A simple problem of conservation of momentum

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SUMMARY

The discussion centers on the conservation of relativistic momentum during an inelastic collision between two particles of identical mass, each moving at speeds v and -v. The initial momentum in frame S is zero, while in frame S', the momentum is calculated using the relativistic momentum formula. The key error identified was neglecting the increase in rest mass post-collision, which is correctly addressed through the conservation of four-momentum. The final conclusion emphasizes the importance of using four-vectors for accurate momentum calculations in relativistic scenarios.

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Trying to understand the conservation of relativistic momentum, I thought of this problem.
It's very simple, and possibly my mistake is embarrassing so I apologize in advance :blushing:

Two particles of identical mass m and speeds v and -v in some frame S collide inelastically and they stop. The relativistic momentum in S is zero all the time.
Now, in system S' moving at -v, the initial momentum should be

p = m v' / √(1-v'2/c2) [1]

where v' is the velocity of the other particle in S'

v' = 2v / (1+v2/c2) [2]

if I substitute [2] into [1] I can get the momentum before impact as a function of v. After some manipulation I get

p = 2mv / (1 - v2/c2)

But the momentum in S' after the impact is that of a particle of mass 2m moving with speed v. So its momentum should be

p = 2mv / √(1 - v2/c2))

so - where did I go wrong?
Thank you
 
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The easiest way to handle these types of problems is through the conservation of the four-momentum. Here is a wikipedia page on four-momentum: http://en.wikipedia.org/wiki/Four-momentum

Using units where c=1 the four-momentum of the particles in S are:
##P_1=(\gamma m, \gamma m v, 0, 0)##
##P_2=(\gamma m, -\gamma m v, 0, 0)##
where ##\gamma=1/\sqrt{1-v^2}##

The four-momentum is conserved, so after the collision we have:
##P_3=P_1+P_2=(\gamma 2 m,0,0,0)##
Note that the mass after the collision is ##m_3=|P_3|=\gamma 2 m## so the mass of the combined particle is greater than the sum of the masses of the original particles! This is the key part that you missed by doing it the long way rather than using four-vectors.

Now, in frame S' we have:
##P_1=((2\gamma^2-1)m, 2mv\gamma^2,0,0)##
##P_2=(m,0,0,0)##
##P_3=(2 m \gamma^2, 2 m v \gamma^2,0,0)##
Note that the rest mass after the collision is again ##m_3=|P_3|=\gamma 2 m##
 
Of course! I forgot about the increased rest mass because of the energy-mass conservation principle... That's where my missing γ went!
I will start using four-momentum but actually this mistake has helped me understand things better.
Thank you very much
 

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