Forces in relativistic rolling motion

  • Thread starter jartsa
  • Start date
  • #51
3,962
20
Analysing the relativistic angular velocity and momentum of a individual particle at any given instant on its cycle can be very complex. The gory details involving a second order antisymmetric tensor and a time dependent quantity called the dynamic mass moment are given in this Wikipedia article. However, if we look at the bigger picture in terms of complete revolutions, then considerable simplifications can be made.

For a flywheel with angular velocity ##\omega_0## as measured in a reference frame in which the rotating flywheel is linearly at rest, then the angular velocity as measured in any other frame with relative motion v is ##\omega = \omega_0/\gamma(v)##. This is true no matter what the orientation of the flywheel is, because ##\omega## is proportional to 1/dt and dt is a scalar. Put another way angular velocity can be expressed in terms of rotations per unit time, so that ##\omega =2\pi r/dt##. Under a Lorentz transformation ##\omega' = 2\pi r/dt'##. More generally, if the flywheel as a whole has linear motion with a velocity component u parallel to the x axis, then after a Lorentz boost of v in the x direction, the transformed angular velocity is given by:

##\omega ' = \omega \frac{\sqrt{1-v^2}}{(1-vu)} = \frac{2 \pi r}{dt}\frac{\sqrt{1-v^2}}{(1-vu)}##

Any component of the linear motion of the flywheel that is not parallel to the direction of the boost has no effect on how the angular velocity transforms. For completeness we can represent the above in the form of a four-position vector as ##(dt, dx, dy, dz) = (2\pi r/\omega, dx, dy, dz)## where the linear 3 velocity of the flywheel is ##(u_x,u_y,u_z) = (dx/dt, dy/dt, dz/dt)##.

Now for the angular momentum. The linear version of momentum is given by ##p = m_0 v \gamma(v)##. The angular momentum analogue is ##L = I\omega_0\gamma(v)## where I is the moment of inertia and ##\omega_0## is the angular velocity in the irf where the flywheel has no linear motion. For a simple flywheel with the bulk of its rest mass ##m_0## at the rim, ##I=m_0 r^2##. When the flywheel is rotating, ##I = m_0 \gamma(\omega_0) r^2 = m_{\omega}r^2## where ##m_\omega## represents the total effective mass of the rotating flywheel in an inertial reference frame where it is linearly at rest. This means that ##m_{\omega}## is the new effective rest mass of the rotating flywheel and so ##m_{\omega}## is invariant under a Lorentz transformation. Putting this together we have ##L= m_{\omega}r^2\omega \gamma(v)##. As mentioned above ##\omega =\omega_0/\gamma(v)## so we end up with ##L = m_{\omega}r^2\omega_0## and since all the components are Lorentz invariant, angular momentum must be Lorentz invariant and L=L' for any orientation of the flywheel rotation axis.

I am not totally sure of the above analysis so any criticisms and corrections are welcome.
 
  • #52
1,447
113
I think this is the crux of where you are going wrong. In relativity there is no such thing a rigid material that transmits signals or impulses instantly. See this FAQ. https://www.physicsforums.com/showthread.php?t=536289 [Broken]
No I do not have that misconception.



Center of mass can react to distant events instantaneously because center of mass is imaginary. If you imagine a delay, then you imagine the wrong way.
 
Last edited by a moderator:
  • #53
Nugatory
Mentor
12,988
5,699
No I do not have that misconception [about rigid objects]....
Center of mass can react to distant events instantaneously because center of mass is imaginary. If you imagine a delay, then you imagine the wrong way.
The calculated center of mass of an object does indeed move around as distant parts of the object change shape and/or density. However, there is a simultaneity convention hidden in that calculation - you have to evaluate the position and density of all parts of the object at the same time - so the result of that calculation is frame-dependent. Thus, the answer to your question in #49:
What is the momentum of the COM of the rod? Same as the momentum of the rod. I don't see how it could be different.
is that the momentum of the COM of the rod isn't even rigorously defined.
 
  • #54
3,962
20
Center of mass can react to distant events instantaneously because center of mass is imaginary. If you imagine a delay, then you imagine the wrong way.
I regret bringing the COM into this conversation. As a mathematical object it is not really the quantity we are interested in, as the real point is about whether the physical centre of the object accelerates or not. The COM has the property that for a closed system (eg lab,rods,guns,hammers) the COM cannot accelerate in any reference frame no matter what happens inside the lab. This was intended to be used as a reference to determine if the physical centre of a rod accelerated at any point, but the intention got lost along the way. A better reference is the third rod C that receives no impulses and initially co-moves with the other rods. A mark at its physical centre can be easily compared to the location of similar marks on the other rods, but you chosen to ignore that option.

Let's have a quick recap and see what we agree on.

...
A still standing observer sees two bullets fired from the longitudinal arm at different times. So therefore the observer sees the lever receiving an impulse to left, then later an opposite impulse to the right.

Why does the observer not see the lever to change its position to the left?

(We know the lever does not change its position to the left, because we know that in the lever frame there is no time between the impulses)

(And we know the lever does not start to rotate, so there is no rolling motion, which possibly may sometimes cause a shift of the center of mass of the rolling thing)
Do you still maintain that the lever should always rotate in any reference frame where the impulses are not seen to occur simultaneously, even if the impulses occur simultaneously in the rest fame of the lever?

You guys are not getting my point.

(1) An object is given an impulse to the left, later an opposite impulse to the right -> object is moved to the left.
(2) An object is given an impulse to the right, later an opposite impulse to the left -> object is moved to the right.
(3) An object is given an impulse to the left, and simultaneously an opposite impulse to the right -> object is not moved to the left and object is not moved to the right.
We had not introduced the COM at this point. Here you are presumably talking about the physical acceleration of the object as a whole that displaces it relative to an initially comovng inertial object.

Do you agree that statement:
(1) is not necessarily true in an inertial ref frame where the object is moving to the left.
(2) is not necessarily true in an inertial ref frame where the object is moving to the right.
(3) is not necessarily true in any inertial ref frame where the object is moving and the simultaneous impulses are spatially separated.
 
  • #55
1,447
113
Let's have a quick recap and see what we agree on.


Do you still maintain that the lever should always rotate in any reference frame where the impulses are not seen to occur simultaneously, even if the impulses occur simultaneously in the rest fame of the lever?

No. Why do you think I have said that? Impulses cause two torques on the two arms, torques cancel. Timing of impulses affects how much torgue the impulses cause, by affecting the distance between the impulses.

We had not introduced the COM at this point. Here you are presumably talking about the physical acceleration of the object as a whole that displaces it relative to an initially comovng inertial object.

Do you agree that statement:
(1) is not necessarily true in an inertial ref frame where the object is moving to the left.
(2) is not necessarily true in an inertial ref frame where the object is moving to the right.
(3) is not necessarily true in any inertial ref frame where the object is moving and the simultaneous impulses are spatially separated.
These were the statements:
(1) An object is given an impulse to the left, later an opposite impulse to the right -> object is moved to the left.
(2) An object is given an impulse to the right, later an opposite impulse to the left -> object is moved to the right.
(3) An object is given an impulse to the left, and simultaneously an opposite impulse to the right -> object is not moved to the left and object is not moved to the right.

All statements are true in all frames, except for the fact that there exists a special shift of COM in rolling motion. Most people, including me, think the lever in the Right-angle lever paradox does not roll.

Oh yes, we are not considering COM. So all statements are just plain true. :smile:

(In cases were the impulses cause a spinning motion of an object, a mark painted on the middle of the object is the mark that shifts left or right or does not shift)

Now I may finally start to understand what the talk about compression waves was about. If two compression waves collide at the middle of an object, then that's same in all reference frames. To this I would say someting about opposite impulses and opposite waves they cause not being opposite ones in a different frame.

Are opposite impulses really not opposite ones in a different frame? Let's see ... No. I made an error :eek:


We have a paradox here! Compression wave analysis says one thing, common sense about transitory momentum change says another thing.
 
Last edited:
  • #56
30,143
6,603
We have a paradox here! Compression wave analysis says one thing, common sense about transitory momentum change says another thing.
We don't have a paradox, we have a lazy poster who isn't doing the math.
 
  • #57
Nugatory
Mentor
12,988
5,699
(3) An object is given an impulse to the left, and simultaneously an opposite impulse to the right ......

We have a paradox here! Compression wave analysis says one thing, common sense about transitory momentum change says another thing.
When describing a relativity paradox, a good rule of thumb is that if you ever find yourself typing the word "simultaneously", you should immediately stop typing and start thinking in the language of local space-time events.

In this case, the "common sense" that you're relying on is based on assumptions about the behavior of solid objects. These assumptions are only valid in the non-relativistic approximation in which the relativistic effects of differences in velocity between different parts of the body are small. Thus, your "common sense" is misleadng you.
 
  • #58
3,962
20
Now I may finally start to understand what the talk about compression waves was about. If two compression waves collide at the middle of an object, then that's same in all reference frames. To this I would say someting about opposite impulses and opposite waves they cause not being opposite ones in a different frame.
Are opposite impulses really not opposite ones in a different frame? Let's see ... No. I made an error :eek:
I think you have finally seen the light which is good news. As for how impulses of the guns transform, we can note that an impulse J is defined as ##J = F*\Delta T## in the rest frame. Under transformation to a frame parallel to the impulse, this becomes ##J' = F'*\Delta T' = F*(\Delta T \gamma)##. For two opposite facing guns, the impulses are still equal in magnitude, but opposite in sign, in the transformed frame.

We have a paradox here! Compression wave analysis says one thing, common sense about transitory momentum change says another thing.
It is less controversial to say "apparent paradox" or "seeming paradox". I am fairly sure you do not think there is any real paradox here now.
Impulses transform too. Impulse that a gun gives to a bullet is larger in the frame where the gun moves forward. I said this in post #42. (and I expected you to disagree)

IF the impulses REALLY are opposite and simultaneus .... and so forth and so on.
As mentioned above, the impulses remain equal and opposite for the opposite facing guns. We can remove any need to even consider the impulse transformation by considering the following special set up. Consider two rotor arms that are parallel to each other and free to rotate relative to a massive bar connecting their centres like this:
attachment.php?attachmentid=63308&stc=1&d=1382803195.jpg


Guns are attached at right angle to the ends of the rotor arms so that they fire outwards. In the rest frame of the apparatus, all four guns are fired simultaneously and there is no nett rotation of either rotor arm. When viewed in a irf where the apparatus is moving to the right, the guns on the left fire first and after a delay the guns on the right fire. Again there is no net rotation of the rotor arms in this reference frame. This is because the "torque waves" from both directions arrive simultaneously at the centres of the their respective rotor arms. In slow motion, the arms will be seen to bend during the transient phase before everything returns to equilibrium. It should be clear with this set up, all the guns are parallel to each other and moving at the same velocity, so that the impulses must transform in the same way. In this case the magnitude of the impulses is ##J' = F/\gamma*\Delta T \gamma =J##.
 

Attachments

  • #59
1,447
113
We don't have a paradox, we have a lazy poster who isn't doing the math.

OK then, I'll do some math.

So we have a spacetime diagram with one mostly vertical world line, describing how an object at rest is given two opposite impulses. It looks like this:
Code:
|
|
|
\
 |
 |
 |


My task is to draw a spacetime diagram where that same world line is tilted. Convert the diagram to other frame, or whatever the correct idiom is.

How do I do that? It will be a tilted line with two angles, is there a nice simple formula to convert the angles?
 
Last edited:
  • #60
30,143
6,603
OK then, I'll do some math.
I am waiting with breathless anticipation. :rolleyes:
 
  • Like
Likes 1 person
  • #61
3,962
20
OK then, I'll do some math.

So we have a spacetime diagram with one mostly vertical world line, describing how an object at rest is given two opposite impulses. It looks like this:
Code:
|
|
|
\
 |
 |
 |


My task is to draw a spacetime diagram where that same world line is tilted. Convert the diagram to other frame, or whatever the correct idiom is.

How do I do that? It will be a tilted line with two angles, is there a nice simple formula to convert the angles?
Ok, lets give the events some labels in this inertial reference frame (S). The start of the worldline is e1, the first kink to the left is e2, the second kink is e3 and the top of the worldline is e4. During the impulse the object moves a distance ##\Delta x## in a time interval ##\Delta t##. x must be less that t as this is a physical object. Now use the Lorentz transforms to find ##\Delta x'## and ##\Delta t'## of events e2' and e3' in another reference frame (S') moving to the right with velocity v relative to frame S. Once you have the deltas it is easy enough to find the angles using simple trigonometry.

I think you will better off demonstrate to yourself that if the equal impulses occur simultaneously in frame S' where the rod is moving, that they do not occur simultaneously in the original rest frame S of the rod and so the centre of the rod will be accelerated in inertial reference frames.
 
Last edited:
  • #62
Nugatory
Mentor
12,988
5,699
So we have a spacetime diagram with one mostly vertical world line, describing how an object at rest is given two opposite impulses. It looks like this:
An object at rest does not have a world line.

Each individual point in the object has a world line, and all of these worldlines together form the world sheet of the object; the individual world lines may converge, diverge, and jog independently.
 
  • #63
3,962
20
An object at rest does not have a world line.

Each individual point in the object has a world line, and all of these worldlines together form the world sheet of the object; the individual world lines may converge, diverge, and jog independently.
I think jartsa is talking about a single point marked at the centre of the object to simplify things.
 
  • #64
Nugatory
Mentor
12,988
5,699
I think jartsa is talking about a single point marked at the centre of the object to simplify things.
I'm sure that he is, but that simplification only works if all parts of the body move in unison - and that's not applicable in all of this discussion about impulses being applied to opposite ends of the object at the same or different times.
 
  • #65
3,962
20
I'm sure that he is, but that simplification only works if all parts of the body move in unison - and that's not applicable in all of this discussion about impulses being applied to opposite ends of the object at the same or different times.
True, but we can determine if the centre of the object moves in its original rest frame by considering the compression waves in isolation. If they arrive simultaneously at the centre in the rest frame, the centre does not move (for equal impulses) and if they do not arrive simultaneously, then it does move. Don't want to overcomplicate things at this stage.
 
  • #66
1,447
113
Let's consider a T-shaped object, which we will call T, moving to the right very fast, with two unstable particles on both ends of the horizontal bar. Both particles decay to two photons simulteneously in T's frame. Then two photons travel from the end points of the horizontal bar to the vertical bar, into which the photons are absorbed.

Now we observe these events from that frame where the T was moving to the right very fast.

We see the left side particle decaying first, then one of the decay products travels to the right as a high energy photon.

Then we see the right side particle decaying, then one of the decay products travels to the left as a low energy photon.


Some of T's stuff traveled from the left to the right, smaller amount of T's stuff travelled from the right to the left.

Conclusion: In this case where opposite impulses pushed the T, a shift to the right happened, as we'd expect when the right pushing impulse preceeds the left pushing impulse.


Clarification:

We consider the photons that leave the T to not be part of the the T, while the other two photons we consider to be part of the T, also the two unstable particles we consider to be part of the T.

So when the left side particle decays, T loses a small part of itself, when the right side particle decays, T loses a larger part of itself, later the balance between left and right is restored by a net mass-energy flow from left to right.
 
Last edited:
  • #67
30,143
6,603
This is pointless. Until you actually go through the effort to mathematically analyze a simple situation using the correct mathematical tools which have been identified there is no point in analyzing progressively more complicated scenarios.

Please try to do some actual work on your own. If you get stuck then post it to a new thread, this one is closed.
 

Related Threads on Forces in relativistic rolling motion

Replies
2
Views
3K
  • Last Post
Replies
5
Views
2K
Replies
125
Views
21K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
6
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
2
Replies
32
Views
12K
  • Last Post
8
Replies
195
Views
24K
  • Last Post
Replies
5
Views
2K
  • Last Post
3
Replies
59
Views
3K
Top