Forces in relativistic rolling motion

[jartsa]

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.

 

[yuiop]

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.

 

[yuiop]

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 traveling 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.

 

[Dale]

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.

 

[jartsa]

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 traveling 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.

 

[yuiop]

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.

 

[jartsa]

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:

 ____|          
 |                   ------>

 

[yuiop]

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.

 

[yuiop]

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:

 ____|          
 |                   ------>

Yep, but in this case the stick rotates in all inertial reference frames.

 

[jartsa]

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?

 

[yuiop]

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.

 

[jartsa]

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.

 

[yuiop]

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.

 

[jartsa]

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

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,

 

[yuiop]

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

 

[yuiop]

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.

 

[jartsa]

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

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.

 

[Nugatory]

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.

 

[yuiop]

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.

 

[jartsa]

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.

(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 something 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 We have a paradox here! Compression wave analysis says one thing, common sense about transitory momentum change says another thing.

 

[Dale]

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.

 

[Nugatory]

(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.

 

[yuiop]

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 something 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

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:

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 $J' = F/\gamma*\Delta T \gamma =J$.

 

[jartsa]

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:

|
|
|
\
 |
 |
 |

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?

 

[Dale]

OK then, I'll do some math.

I am waiting with breathless anticipation.

 

[yuiop]

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:

|
|
|
\
 |
 |
 |

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, let's 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.

 

[Nugatory]

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.

 

[yuiop]

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.

 

[Nugatory]

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.

 

[yuiop]

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.

 

[jartsa]

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 traveled 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.

 

[Dale]

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.