# I A question about invariant mass

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1. Aug 7, 2018

### liuxinhua

A question of invariable mass.

In a inertial system, the invariable mass of a system never change with time. This system may not be an isolated system.

Whether in any inertial system, the invariant mass of the system remains unchanged.

Or, in a certain inertial system, what is the necessary and sufficient condition for the invariant mass of the system to be constant?

The same time
https://en.wikipedia.org/wiki/Mass_in_special_relativity
proposed:
Conservation of invariant mass also requires the system to be enclosed so that no heat and radiation (and thus invariant mass) can escape. As in the example above, a physically enclosed or bound system does not need to be completely isolated from external forces for its mass to remain constant, because for bound systems these merely act to change the inertial frame of the system or the observer.

Where can I find the strict definition of bound-system?

Thank you .

2. Aug 8, 2018

### Staff: Mentor

If the system is isolated then energy and momentum are conserved, therefore the invariant mass is conserved as well.
If the system is not isolated then everything can change. The invariant mass can change if energy or momentum change, if they don't change (or change just in the right way) it doesn't change. "Just the right way" here means the invariant mass stays the same despite energy and/or momentum changing - there is no other way to express that.
I don't think it would help to go more into detail here. What the quote means: If you e.g. accelerate a whole hydrogen atom and it stays a hydrogen atom, then its invariant mass stays the same.

3. Aug 8, 2018

### vanhees71

Not necessarily! If you excite the hydrogen atom in the acceleration process, its invariant mass will change, but that's of course, because you have to put in the excitation energy, i.e., the invariant mass of the hydrogen atom will be a bit larger than before.

4. Aug 8, 2018

### liuxinhua

Mr. mfb , You mean: If the system is not isolated, there is no way to decide its invariant mass except to use the definition of invariant mass.
Example of my question,:

In the inertial reference frame K, an isolated rod AB rotates around its mid point O at uniform angular velocity, keeps the linear state.

For actual rod AB, there is micro motion on molecular level. In the process of simplifying and establishing a mathematical model, we do not consider this micro motion. The rod AB is an ideal system without any relative movement between the components relative to K. The distribution of material and potential energy on the rod AB along the length is symmetry about mid point O and it does not change with time, measured in K.

Rod AB has no thickness along theta direction, if transform K from rectangular coordinate system to polar coordinate system. That is to say consider rod AB as a one-dimensional rod on x-y plane relative to K.

In K, static auxiliary circle D and M, take O as their center of their circle, intersecting rod OA at point D and M.

For section DM, measured in K , its total energy doesn’t change with time, and the absolute value of total momentum of its all components doesn’t change with time. So the invariant mass of section MD is a constant that does not change with time measured in K.

The invariant mass of section DM is a constant in K.

Is the invariant mass of section DM is a constant in K’ ?

Is section DM a bound-system? What is the basis for making the judgment?

Last edited by a moderator: Aug 8, 2018
5. Aug 8, 2018

### liuxinhua

Maybe when you excite the hydrogen atom in the acceleration process, you change its bound state.

6. Aug 8, 2018

### Ibix

A bound system is one for which nothing enters or leaves (at least to the precision you care about). As a consequence mass is constant: there's nothing to carry it away or bring it in.

However, one can imagine parts of a system which are not themselves bound but nevertheless have constant invariant mass. For example, one half of a rotating wheel is not bound - matter is constantly rotating in and out of it. Nonetheless its mass is constant because the inflow exactly balances the outflow.

I'm not sure about your example - I can only see about two lines at a time on my phone screen, so I'm having trouble following it.

Last edited: Aug 8, 2018
7. Aug 8, 2018

### liuxinhua

If measured in the inertial reference frame K, the invariant mass of a system is a constant.
But measured in another inertial reference frame K’ , whether the invariant mass of the system is a constant ?
Whether someone has mentioned this conclusion in any literature?

In K’ , it is difficult to calculate the system’s invariant mass, if using the definition of invariant mass.

8. Aug 8, 2018

### Ibix

"Invariant" means that it does not change between frames. So either it's the same in both frames or its not invariant.

9. Aug 8, 2018

### Staff: Mentor

Not only is it constant, it is the same as in K.

The most interesting question here might be how you would measure that physical property that we call "invariant mass" under two different circumstances:
1) The object in question is at rest relative to you (this is equivalent to measuring its invariant mass in frame K).
2) The object in question is moving relative to you (this is equivalent to measuring its invariant mass in frame K').
#2 is likely to be more difficult than #1, which is why we generally try to work in frames in which whatever studying is at rest, but if we're doing it right we have to get the same answer either way - as @Ibix says, that's what "invariant" means.

10. Aug 8, 2018

### pervect

Staff Emeritus
If the non-isolated system is not pointlike, much caution is needed. I would not use the invariant mass to characterize a non-pointlike, non-isolated system. I would use the stress-energy tensor instead.

An example would be the mass of a section of a stressed rod, or the mass of a box of light ignoring the walls. The intergal of $T^{00}$ over the volume of the rod at some particular time seems like a logical candidate for the invariant mass. But if the system is not isolated, this quantity is not necessarily the same in different frames.

My recollection (which isn't as good as it used to be) is that the integral of $T^{00}$ of a section of a stressed rod depends on the frame of reference. I'm not sure that I recall a solid reference on the topic, though. Rindler has some discussion, but not using relativistic mass. Most other authors introduce the stress energy tensor, but don't address the issue of exactly why it's needed.

If you look at Taylor and Wheeler "Space-time physics", for instance, and read the fine print, I believe they usually make a point of saying that the system is isolated. In my recollection at least, it'd be good to do a double check on this.

11. Aug 8, 2018

### pervect

Staff Emeritus
To add a bit to my previous post, I worked on the box of light in https://www.physicsforums.com/threads/energy-momentum-mass-of-a-box-of-moving-particles.117773/

This is a detailed working out of a mass of bouncing particles (which could be light, but don't have to be).

However, some of the diagrams are no longer legible, and I have a recollection that I made some typos that I found latter that I could not correct. But I did find that if you ignore the walls of the box, the invariant mass of the box of photons excluding the walls of the box is not invariant. This can be traced to the system not being isolated, if you compute the invariant mass of the box including the walls, the invariant mass is invariant as expected.

If I were to get motivated to work the problem again, I'd work it in a similar manner, but I'd use light rather than particles. The energy density winds up transforming as the square of the doppler factor, k = $\gamma(1-\beta)$, as was mentioned in a recent post.

If we take $\beta=3/5$, working it out quickly I get a doppler shift factor of k = 2 in one direciton, 1/2 in the other.

So in the rest frame we have (energy, momentum) = (1,1) for one light beam, and (1,-1) for the other, so in a box of unit length we have
E=2
p=0

In the moving frame, we have energy and momentum densities multipled by k^2, so the energy/momentum per unit length becomes (4, 4) and (1/4, -1/4) modulo sign issues, with the box length contracting to 4/5. So the total energy / momentum is, if I haven't made an error

E=17/5 (3.4)
p = 3

The sign of p depends on the sign convention and the diagram, but it doesn't matter, all we want to compute is p^2.

E^2 - p^2 is not the same, it was 4 in the unboosted frame, and 2.56 afer the boost. The invariant mass of the non-isolated system (the box of light without including the walls) is not invariant!

Hopefully this is right, but it's not impossible I made an error, I did it really fast.

12. Aug 11, 2018

### liuxinhua

I find that the invariant mass of the bound system may not be constant.
Is it possible to write an article, but my English is not good enough， and I don't have enough literature.
Is anyone willing to cooperate with me?

removed private data - mfb

Last edited by a moderator: Aug 11, 2018
13. Aug 11, 2018

### Staff: Mentor

Of an isolated system or a non-isolated system?

14. Aug 11, 2018

### pervect

Staff Emeritus
There's a paper that talks about some of this, though the focus of the paper is on thermodynamics, the point about the invarinat mass is peripheral. It's pretty easy to follow, I think.

Nakumura, "Covariant thermodynamics of an object with finite volume", https://arxiv.org/abs/physics/0505004

There are some other papers references in this paper that may be more directly to the point, but they're not as easily obtainable. Nakumura's paper is on arxiv, so it's easy to get the complete text. Here's a quote from the abstract, and a more detailed quote from the body of the paper.

15. Aug 11, 2018

### liuxinhua

Of a non-isolated system
https://en.wikipedia.org/wiki/Mass_in_special_relativity
proposed:
Conservation of invariant mass also requires the system to be enclosed so that no heat and radiation (and thus invariant mass) can escape. As in the example above, a physically enclosed or bound system does not need to be completely isolated from external forces for its mass to remain constant, because for bound systems these merely act to change the inertial frame of the system or the observer.
It means, Conservation of invariant mass for bound systems don’t need it is an isolated system.

For an object rest in K, if there is heat flux at the boundary of the object, it invariant mass will change.

As of floor #5 and #7
You May have changed its bound state, when you excite the hydrogen atom.
I don’t confirm: for an object rest in K, if there is heat flux at the boundary of the object, the object can still keep its bound state, even if the boundary remains unchanged.

But when you only accelerate a hydrogen atom, the hydrogen atom can keep its bound state.

So we need the strict definition of bound-system. And whether people think as invariant mass of bound-system will change.

Last edited: Aug 11, 2018
16. Aug 12, 2018

### Staff: Mentor

Ok, that is fine and is not a problem.

17. Aug 12, 2018

### liuxinhua

18. Aug 12, 2018

### Ibix

It's not the greatest pieve of writing ever, but I don't think the bit of the article you quoted is claiming that the mass of a bound system must be constant. The bit you quoted says:
That seems to me to be saying that the mass of a bound system can be constant even when not isolated from external forces, not that it must be constant even when not isolated.

19. Aug 12, 2018

### Staff: Mentor

Invariant mass of a non-isolated system may change.
I didn’t see any argument about bound systems in that page, but it correctly describes isolated systems:

“Note that the invariant mass of an isolated system (i.e., one closed to both mass and energy) is also independent of observer or inertial frame, and is a constant, conserved quantity for isolated systems and single observers, even during chemical and nuclear reactions.”

I don’t think being bound is relevant to the conservation of mass. I think it is just about being isolated or not.

20. Aug 12, 2018

### vanhees71

It's quite simple and the true physics content of Einstein's most famous formula $E=m c^2$, which however should rather read $E_0=m c^2$, where $m$ is the invariant mass of the system.

If you have a composite system like a bound state of particles, the invariant mass is not necessarily conserved. Let's first take a very simple case, namely a rather weakly bound state like a hydrogen atom, consisting of a proton and an elektron. Forget for this discussion for a moment that the proton is itself a composite object (but much more complicated and not completely understood today, see below). The bound states are very well known from quantum theory (even including radiative reactions due to qft, but that's not so important here). To define the invariant mass of the composite object, it's most easy to got into the center-momentum frame of the entire system, i.e., the inertial frame, where the total momentum of the hydrogen atom is 0 and suppose the hydrogen atom is in its ground state. Then the total energy is $E=m c^2$. If this hydrogen atom is isolated well enough nothing happens to it, and it's invariant mass stays constant. If you look at it from an arbitrary reference frame, the invariant mass is given by the usual formula $E^2-p^2 c^2=m^2 c^4$, which manifestly covariant description makes it evident that $m$ is a Lorentz scalar and thus called the "invariant mass". In fact in modern physics nobody uses any other kind of "mass" anymore. Unfortunately particularly "didactical books" tend to use old-fashioned notions of "relativistic masses", but that's off-topic here.

Now suppose a physicist is shooting with a laser at the hydrogen atom, bringing it in an excited state. Usually that happens by a single-photon absorption process, i.e., you have effectively a reaction $\text{H}+\gamma \rightarrow \text{H}^*$. The energy-momentum conservation tells you that the four-momentum of the hydrogen atom (starting from the center-momentum frame of the original hydrogen atom) after the reaction is
$$E^*=m c^2+\hbar \omega, \quad \vec{p}^*=\hbar \omega/c \vec{e}_1.$$
The invariant mass of the excited hydrogen atom thus is given by
$$m^{*2}c^4 = E^{*2}-\vec{p}^{*2}c^2 = (m c^2 + \hbar \omega)^2-\hbar \omega^2 = m^2 c^4 +2 \hbar \omega m c^2,$$
i.e., the invariant mass of the excited hydrogen atom is larger than the original invariant mass of the hydrogen atom in the ground state.

An extreme example are hadrons, whose mass is so far only poorly understood on a intuitive level. It's very well understood computationally from lattice quantum chromodynamics. The proton is a very complicated bound state of predominately light quarks and gluons. The light quarks have a mass of a few $\text{MeV}/c^2$, the gluons are massless (how to define these "masses" is itself very tricky). The proton itself has an invariant mass of about $940 \; \text{MeV}/c^2$. In a naive "bag model" one can think of this mass being generated by "confining" the quarks and gluons in some finite volume (the proton has a radius of about 1 fm) hold together by the strong interaction. Thus almost the entire proton mass is dynamically generated somehow by the strong interaction. That this picture is correct also quantitatively can seen today only by numerical calculations, using lattice-QCD, which reproduce the masses of the hadrons within a few percent.