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About tidal forces

  1. Mar 8, 2010 #1
    Recently while searching through some old threads in this section of the forums, I came across a few statements that do not seem to me to be entirely self-consistent. I was wondering if someone could tell me which - if any - of these declarations is not correct:

    -- in a uniformly accelerating frame of reference - like inside a rocket moving with constant proper acceleration - points "lower" in the induced gravitational field experience greater acceleration than ones at "higher" locations

    -- such a frame of reference is described by the Rindler metric

    -- tidal forces are equivalent to curved spacetime

    -- the curvature tensor of the Rindler metric is zero

    Thanks.
     
  2. jcsd
  3. Mar 8, 2010 #2

    bcrowell

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    In these first two statements, you should keep in mind that a uniform gravitational field is not something that really exists in GR. By the equivalence principle, we expect that a uniform gravitational field is equivalent to a uniformly accelerating frame. Therefore there is an ambiguity in the notion of a uniformly accelerated frame in GR. These ambiguities are the subject of Bell's spaceship paradox http://en.wikipedia.org/wiki/Bell's_spaceship_paradox . There are other metrics that are candidates for playing the role of "the" (really "a") uniform field in GR, e.g., the Petrov metric http://arxiv.org/abs/0802.4082 .

    Tidal forces are one type of curved spacetime, measured by one part of the Riemann tensor. There is another part of the Riemann tensor, expressed by the Ricci tensor, which measures non-tidal curvature. If we're looking for something that is going to play the role of a uniform field in GR, we probably want a vacuum solution, so in that special case it makes sense to focus on tidal forces.

    The gravitational field is not a tensor in GR. At a given point in space, an observer can make the gravitational field be anything he likes, including zero, by a choice of coordinates. Therefore we should not expect that a uniformly accelerating frame, or a frame with a uniform gravitational field, should correspond in any particular way to intrinsic properties of spacetime as expressed by tensors like the Riemann tensor. The curvature tensor of the Petrov metric is not zero, but I don't think that has anything in particular to do with whether one or the other (Petrov or Rindler) is a better candidate for a uniformly accelerating frame/uniform field.
     
    Last edited by a moderator: Apr 24, 2017
  4. Mar 8, 2010 #3

    Mentz114

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    I don't know where you found that but it begs the question - which curved spacetime ? As it happens there are no tidal forces in the Rindler spacetime because the Riemann ( and hence Weyl ) tensor is all zero.

    The Rindler chart is a set of coordinates in Minkowski space-time used by an observer with constant proper acceleration. It is a trivial solution of the Einstein field equations but it seems not to to correspond to a ST created by a source.
     
  5. Mar 8, 2010 #4
    I just found the thread where I got that idea, although I vaguely remember seeing it elsewhere, long ago:

    https://www.physicsforums.com/showthread.php?t=156168#

    Entry #62 quotes Kip Thorne, "tidal gravity is a manifestation of spacetime curvature," after which the PF member says,

    "thus it is impossible to have tidal forces without spacetime curvature".

    Looking at it again I notice that these two statements are actually not logically equivalent.
     
  6. Mar 8, 2010 #5
    Wait, does this mean that my first statement about the differences in acceleration at different "heights" in the spaceship is not correct?
     
  7. Mar 8, 2010 #6

    bcrowell

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    Your original statement was correct. This is analyzed in more detail here: http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken] In the Rindler coordinates, the proper acceleration of a particle with an instantaneous coordinate velocity of zero varies with height. This is a good argument against the Rindler coordinates as GR's version of a uniform field. The Petrov metric is better by this criterion. In the Petrov spacetime, the proper acceleration of a particle with an instantaneous coordinate velocity of zero is independent of the "height" r.
     
    Last edited by a moderator: May 4, 2017
  8. Mar 8, 2010 #7

    Mentz114

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    Of course there are tidal effects. The 'ball of coffee grounds' would stretch in the direction of the acceleration. I keep applying GR concepts to the flat Rindler space and coming up with no-no's. There must be a lesson here but I'm not sure what it is ... :biggrin:

    Ben, I can't find a Petrov metric. Do you have a reference ?
     
  9. Mar 8, 2010 #8

    bcrowell

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    The one I gave in #2 is the reference on the Petrov metric that I think is the easiest to understand and most accessible online.
     
  10. Mar 8, 2010 #9

    bcrowell

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    Here is a FAQ entry I wrote up on this topic.

    FAQ: In general relativity, what does a uniform gravitational field look like? What about a uniformly accelerating frame of reference?

    The equivalence principle tells us that (a) these two are really the same question, and (b) both questions are frame-dependent.

    There is no metric and set of coordinates that gives a completely satisfactory uniform gravitational field. The two possibilities that are generally regarded as coming the closest are (1) flat spacetime in Rindler coordinates, and (2) the Petrov metric.

    Rindler coordinates are the coordinates defined by an observer in flat space with a constant proper acceleration a. The line element is (1+ax)^2dt^2-dx^2. This metric describes a flat spacetime, since it is derived from the Minkowski metric by a change of coordinates. It is not a perfect embodiment of our concept of a uniform gravitational field, because a test particle released with zero coordinate velocity at some height x has a proper acceleration that depends on x; only if it is released at x=0 does it have acceleration a.

    The Petrov metric was first found by Lewis in 1932, but is now named after Petrov, who rediscovered it in 1962. It is a vacuum solution of the Einstein field equations, and it is unique because of its high degree of translational symmetry; in technical terms, it is "[t]he only vacuum solution of Einstein’s equations admitting a simply-transitive four-dimensional maximal group of motions" (Gibbons 2008). This translational symmetry is what we want in a uniform field. Unlike the Rindler coordinates, the Petrov metric has the property that the proper acceleration of a test particle released with zero coordinate velocity is the same no matter where it is released. However, the Petrov metric has some strange physical properties, including closed timelike curves, which make it also not satisfactory as "the" uniform gravitational field in general relativity. The most accessible reference for information on this spacetime is Gibbons 2008.

    T. Lewis, Proc. Roy. Soc. Lond. A136 (1932) 176

    Petrov, in "Recent Developments in General Relativity," 1962, Pergamon, p. 371

    Gibbons and Gielen, "The Petrov and Kaigorodov-Ozsváth Solutions: Spacetime as a Group Manifold," http://arxiv.org/abs/0802.4082
     
    Last edited: Mar 8, 2010
  11. Mar 8, 2010 #10

    Mentz114

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    Ben,
    thanks a lot. That's helpful. I gathered the paper in a sweep of Gibbons' papers in the arXiv earlier, but I hadn't seen the ref in #2. I doubt if I'll be able to follow it all but it looks interesting.
     
  12. Mar 8, 2010 #11

    bcrowell

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    Yeah, a lot of it is technical stuff I don't understand either, but for me it was relatively intelligible if I didn't obsess over the details.
     
  13. Mar 8, 2010 #12

    JesseM

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    Wait, are you saying a freefalling "ball of coffee grounds" would stretch? That doesn't seem right to me, since a freefalling ball in Rindler coordinates can be transformed into an inertially-moving ball in an inertial coordinate system on the exact same spacetime.
     
  14. Mar 9, 2010 #13

    Mentz114

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    I don't know. Can there be free-fall in this scenario ? The coffee-grounds are being individually rocket propelled.
     
  15. Mar 9, 2010 #14
    1. Since the coffee grounds at the "lower" end of the sphere accelerate at a higher rate then those at the "upper" end, the sphere must stretch

    2. Since the Wely tensor and the Ricci tensor of the Rindler metric are both zero, the sphere must maintain its shape and size

    3. Since the observer's speed with respect to the sphere is increasing, from his/her perspective the sphere must squash

    This sounds like one of those SR paradoxes that can only be resolved through a detailed mathematical analysis using spacetime diagrams.
     
  16. Mar 9, 2010 #15

    JesseM

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    But that's what I'm asking, what's the scenario? From the perspective of the accelerating observers at rest in Rindler coordinates, are the coffee grounds also supposed to be held at rest (and if so how? Individual rockets? But in that case wouldn't each individual particle be at rest too, so the shape wouldn't change at all?) or do the Rindler observers see the coffee grounds free-falling in the pseudo-gravitational field, so that they're moving towards the Rindler horizon?
     
  17. Mar 9, 2010 #16

    Mentz114

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    In this picture the test bodies would be seen from the Rindler frame to be falling away.
    In GR I would calculate the tidal forces in a frame by transforming the Riemann tensor to the frame, then projecting out with the 4-velocity of the observer, to give a second rank tensor of the tidal forces. So, because the Riemann tensor is zero, the tidal forces will also be zero.

    But I'm almost certainly wrong as usual.

    I don't know how the grounds would look to a Minkowski observer, assuming they are individually propelled.
     
  18. Mar 9, 2010 #17
    Wait a minute! What do you mean by "accelerating observers at rest"? This is only correct if we take the frame of comoving observers to be instantaneously at rest which means their proper velocity is zero momentarily so in this case one leads to [tex]\ddot{t}=\frac{-2a\dot{x}\dot{t}}{1+ax}=0[/tex] for the Rindler metric

    [tex]ds^2=(1+ax)^2dt^2-dx^2-dy^2-dz^2[/tex]

    and from this point it is obvious that [tex]\dot{t}= const.=T[/tex] so that

    [tex]\ddot{x}=-aT^2(1+ax).[/tex]

    But since the proper acceleration of particles is still dependent on the position, so the coffee grounds are not going to retain their initial shape while staying at rest by definition. This strongly suggests that the Rindler metric does not generally admit a uniform gravitational field unless assuming the local discussion and the EP.

    AB
     
  19. Mar 9, 2010 #18

    JesseM

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    Well, the "free-falling in the pseudo-gravitational field seen in Rindler coordinates" scenario is equivalent via a coordinate transformation to having the grounds all moving inertially in an inertial frame, so at least in the inertial frame there will be no change in shape and no tidal forces. It's possible that there is some coordinate-based change in shape seen in Rindler coordinates, although this wouldn't imply any actual physical stresses if the ball were semi-solid rather than just a cloud of noninteracting coffee grounds.
    Again, when you talk about them being "individually propelled", do you mean they are propelled in such a way that each ground remains at rest in Rindler coordinates? In that case the ends of the cloud would get closer together just like the family of accelerating observers at rest in Rindler coordinates. This page shows a diagram of how the worldlines of these Rindler observers look in Minkoski coordinates (along with the surfaces of simultaneity from Rindler coordinates drawn in gray, and the Rindler horizon drawn as a dotted line):

    Coords.gif
     
  20. Mar 9, 2010 #19

    JesseM

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    The full phrase was "accelerating observers at rest in Rindler coordinates"...but I was speaking sloppily, I meant that they are accelerating as seen in any inertial frame (and thus have nonzero proper acceleration), but they are observers "at rest in Rindler coordinates". See the diagram I posted above, which shows the worldlines of observers at rest in Rindler coordinates as seen in an inertial frame.
    I think "uniform gravitational field" has a confusing meaning in relativity, it doesn't actually mean a field where the proper acceleration is the same everywhere--see this page from John Baez's twin paradox page, which says that the time dilation of clocks at different heights in a "uniform" field would actually be different:
    This implies that clocks at rest in this uniform field have different velocities as seen in an inertial frame (which is true for observers at rest in Rindler coordinates who are undergoing Born rigid acceleration, which is designed to ensure that the distance between them remains constant in their own instantaneous inertial rest frame at each moment...again, see the diagram I posted above for what this should look like).

    By the way, I had an idea about the ball of coffee grounds. Although it's true that if they are free-falling in Rindler coordinates they must be moving inertially in an inertial frame, individual grounds at different ends of the ball might still have different velocities in the inertial frame so the ball as a whole could change shape in the inertial frame. And if you require that the grounds were all initially at rest in Rindler coordinates before beginning to fall (held up by individual rockets which are all turned off simultaneously in Rindler coordinates, for example), wouldn't this in fact imply different velocities in the inertial frame? (edit: actually looking at the gray lines of simultaneity in the diagram above, it looks like q=0 in Rindler coordinates would give a line of simultaneity parallel to the t=0 plane (i.e. the x-axis) in the inertial Minkowski frame, and if the worldlines of the grounds at that moment are tangent to the worldlines of the accelerating observers at t=0, then all the worldlines of the grounds should just be vertical in the inertial frame, meaning they are all moving at the same velocity, so that idea of mine above probably isn't the right answer.)
     
    Last edited: Mar 9, 2010
  21. Mar 9, 2010 #20
    This means that the observers have zero proper velocities momentarily.

    I don't see anything in contrast to my understanding of a uniform gravitational field: There it is also said that "...by "uniform" we mean that the force felt by each object is independent of its position." Or, equivalently, the proper acceleration is independent of position. This is a funadamental property of spacetimes if the EP and thus local inertia are taken into account.

    If the grounds were all initially at rest, they all would stay at the same level locally as the instantaneous observer being at rest does only keep being at rest just in a very short time dt and there everything will look like the Minkowski metric. But as long as time elapses much, the grounds start having their own levels as they experience different proper accelerations. If a bus hits a bump in the street, due to the rapid change of acceleration, your body may jump upwards and your handbag would then be in a different shape for moments of being in jump.

    AB
     
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