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B How does GR explain increase/decrease in speed?

  1. Nov 12, 2015 #1
    OK, so my basic understanding is that GR is all about geometry of space-time. It's all geometry, no other mechanism.

    This explains why objects change direction due to gravity. But why does the speed increase?

    How does pure geometry cause a change in speed?
    Also, where does this kinetic energy come from?

    I know people will talk about potential energy, but that's just a hand-waving mathematical trick to balance equations. What does geometry have to do with potential energy?
     
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  3. Nov 12, 2015 #2
    The speed of a falling object does not increase. It is you, the observer, accelerating in the other direction, as the surface of Eaarth pushes your feet.
    The geometry allows you to stay at the same distance from the center of the Earth, while accelerating.
     
  4. Nov 12, 2015 #3
    Forget the observer.
    Are you saying that a falling ball bearing does not increase speed by 9.8 ms-2?
    Sure, we can say that the Earth is increasing its speed towards the ball bearing, but the problem remains.

    Why does geometry cause a change in speed and a change in kinetic energy?
     
  5. Nov 12, 2015 #4
    Kinetic energy is not something absolute. It is always taken relative to some observer. So I can't "forget the observer".

    Geometry does not change speed of objects: acceleration does that. The surface of Earth is accelerating, as viewed from an inertial frame.
    As viewed from the surface of Earth, an inertial ball bearing appears to accelerate in the other direction.
     
  6. Nov 12, 2015 #5

    PeterDonis

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    "Speed" is frame-dependent; so is "direction". These are not good things to focus on if you want to understand how GR models gravity as spacetime curvature.

    Kinetic energy is also frame-dependent; as SlowThinker points out, the kinetic energy of a falling rock does not change in the inertial frame in which the rock is at rest.

    Potential energy is not exactly frame-dependent, but it is dependent on choosing a zero point, and it is only well-defined under certain conditions. But the spacetime around a massive gravitating body like the Earth is one of those conditions, so it's not correct to just discount potential energy as a "hand-waving mathematical trick"; in the conditions where it's well-defined, it is telling you something physically significant. If we choose the zero point to be at infinity, which is the usual convention, then the potential energy at any finite distance from the gravitating body is negative, and it is telling you how much energy it would take to move an object at rest at that distance out to infinity.

    However, that's still not really a geometric viewpoint. Geometrically, the thing to focus on is not "gravity" in general, but specifically tidal gravity. Tidal gravity is what shows the presence of spacetime curvature (in fact, they are really the same thing). For an example of tidal gravity, consider two rocks, which at some instant are both at rest above the Earth at slightly different altitudes. Relative to the Earth, they will both start free-falling downwards, but at slightly different rates, so the separation between them will increase.

    Now shift viewpoints and imagine you are falling along with one of the rocks, and keeping track of your separation to the other rock. You are in free fall, so you feel no acceleration (and you can use an accelerometer to verify that you are experiencing no acceleration)--as far as you can tell, you are at rest in an inertial frame. You can also verify that the other rock is in free fall (you can attach another accelerometer to it to verify this). But the other rock started out at rest relative to you; yet over time, it moves away from you, as viewed from the "inertial" frame in which you are at rest. That is not possible in a "true" inertial frame like the ones we have in Special Relativity; in such an inertial frame, freely falling objects can never change their speed relative to one another, so if they start out at rest relative to each other, they will always be at rest relative to each other. That isn't happening with the two rocks, so clearly SR is not exactly correct in this case.

    What aspect of SR fails in this case? Geometrically, the worldlines of freely falling objects can be viewed as "grid lines" (the technical term is "geodesics") that mark out the geometry of the spacetime. In SR, this geometry is flat: grid lines that start out parallel (i.e., the objects are at rest relative to each other) always stay parallel. But in the presence of gravity, that no longer happens; grid lines that start out parallel (like the worldlines of the two rocks) don't stay parallel, because of tidal gravity. Initially parallel grid lines not staying parallel is the definition of curvature, geometrically speaking; so tidal gravity is spacetime curvature.

    Finally, what about the "change in speed" of a falling rock relative to the Earth? As SlowThinker has pointed out, if we just look at one rock, we can account for its change in speed relative to the Earth by observing that the Earth's surface itself is accelerated; if you are standing on the Earth's surface, you feel acceleration (weight), so you are being pushed upward relative to the rock, which is in free fall. (Once again, you can attach accelerometers to you and the rock to verify that you are the one who is accelerated.) Where spacetime curvature comes in is when you look at rocks falling in different places, and realize that they fall at different rates (depending on height) and in different directions (depending on location on the Earth's surface--rocks in Australia fall in a different direction from rocks in Europe). Spacetime curvature is what makes all those freely falling rocks fall in different directions, and what allows different pieces of the Earth's surface to be accelerated in different directions while the Earth as a whole stays the same size and shape.
     
  7. Nov 12, 2015 #6
    There is no third observer. It is a two body problem.
    Think of a universe with only two particles P1 and P2.
    Take P1's frame of reference. It sees P2 moving towards it and P2's radial speed and KE is increasing.
    Where is P2 getting all this extra KE?

    Your second phrase is a tautology.
    GR purports to explain gravity as a purely geometrical effect, so the question arises?
    From P1's frame of reference, where is P2 getting its KE?
     
  8. Nov 12, 2015 #7
    No energy is created but part of P2's rest energy looks like kinetic energy.
    Just like the Leaning Tower of Pisa is not getting any shorter but it looks wider.
     
  9. Nov 12, 2015 #8
    That's what I mean. PE is a bookkeeping trick which has no physical basis in reality.
    Consider the case of the ball bearing falling to Earth. Normally, we are told that the ball bearing is getting its downward KE from the PE stored into it when I lifted it up. The molecular bonds' energy in ATP -> KE in my muscle fibers -> KE in ball bearing -> PE in ball bearing -> KE in ball bearing.

    But that explanation is nonsense.

    If I lift the ball bearing high enough, it will fall towards the Sun instead of the Earth, and it will pick up a whole lot of KE along the way. If it falls towards a quasar 10 billion light years away, it will pick up even more. How can one claim that the ATP in my muscles gave it all that PE?

    I appreciate the example, but could you please explain the (surely simpler) case of the two-body situation.

    Take P1's frame of reference. P2 starts 'falling' towards it, so P1 sees P2's radial speed and KE increase. I understand that the geodesics explain the directional change in velocity, but what is causing the change in speed and KE?

    Remember we are using P1's frame of reference, so we (P1) are stationary. Only P2 is moving and increasing its speed/KE.
     
  10. Nov 12, 2015 #9
    So you are saying that P1 sees P2's radial speed increase, but P2's total mass/energy remains constant?
    That doesn't sound right. What is there to stop P2 achieving and exceeding the speed of light in that case, if its relativistic mass doesn't increase as seen by P1?
     
  11. Nov 12, 2015 #10

    PeterDonis

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    That's because you lifted it up through the same distance that it fell down. If you let it fall down through a hole in the Earth, all the way to the center, obviously it's going to have more KE than the PE you gave it--because you let it fall through a longer distance than you lifted it. Similar remarks apply to your other cases.

    Before you even tackle relativity, it looks to me like you need to first understand basic Newtonian physics.

    Relative to him, yes.

    The geodesics also explain the change in P2's speed and KE relative to P1. P1 and P2 are both following geodesics of the spacetime they are in (note that this spacetime is not the same as the spacetime around a single gravitating body that I was using earlier in this thread--by hypothesis you have two gravitating bodies in this scenario). Those geodesics are such that they start out parallel (at the instant that P1 and P2 are at rest relative to each other) but don't stay parallel--they approach each other, then pass each other, recede from each other, then become parallel again for an instant (the instant when P1 and P2 have passed each other, decelerated, and are now at rest relative to each other for an instant), then start another cycle of approach, pass, recede. To each of P1 and P2, this behavior of the geodesics appears as the speed and KE of the other object changing.

    The properties of geodesics are independent of the coordinates we choose. Which object we label as having "speed" and "KE" depend on the coordinates, but the geometry of the geodesics does not.
     
  12. Nov 12, 2015 #11
    When all of P2's rest energy has been converted to kinetic energy *, there is no more speed to be gained.
    When the Tower falls, it won't get any longer than it already is.

    * in fact, P2 will always keep a bit of rest energy, so it can always move a bit faster. As if the Tower never falls perfectly horizontal.
     
  13. Nov 12, 2015 #12

    Matterwave

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    In GR, the quantity "velocity" can be more closely associated with the the 4-velocity. The 4-velocity is parallel transported along geodesics (the paths that freely falling objects follow). As such, the 4-velocity does the GR equivalent of "not changing" along free fall paths. There is no 4-acceleration. As there is no 4-acceleration, there is also no change in the 4-momentum (no change here again means parallel transported). The 4-momentum, the quantity that encapsulates Newtonian ideas of energy and momentum into a relativistic theory, does not change along free fall paths.

    What the ball and the surface of the Earth are doing is that they are both following free fall paths, and these free fall paths happen to meet at some point in the future. Just like if you drew two lines which started parallel, but which eventually converged. The "curvature" effect of GR is to make these once parallel lines converge.
     
  14. Nov 12, 2015 #13

    PeterDonis

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    No, this is not correct. First of all, you are trying to apply SR in the presence of gravity, but SR doesn't work in the presence of gravity. Second, even in an SR context, it is not true that, in a frame in which an object is moving, part of its rest energy looks like kinetic energy. An object has the same rest energy in every frame; in a frame in which it is moving, it also has kinetic energy due to that motion, in addition to its rest energy. Total energy is frame-dependent; that's all there is to it.

    "Relativistic mass" is just another term for "total energy", and is frame-dependent. Hence, it is not a good tool for analysis.

    As for P2 somehow exceeding the speed of light, that is impossible because of the geometry of spacetime. The speed of light limitation is not a property of P2, or any other object; it's a property of spacetime itself. A better way to describe that property in curved spacetime is that the worldlines of all objects must lie within the light cone at every event they pass through.

    This is not correct; P2's rest energy is not being converted to kinetic energy. See above.
     
  15. Nov 13, 2015 #14
    Let's drop PE since no one is invoking it as a deus ex machina to conjure up the needed KE in GR.

    Newtonians would probably argue that the Big Bang gave the ball bearing a huge amount of PE which it now converts to KE as it falls into the quasar 10 billion light years away.

    I am confused.

    I was told that we can consider the apple/ball bearing to be at rest and the Earth to be accelerating towards it.

    In our two-body example, I (P1) am at rest and P2 is the only thing moving in my universe. Since GR only works in 3D, and I don't think anyone is invoking time dilation here, I don't see how P2 moving at a steady speed along the 3D space can be perceived as an increase or decrease in speed, no matter how convoluted the geometry.

    By coordinates, do you mean frame of reference?
    Are you saying the warping of spacetime (geodesics) are an intrinsic property of an object regardless of any frame of reference?

    Sorry if I misunderstand but that can't possibly be. Since KE is a relative value, then relativistic mass is a relative value and the resulting warping of space-time by that object must be a relative effect, no?
     
  16. Nov 13, 2015 #15
    Sorry, but isn't that just a restatement of GR's mathematical model?
    I am asking how to reconcile that with the physical reality where the ball does indeed increase speed and momentum as it races towards its rendezvous with the Earth.

    No problem with the trajectory side of things.
    My question is about the fact that the ball covers a greater and greater distance along the geodesic in a given amount of time as it gets closer to its rendezvous point.

    What is there in pure geometry that compels it to do so?
     
  17. Nov 13, 2015 #16

    Orodruin

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    I think it is worth underlining that this fact is not particular of relativity. It is the case also in classical mechanics.
     
  18. Nov 13, 2015 #17

    A.T.

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    Look at the geometry again. The "straight" (geodesic) world-line of a free faller deviates more and more from the time axis, towards the space axis. This corresponds to moving faster and faster through space:

     
  19. Nov 13, 2015 #18
    Thank you.
    Since the answer must be purely geometric, the diagram and animation help.

    This also answers my question above that the curvature cannot be an intrinsic property of an object, but is a property of interaction between two particles.
    The curvature will be custom-tailored for every particle that interacts with the gravitational "field", and each upcoming segment of the curvature will be instantaneously created as the particle moves within that "field". This newly created curvature must occur from local information, since there can be no communication with the other particle, which might be light years away.

    This still leaves the question of where the KE comes from, and we can know that the KE is not just an apparent attribute, but there are absolute constraints on the apple's KE/momentum which every observer must validate.

    For e.g. if we place a sheet of wood at ground level with a known strength that will only be punched through with at least a minimum amount of momentum/KE, and the apple punches through it, then every observer must measure the apple (or wood) to have at least that amount of momentum and KE.
     
  20. Nov 13, 2015 #19

    Dale

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    Don't forget that GR is about the geometry of spacetime, not just space. A change in speed is the same thing as a change in direction in spacetime.
     
  21. Nov 13, 2015 #20

    A.T.

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    KE comes from defining a reference frame. But that is not specific to GR.
     
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