Why is the relativistic mass a rejected concept?

[kmarinas86]

Note: Hurkyl's post above was a response to a previous version of this post which is now deleted. Formatting has been fixed, and I have added a further argument at the end.

I think I can now reconcile the differences between both sides.

Going back to the formula:

m_0 c^2 = \left( E^2 - \|\mathbf{p}c\|^2\right)^{1/2}

1) We know that different inertial frames will see different values for E and \|\mathbf{p}c\|^2 and yet will see the same value for m_0 c^2.
2) I said that E is not something of the object itself, but rather the maximum value of energy change that may occur of that object when colliding with a body in the same inertial frame of reference as the observer. In other words, I claimed that relativistic mass is not tied to anyone body.
3) So the E in this equation, when distorted by a Lorentz boost, is not the energy content of the object.
4) The relativistic mass is often used in consideration of situations where one accelerates a particle. Such cannot allow the particle to be described as a closed system.
5) Therefore, while those against the validity of relativistic mass focus on the Lorentz boosts of a particle, those for the validity of relativistic mass focus on the proper acceleration of a particle.
6) If one stays away from Lorentz boosts and considers the effect that proper acceleration has on the mass of a charge particle (an open system effect), one realizes that E^2 - \|\mathbf{p}c\|^2 is not constant for this particle. Thus, we have the following relations:
E = m_{accelerated} c^2 = \gamma m_{lab} c^2
\|\mathbf{p}c\| = \left(\left(m_{accelerated}c^2\right)^2 - \left(m_{lab}c^2\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\left(m_{accelerated}c\right)^2 - \left(m_{lab}c\right)^2\right)^{1/2}
Where the lab frame is simply the laboratory frame of reference. One such laboratory frame of reference is the foundation of the CERN Hadron Collider facility. It follows that \|\mathbf{p}\| is simply the relativistic momentum with respect to the lab frame:
\|\mathbf{p}\| = \left(\left(\gamma m_{lab}c\right)^2 - \left(m_{lab}c\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\left(\gamma^2 - 1\right)\left(m_{lab}c\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\frac{1}{1-\frac{v^2}{c^2}} - \frac{1-\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{1-\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\gamma\frac{v}{c}\right)m_{lab}c
\|\mathbf{p}\| = \gamma m_{lab}v

Q.E.D.

 

[kmarinas86]

Note: Hurkyl's post above was a response to a previous version of this post which is now deleted. Formatting has been fixed, and I have added a further argument at the end.

I think I can now reconcile the differences between both sides.

Going back to the formula:

m_0 c^2 = \left( E^2 - \|\mathbf{p}c\|^2\right)^{1/2}

1) We know that different inertial frames will see different values for E and \|\mathbf{p}c\|^2 and yet will see the same value for m_0 c^2.
2) I said that E is not something of the object itself, but rather the maximum value of energy change that may occur of that object when colliding with a body in the same inertial frame of reference as the observer. In other words, I claimed that relativistic mass is not tied to anyone body.
3) So the E in this equation, when distorted by a Lorentz boost, is not the energy content of the object.
4) The relativistic mass is often used in consideration of situations where one accelerates a particle. Such cannot allow the particle to be described as a closed system.
5) Therefore, while those against the validity of relativistic mass focus on the Lorentz boosts of a particle, those for the validity of relativistic mass focus on the proper acceleration of a particle.
6) If one stays away from Lorentz boosts and considers the effect that proper acceleration has on the mass of a charge particle (an open system effect), one realizes that E^2 - \|\mathbf{p}c\|^2 is not constant for this particle. Thus, we have the following relations:
E = m_{accelerated} c^2 = \gamma m_{lab} c^2
\|\mathbf{p}c\| = \left(\left(m_{accelerated}c^2\right)^2 - \left(m_{lab}c^2\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\left(m_{accelerated}c\right)^2 - \left(m_{lab}c\right)^2\right)^{1/2}
Where the lab frame is simply the laboratory frame of reference. One such laboratory frame of reference is the foundation of the CERN Hadron Collider facility. It follows that \|\mathbf{p}\| is simply the relativistic momentum with respect to the lab frame:
\|\mathbf{p}\| = \left(\left(\gamma m_{lab}c\right)^2 - \left(m_{lab}c\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\left(\gamma^2 - 1\right)\left(m_{lab}c\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\frac{1}{1-\frac{v^2}{c^2}} - \frac{1-\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{1-\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\gamma\frac{v}{c}\right)m_{lab}c
\|\mathbf{p}\| = \gamma m_{lab}v

Q.E.D.

Considering the above relations, we have the following equation forms:

x=m_{lab}c^2
y=\left(\gamma-1\right) m_{tab}c^2
z=\gamma m_{lab}c^2

y = \left(z^2 - x^2\right)^{1/2}
x^2 + y^2 = z^2
x + y = z

At first, this would require that, for an indivisible object, that is to say one that lacks any degrees of freedom (i.e. a truly fundamental particle), only one of these conditions may apply:

• z=x, v=0
• z=y, v=c

Any particle that travels neither at v=0 nor v=c is a mass consisting of multiple particles of the fundamental kind. This result is also consistent with the following alternate approach:

\gamma-1 = \left(\gamma^2-1\right)^{1/2}
\left(\gamma^2-1\right)^{1/2} = \gamma\frac{v}{c}
\gamma-1 = \gamma\frac{v}{c}
\gamma\left(1-\frac{1}{\gamma}\right) = \gamma\frac{v}{c}
1-\frac{1}{\gamma} = \frac{v}{c}
1-\frac{v}{c} = \frac{1}{\gamma}
\frac{1}{1-\frac{v}{c}} = \gamma
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
1-\frac{v}{c} = \sqrt{1-\frac{v^2}{c^2}}

As with the above, the only solutions for this are v=0 and v=c.

 

[PAllen]

pervect suggested that the rest mass captured the idea of "amount of stuff". If we compare hot and cold boxes in which the molecules have more and less kinetic energy, will the scale still give only the rest masses?

The invariant mass of the system goes up with particles moving faster. This actually shows the superiority of the invariant mass. For a single particle, invariant mass=rest mass, while relativistic mass is a strictly frame dependent quantity that, for example, has no impact on amount of curvature created. However, for a system of particles, invariant mass comes out higher for the faster moving particles, thus showing, for example, that they do generate more curvature. Invariant mass wins again.

Note: higher invariant mass correlates with higher curvature. Higher relativistic mass correlates with .. nothing in this context.

 

[pervect]

One of the most successful, but most abstract, ideas for defining mass, IMO, is the conserved quantity associated with some time translation symmetry with the additional provision that your physics can be expressed in terms of a "least action" principle.

This is broad enough to include the ADM, Komar, and Bondi masses, with the details of "some sort of time translation symmetry" being different.

I'm not sure if the defintion is broad enough to include the other sorts of mentioned - I'm not terribly familiar with them.

While it's a pretty good definition, I have to admit it's not much of a"motivator".

 

[atyy]

The invariant mass of the system goes up with particles moving faster. This actually shows the superiority of the invariant mass. For a single particle, invariant mass=rest mass, while relativistic mass is a strictly frame dependent quantity that, for example, has no impact on amount of curvature created. However, for a system of particles, invariant mass comes out higher for the faster moving particles, thus showing, for example, that they do generate more curvature. Invariant mass wins again.

Note: higher invariant mass correlates with higher curvature. Higher relativistic mass correlates with .. nothing in this context.

Really? How do you get that?

 

[PAllen]

Really? How do you get that?

Simple. Imagine a fast moving single particle with high relativistic mass (in some frame, obviously). Now transform to a frame where it is at rest. Compute curvature scalar and tensor. The former is invariant, the latter covariant. Thus the fast motion contributes nothing to curvature - as long as it can transformed away. The invariant mass, meanhwhile is unchanged from rest mass in this case - correctly correlating with the irrelavance of the relativistic mass.

Meanwhile, once you have a system of particles, their momentum *does* contribute to the invariant mass, and to the curvature. A key point is that for a system, of particles you can't transform away the relative motion. Maybe a further question is what if you have a system of comoving particles? Now, the invariant mass is unaffected by the motion and by the transform argument, only this invariant mass and not the relativistic mass contributes to curvature.

The upshot: invariant mass correlates meaningfually with curvature generated; relativistic mass is meaningless is this context.

 

[atyy]

Simple. Imagine a fast moving single particle with high relativistic mass (in some frame, obviously). Now transform to a frame where it is at rest. Compute curvature scalar and tensor. The former is invariant, the latter covariant. Thus the fast motion contributes nothing to curvature - as long as it can transformed away. The invariant mass, meanhwhile is unchanged from rest mass in this case - correctly correlating with the irrelavance of the relativistic mass.

Meanwhile, once you have a system of particles, their momentum *does* contribute to the invariant mass, and to the curvature. A key point is that for a system, of particles you can't transform away the relative motion. Maybe a further question is what if you have a system of comoving particles? Now, the invariant mass is unaffected by the motion and by the transform argument, only this invariant mass and not the relativistic mass contributes to curvature.

The upshot: invariant mass correlates meaningfually with curvature generated; relativistic mass is meaningless is this context.

Hmm, interesting argument. If you transform away the motion, then the relativistic mass should be the rest mass, which should contribute to the invariant mass. So can we not say the relativistic mass still contributes to the curvature?

 

[PAllen]

Hmm, interesting argument. If you transform away the motion, then the relativistic mass should be the rest mass, which should contribute to the invariant mass. So can we not say the relativistic mass still contributes to the curvature?

If there is any point in used relativistic mass it is when it is different from rest mass. Then my argument adds another significant case where relativistic mass is misleading. (If you only use relativistic mass when it is the same as rest mass, why on Earth introduce it?)

Meanwhile, invariant mass, computed in any frame, and without any special rules (e.g. comoving particles versus random orientations), correctly specifies when KE contributes to curvature generation, and when it does not.

 

[atyy]

If there is any point in used relativistic mass it is when it is different from rest mass. Then my argument adds another significant case where relativistic mass is misleading. (If you only use relativistic mass when it is the same as rest mass, why on Earth introduce it?)

Meanwhile, invariant mass, computed in any frame, and without any special rules (e.g. comoving particles versus random orientations), correctly specifies when KE contributes to curvature generation, and when it does not.

So your claim is that there is a situation in which KE does not contribute to curvatre?

 

[PeterDonis]

Considering the above relations, we have the following equation forms:

x=m_{lab}c^2
y=\left(\gamma-1\right) m_{lab}c^2
z=\gamma m_{lab}c^2

y = \left(z^2 - x^2\right)^{1/2}
x^2 + y^2 = z^2
x + y = z

As you've defined these variables, x is the "rest energy", z is the "total energy", and y is the "kinetic energy" as measured in the lab frame. With these definitions, x + y = z is correct, but x^2 + y^2 = z^2 is *not* correct. The latter equation would only be correct if y were the momentum as measured in the lab frame, but as you've defined it, it isn't; it's the kinetic energy. They're not the same; the kinetic energy is y=\left(\gamma-1\right) m_{lab}c^2, which is how you've defined y, but the momentum is p = \gamma m_{lab} v, where v is the velocity as measured in the lab frame. So it is true that, as you've defined x and z, x^2 + p^2 = z^2; but that equation is *not* true for y as you've defined it. (And of course, the equations involving p permit any value for v from 0 to c.)

 

[PAllen]

So your claim is that there is a situation in which KE does not contribute to curvatre?

Yes, I thought I already explained a few times. The KE of a single body (or a single body plus test bodies) does not contribute to curvature. This follows directly from the ability to transform such KE away.

 

[atyy]

Yes, I thought I already explained a few times. The KE of a single body (or a single body plus test bodies) does not contribute to curvature. This follows directly from the ability to transform such KE away.

Isn't this the same as saying that if the KE is 0, it does not contribute to curvature?

It doesn't show that if the KE is non-zero, it doesn't contribute to curvature.

In any case, if the KE is transformed away, the relativistic mass is the invariant mass.

Fundamentally, the relativistic mass is energy. And somehow one has to argue that energy doesn't contribute to curvature to show the relativistic mass doesn't contribute to curvature. The problem with the relativistic mass is it isn't covariant, but the generalization is to include the full stress-energy tensor.

 

[atyy]

One of the most successful, but most abstract, ideas for defining mass, IMO, is the conserved quantity associated with some time translation symmetry with the additional provision that your physics can be expressed in terms of a "least action" principle.

This is broad enough to include the ADM, Komar, and Bondi masses, with the details of "some sort of time translation symmetry" being different.

I'm not sure if the defintion is broad enough to include the other sorts of mentioned - I'm not terribly familiar with them.

While it's a pretty good definition, I have to admit it's not much of a"motivator".

Yes. The motivation is only how do we get to such an abstract idea? Well, the idea is not abstract if we say that energy is the conserved quantity associated with time translation symmetry in an action theory, since that holds even in Newtonian physics. So the only question: why do we define energy as mass? And that's where the relativistic mass helps.

The clean route would be to say "ADM energy", which I believe people do use. I haven't yet heard about "Komar energy" or "Bondi energy".

 

[harrylin]

Yes, I thought I already explained a few times. The KE of a single body (or a single body plus test bodies) does not contribute to curvature. This follows directly from the ability to transform such KE away.

Now I wonder, why would the KE of matter have a different effect from the KE of light? Light only has KE, and it does contribute to gravitation, right?

See also an old thread: https://www.physicsforums.com/archive/index.php/t-45067.html.
There pervect appeared to contradict you (let's ignore his debate with pete about words):

Saying that ["invariant mass"] curves space-time is wrong [...]
"What curves space, energy or rest mass" [..] is energy, not rest mass.
Saying "relativistic mass" curves space-time would not be wrong

Harald

 

[tom.stoer]

The clean route would be to say "ADM energy", which I believe people do use. I haven't yet heard about "Komar energy" or "Bondi energy".

Have a look at http://tower.ict.nsc.ru/EMIS/journals/LRG/Articles/lrr-2009-4/

 

[atyy]

Yes, I thought I already explained a few times. The KE of a single body (or a single body plus test bodies) does not contribute to curvature. This follows directly from the ability to transform such KE away.

Now I wonder, why would the KE of matter have a different effect from the KE of light? Light only has KE, and it does contribute to gravitation, right?

This brings to mind an interesting question. Would these statements be true in Nordstrom gravity? That was the first relativistic theory of gravity. In Nordstrom gravity, the tensor that is the source of gravity is the trace of the stress-energy tensor, which in the case of a particle - isn't that the invariant mass? Is the trace in general the invariant mass (restricting to special relativity)? There is no global bending of light in Nordstrom theory.

Have a look at http://tower.ict.nsc.ru/EMIS/journals/LRG/Articles/lrr-2009-4/

Thanks!

 

[PAllen]

Isn't this the same as saying that if the KE is 0, it does not contribute to curvature?

It doesn't show that if the KE is non-zero, it doesn't contribute to curvature.

In any case, if the KE is transformed away, the relativistic mass is the invariant mass.

Fundamentally, the relativistic mass is energy. And somehow one has to argue that energy doesn't contribute to curvature to show the relativistic mass doesn't contribute to curvature. The problem with the relativistic mass is it isn't covariant, but the generalization is to include the full stress-energy tensor.

Since when is frame dependent KE not KE? It is the nature of KE of a single body to be frame dependent; it is the nature of relativistic mass to be frame dependent. Invariant mass gets precisely at the frame independent contribution of KE, thus the portion of KE (of a system) that contributes to curvature.

The stress energy tensor included many other factors than invariant mass (stress, pressure, EM fields, etc.), and is the true source of gravity. But in comparing the concept of relativistic mass versus invariant mass, it is simply a fact that to the extent relativistic mass differs from invariant mass, it does not contribute to curvature.

 

[PAllen]

Now I wonder, why would the KE of matter have a different effect from the KE of light? Light only has KE, and it does contribute to gravitation, right?

See also an old thread: https://www.physicsforums.com/archive/index.php/t-45067.html.
There pervect appeared to contradict you (let's ignore his debate with pete about words):


Harald

Light is tricky, but not really an exception to my point of view, if you look at my wording and arguments carefully. You can't transform to a frame where light is at rest. Further, there is no body light enough to be a test particle relative to light. So my argument and and caveats explain why the energy of light needs to be treated a bit special. However, also note, that there is a result that two beams of light going in the same direction don't attract, while in opposite directions they do. It is definitely worth noting that in the former case, the invariant mass of the light beams is zero, while in the latter case it is non-zero.

As for Pervect's quote:

"Saying that ["invariant mass"] curves space-time is wrong [...]
"What curves space, energy or rest mass" [..] is energy, not rest mass.
Saying "relativistic mass" curves space-time would not be wrong"

I agree with the first part indicating incompleteness of invariant mass - it ignores EM, pressure, stress, etc. In strictly comparing the utility of invariant mass versus relativistic mass I am ignoring other contributions to stress energy. I disagree with has last statement. I think invariant mass is a much 'first order' model of how a system of particles generates curvature than relativistic mass.

 

[atyy]

Since when is frame dependent KE not KE? It is the nature of KE of a single body to be frame dependent; it is the nature of relativistic mass to be frame dependent. Invariant mass gets precisely at the frame independent contribution of KE, thus the portion of KE (of a system) that contributes to curvature.

The stress energy tensor included many other factors than invariant mass (stress, pressure, EM fields, etc.), and is the true source of gravity. But in comparing the concept of relativistic mass versus invariant mass, it is simply a fact that to the extent relativistic mass differs from invariant mass, it does not contribute to curvature.

So you are saying that if the KE is non-zero, then it is the invariant mass and not the relativistic mass that contributes to curvature?

BTW, how is the invariant mass defined from the stress-energy tensor? I know for a single particle it is the trace. But is that true for a system of particles?

 

[atyy]

As for Pervect's quote:

"Saying that ["invariant mass"] curves space-time is wrong [...]
"What curves space, energy or rest mass" [..] is energy, not rest mass.
Saying "relativistic mass" curves space-time would not be wrong"

I agree with the first part indicating incompleteness of invariant mass - it ignores EM, pressure, stress, etc. In strictly comparing the utility of invariant mass versus relativistic mass I am ignoring other contributions to stress energy. I disagree with has last statement. I think invariant mass is a much 'first order' model of how a system of particles generates curvature than relativistic mass.

http://arxiv.org/abs/gr-qc/9909014

Carlip has a formula that may be pretty close here. Eq 15 gives the first order approximation as a sum over the diagonal terms of the stress-energy tensor. Is that the invariant mass? I believe that is not the trace, since the trace should have some minus sign in Lorentz signature.

 

[PAllen]

I found an interesting reference on this. On page 4 of this paper it refers to a result by Moller that for an isolated system, the total relativistic 4 momentum is a true 4 vector. This is obviously the case I have been discussing. Thus suggests that for non-interacting particles, no EM field, the norm of this 4 vector is the effective gravitational mass.

http://arxiv.org/abs/physics/0505004

 

[atyy]

I found an interesting reference on this. On page 4 of this paper it refers to a result by Moller that for an isolated system, the total relativistic 4 momentum is a true 4 vector. This is obviously the case I have been discussing. Thus suggests that for non-interacting particles, no EM field, the norm of this 4 vector is the effective gravitational mass.

http://arxiv.org/abs/physics/0505004

Ok, thanks.

Concretely, how does one get the invariant mass out of the dust stress-energy tensor (http://www.mth.uct.ac.za/omei/gr/chap4/node1.html , Eq 15, 16 )?

 

[PAllen]

Ok, thanks.

Concretely, how does one get the invariant mass out of the dust stress-energy tensor (http://www.mth.uct.ac.za/omei/gr/chap4/node1.html , Eq 15, 16 )?

I think that section is too simple. Looking at the discussion from right after (11) up to (15), it seems their concept of dust is one where the particles have no motion relative to each other. Thus in a co-moving frame (and note the emphais that anything significant about T can be described in such a special frame), you get all components of T vanishing except T00, which just becomes the total rest energy of the particles. Note that this quantity would also be the invariant mass of the system in motion, as derived from its total 4-momentum.

While this case is trivial, it continues to support my point that for an isolated system of non-interacting particles, the system invariant mass is at least a good approximation to system gravitational mass, while relativistic mass fails to describe anything useful in this context.

 

[atyy]

I think that section is too simple. Looking at the discussion from right after (11) up to (15), it seems their concept of dust is one where the particles have no motion relative to each other. Thus in a co-moving frame (and note the emphais that anything significant about T can be described in such a special frame), you get all components of T vanishing except T00, which just becomes the total rest energy of the particles. Note that this quantity would also be the invariant mass of the system in motion, as derived from its total 4-momentum.

While this case is trivial, it continues to support my point that for an isolated system of non-interacting particles, the system invariant mass is at least a good approximation to system gravitational mass, while relativistic mass fails to describe anything useful in this context.

Can we make a bunch of particles moving in different directions by adding together stress-tensors of the form of Eq 16?

 

[Hurkyl]

All this talk of "invariant mass" and "total momentum" and "adding stress-tensors" in the context of gravity -- you are working in a theory that is not GR, Or are in GR but analyzing a system confined to an "infinitesimal neighborhood" or equivalently (I think) a setup where second-order effects are negligible, right? Or is there some construction involved that allows you to make sense of these that I am simply totally uninformed of?

 

[atyy]

All this talk of "invariant mass" and "total momentum" and "adding stress-tensors" in the context of gravity -- you are working in a theory that is not GR, Or are in GR but analyzing a system confined to an "infinitesimal neighborhood" or equivalently (I think) a setup where second-order effects are negligible, right? Or is there some construction involved that allows you to make sense of these that I am simply totally uninformed of?

My statement of adding stress-tensors was in special relativity. The idea was to get a form in special relativity and see if it could be written in generally covariant form.

And yes, I expect if the invariant mass is to make sense in GR, it should be in setup where we write g=Minkowski+small perturbation. Eg. Eq 15 of http://arxiv.org/abs/gr-qc/9909014

 

[edguy99]

Simple. Imagine a fast moving single particle with high relativistic mass (in some frame, obviously). Now transform to a frame where it is at rest. Compute curvature scalar and tensor. The former is invariant, the latter covariant. Thus the fast motion contributes nothing to curvature - as long as it can transformed away. The invariant mass, meanhwhile is unchanged from rest mass in this case - correctly correlating with the irrelavance of the relativistic mass.

Meanwhile, once you have a system of particles, their momentum *does* contribute to the invariant mass, and to the curvature. A key point is that for a system, of particles you can't transform away the relative motion. Maybe a further question is what if you have a system of comoving particles? Now, the invariant mass is unaffected by the motion and by the transform argument, only this invariant mass and not the relativistic mass contributes to curvature.

The upshot: invariant mass correlates meaningfually with curvature generated; relativistic mass is meaningless is this context.

Thank you, very clear and precise. I have spent all day reading about invariant mass.

 

[pervect]

My $.02. When you have a small system where the metric coefficents don't vary significantly, one observes that the invariant mass and the relativistic mass are the same in a frame where the momentum is zero. So one simply chooses a frame where the momentum is zero, and integrates T_00, and one gets both. If one choses some other frame, the relativistic mass changes, the invariant mass does not.

When you have a large system where the metric coefficients, in particular, g_00 varies, neither the concept of invariant mass nor the concept of relativistic mass is well-defined, as Hurkyl observes. One needs to use another sort of mass, one that IS defined in GR. If it's a static system, the easiest to understand useful quantity is the Komar mass.

The trace of the stress energy tensor T is T_00 - T_11 - T_22 - T_33, so integrating the trace is wrong.

If one has a small patch of space-time that is normalized to have locally unit metric coefficeints, so that it is Minkowskiian, one can ask what the contribution of that small patch of space-time is to the total Komar mass.

In this case, one integrates not the trace of the stress energy tensor, but 2K(T_00 - T), where K is the "redshift factor" to infinity. This can also be expressed as K * (T_00 + T_11 + T_22 + T_33), or K(rho + 3P) in an isotropic system, where P is the pressure.

The full expression in general is (Wald, pg 289)

<br /> 2 \int \left( T_{ab} - \frac{1}{2} T g_{ab} \right) \n^a \xi^b dV<br />

here n^a is a unit future, and \xi^b is a Killing vector.

A possibly simpler to understand expression, from the wiki article (originally written by yours truly, since then edited) which requires that coordinates be used where none of the metric coefficients are functions of time

http://en.wikipedia.org/w/index.php?title=Komar_mass&oldid=387629121

<br /> 2 \int ( \sqrt{|g_00|} \left( T_{ab} - \frac{1}{2} T g_{ab} \right) n^a n^b dV<br />

Choosing coordiantes so that g_{ij} = \delta_{ij} makes most of this manifest, a vital but not-well presented part is the need for the inclusion of

K = \sqrt |g_{00}| = \sqrt \left( -\xi^a \xi_a\right)

 

[atyy]

My $.02. When you have a small system where the metric coefficents don't vary significantly, one observes that the invariant mass and the relativistic mass are the same in a frame where the momentum is zero. So one simply chooses a frame where the momentum is zero, and integrates T_00, and one gets both. If one choses some other frame, the relativistic mass changes, the invariant mass does not.

When you have a large system where the metric coefficeints, in particular, g_00 varies, neither the concept of invariant mass nor the concept of relativistic mass is defined. One needs to start expanding one's horizons, and use another sort of mass, one that is defined in GR. If it's a static system, the useful quantity is the Komar mass.

The trace of the stress energy tensor T is T_00 - T_11 - T_22 - T_33, so integrating the trace is wrong.

If one has a small patch of space-time that is normalized to have locally unit metric coefficeints, so that it is Minkowskiian, one can ask what the contribution of that small patch of space-time is to the total Komar mass.

In this case, one integrates not the trace of the stress energy tensor, but 2K(T_00 - T), where K is the "redshift factor" to infinity. This can also be expressed as K * (T_00 + T_11 + T_22 + T_33), or K(rho + 3P) in an isotropic system, where P is the pressure.

The full expression in general is (Wald, pg 289)

<br /> 2 \Int \left( T_{ab} - \frac{1}{2} T g_{ab} \right) \n^a \xi^b dV<br />

here n^a is a unit future, and \xi^b is a Killing vector.

Choosing coordiantes so that g_{ij} = \delta_{ij} makes most of this manifest from the above equation, it's easy to miss the redshift-factor
K = \sqrt \left( -\xi^a \xi_a\right) but it's essential to include if your metric coefficients vary.

I agree. What I haven't been able to figure out is: how does one write the invariant mass from the stress energy tensor? For a single particle it is the trace, I think. What is it for a system of particles?

 

[pervect]

I agree. What I haven't been able to figure out is: how does one write the invariant mass from the stress energy tensor? For a single particle it is the trace, I think. What is it for a system of particles?

E = integral T_00, P = integral T_01, and m = sqrt(E^2 -P^2) , in general, in flat space-time.

Watch the signs on the trace! T^j_k is g^ij T_ik, and g_00 has the opposite sign from g_11, g_22, and g_33. So the trace isn't what you want, aside from the need to do two seprate integrals for E and P.

 

[kmarinas86]

Note: Hurkyl's post above was a response to a previous version of this post which is now deleted. Formatting has been fixed, and I have added a further argument at the end.

I think I can now reconcile the differences between both sides.

Going back to the formula:

m_0 c^2 = \left( E^2 - \|\mathbf{p}c\|^2\right)^{1/2}

1) We know that different inertial frames will see different values for E and \|\mathbf{p}c\|^2 and yet will see the same value for m_0 c^2.
2) I said that E is not something of the object itself, but rather the maximum value of energy change that may occur of that object when colliding with a body in the same inertial frame of reference as the observer. In other words, I claimed that relativistic mass is not tied to anyone body.
3) So the E in this equation, when distorted by a Lorentz boost, is not the energy content of the object.
4) The relativistic mass is often used in consideration of situations where one accelerates a particle. Such cannot allow the particle to be described as a closed system.
5) Therefore, while those against the validity of relativistic mass focus on the Lorentz boosts of a particle, those for the validity of relativistic mass focus on the proper acceleration of a particle.
6) If one stays away from Lorentz boosts and considers the effect that proper acceleration has on the mass of a charge particle (an open system effect), one realizes that E^2 - \|\mathbf{p}c\|^2 is not constant for this particle. Thus, we have the following relations:
E = m_{accelerated} c^2 = \gamma m_{lab} c^2
\|\mathbf{p}c\| = \left(\left(m_{accelerated}c^2\right)^2 - \left(m_{lab}c^2\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\left(m_{accelerated}c\right)^2 - \left(m_{lab}c\right)^2\right)^{1/2}
Where the lab frame is simply the laboratory frame of reference. One such laboratory frame of reference is the foundation of the CERN Hadron Collider facility. It follows that \|\mathbf{p}\| is simply the relativistic momentum with respect to the lab frame:
\|\mathbf{p}\| = \left(\left(\gamma m_{lab}c\right)^2 - \left(m_{lab}c\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\left(\gamma^2 - 1\right)\left(m_{lab}c\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\frac{1}{1-\frac{v^2}{c^2}} - \frac{1-\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{1-\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\gamma\frac{v}{c}\right)m_{lab}c
\|\mathbf{p}\| = \gamma m_{lab}v

Q.E.D.

Considering the above relations, we have the following equation forms:

x=m_{lab}c^2
y=\left(\gamma-1\right) m_{tab}c^2
z=\gamma m_{lab}c^2

y = \left(z^2 - x^2\right)^{1/2}
x^2 + y^2 = z^2
x + y = z

At first, this would require that, for an indivisible object, that is to say one that lacks any degrees of freedom (i.e. a truly fundamental particle), only one of these conditions may apply:

• z=x, v=0
• z=y, v=c

Any particle that travels neither at v=0 nor v=c is a mass consisting of multiple particles of the fundamental kind. This result is also consistent with the following alternate approach:

\gamma-1 = \left(\gamma^2-1\right)^{1/2}
\left(\gamma^2-1\right)^{1/2} = \gamma\frac{v}{c}
\gamma-1 = \gamma\frac{v}{c}
\gamma\left(1-\frac{1}{\gamma}\right) = \gamma\frac{v}{c}
1-\frac{1}{\gamma} = \frac{v}{c}
1-\frac{v}{c} = \frac{1}{\gamma}
\frac{1}{1-\frac{v}{c}} = \gamma
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
1-\frac{v}{c} = \sqrt{1-\frac{v^2}{c^2}}

As with the above, the only solutions for this are v=0 and v=c.

As you've defined these variables, x is the "rest energy", z is the "total energy", and y is the "kinetic energy" as measured in the lab frame. With these definitions, x + y = z is correct, but x^2 + y^2 = z^2 is *not* correct. The latter equation would only be correct if y were the momentum as measured in the lab frame, but as you've defined it, it isn't; it's the kinetic energy. They're not the same; the kinetic energy is y=\left(\gamma-1\right) m_{lab}c^2, which is how you've defined y, but the momentum is p = \gamma m_{lab} v, where v is the velocity as measured in the lab frame. So it is true that, as you've defined x and z, x^2 + p^2 = z^2; but that equation is *not* true for y as you've defined it. (And of course, the equations involving p permit any value for v from 0 to c.)

I'm not using the "natural units" where you have m_0=\left(E^2 + p^2\right)^{1/2}.

Notice how I use pure algebra to derive:

\|\mathbf{p}\| = \gamma m_{lab}v

From the assumption that:

E = m_{accelerated} c^2 = \gamma m_{lab} c^2

Notice how I must match all the dimensions. The dimensions for x, y, and z must therefore be all the same.

Now consider what I have above, but to convince you better that the first post is right, I will run the last several equations in reverse.

\|\mathbf{p}\| = \gamma m_{lab}v
\|\mathbf{p}\| = \left(\gamma\frac{v}{c}\right)m_{lab}c
\|\mathbf{p}\| = \left(\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{1-\left(1-\frac{v^2}{c^2}\right)}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\frac{1}{1-\frac{v^2}{c^2}} - \frac{1-\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)^{1/2}\left(m_{lab}c\right)
\|\mathbf{p}\| = \left(\left(\gamma^2 - 1\right)\left(m_{lab}c\right)^2\right)^{1/2}
\|\mathbf{p}\| = \left(\left(\gamma m_{lab}c\right)^2 - \left(m_{lab}c\right)^2\right)^{1/2}

This is equivalent to:

m_{lab}c = \left(\left(\gamma m_{lab}c\right)^2 - \left(\|\mathbf{p}\|^2\right)\right)^{1/2}

And finally, it is also equivalent to:

m_{lab}c^2 = \left(\left(\gamma m_{lab}c^2\right)^2 - \left(\|\mathbf{p}c\|^2\right)\right)^{1/2}

The solution E, going down line by line, is therefore deduced from \gamma m_{lab}c^2.

As you can see, my base assumption was that:

E = m_{accelerated} c^2 = \gamma m_{lab} c^2

Therefore, per the above algebra, this equality E = \gamma m_{lab} c^2 can be derived simply from the fact that \|\mathbf{p}\| = \gamma m_{lab}v.

This E is obviously the relativistic energy. The relativistic kinetic energy must therefore between the difference E and m_0 c^2.

However, I did make a mistake in my second (most recent) post. It turns out that I have not proved that \|\mathbf{p}c\| is indeed equal to y. However, what I did prove is that when \|\mathbf{p}c\| is equal to the y in my Pythagorean equation, either v=0 or v=c. That's not as interesting as a result, but it's better to be correct nonetheless. I hope that clarifies it.

 

[atyy]

E = integral T_00, P = integral T_01, and m = sqrt(E^2 -P^2) , in general, in flat space-time.

Watch the signs on the trace! T^j_k is g^ij T_ik, and g_00 has the opposite sign from g_11, g_22, and g_33. So the trace isn't what you want, aside from the need to do two seprate integrals for E and P.

Well, actually, I can't seem to get the invariant mass of a single particle even with the wrong signs for the diagonal elements. And yes, of course E and P and m are as you gave. I am simply trying to figure out if it is possible that the invariant mass is a better source of gravity than the energy in some situations.

I have had in mind Eq 15 of http://arxiv.org/abs/gr-qc/9909014v1 , which I pointed out in post #70 was probably not the trace in Lorentzian signature, which as far as I understand agrees with what you wrote in post #78.

 

[atyy]

The point of http://arxiv.org/abs/gr-qc/9909014v1 is that in the weak field, slow motion approximation, for a box of stuff, it is the "energy" or "relativistic mass" that couples to gravity. Eq 15 sets up the scenario, and the following discussion shows that the virial theorem for bound systems results in the coupling being given by "rest mass+kinetic+potential energy" (so we get the result kinda by accident). Interestingly, in Eq 20-22 he gives an argument that the cancellations are not accidental but result from general covariance.

 

[PAllen]

The point of http://arxiv.org/abs/gr-qc/9909014v1 is that in the weak field, slow motion approximation, for a box of stuff, it is the "energy" or "relativistic mass" that couples to gravity. Eq 15 sets up the scenario, and the following discussion shows that the Virial theorem for bound systems results in the coupling being given by "rest mass+kinetic+potential energy" (so we get the result kinda by accident). Interestingly, in Eq 20-22 he gives an argument that the cancellations are not accidental but result from general covariance.

But this is all in a frame where the box as a whole is at rest. There, the rest mass + kinetic energy is analogous to invariant mass (which is exactly this: rest mass plus kinetic energy in a frame where the 'system' is at rest - except it can be computed in any frame). Thus, precisely where relativistic mass differs from invariant mass (the case of the box moving), the relativistic mass is irrelevant, and the invariant mass is relevant.

Let's take a simple example. You have a bunch particle moving in different directions. In the frame where total P is zero, the invariant mass is the total E (rest energy + kinetic energy of each particle) / c^2, precisely because total P=0. In any frame where the system as a whole is moving, total E gets larger (relativistic mass), and total P is nonzero. In this frame, total E should *not* used as a simple case approximation for gravitational mass. However, invariant mass comes out the same in this frame. Thus, again, whenever, relativistic mass and invariant mass differ, it is the latter that is meaningful to talk about the the effective mass of the system *including* the irremovable contribution of kinetic energy.

 

[atyy]

The box is moving.

But yes this is a frame dependent calculation. The choice of frame is taken when he does the split g=Minkowski+h, and h is related to the Newtonian potential. I'm not sure how to do a frame-invariant Newtonian limit - is there such a thing - the Newtonian potential is not gauge invariant even in Newtonian gravity, let alone general relativity?

Let's take a simple example. You have a bunch particle moving in different directions. In the frame where total P is zero, the invariant mass is the total E (rest energy + kinetic energy of each particle) / c^2, precisely because total P=0. In any frame where the system as a whole is moving, total E gets larger (relativistic mass), and total P is nonzero. In this frame, total E should *not* used as a simple case approximation for gravitational mass. However, invariant mass comes out the same in this frame. Thus, again, whenever, relativistic mass and invariant mass differ, it is the latter that is meaningful to talk about the the effective mass of the system *including* the irremovable contribution of kinetic energy.

What happens is that in the weak field limit, the Newtonian potential is related to the diagonal terms of the perturbation to Minkowski. Then terms that are not E get included, as one expects. Surprisingly, when the system is bound, the virial theorem results in cancellations that result in the coupling being related to E again. So although it is wrong to only include E, the mistakes cancel out.

 

[nonequilibrium]

I haven't read through all the posts, but on the first page the user ZealScience asked a sensible question I've often wondered about and seems very important for this discussion, yet I can't see an answer on page 1 or 2 and it seems doubtful there was ever any reply to ZealScience's question, so let me rephrase it succintly:

Is invariant mass well-defined? Let's say a particle A has invariant mass m, but later we find out that the particle itself is rather a shell with three smaller particles (B,C,D) in it, bouncing around at high speeds. We then conclude that the "invariant mass" of A is actually the sum of the invariant masses of B, C & D and their kinetic energies (i.e. the sum of the relativistic masses of B, C & D).

Okay maybe we can still make the concept well-defined, but at least it's not fundamental and one could sensibly wonder if there maybe is no "invariant mass" strictly speaking, and all "invariant mass" is purely "relativistic". (I'm not saying the concept of invariant mass is useless, but it's certainly as arbitrary/unfundamental as the distinction between kinetic energy and temperature)

EDIT: okay i see two posts above is pretty much adressing this issue

 

[PAllen]

A few more points on what form of SR mass corresponds roughly to the active gravitational mass for suitably simple system.

MTW section 5.4 gives the derivation of stress energy tensor T for a swarm of particles that has been sought in these discussions.

MTW section section 19.1 discusses active gravitational mass of a suitably simple system. It concludes that the answer is the integral of T00 *done in a frame where total momentum is zero*. Combining this with the results of section 5.4, establishes, indeed, that SR invariant mass gives the active gravitational mass of system of particles under appropriate simplifying assumptions.

 

[PAllen]

To sum up, given my prior post, the reasons I have abandoned relativistic mass (I used to use it) are:

1) Because, for momentum, you can get a valid formula using relativistic mass in the Newtonian formula, there is great temptation to try to use it in other Newtonian formulas. Unfortunately, virtually every other case fails. Attempts to repair these formulas leads to nothing but strange forms of the normal SR formulas, defeating any value to relativistic mass.

2) It is conceptually cleaner to have non-overlapping meanings for mass and energy. Mass is an invariant quantity (rest mass for a single particle). Energy includes the total contributions due to mass, fields, and motion. Kinetic energy is the part if energy due to matter in motion. Energy is frame dependent.

3) Relativistic mass leads to the plethora of questions about: does a rapidly moving body form a black hole? Realizing that geometry (curvature = gravity) is coordinate independent suggests analyzing in the center of momentum frame, in which case relativistic mass becomes invariant mass. Thus, relativistic mass is misleading in any frame where it differs from invariant mass. Under simplifying assumptions, gravity comes from invariant mass. More generally, all intrinsic features of curvature must be the same as those analyzed in a center of momentum frame; I consider such features to be generalizations of invariant mass.

[Slight caveat, as Harrylin pointed out: there is no center of momentum frame for light. However, there is still invariant mass for systems including light, and, to a first approximation, these correctly incorporate light for the purposes of (3).]

 

[harrylin]

But this is all in a frame where the box as a whole is at rest. There, the rest mass + kinetic energy is analogous to invariant mass (which is exactly this: rest mass plus kinetic energy in a frame where the 'system' is at rest - except it can be computed in any frame). Thus, precisely where relativistic mass differs from invariant mass (the case of the box moving), the relativistic mass is irrelevant, and the invariant mass is relevant.

Let's take a simple example. You have a bunch particle moving in different directions. In the frame where total P is zero, the invariant mass is the total E (rest energy + kinetic energy of each particle) / c^2, precisely because total P=0. In any frame where the system as a whole is moving, total E gets larger (relativistic mass), and total P is nonzero. In this frame, total E should *not* used as a simple case approximation for gravitational mass. However, invariant mass comes out the same in this frame. Thus, again, whenever, relativistic mass and invariant mass differ, it is the latter that is meaningful to talk about the the effective mass of the system *including* the irremovable contribution of kinetic energy.

Surely here you relate to equations for invariant mass; logically it would be erroneous to abuse them for relativistic mass. Can any of those equations not be adapted to relativistic mass? Or is your argument basically that it complicates such equations without any benefit?

Anyone understands since the time of Newton that in mechanics problems it is useful to calculate with a centre-of-mass frame; and for such a frame, there is no advantage of invariant mass over relativistic mass as they are the same value.

 

[PAllen]

Surely here you relate to equations for invariant mass; logically it would be erroneous to abuse them for relativistic mass. Can any of those equations not be adapted to relativistic mass? Or is your argument basically that it complicates such equations without any benefit?

Anyone understands since the time of Newton that in mechanics problems it is useful to calculate with a centre-of-mass frame; and for such a frame, there is no advantage of invariant mass over relativistic mass as they are the same value.

The point of this discussion is the utility of relativistic mass for a fast moving body. My point is that as a source of gravity, this is a useless mass to use.

 

[atyy]

A fast moving object will length contract. Its density will increase If its density is given by its invariant mass divided by volume. Will it turn into a black hole?

 

[PAllen]

A fast moving object will length contract. Its density will increase If its density is given by its invariant mass divided by volume. Will it turn into a black hole?

Nah, more like a black pancake. I have to be done with this thread. Several major reasons why relativistic mass leads to confusion have been presented. As I see, not a single reason for its value has been presented. At best, strained defenses that it can be used without error - which is, of course true.

 

[harrylin]

[..] on the first page the user ZealScience asked a sensible question I've often wondered about and seems very important for this discussion, [..]
let me rephrase it succintly:
[..] Let's say a particle A has invariant mass m, but later we find out that the particle itself is rather a shell with three smaller particles [..] in it, bouncing around at high speeds. [..] the "invariant mass" of A is actually [..] the sum of the relativistic masses [..].

Okay maybe we can still make the concept well-defined, but at least it's not fundamental [..]

Good observations, you hit the nail on its head
Indeed, "invariant" mass is simply a standardised relativistic mass.

 

[pervect]

The point of http://arxiv.org/abs/gr-qc/9909014v1 is that in the weak field, slow motion approximation, for a box of stuff, it is the "energy" or "relativistic mass" that couples to gravity. Eq 15 sets up the scenario, and the following discussion shows that the virial theorem for bound systems results in the coupling being given by "rest mass+kinetic+potential energy" (so we get the result kinda by accident). Interestingly, in Eq 20-22 he gives an argument that the cancellations are not accidental but result from general covariance.

I've been giving this paper a careful re-read, and I want to see if you agree with my understanding of eq 15.

My interpretation of what Carlip is saying here is that if we look at the contribution of the presence of matter (I use the term loosely, to include light as well as solid matter) to the total Lagrangian that said contribution due to matter, which he calls the coupling, is going to be proportional to the "gravitational mass" of the matter and can be used to define said gravitational mass.

He talks about inertial mass, but I think all he really does is assume that that's equal to the total energy E, without justifying it explicityly.

Is this close to your interpretation?

 

[atyy]

I've been giving this paper a careful re-read, and I want to see if you agree with my understanding of eq 15.

My interpretation of what Carlip is saying here is that if we look at the contribution of the presence of matter (I use the term loosely, to include light as well as solid matter) to the total Lagrangian that said contribution due to matter, which he calls the coupling, is going to be proportional to the "gravitational mass" of the matter and can be used to define said gravitational mass.

He talks about inertial mass, but I think all he really does is assume that that's equal to the total energy E, without justifying it explicityly.

Is this close to your interpretation?

Yes, that's my understanding.

I don't have anything definite worked out for the inertial mass, but I would suggest trying http://arxiv.org/abs/astro-ph/0006423, Eq 21, with the first term producing the inertial mass and the second term being related to Carlip's terms.

 

[atyy]

Carlip, http://arxiv.org/abs/gr-qc/9909014
Straumann, http://arxiv.org/abs/astro-ph/0006423

It seems that Carlip takes phi constant in going from his Eq 15 to 16.

But in trying to get to something like Carlip's Eq 16 from Straumann's Eq 21, it seems I need to take grad(phi) constant, which makes sense, but doesn't quite seem to match up with Carlip's steps.

 

[pervect]

Yes, that's my understanding.

I don't have anything definite worked out for the inertial mass, but I would suggest trying http://arxiv.org/abs/astro-ph/0006423, Eq 21, with the first term producing the inertial mass and the second term being related to Carlip's terms.

I was thinking that the inertial mass would probably come from the pure-matter terms in the Lagrangian, the gravitational mass from the coupling terms. Though I haven't worked it out in any detail.

This would leave the pure field terms, which would persumably give the "mass" or energy of the field. But of course only for the linear model, not in the general theory!

 

[PAllen]

On the topic of the utility of relativistic mass, I wanted, and still want to say no more. However, this thread has gone in a different direction. I wonder if you (Pervect or Atyy) have any comments on my outline of derivation in post #87? Under the simplifying assumptions of those two MTW sections, it seems to follow almost immediately that invariant mass of a swarm of partices is the same as its active gravitational mass (the thing called M in section 19.1).

 

[atyy]

On the topic of the utility of relativistic mass, I wanted, and still want to say no more. However, this thread has gone in a different direction. I wonder if you (Pervect or Atyy) have any comments on my outline of derivation in post #87? Under the simplifying assumptions of those two MTW sections, it seems to follow almost immediately that invariant mass of a swarm of partices is the same as its active gravitational mass (the thing called M in section 19.1).

I don't have a direct comment off the top of my head. But the funny thing is that we know that in some exact solutions, the "mass" measured at infinity is certainly not the relativistic mass (eg. http://books.google.com/books?id=qhDFuWbLlgQC&source=gbs_navlinks_s, p259).

 

[PAllen]

I don't have a direct comment off the top of my head. But the funny thing is that we know that in some exact solutions, the "mass" measured at infinity is certainly not the relativistic mass.

No, it isn't the relativistic mass; it is (given the simplifying assumptions) the invariant mass, which includes only kinetic energy in the center of momentum frame, plus rest mass of the particles (with the feature that it can be computed directly in any frame). Really, the system rest mass of the swarm of particles.