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Equivalence Principle & Tidal Effects

  1. Dec 28, 2008 #1
    It is my understanding that the Equivalence Principle postulates that if I were standing in a closed room, I would not be able to distinguish whether the downward force that I felt was caused by (1) the presence of a massive body such as the Earth exerting a downward gravitational force or (2) the upwards acceleration of the entire room. Would I not in principle be able to measure a tidal effect (slightly less downward force measured at the ceiling than at the floor) in the gravitational case that would not be present in the acceleration case?
     
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  3. Dec 28, 2008 #2

    atyy

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    I don't think know if it's possible to produce a uniform gravitational field by acceleration. The most common accelerated observer is the Rindler mouse, but it doesn't experience a uniform gravitational field.

    Nonetheless, tidal effects do go beyond the equivalence principle. Some ways to think of the EP are that in Newtonian gravity, an accelerating free falling accelerometer will read zero. Hence we should redefine acceleration to be what an accelerometer reads. This leads to the Newton-Cartan formulation of Newtonian gravity as curved spacetime. From this point of view, Einsteinian gravity is a reworking of Newton-Cartan gravity to include special relativity. (Historically Einstein discovered GR first, and inspired Cartan's reworking of Newton.) In GR, the EP is the statement that (i) at every point in spacetime we can set up coordinates so that the metric is exactly flat at the origin, and deviates from flatness away from the origin at second order or more in Taylor series (ii) the laws of physics take the same form at the origin as in flat spacetime. In GR (i) is always true, but (ii) is true only for "first derivative" laws such as Maxwell's equations in GR, but it is not generally true for "second derivative" consequences which can detect spacetime curvature or tidal effects at an event. Ohanian's textbook, and Carroll's and Blandford and Thorne's online texts have more discussion of this.
     
  4. Dec 29, 2008 #3
    The simple answer is that the equivalence principle equates acceleration of the "room" with a uniform gravitational field. Tidal effects are due to non-uniformity of a gravitational field.

    Al
     
  5. Dec 29, 2008 #4

    JesseM

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    Can't we also say that the equivalence principle holds in curved spacetime when you take the limit as the size of your "room" (and the time interval in which you are making your measurements) approaches zero, and that in this limit the tidal forces measurable in this patch of spacetime approach zero as well?
     
  6. Dec 29, 2008 #5

    atyy

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    Perhaps related: is a non-zero second derivative local or nonlocal? It is local in the sense that it is a well-defined limit at every point. It is non-local in the sense that a derivative, and even more so a second derivative, always involves taking a difference in space or time.
     
  7. Jan 3, 2009 #6
    The reason Einstein was so happy when he found the "Equivalence Principle", because its predicting power. He used it to predict that light must fall/bend in gravitational field, since he deduced that light must bend in an accelerating elevator

    True, tidal effect would not be present in the acceleration case. But can you use tidal effect to invalidate Einstein's prediction of light bending? If not, then Equivalence Principle still hold regardless of tidal effect
     
    Last edited: Jan 4, 2009
  8. Jan 4, 2009 #7
    There are a number of different formulations of the "equivalence principle (EP)" with slightly different implications. Wikipedia discusses some. Post #3 here reflects Einstein's version, I believe.

    The distinctions can be subtle and are not self evident to me. Peter Bergmann, a student of Einstein's writes in one definition of EP (THE RIDDLE OF GRAVITATION):

     
    Last edited: Jan 5, 2009
  9. Jan 4, 2009 #8
    So equivalence principle only work in uniform gravitational field? Einstein's prediction of bending of light beam deduced from equivalence principle. Is it valid only in uniform gravitational field too?
     
  10. Jan 4, 2009 #9

    JesseM

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    I recommend reading through this thread from December, apparently defining the equivalence principle is fairly subtle and there are a few different definitions which apply to different cases.
     
  11. Jan 4, 2009 #10
    Totally Useless, waste of my time and yours
     
  12. Jan 5, 2009 #11

    JesseM

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    If you say so. But unless you think the info on that thread is wrong, then you see that your question about how the equivalence principle works in curved spacetime has no simple answer. For instance, in post #34 on p. 3 of that thread atyy gives two mathematical ways of defining the equivalence principle in curved spacetime:
    atyy also started another thread more focused on definitions of the equivalence principle here. The main subtlety seems to be the issue of the principle only working locally up to the first-order terms in the Taylor (no second or higher derivatives)--if those higher-order derivatives are taken into account, then apparently GR can't even be said to reduce to SR locally.
     
  13. Jan 5, 2009 #12
    feynmann asks:
    If you are referring to Einsteins original insight, I think the general answer is "yes"...but (a) I am unsure just how far Einstein carried that "equivalence" and (b) there are now so many definitions of EP that inferences vary from definition to definition...

    Einstein found enough of a link between acceleration and gravity to make a selection from among different formulations for GR. But it would be wrong to conclude that acceleration and gravity are equal in every respect: Peter Bergmann, a student of Einsteins, says in THE RIDDLE OF GRAVITATION, (page 9)
    (I'm unsure exactly what the distinction is!!)

    But apparently a distinction between acceleration and gravity can (maybe) be made via the Unruh Effect, of which Wikipedia says:
    In turn, this is explained via an apparent event horizon which forms for all accelerating observers. I don't believe it has been experimentally verified.



    And Jesse posts:
    I have not looked at all the math involved, but this has been a point I have also seen elsewhere. Part of the issue is how closely you take a portion of curved space to be "flat"....in the simple case of acceleration versus free fall in an elevator, it makes little difference. If you drop two balls separated by a distance and ask "do they come together as they fall" the whole focus is different.

    I have read elsewhere that Einstein chose a formulation of GR that reduces to SR without gravity....which implies he had considered other formulations that did not so simplify.
     
    Last edited: Jan 5, 2009
  14. Jan 5, 2009 #13
    atyy posts:
    That's also my understanding, if I understand the statement correctly, but via explanations from others rather than first hand mathematical analysis.
    In another thread, DrGreg explained there is an "observed" curvature of spacetime relative to an accelerating observer...this is what I have read elsewhere but precisely how it relates to equivalence and gravitational curvature is not clear to me. For the time being I am taking them to be similar but distinct phenomena.
     
  15. Jan 5, 2009 #14

    HallsofIvy

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    The derivative is the limit of a difference. Any derivative is purely local.
     
  16. Jan 5, 2009 #15

    atyy

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  17. Jan 5, 2009 #16
    glad to see that expressly made!!!! I thought maybe I was missing something...and ditto for subsequent derivatives!!
     
  18. Jan 5, 2009 #17

    JesseM

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    That's true, but in the case of knowing all the infinite higher-order derivatives of a function at a point, you can use the Taylor series to get "nonlocal" information about the exact shape of the function out to some finite distance (the radius of convergence, which may be infinite)...is there anything analogous in GR? Is there any quantity where a local observer could in principle measure an infinite number of nonzero higher-order derivatives of that quantity, and use this to determine the curvature of spacetime out to some finite (or infinite) distance?
     
  19. Jan 5, 2009 #18

    atyy

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  20. Jan 5, 2009 #19

    DrGreg

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    It would help if you gave a link to any other thread you mention. In my (relatively) old age I can't remember everything I ever said in every thread!

    It is true that an accelerating observer (in a Born-rigid rocket in the absence of gravity -- the "Rindler mouse" that atyy refers to) will see a number of effects that a "stationary" observer would see near a black hole e.g. an event horizon and (pseudo-)"gravitational" time dilation. But I would not describe these effects as "curvature of spacetime". Spacetime curvature is, loosely speaking, another name for tidal effects, and occurs only with a "real" gravitational source.
     
  21. Jan 6, 2009 #20
    Suck it up, DrGreg!!!

    I can't remember all the threads I respond to!!!!

    I link when I remember threads and find them....
     
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