Forces in relativistic rolling motion

  • Thread starter jartsa
  • Start date
  • #26
30,152
6,616
This seems to be a good introductory reference for relativistic angular momentum.
http://panda.unm.edu/Courses/Finley/P495/TermPapers/relangmom.pdf [Broken]

You may want to go through that before just randomly proposing another more complicated scenario.
 
Last edited by a moderator:
  • #27
3,962
20
I would like to expand on the four vector approach to the right angle lever paradox.

Consider a right angle lever in its rest frame, that has one arm parallel to the x axis with a force fy=1 acting in the y direction on the free end. The other arm is parallel to the y axis that has a force fx=1 acting in the x direction on its free end, where fx and fy are equal in magnitude and the length of the arms are of equal length r=2.

Using the regular 4 force vector ##\gamma##(P,fx,fy,fz) where P is the power, we can represent the force acting on the y arm as ##\gamma##(0,1,0,0) where P and ##\gamma## are set to zero because the lever is stationary and not rotating. Under a lorentz boost of velocity u=0.6 in the x direction, the 4 force on this y arm is now (-0.75,1.25,0,0). The x component of the 3 force is 1.25/##\gamma##(0.6) = 1 and the torque is 1*r = 2.

Now we consider the x arm. The four force acting on this arm in the rest frame of the arm is (0,0,1,0) and after the boost is still (0,0,1,0), but now the 3 force is 1/##\gamma##(0.6) = 0.8. The torque is 0.8*r/##\gamma## = 1.28 where we take account of the length contracted length of the x arm. It is now clear from this naive analysis, that the torques in the two arms are unequal and the lever should rotate in the boosted frame and this gives rise to the paradox.

We can note however, that the power component of the boosted 4 vector for the y arm was non zero. This hints at the internal energy transfer or internal counter torque that some texts use to explain why the lever does not rotate in the boosted frame (as in the link given by Dalespam).

Now lets try another approach, that uses the cross product form of torque force. Now the 4 force for the y arm in the rest frame of the lever is (0,0,0,2) where we choose the fz component because this is the axis of rotation and is the direction the cross product torque vector points in and the length of the arm is already included in the 3 torque force. Under the boost in the x direction of 0.6c, the 4 force remains (0,0,0,2) and the 3 torque is 2/##\gamma## = 1.6.

For the y arm, the torque is around the same axis but in the opposite direction, so in the rest frame the 4 torque is (0,0,0,-2) and the boosted 4 torque is also (0,0,0,-2) and so the 3 torque is -2/##\gamma## = -1.6 which is equal and opposite to the torque on the x arm, so no rotation in the boosted frame and internal energy transfer or internal counter torque is not required when using 4 torque, with torque defined as a cross product.
 
Last edited:
  • #28
pervect
Staff Emeritus
Science Advisor
Insights Author
9,862
1,057
I don't know. WBN posted the mathematical quantity which corresponds to angular momentum in relativity, but I have no experience with it. Unless someone with some familiarity with it responds you will probably have to learn it on your own.
ed:
((I think I'll snip some rather speculative ideas that may be useful but that I haven't fully worked out yet))

For a textbook discussion of a related problem, I'd suggest MTW's exposition on pg 175 of Thomas precession, which gives the results for the inertial-frame spin components of an electron rotating around the nucleus in the (x,y) plane in the lab frame. (The lab frame is presumed to be accelerating).

Unfortunately, MTW presents the results, without giving a detailed derviatio.
 
Last edited:
  • #29
1,447
113
Let's try to solve the right-angled lever paradox.

http://en.wikipedia.org/wiki/Trouton–Noble_experiment#Right-angle_lever_paradox

A right-angled lever is moving at velocity v, witch is 0.99999999 c.
The two arms of the lever have the same rest length.

From both ends of both arms two bullets are fired, at the moments when light signals sent from the middles of the arms arrive at the ends of the arms. (Total number of bullets is four)


Here's a picture of two guns and one horizontal lever arm:
______|
|


Now we are looking from a viewpoint of a still standing observer:

The aforementioned signals travel at velocities:

Transverse arm: c / gamma (That's the transverse speed)
Longitudinal arm: c


The travel times of signals are:

Transverse arm: t
Longitudinal arm: 2 * t for one signal, 0 for the other (this is very high velocity approximation )

Distances between the two firing events are:

Transverse arm: 2 * t * c / gamma
Longitudinal arm: 2 * t * c


Forces of the two firing events are:

Transverse arm: F
Longitudinal arm: F / gamma



Torques are:

Transverse arm: 2 * t * ( c / gamma ) * F
Longitudinal arm: 2 * t * c * ( F / gamma )


There is no problem! Torques are equal.


Here's a non-mathematical version of the scenario above:
From each of the two arms of a high speed right-angled lever two bullets are fired, so that opposite torques are caused on the arms by the firings, in the lever frame.


Here's a question about that scenario:

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)
 
Last edited:
  • #30
30,152
6,616
(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)
No, the lever cannot be perfectly rigid for exactly this reason.
 
  • #31
3,962
20
(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)
In the lever frame all 4 bullets are fired simultaneously. In this frame, the two bullets at the fulcrum fire simultaneously, tending to push the fulcrum up and to the right and after a delay the opposing impulses from firing the bullets at the free ends of the arms, arrive simultaneously with each other at the fulcrum restoring the position of the fulcrum. This means there might slight motion of the fulcrum visible in both frames due to this delay. Similarly, when the bullets at the free ends of the arms are fired, the reaction forces will bend both arms in opposite directions and this is again followed later by restoring forces. The main point is that although the lever is presented as a static situation in the lever frame, it is actually quite dynamic with a lot of bending going on (visible in both frames). Of course if the lever was perfectly rigid, none of this bending could happen and there would be no delay between the initial impulses and the restoring forces, but perfect rigidity is not allowed in relativity (as Dalespam mentioned).
 
Last edited:
  • #32
1,447
113
but perfect rigidity is not allowed in relativity (as Dalespam mentioned).
You guys are not getting my point.

An object is given an impulse to the left, later an opposite impulse to the right -> object is moved to the left.
An object is given an impulse to the right, later an opposite impulse to the left -> object is moved to the right.
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.

That was the idea.

"No time between impulses" = impulses are simultaneous.
 
  • #33
3,962
20
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.
Assuming this is the rest frame of the centre of mass of the object, when the impulses are applied to the ends equidistant from the COM, the object is compressed and each end travels towards the COM and later restores to its original length when it reaches equilibrium.

An object is given an impulse to the right, later an opposite impulse to the left -> object is moved to the right.
Let us say this is the same set of events as above, but as seen in a reference frame where the object is moving from left to right. When the impulse occurs at the left end of the object, pushing the object to the right, the whole object does not instantly accelerate to the right. Instead, the left end accelerates to the right and the left end of the object compresses towards the COM. Similarly when the impulse on the right end of the object occurs, the right end compresses towards the COM. These compression waves meet at the COM simultaneously, so there is no acceleration of the COM. Later restoring forces expand the object to its original length. At no point does the COM accelerate in any reference frame. The non rigid nature of materials allows the ends to temporarily move independently of each other.

Alternatively, let us say there is a reference frame where the COM of the object is initially stationary and the delay between applying the left and right impulses is large enough that the COM is displaced from its original rest position. Now it will be impossible to find an inertial reference frame in which the COM does not accelerate.
 
Last edited:
  • #34
30,152
6,616
You guys are not getting my point.

An object is given an impulse to the left, later an opposite impulse to the right -> object is moved to the left.
An object is given an impulse to the right, later an opposite impulse to the left -> object is moved to the right.
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.

That was the idea.

"No time between impulses" = impulses are simultaneous.
I get your point, but your point is wrong. It relies on the object being rigid.

If you give an impulse to the left a compression wave travels through the object displacing various parts of the object to the left at various times depending on the speed of sound in the material. If you give an impules to the right a compression wave travels through the object displaing various parts of the object to the right at various times depending on the speed of sound in the material.

Because the object is not rigid, when the impulses are simultaneous then both ends move in their respective directions until the opposite compression wave pushes them the opposite direction. Each end is displaced in all frames.
 
  • #35
1,447
113
Assuming this is the rest frame of the centre of mass of the object, when the impulses are applied to the ends equidistant from the COM, the object is compressed and each end travels towards the COM and later restores to its original length when it reaches equilibrium.

Let us say this is the same set of events as above, but as seen in a reference frame where the object is moving from left to right. When the impulse occurs at the left end of the object, pushing the object to the right, the whole object does not instantly accelerate to the right. Instead, the left end accelerates to the right and the left end of the object compresses towards the COM. Similarly when the impulse on the right end of the object occurs, the right end compresses towards the COM. These compression waves meet at the COM simultaneously, so there is no acceleration of the COM. Later restoring forces expand the object to its original length. At no point does the COM accelerate in any reference frame. The non rigid nature of materials allows the ends to temporarily move independently of each other.

Alternatively, let us say there is a reference frame where the COM of the object is initially stationary and the delay between applying the left and right impulses is large enough that the COM is displaced from its original rest position. Now it will be impossible to find a reference frame in which the COM does not accelerate.

:confused::confused::confused:

It's silly to say "These compression waves meet at the COM simultaneously, so there is no acceleration of the COM."

Here are two reasons:

1: COM may very well move when some matter in the position where the COM is does not move
2: It may be possible to push a COM, by poking the position where the COM is with a stick, but ... well I don't know
 
  • #36
1,447
113
I get your point, but your point is wrong. It relies on the object being rigid.

Well that's easy to refute: My idea does not rely on the object being rigid.

Here's the idea again: Opposite impulses at different times cause an object to be shifted to another position.


Let's try some mathematical analysis:

A very short time Impulse gives a very large object a momentum mv. Immediately after the impulse the COM (center of mass) of the object moves at velocity v. (Object was at rest at the beginning)

Opposite impulse stops the object after time t. The COM of the object moved a distance: v * t.
 
  • #37
3,962
20
Let's try some mathematical analysis:

A very short time Impulse gives a very large object a momentum mv. Immediately after the impulse the COM (center of mass) of the object moves at velocity v. (Object was at rest at the beginning)

Opposite impulse stops the object after time t. The COM of the object moved a distance: v * t.
In this case the COM moves in all reference frames, so no issue. As I mentioned before:
Alternatively, let us say there is a reference frame where the COM of the object is initially stationary and the delay between applying the left and right impulses is large enough that the COM is displaced from its original rest position. Now it will be impossible to find an (inertial) reference frame in which the COM does not accelerate.
 
  • #38
3,962
20
Consider the following experiment:

We have a lab where two guns at opposite ends of the lab are lined up with each other and fired simultaneously, such that the two bullets collide inelastically at the centre of the lab and stop dead. Do you agree that from the point of view of an observer travelling at high speed relative to the lab that sees one gun fired much earlier than the other, that he still sees both bullets arriving at the centre of the lab simultaneously? If you do, think of the bullets as analogous to the compression waves.
 
  • #39
30,152
6,616
Let's try some mathematical analysis
Please go back to my first response to you and read up about 4-vectors. Then try the correct mathematical analysis.
 
  • #40
1,447
113
Consider the following experiment:

We have a lab where two guns at opposite ends of the lab are lined up with each other and fired simultaneously, such that the two bullets collide inelastically at the centre of the lab and stop dead. Do you agree that from the point of view of an observer travelling at high speed relative to the lab that sees one gun fired much earlier than the other, that he still sees both bullets arriving at the centre of the lab simultaneously? If you do, think of the bullets as analogous to the compression waves.
Ok I thought. Seems I don't disagree with anything. But I did not think about any implications.




We have two right moving sticks A and B, with guns, guns are the short lines, guns point away from the sticks:

A:
_ _____ _ ->

B:
_ _____ _ ->



Guns of stick A fire simultaneously. Effect: No change at all on velocity of COM of stick.

Guns of stick B fire, first the right one then the left one. Effect: A transient slowing down of COM of stick. COM of B will be shifted to the left of COM of A.
 
Last edited:
  • #41
3,962
20
We have two right moving sticks A and B, with guns, guns are the short lines, guns point away from the sticks:

A:
_ _____ _ ->

B:
_ _____ _ ->


Guns of stick A fire simultaneously. Effect: No change at all on velocity of COM of stick.

Guns of stick B fire, first the right one then the left one. Effect: A transient slowing down of COM of stick. COM of B will be shifted to the left of COM of A.
There will be a change of velocity of stick A if A's guns fire simultaneously in this reference frame where both sticks are moving to the right.

As a general rule, if there is reference frame where the COM of a system is at rest, then in any other inertial reference frame with relative velocity v, the COM has a constant velocity of -v and due to conservation of momentum there is nothing that can occur internally within that system that can change the velocity of the COM. However parts of the system can move relative to the COM of the system. In your example, the centre of stick A will move relative to the COM of system A (stick + guns + bullets) because the impulses do not arrive simultaneously at the COM of the stick. It might be worth considering a third system C, which is similar to A and B but the guns never fire for use as a reference. Both sticks A and B will move relative to stick C after the guns are fired, in the particular example you gave above.
 
Last edited:
  • #42
1,447
113
There will be a change of velocity of stick A if A's guns fire simultaneously in this reference frame where both sticks are moving to the right.

Oh yes. That happens if the stick is relativistic, because the right flying bullet experiences a larger change of momentum.

In this stick-gun configuration such thing does not happen:

Code:
 ____|          
 |                   ------>
 
Last edited:
  • #43
3,962
20
Here is an adaptation of your experiment, with the addition of rference system C that does not fire its guns. The (:) at the centre of each stick marks the COM of each stick (not each system). All the sticks are initially moving inertially to the right in inertial reference frame S.

A:
_ ___:___ _ ->

B:
_ ___:___ _ ->

C:
_ ___:___ _ ->

This time guns of stick A fire simultaneously in system A's own rest frame (S'). Effect: No change at all on velocity of COM of stick A. (It keeps level with the COM of stick C). (Note that the guns of A do not fire simultaneously in the reference frame (S) where the sticks are moving.) In S, the gun at the rear of stick A fires first and the stick compresses to the right. This is followed by the gun at the front firing and compressing the leading end back towards the COM, but at no time does the COM of stick A move relative to the COM of stick C.

Guns of stick B fire, first the right one then the left one as seen in the reference frame S. Effect: A transient slowing down of COM of stick B. COM of B will be shifted to the left of COM of A.

Note that there is no inertial reference frame where the COM of stick B remains at rest.
 
Last edited:
  • #44
3,962
20
Oh yes. That happens if the stick is relativistic, because the right flying bullet experiences a larger change of momentum.

In this stick-gun configuration such thing does not happen:

Code:
 ____|          
 |                   ------>
Yep, but in this case the stick rotates in all inertial reference frames.
 
  • #45
1,447
113
It has not in any way become clear to me how I am wrong when saying for example this:

Impulses that are opposite and simultaneous in some frame do not shift the COM of an object, nonexistence of the shift is true in all frames.

Impulses that are opposite and non-simultaneous in some frame shift the COM of an object, the existence of the shift is true in all frames.


Any counter examples?
 
  • #46
3,962
20
It has not in any way become clear to me how I am wrong when saying for example this:

Impulses that are opposite and simultaneous in some frame do not shift the COM of an object, nonexistence of the shift is true in all frames.

Impulses that are opposite and non-simultaneous in some frame shift the COM of an object, the existence of the shift is true in all frames.


Any counter examples?
If you are suggesting that impulses that are opposite and simultaneous in some frame will never cause the COM to shift, then a counter example is this: If a rod is moving wrt frame S, then spatially separated and simultaneous impulses that are equal and opposite in frame S will accelerate the COM. This is because the simultaneous impulses at the ends of the rod in frame S are not simultaneous in the rest frame of the rod.

If your intended meaning was essentially:

If the COM shift of an object is non existent in one frame, the non existence of the shift is true in all frames.

If the COM shifts in one frame, then the existence of the shift is true in all frames.


then I would agree with that. That does however answer this question that started this series of posts:

Here's a non-mathematical version of the scenario above:
From each of the two arms of a high speed right-angled lever two bullets are fired, so that opposite torques are caused on the arms by the firings, in the lever frame.


Here's a question about that scenario:

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 centre of mass of the rolling thing)
Here you seem to be suggesting that the lever should rotate when it is moving relative to an observer (and the impulses are not simultaneous) when the lever does not rotate in its rest frame.
 
Last edited:
  • #47
1,447
113
A counter example is that if a rod is moving wrt frame S, then impulses that are equal and opposite in frame S will accelerate the COM. This is because the simultaneous impulses at the ends of the rod in frame S are not simultaneous in the rest frame of the rod.
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.
 
  • #48
3,962
20
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.
OK, if we adjust the impulses at the ends of the rod, in such a way that they are really are equal and opposite in frame S where the rod has relative motion, then if the impulses occur simultaneously in frame S, the compression waves will not arrive simultaneously at the COM of the rod. The compression waves from the front end arrive first, temporarily slowing the COM of the rod.
 
  • #49
1,447
113
OK, if we adjust the impulses at the ends of the rod, in such a way that they are really are equal and opposite in frame S where the rod has relative motion, then if the impulses occur simultaneously in frame S, the compression waves will not arrive simultaneously at the COM of the rod. The compression waves from the front end arrive first, temporarily slowing the COM of the rod.

Pay attention. I try to explain something :smile:

When a gun attched to a rod is fired, the gun imparts some momentum on the rod on that place where the gun is attched. The rod has that extra momentum as soon as the momentum is in the rod.

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.

So when we hit one end of a 100 km long rail with a hammer, the COM of the rail has gained the momentum of the hammer as soon as the hammer has stopped.

If another hammer is hitting the other end of the rail, diffrent observers may have different opinions about which hammer gives its momentum to the COM of the rail first,
 
Last edited:
  • #50
3,962
20
When a gun attched to a rod is fired, the gun imparts some momentum on the rod on that place where the gun is attched. The rod has that extra momentum as soon as the momentum is in the rod.

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.

So when we hit one end of a 100 km long rail with a hammer, the COM of the rail has gained the momentum of the hammer as soon as the hammer has stopped.
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]
 
Last edited by a moderator:

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