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The (non-?) flatness of a uniform gravitational field and the equivalence principle

  1. Oct 31, 2014 #1
    Desloge (1989) published the article: ## '##Non-equivalence of a uniformly accelerating reference frame and a frame at rest in a uniform gravitational field## '##. His result, briefly resumed: a uniform gravitational field is not flat, is quite interesting. But the way he proves the result is usable for any gravitational field; I will start with a description of his ‘method’, though in a slightly modified way; the result will be the same.
    So, first a definition of a rigid reference system: observers, for example space-ships, provided with an ideal clock, are positioned in an arbitrary gravitational field, in the following way:

    - the power of the observers (space-ships) is constant.
    - the light-distance, measured by any observer to any other one is constant.
    - one of the observers is designed as head observer H; his spatial coordinate will be 0; the spatial coordinate of observer A will be the light-distance, measured by H of A: x##_A##
    - the standard clocks of the observers cannot be synchronized; the clock of H will be appointed as coordinate clock. All other observers are supplied with a coordinate clock in the following way: suppose the light-distance of observer A, measured by H, is a; the light-distance of H, measured by A, is b.

    Then, b/a determines the ratio between standard clock and coordinate clock; the ratio will be ##\alpha##(x), depending of the spatial distance x of the observer. All coordinate clocks can be synchronized.
    We consider the one-direction case.

    The frame is rigid; the corresponding coordinate system will be represented by ds##^2 = \alpha(x)^2##(dt##^2## - dx##^2##),(the so-called Weyl -form) where ##\alpha##(t=0) = 1. So local velocity=dx/dt, coordinate velocity. The relationship between local acceleration (this is what every observer really‘feels’, and is able to measure in his space-ship lab, by releasing a free falling particle alongside a rod), and coordinate acceleration, is more complex.
    We have to consider the equations for geodesics. We start with the more general form:
    $$
    \frac{d^2x^a}{du^2}+ \Gamma^a_{bc}\frac{dx^b}{du} \frac{dx^c}{du} =\lambda(u)\frac{dx^a}{du}

    $$
    In our special case, we have ##\Gamma^a_{bc}##=0, except ##\Gamma^0_{01}=\Gamma^0_{10}=\Gamma^1_{00}=\Gamma^1_{11} = \frac{\alpha '}{\alpha}, \alpha##’ being d##\alpha##(x)/dx. Given that x##_0##=t, x##_1##=x, we find for the first equation:
    $$
    \frac{d^2t}{du^2}+\frac{2\alpha '}{ \alpha} \frac{dt}{du} \frac{dx}{du} = \lambda(u) \frac{dt}{du}
    $$
    This holds for any parameter, with corresponding##\lambda##(u). So, also for t(u). Result:
    $$
    \frac{2 \alpha '}{\alpha} \frac{dx}{dt} = \lambda (t)
    $$
    The second equation will become:
    $$
    \frac{d^2x}{dt^2}+\frac{\alpha '}{\alpha} + \frac{\alpha '}{\alpha} \left( \frac{dx}{dt}\right)^2 = \frac{2 \alpha '}{\alpha}\left( \frac{dx}{dt} \right)^2
    $$
    And therefore
    $$
    \frac{d^2x}{dt^2} + \frac{\alpha '}{\alpha} - \frac{\alpha '}{\alpha} \left( \frac{dx}{dt}\right)^2 = 0
    $$
    Suppose, we release a free falling particle at moment t, so dx/dt =0 at that moment: d##^2##x/dt##^2##= ##-\alpha '/\alpha##. In this case, the relation between local and coordinate acceleration is: localacc.=1/##\alpha##. Coordinate acc.= ##\alpha '/\alpha^2##. This holds for every gravitation field. Desloge uses the result for a uniform field; we investigate what comes out for natural gravity. Suppose there is a rigid reference frame, radially positioned with respect to a heavy body. The one-direction form of the Schwarzschild metric is:
    $$ds^2 = \left(1-\frac{2m}{x}\right)dt^2 - \left(1-\frac{2m}{x}\right)^{-1}dx^2; \; \; \; x>2m
    $$
    First, we have to transform the metric in the Weyl-form.
    The transformation ##\bar{t}##=t, ##\bar{x}##=x+2m*ln(x - 2m) will do the job.
    The result: ds##^2##={1 – 2m/x(##\bar{x}##)(d##\bar{t}^2## - d##\bar{x}^2##).
    (There is no mathematical expression for the inverse of ##\bar{x}##(x) ,so we leave it: x(##\bar{x}##).
    There is one final task: 1 - 2m/x ##\rightarrow## 1 for ##\bar{x} \rightarrow \infty##. We want the head observer at some finite ##\bar{x}##. We take A, with coordinate ##\bar{x}_A##, and ##\bar{t}##=(1/k). ##\bar{t}##, ##\bar{x}##=(1/k). (x(##\bar{x}##)–x(##\bar{x}_A##)); k = (1 - (2m/x(##\bar{x}_A##)))##^2## will have as result: ##\hat{\alpha}##=(1/k)(1 - 2m/x)##^\frac{1}{2}##; x = x(##\bar{x}(\bar{x}##))and for ##\bar{x}=0##, ##\alpha_A##=1. Now we are able to use the previous result:
    \begin{equation}
    \frac{d^2\hat{x}}{d\hat{t}^2} = \frac{\alpha '}{\alpha} = \left( \left[1 - \frac{2m}{x}\right]^{-\frac{1}{2}} \frac{2m}{x^2}\frac{dx}{d\bar{x}} \frac{d\bar{x}}{d\hat{x}}\right)\left[1-\frac{2m}{x}\right]^{-\frac{1}{2}} = \frac{2km}{x^2}
    \end{equation}
    This is the result we expect (k only dependent on the choice of A as head observer), if all observers are measuring with their coordinate clocks. However, an observer, eventually not being member of the reference system, and so measuring with his standard clock, will find:
    \begin{equation}
    \frac{1}{\alpha}\frac{2km}{x^2} = \left(1-\frac{2m}{x}\right)^{-1}\frac{2km}{x^2}
    \end{equation}
    The result is classical gravity multiplied with a ##‘##relativistic correction##’##, neglegible in normal circumstances, but considerable in the neighbourhood of the Schwarzschild radius.

    P.S. To my opinion, the way Desloge uses the concept of a rigid reference frame, together with the link with a coordinate system is an important contribution to the theory. Still, there are several questions, open to me, which I would like to discuss with those who are interested in the matter.
     
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  3. Oct 31, 2014 #2

    atyy

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    In general, accelerated frames and spacetime curvature both contribute to deviations from a "Minkowski metric". You can see that in Eq 9.14-9.16 pf http://relativity.livingreviews.org/Articles/lrr-2011-7/fulltext.html [Broken]. If the observer is non-accelerating, the curve is a geodesic, and the Christoffel symbols vanish on the curve. Even for a non-accelerating observer, as one moves away off the geodesic, the Christoffel symbols will no longer vanish if spacetime is curved. (In all cases, spacetime curvature cannot be hidden if one looks at derivatives of the Christoffel symbols - ie. curvature.) There's a great discussion also in sections 24.3 and 24.6 of Blandford and Thorne's notes.
     
    Last edited by a moderator: May 7, 2017
  4. Oct 31, 2014 #3

    PAllen

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  5. Nov 1, 2014 #4

    stevendaryl

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    I wouldn't say that it was refuted. I would say that the two authors are disagreeing about what's a reasonable definition of "uniform gravitational field".

    To me, it seems that a "gravitational field" is only meaningful relative to a coordinate system in which the metric components are time-independent. In such a case, you can define the gravitational field at a point to be proper acceleration of a point mass that is "stationary" (all spatial coordinates are constant in time). According to this definition of "gravitational field", the gravitational field for Rindler coordinates is not uniform. Or contrariwise, if there is some coordinate system for some spacetime in which the "gravitational field" has the same magnitude everywhere, then that spacetime is certainly not flat spacetime.

    That paper is arguing that that is not a reasonable definition of "uniform gravitational field".
     
  6. Nov 1, 2014 #5

    PAllen

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    For the purpose of the equivalence principle as discussed by Einstein, the definition in the paper I give is the reasonable one. A long rocket
    Well, they show a lot more than that. Specifically, that such an alternative does not exist unless you let go of other properties expected of a uniform field , specifically that the spatial geometry is homogeneous and isotropic in planar slices orthogonal to the field. Since this seems so fundamental to what you would want for a uniform field, I take this as proof that a uniform gravitational field inconsistent with the EP cannot exist.
     
  7. Nov 1, 2014 #6

    stevendaryl

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    It seems to me that the conclusion is that there are no uniform gravitational fields, in the intuitive sense.
     
  8. Nov 1, 2014 #7

    PeterDonis

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    Has anyone (Desloge, for example) actually given an explicit example of a spacetime in which the "gravitational field" has the same magnitude everywhere, which is a valid solution of the Einstein Field Equation?
     
  9. Nov 1, 2014 #8

    PeterDonis

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    One item in the introduction of this paper makes me wonder: they say that the "acceleration" ##d^2 x^{\mu} / d \tau## is not a 4-vector. I don't understand what they mean by that. The proper acceleration ##u^{\nu} \nabla_{\nu} u^{\mu}## is certainly a 4-vector, and that's usually what is meant by ##d u^{\mu} / d \tau## (where ##u^{\mu} = d x^{\mu} / d \tau##).
     
  10. Nov 1, 2014 #9

    DrGreg

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    I would interpret ##d u^{\mu} / d \tau## as the coordinate derivative and would denote the "absolute derivative" (a.k.a. "intrinsic derivative" or "covariant derivative") as ##D u^{\mu} / D \tau = u^{\nu} \nabla_{\nu} u^{\mu}##
     
  11. Nov 1, 2014 #10

    PeterDonis

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    I agree that small "d" can mean a coordinate derivative, but ##\tau## isn't a coordinate, so I'm not sure it can be that simple.
     
    Last edited: Nov 1, 2014
  12. Nov 1, 2014 #11

    PAllen

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    I momentarily stumbled on the same point, but reached the same conclusion as Dr. Greg - that they were pointing out that the simple first derivative by proper time is a vector, but the simple second derivative is not - you need an absolute derivative, involving the connection.
     
  13. Nov 8, 2014 #12

    stevendaryl

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    I think "coordinate derivative" means in this context the derivative OF the coordinate, rather than a derivative with respect to a coordinate.

    To me, [itex]\frac{d^2}{d\tau^2} x^\mu[/itex] is completely unambiguous. [itex]x^\mu[/itex] is a coordinate for some spacetime path parameterized by proper time [itex]\tau[/itex], and that expression just means the second derivative of [itex]x^\mu[/itex] with respect to [itex]\tau[/itex]. [itex]x^\mu(\tau)[/itex] is just an ordinary real-valued function of a real-valued parameter, and there is only one kind of derivative.

    However, if [itex]e_\mu[/itex] is the coordinate basis for coordinate system [itex]x^\mu[/itex], then the combination [itex]U = \sum_\mu \frac{dx^\mu}{d\tau} e_\mu[/itex] is a 4-vector, with components [itex]U^\mu = \frac{dx^\mu}{d\tau}[/itex]. To me, the confusion comes in when people use [itex]U^\mu[/itex] to mean, ambiguously, the 4-vector [itex]U[/itex] and one of its components. Then [itex]\frac{d}{d\tau} U^\mu[/itex] can either mean:

    • A component [itex]\mu[/itex] of the derivative of [itex]U[/itex].
    • The derivative of the [itex]\mu[/itex] component of [itex]U[/itex].
    If you know which of those is meant, then there is only one interpretation of [itex[\frac{d}{d\tau}[/itex]. There is no way to take the derivative of a 4-vector other than using parallel transport. Since components of a vector are scalars, there is no way to take a derivative of a scalar other than to use the ordinary derivative.

    Although it requires extra parentheses, I think it's clearer to use

    [itex](\frac{d}{d \tau} U)^\mu[/itex]

    and

    [itex]\frac{d}{d \tau} U^\mu[/itex]

    but people don't tend to notice the distinction between the two.
     
    Last edited: Nov 8, 2014
  14. Nov 8, 2014 #13

    DrGreg

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    stevendaryl has explained it a bit better than I did. For any vector V I would denote
    • the [itex]\mu[/itex] component of the derivative of V by [itex]DV^\mu/D\tau[/itex]
    • The derivative of the [itex]\mu[/itex] component of V by [itex]dV^\mu/d\tau[/itex]
    The two are related, of course, by[tex]
    \frac{DV^\mu}{D\tau} = \frac{dV^\mu}{d\tau} + \frac{dx^\nu}{d\tau} \Gamma^\mu_{\nu\lambda}V^\lambda
    [/tex]
    I suppose, if you wanted to be completely unambiguous, you could write that as[tex]
    \left( \frac{DV}{D\tau} \right)^\mu = \frac{d}{d\tau} \left( V^\mu \right) + \frac{dx^\nu}{d\tau} \Gamma^\mu_{\nu\lambda}V^\lambda[/tex]
     
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