to a explaination of the gravity or the fact every thing around in rocket is falling, how could I explain?

    For this statement "In a reference frame that is in free fall, the laws of physics are the same as if there were no gravity at all ", I can understand, you can say there are no gravity at all , all body is go its geodesics, for their own frame, the world are minkowski spacetime.
    However i can't understand the inverse statement, "in a gravity-free region of space, objects fall towards the floor if the room we are in is being accelerated"( my situation A). how could I give this a reasonble explaination, i mean what is the cause of this phenomenon?
    Last edited: Aug 6, 2011
  13. Aug 6, 2011 #12


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    If the man in the rocket does only "local" experiments, he could imagine that he is either accelerating or in a gravitational field. He cannot tell the difference.

    If he does "non-local experiments" (let's say he has a slightly bigger rocket), then he can distinguish between the two possibilities.
  14. Aug 7, 2011 #13
    But I think that equivalence of gravity and acceleration is not general to global scale. There is never a uniform gravitational field which corresponds to uniform acceleration because there is curvature...
  15. Aug 7, 2011 #14
    It appears to me you are comparing apples and oranges.

    An observer undergoing constant proper acceleration is not equivalent with an observer free falling, it is equivalent with an observer trying to be stationary in a gravitational field.
  16. Aug 9, 2011 #15
    The effect of gravity on an object is it pulls down on every atom, electron and particle. The floor pushing up on an object giving the illusion of gravity (like an elevator that starts to go up) seems to me to be a different phenomenon and not indishtinguishable from gravity. Any serious spaceman with sensitive equipment would be able to measure the difference, unless someone smarter was able to attach miniature booster rockets to every atom in order to mimic gravitation perfectly.:smile:
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