Applying Velocity Addition in Rotating Frame: Is It Correct?

[lindberg]

TL;DR Summary

How can we apply relativistic velocity addition to a rotating disk? Imagine we launch two objects in opposite direction from point P on a disk set in rotation. The launching point applies the same force to launch two objects with velocity v in opposite directions. If I am an observer in a non-rotating laboratory frame, can I apply the relativistic velocity addition formula to the bullets adding and subtracting (respectively) the angular velocity of the disk to each bullet's velocity?

From the top of my head, I would say that yes, the very moment our clocks are aligned, and the two bullets are launched it is perfectly ok to use the relativistic velocity addition formula to determine the speed of the bullets from my reference frame. But the more the disk keeps rotating, the more our clocks get out of sync. So then no, I cannot apply this operation anymore, except for one single moment to calculate the initial velocity in very special conditions (clocks perfectly aligned and synchronized and bullets launched with identical velocities in opposite directions).

That brings me to the conclusion that it is not correct to apply relativistic velocity addition to objects moving on the rim of a rotating disk.
Am I getting something wrong here? If yes, could someone disentangle my thinking?
Thank you.

 

[jbriggs444]

Am I getting something wrong here? If yes, could someone disentangle my thinking?

You have a possible difficulty with the notion that there is such a thing as "the" reference frame associated with a rotating disk and, in particular, that a rotating coordinate system tied to a rotating disk may not match the tangent inertial frame for a particular point on the rim.

In order to make the rotating frame work at all you will have to play some games with the time coordinate. You need to specify those games before trying to reason further.

 

[Sagittarius A-Star]

That brings me to the conclusion that it is not correct to apply relativistic velocity addition to objects moving on the rim of a rotating disk.
Am I getting something wrong here? If yes, could someone disentangle my thinking?

Maybe, the following link helps:
http://www.physicsinsights.org/sagnac_1.html

 

[Vanadium 50]

Let's simplify things. Ignore relativity for now, and ask your question classically. I suspect you will have difficulty formulating it, because of what you are measuring when. If I measure v1 and v2 at two different times and/or places, do I even expect the classical velocity addition rules to hold?

 

[PeterDonis]

Imagine we launch two objects in opposite direction from point P on a disk set in rotation. The launching point applies the same force to launch two objects with velocity v in opposite directions.

The first question you need to answer carefully is, what is this velocity $v$ relative to? Obviously it's not the lab frame. But if you are going to use the relativistic velocity addition formula, it must be some inertial frame (although it could be two different inertial frames for the two objects). So which inertial frame (or frames) is it, that the velocity $v$ is relative to?

If I am an observer in a non-rotating laboratory frame, can I apply the relativistic velocity addition formula to the bullets adding and subtracting (respectively) the angular velocity of the disk to each bullet's velocity?

Not the angular velocity of the disk, but the instantaneous velocity of each of the launch points on the disk, relative to the lab frame (which is assumed to be inertial). If you do that, and if you correctly answer my question above regarding what inertial frame or frames the velocity $v$ is relative to, then yes, you can use the relativistic velocity addition formula to get the velocity of each object relative to the lab frame.

 

[lindberg]

The first question you need to answer carefully is, what is this velocity v relative to? Obviously it's not the lab frame. But if you are going to use the relativistic velocity addition formula, it must be some inertial frame (although it could be two different inertial frames for the two objects). So which inertial frame (or frames) is it, that the velocity v is relative to?

Well, I will tentatively reply that v is the velocity at the moment of launching the two bullets (or whatever objects you wish) in the moment of its launching at the launching point k0, given the force applied on the bullets is identical. It is in the infinitesimal inertial (and tangential?) frame of the launching device.
What I wonder is whether I could use the clock k1, which is just one centimeter away on the disk rim from the launching device, co-rotating with k0, to check this velocity several instants later and say "yeah, the two objects are still keeping the same speed". Correct me if whatever I have stated makes no sense for some reason or is just outright wrong/confused.

Please note: I am only trying to determine the velocity of the two objects for now and I explicitly stay (for now) on the disk of the rim, not considering yet the lab frame. So velocity addition in the non-rotating lab frame comes later for me, once I am cleared on this one.
Thanks @PeterDonis

 

[lindberg]

Let's simplify things. Ignore relativity for now, and ask your question classically. I suspect you will have difficulty formulating it, because of what you are measuring when. If I measure v1 and v2 at two different times and/or places, do I even expect the classical velocity addition rules to hold?

Good point. Should I? Should the classical velocity addition hold? Not sure about it...

 

[PeterDonis]

It is in the infinitesimal inertial (and tangential?) frame of the launching device.

The standard term for this is the "momentarily comoving inertial frame": the inertial frame in which the launching device is momentarily at rest at the instant it launches the object. Yes, that's the correct answer.

What I wonder is whether I could use the clock k1, which is just one centimeter away on the disk rim from the launching device, co-rotating with k0, to check this velocity several instants later and say "yeah, the two objects are still keeping the same speed".

If you wait a few instants and then pick a clock on the disk rim which is momentarily at rest in the same inertial frame that the launch device was a few instants before, then that new clock will momentarily see both objects moving at speed $v$ relative to it (in opposite directions). However, you would have to pick a method of measuring speed that is instantaneous (such as looking at the Doppler shift of light emitted by the objects and received at the appropriate point on the disk rim).

 

[lindberg]

Maybe, the following link helps:
http://www.physicsinsights.org/sagnac_1.html

Ah thanks. Yes, I see they use the relativistic velocity addition to deduce the Sagnac effect from the point of view of the observer in the lab, a non-rotating guy.
Although, instead of the author of the blog, I guess I would use [2πr +/- (angular velocity x time that has passed since the observer got hit)] to calculate the time needed for each beam...
I guess they give the same results, though.
He does apply relativistic velocity addition, looking from the lab. Again, I am not quite sure whether that can be done correctly, but hey, maybe he's right, and I am missing something.

 

[Ibix]

Good point. Should I? Should the classical velocity addition hold? Not sure about it...

I think you need to specify your question carefully before you can answer it. What exactly are you trying to calculate? The velocities in the lab frame at launch? Or at some general time $t$? What measurements will you have available in either case?

 

[lindberg]

I think you need to specify your question carefully before you can answer it. What exactly are you trying to calculate? The velocities in the lab frame at launch? Or at some general time $t$? What measurements will you have available in either case?

At some point $t$ after the launch, ONLY having available as data the initial velocity $v$ of the launched bullets (identical in opposite direction) at the moment of launching, which is when my clock (non-rotating) and the clock of the launcher are aligned and synced at the very moment of alignment.
I also have $w$, which is the angular velocity of the rotating disk as seen from the lab.
My question is whether whatever total velocity I have deduced at that very moment still holds as the disk keeps rotating. In other words, is my relativistic addition formula naive, given I expect it to hold for the entire duration of rotation of the disk.

 

[Ibix]

So you are launching the bullets and then they fly inertially? In that case I'm not clear why do you expect the lab frame velocities to be different at time $t$ from when they were launched.

 

[lindberg]

The standard term for this is the "momentarily comoving inertial frame": the inertial frame in which the launching device is momentarily at rest at the instant it launches the object. Yes, that's the correct answer.

Thanks, I did not know the exact term.

If you wait a few instants and then pick a clock on the disk rim which is momentarily at rest in the same inertial frame that the launch device was a few instants before, then that new clock will momentarily see both objects moving at speed v relative to it (in opposite directions). However, you would have to pick a method of measuring speed that is instantaneous (such as looking at the Doppler shift of light emitted by the objects and received at the appropriate point on the disk rim).

But is there such a clock on a rotating disk? Or is each clock in its own inertial frame?
What I mean by that is... the clock k1 which is several centimeters away from my first clock k0 is on a slightly curved surface compared to it, right? So unless we sync them by some method... hm...
Is it theoretically possible to force my second clock k1 to be in the same inertial frame?
When you say momentarily, I suspect you say this because these clocks, left rotating on the rim, will desync quite fast.
What I was thinking is a constant GPS connection of each clock so that they all stay synced. But that won't work, because I would sync them with a central GPS controller, which is non-rotating, so the whole things will fall apart logically.

 

[lindberg]

So you are launching the bullets and then they fly inertially? In that case I'm not clear why do you expect the lab frame velocities to be different at time $t$ from when they were launched.

They fly around the rim. I do realize they cannot fly on a round path just by themselves. Obviously, if they fly away from the disk, the situation is different altogether, but I am not implying that.
So I am thinking more of a way to make them follow the rim of the disk and stay in the disk's frame. I ignore the technical details of that, sorry (I suppose for some small particles they use mirrors or other guiding devices). Sorry, maybe my wording was wrong and lead to confusion.
I am not interested in situations in which there is no meaning in talking about velocity of the bullets compared to the disk past the moment of launching. What I am trying to figure out is a way to compose velocities for bullets (but hey, you can take guys running along a disk or some other objects that stay conceptually attached to disk, still moving in opposite directions).

 

[Sagittarius A-Star]

He does apply relativistic velocity addition, looking from the lab. Again, I am not quite sure whether that can be done correctly, but hey, maybe he's right, and I am missing something.

The Lorentz transformation can only be used, if in both frames the time-axis is defined based on Einstein-synchronization. You can use a 2D (ct,x)-LT under this condition.

Imagine, you would curl the x' axis around the rotating disk and have placed many clock along this curled x'-axis. In the rotating frame, you can do locally Einstein-synchronizations between neighboring clocks. But this works only without contradictions, if the used range on the x'-axis is limited to for example $-\pi R \gamma \leq x' < +\pi R\gamma$. You must avoid, that one end of the used range overlaps with the other end. At a "normal" straight x'-axis, clocks located at x' locations $-U/2$ and $+U/2$ cannot be at the same location in the primed frame.

 

[lindberg]

The Lorentz transformation can only be used, if in both frames the time-axis is defined based on Einstein-synchronization. You can use a 2D (ct,x)-LT under this condition.

Thanks @Sagittarius A-Star if I got it right from my readings, the time-axis is not defined in the same way in a rotating frame, since it is non-Euclidean. Time is not orthogonal on space in such a framework. Hence, the headache...

 

[Sagittarius A-Star]

Thanks @Sagittarius A-Star if I got it right from my readings, the time-axis is not defined in the same way in a rotating frame, since it is non-Euclidean. Time is not orthogonal on space in such a framework. Hence, the headache...

If you limit yourself to the range on the x'-axis, that I described in posting #15, then you can get a 2D-coordinate chart, were the time axis ct' is orthogonal to the x' axis in the rotating 2D-frame. The Einstein synchronizations should be really locally, or, with a finite small distance between the clocks, you should guide the light via an optical fiber cable exactly along the rim of the disk to get the same path as your moving objects.

The formula for the Sagnac $\Delta t$ is nothing else than the term for "relativity of simultaneity" in the inverse Lorentz transformation, multiplied with $2$ (and with $\require {color} \color{orange} \gamma \color{black}$ for the stationary frame).

$\require {color} \Delta t = \color{orange} \gamma \color{black} (\Delta t' + \color{red} {\frac {v} {c^2} \Delta x'} \color{black})$​

You need only to plug-in:
$\Delta t' := 0$ (measured with 2 clocks at both end of the cable, that were synchronized on the rotating disk with light in both directions via the cable)
$\require {color} \color{red} \Delta x' \color{black} := \ 2 \pi r \gamma$
Multiply $\Delta t$ of the LT with $2$ because there are 2 signals in opposite direction involved.

 

[Nugatory]

the time-axis is not defined in the same way in a rotating frame, since it is non-Euclidean.

Spacetime is non-Euclidean, and this has nothing to do with the frame we use to describe it.

In this problem we are working with a flat non-Euclidean spacetime, and we can choose to use coordinates in which the time axis is perpendicular to the three spatial axes (the x,y,z,t coordinates of an inertial frame) or more complex coordinates in which something not moving inertially (in this case, a point on the rim of the disk) is at rest.

 

[PeterDonis]

is there such a clock on a rotating disk?

Only momentarily. Any clock momentarily reads the same time as a clock in its momentarily comoving inertial frame. But only momentarily. Each clock on the rim of the disk is momentarily at rest in the inertial frame in which the two objects have speed $v$ in opposite directions--but each such clock is at rest in that frame at a different instant.

If what you want is to find the speeds of the two objects in the non-rotating lab frame, all you need is their speeds in the inertial frame described above. Then it's just a matter of using the relativistic velocity addition formula to find their speeds in the lab frame. Everything else you have been talking about--whether you can synchronize different clocks on the rotating disk, etc., etc.--is completely irrelevant to that.