# The Relativistic Figure Skater

1. Sep 1, 2015

### AVentura

Picture a large ring floating in space. It is rotating with angular momentum L. However, its radius is so large that it has essentially no tangential velocity or momentum and therefore no kinetic energy. The total energy is mc^2, with m being the invariant mass.

It then begins to draw itself in by applying a force over a distance along its circumference. (picture strong stretched springs being allowed to pull in) Being a ring, the net force is inwards towards its center. As it reduces its radius the tangential momentum increases to conserve angular momentum, like a figure skater.

It now seems to have kinetic energy, but this is a closed system. The relativistic energy/momentum equation seems to have a problem:

E^2 = (mc^2)^2 + (pc)^2

There is only one variable, p. Unless we want to entertain the idea that m can change.

What gives?

2. Sep 1, 2015

### Geofleur

Where is the energy that was stored in the springs included?

3. Sep 1, 2015

### AVentura

I want to say that when they were stretched they had more mass. And that this mass was converted into kinetic energy. It seems logical.

4. Sep 1, 2015

### Geofleur

It's logical, but the problem is that the equation only contains invariant rest masses. I have always seen that equation derived assuming the absence of external forces. But each element of the loop has an external force on it, which can be lumped into a potential function. It seems the final equation really does need an extra term in it.

5. Sep 1, 2015

### AVentura

Just to be clear, there are no external forces. I realize you are talking about what the individual elements feel from the springs between them.

The potential function is pv/r, integrated over r inwards. Using L=pr this can be converted to v, integrated over p increasing (just like an object being pushed linearly)

6. Sep 1, 2015

### Geofleur

OK, so we enlarge the system boundary to include both the loop and the springs. Since we can write $E = mc^2$, the spring energy must be accounted for in increased $m$, like we want to say. Then we can combine this equation with our earlier one and get $m^2 c^4 = m_0^2 c^4 + p^2 c^2$ or, rearranging, $c^4(m^2-m_0^2) = p^2 c^2$. In the beginning the energy is on the left, and in the end, on the right. Well, it's not entirely clear, since $m$ can be large due to the increased velocities...

7. Sep 1, 2015

### AVentura

Before moving, the elements with stretched springs were the invariant mass. It was all at rest. It seems like you just have a new "invariant" mass, that is missing the energy from the springs.

8. Sep 1, 2015

### Geofleur

I'm not sure I understand. If we use $E = mc^2$ to describe the energy of a spring then, if you stretch the spring, the $m$ in this equation increases. It's bigger than it was, even if the spring is not moving. I am suggesting that the stretching energy is included in the $E^2$ term of the energy-momentum relation that we started with.

9. Sep 1, 2015

### AVentura

We are saying the same thing. (m1c^2)^2 = (m2c^2)^2 + (pc)^2

m1 > m2, right?

10. Sep 1, 2015

### Geofleur

Yes. I am saying, further, that as the spring becomes unstretched, m1 decreases. This decrease in the spring's mass goes into increasing the velocity of the loop.

11. Sep 1, 2015

### Orodruin

Staff Emeritus

Note that the invariant mass of the system as a whole does not change. The overall momentum is still zero.

12. Sep 1, 2015

### AVentura

Orodruin, is there kinetic energy?

13. Sep 1, 2015

### Orodruin

Staff Emeritus
This depends what you mean by kinetic energy. If you mean energy due to the motion of the system as a whole, then no. If you mean energy due to the motion of individual pieces of the system, then yes - this contributes to the total invariant mass of the system.

14. Sep 1, 2015

### AVentura

So looking at the individual pieces of the system, is this correct?:

(mc^2)^2 = (mc^2 - KE)^2 + (pc)^2

with m being the mass of the ring before the motion started.

I am asking because when I use this relationship between p and KE to integrate the potential function KE=∫ v dp I get nonsense.

Last edited: Sep 1, 2015
15. Sep 1, 2015

### Orodruin

Staff Emeritus
No, you are now looking at the entire system. The overall momentum of the entire system is zero. If you are looking at individual parts of the system, there are forces being exchanged between the different parts of the system and energy and momentum in each part is not conserved.

16. Sep 1, 2015

### AVentura

OK. So the energy/momentum relation cannot be used in this way. Thanks

17. Sep 1, 2015

### Staff: Mentor

It can, but you have to include potential energy in the rest mass of the system. The initial state is a ring with total energy $E$ composed of ring rest mass $m$ and potential energy $U$ stored in the springs, i.e., $E = m + U$. The final state is a ring with total energy $E$ composed of ring rest mass $m$ and kinetic energy $K$, i.e., $E = m + K$. So what has happened is that the potential energy $U$ stored in the springs has been converted into kinetic energy $K$ of the speeded up rotation. Total linear momentum is still zero and angular momentum is still $L$.

If you want to look at the springs and the ring as separate systems, then you start out with a ring of total energy $m$ and no kinetic energy, and springs with total energy $U$ and no kinetic energy. You end up with a ring of total energy $E = m + K = m + U$ and springs with total energy zero (we are assuming they have negligible rest mass). You could then calculate the final rotational velocity of the ring using $E^2 - p^2 = ( m + K )^2 - p^2 = m^2$.

Last edited: Sep 1, 2015
18. Sep 1, 2015

### pervect

Staff Emeritus
Let's replace the large ring with a couple of balls on a string, the balls being 180 degrees apart in their circular orbit, and held in orbit by tension in the string.

Lets tug on the string to pull the balls in closer to the center and think about what happens. Additionally, we'll gloss over the question of the contribution of the string towards the linear and angular momentum of the system by assuming it's negligible, because it's much easier that way, even though this assumption is a bit more suspect than it initially appears, especially if we want to do a really good relativistic analysis.

Now, for each ball, L = 2 p x r, L being the total angular momentum, p being the linear momentum of each ball, r being the radius of the orbit, and the factor of 2 coming from the fact that we have two balls. This works for special relativity and newtonian theory.

Since L is constant p * r is constant, but as r decreases, p must go up to keep L constant. This contradicts the notion that the force is purely central, at least I think it does, because a force at right angles to the motion shouldn't make the linear momentum in a direction perpendicular to the radius go up. So the assumption of a purely central forece is probably wrong, though I don't have any really simple, clear explanation for why it's wrong at this point. The simplest approach I can think of is to introduce a co-rotating frame and consider the coriolis force as being the suspect for how p goes up, but to do this relativistically we'd need to introduce a relativistic rotating frame with an appropriate metric, which is both beyond what I would want to do, and most likely not helpful to the OP.

Last edited: Sep 1, 2015
19. Sep 2, 2015

### Orodruin

Staff Emeritus
This is not correct. The point is that the direction which is perpendicular to the string changes all the time. In the case where you are reeling in the masses, the force is larger than required to simply keep the masses moving in a circle. As a result, the acceleration in the inward radial direction will result in a larger acceleration than for the circular movement. As the masses go on moving, this larger acceleration will result in a larger amount of radial momentum transfered to the angular direction due to the change in what is the radial direction as the masses progress.

Mathematically, this is captured by the Christoffel symbols in polar coordinates.

20. Sep 2, 2015

### harrylin

As clarified by others, potential energy of the springs has been converted into kinetic energy of the ring.
Consequently also the ring's mass has increased (as the ring's linear speed is zero, there is no reason to specify "invariant mass").

21. Sep 2, 2015

### A.T.

If r decreases, then a central (or radial) force isn't at right angles to the motion.

22. Sep 2, 2015

### pervect

Staff Emeritus
That's a good observation, and leads to a simple explanation as well. Now that you've mentioned it, it's obvious. It also leads to a much simpler analysis than my initial attempt. If we keep the force purely central, rather than using a "string" model which does not guarantee a central force (as the string won't necessarily be "straight"), we get a much simpler model. We can consider the orbit that the mass follows under this purely central force by the well known form of Newton's equations in polar coordinates:

$r^2 \dot{\theta} = L/m \quad \ddot{r} - r \dot{\theta}^2 = f(r)/m$

where f(r) is some radial force. We could derive the above equations by Christoffel symbols too, but it's more elementary to appeal to Newton's equations for a central force. Then we can observe what you said - the force always remains central, but does not always remain at right angles to the motion of the particle, hence the central force does do work on the particle, causing it to gain/loose energy.

Remaining to do: write the relativistic form of the above equations. I'd tend to start by writing the (relativistic) Lagrangian. I'm not sure if the OP would be interested in a fuller exposition using this approach though.

23. Sep 2, 2015

### AVentura

Thank you everyone. I was just putting the KE in the wrong place in the energy/momentum relation.

With (mc^2+KE)^2 = (mc^2)^2 + (pc)^2
L = p*r
KE = ∫ pv/r (-dr) = ∫ v dp
Solves for the correct KE(v)

Seems obvious now. E=mc^2+KE. It's just that the KE is a KE to be.

Last edited: Sep 2, 2015
24. Sep 5, 2015

### Jess H. Brewer

Your second sentence makes a false assertion that because the radius is large there is no kinetic energy. If there is angular momentum, there is kinetic energy. I didn't get past that error.

25. Sep 5, 2015

### Staff: Mentor

Strictly speaking, yes, this is true, but if the radius is large enough, the kinetic energy can be so small that it is negligible. That's what the OP was assuming.