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Counter rotating frames

  1. Jun 15, 2004 #1


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    A question that led to an interesting answer was posed in a newsgroup recently that I thought I'd share here. Someone proposed a scenario where there were two rocket ships that would travel at constant speed along a circle in opposite directions. They would pass each other twice per revolution. The times of the onboard crew's watches are easy to express as reckoned from an inertial frame still with respect to the circle. According to the inertial frame observer both accelerated frame watch times are simply given by special relativistic time dilation
    [tex]t = \frac{t'}{\sqrt{1 - \frac{v^2}{c^2}}} [/tex]
    [tex]t = \frac{t''}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex]
    This means that the accelerated frame observers must agree that their watches should read the same time each time they pass nearby. However, while they are nearby eachother each will observe the others local motion yielding special relativistic time dilation as they pass. This means that while they are far apart each must find that the other is "time contracted". This has been understood qualitatively in terms of the curvalinear nature of accelerated frame coordinates which are composed of a continuum of inertial frames with different standards of simultaneity. In other words at each point in the trip the rocket is instantaneously at rest with respect to some inertial frame, but in the next moment it is at rest with respect to another inertial frame and so the crews standards of simultaneity are continuously adjusted. The question that was posed was at what time does one crew reckon that the other transitions from a state of time dilation to time contraction with respect to their own accelerated frame. For the simplest choice of accelerated frame coordinates the A' frame observers find that the A" frame observers watch runs according to
    [tex]\gamma t' - r_{0}\frac{v}{c^2}sin[\frac{v}{r_{0}}\gamma (t' + t'')] - \gamma t'' = 0[/tex]
    [tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex]
    and v is the speed of one with respect to the inertial frame and [tex]r_0[/tex] is the radius of the circle according to the inertial frame.
    This equation implies that the watches do synch up according to the accelerated frames not only upon passing, but also at opposite ends of the circle. Differentiation of this does also yield that "upon passing" eachother the A' frame observers do observe the A" frame observer's watch as time dilated according to
    [tex]dt' = \frac{dt''}{\sqrt{1 - \frac{u^2}{c^2}}}[/tex]
    [tex]u = \frac{2v}{1 + \frac{v^{2}}{c^{2}}}[/tex]
    and at opposite ends where the clocks resynch as well to be time contracted maximally by
    [tex]dt'' = \frac{dt'}{\sqrt{1 - \frac{u^2}{c^2}}}[/tex]
    These were derived as the answer to problem 5.4.5 from equation 5.4.4 at
    Last edited: Jun 15, 2004
  2. jcsd
  3. Jun 15, 2004 #2


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    I had some trouble getting equations to display. The Latex generator didn't like the double primes. Anyone not seeing them still should be able to by refreshing their browsers.
  4. Jun 18, 2004 #3
    [tex]\gamma dt \prime=dt\prime\prime[/tex]

    can I write

    [tex]\gamma dt \prime=dt\prime\prime[/tex]

    Last edited: Jun 18, 2004
  5. Jun 18, 2004 #4


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    I used [tex]\gamma[/tex] to meen [tex]\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/tex] where v is the speed of each with respect to the inertial frame according to which the circle does not translate. But, if you want to use it to instead represent [tex]\frac{1}{\sqrt{1 - \frac{u^2}{c^2}}}[/tex] where u is the speed one finds the other to pass at then that equation would be the correct time dilation of A' as observed by A" as they pass, or the correct time contraction of A" as observed by A' as they are on opposite sides of the circle.
  6. Jun 21, 2004 #5
    I am curious about a couple of things, as I try to read these equations.

    Perhaps you would explain a little more about the use of "u"? I think I see that from the frame of reference of the circle, the ships each travel at v, so they pass each other at 2v. But from the rotating frame onboard the ship, the other ship's instantaneous velocity as they pass must be calculated using two gammas, one for the effect of the observer's velocity and one for the effect of the observed's velocity. Is this a correct interpretation?

    Also, I see that you use r_0 for the radius of the orbits. Does the value of r vary so that you must indicate a rest frame for r? Is r affected by the contraction? I was thinking it would not be affected because it is orthagonal to the direction of the velocity vectors. This led me onto the thought that the frame of reference of the circle could also be moving, so that r would vary according to the direction of movement of the circle vector. Is this the reason for r_0?

    It has also occurred to me while thinking on this problem that the ships are actually moving instantaneously in the same direction when in opposition on the circle, so their relitive velocity at that point is actually zero, and for that instant their frames of reference converge. Their common frame of reference at that point is still moving at v with respect to the frame of the circle. Any comments on this approach?


  7. Jun 22, 2004 #6


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    Actually they pass eachother at [tex]\frac{2v}{1 + \frac{v^2}{c^2}}[/tex] which is what I called u.

    You do have to use relativistic velocity addition which is in the expression above. It isn't 2v multiplied by to factors of gamma though if that is what you mean.

    I specifically said that this was the radius of the circle according to a frame for which the circle was at rest, yes.

    According to a ship frame that ship is not moving in a circle at all. Everyone is always at rest according to their own coordinate frame. It is the other ship that would be in motion according to the one ship's coordinate frame.

    Yes. According to another inertial frame the circle would become flattened in the direction of motion.

    Actually even though they are instantaneously at rest then, they do not share a common frame then. This is due in part to their acceleration vectors pointing in opposite directions and in part to being some distance apart along the acceleration. At that instance each observes the others watch to instantaneously be synched up to their own, but to be running faster than his own. This is like when clocks at a higher location in a gravity field run fast as compared to your own even though they are at rest with respect to each other.
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