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Question on relativity of movement

  1. Apr 29, 2013 #1
    On a layman level I'm pretty familiar with the principles of modern physics but there's something about relativity which I'm afraid I do not understand properly.
    Relativity tells us that non-accelerating motion is relative to the observer. Saying that I am travelling at 100 Km/h relative to you who are sitting still motionless is indistinguishable from saying that it's me who is motionless and you who are travelling at 100 Km/h relative to me.
    Unless we define the reference frame, one description is physically equivalent and (as I have read frequently and this is the crucial word) indistinguishable from the other.

    This is pretty straightforward when referring to linear motion, but I'm not sure I get it properly when it comes to gravitational orbital motion (for the time being I'm not going to complicate things by arguing that 'gravitational orbital motion' is 'linear', geodesically speaking).

    Let's say, Wikipedia says that the ISS is orbiting the Earth at 27,000 Km/h. Of course this is relative to the Earth's close environment (we can not even say 'relative to the Earth's surface' since the Earth is also spinning). An astronaut out on a spacewalk at the ISS feels like floating motionless, due to the absence of aerodynamic drag and acceleration.
    But can the astronaut really say that he is motionless and it's the Earth (not meaning some particular point in the Earth's surface) which is rotating around him at 27,000 Km/h? Is this description really indistinguishable? By extension the same goes to the fact that both the Earth and the ISS are travelling much faster than that in their orbit around the Sun, not to mention the movement of the Solar system around the Milky Way and the Milky Way's relative to other galaxies.

    My point is that even if the physical behavior is equivalent, I don't see the two descriptions as indistinguishable. We have a physics theory explaining why the ISS and its astronaut orbit the Earth, why the Earth orbits the Sun, why the Sun orbits the Milky Way and so forth.
    But if the astronaut is the one considered to be motionless and everything else moving relative to him, we need a physics theory to explain why a hugely massive Earth orbits a tiny astronaut, why a hugely massive Sun orbits a tiny Earth and so forth.

    Obviously we do not have such a theory. General relativity tells us that both descriptions are equivalent in terms of physical behavior, but surely not indistinguishable from each other. By observing the facts we can deduce that it's the ISS and its astronaut which are orbiting the Earth and not the other way around, can't we? For one description we have a theoretical explanation but for the other one we don't. So the two descriptions are equivalent but surely distinguishable which would contradict the standard interpretations about 'the relativity of motion' which I usually find in popular science books.

    Or am I missing something here?
    Last edited: Apr 29, 2013
  2. jcsd
  3. Apr 29, 2013 #2


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    You are missing a very fundaemental point, confusing "speed" and "velocity". Velocity and acceleration are vector quantities having magnitude and direction. You can only move in a straight line with constant velocity. Any variation from a straight line, in particular an elliptic orbit, is accelerated motion. And accelertation, necessarily involving a force, is NOT "relative".
  4. Apr 29, 2013 #3
    Thanks! So the astronaut, even if he feels motionless, can not say that from his reference frame he is motionless and the Earth is orbiting him, is that right?

    And sorry to complicate things, but 'straight line' is a tricky concept in GR, isn't it? An elliptic orbit is a straight line after all, isn't it?
    Last edited: Apr 29, 2013
  5. Apr 29, 2013 #4
    Well problem is you can't "feel" acceleration nor motion at least not until you bump against something which resists the same acceleration or something just crashes into you because of its different motion. You could (with high sensibility) feel a difference of acceleration in different parts of you body as its parts have a different distance to the centre of gravity they would experience a slighter different acceleration. The electromagnetic forces which bind the atoms of your body would resist the different acceleration and you would feel some stretching.

    Last edited: Apr 29, 2013
  6. Apr 29, 2013 #5


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    True for motion, not true for acceleration. Acceleration (there's a GR versus SR subtlety about gravitational acceleration, which I'll get to in a moment) can be felt and interpreted in a non-relative way.

    Consider an experimenter in a sealed, windowless room. He cannot feel or otherwise detect motion. For all he knows the entire universe is moving past him at some huge velocity while he's motionless, or maybe he's rushing through the universe instead, or maybe there's nothing moving out there at all; he can't tell.

    But acceleration he can detect. Attach a weight to each wall, floor, and ceiling with springs; and any time the springs stretch he is undergoing acceleration.

    Gravity is a bit weird because gravitational acceleration cannot be detected in this way. SR punts on this problem, GR treats movement under the influence of gravity as unaccelerated motion.
  7. Apr 29, 2013 #6
    :confused: so I'm left with contradictory responses, sorry.:frown:
  8. Apr 29, 2013 #7
    Gravity is not weirder than lets say EM acceleration. If for example a body would consist of only positive or negative charges it would have the same problems to "feel" the acceleration which source would be an electric field. The detection of the springs is a misleading thing because it is not the weights which are accelerated but the mounts on the other end of the spring. What you actually measure is a difference of acceleration.

  9. Apr 29, 2013 #8


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    No contradiction, just different mathematical treatments of the same phenomenon. HallsOf Ivy is using teh classical and special relativistic math; GR uses a different treatment. The trick in GR is that an observer in free-fall cannot detect acceleration due to gravity... so if we always do our math from that observer's point of view there won't be any acceleration due to gravity. (There's also no way for the observer to know that he's in a circular or elliptical orbit).
  10. Apr 29, 2013 #9
    So coming back to my OP example, from the point of view of the astronaut in a spacewalk next to the ISS, it's the Earth which is orbiting around him at 27,000 Km/h. Is this what you mean? Which theory would he use to explain that a hugely massive Earth is orbiting his tiny mass?
  11. Apr 29, 2013 #10


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    Since according to Newton he and the earth are orbiting their common center of mass we do not need a theory, we have one. From his observational point of view he cannot tell. Just like you cannot tell that we are orbiting the sun.
  12. Apr 29, 2013 #11


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    Not quite -more that if we blindfolded him (no way of seeing the earth going around him) he would have no way of telling whether he was in free-fall around a massive planet or in free-fall in empty space. So the same math will work for him and his immediate neighborhood whether the earth is there or not. It's only when he looks outside of his immediate neighborhood that he has to worry about....

    Using GR, he would say that both he and the earth are in free-fall through space, undergoing no acceleration; but the mass of the earth has curved spacetime so that his straight-line unaccelerated free-fall path curves around the earth's straight-line unaccelerated free-fall path. In the GR picture, neither he nor the earth is accelerating (that is, responding to a force as predicted by Newton's F=ma) but the curvature of space-time is still conspiring to change his position and velocity relative to that of the earth.

    You will sometimes see the terms "proper acceleration" and "coordinate acceleration" being used. Proper acceleration is the acceleration that can be detected by the weight-and-springs accelerometer mentioned above. Coordinate acceleration is the changes in relative velocity that we can calculate based on what we know about the curvature of spacetime and any forces that may be producing proper acceleration.
    Last edited: Apr 29, 2013
  13. Apr 29, 2013 #12


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    Let me start out by saying that usually, the "relativity of motion" is a concept used in SR, and not in GR.

    If the Earth was just a spec in a telescope, along with a bunch of other specs, I think you wouldn't have the same issues?

    So I think a lot of your issue is applying SR philosophy to GR.

    But the good news is you _can_ make it work in the way you describe, the one that seems odd to you. If you define a metric, from the observer in the ISS, you can predict the natural motion of the galaxies from just the metric, with the same basic physics.

    A few stumbling blocks:

    1) Constructing this metric isn't going to be easy. There's some technical issues out there waiting for you. There's a pretty good chance that the coordinate construction issue is a lot thornier than you think it is. People tend to underestimate the issue.

    2) You MAY not be able to rely on using Newtonian approximations if you go that way. (See below). And Newtonian approximations are MUCH easier than trying to due everythign with GR.

    You can solve both problems with one fell swoop by choosing a suitably standard coordinate system.

    If your problem is on the Earth's, you probably just want to use lattitude and longitude.

    You really don't want to (and don't have to) worry about the Earth's rotation, the Earth's motion around the sun, the sun's motion in the galaxy, if you're walking to the corner store.

    Various small effects do arise (such as Coriolis forces, and tidal forces) due to the Earth's rotation, the presence of the Sun and moon, etc, or the curvature of the Earth's surface can be noticable (maybe not going to the store, but on a longer trip).

    You can incorporate them as needed, if you need them.

    If your problem is near Earth, but not on Earth, an Earth satellite, you'd probably want to use a geocentric coordinate system. Last I recall, the IAU recommended one based on the Earth-moon barycenter. (But I didn't double check this). At this level, the behavior using coordinates fixed on the Earth's surface becomes too non-Newtonian.

    If your problem is of a bit broader scope, say interplanetary, you'd want a coordinate system based on the solar system barycenter.

    The point of this is that you choose coordinates to make your problem easier. The other point of this is that with enough effort, and knowledge, you can do physics using *any* coordinates you like. You can't do this with Newtonian mechanics, that requires inertial frames. You *CAN* do this with General relativity.

    Just because you CAN do this, doesn't mean you SHOULD. You still want to choose your coordinates wisely, to make the problem as simple as possible.

    The point is that with a sufficiently advanced approach (this rules out high-school Newtonian physics!) you don't NEED to set your coordinates up in any special way. You CAN do it any-old-way. It'll just take weeks and months of effort, you'll still wind up with the right answers :-).

    At least that's the ideal. We frequently use approximations in GR too, like test bodies that follow geodesics, rather than solving everything totally.

    Still, using coordinate independent methods, I believe you can use the same approximations we use now, it might take some thought as to how to apply them properly in the new coordinates though, especially if they're unusual ones.

    The first step towards coordinate independent physics would be Lagrangian mechanics, and its "generalized coordinates" and virtual work. This is basically a re-casting of Newtonian mechanics (you can put SR in this form as well).
    Last edited: Apr 29, 2013
  14. Apr 29, 2013 #13


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    The theory is GR. It explains all of the motion and gravitational effects regardless of whatever strange coordinates you might use. In GR both the earth and the ISS are traveling along geodesics, and the equations to determine geodesics are coordinate independent.
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