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B Is it that hard to do Relativity in accelerating frames?

  1. Apr 13, 2017 #1
    According to the Lorentz transformations, in the absence of gravity, the relations between coordinates of a primed system to those of a unprimed system are $$x'^{\ \alpha} = \Lambda^{\alpha}{}_{\beta}x^{\beta}$$ For the Lorentz invariance to be satisfied we must have ##\Lambda^{\alpha}{}_{\beta}## constant, because then $$dx'^{\ \alpha} = \Lambda^{\alpha}{}_{\beta}dx^{\beta}$$
    My questions are:
    - is it always possible to "adjuste" a accelerating frame so that it becomes unaccelerated?
    - the equations get much more complicated if we assume the accelerating frames without making any adjustment or do they become just a little more complicated?
     
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  3. Apr 13, 2017 #2

    Orodruin

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    Using accelerated frames in special relativity is very much like using curvilinear (and not necessarily orthonormal) coordinates in regular calculus. Depending on whether on not you find that difficult (along with seeing time as just another coordinate), you may or may not find relativity in accelerated frames difficult.
     
  4. Apr 13, 2017 #3

    Dale

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    It also depends strongly on the problem. For certain problems picking a good non inertial coordinate system dramatically simplifies the problem. Picking a bad coordinate system makes things even worse, of course.
     
  5. Apr 13, 2017 #4

    Orodruin

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    Just like when using curvilinear coordinates in regular calculus! :rolleyes:
     
  6. Apr 13, 2017 #5

    vanhees71

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    It's in fact pretty simple. Just take arbitrary coordinates ##q^{\mu}## to parametrize the components of the spacetime four-vector in Minkowski coordinates. Then you have for the invariant increment
    $$\mathrm{d} s^2=\eta{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} = \eta_{\mu \nu} \frac{\partial x^{\mu}}{\partial q^{\rho}} \frac{\partial x^{\nu}}{\partial q^{\sigma}} \mathrm{d} q^{\rho} \mathrm{d} q^{\sigma},$$
    i.e., the pseudometric components with respect to the holonomous (!) curvilinear coordinates are given by
    $$g_{\rho \sigma} = \eta_{\mu \nu} \frac{\partial x^{\mu}}{\partial q^{\rho}} \frac{\partial x^{\nu}}{\partial q^{\sigma}}.$$
    BTW: You are now already very close to the general tensor framework needed for general relativity. The only thing one must learn is that there is no spacetime four-vector anymore but only a differential manifold with tangent vectors and tangent co-vectors and a pseudo-Riemannian (Lorentzian) fundamental form, which physicists like to simply call "the metric" although it's not positive definite ;-).
     
  7. Apr 13, 2017 #6
    Thank you all.
    It's interesting to know this.

    Is this the same as working on the problem in the intertial frame and after changing to the accelerating frame?
     
  8. Apr 13, 2017 #7

    robphy

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    One aspect that [to me] isn't so clear, however, is making and interpreting measurements by an accelerating observer (e.g. Piecewise inertial, uniformly accelerating, piecewise uniformly accelerating, rotating, arbitrary, etc) ... they don't just carry over from the inertial case. For example, as I currently understand it, radar measurements yield different results from those associated with clocks and "rods" (somehow defined).
     
  9. Apr 13, 2017 #8

    pervect

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    I'm not sure of the context of the question. If one is in the flat space-time of special relativity, one can always find a global inertial frame of reference. If one is in the curved space-time of General Relativity, in general there isn't any such thing, nor is there any general global Lorentz transform.

    The key point I think is defining what one means by "frame". Is it a global coordinate system? Is it a purely local thing (a tangent space, and/or a set of basis vectors for the tangent space near some particular point). Or perhaps it's a congruence of time-like worldlines, each worldline representing an "observer" in the frame.

    Once one has an agreed on definition of what a "frame" really is, it becomes much easier to answer questions like the one you posed.
     
  10. Apr 13, 2017 #9
    What would you personally define to be a frame from these that you mentioned?
     
  11. Apr 13, 2017 #10

    pervect

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    I'm not totally consistent, but usually I regard a frame as being defined by a set of basis vectors, one set at every point along the worldline of some particular "observer". One thinks of the observer as "carrying along" a set of basis vectors, or transporting them.

    If the worldline is a geodesic, the frame is "not accelerating". If the basis vectors are fermi-walker transported, the frame is "not rotating". (I believe that in the case where the observer is not accelerating, i.e. is a geodesic observer, fermi-walker transport is equilvalent to parallel transport. But don't shoot me if I'm wrong.)

    So by this definition, a non-accelerated frame of reference exists if geodesics exist. This is nearly always, though singularities (where geodesics are incomplete, i.e. they just stop) would be an exception, I suppose.
     
  12. Apr 13, 2017 #11

    PeterDonis

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    This is correct.
     
  13. Apr 13, 2017 #12
    Just to know if I'm getting things right, I have constructed an example. The coordinates of one system are labeled ##y##, while those of the other system are ##x##.

    $$y^1 = x^1 \text{Cos}(wt') \\ y^2 = x^2 \text{Sin}(wt') \\ y^3 = x^3 \\ y^0 = t' = x^0 = t$$ where ##w## is the angular velocity.
    Because ##dy^{\alpha} = (\partial y^{\alpha} / \partial x^{\beta}) dx^{\beta}## and ##\eta_{\sigma \kappa} = (\partial y^{\alpha} / \partial x^{\sigma}) (\partial y^{\beta} / \partial x^{\kappa}) \eta_{\alpha \beta}##, we have for the space time interval $$d \tau '{\ }^2 = dt^2 - (\text {Cos}^2(wt)(dx^1)^2 + \text{Sin}^2(wt)(dx^2)^2 + (dx^3)^2)$$ Does it make sense?

    At a constant time ##t = n \pi / 2 ##, ##n \in \mathbb{N}## the proper time in the primed (##y##) system is equal to ##dx^3##!
     
    Last edited: Apr 13, 2017
  14. Apr 13, 2017 #13

    DrGreg

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    1. You have forgotten to include ##\partial y^1 / \partial x^0## and ##\partial y^2 / \partial x^0##.
    2. I assume you intended this to be a "rigidly rotating" coordinate system, so $$y^1 = x^1 \cos(\omega x^0) - x^2\sin(\omega x^0) \\y^2 = x^1 \sin(\omega x^0) +x^2\cos(\omega x^0)$$
     
    Last edited: Apr 13, 2017
  15. Apr 13, 2017 #14
    Thanks Dr Greg. So would the proper time become $$d \tau '{\ }^2 = dt^2 - (\text {Cos}^2(wt)(dx^1)^2 + \text{Sin}^2(wt)(dx^2)^2 + w( \text{Cos}^2(wt) - \text{Sin}^2(wt))(dt)^2 + (dx^3)^2)$$
     
  16. Apr 14, 2017 #15

    haushofer

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    It is similar to doing Newtonian mechanics in accelerating frames. Only the coordinate transformations differ.
     
  17. Apr 14, 2017 #16

    vanhees71

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    Nearly correct! Fermi-Walker transport is parallel transport if the (timelike) worldline along which you transport is a geodesic. In Minkowski space it's only the same, if the worldline is given by
    $$x^{\mu}(s)=u^{\mu} s+x_0^{\mu},$$
    where ##u^{\mu}=\text{const}## with ##u^{\mu} u_{\mu}=1## and ##x_0^{\mu}=\text{const}##, where the ##x^{\mu}## are the components of the spacetime vector wrt. a Minkowski basis. Then indeed ##a^{\mu}=\ddot{x}^{\mu}=0##, and Fermi-Walker transport is just parallel transport (which is trivial in a flat space).

    Fermi-Walker transport is more general. It works along any time-like worldline and transports some Minkowski basis (tetrad) from one point to another such that it is locally non-rotating. For details, see the new Section on Fermi-Walker transport in my SRT FAQ, which is unfortunately down right now :-(. I hope our sysadmins bring the server up soon (although we have holidays till including Monday of course):

    http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf
     
  18. Apr 14, 2017 #17

    DrGreg

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    You've made a mistake somewhere in your working. Think about it. Do you really expect the answer to depend on the angle you have rotated through? You should find all the angles cancel out, either via ##\cos^2 \omega t + \sin^2 \omega t = 1## in some places, or ##2\cos \omega t \sin \omega t - 2\cos \omega t \sin \omega t = 0## in other places.
     
  19. Apr 14, 2017 #18

    martinbn

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    Why nearly? You are saying exactly the same thing.
     
  20. Apr 14, 2017 #19

    PeterDonis

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    Yes, that's what pervect said and I said was correct.
     
  21. Apr 14, 2017 #20

    vanhees71

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    Argh! Sorry, I misread the statement. So it's indeed fully correct. Sorry again!
     
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