I Weight of a relativistic particle

[Sorcerer]

So I can agree with your statement "if you accelerate away from an object, creating a relative velocity between you and the object, how could that create more "stuff""

except that we are still dealing with inertial frames, not acceleration, but I get your meaning.

Now gamma is the Lorentz factor which contains the velocity v. When v gets large then gamma gets large so we have to conclude that defining mass as gamma.mass implies that it's mass varies according to its velocity relative to something.

Now since its gamma.mass is dependent on its relative velocity (to something) why do we "deprecate" (disapprove of) the term relativistic mass. Doesn't the Lorentz factor tell us that it is relative?

Free two cents:

There is one "new" definition that is still used today, and it's momentum. Momentum was redefined by Einstein so that it would still be conserved in special relativity. Instead of wasting time with archaic definitions of mass, just call mass m, and when speed comes into play, consider the relativistic momentum γmv instead. Then you don't have to deal with all this confusion.

 

[vanhees71]

I am not confusing mass with weight. Weight is a measure of the force of gravity. Mass is a measure of the amount of "stuff" in the object, as we know, so gravity is not a concern.

So help me to understand why I can define mass as "anything I want it to be". I understand i can use various units, of course, but stuff is still stuff, isn't it?

First of all, if you want to really discuss gravity in relativity you must use General Relativity as the best established theory describing the gravitational interaction yet.

For mass, of course SR is sufficient, and it's "defined" very clearly as a Casimir operator of the Poincare group. With the total four-momentum of your system, it's given by
$$m^2 = \frac{1}{c^2} P_{\mu} P^{\mu},$$
and since $(P^{\mu})$ is a four-vector it is a scalar. In the center-momentum frame, where $\vec{P}=0$ you have, because of $P^0=P_0=E_{\text{CM}}/c$, you get the famous formula
$$E_{\text{CM}}=m c^2,$$
in its modern form, which is due to Einstein from the very beginning, where $E_{\text{CM}}$ is the energy of the system in the center-momentum frame. The less precise even more famous formula $E=m c^2$ is misleading, because it's less precise ;-)).

To answer your question in #12 one last time: Since 1908, when Minkowski gave the most convenient mathematical formulation of SR, we rather use quantities that have a well-defined and simple behavior under Lorentz transformations, and thus we define mass to be a scalar (as by the way we also define temperature and chemical potential in thermodynamics/statistical physics to be scalars).

 

[bhobba]

There is one "new" definition that is still used today, and it's momentum. Momentum was redefined by Einstein so that it would still be conserved in special relativity. Instead of wasting time with archaic definitions of mass, just call mass m, and when speed comes into play, consider the relativistic momentum γmv instead. Then you don't have to deal with all this confusion.

Actually there is this thing called Noethers theorem that defines momentum, energy etc etc, and not only defines it but shows why its conserved. The definition is the same in classical or relativistic mechanics, EM, - even quantum mechanics. Unfortunately not in GR - but that is a whole new thread. Its what the greatest mathematician you probably never heard of, Emmy Noether, sorted out for Hilbert and Einstein that they could not:
https://www.physicsforums.com/threa...00-years-since-emmy-noethers-theorems.939388/

Thanks
Bill

 

[vanhees71]

What do you mean by "Unfortunately not in GR"? Also there Noether's theorem of course applies, and her papers about the relation between symmetries and conservation laws were motivated by understand the question under which circumstances energy and momentum and the other conservation laws known from Galilei and Minkowski spacetime are definable in GR too. Indeed, in GR in general energy and momentum are not conserved, but thanks to Noether it's very clear why!

 

[bhobba]

Indeed, in GR in general energy and momentum are not conserved, but thanks to Noether it's very clear why!

That's exactly what I meant - sorry for any confusion.

Gravity, being space-time curvature, means, for example you don't have time translation symmetry hence the theorem does not apply - you can't define energy as the conserved quantity related to time translation invarience.

I am sure Vanhees knows this, but for those that do not John Baez wrote a nice article about the issue:
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html

We had a nice discussion a while ago now on what is energy. I maintained it was the conserved charge related to time translational symmetry as per Noether. Some agreed with me, others were eventually converted, but others were not convinced. It explained exactly what this thing is when physicists, engineers etc talk about and use it. It makes E=MC^2 trivial from the free particle Lagrangian (an elegant approach - but yours related to the Ponicare group in the most elegant of all - strangely though I still go for the Lagrangian view probably Landau's influence since that's how he does it). Feynman talked about it like a game where you search about to find something you think is lost. Of course true, and very insightful as you would expect from Feynman - but avoids - what exactly is it. That's where the definition as Noether charge from time translation symmetry comes in. But you face the issue of GR and it's lack of global time transnational symmetry - you can't define it that way. Locally, since you can always find local inertial coordinates no problem, but globally its a mess from this viewpoint.

Thanks
Bill

 

[pervect]

Relativistic mass is relative, as are the longitudinal mass and transverse mass. Rest mass (aka invariant mass) is not. They are different concepts. The word "mass", unqualified could mean any of these.

Having a plethora of things all called some variant on "mass" turns out to confuse people.

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.

 

[Herman Trivilino]

The point I'm trying to make is that the response of such a scale would depend on the velocity of the truck.

Yes, but it's an issue of what to do with that dependence.

The sort of weight one measures with a scale isn't the sort of weight we use in commerce.

Sure it is! Still the most precise way to legally weigh things is with a mechanical balance. Take a case where the legal weight really matters, the buying and selling of gold. They're not so silly as to buy and sell by a quantity that changes with location, they measure and use mass. Calling it weight, as is the legal practice.

The motto of the Toledo Scale Company: NO SPRINGS, HONEST WEIGHT. They're not measuring a force. They're measuring mass.

I'm not sure why the physics definition is what it is, but that is my understanding of the common usage of weight as it is used in physics and engineering, where we measure weight (and force) in Newtons. I'm not sure what papers you refer to that argue about issues about how to define the force. At the most basic level, I would treat the problem as a point particle

By far the two most common definitions of weight as a force each define a particle of mass $m$ as having a weight $mg$. Where they differ is in how they define $g$. I shall use the terminology "gravitational field strength" to refer to one, and "free fall acceleration" to refer to the other. In the literature and in the introductory textbooks the phrase "acceleration due to gravity" is used instead of one or the other of the above, inconsistently. While the terminology used to describe each of these two definitions differ, they are mentioned in virtually every introductory physics textbook.

I'm not aware of any issues with defining the force on a point particle, if you think you have some references that call this into question, I'd like to see them if you think the point is worth discussing.

I believe it is worth discussing. I think the pedagogical confusion is on par with the one surrounding the relativistic mass concept. Just as a pedagogical reform addressed that issue in recent decades, a similar reform is needed here.

The implications of this discrepancy in definitions is well-explored by these two letters to the editor, one by Mario Iona and the other Anthony P. French, each appearing in the Feb 1995 issue of AJP. (Am. J. Phys, Vol 63, pp.105 - 106). These were letters written in response to a previous letter that argued about objects in free fall, and whether or not they have weight. One of the two force definitions mentioned above assigns a nonzero weight to objects in free fall, the other assigns a weight of zero. The objects could be a common toy where a penny in free fall is caught in a tube, or an astronaut aboard the ISS.

Using the force definition where I refer to $g$ as the free fall acceleration, the value varies at sea level from about 9.78 m/s² at the equator to 9.83 m/s² at the poles. Two-thirds of that difference, 0.034 m/s², is due to Earth's spin, the remaining one-third or 0.017 m/s² is due to the fact that Earth is not a sphere and is wider at the equator. In this case the quantity $mg$ is called the weight, or as some textbook authors call it, the apparent weight.

In the other force definition, the one where I refer to $g$ as the gravitational field strength, its value only varies from about 9.81 m/s² at the equator to 9.83 m/s² at the poles. In this case the quantity $mg$ is the gravitational force, also called true weight (by those who use apparent weight in the way mentioned above). Among those authors who call it the true weight are those who use a value of 9.80 m/s² for $g$. Befuddling!

For those considering this issue solved merely by a "clear" explanation of the distinction, witness the difficulty students have with the concept of weight. I've come to believe that if we want students to understand weight as a force (as opposed to being equivalent to mass) we must have coherent terminology that's consistent and clear, with one definition of weight as "a quantity of the same nature as a force" where $g$ is called the free fall acceleration. This distinguishes it from the gravitational force $mg$ where $g$ is the gravitational field strength.

 

[Herman Trivilino]

Mass is a measure of the amount of "stuff" in the object, as we know,

No, it's not.

So help me to understand why I can define mass as "anything I want it to be". I understand i can use various units, of course, but stuff is still stuff, isn't it?

You can't use mass as a measure of the amount of stuff. Because the energy of the constituents of a piece of stuff contributes to the mass of that piece of stuff.

 

[pervect]

Newton famously defined mass as "the quantity of material" in his Principa, which is where I believe the idea that mass is a measure of the "amount of stuff" comes from. This is not entirely true in relativity, but the concept in special relativity that comes the closest to respecting this essentially philsophical idea is the invariant, or rest, mass. If one changes the motion of a single particle, one doesn't change the amount of "stuff" in it, it does not change the internal structure of the particle, so one shouldn't expect the mass to change. This is particularly clear when one changes the motion of a particle by changing one's viewpoint, i.e. changing the frame of reference of the observer. According to one observer the particle is stationary. According to a different observer, the exact same particle is moving. If mass is to be a measure on the amount of "stuff", then the mass of the particle should be the same for both observers.

But as a logical consequence, one has to abandon the idea that F=ma. Which also comes from Newtonian mechanics, and is not a relativistic equation. In order to learn relativity, one has to at some point realize that it's a different theory than Newtonian mechanics, a theory that one has to learn, which takes study. There are multiple ways of teaching SR, the modern way is to keep (as much as one can) the idea that the mass of a point particle is "the amount of stuff", and doesn't depend on how the particle moves. Then one winds up modifying the famous Newtonian equation F=ma to $F=\gamma m a$. The Newtonian equation F = dp/dt survives the change in paradigm, so if one wants to stress the parallels between SR and Newtonian theory rather than the differences, it might be useful to note that F=dp/dt remains the same.

There are more adjustments that need to be made further down the road when one deals with objects that aren't just point particles. This gets rather involved, though it leads to the eventual weakening of the idea that mass can be just the "amount of stuff". I know that Max Jammer discussed some of this in his book(s), of which I've read one, (https://www.amazon.com/dp/0486299988/?tag=pfamazon01-20) and not the other ( https://press.princeton.edu/titles/6885.html) but I don't recall what I did read well enough to give a really good summary. I'll just note that there's enough material there for a book, even without trying to tackle GR. (And I wouldn't recommend trying to tackle the concept of mass in GR without first understanding the basics at a mathematical level.)

 

[SiennaTheGr8]

Newton famously defined mass as "the quantity of material" in his Principa, which is where I believe the idea that mass is a measure of the "amount of stuff" comes from. This is not entirely true in relativity, but the concept in special relativity that comes the closest to respecting this essentially philsophical idea is the invariant, or rest, mass. If one changes the motion of a single particle, one doesn't change the amount of "stuff" in it, it does not change the internal structure of the particle, so one shouldn't expect the mass to change. This is particularly clear when one changes the motion of a particle by changing one's viewpoint, i.e. changing the frame of reference of the observer. According to one observer the particle is stationary. According to a different observer, the exact same particle is moving. If mass is to be a measure on the amount of "stuff", then the mass of the particle should be the same for both observers.

But as a logical consequence, one has to abandon the idea that F=ma. Which also comes from Newtonian mechanics, and is not a relativistic equation. In order to learn relativity, one has to at some point realize that it's a different theory than Newtonian mechanics, a theory that one has to learn, which takes study. There are multiple ways of teaching SR, the modern way is to keep (as much as one can) the idea that the mass of a point particle is "the amount of stuff", and doesn't depend on how the particle moves. Then one winds up modifying the famous Newtonian equation F=ma to $F=\gamma m a$. The Newtonian equation F = dp/dt survives the change in paradigm, so if one wants to stress the parallels between SR and Newtonian theory rather than the differences, it might be useful to note that F=dp/dt remains the same.

There are more adjustments that need to be made further down the road when one deals with objects that aren't just point particles. This gets rather involved, though it leads to the eventual weakening of the idea that mass can be just the "amount of stuff". I know that Max Jammer discussed some of this in his book(s), of which I've read one, (https://www.amazon.com/dp/0486299988/?tag=pfamazon01-20) and not the other ( https://press.princeton.edu/titles/6885.html) but I don't recall what I did read well enough to give a really good summary. I'll just note that there's enough material there for a book, even without trying to tackle GR. (And I wouldn't recommend trying to tackle the concept of mass in GR without first understanding the basics at a mathematical level.)

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 $f=\gamma^3 m a$.)

 

[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).

--
lightarrow

 

[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).

--
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!

 

[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?

--

 

[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.

--
lightarrow

 

[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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