Relativity - proton-proton collision

[alfredbester]

A Proton, m_{1} with Kinetic energy T = 200MeV strikes a stationary proton in the lab frame.
p + p -> p + p + X

what is the maximum mass of X, which can be produced.

I think I need to make use of E^2 - P^2 c^2 is invariant.
and
In S: E = (m_{1} + m_{0})c^2 , p = p_{1}
In S ' :

E^2 - P^2 c^2 = E^2_{1}+ 2m_{0}E_{1}c^2 + m^2_{0}c^4 -T^2 (1) where E_{1} = T + m_{0}c^2
I'm not sure what is happening in the centre of mass frame, I thought that the particle would have maximum mass when there was zero K.E i.e E' = (2m_{0} + m_{x})c^2, p' = 0 in the lab frame but I got lost when I tried to equate this with (1).

 

[Meir Achuz]

Find the center of mass energy W, using W^2=(T+M+M)^2-p^2.
This can be solved to T in terms of W. Then W=2M+X will give T.

 

[alfredbester]

Thanks.

I have a similar question, for electron-proton collision and I'll try to post my full working as I'm not really sure on these problems at all. Could someone check my answer please.

An electron with with total energy E collides with a proton at rest
e + p -> e + p + X, find the maximum possible mass of X making use of invariants.
Masses quoted are all rest masses.

P_{0} = E / c, p^2 = -m^2 c^2
E_{t} = E + m_{p}c^2 , |p_{t}| = sqrt[{(E^2 / c^2) + m^2_{e}c^2}]

E^2_{t} - |p_{t}|^2 c^2 = (m_p + m_e + m_x)^2 c^4
Therefore
E^2 + 2Em_{p}c^2 + m^2_{p}c^4 - (E^2 - m^2_{e}c^4) = <br /> (m^2_{e} + m^2_{p} + m^2_{x} + 2m_{e}m_{p} + 2m_{e}m_{x} + 2m_{p}m_{x})c^4

Factoring:
m^2_{x} + 2m_{x}(m_{e}+m_{x}) + 2m_{e}m_{p} = 2Em_{p}c^{-2}
E = T + m_{e}c^2 <br /> =&gt; m^2_{x} + 2m_{x}(m_{e}+m_{p}) + 2m_{e}m_{p} = 2[Tm_{p}c^{-2} + m_{e}m_{p}]<br /> =&gt; m^2_{x} + 2m_{x}(m_{e}+m_{p}) - 2Tm_{p} = 0
multipled the above by c^2, so I can use masses in MeV.

If the initial kinetic energy, T, of the electon is T = 5000meV, m_{e} = 0.511 MeV, m_{p}. Plugging into a quadratic I get m_{x} = 2265MeV. Is this realistic?