I have two identical massess with rest mass $m_0$. One of them is made to move with very high velocity $v$, while other is at rest. At some point in time the moving mass collides with the stationary mass and gets stuck to it, now they both move as 1 body with some unknown velocity. My Doubt- I want to know the rest mass of this new body formed by collision of the old 2 massess. I am seriously surprised to find out that it is not $2m_0$. Please explain what's happening here and why $2m_0$ is not the right answer.
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In relativity, mass is the length of a vector. Specifically, it is $$m^2 c^2=E^2/c^2-p^2=-P_\mu P^\mu$$ where $P^\mu=(E/c,\vec p)$ is the four-momentum vector.
Because mass is the length of a vector, it is not additive. This is exactly the same as other lengths of vectors, like speed. If you have two velocity vectors $$\vec v=(3,0,0)$$$$\vec u=(0,4,0)$$ then the speeds are $v=3$ and $u=4$. When you add $\vec u$ and $\vec v$ the resulting speed is $5$, not $7$. To find it you have to add $\vec u+\vec v$ first, and then take the magnitude second to get the combined speed.
Since mass is the magnitude of the four-momentum vector, the same rule applies. You have to add the four-momenta first and take the magnitude second. So for the mass at rest $P^\mu=(m_0 c,0,0,0)$ and for the moving mass $Q^\mu=(\gamma m_0 c, \gamma m_0 v, 0,0 )$. So to get the total mass we first add the momenta $R^\mu=P^\mu+Q^\mu$ and then take the magnitude $$m^2 c^2 = -R^\mu R_\mu = 2\ (1+\gamma)\ m_0{}^2 c^2$$
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1$\begingroup$ Nice answer Dale. I was wondering, so I know 4-momentum is a conserved quantity. Is this how you found the equation with $R^\mu = P^\mu +Q^\mu$? $\endgroup$– JohnA.Commented Dec 28, 2023 at 14:10
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1$\begingroup$ @JohnA. no, that additivity is just a property of vectors. It applies to vectors regardless of whether they are conserved or not. If you have some physical quantity that doesn’t add, then you would not represent it by a vector $\endgroup$– DaleCommented Dec 28, 2023 at 14:14
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$\begingroup$ So you are saying that nowhere in your answer you used that energy and momentum are conserved? $\endgroup$– JohnA.Commented Dec 28, 2023 at 14:16
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$\begingroup$ @JohnA. that is correct. Energy and momentum are conserved, but I did not use that fact anywhere in the answer. For example, you could add a pair of four-force to get a net force. If you were interested in the magnitude of the net force you would use the same math as here. Regardless of the fact that force is not conserved. All that is needed is for them to be colocated vectors $\endgroup$– DaleCommented Dec 28, 2023 at 14:24
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$\begingroup$ I guess how do you know that the specific 4-vector of the resulting particle after the collision is equal to the sum of the 4-vectors of the incoming particles without using conservation. I agree vectors add like vectors but it seems to me you need to consider these conservation laws (and assume elasticity of collision) to add these and claim they result in the momentum 4-vector of the resulting particle $\endgroup$– JohnA.Commented Dec 28, 2023 at 14:29