Forces in relativistic rolling motion

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  • #1
jartsa
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Let's tie a rope on an object, then we swing the object around in a circle, then we start running forwards while continuing the swinging. (All velocities are relativistic)

An assistant takes lot of pictures of us doing this stunt.

Then finally we study the pictures.

We will notice that the object is more probable to be found on the side where the velocity of the object is larger.
(This effect is a result of relativistic addition of the linear and the circular velocities of the object)

So we conclude the object spends a longer time on one side. From this we conclude that the force exerted
on the object is smaller on this same side. (So that the the impulse (F*t) does not become too large)

From the principle of relativity we can see that the rope does not feel any decrease of stress at any time.

Question1: Why does the part of the rope that moves very fast not feel any decrease of stress? (Because of the high velocity, I guess)

Question2: Why does the part of the rope near the center of the circle that does not move so fast not feel any decrease of stress?
 

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  • #2
Dale
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The best way to deal with forces in problems like this is to use the four-force:
http://en.wikipedia.org/wiki/Four-force

That includes all of the relativistic corrections needed.

The Minkowski norm of the four-force gives the force "felt" by the particle. It is invariant.
 
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  • #3
Bill_K
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Let's tie a rope on an object, then we swing the object around in a circle, then we start running forwards while continuing the swinging. (All velocities are relativistic)
Arg, and also there's a guy sitting on the object who stirs his coffee while flipping a coin (all relativistic of course.)

I don't really see how pondering such complicated problems like this helps one understand relativity!
 
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  • #4
yuiop
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So we conclude the object spends a longer time on one side. From this we conclude that the force exerted
on the object is smaller on this same side. (So that the the impulse (F*t) does not become too large)

From the principle of relativity we can see that the rope does not feel any decrease of stress at any time.

Forces in relativity can get really messy. Look up the right angled lever paradox for example. Circular motion makes thing messy ^2.

There are various formulas for relativistic force quoted in the literature and the tricky part is knowing which ones to use in a given situation. I am using the transformation equations in the last paragraph of this link. Note that when a force acts on an object with a relative velocity, the tangential force is greater than m*a by a factor of gamma. When a force acts on an object that is momentarily at rest and then transform to a frame where the object has a tangential velocity relative to the force, the transformed force is reduced by a factor of gamma. The difference is subtle, but you have to pay attention to it.

Lets say in the rest frame of the circle, the tangential velocity of the object is v. You are running at v relative to the road, in the +x direction and the rotation is such that when the object is on the -y side of the road the tangential velocity is momentarily zero and on the +y side of the road the tangential velocity is w, where w is the relativistic addition of v with itself.

In the frame where the centre of the circle is at rest, let us say the magnitude of the force is ##F_y## in both y directions. When transformed to the rest frame of the road, the force on the fast +y side is ##F_y*\gamma(v)/\gamma(w)##.

The tangential velocity of the object on the -y side of the road is zero, so the transformed force is is ##-F_y*\gamma(v) ##.

So in the road frame the magnitude of the force acting on the object when it is on the slow side of the road is ##\gamma(w) ## greater than the force on the fast side.

Of course, like I said, it is tricky and there is plenty of room for error, so that is why it is probably better to use four force as Dalespam suggested.
 
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  • #5
dauto
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In addition to what was explained above, the stresses in the rope also must be corrected because they are not relativistic invariants either. The stress is part of the Stress–energy tensor. One cannot over-emphasize the fact that it can become very confusing trying to understand the mess of relativistic corrections. The easiest way to deal with that is - as already mentioned above - to learn how to use four-vectors and other four-tensors.
 
  • #6
jartsa
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Thank you for the good anwers. None of which was an answer to my questions. :smile:

Let me try to answer:

When we combine the high linear speed and the low circular speed of a short part of the rope near the center, we would get a nearly constant angular velocity, if the changing Lorentz-contraction was not there making things complicated.

When we combine the high linear speed and the high circular speed of a short part of the rope near the outer end, we would get a changing angular velocity, even if the changing Lorentz-contraction was not there.

If we use some fancy rope contraction elimination tecnology, we observe that the rope curves. (it curves because of the differing changes of angular velocity at different parts of the rope)

So:
The rope contracts and extends and curves and straightens. The result of these changes is a constant proper stress on the rope.
 
  • #7
jartsa
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Question 3:

If a force is a Coulomb force, are the force formulas in post #4 valid?

Are force formulas in post #4 equivalent to non-relativistic force formulas + magnetic force formulas?
 
  • #8
Dale
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Thank you for the good anwers. None of which was an answer to my questions.
Huh?

I gave some background on the four-force and said:
The Minkowski norm of the four-force gives the force "felt" by the particle.

Your question 1:
Why does the part of the rope that moves very fast not feel any decrease of stress?
My answer, referring to the Minkowski norm of the four-force:
It is invariant.

Your question 2:
Why does the part of the rope near the center of the circle that does not move so fast not feel any decrease of stress?
Same answer, also referring to the Minkowski norm of the four-force:
It is invariant.

Your questions were answered very directly. The Minkowski norm of the four-force is the quantity that determines the stress felt by the rope. That quantity is invariant, so no decrease is felt in any frame. How is that not an answer your question?
 
  • #9
jartsa
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Huh?

That quantity is invariant, so no decrease is felt in any frame. How is that not an answer your question?


Because I don't get the feeling that this is an answer to the question.

But maybe I can make something out of the four-vectors.


I have been a fan of this kind of answers:

Why the stress is largest were it is largest? Because forces that keep the rope together are weakened at large speeds.
 
  • #10
Dale
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Because I don't get the feeling that this is an answer to the question.
Well, maybe you should re-examine your feelings :smile:

You have a quantity of interest: the force felt by the rope. Physics has a mathematical object which is used to represent that specific quantity of interest: the Minkowski norm of the four-force. From the standpoint of theory, all answers about the physical quantity of interest are given in terms of the mathematical object which is used to represent it. I don't know how else a question about physics can be answered.

I suspect that your feeling is simply based on unfamiliarity with four-vectors. If so, then I would strongly encourage you to become familiar with them. For me personally, they were the key to understanding SR. Once I was exposed to them, SR suddenly "clicked".
 
  • #11
WannabeNewton
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Perhaps this may ameliorate possible confusion: keep in mind that the 4-force ##f^{\mu}## is a purely space-like 4-vector i.e. it is always orthogonal to the 4-velocity (which is purely time-like). This is simply because the 4-velocity can be parametrized by proper time so that ##u^{\mu}u_{\mu} = -1## hence ##u_{\mu}\partial_{\nu}u^{\mu} = 0## which implies that ##f^{\mu}u_{\mu} = ma^{\mu}u_{\mu} = mu^{\nu}u_{\mu}\partial_{\nu}u^{\mu} = 0##.

Therefore if you go to an inertial frame instantaneously comoving with the particle at some event on its worldline, you will have ##f^{\mu} = (0,\vec{f})## i.e. ##f^{\mu}f_{\mu} = \left \| \vec{f} \right \|^2## so the norm squared of the 4-force, which is a relativistic invariant, is just the magnitude squared of the usual force.
 
  • #12
yuiop
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Question 3: If a force is a Coulomb force, are the force formulas in post #4 valid?
All forces transform in exactly the same way.

Consider a test mass suspended by a spring such that the spring force opposes a gravitational or electromagnetic force acting in the opposite direction. The system is in equilibrium and the test mass is stationary in the lab frame. Let us say the opposing forces are parallel to the y-axis of the lab. Now consider the point view of an observer whose velocity in the lab frame is parallel to the mutual x axis. We would not expect this observer to see any motion of the test particle in the y direction and so the system is still in equilibrium from his point of view. This means all static transverse forces must transform in the same way.

Are force formulas in post #4 equivalent to non-relativistic force formulas + magnetic force formulas?
Don't think so.
 
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  • #13
jartsa
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Lets say in the rest frame of the circle, the tangential velocity of the object is v. You are running at v relative to the road, in the +x direction and the rotation is such that when the object is on the -y side of the road the tangential velocity is momentarily zero and on the +y side of the road the tangential velocity is w, where w is the relativistic addition of v with itself.

In the frame where the centre of the circle is at rest, let us say the magnitude of the force is ##F_y## in both y directions. When transformed to the rest frame of the road, the force on the fast +y side is ##F_y*\gamma(v)/\gamma(w)##.

The tangential velocity of the object on the -y side of the road is zero, so the transformed force is is ##-F_y*\gamma(v) ##.

So in the road frame the magnitude of the force acting on the object when it is on the slow side of the road is ##\gamma(w) ## greater than the force on the fast side.


I would say that:

force in road frame = what a force meter between the rope and the object reads / gamma(v)

where v is the two dimensional relativistic sum of the linear and circular velocities


... which is wrong, except at the two positions that you were considering, because the force meter is transverse to the velocity at those positions.


Edit: Oh yes, this sentence is in the page that you gave the link of: "The transverse force is a factor of γ larger in the particle's frame"
 
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  • #14
jartsa
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Are force formulas in post #4 equivalent to non-relativistic force formulas + magnetic force formulas?

Don't think so.


Well I'm disappointed. But at least one such case exists:

Coulomb force between two charges at rest can be calculated with Coulombs law.
The result can be transformed to other frame by the law F'= F/gamma(v), in the simplest case, with nice directions of the force and the velocity.

Alternatively a magnetic force can be added to the non-transformed Coulomb-force. Isn't this right?
 
  • #15
yuiop
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Well I'm disappointed. But at least one such case exists:

Coulomb force between two charges at rest can be calculated with Coulombs law.
The result can be transformed to other frame by the law F'= F/gamma(v), in the simplest case, with nice directions of the force and the velocity.

Alternatively a magnetic force can be added to the non-transformed Coulomb-force. Isn't this right?

A counter example is given in the ongoing EM thread. In the rest frame of a charge moving relative to a wire, the force (with e.g magnitude x Newtons) is entirely electric (in one particular example) and when transformed to the rest frame of the wire the force is entirely magnetic (with magnitude x*gamma Newtons). Obviously transforming back the other way the electric force reappears and the magnetic force disappears. (This is not the general case, just a carefully chosen example). It is not simply a case of adding the magnetic force to the electric force.
 
  • #16
jartsa
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Angular momentum in relativistic rolling motion

We had this person swinging a rock on the end of a rope. Then the person started running and we observed the averege position of the rock shifting to to the fast side. (See post #1)

If a runner running on the right side of a straight road changes to the left side, then the change of angular momentum is:
momentum of the runner * width of the road

Does similar change of angular momentum happen when the rock swinging guy increases his linear velocity?

I mean, does center of mass of the system shift, and does angular momentum change as a result of that, and is there an opposite change of angular momentum somewhere?
 
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  • #17
Dale
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I actually don't know how to treat angular momentum in SR. I am not aware of an angular momentum four-vector. That is a good question.
 
  • #18
WannabeNewton
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There is no such quantity. One instead has the 2-form ##L = r\wedge p##.
 
  • #19
jartsa
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Here's an alternative thought experiment for people who dislike the silly running and swinging:

A fast moving spaceship shoots simultaneously in its frame two bullets, one from the rear to the left, one from the front to the right.

The spaceship will rotate with the appropriate normal angular momentum according to people onboard, but very small angular momentum according to still standing observers.

But when the spaceship stops its linear motion, it will have the normal angular momentum according to everyone.


Right?
 
  • #20
Dale
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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.
 
  • #21
jartsa
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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.
 
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  • #22
yuiop
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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
Agree.

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 )
Agree. In fact we can show that for the longitudinal arm, the travel time going forward is t(1+v/c) and going backward is t(1-v/c) so that the total time is exactly 2*t for any velocity of the lever.

Distances between the two firing events are:

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

This is also correct, but the events are not simultaneous in the reference frame under consideration. You seem to be defining the length of the longitudinal arm (for the purposes of measuring torque) as the distance between the free end of the arm when the bullet was fired relative to where the fulcrum was, when the bullet was fired at the free end of the transverse arm.

It is a nice piece of mathematical work, but I am not sure redefining how length is measured will be readily accepted.
 
  • #23
pervect
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One important factor that hasn't been mentioned is that the circular orbit will become an elliptical one due to length contraction when viewed from the ground, due to length contraction in the direction of motion. I rather suspect that including this effect in the analysis will answer the OP's question, but I don't have the time to work it out in detail to confirm this suspicion.

I concur that the easiest way to do a detailed analysis would be with four-forces as other posters have mentioned. Convincing someone to start using four vectors seems to be about as hard as trying to follow some funky 3-vector derivation, on the other hand it's probably more productive then wasting a lot of time trying to debug a 3-vector formalism.

I'm a bit confused on the representation of angular momentum.

For instance, MTW, on pg 175, talks about the Thomas precession of "the intrinsic angular momentum vector S". I believe there are some circumstances where people can and do "get away" with representing angular momentum as a vector (as per this textbook example), but I don't recall the details. I'd agree in general that it's better to treat angular momentum as a two form, however.
 
  • #24
yuiop
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One important factor that hasn't been mentioned is that the circular orbit will become an elliptical one due to length contraction when viewed from the ground, due to length contraction in the direction of motion. I rather suspect that including this effect in the analysis will answer the OP's question, but I don't have the time to work it out in detail to confirm this suspicion.
In relation to the original question, the OP was particularly interested in the forces acting when the string connecting the object to the centre was transverse to the motion of the centre relative to the road. As such length contraction of the string when it was parallel to the motion was not a big factor.
I concur that the easiest way to do a detailed analysis would be with four-forces as other posters have mentioned. Convincing someone to start using four vectors seems to be about as hard as trying to follow some funky 3-vector derivation, on the other hand it's probably more productive then wasting a lot of time trying to debug a 3-vector formalism.
Any 4 vector analysis is welcome here and may be educational.
I'm a bit confused on the representation of angular momentum.

For instance, MTW, on pg 175, talks about the Thomas precession of "the intrinsic angular momentum vector S". I believe there are some circumstances where people can and do "get away" with representing angular momentum as a vector (as per this textbook example), but I don't recall the details. I'd agree in general that it's better to treat angular momentum as a two form, however.
This Wikipedia article defines angular momentum as a vector, which is the cross product of the linear momentum vector and the displacement vector. It seems more natural for angular momentum to be a vector as the direction of rotation as well as the orientation of the rotation axis have to be defined. It also seems that the OP has moved on to analysing the "right angle lever paradox", which is more about torque than angular momentum, although the two are closely related as demonstrated on the Wikipedia page.

When torque is considered as a cross product, the torque vector points along the axis of rotation. As such, thinking in terms of the cross product would seem to make the paradoxical nature of the right angled lever go away. The torque vectors of both arms are parallel to each other (but pointing in opposite directions) and so are subject to the same change during the Lorentz transformation.

I am fairly sure that if torque force is considered a vector parallel to the rotation axis, that a suitable four vector form could be constructed, based on the regular four-force. All we have to do is remember that the x,y and z three force vectors refer to the direction of the torque vector along the rotation axis and not to the direction of the linear force acting on a given arm. Once we do that, torque can be treated like any other force rather than requiring a special four-torque vector.
 
  • #25
jartsa
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This is also correct, but the events are not simultaneous in the reference frame under consideration.

Yes. Exactly. Thank you.:smile:

The transformation of simultaneity increases the lever arm length of a moving arm.

A fire truck is driving at speed 100 km/h. A water stream is spouted from the rear to the right. After a long time an other water stream is spouted from the front to the left. The water was given a huge angular momentum, because of the timing of the spoutings. Angular momentum of the same magnitude was given to the fire truck.

In the frame where the aforementioned fire truck moves million times faster, the two angular momentums are million times larger, if the timings of spoutings are timed by clocks that are synchronized in the truck frame and are co-moving with the truck.
 
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  • #26
Dale
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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.
 
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  • #27
yuiop
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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 let's 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.
 
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  • #28
pervect
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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.
 
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  • #29
jartsa
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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)
 
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  • #30
Dale
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(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
yuiop
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(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).
 
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  • #32
jartsa
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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
yuiop
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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.
 
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  • #34
Dale
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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
jartsa
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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
 

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