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

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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?

Qmechanic
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jamie
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  • The concept of "relativistic mass" is in decline. It is better IMO to speak solely of the rest mass (which is frame invariant, but can still change e.g. if you heat the object up). You can then formulate all of the necessary equations for total energy/kinetic energy etc. in terms of this rest mass. – 13509 May 14 '20 at 08:41
  • @JamesWirth I agree completely, but the course I'm doing is introductory to nuclear physics so no distinction is made between relativistic and rest mass. I Just wondered how true the cases they presented and using E=mc^2 for a body not at rest would be when taking the effects of relativity seriously. – jamie May 14 '20 at 08:55
  • It's unfortunate that you're using a textbook that uses relativistic mass. Modern treatments of special relativity avoid it because it can lead to confusion. See https://physics.stackexchange.com/a/133395/123208 – PM 2Ring May 14 '20 at 08:57
  • @PM2Ring Well, I don't know whether if it uses relativistic mass, that's kind of the question here, it just says "mass" in the examples above and claims that all energy changes can be governed by e=mc^2. I have a little more SR experience than the book anticipates, so I want to know whether it is talking about relativistic mass(but just calls it mass) or if this equation applies to moving bodies also(without gamma) – jamie May 14 '20 at 09:59
  • Oh, it's definitely using relativistic mass in example 2. That example is a bit silly though, since for the car to increase its KE it burns fuel and air and produces exhaust gases, and the masses involved in that dwarf the mass equivalence of the KE. (But if you could track all of that energy, including the rest masses of everything, you'd find that energy is conserved). – PM 2Ring May 14 '20 at 10:17

2 Answers2

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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.

Umaxo
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  • This answer seems to fit a related question thats asks if water "in a box" being heated gains mass. Rest mass colloquially seems to be opposed to the relativistic mass gained by acceleration of some "system". Why does heating up not increase weight of water is parallel to why does emitting light decrease weight of torch. I wouldn't use the concept of "rest" mass to explain, "just mass" - the torch is a jar, it's not the car. What about the second example given ... – Peter Bernhard Nov 11 '22 at 19:18
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'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.

RogerJBarlow
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