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General relativity and accelerated frames

  1. May 5, 2014 #1
    Hi,
    I'm going deeper in basics points on general relativity but, instead of swimming directly between the differential geometry, I'm trying to base my knowledge on strong physics bases first. I'm studying both on Wheeler's stuff ( I'm collecting almost all his books ), directly on Einstein's papers and on Thorne's books as well.

    The most 2 important things I think to have grasped are:

    1) "in a gravitational fields ( of small spatial extension ) things behave as they do in space free of gravity". After having thought a while, it seems soooo simple but soooo strong as principle. If the frame is sufficient small ( i.e. you can neglect tidal forced ) particles free falling within this frame cannot notice any gravity effect on them, and so gravity can be removed from the picture.
    This also mean these particle can be seen as inertial and so we can apply special relativity principle to this situation ( turning free falling into free floating).

    2) when we are standing still on Earth surface, we receive a repulsive force which is opposite to the gravitational pull in that point. This force is not intertial at all, since it's due to columbian and nuclear forces. This makes our "true" frame of reference as not inertial indeed. This is why we usually see bullets following a parabolic trajectory. It's not because the bullet is not inertial, is because WE are not in a inertial frame. If we'd remove the Earth below us, we would float together the bullet, seeing its trajectory being a geodesic in flat space ( i.e. a straight line ).

    It seems something is missing in my picture to me and I will try to explain it. so far we are able to get rid of gravity in a sufficient small "local inertial free float frame" and apply special relativity rules. I know, and it makes a lot of sense to me, a rocket with engines ignited producing a 1g upward acceleration in a gravity free far space is not distinguishable from a man standing still on Earth, experiencing costant downward acceleration. This is another aspect of the equivalence principle ( it should be the strong version of it ).

    But now my doubt and missing brick is asking me: "does this mean we can consider the physics on the rocket as there would be gravity? If yes, it means I would standing still experiencing a sort of non inertial force on my feet, exactly if I were on Earth."
    So this would mean I could not get rid of gravity, since my frame is not intertial?
    And finally "how I could apply special relativity to accelerated frames if I'm not able to get rid of gravity?".
    This sounds also like asking "are these "special" local inertial free float frames still appliable to "pure" accelerated rockets and frames? ( or at least "how are they related?" ).

    I mean: it's clear to me when the I'm free float I can get rid of gravity and I can apply special relativity, but I cannot see how general relativity applies to an accelerated frame.
    It sounds like "we can replace the engine acceleration with a place holder gravitational field" ( and it sound good to me ) but I cannot see how we can apply special relativity to those frames which are not inertial...

    I remember also from an old post of mine the accelerated frames are already encompassed somehow in special relativity, but no one is usually talking about that ( eventually please depict them a bit, if you are going to ). Since it's true, it seems the general relativity is really a theory of mainly gravity, since accelerated frames take already a good place in special relativity framework.

    I know I've been a bit not enough precise in my sentences above, so I apologize in advance for that. These idea are so "simple" but also so difficult to put together coherently.
    this is probably their strength.

    Thanks
     
  2. jcsd
  3. May 5, 2014 #2

    mfb

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    Right.
    You do not have gravity in the rocket. You feel the same non-inertial force, but due to a different reason (rocket engines instead of a solid, non-falling floor). You can use special relativity without any problems.
     
  4. May 5, 2014 #3
    Thanks for you quick answer. The main problem is that I cannot see how to use special relativity in that situation. The main point ( or at least the one I'm confusing ) is the rocket frame, which is accelerated, is not intertial at all ( exactly as the one with gravity , so how can I apply special relativity?

    Of course, you can see easily from here, my main problem is how accelerated frames are encompassed in special relativity. People always claim about GR and how it extends special relativity encompassing accelerated frames, but it turns out to be not true to me, since it encompass "only" gravity and not all accelerated frames.


    So the point is: I've 2 frames with a relative acceleration. How SR can model them?
     
  5. May 5, 2014 #4

    DrGreg

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    Rindler coordinates cover a rocket rigidly accelerating with constant proper acceleration in a straight line.

    Born coordinates cover an object rigidly rotating with constant angular velocity.
     
  6. May 5, 2014 #5

    WannabeNewton

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    In SR, the rest frame of an accelerating observer can be idealized as a continuous sequence of inertial frames instantaneously at rest with respect to the observer (the spatial orientations of the instantaneous inertial frames must be fixed separately).

    In GR the same holds but with local inertial frames; that's all there is to it.
     
  7. May 5, 2014 #6

    jtbell

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    By using calculus. In an analogous situation, in elementary classical mechanics, students learn that work is force times distance: W = FΔx. However, this works only if F is constant. If the force varies at different points, you have to use integral calculus:
    $$W = \int_{x_1}^{x_2} {F(x)dx}$$
     
  8. May 5, 2014 #7
    Hi WannabeNewton, it's always a pleasure to meet your explanations. I was totally lacking about the part of sequence of inertial frames in SR ( I has a SR course in Politecnico of Tourin, but no mention was done about it ). Now you are depicting it it sounds really really like the sequence of local frames in GR.
    So, again, it seems to me GR is about gravity and not about "generic" accelerated frames, since they are already encompassed in SR. Don't know why some textbooks report that "general" refers to accelerated Vs uniform frames. this now sounds really odd to me.

    More than this, I cannot see why the rocket Vs Earth non-inertial frame example is so widely used in GR ( or at least again on some texts ). In that accelerated rocket frame the SR is still enough to describe things, so no GR is needed at all.
    So why it's so used as main example? Probably because it's the Einstein version of equivalence principle ( Galileo weak one ) which states the equivalence between gravitational and inertial mass?

    From a mere historical point of view, was Einstein going to encompass only gravity in order to move from SR to GR? I mean, there were no need to model accelerated frames since they were already there in SR, isn'it?
     
    Last edited: May 5, 2014
  9. May 5, 2014 #8
    OK this makes a lot sense as well, thanks
     
  10. May 5, 2014 #9
    Yes I've seen Rindler coordinates within a Susskind GR lecture ( when talking about observers falling or pasting near a black hole horizon ) but I dind't immediately realized they were good within SR as well. Thinking of it now, it was straighforward, and Susskind described accelerated wordlines as hyperbolas in Minkowsky space.
     
  11. May 5, 2014 #10

    WannabeNewton

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    Yes that is correct.

    They might be outdated books. Regardless, it's just flat out wrong so definitely don't waste your time with books that make such claims.

    People were already working with rotating systems in SR well before Einstein formulated GR (c.f. Ehrenfest's paradox).
     
  12. May 5, 2014 #11
    A bit confused. How does this fit in with the Equivalence Principle? Does it mean that a non-gravitational accelerated frame (even if tidal/central) is not equivalent to a true gravitational field? (If not, in what way?)
     
  13. May 5, 2014 #12

    WannabeNewton

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    What is a "gravitational accelerated frame" or a "non-gravitational accelerated frame"? I suppose you mean instead accelerated frames that are in curved space-time as opposed to those in flat space-time? A rigidly rotating frame in flat space-time defined up to the light barrier can, through the equivalence principle, be considered instead as a stationary (but not static!) gravitational field i.e. one which possesses angular momentum. The equivalence principle simply says we can do physics using either paradigm: that of a rigidly rotating frame in free space or that of a rotating gravitational field. Of course the gravitational field so obtained will not correspond to a curved space-time geometry i.e. there will be no tidal forces.
     
  14. May 5, 2014 #13

    king vitamin

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    An accelerated frame in Minkowski space still has a zero Reimann tensor everywhere, and thus must correspond to a trivial stress-energy tensor everywhere by Einstein's equations. Also, there can be no tidal forces (since only the Reimann tensor enters the geodesic deviation equation). Therefore, no physical (produced by sources) gravitational field corresponds to an accelerated Minkowski frame.

    In a small enough region that there are no tidal forces (we are measuring zero Reimann tensor), we are equivalent to Minkowski space, possibly with acceleration. We can go to geodesic normal coordinates, where the metric is Minkowski; this is the equivalence principle.

    In this way, "true gravity" is distinguished by the existence of tidal forces.
     
  15. May 5, 2014 #14
    OK, I am trying to get a simple answer, and an example may help.

    Situation (a): A pebble is moving in a circular orbit around Earth at a distance R from center of Earth, under Earth's gravitational influence.

    Situation (b): An alien sitting in a rocket in deep space is whirling about a pebble attached to a longish string of length R in a circular trajectory.

    The orbital velocity of both pebbles are the same (i.e. the accelerations on both are the same).

    Would there be a difference between the two pebbles in any relativistic way - e.g. would their time dilations turn out to be different?
     
  16. May 5, 2014 #15

    WannabeNewton

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    (A) and (B) are not equivalent scenarios. Yes their time dilations will turn out to be different. Situation (B) involves only kinematical time dilation due to orbital rotation in free space whereas (A) involves both kinematical and gravitational time dilation due to both orbital rotation and the intrinsic gravitational field of the Earth. If you apply the equivalence principle to (B) then you are simply reinterpreting the kinematical time dilation as a pseudo-gravitational time dilation but you are not introducing anything new.
     
  17. May 5, 2014 #16

    PeterDonis

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    Assuming the alien adjusts his rate of whirling appropriately, yes. There's no requirement for him to do that, however; he could whirl the pebble at any angular velocity he likes, so long as its tangential velocity is less than the speed of light.

    No, they're not, even if the orbital velocities match. The pebble in (b) feels a force; the pebble in (a) does not, it's in free fall. This means the *proper* accelerations of the two pebbles are different, and in relativity, the most useful definition of acceleration is proper acceleration.

    There is something called "coordinate acceleration", which is what you are thinking of here, and which is the same for both pebbles (assuming we choose coordinates appropriately). But in relativity, coordinate acceleration isn't very useful because it depends on how you choose coordinates; you could choose coordinates centered on the pebble and its coordinate acceleration would be zero. Proper acceleration, OTOH, is a direct observable; just attach an accelerometer to the pebble. It will read zero in (a) and nonzero in (b).

    Definitely. See above.

    Yes, because gravity is present in (a) and not in (b).
     
  18. May 5, 2014 #17
    Then let us consider the 'pebbles' to be 'satellites' with human beings inside. Would they be able to tell the difference in the two scenarios (without any reference to the outside)? I feel in the second case also the centrifugal force will match the pull on the string, so inside observers would float freely and not be able to tell the difference... may be I am wrong.

    In other words... are we saying that the following WEP statement (Wikipedia) is wrong?
    "The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception."
     
  19. May 5, 2014 #18

    WannabeNewton

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    Yes because in one case you have freely falling observers in curved space-time, thus observers who are locally inertial, and in the other you have accelerated observers in flat space-time. Accelerometers on board the satellites will allow observers in the satellites to tell apart the two scenarios.

    Nothing stated thus far contradicts Wikipedia's statement.
     
  20. May 5, 2014 #19

    PeterDonis

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    Certainly. In case (a) the humans will be free-floating inside the satellites, just as astronauts do inside the Space Shuttle or the International Space Station.

    In case (b), the humans will feel a force pressing them against the outer wall of the satellite; they won't be free-floating.

    What "centrifugal force"? In the global frame you were implicitly using (the one where "acceleration" in the sense of coordinate acceleration is present), there is no such thing as "centrifugal force". In case (a), there is the "force" of gravity, and that is the only "force" present; but "force" is in scare-quotes because objects moving solely under the influence of gravity feel no force; they are in free fall, weightless.

    In case (b), the only force present is the pull on the string; and since there are no scare-quotes this time, this force is actually felt; it is transmitted by the outer wall of the satellite, which pushes radially inward on the humans, who therefore feel as though they are being pressed against the wall.

    If you want to analyze these scenarios using "centrifugal force", you have to pick a frame where that force is present, i.e., a frame rotating with the "satellite". In this frame, the satellite is at rest. In case (b), it's at rest because there are two forces on it which balance: the pull on the string, and centrifugal force. But the latter is a "force" with scare-quotes, because it's like gravity: objects moving solely under the influence of centrifugal force don't feel any force. So in this case the humans inside the satellite feel the force exerted by the string and transmitted to them by the wall of the satellite.

    In case (a), in the rotating frame, the satellite is at rest because there are two "forces" on it which balance: gravity and centrifugal force. But here *both* "forces" are in scare-quotes; no force is actually felt. So both the satellite and the humans inside it are just freely floating.

    If you're wondering how the satellite and the humans inside it can be following a curved path while feeling no force in case (a), see below.

    No, you're just mistaken about what "the local effects of motion in a curved space (gravitation)" are. You, standing on the surface of the Earth, feel the same local effects as you would if you were in a rocket in deep space accelerating at 1 g. The local effect of feeling 1 g acceleration is the "local effect of motion in a curved space (gravitation)" that the Wikipedia statement is referring to. (Granted, the Wikipedia statement doesn't make that very clear; that's why Wikipedia is not a good authoritative source. :wink:)

    In other words, the reason case (a) is different from case (b) is that spacetime in case (a) is curved, so the path followed by the satellite, even though it looks "circular" (in space, not spacetime), just like the circular path in case (b), is actually straight (but straight in spacetime, not space), so an object following it feels no force. In case (a), spacetime is flat, so the path that looks curved *is* curved (circular in space, but a helix in spacetime because the time dimension has to be included), and an object following it must be feeling a force that is pushing it out of its natural straight line path.
     
  21. May 6, 2014 #20
    I think what you are saying is right, and I am getting a sense of it, but there's something I am still not quite able to understand...

    First, how are the two scenarios different from Einstein's elevator example... and maybe I am missing the understanding of the word 'tidal' ... will help if you can explain how that might make a difference...

    Secondly, you are saying 'freely falling observers in curved space-time' will have a different experience than 'accelerated observers in flat space-time'. That seems to contradict the statement 'local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception', which you are saying is still correct. I don't get the difference.

    Perhaps I am understanding something wrongly or missing a nuance... is it 'space' vs. 'space-time'...?
     
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