# Kinetic energy of a spinning body

1. Nov 20, 2015

### lightarrow

I know that, at least for a point mass or a rigid body travelling at constant velocity, in special relativity kinetic energy T is defined as: T = E - Mc2 where E is the body's total energy and M its invariant mass.

I have a problem in undestanding how to define T in the case of a spinning body. I assume the body has cylindrical simmetry around its axis of rotation, that its centre of mass is still, that it has uniform density even during spinning and that we could neglect internal tensions.

I have problems in doing that because M now depends on its spinning speed (it also depends on "how" it speeds because we can't simply make the assumption of rigid body, in general, but let's forget these complications for the moment). Infact E = Mc2 in this case, and this means kinetic energy would be zero according to that definition. Presumably we should instead define kinetic energy as: T = E - mc2 where m is the body's mass when it's not spinning at all. Does it make sense?

If then we don't neglect the energy due to internal elastic tensions Eel, should we count it as kinetic energy or not? If not, should we then say that T = E - mc2 - Eel?

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lightarrow

Last edited: Nov 20, 2015
2. Nov 20, 2015

### Staff: Mentor

The kinetic energy of the object in the center of mass frame would be zero. Kinetic energy (and total energy, since it includes kinetic energy) is frame-dependent. In the CoM frame, the rotation of the object about an axis passing through its center of mass produces zero net motion, and therefore zero kinetic energy.

More generally, the energy associated with the rotation of an object about an axis passing through its center of mass is "internal energy", and is part of the invariant mass of the object. You are implicitly acknowledging this when you say that "M would depend on its spinning speed".

The same idea applies here; energy due to internal elastic tensions is internal energy, which means it's part of the invariant mass of the object.

3. Nov 21, 2015

### lightarrow

Thanks.
However, can I say "what a pity!" My definition of kinetic energy seemed more useful to me

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lightarrow

4. Nov 21, 2015

### DrGreg

It's perhaps worth pointing out that within relativity, the term "kinetic energy" isn't used all that much; people usually talk about the total energy (or just "energy"). There's usually no need to decompose the energy as kinetic energy + rest energy.

A more useful decomposition is$$E = \sqrt{(mc^2)^2 + |\textbf{p}c|^2}$$

5. Nov 21, 2015

### Staff: Mentor

In what way? Note that, if you have a rotating body, you have no way to directly measure "what its invariant mass would be if it were not rotating".

6. Nov 21, 2015

### lightarrow

Stopping the body? In physics we always are faced with quantities which are not directly measurable but that we infer from other observations, e.g. in astrophysics. Anyway, even for a body travelling in a straight line in an inertial frame would be difficult to directly measure its invariant mass (= what its invariant mass would be if it were at rest, because in this case they are the same).

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lightarrow

7. Nov 21, 2015

### lightarrow

It can be useful if I want to compute the energy I could in theory get from a fast spinning body or from a system of bodies rotating around their centre of mass at high speeds, as in a flying wheel or similar systems. Certainly, it's difficult for the present technologies that we had to use relativity for such objects, but in theory...
But it's not useful for spinning/rotating bodies in their centre of mass frame because p = 0 in this case.

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lightarrow

8. Nov 21, 2015

### Staff: Mentor

It's less useful than it seems. What you're doing is calculating the kinetic energy of each tiny volume element of the wheel, each one moving in its own direction at its own speed, and then adding the contributions from each to get what seems to be total kinetic energy of the wheel. This is a valid approach in classical physics (it's one way to derive the $E=I\omega^2$ law for the rotational energy of a classical mass) but there's a serious problem in relativistic physics: For that sum to be any sort of total energy at a point in time, you have to have calculated the contribution from each element at the same time and that's not easily defined on the rotating disk.

9. Nov 21, 2015

### Staff: Mentor

Well, sure, that would let you directly measure how much energy you could get by stopping the body. But it doesn't help if you want to try to predict that result in advance, without actually doing it.

Not really. Measure its energy and momentum and use the formula $E^2 - p^2 c^2 = m^2 c^4$. (This is just the formula DaleSpam gave, squared and rearranged.)

10. Nov 21, 2015

### pervect

Staff Emeritus
Was there any interest by the OP in the "gravitational field" of the spinning body, or is this just about the energy?

11. Nov 22, 2015

### lightarrow

Just the Energy. I was trying to generalize (1/2)Iw^2 in some simple model of a spinning disk at relativistic speeds.

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lightarrow

12. Nov 22, 2015

### lightarrow

How do you measure E and p without altering its motion?

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lightarrow

13. Nov 22, 2015

### Staff: Mentor

Strictly speaking, you can't. But strictly speaking, you can't measure anything without altering the measured object in some way. But you could, in principle, make a measurement that changed the object's motion very little--you wouldn't have to go to the extreme of stopping it.

(Also, measuring E and p isn't the only option; you could, for example, measure v and p, velocity and momentum, and use $p = m \gamma v$ to calculate $m$.)

14. Nov 22, 2015

### lightarrow

Ok concerning measuring v without altering (significantly) the motion of a macroscopic object (we are talking of this). And about p?

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lightarrow

15. Nov 22, 2015

### Staff: Mentor

More precisely: measure the change in $p$ when bouncing some light off the object, and also measure the change in $v$. You can measure the change in $p$ by measuring the change in momentum of the light and applying conservation of momentum.

16. Nov 23, 2015

### lightarrow

Right. Does an equation similar similar to E2 = p2c2 + m2c4 but involving angular momentum instead of p, exists in SR?

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lightarrow

17. Nov 23, 2015

### Staff: Mentor

Not to my knowledge. Angular momentum is properly represented by an antisymmetric 2-index tensor, not a vector, and I don't know of any simple scalar invariant that corresponds to it. (In ordinary 3-space, we can represent angular momentum as a vector--more precisely, as a pseudovector--because there is a 1-to-1 correspondence between antisymmetric tensors and pseudovectors, since both have three independent components. But this doesn't work in 4-d spacetime, because a pseudovector has 4 independent components, but an antisymmetric tensor has 6.)

18. Nov 23, 2015

### Ibix

What are the degrees of freedom here? In non-relativistic mechanics there are three independent components, two of which correspond to my choice of coordinates and the third to the magnitude of the angular momentum. Obviously the relativistic version needs me to choose a simultaneity convention too. Does that require three parameters (directional derivatives?) to fix it?

19. Nov 23, 2015

### lightarrow

Thanks, anyway I realize now that such a "similar equation" would have little sense, since E = mc2 here.

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lightarrow

20. Nov 23, 2015

### lightarrow

Isn't possible to use suitable synchronizations, e.g. using a cylindrical pulse of light from the centre?

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lightarrow