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If we have two particles having same rest mass say $m$ and equal velocity approaching $c$ and they do a head on collision and stick together, then the mass of both particles after collision should be equal to 2(relativistic mass) . My view is it should be 2(rest mass) because they are no longer in motion so their masses should be again equal to rest mass. But by energy conservation it should be equal to relativistic mass. What do actually happen here?

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  • $\begingroup$ They lost all their kinetic energy in the form of EM waves $\endgroup$ – Aniansh Mar 5 '18 at 2:15
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The reverse of this process happens in radioactive decay: a nucleus with some mass (and hence weight) decays to some number of products. The sum of the mass of those products is less than the original nucleus’s mass. The difference makes up the kinetic energy of the products.

Now imagine it in reverse: the total energy, including kinetic, makes up the final mass.

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The energy of the system won't change. The energy of the particles alone might, so their relativistic mass might too. As was pointed out in a reply the particles might have emitted radiation that would balance out this difference in relativistic masses, though the emitted radiation would still be a part of the system, so the system's energy wont change. But if we say they didn't emit radiation we can still find places where the difference in mass was sunk in.

First there is heat energy. The kinetic energy of the "obvious" movement might have been converted to heat, which is a mixture of kinetic and potential energy. Which brings me to the second point - potential energy. Potential energy is a characteristic of the system, rather than the particles. So it is possible that part of the relativistic mass difference of the particles was transferred to potential energy in fields corresponding to the fundamental interactions that the particles can participate in.

Finally something that might be worth mentioning is that two particles can not just stick together and stop moving. This implies that both their position and momentum are perfectly defined, which violates the Heisenberg's uncertainty principle.

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  • $\begingroup$ Thanks for answering but I was thinking if I take weight of both particles after collision what will it be? $\endgroup$ – Himanshu Tyagi Mar 5 '18 at 3:40
  • $\begingroup$ Measuring rest masses would give the same result at the start and at the end, measuring the relativistic masses - wont, because the relativistic mass doesn't take in account potential energy. So it goes: mass of the whole system - unchanged; relativistic mass - reduced; rest mass - unchanged; mass stored in potential energy - increased by the amount that relativistic mass was reduced. $\endgroup$ – K. Kirilov Mar 5 '18 at 4:02
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The relativistic mass after the inelastic collision is defined by the total energy of the particles via $$m_{rel}=\frac{E_{tot}}{c^2}$$ The total energy is just the sum of the energies of the two particles before collision $$E_{tot}=E_1+E_2=m_{1rel}c^2+m_{2rel}c^2$$ with $$m_{1rel}=m_{2rel}=\frac {m_0}{\sqrt{1-v^2/c^2}}$$ where $m_0$ is the rest mass of the particles. Thus the relativistic mass of the two particles after collision is just the sum of the relativistic masses of the two particles $$m_{rel}=m_{1rel}+m_{2rel}=\frac {2m_0}{\sqrt{1-v^2/c^2}}$$ The kinetic energies of the particles after the collision is zero and has converted into an equal amount of internal energy (e.g. heat). Thus the mass of the combined particles after collision is larger than the sum of their rest masses.

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  • $\begingroup$ Well then the total weight on earth would be relativistic mass times g? $\endgroup$ – Himanshu Tyagi Mar 5 '18 at 3:42
  • $\begingroup$ @HimanshuTyagi - Yes, the total gravitational force of the combined particles after the inelastic collision would be $g m_{rel}$. $\endgroup$ – freecharly Mar 5 '18 at 4:55
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When working with velocities close to c, one should be working with four vectors,

In special relativity mass is not a conserved quantity. It uniquely defines quantum mechanical particles by the length of the four vector describing them called the rest mass, or the invariant mass, characterizing each particle, invariant under Lorentz transformations.

The term relativistic mass and its algebra has fallen by the way side , exactly because it causes confusions as the one in the question. It connects Newtonian inertial mass concepts with the relativistic energy concepts and is useful only when thinking of spaceships and the fuel they will need to reach relativistic energies.It is not used in particle physics.

Suppose the two particles you are assuming are elementary particles of the standard model of particle physics.They cannot "stick together" inelastially, because energy would not be conserved as you state in the question. The experiment has been done several times, in e+e- scattering, and there is no channel where at the center of mass a single entity forms/sticks together, except within the Heisenberg uncertainty principle, resonances as seen here:

e+e-

These are hadronic crossections, i.e. e+e- scattering into various hadrons.

Note the peaks as the energy of the beams increases, these are resonances, with the necessary mass for conservation of energy but decaying very fast into their constituent elementary particles, within the HUP. For a Δ(t) they make up a new massive entity , which has to decay into standard model particles because they are resonances. This is the experimental fact, and lorentz transormations describe it exactly.

Now in nuclear physics, if a stable state could be reached in the passing over the resonant part of the potential, it would be stable but not completely inelastic because there is no just "sticking" in quantum physics. There will be other particles taking away kinetic and binding energy ( the sticking part in quantum regime) as in this example of fusion in the sun. Proton proton fusion will always give a deuteron which has a mass lower than two protons, and thus at least a positron and an electron neutrino leave the interaction region balancing the energy budget. Note that also quantum numbers have to be conserved, hence the positron electron neutrino pair.

In general , in the quantum regime, stability ( sticking) depends on stable energy levels in nuclear matter otherwise it is just resonances within the HUP, and quantum number conservation has to be obeyed.

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