Why is the relativistic mass a rejected concept?

[ZapperZ]

Your analogy is an inaccurate reflection of what I stated. The "invariant mass" of your "cake" is not changed by biting into it, but rather it is simply split into two kinds of pieces: 1) the cake pieces that come off 2) the cake that remains. Only when you can get that cake pieces' atoms and molecules to lose some mass through the metabolism of one's body, in the form of radiative heat, would I question the time-invariance of this so-called "invariant mass".

You are applying an incomplete conservation law, i.e. your accounting process is flawed. You are only using the conservation of the invariant mass, when the actual conservation law is the conservation of mass/energy. But that is still besides the point because this is NOT what is being discussed here, i.e. we 're not talking about a conversation of mass-energy, but rather the accounting of mass ONLY. Considering that, in high energy physics experiments, where mass-energy conversion happens ALL THE TIME, only invariant mass is used, and it is the ONLY thing that actually has any meaning.

So go ahead and submit your "questions" as rebuttals to all those high energy physics papers.

Zz.

 

[kmarinas86]

You are applying an incomplete conservation law, i.e. your accounting process is flawed. You are only using the conservation of the invariant mass, when the actual conservation law is the conservation of mass/energy. But that is still besides the point because this is NOT what is being discussed here, i.e. we 're not talking about a conversation of mass-energy, but rather the accounting of mass ONLY. Considering that, in high energy physics experiments, where mass-energy conversion happens ALL THE TIME, only invariant mass is used, and it is the ONLY thing that actually has any meaning.

So go ahead and submit your "questions" as rebuttals to all those high energy physics papers.

Zz.

In two contributions to this thread, you think that you have a monopoly over what others are discussing? Mind you, not even most posts here are the same things I discussing, so I do not claim such a monopoly.

To my surprise in fact, only one person in this thread so far, you, seem to be direct about the rejection of relativistic mass, whereas most here either wondered about its existence and/or have discussed the pedagogy of this subject.

All I glean from this so far is an absolute total lack of evidence of scientific consensus on the part of participants in this thread regarding the subject of relativistic mass.

 

[atyy]

I don't think this is the right idea. It's not a closed system; the hot box has gained more gravitational stuff.

$$stuff = \int_\Sigma P_{\mu\nu\sigma}d^{3}x$$

I agree. I should have made the scenario two closed boxes that have equal weight on the balance. In one box, a stationary particle decays into moving parts.

 

[Hurkyl]

All I glean from this so far is an absolute total lack of evidence of scientific consensus on the part of participants in this thread regarding the subject of relativistic mass.

Why not glean "there are so many problems with using the notion of relativistic mass instead of invariant mass and 4-momentum that everyone can pick a different reason to reject it when asked"?

I mentioned pedagogy simply because everything boils down to it -- relativistic mass in general simply isn't as good a tool for understanding and explaining things as invariant mass and 4-momentum. All the comments are either explaining why it's not as good, or demonstrating the community has accepted that it isn't as good.

 

[ZapperZ]

In two contributions to this thread, you think that you have a monopoly over what others are discussing? Mind you, not even most posts here are the same things I discussing, so I do not claim such a monopoly.

This is because you were using what I noticed to be something beyond what the actual topic is.

To my surprise in fact, only one person in this thread so far, you, seem to be direct about the rejection of relativistic mass, whereas most here either wondered about its existence and/or have discussed the pedagogy of this subject.

All I glean from this so far is an absolute total lack of evidence of scientific consensus on the part of participants in this thread regarding the subject of relativistic mass.

So where is the "total lack of evidence of scientific consensus" when you read the mass of the top quark, the mass electron neutrino, etc.? After all, many of the high energy collider experiments continually have mass-energy conversion. This, you did not address. Rather, you focused on the fact that I have only posted twice in this thread, as if that in itself is a point in your favor.

I pointed out two AJP papers addressing why the use of the term "relativistic mass" is faulty. In fact, Lev Okun even has a stronger opinion on this than I do (read his "Relativistic Mug" preprint). I don't believe you've supported your assertion with any valid citation.

Zz.

 

[kmarinas86]

You are only using the conservation of the invariant mass, when the actual conservation law is the conservation of mass/energy.

http://en.wikipedia.org/wiki/Invariant_mass

The invariant mass, rest mass, intrinsic mass, proper mass or just mass is a characteristic of the total energy and momentum of an object or a system of objects that is the same in all frames of reference related by Lorentz transformations.
[...]
This equation says that the invariant mass is the pseudo-Euclidean length of the four-vector (E, p), calculated using the relativistic version of the pythagorean theorem which has a different sign for the space and time dimensions. This length is preserved under any Lorentz boost or rotation in four dimensions, just like the ordinary length of a vector is preserved under rotations.
[...]
Since the invariant mass is determined from quantities which are conserved during a decay, the invariant mass calculated using the energy and momentum of the decay products of a single particle is equal to the mass of the particle that decayed.

Now that you know that Wikipedia's description of "invariant mass" is wrong, then why don't you go fix it?

 

[harrylin]

Oh sorry, were you being serious? I thought you were joking. OK, I did misunderstand then. Would you mind explaining again?

Misunderstanding (as we know from elsewhere) due to a poor explanation of "relativistic mass":
"do particles turn into black holes b/c of increasing mass near speed of light?"

Misunderstanding (as we know from this thread) due to a poor explanation of "invariant mass":
"if we compare hot and cold boxes in which the molecules have more and less kinetic energy, will the scale still give only the invariant masses b/c invariant mass is the amount of stuff?"

Cheers,
Harald

 

[ZapperZ]

http://en.wikipedia.org/wiki/Invariant_mass
Now that you know that Wikipedia's description of "invariant mass" is wrong, then why don't you go fix it?

That's it? That's all you got? Asking me to go correct something that I don't care for in the first place? And this is what you use as a source?

I could ask you the same thing. Since I've given you published sources, if you don't think they are correct in their assertion to drop the idea of relativistic mass, why don't you write a rebuttal?

I noticed you still refuse to address my point about high energy physics experiments.

Zz.

 

[harrylin]

[..]
All I glean from this so far is an absolute total lack of evidence of scientific consensus on the part of participants in this thread regarding the subject of relativistic mass.

The consensus or not of the few people here is not relevant!

What matters is the quality references that we provided. And if you checked out the physics FAQ which gives a summary overview, you would now understand why relativistic mass is disliked by many but appreciated by others, and not generally rejected.

 

[kmarinas86]

I can see why $m_0$ is invariant, but let's look at what we are actually seeing:

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

So here is the thing. $E$ does not contain a norm, and yet $|pc|$ does. If the momenta which make up $|pc|$ are not aligned, then $E$ is greater than $|pc|$. So even if we converted all mass into photons, we would still preserve this difference.

As far as I know, $\frac{E}{c^2}$ is the relativistic mass. If this is not invariant, then neither is $|pc|$. However, if the momentum of the whole system were a constant with time, implying a system closed to any other environment, I would fail to see how $|pc|$ would vary with time. This would make the system's $m_0 c^2$, $E$, and $|pc|$ not a function with time.

However, then it is claimed that one could increase this $E$ by choosing a different reference frame. Now realizing that this is the "relativistic mass" and considering a recent argument of mine against the "physicality" of this $E$, I now realize that I myself rejected the relativistic mass.

Wouldn't $E$ simply be the maximum amount of energy that may be transferred to a separate body in that given reference frame? The actual amount of energy transfer would seem to be a function of the elasticity of the collision with the separate body. The more elastic the collision, the greater the energy change $\Delta E$ would be observed of the separate body on impact. This seems to be the direct result of having $\Delta E$ include the part of the initial energy of the affected separate body that is impedance-matched to the incoming object. It makes no sense to me that the motions of this separate body, which is a receiver (measurer) of $\Delta E$, would not somehow contribute some of its own energy into $\Delta E$ in the form of $\Delta pc$, through gauge bosons, such as photons, which travel at $c$ with a momentum transfer of $\Delta p$. It seems that there would be a scalar product involved in such a collision to determine the limits of the amount energy exchanged if there are additional degrees of freedom, would there not?

In the quote above, here 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.

 

[pervect]

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?

In Newtonian physics, I'd just use a balance, as others have remarked. And in Newtonian physics you wouldn't expect the answer to change when you add heat to the box, so this is something new that arises in relativistic physics. We've subtly changed our philosophical idea of "stuff" - from being just matter, to being energy. So now, the amount of "stuff" in the box is the amount of energy in the box, not just the amount of gross matter. A subtle, but important, evolution from the original Newtonian idea.

In special relativity, we can define a satisfactory answer for the "amount of stuff", which we have newly interpreted as the total energy, of a system, IF the system is isolated. Due to the relativity of simultaneity, the amount of energy in a non-isolated system depends on the observer when the system is interacting with the environment. In GR, the situation is even worse - there isn't any general answer for total energy unless one has some preconditions.

Even in special relativity, the problem can be tricky. The surest way of getting the correct answer is also the most technical. The sure way is to use another geometric object the stress-energy tensor, to compute the energy, and momentum, of the system. Then, for isolated systems, one can show that E^2 - p^2 is invariant, it's independent of the "view" or coordinates or frame one takes. We say that the energy-momentum of the box transforms as a four vector, even for an extended object - but ONLY when the extended object is isolated. So E^2 - p^2, or rather its square root, turns out to be a good way to "weigh" a box.

However, the general reason why we use the stress-energy tensor is, I think, a mystery to most students. The answer goes back to geometric objects again. The stress energy tensor is a geometric object, it's a frame-independent way of thinking about the distribution, or density, of energy and momentum.

The easy way of getting the correct answer, without using the stress-energy tensor, is to compute E^2 - p^2 in the rest frame of the box. In this case, one can omit the walls of the box. But one has to presuppose the result that E^2-p^2 is invariant.

If you compute the total energy and momentum of the box in some other frame, you have to be sure to include the walls of the box. This is part of having a closed system, without a box, some collection of bouncing non-interacting particles would fly apart. There must be tension in the box to hold it together.

This tension in the walls doesn't contribute to either the total energy or the total momentum in the rest frame of the box. But in other frames (or views, as I called them eariler), it does! So to have frame independent physics, one has to include these contributions from the walls of the box.

I have posted a worked example somewhere of a "box of bouncing particles" using SR that shows that if you omit the walls and include only the energy and momentum of the particles, E^2 - p^2 of the sub-assembly (the contents of the box excluding the walls) is NOT constant, a consequence of the system being NOT isolated unless you include the walls.

 

[atyy]

In Newtonian physics, I'd just use a balance, as others have remarked. And in Newtonian physics you wouldn't expect the answer to change when you add heat to the box, so this is something new that arises in relativistic physics. We've subtly changed our philosophical idea of "stuff" - from being just matter, to being energy. So now, the amount of "stuff" in the box is the amount of energy in the box, not just the amount of gross matter. A subtle, but important, evolution from the original Newtonian idea.

In special relativity, we can define a satisfactory answer for the "amount of stuff", which we have newly interpreted as the total energy, of a system, IF the system is isolated. Due to the relativity of simultaneity, the amount of energy in a non-isolated system depends on the observer when the system is interacting with the environment. In GR, the situation is even worse - there isn't any general answer for total energy unless one has some preconditions.

Even in special relativity, the problem can be tricky. The surest way of getting the correct answer is also the most technical. The sure way is to use another geometric object the stress-energy tensor, to compute the energy, and momentum, of the system. Then, for isolated systems, one can show that E^2 - p^2 is invariant, it's independent of the "view" or coordinates or frame one takes. We say that the energy-momentum of the box transforms as a four vector, even for an extended object - but ONLY when the extended object is isolated. So E^2 - p^2, or rather its square root, turns out to be a good way to "weigh" a box.

However, the general reason why we use the stress-energy tensor is, I think, a mystery to most students. The answer goes back to geometric objects again. The stress energy tensor is a geometric object, it's a frame-independent way of thinking about the distribution, or density, of energy and momentum.

The easy way of getting the correct answer, without using the stress-energy tensor, is to compute E^2 - p^2 in the rest frame of the box. In this case, one can omit the walls of the box. But one has to presuppose the result that E^2-p^2 is invariant.

If you compute the total energy and momentum of the box in some other frame, you have to be sure to include the walls of the box. This is part of having a closed system, without a box, some collection of bouncing non-interacting particles would fly apart. There must be tension in the box to hold it together.

This tension in the walls doesn't contribute to either the total energy or the total momentum in the rest frame of the box. But in other frames (or views, as I called them eariler), it does! So to have frame independent physics, one has to include these contributions from the walls of the box.

I have posted a worked example somewhere of a "box of bouncing particles" using SR that shows that if you omit the walls and include only the energy and momentum of the particles, E^2 - p^2 of the sub-assembly (the contents of the box excluding the walls) is NOT constant, a consequence of the system being NOT isolated unless you include the walls.

Yes. The only question, and it is a matter of taste, is how does one motivate the change in the definition of "amount of stuff"? And really, doesn't one need GR to appreciate this definition of "amount of stuff"? (Well, I suppose generalizing Newtonian conservation laws applied to fields would be enough - but maybe not - I think there's this issue of being able to choose non-symmetric stress-energy tensors) If so, wouldn't part of the motivation be a heuristic for guessing the stress-energy tensor as the source of gravity in a relativistic theory of gravity? It can be done without relativistic mass, but would you prefer to do it that way?

BTW, I happened to learn SR on my own from WGV Rosser's text that advocated using only rest mass. When I went to university, I was converted, reluctantly, to accepting relativistic mass - I believe we used Purcell, but am not sure what Purcell actually does, since that book was so impenetrable to me. Does anyone know whether Purcell used relativistic mass? Or was that worse than SI units;) Jackson the betrayer (see his latest edition)!

 

[pervect]

Well, the idea of having gravity depend on a single scalar quantity - "relativistic mass", or "quantity X" - would be nice. But in the end, it winds up depending on a rank 2 tensor, so it never really seems to happen.

And it seems that the well-meaning shortcuts wind up biting one in the back eventually. A lesser but common example of this is when students start asking why you don't turn into a black hole if you move too fast - after all, your "relativistic mass" goes up.

Their logic is sound, the problem is taking a flawed concept (relativistic mass) too seriously. It may be a motivator, but in the end it doesn't quite work right.

Having gravity depend on the scalar rho+3P works better than having it depend on "relativisitic mass", in my opinion. But that's not perfect either, though it works reasonably well for static systems. And it's a relatively advanced concept (the Komar mass concept) anyway.

Also, if we take the energy as mass concept too seriously, we wind up with having 1kg on a mountaintop being different than 1kg in a valley, due to the difference in gravitational potential energy. (We had a recent thread, I was more or less uncessesful in cautioning an enthusiastic but rather misguided poster about this idea). And that's not good. Especially when we start trying to push that idea to it's logical conclusion. Where is the place that 1kg is "really' one kg? The Earth's gravitational field has some effect, but the Suns' gravitational field has more. And the galaxy's gravitational field has more than that. And so on, and so on. In the end, we can't find any place that 1kg is really 1kg. Eliminating gravitational self energy is tempting, but it isn't a good idea, either.

In the end, we are left with a bunch of competing ideas for mass, all of them different (the Komar formula, rho + 3P, for instance, is NOT exactly the same as either the relativistic mass OR the invariant mass), none of which is fully satisfactory.

 

[Hurkyl]

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.

There's already a notion of relativistic mass. If you're going to start inventing new concepts, you should give them different names.

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.

Eh? The way I've seen it presented, accelerated particles are pretty much the biggest strike against relativistic mass. With 4-vectors and invariant mass, you get an equation F=ma. When 3-vectors and relativistic mass, you don't get any simple relationship (except in the special case where force and velocity are in the same direction).

 

[atyy]

Well, the idea of having gravity depend on a single scalar quantity - "relativistic mass", or "quantity X" - would be nice. But in the end, it winds up depending on a rank 2 tensor, so it never really seems to happen.

And it seems that the well-meaning shortcuts wind up biting one in the back eventually. A lesser but common example of this is when students start asking why you don't turn into a black hole if you move too fast - after all, your "relativistic mass" goes up.

Their logic is sound, the problem is taking a flawed concept (relativistic mass) too seriously. It may be a motivator, but in the end it doesn't quite work right.

Having gravity depend on the scalar rho+3P works better than having it depend on "relativisitic mass", in my opinion. But that's not perfect either, though it works reasonably well for static systems. And it's a relatively advanced concept (the Komar mass concept) anyway.

Also, if we take the energy as mass concept too seriously, we wind up with having 1kg on a mountaintop being different than 1kg in a valley, due to the difference in gravitational potential energy. (We had a recent thread, I was more or less uncessesful in cautioning an enthusiastic but rather misguided poster about this idea). And that's not good. Especially when we start trying to push that idea to it's logical conclusion. Where is the place that 1kg is "really' one kg? The Earth's gravitational field has some effect, but the Suns' gravitational field has more. And the galaxy's gravitational field has more than that. And so on, and so on. In the end, we can't find any place that 1kg is really 1kg. Eliminating gravitational self energy is tempting, but it isn't a good idea, either.

In the end, we are left with a bunch of competing ideas for mass, all of them different (the Komar formula, rho + 3P, for instance, is NOT exactly the same as either the relativistic mass OR the invariant mass), none of which is fully satisfactory.

Well, let's exclude the more Komar, AdM, Yau, Yorke etc proposals for the moment. My own thinking is that while the relativistic mass is an important motivator, it should also be shown to be limited - ie. we search for a covariant analogue of the relativistic mass = energy - which for particles would be the invariant mass or the energy-momentum 4-vector. However, we also wish to define it for fields, which leads to the stress-energy tensor, which is good for particles and fields. And this is what we expect to take the place of "gravitational mass", since mass=energy heuristically.

An analogous paedagogical discussion might involve Bohr-Sommerfeld quantization. I know ZapperZ dislikes it, so he's being consistent in having everything very clean. But then this comes back in the field of quantum chaos, so perhaps it's best to learn these restricted ideas as advanced ideas. But don't we want students to have the joy of being confused by things like the EP (now can one argue that the relativistic mass is a worse heuristic than the EP)?

 

[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.