A question about invariant mass

[liuxinhua]

Now, we can sure: The invariant mass of a “Rigid body or Born rigid body" can’t keep constant when it is accelerated.
Any rigid body is bound systems.
The invariant mass of bound systems can’t keep constant when it is accelerated.

Wikipedia article may need to be modified.
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.Ibix said: ↑
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.
Ibix think :
the (invariant) mass of a bound system can be constant even when not isolated from external forces.

I think it means: sure, not can.
Because if it is “can”,then, for any system , its invariant mass can be constant for some process.
So if it is “can”, they say this here “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. ”has no means.

 

[PeterDonis]

Now, we can sure: The invariant mass of a “Rigid body or Born rigid body" can’t keep constant when it is accelerated.

That's not quite what has been shown in this thread. What has been shown is that the invariant mass of a spatially extended body is frame-dependent.

The invariant mass of a body undergoing Born rigid acceleration, relative to a fixed inertial frame, does indeed increase as the body accelerates; as @PAllen noted some posts back, this is fine because the body by itself is not a closed system, there is also whatever is producing the acceleration.

However, as @PAllen also noted some posts back, the invariant mass of a body undergoing Born rigid acceleration, in an inertial frame momentarily comoving with the body, stays the same as it accelerates; it is just the sum of the rest masses of the constituents. Intuitively, this is what most people probably think of when they are asked what happens to the invariant mass of a body as it accelerates. But of course this requires changing the inertial frame you are evaluating the invariant mass relative to, as the body accelerates.

So before one can even make statements about the invariant mass of a spatially extended body, one has to specify what inertial frame is being used.

 

[PAllen]

Let me emphasize this: the very notion of what a body is is frame dependent, to the extent you want treat it as a 3 d spatial object. This is simply because different frames use completely different sequences of sets of events to describe the body’s history due to relativity of simultaneity. The invariant notion of an extended body is world tube containing worldlines of constituent elements. In the general case, you cannot define any meaningful notion of invariant mass, even in SR. Only in the special case of Born rigid object do you have the feature that an MCIF of one world line is also an MCIF for all the world lines, and thus you can give a mass for the sequence of different MCIFs, and this mass does not change, and is the sum of rest masses. However, for any frame in which the rigid body is not at rest [edit: and not moving inertially], the invariant mass per that frame is greater than the sum of rest masses.

This, of course, suggests a fundamental limitation of the notion of invariant mass for extended bodies. The mass is invariant given a specified slice through the world tube. But in general, there is no natural slice through the world tube. It is entirely frame dependent.

 

[PAllen]

I think there are clearly inaccuracies in the Wikipedia description, not just bad wordings. This is not surprising and I long ago gave up on trying to maintain wiki accuracy on areas I know.

[edit: perhaps all that’s needed is a statement that an accelerating particle or body is not a closed or isolated system; you must include the sources of acceleration]

 

[PAllen]

I guess another point worth mentioning is that invariant mass for systems of particles in particle physics is well defined because the extent of the interaction region is tiny. In contrast, the problematic case of realistic extended bodies accelerating to substantial fractions of c per some observer have never been achieved.

 

[vanhees71]

I'd also like to point out some of the usual sign conventions that are different than what the OP used

Modifying wiki's sign conventions to use "i c t", https://en.wikipedia.org/wiki/Four-momentum, we'd write the following for the 4-momentum, which we'll calll $\vec{P}$, where E is the energy, p_x, p_y, and p_z are the x, y, and z components of the 3-momentum.

We've basically just added the "i" to wiki's approach, wiki doesn't use the i.

Wikipedia is better than many people think! At least they seem not to commit the sin to use the outdated "$\mathrm{i}c t$" formulation of SR. It's not plain wrong but very confusing, particularly for me as I'm involved in relativistic many-body QFT. There time is real as it must be. It's real since the very beginning of modern physics! So there's a real-time formalism, applicable to all of physics and a mathematical trick, called imaginary-time or Matsubara formalism, which simplifies calculations tremendously in thermal equilibrium, where you need the vertical part of the extended Schwinger-Keldysh time contour only. Another use of imaginary-time formulations is also vacuum QFT (which is of coarse also thermal equilibrium at 0 temperature and 0 chemical potentials), where the Euclidean formulation helps to do some calculations. Of course you have to pay a prize, because at the end you have to analytically contnue back to real time quantities, which is everything than straight forward. This holds the more true for lattice calculations which also are done in the imainary-time formulation for obvious reasons.

 

[liuxinhua]

In the general case, you cannot define any meaningful notion of invariant mass, even in SR. Only in the special case of Born rigid object do you have the feature that an MCIF of one world line is also an MCIF for all the world lines, and thus you can give a mass for the sequence of different MCIFs, and this mass does not change, and is the sum of rest masses. However,

Is there have any detailed information ?
What term is MCIF abbreviated?
Many of the information I saw was not in English.
So I may have some low-level problems. Excuse me.

 

[liuxinhua]

So before one can even make statements about the invariant mass of a spatially extended body, one has to specify what inertial frame is being used.

Always measure the invariant mass of this born rigid body in $K$, the invariant mass of this born rigid body is changed.

 

[PAllen]

Is there have any detailed information ?
What term is MCIF abbreviated?
Many of the information I saw was not in English.
So I may have some low-level problems. Excuse me.

MCIF is momentarily comoving inertial frame.

Not sure of a good reference for relativistic treatment of extended bodies, other than the special case of Born rigid motion.

 

[vanhees71]

[edit: perhaps all that’s needed is a statement that an accelerating particle or body is not a closed or isolated system; you must include the sources of acceleration]

That's indeed a very important point. Many misunderstandings (at least I had them, when I started to learn relativity) comes from not carefully keeping in mind that defining the total energy and momentum of extended systems is difficult. The only exception are indeed closed systems.

Math is, as always, your friend, and it's easy to see, where this problem comes from. First of all the natural description of relativistic physics are local laws for fields (classical and quantum, but I'll stay here within the realm of classical physics), i.e., you usually deal with local field equations of motion like, e.g., classical electrodynamics, where you describe matter naturally in terms of continuum mechanics, i.e., in terms of the energy-momentum tensor and electric charge and current densities.

Now, if you want to consider extended objects, you have to somehow integrate in a meaningful way the corresponding densities, well defined as relativistic field quantities. The mathematical difficulty is that naively taking spatial integrals in a given (inertial) reference frames in general do not lead to tensor quantities, but only when you have a conserved quantity in a closed system.

E.g. if you have electric charges and currents, it's described by the four-current density $j^{\mu}$, which fulfills (from the Maxwell equations alone due to Noether's 2nd theorem/gauge invariance) the continuity equation,
$$\partial_{\mu} j^{\mu}=0.$$
Then the total charge of the system is given by
$$Q=\int_{\mathbb{R}^3} \mathrm{d}^3 r j^0(t,\vec{r}).$$
The equation of continuity immediately guarantees that this is a conserved quantity, i.e.,
$$\dot{Q}=\int_{\mathbb{R}^3} \mathrm{d}^3 r \partial_t j^0(t,\vec{r})=-\int_{\mathbb{R}^3} \mathrm{d}^3 r \vec{\nabla} \cdot \vec{j}(t,\vec{r}=0,$$
because for realistic situation the current goes to 0 at infinity sufficiently large so that the surface integral at infinity from Gauss's theorem vanishes.

Nothing in the above calculation, however ensures that $Q$ is a scalar. But that's also easy to prove from the equation of continuity! Just use the 4D Gauss's theorem. As integration four-volume take a "cylindrical" volume with two time-like bottom and top non-overlapping hyper-surfaces and let the space-like "rim" of the cylinder go to infinity. Then the continuity equation shows
$$\int_{V^{(4)}} \mathrm{d}^4 x \partial_{\mu} j^{\mu}(t,\vec{x})=\int_{\partial V^{(4)}} \mathrm{d}^3 \sigma_{\mu} j^{\mu}(t,\vec{x})=0.$$
This shows that the integral over the top and bottom time-like hypersurface exactly cancel. As one of the hypersurfaces you can choose $t=\text{t}_0=\text{const}$ of the "lab frame" of the observer, which shows that the quantity which you get from integrating over either the top or the bottom is simply $Q$, and this value doesn't depend on the chosen time-like hypersurface. That's why $Q$ is a Lorentz scalar, and it's only a Lorentz scalar, because the equation of continuity holds.

In the same way it's clear that energy and momentum are field-theoretically described by the (symmetric) energy-momentum tensor $T^{\mu \nu}$, and the total momentum
$$P^{\mu}=\int_{\mathbb{R}^3} \mathrm{d}^3 r T^{\mu 0}(t,\vec{r})$$
is in general only a four-vector, if $\partial_{\nu} T^{\mu \nu}=0$, i.e., if you consider the total energy-momentum tensor of a closed system. Then the total energy and momentum defined by the non-covariantly written integral is nevertheless a four-vector, and it's conserved: $\dot{P}^{\mu}=0$.

 

[vanhees71]

MCIF is momentarily comoving inertial frame.

Not sure of a good reference for relativistic treatment of extended bodies, other than the special case of Born rigid motion.

A nice example is the following paper on the problem of radiation reaction of classical "point charges". Of course there's no such thing as a classical point charge, but the paper treats the case of an extended charged quasi-rigid body as a model of a classical "point charge", including the necessary Poincare stresses with a very careful analysis also of the issue of self-energy and momentum:

https://arxiv.org/abs/physics/0508031v3

 

[PeterDonis]

Always measure the invariant mass of this born rigid body in $K$, the invariant mass of this born rigid body is changed.

Yes, I've already said that we agree on that now. But measuring everything relative to one fixed $K$ is not the only possible choice.

 

[liuxinhua]

In my previous remarks, the Bound system I mentioned refers to Born rigid system.
Here, this should be a more formal definition of Born rigid body on three-dimensional

Under Newton's classical space-time, the Born rigid system is a system that the distance between any components not changes over time. It can also be understood as: there is no relative movement between any components.

Born rigid system is actually a rigid body. In the special relativity, rigid bodies do not exist.

And The distance between objects needs the coordinates of two positions measuring at the same time. But the simultaneity is no longer absolute. Therefore, we must redefine Born rigid system in the special theory of relativity.

There is no transformation between matter and energy in the Born rigid system.

Defined Born rigid system as: Give a unique markup for every component of the system, such as Ma, Mb…. For every component of the system, such as Ma, it may be non inertial motion and it has its own original timeτ. For every component and its every original time, such as Ma at time τ1, there exists an inertial reference frameK1. And at time t1 of K1 the component of the system Ma is rest in K1. Measured in K1 at time t1, the system has a material distribution and shape. For Ma at timeτ2, there exists another inertial reference frame K2. And at time t2 of K2 the component of the system Ma is rest inK2. Measured in K2 at timet2, the system has another material distribution and shape. If the material distribution and shape of the system don’t change with time τ(τ1,τ2…), the system is a Born rigid system.

 

[PeterDonis]

Born rigid system is actually a rigid body.

No, it is a body undergoing rigid motion. There is a difference.

In the special relativity, rigid bodies do not exist.

In SR, bodies which respond instantly to applied forces all throughout the body do not exist; effects of applied forces, like anything else, can only propagate through the body at the speed of light (or much slower, in the case of most actual bodies).

But in SR, it is perfectly possible for Born rigid motions of bodies to exist. You just have to arrange the applied forces appropriately on the different parts of the body. (Or, you can apply a constant force at one part and wait for the body to reach equilibrium; for an elastic body where the applied force is not above the elastic limit, the equilibrium motion of the body will in general be Born rigid.)

Defined Born rigid system as

Your definition is much too cumbersome. See here:

https://en.wikipedia.org/wiki/Born_rigidity#Definition