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Accelerated coordinate systems

  1. Aug 13, 2008 #1

    DrGreg

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    This is in response to the following in another thread:
    I answer it here to avoid diverting that other thread's course.

    I'm assuming that Fredrik refers to using the plane of simultaneity of the co-moving inertial observer as a means of assigning coordinates to events in an accelerating frame. This works for Rindler coordinates in the case of uniform acceleration in flat spacetime, but I see no reason why it shouldn't work for arbitrary accelerating observers.

    You will have to impose some restrictions on the smoothness of the observer's worldline. ([itex]C^\infty[/itex] would certainly be good enough, in my guess, and probably much worse would do.) And you are guaranteed to get a one-one correspondence between coordinates and events only locally near the observer. (I would guess within a distance of [itex]c^2/a[/itex], the "radius of curvature" of the worldline, where a is the instantaneous proper acceleration; this radius is pretty large for most physically realistic accelerations, e.g. about one light year for 1g. But even beyond that radius, coordinates are still defined, just no longer one-one. That being said, there may nevertheless be some events lying behind an event horizon with no coordinates.) But apart from that I don't see why it shouldn't work, and why you shouldn't regard this as "natural" -- or to be more specific, I don't see why uniform acceleration should be singled out as being different. (I'm using my geometric intuition here, I have no proof.)
     
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  3. Aug 13, 2008 #2
    I don't see why it would ever be a problem in principle, only as a practical matter it could be very complicated. As long as the acceleration is well defined, you could always divide it into arbitrarily small segments, and consider a co-moving inertial frame for each segment. And make the segments as small as necessary to get the desired accuracy, even at the limit as each period of elapsed time approaches zero. Hey that sounds suspiciously like the reason Newton invented calculus :smile:.

    Al
     
  4. Aug 13, 2008 #3

    Fredrik

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    OK, I agree that there's no essential difference between constant acceleration and arbitrary acceleration. If we define the time coordinate using the simultaneity lines of co-moving inertial observers as you suggested, we get a coordinate system that's well defined on some proper subset of spacetime that contains the world line of the rocket. This is true regardless of whether the acceleration is constant.

    Some caution is required however. Consider e.g. what would happen if we have the rocket (in the standard twin paradox scenario) change its direction three times before it finally returns. In this case, this "natural" coordinate system isn't going to extend very far from the world line of the rocket. It can't extend to a point where two of the simultaneity lines intersect, because such a point would be an event with two different sets of coordinates in the same coordinate system, and that contradicts the definition of a coordinate system.

    There are of course ways around this. We can e.g. cut the world line in three pieces so that the rocket only changes direction once in each piece. Now we have three of these "natural" coordinate systems that overlap each other a bit, and the first and third can be extended all the way to Earth. The region where the second coordinate system is well defined may or may not include Earth.

    I still feel a bit uneasy about describing these coordinate systems as "natural". I'm not sure why though. I haven't been able to identify the difference between a coordinate system I would describe as "natural" and one that I wouldn't.
     
  5. Aug 14, 2008 #4

    DrGreg

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    Of course this is why, in GR, we can't insist on there being a single coordinate chart to cover the whole of the manifold and have to make do with multiple overlapping charts.

    As I said earlier, you may still be able to assign coordinates to some events beyond the "centre of curvature", but they won't be unique and you may find proper time running backwards (or even stopped!) relative to coordinate time.

    So, for a given particle you may still be able to find a continuous function mapping the accelerating frame's coordinate time t to the particle's proper time [itex]\tau[/itex] but that function need not be injective (one-to-one), or, to put it another way, it need not be monotonically increasing. Which all sounds rather nasty, but you could use such a function to account for dilation [itex]d\tau/dt[/itex] in a twins' paradox scenario, and that function would still give the expected elapsed time at either end of the worldline when the twins meet, and it would give the "usual" dilation along any segments where the observer moves inertially.

    However there's no guarantee that such a function [itex]\tau(t)[/itex] exists for every worldline -- some worldlines may be behind an event horizon.
     
  6. Aug 15, 2008 #5
    Also, if the proper acceleration is uniform, we don't need to do much, we can just consider ourselves at rest. Hovering helicopter pilots do it all the time. We just have to use the concept of pseudo-forces to explain the movements of inertial masses, just like we consider gravity a force to explain the motion of inertial masses near earth.

    And with a twins scenario involving uniform 1G acceleration, we normally are not given enough information to rule out the possibility that there is a vast uniform 1G gravitational field present. If so, the ship would be at rest (or uniform motion) relative to it's source during the "turnaround", and the earth would turn around due to this field's "force". And as far as I can tell, not only are the results the same, there is no way to determine that this isn't exactly what happened with the info given.

    And the only difference anyway is the choice of reference frames. If we choose to consider the uniformly accelerated frame to be "at rest", does it matter whether the pseudo-force which causes the earth to turnaround is gravitational or not?

    Even if this gravity field existed, acting on earth and the ship equally, would it make any difference?

    Al
     
  7. Sep 28, 2008 #6

    Fredrik

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    I'm bumping this because of something that came up in a discussion in another thread:
    Does anyone have any comments about what I said there, especially the last part. Is it really possible to define a coordinate system this way?
     
  8. Sep 28, 2008 #7

    George Jones

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    Is the stuff below a special case of what you mean?

    The radar method gives one natural way of constructing Rindler coordinates for an observer who has constant 4-acceleration. I outlined such a construction,

    http://groups.google.ca/group/sci.physics/msg/91d8704c115d8899?dmode=source&hl=en

    Note the mistake in my construction:

    [tex]t = \frac{e^{2at'} + 1}{e^{2at'} - 1}x[/tex]

    should be

    [tex]t = \frac{e^{2at'} - 1}{e^{2at'} + 1}x.[/tex]
     
  9. Sep 28, 2008 #8

    Fredrik

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    I haven't tried to understand every step yet, but yes, it appears to be a special case of what I'm talking about.

    One thought that occurs to me when you mention Rindler coordinates is that they are well-defined no matter how high the acceleration is. (They actually describe all possible constant proper accelerations. The hyperbolas that are closer to the origin represent higher proper accelerations). So when the acceleration is constant, we never run into the problem that I was worried about when I bumped this thread. My next thought is that when we look at a short enough segment of an arbitrary* world line, it should look like a world line describing constant acceleration. This suggests that we will never run into the problem I've been worrying about. It isn't a rigorous proof, but I find it convincing enough, at least for the moment.

    *) We probably have to impose some technical requirements, e.g. that the spatial coordinates as functions of proper time are smooth, i.e. differentiable an infinite number of times.
     
  10. Sep 29, 2008 #9

    George Jones

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