Is energy mass equivalence ficticious when not at rest?

Question

Here are some examples from my textbook I will refer to, which I could not copy paste,

I understand that $E=mc^2$ applies fine for thinking about rest energy, but when the body is not at rest, would this be the change in fictitious "relativistic mass", or is this a legitimate phenomenon? In example 1, this would then be true, because the torch is at rest. In example 2, this would be fictitious because the movement of the car only affects its inertia, not its mass. In example 3 is where I would struggle to draw the line, because I would assume the nucleus to be at rest, but I think this would be used either way? Motion was not considered at all. I understand that the binding energy of a nucleus decreases some kind of mass than if the nucleons were considered separately, is this the true mass, or a "relativistic" mass? In other words, is the only reason the measured mass of a nucleus is different to its individual nucleons due to a relativistic effect?

Answer 1

The rest mass is mass as measured in rest frame. But the rest frame of composite system is not equivalent to rest masses of its constituents. The rest frame of composite system is frame in which center of mass has zero velocity.

So when you have jar of some gass its rest mass includes inner kinetic energy of all the molecules. So when the gass gets cooled down, the rest mass of the jar decreases, even though the rest mass of the constituents remains the same. Similarly with nucleus.

This works, because rest mass is given by norm of 4-momentum $p\cdot p=-m^2c^2$. For system of two particles however you get: $$m^2c^2=(p_1+p_2) \cdot (p_1+p_2)=-2m^2_0c^2+2 p_1 \cdot p_2,$$ where the last term is given by inner kinetic energy of the constituents. This can be demonstrated for two particles with same mass in center of mass frame (thus they have opposite velocities): $$p_1 \cdot p_2=-\gamma^2m_0^2(c^2+v^2_0),$$ where quantities with index zero belong to constituent particles. Thus: $$-m^2c^2=-2m^2_0c^2 -2\gamma^2m_0^2(c^2+v^2_0)=-4m^2_0\gamma^2c^2=-4\frac{E_0^2}{c^2},$$ so it holds $mc^2=2E_0$, where $E_0$ is relativistic energy of constituent particles which includes both rest mass and kinetic energy of particles in center of mass system.

The point is that rest mass of system includes rest masses of constituent particles plus addition due to inner energy of the system. When system looses inner energy, the rest mass of the system also decreases.

Answer 2

'Relativistic mass' is often unhelpful, but it is not fictitious. It is legitimate.

In your second example, the moving car, although it is old-fashioned to say that 'the mass increases' it is still the case that the moving car has more inertia than a stationary one. If you apply a specific impulse to it, the change in velocity will be less. Whether you ascribe this to an increase in $m=m_0 \gamma$ or the gamma factor separately is just convention. Not a big deal, don't worry about it.

Consider a closed 'black box' system containing one or several particles, say $\pi^0$ mesons, which can decay to photons and also be formed from photons. Sometimes the box contains mostly mesons, which have rest mass; sometimes it contains mostly photons, which don't. But from the outside we don't know whether the energy inside is the former or the latter or somewhere in between: all we know is the total (inertial) mass - we give it a kick and see how fast it moves. That must be the same whatever the internal state. So don't try and distinguish between 'fictitious' and 'legitimate' mass/energy effects.