Weight of a relativistic particle

[lightarrow]

A box with a hot gas will weigh more than a box with the same gas but cold.

But in this case you assume the box at rest in the FOR? If so, it's simply because the mass' box has increased (rest mass = invariant mass = mass).

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[Dale]

But in this case you assume the box at rest in the FOR?

I assume that it is at rest with respect to the scale or balance.

 

[vanhees71]

But in this case you assume the box at rest in the FOR? If so, it's simply because the mass' box has increased (rest mass = invariant mass = mass).

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The mass of a gas is independent of its state of motion. Mass is a Lorentz scalar by definition. You find tons of debates about the fact that old-fashioned ideas on "relativistic velocity-dependent masses" are outdated for about 110 years now!

 

[lightarrow]

Coming back to the OP's question, or at least to some near to it, is what is following correct?
1) SR context.
A point mass of mass m, rotates at constant relativ. speed v around a point at distance r in an inertial frame; to make this motion possible, a constant centripetal force F is applied on the point mass. We know from SR that
F = γma
where a is the centripetal acceleration a = v^2/r. Can I say, as in Newtonian mechanics, that F does a null work on the point mass and so this system's energy is simply γmc^2? So its mass is γm?

2) GR context.
The point mass rotates around the Earth and so there is gravity now. Has the previous result in 1) any relation with the effect here? What could we say?

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[lightarrow]

The mass of a gas is independent of its state of motion. Mass is a Lorentz scalar by definition. You find tons of debates about the fact that old-fashioned ideas on "relativistic velocity-dependent masses" are outdated for about 110 years now!

Certainly. I was just saying (actually I pointed this out to Dale's interlocutor) that there isn't any mistery in the fact the box' weight has increased after heating it: it's simply because it's mass (invariant mass) has increased; no need of relativistic mass sort of things...
And, in case: no, I'm not advocating any "relativistic mass" in any context, on the contrary I could tell you a lot of reasons NOT to use it or at least supporting the fact it's a useless concept.

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[jbriggs444]

So its mass is γm?

This system (the moving particle) has mass m. Period. This mass is equal to its total energy in the frame of reference where it is momentarily at rest. In this frame it is not moving. [Assume units where c=1]

The presence of an external force does not change this.

If you want to close the system by including a central body which is responsible for the centripetal force on the moving particle then we can talk about the mass of that system and include the kinetic energy of the moving particle in the computation.

 

[pervect]

What do you mean by "a plethora of masses happens anyway"? As you correctly say in the next sentence, there's one notion of mass in GR, which is invariant mass, and that's it. It's just a term in the matter-field Lagrangian and appears thus as part of the sources of the gravitational field in the Einstein equation, i.e., the energy-momentum tensor of matter (matter meaning all fields except the gravitational field, which in the Einstein equation occurs on the left-hand side in terms of the Einstein tensor of gravity).

In GR, we have as the most frequently used possibilities for "mass" in the sense I was talking about it, the ADM mass, the Bondi mass, and the Komarr mass. These are not tensors, sometimes they are introduced by pseudotensor methods. Using these pseudotensor methods, mass is sometimes described as the result of integrating some pseudotensor, usually the Landu-Lifschitz pseudotensor, with the appropriate conditions as required to ensure that the result is independent of gauge choice.

I don't think this integral winds up as being a fourth alternative to the above three, but I don't have a solid reference on that point, which would entail demonstrating that it's equal to one of the three.

Ignoring pseudotensors and consdering only tensor quantites, we have the matter field Lagrangian, and the stress-energy tensor as you say. Interpreting the Lagrangian density as some sort of "mass density" is somewhat problematical. Consider the relativistic Lagrangian of a free particle, $-m_0 \,c^2 \, \sqrt{1-v^2/c^2}$, the fact that it's absolute value decreases when we increase velocity (it's a negative quantity which becomes less negative) says to me that we should call it the Lagrangian, not the mass, to avoid confusion. Because of it's sign and it's behavior with velocity, it can't be equivalent either to invariant mass or relativistic mass. Similar issues arise if we consider the Lagrangian density rather than the Lagrangian.

 

[vanhees71]

In GR a free massive particle has the Lagrangian
$$L=-m c^2 \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}{\nu}},$$
where the worldline is parametrized with an arbitrary scalar world parameter. The action is parameter independent, and of course $m$ is the invariant mass of the particle also in GR. In GR there's no way to get along with any kind of "velocity dependent" mass.

All this has nothing to do with the much more subtle problem of the total mass of a composite object like a star, but it's also not needed, because all that counts is the energy-momentum tensor of matter appearing on the right-hand-side of the Einstein-Hilbert equations.

 

[lightarrow]

This system (the moving particle) has mass m. Period.

A disk (or a ring) which has mass m when at rest has mass m'>m (invariant mass) when rotating about its centre. Clearly the system now includes the forces (only the disk's elastic forces if it's homogeneous) needed to maintain it in rotation, it's not only "the set of point masses which are moving about a centre". Did you mean this, isnt'it?

This mass is equal to its total energy in the frame of reference where it is momentarily at rest. In this frame it is not moving. [Assume units where c=1]

But if I write that the body "is rotating" it obviously means that the frame of reference we have to consider is not the one in which the body is not moving...
Anyway, if a point mass of mass = m is rotating at a constant relativistic speed v around a fixed point at distance r, how much it is the centripetal force needed to keep it at that distance?

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[jbriggs444]

A disk (or a ring) which has mass m when at rest has mass m'>m (invariant mass) when rotating about its centre.

There is a difference between a disk and a particle. The disk has significant extent. The particle does not. The pieces of a rotating disk are all in motion relative to any inertial frame in which another piece is at rest. The relative motion means that a rotating disk has more mass than one which is not rotating. And more mass the the sum of its pieces.

Clearly the system now includes the forces (only the disk's elastic forces if it's homogeneous) needed to maintain it in rotation, it's not only "the set of point masses which are moving about a centre".

Forces do not contribute to mass. It is far from clear what relevance there is in lumping such forces in as part of the system.

Anyway, if a point mass of mass = m is rotating at a constant relativistic speed v around a fixed point at distance r, how much it is the centripetal force needed to keep it at that distance?

Why do you ask?

 

[pervect]

A disk (or a ring) which has mass m when at rest has mass m'>m (invariant mass) when rotating about its centre.

This is complicated by the fact that there can't be a rigid rotating disk in special relativity, due to the Ehrenfest paradox. But it is true that if you start out with a disk at rest, and spin it up, this requires energy, and adding energy to the system increases the invariant mass of the system.

In general, what actually happens is the disk deforms as you spin it up, and breaks well before anything approaching relativistic speeds can be reached. We had a thread on this a while back, there is a characteristic velocity at which a tether or hoop will break when you spin it up, it's under 10km/sec for the strongest known materials.

 

[PAllen]

This is complicated by the fact that there can't be a rigid rotating disk in special relativity, due to the Ehrenfest paradox.

I don’t think that is right. There can’t be rigidity while it is increasing its spin, but then it can settle into a Born rigid constant spin state. Any energy added to spin it up will contribute to an increased mass.

 

[pervect]

I don’t think that is right. There can’t be rigidity while it is increasing its spin, but then it can settle into a Born rigid constant spin state. Any energy added to spin it up will contribute to an increased mass.

Yes, it's better to say that a rigid disk cannot change it's rate of rotation.

 

[lightarrow]

There is a difference between a disk and a particle. The disk has significant extent. The particle does not. The pieces of a rotating disk are all in motion relative to any inertial frame in which another piece is at rest.

Ok.

The relative motion means that a rotating disk has more mass than one which is not rotating.

Not only because of relative motion, but also because being its centre of mass stationary and so its momentum p zero, we have:
E² = (cp)² + (mc²)²⇒E = mc².
So its mass is its total energy (divided c²).

And more mass the the sum of its pieces.

Yes, the sum of its "pieces" mass is m, the disk's mass when stationary; its mass when in rotation is m' = E/c² = mc² + Ek + El
El = elastic energy due to disk's deformation
Ek = kinetic energy
Is this correct?

Forces do not contribute to mass.

Yes, I was referring to the elastic energies, I took it for granted, but better being precise, thank you.

It is far from clear what relevance there is in lumping such forces in as part of the system.
...
Why do you ask?

To know your answer, I still haven't understood which is it (my problem,).

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