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A perfectly inelastic central collision of two equal relativistic particles whose kinetic energies are equal to their resting energies results a single relativistic particle (and nothing but it). The mass of the resulting particle is greater than the sum of the masses of the particles that collided, but why and how?

pat._
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2 Answers2

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In relativity the mass of a system is $$m^2 c^2 = E^2/c^2 - p^2$$ For a system of multiple parts you add the $E$ for each part and the $\vec p$ for each part to get the $E$ and $\vec p$ for the whole system. Then you find the resulting $m$ for the system using the formula above.

So for $(E,\vec p) = (E_1 + E_2,\vec p_1+ \vec p_2)$ we have $$m^2 c^2 = E^2/c^2 - p^2 = (E_1+E_2)^2/c^2-|\vec p_1 + \vec p_2|^2$$$$>(E_1{}^2/c^2-p_1{}^2)+(E_2{}^2/c^2-p_2{}^2)=m_1{}^2 c^2 + m_2{}^2 c^2$$

So the mass of a system is always greater than the sum of the masses of the parts. This is just the triangle inequality $A^2+B^2>C^2$ from the Pythagorean theorem, but modified for spacetime.

Dale
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(An extension of my original comment)

If you draw an energy-momentum diagram of the totally-inelastic collision (using the conservation of total 4-momentum), you get a triangle with future-timelike legs.

This is similar to the spacetime diagram of the clock-effect/twin-paradox, where the total proper time of the traveling twin between the separation and reunion events is less than that of the inertial twin.

Possibly useful: my answer https://physics.stackexchange.com/a/318858/148184 to Mass Addition in Special Relativity

Ultimately, it's due to the "reverse-triangle inequality" property of Minkowski spacetime. See, for instance, my answer to Is there a space-like equivalent of the Twin Paradox? (I see @Dale made a similar comment.)

robphy
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