What is the four-vector momentum of the fused particle X in the lab frame?

[DanAbnormal]

Homework Statement

Hey, here's a question I am not too sure about.

An electron and a positron travel on opposite directions along the z-axis (with the electron traveling in the +z direction) and collide head-on. Their energies are Ee and Ep respectively, and they both have mass m.

a) Write down their four vector momenta in the lab frame in terms of Ee, Ep and m.

The electron and positron fuse briefly to form a single particle, labelled X. For the case when Ee = 50 GeV, Ep = 100 GeV and m is negligible, calculate the four-vector momentum of X i the lab frame, and its rest mass.

Homework Equations

E = Mc^2

M = relitivistic mass

M = m.gamma

m = rest mass

The Attempt at a Solution

This is what I have done, but I do not know if its right.
Well for the four vector momenta of Ee and Ep I have, respectively:

0
0
mc
Ee/c

and

0
0
-mc
Ep/c

Which, with Ee = 50 GeV and Ep = 100 GeV, makes the four vector momentum of particle X:

0
0
0
150 GeV/c

Using E = Mc^2 the relitivistic mass is 8 x 10^-25 kg.

With the M = m.gamma equation I get stuck, because according to their four vector momenta, their speeds are c, which would make gamma zero.
Unless the four vector momentum of Ee (and Ep as negative) should be this instead:

0
0
mv
Ee/c

Im not sure, could you point me in the right direction please?
Thanks
Dan

 

[Final]

Hi,,,
It's better to use always the rest mass and the equation $$m^2=E^2-|\vec{p}|^2$$ (the norm of the four vector momenta).
In your case the 4-vectors are $$P_e=(E_e,\sqrt{(E_e^2-m^2)}\hat{z})$$ and $$P_p=(E_p,-\sqrt{(E_p^2-m^2)}\hat{z})$$.
$$P_X=P_p+P_e=(E_e+E_p,(E_e-E_p)\hat{z})$$ where I have negletted the mass. $$M_X^2=(E_e+E_p)^2-(E_e-E_p)^2$$
The norm of the 2 momentums are not equal... You are in the lab frame not in the centre of mass...

 

[DanAbnormal]

Hi,,,
It's better to use always the rest mass and the equation $$m^2=E^2-|\vec{p}|^2$$ (the norm of the four vector momenta).
In your case the 4-vectors are $$P_e=(E_e,\sqrt{(E_e^2-m^2)}\hat{z})$$ and $$P_p=(E_p,-\sqrt{(E_p^2-m^2)}\hat{z})$$.
$$P_X=P_p+P_e=(E_e+E_p,(E_e-E_p)\hat{z})$$ where I have negletted the mass. $$M_X^2=(E_e+E_p)^2-(E_e-E_p)^2$$
The norm of the 2 momentums are not equal... You are in the lab frame not in the centre of mass...

Sorry, I am not too sure what most of that means, could you explain?