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Determining location along a rotating rim

  1. Feb 21, 2014 #1
    1. The problem statement, all variables and given/known data

    Is it possible to determine the relative location of two points on a rotating rim by sending a signal that goes from one to the other, reflects, and then returns to the sender? Assume a rim, like a bicycle wheel rim without the tire or spokes; the interior surface of the rim is mirrored, so that light signals can travel around the circumference (the interior surface) without the need for fiber optics or other means.

    2. Relevant equations

    The typical way to analyze light signals moving around a rotating rim is that the rim is at rest, a counter-rotating signal travels at c + v, and a co-rotating signal travels at c - v.

    3. The attempt at a solution

    Start with a single case of determining two points that are at opposite ends of a diameter (one is half way around the rim from the other). Set a recording clock and signaling device (the Device) at an arbitrary point marked 0 degrees on the rim. Send a signal around the rim in the co-rotating direction and record the time that it takes for the signal to go around and return to 0 degrees. Next, do the same with a counter-rotating signal. Average the two times.

    As a check, send a single signal around the rim in one direction until it makes a complete circuit, reflects off a mirror placed at 0 degrees, and then returns around the rim in the opposite direction back to the Device. The total time for this out and back trip should be the same as the total time for the two separate signal trips described above. In both cases, you have signals traveling the same distance in opposite directions.

    Next, set a mirror at an arbitrary other point on the rim perpendicular to its surface. Send a signal from the device in either direction (it should not matter); assume the counter-rotating direction. The signal travels around the rim in the counter-rotating direction until it strikes the mirror; then it reflects; then it travels around the rim in the co-rotating direction until it returns to the device. Note the total time for this out and back trip. Move the mirror until the total time for the out and back trip equals half of the two complete circular trips determined above (that is, the average of the co-rotating trip time and the counter-rotating trip time, as described above). The mirror should be at 180 degrees on the rim, half way around the rim from the Device.

    Why should this work? In inertial frames one can determine distance as half the time that a signal takes to get from one point to another, reflect, and return. But that is because light travels at c in both directions (or at least it is assumed to travel isotropically at c). One cannot use that method on a rotating rim because light travels at c + v or c - v depending upon the direction of the rotation. However, if you only want to determine the proportionate distance around the rim then you should be able to use the method outlined above. It should work whether you send the initial signal in the co-rotating or counter-rotating direction, again because the trip will be out and back in both cases, with the signal traveling the same distance to the mirror as back from it.

    This seems logical at least at nonrelativistic speed of rotation. I do not know what effect would occur if the rim were spinning at relativistic speeds. However, I read something about the Sagnac Effect that suggests that higher speeds would not change the result. I really do not understand it, but I will quote it here with some relevant parts in bold. It seems to say that there are two offsetting effects, at least for a signal traveling a complete circuit.

    Clearly the pulse traveling in the same direction as the rotation of the loop must travel a slightly greater distance than the pulse traveling in the opposite direction, due to the angular displacement of the loop during the transit. As a result, if the pulses are emitted simultaneously from the “start” position, the counter-rotating pulse will arrive at the "end" point slightly earlier than the co-rotating pulse. Conversely, if two pulses arrive simultaneously at the end point, the co-rotating pulse must have been emitted from the starting point earlier than the counter-rotating pulse. . .

    This analysis is perfectly valid in both the classical and the relativistic contexts. Of course, the result represents the time difference with respect to the axis-centered inertial frame. A clock attached to the perimeter of the ring would, according to special relativity, record a lesser time, . . . [h]owever, the characteristic frequency of a given light source co-moving with this clock would be greater, compared to its reduced value in terms of the axis-centered frame, by precisely the same factor, so the actual phase difference of the beams arriving at the receiver is invariant. (It's also worth noting that there is no Doppler shift involved in a Sagnac device, because each successive wave crest in a given direction travels the same distance from transmitter to receiver, and clocks at those points show the same lapse of proper time, both classically and in the context of special relativity.) http://mathpages.com/rr/s2-07/2-07.htm

    Although my proposed answer seems logical, I am somewhat skeptical, because it seems a bit too simple and lacks the complexity that always seems to accompany rotation in relativity. In fact I started to use this analysis in a thread on the regular forum for a broader purpose but had to drop it because of the issues involved in that purpose (synchronizing clocks).

    Thanks in advance.
     
  2. jcsd
  3. Mar 10, 2014 #2
    Please ignore this; the prior thread answered the question that yes, the method works.
     
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