# Speed of an object in a rotating frame

1. Nov 14, 2015

### JVNY

Is it meaningful to measure the speed of an object in a rotating frame, and if so how do you do it in the following case? Consider a rim of circumference 100 rotating at 0.8c, both measurements with respect to the inertial lab frame. An object is at a point on the rim with a standard clock affixed there (the origin). The object begins to move around the rim in a counter-rotating direction at 0.8c relative to the rim (as measured in the lab frame). The result is that the object is at rest in the lab frame (like a person running on a treadmill). After lab time 125 the rim will complete one rotation in the lab, and the clock and object will be together again at the origin on the rim.

Is it possible to transform this to determine the object's speed in the rim frame? In the rim frame, the rim and the clock are at rest. The object is in motion, moving around the rim. The elapsed time on the affixed clock when the object completes one circuit around the rim and returns to the origin will be 75 because of time dilation (gamma = 1.67).

I can think of at least three possibilities. Unlike a Born rigidly accelerating rod, standard clocks affixed to all points on the rim will run at the same rate, so it seems reasonable to use the clock at the origin to measure the total rim time for the object's circuit.

1. If the rim's own circumference equals its ground circumference times gamma, then the object will have traveled rim distance 167 in rim time 75 (greater than c).

2. If the rim's own circumference equals its ground circumference, the object will have traveled rim distance 100 in rim time 75 (greater than c).

3. If the rim's own circumference equals its ground circumference divided by gamma, then the object will have traveled rim distance 60 in rim time 75 (0.8c).

Thanks for any help.

2. Nov 14, 2015

### Staff: Mentor

The implicit comparison you're making here isn't really valid. In the case of an accelerating rod, the reason clocks at different points on the rod run at different rates is that the direction of their separation is the same as the direction of acceleration. But in the case of the clocks all around the rim, the direction of their separation is perpendicular to their acceleration (which is inward towards the center). That's why they all run at the same rate.

By "clock at the origin", do you mean the clock on the rim that is co-located with the counter-rotating object at the start and end of one revolution? If so, that location is not the usual origin of a rotating frame; the usual origin is at the center of rotation, which is not anywhere on the rim; it's at the center of the circle. A clock at the center of the circle does not go at the same rate as clocks on the rim.

The rest of your post opens a can of worms that I don't think you realize is there. In fact it is highly nontrivial to even define a "rotating frame" and give a physical meaning to the "space" in this frame, in which you are implicitly trying to define the circumference of the rim. See these previous threads for a start:

This is the closest you can get to a correct answer without getting into the can of worms I mentioned above. It's not fully correct (because of the issues in that can about defining the "space" in the rotating frame in which the rim's circumference is to be measured), but it does give you the right answer for the velocity of the object in the "rim frame" (0.8c).

3. Nov 15, 2015

### JVNY

The question implicates the Ehrenfest Paradox but I have in mind the Sagnac effect. Rizzi and Ruggiero imply that the effect also occurs with co- and counter-rotating subluminal objects unless they maintain the "same" velocity in opposite directions by dividing the distance traveled along the rim by elapsed time on clocks affixed to the rim that were initially synchronized in, e.g., the inertial ground frame. In this case the objects return to their starting point at the same time. See http://arxiv.org/pdf/gr-qc/0305084v4.pdf pages 7-12.

This implies that there is a physical meaning to space on the rim because each of the co- and counter-rotating objects, say two people running in opposite directions around the rim, is able to regulate his effort expended in order to maintain an agreed constant speed with respect to the rim and to travel the same distance (physically meaningful distance measured incrementally on the rim) in the same amount of elapsed time (physically meaningful incrementally elapsed time on clocks affixed at points all around the rim, initially synchronized in the inertial ground frame) and return to the starting point simultaneously with the other.

Discussions of the Ehrenfest Paradox (like the one in the second link above) get tangled up with the question of what happens to a disk as it spins up. I agree with Koks, who says that this is irrelevant and distracting -- we should simply assume that the rim is already in constant rotation and analyze it from there. As Koks writes,

The . . . discussion of how the disk rotates, how it gets accelerated and so on, are completely irrelevant to the analysis of a rotating frame within the context of special relativity . . . In the standard Lorentz transform . . . we don't concern ourselves . . . with how the "primed frame" was ever made to move. It is simply taken as having always been moving, forever . . . We are content to treat a constant-velocity primed frame as having had its state of motion forever, so we should do likewise for the rotating frame and not allow discussion of how the disk was set into motion derail the core issue, which is the analysis of events in a rotating frame.

http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

So let's assume that the rim has always been rotating, that an inertial primed object has always been moving, and that a Born rigid linearly accelerating rod has always been accelerating. We can give clearly defined physical meanings to the proper length of the inertial object and to the proper length of the Born rigid linearly accelerating rod. The ability for moving objects to eliminate the Sagnac effect by maintaining the same speed on the rim seems to require a physically meaningful measure of distance on the rim in order for the two runners to exert exactly the right effort at all times to move the same incremental rim distance in the same incremental elapsed rim time and return to the starting point at the same time.

If (as you suggest) we cannot give a clearly defined physical meaning to the length of the rim in its own frame, that would be highly interesting. It would make the rotating object qualitatively different from inertial objects and Born rigid linearly accelerating objects in SR.

It would also seem to dismiss as irrelevant all of the historical arguments over whether the length of the rim in its own frame is greater than in the ground frame (Einstein's view), the same, or lesser, because there isn't any physically meaningful length in the rim's own frame. For a contrary description, see your statement in the second linked thread referring to "the actual, physical circumference of the disk." But in a view like yours in the same thread, pervect writes that "If you can't agree on a concept of 'physical space', it becomes obvious that it's confusing to talk about it's circumference."

Last edited: Nov 15, 2015
4. Nov 15, 2015

### SlowThinker

The length of the rim measured by rods would be 167.
The radar length in the rotating direction is 375.
The radar length in the counter-rotating direction is 41.67 (unless I'm mistaken... again).

This tells you that something is very wrong with the distances on a rotating disc, and it's not only because the outward acceleration is different at different speeds. Generally, in a looped universe, strange things can happen when you take a trip around the universe, or in your case, around the disc.

These 2 documents discuss the twin paradox in a looped universe. It's not really your question, but it might give you some insights.
http://arxiv.org/abs/gr-qc/0101014
http://arxiv.org/abs/astro-ph/0606559

5. Nov 15, 2015

### Staff: Mentor

But note that this "elapsed time on clocks affixed at points all around the rim" requires imposing a clock synchronization for those clocks that is done with respect to an inertial frame--not with respect to the non-inertial frame defined by the rotating rim. (In fact it is not possible to define a clock synchronization with respect to that non-inertial frame, if by that we mean one that matches the natural notion of simultaneity for observers at rest on the rotating rim.) So the "space on the rim" here is being defined with respect to a clock synchronization in one frame (the "inertial ground frame"), yet it is supposed to represent "space" for objects in motion in that frame (the observers on the rim). At the very least, this leaves something out.

I agree that for purposes of this discussion, how the rim got to be rotating is irrelevant; we can treat it as having always been rotating at the same rate.

Agreed.

And there is one; but it has some properties that are different from the notion of distance in the other two cases (inertial object and linearly accelerating Born rigid rod).

That's not what I'm suggesting. What I'm suggesting is that this notion of distance, and in fact the notion of "the rim in its own frame" in general, has properties that are different from the other two cases mentioned, and those differences are significant. At the very least, they mean that intuitive reasoning of the sort that has been advanced in this thread cannot be used without more justification.

6. Nov 15, 2015

### Staff: Mentor

How are you deriving these results?

7. Nov 15, 2015

### SlowThinker

Nothing complicated...
a) The measuring rod is $\gamma$-times shorter, so we have 100*1.67.
b) If light signal is sent along the rim forward, I think it takes 5 revolutions to arrive back at the source. 5 revolutions take 5*75 rim-seconds.
c) If light signal is sent backwards, it takes 1/(1+0.8)=0.556 revolutions, or 75/1.8=41.67 rim-seconds.

Is this a prime example of misuse of time dilation & length contraction?

8. Nov 15, 2015

### Staff: Mentor

If the rim were moving linearly, this reasoning would be correct. But it isn't; it's moving in a circle. So each rod is moving in a different direction, which means the Lorentz boost required to determine the length contraction of each rod is in a different direction. That means you can't just add all the length contracted lengths together; they're not all length contracted lengths in the same inertial frame.

I'll defer further comment on this until we've looked at the "radar distance" below.

Not quite. I see how you arrived at this--0.8 revolutions of the rim for every 1 revolution of the light, so the "closure speed" is 0.2 revolutions per revolutions, so to speak, and 1/0.2 = 5--but it's wrong. To see why, ask: whose revolutions? The light makes 5 revolutions, yes; but the rim itself only makes 4. (Think about it--how many times will the rim--not the light--pass the clock at rest in the inertial frame, assuming that the clock and rim emitter/receiver are co-located when the light is emitted?) And it's rim-revolutions we are concerned with. So the correct answer is that it takes 4 revolutions--of the rim--for the co-rotating light signal to arrive back at the source.

The general formula is that, for a rim moving at velocity v, it takes v / (1 - v) rim revolutions for a co-rotating light signal to return. It takes 1 / (1 - v) light revolutions.

The time dilation factor of 1.67 is correct, but as above, the number of revolutions is 4, so the time is 4*75 = 300 rim-seconds.

Same problem here as above, except that now the factor in the denominator is 1 + v instead of 1 - v. The light makes 1 / (1 + v) = 5/9 = 0.545 revolutions, but the rim makes only v / (1 + v) = 4/9 = 0.455 revolutions, for a time of 75 * 4/9 = 100/3 = 33.33 rim-seconds.

However, neither of these last two calculations are of "radar distance", since that involves a light signal going out and being reflected back. So the correct "radar length" of the rim would be found by combining these last two results--in other words, we imagine a light signal going out in one direction around the rim, being reflected at the point on the rim it started from and going in the other direction back around the rim, and finally returning to its starting point. The total time required for this is 4.455 revolutions, or 333.33 rim-seconds; the "radar distance" around the rim is then half that, or 166.67 [Corrected--was 116.67].

Note that this number still presents a problem if we interpret it "naively": it tells us that the clock at rest in the inertial frame covers a distance of 116.67 around the rim in a time of 75 rim-seconds, for a speed of 166.67/75 = 2.333 c [Corrected--was116.67 / 75 = 1.455 c]. What gives?

The simple answer is that "speed" in a non-inertial frame is not limited to c; it's only limited to c in inertial frames. However, there's more to it than that. Consider this: suppose we want to run the radar distance measurement above, and we first emit the light signal in the counter-rotating direction. We time it so that the signal is emitted at the exact instant that the clock at rest in the inertial frame passes our light emitter (which will also be the reflector and the receiver). We will at once realize that the clock can't be moving faster than light, because the light signal we emitted, which is moving in the same direction the clock is moving, is moving faster than the clock. It must be, because we saw above that it only takes 0.455 revolutions for the counter-rotating light signal to return, whereas it takes exactly one revolution for the clock to return. So if the clock is moving "faster than c" in the "rim frame", then light itself must also be moving "faster than c" by even more in that same frame!

Things get even weirder when we consider the co-rotating direction. Suppose we emit a co-rotating light signal at the same time as the clock is co-located with the emitter/receiver. The signal will take 4 revolutions to return, so the clock, by definition, makes 4 revolutions relative to the emitter/receiver--but the light only makes one. So here it seems like the clock is moving 4 times faster than light!

Obviously what all this is telling us is that our intuitions are not adequate to this case. Not only is "speed" apparently not limited to c, it's not even isotropic. So whatever this "speed" is, it clearly isn't following the same rules as speed in an inertial frame, not just with respect to the "speed limit" of c, but with respect to the whole structure of how it works. The same obviously goes for "distance".

Last edited: Nov 15, 2015
9. Nov 15, 2015

### SlowThinker

In fact, I used radar "length" on purpose. It's something you can't even measure in normal space.
The OP is asking about such a "length" or "half-distance", so I wanted to point out that the main problem here is that the rim is looped.
There are other effects in play, but this one seems to cause the most trouble.

Obviously I made the mistake of using the wrong counter, sorry about that

10. Nov 15, 2015

### JVNY

I think that the radar distance is 166.67, as follows.

A co-rotating circuit for a light flash takes ground time 100 / (1 - 0.8) = 500, for rim time of 300 (ground time dilated by gamma 1.67).

The counter-rotating circuit takes ground time 100 / (1 + 0.8) = 55.56, or rim time of 33.33.

The radar distance on the rim is the sum of the out and back trip times divided by 2, so the radar measured circumference of the rim in its own frame is 333.33 / 2 = 166.67.

This is the same as Einstein's conclusion based on the additional number of length-contracted rods that fit around the circumference in the inertial frame.

This suggests that there is a physically meaningful own length of the rim's circumference, and that it is 166.67.

It is important to note, as in the OP, that clocks affixed anywhere on the rim run at the same rate. The radar method does not work on a Born rigidly accelerating rod, and Misner Thorne Wheeler Gravitation points out in exercise 6.5. This is presumably because clocks run at different rates along the rod, so the elapsed time on a single clock does not reflect the sum of the local times of the light flash as it moves across infinitely small segments along the rod.

In the case of clocks affixed to the rotating rim, all of the clocks run at the same rate, so the radar distance measured along the whole circumference using one clock will be the same as the sum of the radar distances of many small segments making up the entire rim. It does not matter whether the clocks at each of those points are synchronized with each other. It is only necessary that each one runs at the same rate and that for each segment you use a single clock to send a signal to the end of the segment and to measure the elapsed time upon receiving the reflected signal back again.

So it seems that the proper physical length of the circumference in the OP is 166.67, but that then makes it hard to conclude that an observer on the rim measures the counter-rotating object to be traveling at 0.8c.

Is this consistent with your initial response that the object's speed is 0.8c? That response seems to follow the same rules as the inertial frame. This quotation seems instead to be consistent with concluding in the OP that the object travels at greater than c in the rim frame because it travels rim distance 166.67 in rim time 75 (consistent with your statement just quoted that speed is apparently not limited to c).

But the radar method shows that the proper length equals the ground length times gamma, which is exactly the same rule that applies to inertial frames. So distance seems to follow the same rules as in an inertial frame, although speed measured by proper distance divided by elapsed time on the clock at the starting and ending point of a circuit can exceed c.

Is there a good prior thread (albeit involving the full can of worms) that explains those differences?

11. Nov 15, 2015

### Staff: Mentor

I think you miss the point. It isn't that you cannot give a clearly defined meaning to the length of the rim in its own frame. It is that you first have to give a clearly defined meaning for the rim's frame.

If you don't define what you mean by the rim's frame then obviously you cannot even ask what its length is in that frame.

12. Nov 15, 2015

### JVNY

Fair enough. Is there a good prior thread that clearly defines the rim's frame?

13. Nov 15, 2015

### Staff: Mentor

I don't know. I would just define it as:
$t'=t$
$r'=r$
$\theta'=\theta+\omega t$
$z'=z$
With $(t,r,\theta,z)$ being the usual cylindrical coordinates.

That certainly is not the only possible choice, and it has disadvantages, but that is what I would do. You just have to be explicit.

14. Nov 15, 2015

### Staff: Mentor

Oops, you're right, I divided 333.33 in half incorrectly. I've corrected my previous post.

But they are not naturally synchronized--that is, if we use the momentarily comoving inertial frame of a particular clock on the rim to define a synchronization, all the other clocks on the rim will be out of sync with the chosen clock.

Not for that particular measurement, no. But as soon as you try to extend things beyond the rim, i.e., to any radius besides the radius of the rim, it will matter.

It's "consistent" in the sense that it shows that there is not a unique notion of "speed" that can be applied in the non-inertial frame. If any observer on the rim measures the speed of the clock relative to him as it goes by, he will measure it to be 0.8c. The different answers for "speed" come from looking at other, global observations, not from any local measurement.

Which means that distance does not follow all of the same rules as in an inertial frame, since in an inertial frame speed measured by proper distance divided by elapsed time cannot exceed c.

15. Nov 15, 2015

### Staff: Mentor

This is called "Born coordinates" (with one difference, the definition of $\theta'$ should have a minus sign in it), and is discussed briefly here, along with some of the "can of worms" issues:

https://en.wikipedia.org/wiki/Born_coordinates

16. Nov 15, 2015

### Staff: Mentor

Excellent, it even has the metric calculated. Not having to re-do that is worth switching the sign in my definition ☺

17. Nov 15, 2015

### Staff: Mentor

Yes, but don't expect it to resolve all the questions. Spacelike surfaces of constant time in this chart are Euclidean (that's obvious from the line element), which means they do not represent the "space" seen by a rim observer--more precisely, they do not represent the non-Euclidean quotient space which is the best we can do at describing "the space seen by a rim observer". (That quotient space does not correspond to any spacelike slice cut out of Minkowski spacetime.) The surfaces of constant time are determined according to the clock synchronization of an inertial observer at rest at the center of rotation.

18. Nov 15, 2015

### JVNY

This is clear in the case of a Born rigidly accelerating rod. Say an observer is at the rear of the rod, which has proper length 100. If the observer makes a radar measurement of the distance to the front, half the elapsed time does not equal 100. But we know that the proper length is 100 (we specified that). So the observer is using a global observation. If instead he makes a radar measurement of the length from the rear to the center, and another observer makes a radar measurement of the length from the center to the front, the sum gets closer to 100. If you use an infinite series of infinitely small segments, the sum should be exactly 100. This essentially sums the local times that light takes to travel along the rod. So light takes 100 of local time to travel proper length 100.

But the rotating rim is different. The radar measurement of the entire circumference using light sent from a clock at a single point around to reflect off of itself and return in the opposite direction is exactly the same as the sum of the radar measurements of an infinite set of infinitely small segments making up the circumference. That is the same as in an inertial frame -- the time that a single light flash takes from the rear of an inertial object to the front and back again as measured on a clock at the rear is exactly the same as the sum of the times of a series of radar measurements along the entire length of the object. So there should be no difference between local and global measures of speed in the rotating frame any more than there is in an inertial frame.

19. Nov 15, 2015

### Staff: Mentor

How are you defining "proper length"? Obviously it isn't "radar distance", so what is it?

Part of the problem with all scenarios of this sort is that people assume that there is some well-defined notion of "proper length" independent of any measurements. There isn't. In order to define "length" physically, you have to define how it is measured. If there are different measurements that yield different results, then it's doubly important to specify which one you mean.

20. Nov 15, 2015

### Staff: Mentor

This is true, but this...

...does not follow from it. There are ways of measuring speed other than distance/time. Suppose, for example, that an observer on the rim observes light emitted by the clock in the inertial frame; he measures the Doppler shift of the light and determines what relative velocity would produce that Doppler shift. He will observe a shift consistent with a relative velocity of 0.8c. That is a local measurement of speed, and it gives different results from the global measurements we discussed.

One key difference between the rim and an inertial frame is that the global measurements involve light traveling around closed paths. In all of the length measurements in inertial frames that you describe, that is not the case: there is no way to send out light in an inertial frame in one direction and have the same light signal come back from the opposite direction.