Undergrad Weight of a relativistic particle

[PAllen]

I think everything falls into place quite nicely when you abandon the "amount of stuff" heuristic entirely. Mass simply isn't additive, although it's approximately so in the classical limit. No need for a book! (Though Jammer is a nice source for learning about the history of the mass concept.)

(Also, $f=\gamma m a$ isn't right, unless the force is perpendicular to the velocity. If the force is parallel to the velocity, it's $f=\gamma^3 m a$.)

Actually, I think the amount of stuff -> invariant mass is the most natural transition from Newtonian mechanics to SR. Of course, I think only dp/dt definition of force should be used, with p becoming a nonlinear function of velocity for a given mass, consistent with velocity addition and invariant speed. The gamma is related to algebra of velocities not mass. In both Newtonian physics and SR there is no change in mass unless something flows in or out of a body; of course that stuff can be radiation.

A little more on the momentum issue. In Newtonian physics, momentum is additive for both mass (as Newton defined it) and velocity. In SR it can’t be additive for both. So which to give up? Well, velocities themselves are not additive, and there is a limiting speed, so it makes more sense to give up momentum additivity for velocity. This keeps the very useful notion that mass does’t change without a flow of something

 

[SiennaTheGr8]

Actually, I think the amount of stuff -> invariant mass is the most natural transition from Newtonian mechanics to SR.

Agree to disagree, I fear.

When I see a formulation like "mass = amount of [blank]," I infer that I can add up all the [blank] to find the mass. For that to work in SR, the solution is: [blank] = "energy in the rest frame."

"Stuff," like "matter," is a vague word. Unless you define it carefully and unusually broadly, it's likely to conjure thoughts of massive things—precisely the association one must avoid when trying to unlearn the additivity of mass. And anyway, whatever "stuff" is, energy is only a property of it, not the "stuff" itself. An important distinction I think, even though all "stuff" has energy.

 

[PAllen]

Agree to disagree, I fear.

When I see a formulation like "mass = amount of [blank]," I infer that I can add up all the [blank] to find the mass. For that to work in SR, the solution is: [blank] = "energy in the rest frame."

"Stuff," like "matter," is a vague word. Unless you define it carefully and unusually broadly, it's likely to conjure thoughts of massive things—precisely the association one must avoid when trying to unlearn the additivity of mass. And anyway, whatever "stuff" is, energy is only a property of it, not the "stuff" itself. An important distinction I think, even though all "stuff" has energy.

Well, let’s play this game. I add two masses in rest frame, and I always get 2m for the result mass. The case of mass defect arises when the combined mass emits radiation to fall into a ground state. Then mass has changed due to a flow of radiation. Similarly for absorption of radiation. If you use natural units, there is no artificial conversion factor to worry about. The only new notion beyond Newtonian is that captured radiation has mass. [edit: you also need the notion of conversion of KE to mass when captured, which actually subsumes the radiation case]

 

[PeterDonis]

"Stuff," like "matter," is a vague word.

Yes, but you can make it precise by using the stress-energy tensor, which is the proper relativistic generalization of that meaning of the word "mass". In other words, to model "amount of stuff" in full generality in relativity, a scalar is not sufficient; you need a second-rank tensor.

 

[PAllen]

Yet another reason to relate Newton’s mass to invariant mass is that of the quantities energy, momentum, force and mass, the only one that is frame independent per Newton is mass. Similarly in SR if invariant mass is simply called mass.

 

[bhobba]

Peter is correct, although the most elegant definition of rest mass was given by Vanhees using the Poincare group. I am however probably far too influenced by Landau and somewhat old fashioned - I use the Lagrangian. In SR we expect the equations of physics to be the same in all frames - otherwise you can tell what frame you are in in violation of the POR. If we want to write the action of a free particle in Lagrangian form is should be, from the definition of Lagrangian, of the form ∫L dt. Now we impose the POR on it and require it to be invariant. For dt we use dτ where τ is the proper time and so is invariant. Lagrangian's normally depend of position and velocity. So we will use the same trick we did with proper time - write it in the frame its at rest - ie a frame attached to the particle - in the rest frame it has no velocity (the rest frame is the frame attached to the particle - so obliviously it has no velocity, proper time is the time measured by a clock at least conceptually attached to the particle) and since inertial frames are homogeneous and isotropic it can't depend on position or direction - so L is just a constant - defined as -m and m is called, by definition, the rest mass. The rest mass naturally comes out of the POR and trying to find the Lagrangian of a free particle. From that Lagrangian you can derive all sorts of things from Nothers theorem - its relativistic momentum, energy etc. You also get from it's energy the famous E=mc^2 if its at rest.

The point is to get that invariant Lagrangian you have to go to a frame where it is at rest - and from that you get the rest mass - it the natural mass to use in SR.

Thanks
Bill

 

[vanhees71]

Eventually, in GR, a plethora of masses happens anyway, but for SR, one can get away with only using one sort of mass. Invariant or rest mass is an excellent choice, and the one I use because it's the way I learned it and leads to less errors on my part. It's also the choice of most science advisers here.

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).

 

[SiennaTheGr8]

Well, let’s play this game. I add two masses in rest frame, and I always get 2m for the result mass. ...

But only if they're at rest relative to one another and if there's no potential energy associated with their relative positions. If the kinetic- and potential-energy contributions in the rest frame are non-zero but small enough to neglect, then the mass of the system is approximately equal to 2m.

... The case of mass defect arises when the combined mass emits radiation to fall into a ground state. Then mass has changed due to a flow of radiation.

Yes, but the energy radiated away didn't come from the masses of the system's constituents. That's my point. The mass of the system wasn't 2m—it was 2m plus the internal kinetic- and potential-energy contributions (some of which flows out of the system here).

 

[vanhees71]

But as a logical consequence, one has to abandon the idea that F=ma.

I couldn't agree more! The irony is that Newton got it right in the very first attempt, discovering at the same time three most important things:

(a) Newtonian mechanics (guess, why it's called Newtonian!)
(b) calculus as the adequate language to express his ideas (although Leibniz's approach was much more powerful and thus prevailed)
(c) theoretical physics as a way to bring order into the empirical facts, upon which modern physics is based till today

So indeed, he made the statement (put in modern terms) that
$$\frac{\mathrm{d}}{\mathrm{d} t} \vec{p}=\vec{F},$$
which is $\vec{F}=m \vec{a}$ for the special case of a "point particle" (i.e., a finite macroscopic body whose spatial extension is negligible under the circumstances the point-particle approximation is avlid) of constant mass.

What we also know is that Newtonian mechanics is a good approximation to relativistic mechanics whenever the speed of the particles is small compared to the speed of light. That's why it is very clear that the correct covariant description is
$$\frac{\mathrm{d}}{\mathrm{d} \tau} p^{\mu} = K^{\mu},$$
where
$$p^{\mu}=m \frac{\mathrm{d}}{\mathrm{d} \tau} x^{\mu},$$
where $\tau$ is the proper time of the particle (Minkowski scalar), $p^{\mu}$ the four-momentum components (Minkowski-vector components), and $K^{\mu}$ the Minkowski-force (Minkowski-vector) components, and $m$ is the invariant mass (scalar) of the "particle".

Since $\mathrm{d} \tau$ is the local time increment in the local rest frame of the particle, it's very clear that $m$ is the same quantity as appears in Newton's theory, i.e., in special relativity the invariant mass (and imho only the invariant) mass is to be identified with Newton's notion of (inertial) mass.

For more details on relativistic mechanics see my FAQ

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Cpht. 2

 

[pervect]

Yes, but you can make it precise by using the stress-energy tensor, which is the proper relativistic generalization of that meaning of the word "mass". In other words, to model "amount of stuff" in full generality in relativity, a scalar is not sufficient; you need a second-rank tensor.

I was sort of dancing around the point earlier, but I agree. "Stuff" isn't just characterized by a single number in special relativity, it's characterized by the stress-energy tensor. Energy (which may also be familiar to some readers by its other name, relativistic mass) is _part_ of the stress-energy tensor, but one also needs to include the other parts of the tensor as well. The other parts are stress (as one might guess from the name), and momentum.

One needs to understand the SR frame transformation laws and tensors to fully appreciate this point, which is one reason I danced around the issue until now. The transformation laws of SR, the Lorentz transforms, are I-level topics, and tensors are A-level.

The ratio of accleration / force in a stressed rod with a relativistic velocity is a concrete example for the need for the stress energy tensor in SR. Rindler has a treatment of it in his book <<amazon link>>, as I recall. (I don't own the book, but the topic came up in some by now ancient discussions on PF, and I borrowed it from the library).

One might not need an extreme amount of knowledge of tensors to follow Rindler's introductory treatment, but I don't think one could get much out of the book that I haven't said already without knowing how to represent a stressed rod mathematically. It is still useful to cite a source for the skeptical, and/or those few readers who might want to check the original source or find out more information.

Note that knowing the mass of the rod doesn't tell one what the stress in the rod is, so this result shows that in special relativity, finding the ratio of acceleration/applied force can depend on things that are not "mass" when one deals with objects that are not point particles.

When one realizes that "stuff" is represented by the stress-energy tensor, rather than mass, then it's not at all surprising that it's the stress-energy tensor that is the source of gravity in Einstein's field equations.

 

[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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