I If energy is relative, is the rest mass also relative?

[qnt200]

TL;DR Summary

energy;relative;mass

The mass (rest mass) of an atom, for example, depends on the kinetic and potential energy of the particles and their individual masses. Kinetic and potential energy are relative. Why is the mass not relative, but the same for all reference frames?

 

[PeterDonis]

The mass (rest mass) of an atom, for example, depends on the kinetic and potential energy of the particles and their individual masses.

Not just "the kinetic and potential energy" without qualification; the kinetic and potential energy in the center of mass rest frame. Big difference.

Kinetic and potential energy are relative.

Not if you specify the center of mass rest frame of the system; that can be specified in an invariant way.

 

[Orodruin]

Because it is the rest mass and rest mass is not relative. It is a particular combination of energy and momentum that is invariant.

 

[Dale]

Summary:: energy;relative;mass

The mass (rest mass) of an atom, for example, depends on the kinetic and potential energy of the particles and their individual masses. Kinetic and potential energy are relative. Why is the mass not relative, but the same for all reference frames?

The correct formula is $m^2 c^2= E^2/c^2 - p^2$. In that formula both $E$ and $p$ are relative, but their combination is not. In fact, $E$ has the same relationship to $p$ as time has to space. Such quantities form four-vectors and four-vectors have an invariant norm, in this case $m$

 

[PeterDonis]

Because it is the rest mass and rest mass is not relative. It is a particular combination of energy and momentum that is invariant.

The correct formula is $m^2 c^2= E^2/c^2 - p^2$.

I don't think the OP's question is about what rest mass/invariant mass is for a single object. I think it is about the fact that rest mass/invariant mass is not additive for a system consisting of multiple objects that are not all at rest relative to each other; the invariant mass of the system will contain contributions from the kinetic and potential energies of the individual constituents, as well as their invariant masses, and one has to be careful how those energies are defined in order to understand how the invariant mass of the system as a whole can still be invariant. That's why I focused on the center of mass rest frame in post #2.

 

[Dale]

I don't think the OP's question is about what rest mass/invariant mass is for a single object. I think it is about the fact that rest mass/invariant mass is not additive for a system consisting of multiple objects that are not all at rest relative to each other

Nothing there indicates that to me, but it certainly doesn’t hurt to discuss your take on it too.

 

[Orodruin]

I don't think the OP's question is about what rest mass/invariant mass is for a single object. I think it is about the fact that rest mass/invariant mass is not additive for a system consisting of multiple objects that are not all at rest relative to each other; the invariant mass of the system will contain contributions from the kinetic and potential energies of the individual constituents, as well as their invariant masses, and one has to be careful how those energies are defined in order to understand how the invariant mass of the system as a whole can still be invariant. That's why I focused on the center of mass rest frame in post #2.

The answer to the question inthe OP (why is the mass the same in all frames?) is still the same though. It is invariant because mass is an invariant combination of energy and momentum, not just the energies.

 

[PeterDonis]

Nothing there indicates that to me

This indicates it to me:

The mass (rest mass) of an atom, for example, depends on the kinetic and potential energy of the particles and their individual masses.

Note that the OP here is not stating explicitly that rest mass is not additive, but that's what he's describing.

 

[PeterDonis]

It is invariant because mass is an invariant combination of energy and momentum

The invariant mass of a single object is an invariant combination of energy and momentum. But for a composite system it's not so simple. The whole point is that the invariant mass of the system as a whole, if the constituents are in relative motion, is not the sum of the invariant masses of the constituents; so you can't just plug each constituent's energy and momentum into the invariant mass formula and come out with the invariant mass of the system as a whole.

Consider a simple example, positronium--an electron and positron each in a circular orbit about their mutual center of mass. Forget all the quantum stuff involved and just consider the situation classically. We have two objects, each with invariant mass $m$ and velocity $v$ (and thus kinetic energy $K = \left( \gamma(v) - 1 \right) m$) in the center of mass frame. We also have a (negative) potential energy of interaction $- V$ between them because they are bound. So the invariant mass of the system as a whole is not $2 m$; it's $2 \left( m + K \right) - V$ (which by the virial theorem works out to $2 m - V / 2$). But if you take each particle's energy and momentum and plug it into the invariant mass formula, and sum the results, you'll get $2 m$. So doing that does not answer the question of why the invariant mass of the system is something other than $2 m$.

 

[Orodruin]

But for a composite system it's not so simple.

Yes it is. It is the square root of the square of the 4-momentum of the system, which is a particular combination of energies and momenta. It may be a more complicated one than for a single object, but it still is one.

 

[pervect]

Summary:: energy;relative;mass

The mass (rest mass) of an atom, for example, depends on the kinetic and potential energy of the particles and their individual masses. Kinetic and potential energy are relative. Why is the mass not relative, but the same for all reference frames?

Rest mass does not include kinetic energy, as the phrase "rest mass" should imply. By definition, if one has two otherwise identical systems, one moving, one not moving, the rest mass of both systems is the same. The rest mass simply doesn't include kinetic energy, as the name implies.

Potential energy is a bit trickier, but the answer is similar. Strictly speaking, a system has an invariant mass only if it's isolated. (Reference: Taylor & Wheeler's "Spacetime Physics"). So it doesn't make sense to talk about the rest mass of a strongly interacting system, one typically takes a larger system that includes all the significant interactions and finds the rest mass of the larger system.

One would run into this issue if a system has a significant potential energy - this implies that the system is not isolated.

A piece of a strongly interacting system doesn't really have a well-defined rest mass. But it can be described with more advanced mathematical description, namely one can assign a stress-energy tensor to the "piece" of the larger system even when it's strongly interacting with the rest of the system. The stress energy tensor is not usually taught at an undergraduate level, though.

Interactions where the ineteraction is strong enough for this to be necessary are rare, however - usually the interaction energy is negligible, or can be made negligible by considering a different, larger, system that includes the interaction.

An interesting though somewhat advanced example that shows some of the issues is trying to find the "mass" of a box of light excluding the box. The light strongly interacts with the box, so it's not an isolated system until one includes the box.

 

[Orodruin]

the invariant masses of the constituents; so you can't just plug each constituent's energy and momentum into the invariant mass formula and come out with the invariant mass of the system as a whole.

Yes you can, but not separately (I never said to do it separately). The 4-momentum of the system is the sum of the 4-momenta of its components.

 

[PeterDonis]

The 4-momentum of the system is the sum of the 4-momenta of its components.

How does that account for the potential energy in the positronium example I gave?

 

[PeterDonis]

it doesn't make sense to talk about the rest mass of a strongly interacting system, one typically takes a larger system that includes all the significant interactions and finds the rest mass of the larger system.

One would run into this issue if a system has a significant potential energy - this implies that the system is not isolated.

I'm not sure I understand. The positronium example I gave is an interacting system with a potential energy which can be treated as isolated (that's what I was implicitly doing in my example). The electron by itself, and the positron by itself, are not isolated, but the positronium system as a whole is, and that is the system that has potential energy.

 

[Orodruin]

How does that account for the potential energy in the positronium example I gave?

You need to include the 4-momentum of the EM field. This is a part of the system. You would not have positronium without the EM field.

 

[PeterDonis]

You need to include the 4-momentum of the EM field.

Ah, ok.

 

[pervect]

I'll use the box of moving particles as an example as that's the one I'm familiar with. The particles could be light. One can compute and sum E and p of the particles. However, if one works through the details, E^2 - (pc)^2 is not frame invariant. I worked this out once upon a time, see https://www.physicsforums.com/threads/energy-momentum-mass-of-a-box-of-moving-particles.117773/.

E^2 - (pc)^2 for the system of box + particles is frame invariant - but if one excludes the walls of the box from the computation, it's not.

 

[mitochan]

Why is the mass not relative, but the same for all reference frames?

Because we have convention that mass is energy measured in a special frame of reference called COM frame of reference.

 

[pervect]

Here's a calculation of E^2-p^2 for the box of moving particles using the stress-energy tensor approach and geometric units.

In the rest frame, the interior of the box has a stress energy tensor of
$$T^{ab} = \begin{bmatrix} \rho & 0 \\ 0 & P \end{bmatrix}$$

T transforms under a boost as $T'^{ab} =\Lambda^a{}_c \Lambda^b{}_d T^{cd}$ where
$$\Lambda = \begin{bmatrix} \gamma & \beta \gamma \\ \beta \gamma & \gamma \end{bmatrix}$$

In the rest frame the energy is given by $E=T^{00} V$, where V is the volume of the box. Let the volume V of the box in the rest frame be 1, then $E=\rho$. Similarly, the momentum in the rest is given by $p = T^{01} \, V = 0$. So in the rest frame $m^2 = \rho^2$.

Now let's consider what happens when we do a boost.
The boosted stress energy tensor is:
$$T = \begin{bmatrix} \gamma^2 \rho + \beta^2 \gamma^2 P & \beta \gamma^2 \rho + \beta \gamma^2 P \\ \beta \gamma^2 \rho + \beta \gamma^2 P & \beta^2 \gamma^2 \rho + \gamma^2 P \end{bmatrix}$$

The box is length contracted , so it's volume is $1/\gamma$.

Thus we have in the boosted frame
$$E = \gamma \rho + \beta^2 \gamma P \quad p = \beta \gamma \rho + \beta \gamma P$$
m'^2 in the boosted frame is E^2 - p^2, which is
$$m'^2 = \gamma^2(1-\beta^2) \rho^2 - \beta^2 \gamma^2 (1-\beta^2) P^2 = \rho^2 - \beta^2 P^2$$

So we see that when the pressure P is zero, the mass is invariant, but when P is nonzero, it isn't.

Basically, if we consider the whole box (it's pressuried contents and the walls), the integral of the pressure term vanishes, and we get the expected result that E^2-p^2 is unchanged by the boost. But if we don't include the walls of the box, and only compute E^2-p^2 of the interior, E^2-p^2 is changed by the boost.

 

[Vanadium 50]

At this point, 30x as much ink has been spent on figuring out what the OP "must surely have meant" than the original question. The OP has been back, but hasn't clarified.

 

[Dale]

This indicates it to me:Note that the OP here is not stating explicitly that rest mass is not additive, but that's what he's describing.

I can sort of see that. But I will nevertheless leave my response because the way I read it the bigger problem is that he is neglecting $p$ in his description.

 

[Dale]

Rest mass does not include kinetic energy, as the phrase "rest mass" should imply.

Well, you have to be careful. Consider a spinning disk. In the frame where the disk has $p=0$ it also has $KE \ne 0$. That non-zero KE in the center of momentum frame does contribute to the "rest mass", which I find confusing. So I prefer to say "invariant mass" for that reason.

 

[Dale]

At this point, 30x as much ink has been spent on figuring out what the OP "must surely have meant" than the original question.

[tongue in cheek] I would like to remind all participants that printing out PF threads kills trees and wastes ink. We encourage everyone to be environmentally conscious and view PF electronically rather than in hard-copy. [/tongue in cheek]

 

[Vanadium 50]

view PF electronically rather than in hard-copy

That's all 1's and 0's, but doesn't that just waste 1's?

 

[Orodruin]

[tongue in cheek] I would like to remind all participants that printing out PF threads kills trees and wastes ink. We encourage everyone to be environmentally conscious and view PF electronically rather than in hard-copy. [/tongue in cheek]

Are you saying I’ve been doing it wrong?

 

[pervect]

Well, you have to be careful. Consider a spinning disk. In the frame where the disk has $p=0$ it also has $KE \ne 0$. That non-zero KE in the center of momentum frame does contribute to the "rest mass", which I find confusing. So I prefer to say "invariant mass" for that reason.

True, let's go through the whole thing in detail for the OP's benefit, hopefully they're still around. This is in the context of special relativity.

step 1. Ensurer you have an isolated, self-contained system, so the rest mass is defined
step 2. Compute the total energy E of this system. Call it E.
step 3. Compute the total momentum of the system, call it p.
step 4. Compute $\sqrt{(\frac{E}{c^2})^2-(\frac{p}{c})^2}$. This is the rest mass of the system.

This is most simply done in the frame of reference where the momentum is zero, in which case the rest mass is m=E/c^2. However, as long as the system is complete and not strongly interacting with its environment, the rest mass can be computed in any desired frame by the above procedure, if one knows how to compute E and p. Unfortunately, some of the details of precisely computing E and p for a anything more complex than a point particle is rather advanced. Correctly computing in detail the exact mass of a spinning disk including all relativistic effects would not be an undergraduate problem, though it is simple to note that adding energy to the disk by making it rotating does increases its mass. The disk can be thought of as composed of particles, but these particles are interacting, not isolated, so knowing the formula for an isolated particle isn't sufficient. Egan did some work on this problem in https://www.gregegan.net/SCIENCE/Rings/Rings.html. The good news is that any physically reasonable disk would self-destruct long before any relativistic effects became significant.

Rather than going into the detailed calculations, some general observations are more likely to be helpful. If one has a "box" of gas, and one heats up said box, the mass after heating is ever so slightly larger than the mass before heating due to the energy added to the box. Note that we regard the box as isolated before the heating, not isolated during the heating, and isolated again after it's heated. If the box is allowed to cool down again, the energy is lost and the mass of the box decreases. Similarly if one has a spring + screw rod combination, where turning the screw stretches the spring, turning the screw does mechanical work on the system and it's mass increases. In your example of the spinning disk, spinning the disk requires energy, adding this energy makes the mass of the spinning disk very slightly larger. If one imagines a very strong box containing an atomic bomb that is totally sealed, exploding the bomb doesn't change the mass of the box as long as the box doesn't fail and nothing escapes the box. (This is rather unlikely for a real box). Energy is converted from one form to another but the mass of the system doesn't change. The mass changes only when energy enters or leaves the system.

 

[qnt200]

I thank everyone for the answers, which are extensive and in any case focused on the questions I asked. I am very pleased with the interesting answers. I think I got good answers to my questions.
My apologies for the vague questions.

Additional explanations of the question:

The mass (rest mass) of an atom, for example, depends on the kinetic and potential energy of the particles and their individual masses.

Here I wanted to emphasize that mass, kinetic and potential energy have their different characteristics, distribution in space and dynamics. It is questionable how to reduce all this to a rest mass that would also be invariant?

Kinetic and potential energy are relative. Why is the mass not relative, but the same for all reference frames?

For example, in protons and neutrons most energy is the energy of interaction between parts of particles. Since that energy is relative, I asked, perhaps a naive question, why the rest mass that makes up most of that energy is not relative?

 

[PeterDonis]

mass, kinetic and potential energy have their different characteristics, distribution in space and dynamics. It is questionable how to reduce all this to a rest mass that would also be invariant?

In the general case, you can't describe matter by a single "rest mass". You have to describe it using a tensor, the stress-energy tensor.

In this thread, we have been restricting discussion to special cases where matter is being modeled by particles (although as @Orodruin pointed out, in my positronium example you have to also include the energy and momentum in the EM field, which is continuous and should really be described using a stress-energy tensor.)

in protons and neutrons most energy is the energy of interaction between parts of particles.

Yes, and as above, that energy really should be modeled using a stress-energy tensor.

Since that energy is relative

No, if you model it using a stress-energy tensor, it isn't. From the standpoint of this discussion, the stress-energy tensor is just like the 4-momentum vector from which the invariant rest mass is derived: the individual components of the tensor change when you change frames, but they change in just the right way to keep all invariants--i.e., all actual observable quantities--the same. The "energy of interaction" you refer to is one of those observables.

 

[vanhees71]

I agree that the local laws are the fundamental and most tranparent form to express relativistic physical laws, and if it comes to energy and momentum the local description is the energy-momentum-stress tensor $T^{\mu \nu}$. For a closed system it is conserved, which means in special relativity that $\partial_{\mu} T^{\mu \nu}=0$, reflecting the four conservation laws following from the space-time translation invariance of the theory. Then (and only then!)
$$P^{\nu}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} T^{0 \nu}$$
is a four-vector and is the total energy $P^0$ and momentum $\vec{P}$, and it's conserved. Then also $s=P_{\mu} P^{\mu}=M^2$ is conserved, and you can call $M$ the invariant mass of the system. Since this total energy and momentum and thus also the invariant mass includes interaction energy invariant mass is not additive, i.e., the invariant mass of an interacting system is not the sum of the invariant masses of its constituents, and in special relativity there is not an additional conservation law for invariant mass as there is one in Newtonian physics (for pretty subtle reasons concerning the Lie algebra of the respective space-time symmetry groups).

For a positronium in the approximation to treat it within QED (i.e., neglecting strong and weak interactions) the closed system is the one including the quantized Dirac and electromagnetic field.

For open systems energy, momentum and thus also invariant mass can change. The most simple example is some piece of matter, which you heat up, i.e., you add heat energy to the system, i.e., you enhance the total energy of the body in its center-of-momentum frame, and it's invariant mass gets enhanced by the corresponding energy (divided by $c^2$ if you don't set $c=1$ as I did above). Another example is a capacitor whose invariant mass is a bit larger when charged to a voltage $U$ by the corresponding static-electric field energy $E_{\text{field}}=C U^2/2$ etc. etc.

 

[PeterDonis]

For a closed system it is conserved, which means in special relativity that $\partial_{\mu} T^{\mu \nu}=0$

I'm not sure what you mean by "for a closed system". The SET in SR always obeys the vanishing divergence condition you give, at every point of spacetime. In GR that condition becomes $\nabla_{\mu} T^{\mu \nu}=0$, and is again always true. Since it applies at each point of spacetime, there is no such thing as a "closed system" vs. "open system" if you're just looking at the SET; it's just a continuous distribution of stress-energy.

the four conservation laws following from the space-time translation invariance of the theory.

In relativity these conservation laws follow from the presence of Killing vector fields and are different from the divergence condition on the SET.

Then (and only then!)
$$P^{\nu}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} T^{0 \nu}$$
is a four-vector and is the total energy $P^0$ and momentum $\vec{P}$, and it's conserved.

AFAIK the conditions for this working are that the spacetime is stationary (so the quantity is conserved) and asymptotically flat (so the integral is well-defined). As you've written the integral, it also requires a particular choice of coordinates (adapted to the timelike KVF). An invariant formulation would be the Komar energy.

For open systems energy, momentum and thus also invariant mass can change.

An "open system" here would be a region of spacetime that is not isolated, i.e., stress-energy crosses its boundary. But at each point of the spacetime, including those on the boundary where stress-energy is crossing, the SET still obeys the vanishing divergence condition.

 

[Orodruin]

The SET in SR always obeys the vanishing divergence condition you give, at every point of spacetime.

He did mention "for a closed system". While the total SET in SR vanishes, it is not necessarily true for a particular system as it may not be everything that contributes to the total SET. What holds is that $\sum_i \partial_\mu T_i^{\mu\nu} = 0$ where the sum is over all contributions to the SET. In some sense, this is the SR equivalent of Newton's third law. If the divergence of one system (say the EM field) is non-zero, then this must be compensated by the negative of that divergence for another system (say the field of charged matter). This is then essentially summarised as the Lorentz force law.

Edit: In fact, for suitable choices of constants and metric conventions, one finds that
$$
\partial_\mu M^{\mu\nu} = - F^{\nu\mu} J_{\mu},
$$
where $M^{\mu\nu}$ is the SET of the electromagnetic field, $F^{\mu\nu}$ is the EM field tensor, and $J^\mu$ the EM 4-current density. The RHS here is not identically equal to zero. In fact, it is just the negative of the 4-force acting on the 4-current according to the Lorentz force law - just as it should be.

 

[PeterDonis]

He did mention "for a closed system".

But he didn't say what he meant by that. The usual interpretation of "closed system" would be "a region of spacetime such that no stress-energy crosses its boundary". But no such condition is necessary for the vanishing divergence condition to hold.

While the total SET in SR vanishes, it is not necessarily true for a particular system as it may not be everything that contributes to the total SET.

I see what you mean--for example, we might only look at the SET due to matter and not that due to the EM field, and if both are present and the matter is charged, neither SET would obey the divergence condition by itself, only the sum of them would. (You describe this further on in your post.)

However, I don't think the term "closed system" is a good way to say "we're counting all of the SET at this point in spacetime, not just a piece of it". One should always do that anyway; doing the opposite is not "an open system", it's just wrong.

 

[Orodruin]

But he didn't say what he meant by that. The usual interpretation of "closed system" would be "a region of spacetime such that no stress-energy crosses its boundary". But no such condition is necessary for the vanishing divergence condition to hold.I see what you mean--for example, we might only look at the SET due to matter and not that due to the EM field, and if both are present and the matter is charged, neither SET would obey the divergence condition by itself, only the sum of them would. (You describe this further on in your post.)

However, I don't think the term "closed system" is a good way to say "we're counting all of the SET at this point in spacetime, not just a piece of it". One should always do that anyway; doing the opposite is not "an open system", it's just wrong.

I guess I have a more "liberal" view of what a "system" refers to where what you decide to include cannot only be spatially and temporally separated, but also separated by what it is. At the very least I find it useful to talk about things such as the force on a charge exerted by the electromagnetic field, which is in essence drawing a boundary between the two - one system exerting a force on the other and the other exerting an equal but opposite force on the first. We do this in non-relativistic electrostatics too, where we consider the force on an object exerted by the electric field. We do not include the energy of the electric field in the region occupied by the object. A closed system would just be a system not subject to any external forces, whether forces acting across a spatial boundary or within the volume.

 

[Dale]

what you decide to include cannot only be spatially and temporally separated, but also separated by what it is.

I think that is often done. However, if you do that you have to be willing to accept the fact that there may not be a unique separation based on what it is, and that the conservation laws may not apply individually to the separated components.

A good example of this is the famous Abraham Minkowski controversy about the momentum of light in a transparent medium. Abraham and Minkowski disagreed about how to separate the field from the matter. As a result they obtained different expressions for the momentum of the light. But the total momentum is what is conserved and agreed for both approaches.

 

[Orodruin]

I think that is often done. However, if you do that you have to be willing to accept the fact that there may not be a unique separation based on what it is, and that the conservation laws may not apply individually to the separated components.

A good example of this is the famous Abraham Minkowski controversy about the momentum of light in a transparent medium. Abraham and Minkowski disagreed about how to separate the field from the matter. As a result they obtained different expressions for the momentum of the light. But the total momentum is what is conserved and agreed for both approaches.

Sure, I am fine with this. Momentum in classical mechanics also depends on how I choose to define my system. The question is if you can really define anything that is really a closed system. In some sense this will never be the case unless you include everything, at least not in SR as, if you do have two or more completely closed systems, they would just as well be describable on their own.

 

[vanhees71]

I'm not sure what you mean by "for a closed system". The SET in SR always obeys the vanishing divergence condition you give, at every point of spacetime. In GR that condition becomes $\nabla_{\mu} T^{\mu \nu}=0$, and is again always true. Since it applies at each point of spacetime, there is no such thing as a "closed system" vs. "open system" if you're just looking at the SET; it's just a continuous distribution of stress-energy.

For a closed system yes. There's a lot of confusion in the literature by considering only the energy-momentum tensor of the em. field and then wondering, why there is problem with total em. energy and momentum at presence of sources $\rho$ and $\vec{j}$. You have to consider always the total EM tensor of a closed system, obeying the continuity equation, and only then the spatial integral gives an energy-momentum four-vector (for the entire system consisting of the em. field and the charges).

In relativity these conservation laws follow from the presence of Killing vector fields and are different from the divergence condition on the SET.

I didn't want to discuss the GR case here, where it is of course even more complicated. You only have local energy conservation $\nabla_{\mu} T^{\mu \nu}=0$, but you cannot so easily define total energy and momentum in a coordinate independent way. I guess you know this better than I.

 

[PeterDonis]

my own paper

Which is not a valid reference. Personal research is off limits for discussion at PF. (And no, posting it on vixra doesn't mean it isn't personal research. If you get it published in an actual peer-reviewed journal, then we might take a look.)

I'm sure other ones are possible.

I'm afraid no one else shares your belief on this point. Please do not post further about this in this thread; if you do, you will receive a warning and a thread ban.