Kinetic energy of a spinning body

In summary, the problem is that the definition of kinetic energy in special relativity depends on the frame of reference in which the calculation is made, and it's difficult to determine the invariant mass of a rotating body without knowing its spinning speed.
  • #1
lightarrow
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Sorry if this question has already been addressed.
I know that, at least for a point mass or a rigid body traveling 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 symmetry 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). In fact 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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  • #2
lightarrow said:
Infact E = Mc2 in this case, and this means kinetic energy would be zero according to that definition.

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".

lightarrow said:
If then we don't neglect the energy due to internal elastic tensions Eel, should we count it as kinetic energy or not?

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.
 
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  • #3
PeterDonis said:
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.
Thanks.
However, can I say "what a pity!" My definition of kinetic energy seemed more useful to me :smile:

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  • #4
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
lightarrow said:
My definition of kinetic energy seemed more useful to me

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
PeterDonis said:
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".
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 traveling 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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  • #7
DrGreg said:
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.
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...
A more useful decomposition is $$
E = \sqrt{(mc^2)^2 + |\textbf{p}c|^2}
$$
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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  • #8
lightarrow said:
However, can I say "what a pity!" My definition of kinetic energy seemed more useful to me :smile:

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
lightarrow said:
Stopping the body?

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

lightarrow said:
even for a body traveling in a straight line in an inertial frame would be difficult to directly measure its invariant mass

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
Was there any interest by the OP in the "gravitational field" of the spinning body, or is this just about the energy?
 
  • #11
pervect said:
Was there any interest by the OP in the "gravitational field" of the spinning body, or is this just about the energy?
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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  • #12
PeterDonis said:
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.)
How do you measure E and p without altering its motion?

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  • #13
lightarrow said:
How do you measure E and p without altering its motion?

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
PeterDonis said:
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##.)
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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  • #15
lightarrow said:
And about p?

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.
 
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  • #16
PeterDonis said:
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.
Right. Does an equation similar similar to E2 = p2c2 + m2c4 but involving angular momentum instead of p, exists in SR?

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  • #17
lightarrow said:
Does an equation similar similar to E2 = p2c2 + m2c4 but involving angular momentum instead of p, exists in SR?

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
PeterDonis said:
But this doesn't work in 4-d spacetime, because a pseudovector has 4 independent components, but an antisymmetric tensor has 6.)
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
PeterDonis said:
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.)
Thanks, anyway I realize now that such a "similar equation" would have little sense, since E = mc2 here.

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  • #20
Nugatory said:
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.
Isn't possible to use suitable synchronizations, e.g. using a cylindrical pulse of light from the centre?

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  • #22
lightarrow said:
Isn't possible to use suitable synchronizations, e.g. using a cylindrical pulse of light from the centre?

Sure, but what's the physical significance of the quantity that you've calculated that way? Why should it be any more meaningful than the quantity you calculate using some other simultaneity convention?

This is why PeterDonis is telling you that you need a rank-two tensor instead of a scalar here.
 
  • #23
PeterDonis said:
No, it doesn't. It just requires a choice of "spacetime origin", i.e., a point in spacetime about which the relativistic angular momentum is evaluated.
Ah - so the extra independent components come in because two equations that are unrelated in Newtonian terms are entwined in relativistic terms. I was having trouble seeing why a relativistic description required so much more information than a Newonian one, but it's because this one equation is describing all the stuff described by two equations in the pre-relativistic model. c.f. 4-momentum conservation versus Newtonian 3-momentum conservation and energy conservation.
 
  • #24
Nugatory said:
Sure, but what's the physical significance of the quantity that you've calculated that way? Why should it be any more meaningful than the quantity you calculate using some other simultaneity convention?

This is why PeterDonis is telling you that you need a rank-two tensor instead of a scalar here.
Thanks.

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  • #25
Ibix said:
the extra independent components come in because two equations that are unrelated in Newtonian terms are entwined in relativistic terms.

Yes.
 

1. What is the definition of kinetic energy of a spinning body?

The kinetic energy of a spinning body is the energy that an object possesses due to its motion. It is a type of energy that is associated with the movement of an object.

2. How is the kinetic energy of a spinning body calculated?

The kinetic energy of a spinning body can be calculated using the formula: KE = 1/2 * I * w^2, where KE is kinetic energy, I is the moment of inertia, and w is the angular velocity of the body.

3. What factors affect the kinetic energy of a spinning body?

The kinetic energy of a spinning body is affected by its mass, moment of inertia, and angular velocity. The greater the mass and the faster the angular velocity, the higher the kinetic energy will be. The shape and distribution of mass of the object also affect its moment of inertia, thus impacting its kinetic energy.

4. How does the kinetic energy of a spinning body relate to its rotational motion?

The kinetic energy of a spinning body is directly related to its rotational motion. As the body spins faster, its angular velocity increases, resulting in a higher kinetic energy. Similarly, if the body's rotational speed decreases, its kinetic energy will also decrease.

5. Can the kinetic energy of a spinning body be increased or decreased?

Yes, the kinetic energy of a spinning body can be increased or decreased by changing its angular velocity. This can be done by applying an external force, such as pushing or pulling, to the body. The direction of the force will determine whether the kinetic energy increases or decreases.

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