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Translate acceleration to GR gravity

  1. Jun 15, 2015 #1

    ChrisVer

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    Suppose I have a toroidal tube, in which I lock a lucky/unlucky mouse.
    Then I start spinning the torus [of radius ##r##] around its center with some velocity ##\beta##.
    Then the centrifugal acceleration felt by the mouse due to rotation is given by some special relativistic maths:
    [itex]a= \frac{\beta^2}{r (1 - \beta^2)}[/itex]
    This acceleration can be seen as a gravitational acceleration.
    How would I translate this acceleration into gravity of GR?
     
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  3. Jun 15, 2015 #2

    Nugatory

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    You cannot, because you're reading the equivalence principle backwards. The equivalence principle does not say that all inertial pseudo-forces are gravitational forces, it says that all gravitational forces are members of the larger class of inertial pseudo-forces.

    In fact, there is no configuration of matter that will curve spacetime in a way that would produce gravitational effects equivalent to the flat space-time inertial pseudo-force that the unfortunate mouse experiences.
     
    Last edited: Jun 15, 2015
  4. Jun 15, 2015 #3

    WannabeNewton

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    Transform to the rest frame of the tube and accordingly transform the Minkowski metric. The transformed metric will give you the gravitational field in the rotating frame.
     
  5. Jun 15, 2015 #4

    ChrisVer

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    you mean that the metric will become:
    [itex]g_{\rho \sigma} =\frac{\partial x^\mu}{\partial \bar{x}^\rho} \frac{\partial x^\nu}{\partial \bar{x}^\sigma} n_{\mu \nu}[/itex]
    Where [itex]x^\mu[/itex] are the coordinates where the tube is circulating (cartesian coordinates) and [itex]\bar{x}^\nu[/itex] the ones that it looks at rest?
    [itex] \bar{t}= t[/itex]
    [itex] \bar{x} = r \cos \theta(t)[/itex]
    [itex] \bar{y} = r \sin \theta(t)[/itex]
    [itex] \bar{z}= z[/itex]
     
  6. Jun 15, 2015 #5

    A.T.

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    If the tube itself is considered mass-less, that you simply looking for the rotating frame metric:
    http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken]
     
    Last edited by a moderator: May 7, 2017
  7. Jun 15, 2015 #6

    ChrisVer

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    My initial goal was to think about the scenario they propose for time-travel, like getting a very long rod and start rotating it fast enough so that at its endpoints an observer would feel a different clock time than the one on the earth.
    However I think that this is a result of the coordinates and so "time" shouldn't be taken seriously.
     
  8. Jun 15, 2015 #7

    A.T.

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    The ends of a spinning rod will age less than it's center. These proper time intervals are frame invariant.
     
  9. Jun 15, 2015 #8

    ChrisVer

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    That's if you really care about reference frames. Instead, I'm initially suggesting that the Lorentz invariance of frames should not work (as it doesn't in GR).
     
  10. Jun 15, 2015 #9

    Dale

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    Frame invariant in this context means invariant under any coordinate transform. Not just the Lorentz transform.
     
  11. Jun 15, 2015 #10

    Nugatory

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    You've complicated the situation by using non-inertial coordinates, but that's still just a coordinate transformation with no more physical significance than the choice to use polar instead of Cartesian coordinates in a classical central-force problem. You won't find any physics that's not already part of Minkowski's formulation of special relativity as physics in flat spacetime.
     
    Last edited: Jun 15, 2015
  12. Jun 15, 2015 #11

    A.T.

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    No, it's about what intervals the physical clocks will have accumulated, if you bring them back together for comparison after a while.
     
  13. Jun 17, 2015 #12

    pervect

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    The first step is to describe the geometry you're talking about with a metric. If you're looking for a bit of philosophy, try arXiv:gr-qc/9508043


    Anyway,the short summary of this is "the metric is all". A few abbreviated quotes , I'd recommend the original for clarity, but I hope this gives the gist:

    Next up is to figure out which mathematical entity describes "the gravity of GR". One popular choice is to view the field as the metric itself. Note that the above quotes support this interpretation, but I personally am not fond of it though it's quite common. Another popular choice of what you mean by "gravity" is the Riemann curvature tensor. The most Newtonian choice is to use the Christoffel symbols. While this is the most Newtonian interpretation of the "gravitational field", it's not a tensor and hence requires you to specify a coordinate system before it's meaningful.

    On the other hand, my interpretation of your post is that you've essentially started to specifying a coordinate system. If you regard coordinates as totally irrelevant, specifying a flat Minkowskii geometry is simpler and equivalent, there's no need to talk about rotation or rotating frames at all. The underlying geometry is the same regardless of whether or not you choose rotating or non-rotating coordinates. The tensor nature of the metric and the Riemann curvature tensor means that the coordinate choices in some sense "wont matter" - the components are different, but because there is a diffeomorphism between the rotating and non-rotating coordinates, the tensors are equivalent even though their components change. Because you think that the coordinate choice matters, I am thinking that you are thinking of the "gravitational field" as the Christoffel symbols.

    Once you've gotten this far, if you agree with the approach, I'd suggest thinking about what particular components of the Christoffel symbol measure rotation, and how the different components relate to the physical ideas of "centrifugal force" and "coriolis force" in the rotating frame.
     
  14. Jun 17, 2015 #13

    ChrisVer

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    Well that was the first idea I had, but obviously I could not use the metric of a torus, since the torus is rolling and intuitively I have a feeling this motion should be in the metric (as you insert the angular momentum in Kerr's metric vs the Schwarchzild's one).
     
  15. Jun 17, 2015 #14

    PeterDonis

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    Only if the rotating torus has enough stress-energy to significantly curve spacetime. Previous responses to you in this thread have assumed that's not the case, i.e., that spacetime is flat Minkowski spacetime, meaning the stress-energy of the torus is negligible and the only thing of interest is its rotation. Then you can use rotating coordinates, in which the angular velocity of the torus will appear in the line element, but the curvature of spacetime will still be zero--it will still be flat Minkowski spacetime, just in different coordinates. For a look at such coordinates, see here:

    https://en.wikipedia.org/wiki/Born_coordinates

    If, OTOH, you want the stress-energy of the rotating torus to be significant, then nobody knows what the exact spacetime geometry is; it won't be flat (because no spacetime geometry with significant stress-energy can be flat), but nobody knows a closed-form solution for it. The best you could do would be to solve the Einstein Field Equation numerically.
     
  16. Jun 18, 2015 #15

    pervect

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    Yes, fors steady rotation you'd just assume ##\theta = \omega t##

    Also you can do it algebraically, though it would be advisable to write ##\bar{x}## as functions of x,y,t instead of the other way around. The algebraic approach only involves using the chan rule expansion of dx as a function of d##\bar{x}##, d##\bar{y}##, d##\bar{z}##, d##\bar{t}##, doing the same for dy, dz, and dt, then computing dx^2 + dy^2 + dz^2 - dt^2 in terms of the barred coordinates.


    You also need to note that ##r^2 = x^2 + y^2 = \bar{x}^2 + \bar{y}^2##
     
  17. Jun 18, 2015 #16

    pervect

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    Actually, the easiest thing to do is this: use cylindrical coordinates. The non-barred coordinates will be non-rotating, the barred corridnates will be the rotating coordinates you're after.

    Then ##r = \bar{r} \quad \theta = \bar{\theta} - \omega \bar{t} \quad t = \bar{t} \quad z = \bar{z}##

    You know the metric (I probably should have said - line element) is ##dr^2 + r^2 d\theta^2 + dz^2 - dt^2## in the non-rotating coordinates. Applying the chain rule you get the metric in the rotating coordinates as ##d\bar{r}^2 + \bar{r}^2 \left(d\bar{\theta} - \omega d \bar{t} \right)^2 + d\bar{z}^2 - d \bar{t}^2##

    I'll leave the rest of the messy calculations up to you - a program such as Maxima or Maple helps a lot. You eventually probably want to calculate ##u^b \nabla_b u^a## , or the equivalent in whatever notation you're familiar with, to get the 4-acceleration. A more advanced thing to ponder is how the equations ##\nabla_a T^{ab}## play out in your rotating coordinates. This would allow you to answer questions like - "what would be the tension on a string holding an object in place in the rotating frame (or in a circular orbit in the nonrotating frame). Some thought is required how to define the string, to take a trivial example, if the string has weight of its own, some of the tension in the string is needed to support its own weight. Defining a suitable notion of a "weightless" string turns out to be coordinate dependent.

    If you've got a text that talks about orthonormal basis of one-forms, you might want to read up on that section and apply those techniques.

    [add]It wasn't in your original question, but it wouldn't be a bad idea to work out the 4-acceleration in both the rotating and non-rotating coordinates to demonstrate to yourself that the coordinate choice doesn't matter (as should happen when you express everything in tensors). By "doesn't matter" I mean that you can work the problem in any coordinate choices you like, and the tensor result will transform according to the tensor rules, so that you can just transform the solution from one set of coordinates to the new set rather than work the problem out from the start in both sets of coordinates.
     
    Last edited: Jun 18, 2015
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