Forces in relativistic rolling motion

1. Oct 9, 2013

jartsa

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?

2. Oct 9, 2013

Staff: Mentor

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.

Last edited: Oct 9, 2013
3. Oct 9, 2013

Bill_K

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!

4. Oct 9, 2013

yuiop

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.

Last edited: Oct 9, 2013
5. Oct 9, 2013

dauto

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. Oct 10, 2013

jartsa

Thank you for the good anwers. None of which was an answer to my questions.

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. Oct 10, 2013

jartsa

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. Oct 10, 2013

Staff: Mentor

Huh?

I gave some background on the four-force and said:
My answer, refering to the Minkowski norm of the four-force:
Same answer, also refering to the Minkowski norm of the four-force:
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. Oct 11, 2013

jartsa

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. Oct 11, 2013

Staff: Mentor

Well, maybe you should re-examine your feelings

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. Oct 11, 2013

WannabeNewton

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. Oct 11, 2013

yuiop

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.

Don't think so.

Last edited: Oct 11, 2013
13. Oct 11, 2013

jartsa

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"

Last edited: Oct 11, 2013
14. Oct 12, 2013

jartsa

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. Oct 12, 2013

yuiop

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. Oct 18, 2013

jartsa

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?

Last edited: Oct 18, 2013
17. Oct 18, 2013

Staff: Mentor

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. Oct 18, 2013

WannabeNewton

There is no such quantity. One instead has the 2-form $L = r\wedge p$.

19. Oct 19, 2013

jartsa

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. Oct 19, 2013

Staff: Mentor

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.