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Limit of Rindler coordinates

  1. Nov 27, 2012 #1

    zonde

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    It seems that acceleration at some point in Rindler coordinates completely determines it's distance from rindler horizon, right?

    If we have two rockets with equal hight and experiencing equal acceleration at the bottom there are no other parameters we can vary to get different results for two cases. So that means that time dilation at the top of the rocket is the same for both rockets.

    That contrasts with gravitational field where we have two parameters determining acceleration (r and rs) so we don't know time dilation at the top of the rocket given acceleration at the bottom (and given hight) if it stand on the surface of some gravitating body.

    So it seems that there is no limiting case where proper acceleration and gravitational acceleration would tend to become equal.
     
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  3. Nov 28, 2012 #2

    PeterDonis

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    Right.

    Yes.

    Yes; we also need to know the gravitating body's mass (or alternatively its radius, since we know the acceleration, and that plus the radius is enough to give us the body's mass).

    How does that follow from what you said above? Given any combination of acceleration and time dilation in Rindler coordinates, we can always find *some* combination of mass and radius for a gravitating body that will yield the same combination of acceleration and time dilation. And vice versa. Nothing that you said above makes that impossible. It's true that the distance from the Rindler horizon in that case is highly unlikely to be the same as the radius of the gravitating body in that case, for the same acceleration-time dilation combination. But so what? That's not required by the principle of equivalence.
     
  4. Nov 28, 2012 #3

    pervect

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    It seems to me there isn't any huge difference between the two cases. The time dilation for a small height h will always be 1+gh/c^2, you can't really change this fundamental fact.

    There are some second order differences, but to first order you can say that the time dilation is what I said above.
     
  5. Nov 28, 2012 #4

    Matterwave

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    I'm getting the impression that Zonde has figured out that the strong principle of equivalence doesn't apply to spatially varying gravitational fields, or, equivalently, that it only applies in the limit of uniform gravitational fields or infinitesimal displacements.
     
  6. Nov 28, 2012 #5

    zonde

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    Right, it does not follow from above.

    But your proposal that we can find some situation where the equivalence works is not quite satisfactory either. If there is a limit then where is this limit?

    Then from different side.
    proper acceleration in Rindler coordinates is proportional to 1/x
    Newtonian gravitational acceleration is proportional to 1/r^2
    Is this right? Maybe that 1/x acceleration is actually (Rindler) coordinate acceleration and you have to take into account time dilation to get proper acceleration?
    And the same about Newtonian gravitational acceleration. From GR perspective does it approximate coordinate acceleration (as seen by far away observer) or proper acceleration?
     
  7. Nov 28, 2012 #6

    zonde

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    1+gh/c^2 is for accelerated observer.
    I calculated with example values that gravitational time dilation seems to give the same value. But I don't understand that. Gravitational acceleration seems to vary differently than Rindler acceleration (see replay to PeterDonis) and than there is still mass parameter.
     
  8. Nov 28, 2012 #7

    PeterDonis

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    I don't understand. What limit are you talking about?

    Yes.

    Yes; the correct relativistic formula has an extra factor of sqrt(1 - 2m/r) in the denominator.

    No; 1/x is the proper acceleration.

    Proper acceleration.

    None of this contradicts what I said above. Show me an accelerated observer in flat spacetime, feeling proper acceleration 1/x, and I will show you an accelerated observer in Schwarzschild spacetime who feels the same proper acceleration; it's just a matter of setting the two formulas equal:

    [tex]a = \frac{1}{x} = \frac{m}{r^2 \sqrt{1 - 2m / r}}[/tex]

    You should be able to convince yourself that for any x > 0, we can pick any m > 0 that we like, and then find some r > 2m for which the equality above is satisfied. That means that for any Rindler observer (accelerated in flat spacetime), we can find some corresponding Schwarzschild observer (hovering over a black hole in curved spacetime) who feels the same proper acceleration. That's all I was trying to say.
     
  9. Nov 28, 2012 #8

    zonde

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    You are sarcastic, right? :smile:
    But I doubt that strong principle of equivalence applies in the limit of infinitesimal displacements. And I want to either confirm my doubt or get over it.

    And what I just figured out is that strong equivalence principle in GR is the same as relativity principle in SR i.e. it is the statement that gives physical content to GR.
    So I want good understanding of things around it.
    And as a physical statement it is subject to experimental tests. So in perspective I want to understand how (tested quantitative) predictions of GR follow from strong principle of equivalence.
     
  10. Nov 28, 2012 #9

    PeterDonis

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    Why do you doubt this?

    It is one of the principles that gives physical content to GR, yes. It's not the only one.

    Also, are you really intending to talk about the *strong* EP, as opposed to the weak EP or the Einstein EP? I'm using the terminology that's used on the Wikipedia page, which gives a brief overview of the different versions of the EP:

    http://en.wikipedia.org/wiki/Equivalence_principle

    I ask because the statement you appear to be concerned about is, more or less, "proper acceleration in flat spacetime is equivalent to being at rest in a static gravitational field". In the Wikipedia page terminology, this is the weak EP, not the strong EP.
     
  11. Nov 29, 2012 #10

    zonde

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    I'm not saying you are wrong. But that it doesn't really answer my question.

    What is other?
     
  12. Nov 29, 2012 #11

    pervect

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    There seems to be more argument about the EEP than one would expec, even in the literature. BUt my view is that the EEP says that gravity and acceleration are the same to the first order. At higher orders, gravity has tidal effects, and acceleration doesn't really (not in the sense of geodesic deviation at least). But that's not really the point, the point is that it's the same at lower orders.
     
  13. Nov 29, 2012 #12

    zonde

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    How would you argue that tidal effects are higher order effect than acceleration? We can't go down on orders so much that we can't speak about acceleration any more. So there is some level of orders where we should stay. What if tidal effects are still there at this level?

    I would like to add that when speak about tidal effects I usually think about radial effects and not convergence of different angular directions.

    PeterDonis, this answers your question too "Why do you doubt this?". My doubts are that acceleration and tidal effects might be at the same level of orders. And in that case we can't speak about the limit where EP tends to hold better.
     
  14. Nov 30, 2012 #13

    PeterDonis

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    Tidal effects depend on the second derivatives of the metric coefficients. Acceleration depends on the first derivatives of the metric coefficients. That's why tidal effects are higher order.
     
  15. Nov 30, 2012 #14

    PeterDonis

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    Some other principles that give physical content to GR:

    (1) Spacetime is a geometric object. Gravity is curvature of this geometric object.

    (2) Physics is contained in geometric invariants.
     
  16. Nov 30, 2012 #15

    pervect

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    If U is the Newtonian potential, [itex]\partial U / \partial x[/itex] is the force in the x direction, and [itex]\partial^2 U / \partial x^2[/itex] is the tidal acceleration in the x direction.

    I'm using "order" in the calculus sense, first order effects are proportional to the derivative, second order effects are proportional to the second derivative.

    As you take the limit as dx->0, the first order terms will dominate the second. If we expand U in a taylor series we'd get

    U = some constant + force terms * dx + (1/2) tidal force terms * dx^2

    (Expanding U in a taylor series is the same as expanding the "gravitational time dilation" in a series, the two are proportional).

    By choosing a small enough dx, i.e. by limiting the size of your box, you can always guarantee that the second order effects are low enough to ignore. This is the sense in which the EEP says acceleration is the same as gravity. It doesn't mean that tidal forces don't exist, it just means that by taking your box small enough you can ignore them.


    I'm afraid I don't follow this question as written, I hope my answer above answers it.


    I generally mean radial "stretching" effects more often than I mean the "crushing" convergence effects, but "more often" doesn't mean "always". As far as their contributions to U or time dilation goes, the only difference is in the sign of the effect.
     
    Last edited: Nov 30, 2012
  17. Nov 30, 2012 #16
    It is a bit surprising since the EEP is the published opinion of a single person; that should be rather easy to verify.

    The EEP says that we can treat an uniformly accelerated reference system in space that is free from gravitation, as being "at rest" in a homogeneous gravitational field.

    That is quite different from "gravity and acceleration are the same to the first order".
     
    Last edited: Nov 30, 2012
  18. Nov 30, 2012 #17

    PeterDonis

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    You and pervect are using "EEP" to refer to two different things. You mean "the EEP that Einstein stated." He means "the EEP that is actually used, today, in GR." They're not necessarily the same, and the argument he is referring to about the EEP is not about what Einstein said, it's about what the actual principle that is used in GR should be.
     
  19. Nov 30, 2012 #18

    pervect

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    An example of a generally similar opinion to mine about the EEP occurs in the literature:

    http://dx.doi.org/10.1007/BF02450447 "Covariance, invariance, and equivalence: A viewpoint"

    It is clear from Zonde's argument that when you actually try to apply the offered defintion from Harry

    it doesn't actually work as stated if one considers a large enough region of space-time.

    But how small is "small enough?" I think it's sufficient to say that it works in the limit as the box size approaches zero. The remark about the "orders" is simply a helpful way to help understand why the equivalence is exact in the limit, and non-exact over larger regions. It was not my intention to represet is aspart of any formal statement or defintion of the equivalence principle.

    Other proferred statements of the equivalence principle as "The Universality of Free Fall" seem to me to be clearer formulation of the equivalence principle than the original formulation about trying to determine whether or not you're in an elevator or on a planet by the means of physical experiments - as there are known means of building reasonably compact gravity gradient meters (such as the Forward mass detector, variants of which are actually used in prospecting for oil)

    [add]
    Another reference:

    http://dx.doi.org/10.1007/BF00763538

     
    Last edited: Nov 30, 2012
  20. Nov 30, 2012 #19

    zonde

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    Acceleration is second order effect in respect to time. So you want to talk about second order effect in respect to time but first order effect in respect to distance. But then you can't talk about patch of space-time.
     
  21. Nov 30, 2012 #20

    pervect

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    That's a logical interpretation, but not what I meant, alas.

    Meanwhile, there's an interesting paper I ran across (it may disappear from public access today) that takes a different position on the equivalence principle that may be helpful.

    Gravitational redshift and the equivalence principle
    There authors prescribe only one thing to successfully apply the EEP to this problem.

    The thing that is prescribed directly is that "the field must be uniform. This means no differences in the accelerometer readings between h and h+dh. This rules out tidal forces.

    THis is similar to the earlier observation I made that for small enough "h", there isn't any difference between the gravitational field due to the acceleration and the gravitational field due to matter - the time dilation is (1+/-gh) in each case. BTW, this is one of the relations that this paper derives with an effort to be exact and not make approximations.

    The authors don't directly prescribe that there is no matter in the "accelerated spacetime", and I suppose as long as the first condition is met, it's not necessary. You simply have a hybrid case, where you have gravitation due to matter plus gravitation due to acceleration, and they both follow the EP (as long as you don't choose a region so big that g varies over the region).

    However, the authors make a point of analyzing the case where there is no matter present, even though they don't prescribe it as "necessary".
     
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