Proving momentum equation for neutrino/nucleon scattering

[gildomar]

Homework Statement

Prove the relationship between the momentum of the neutrino or nucleon in an elastic scattering of them in the center of mass frame is p'^{2}=m_{1}E_{2}/2, where p' is the momentum of the neutrino or nucleon in the center of mass frame, m_{1} is the mass of the nucleon, and E_{2} is the relativistic energy of the neutrino in the laboratory frame. Assume the nucleon is initially at rest in the laboratory frame.

Homework Equations

p'=\frac{1}{\sqrt{1-(v/c)^{2}}}[p-\frac{vE}{c^{2}}]

p'_{1}: momentum of nucleon in center of mass frame
p_{1}: momentum of nucleon in laboratory frame
p'_{2}: momentum of neutrino in center of mass frame
p_{2}: momentum of neutrino in laboratory frame
E'_{1}: energy of nucleon in center of mass frame
E_{1}: energy of nucleon in laboratory frame
E'_{2}: energy of neutrino in center of mass frame
E_{2}: energy of neutrino in laboratory frame

v: velocity of center of mass frame relative to laboratory frame


E^{2}=p^{2}c^{2}+m^{2}_{0}c^{2}

The Attempt at a Solution

First find the velocity of the center of mass frame:

p'_{1} + p'_{2}=0
\gamma[p_{1}-\frac{vE_{1}}{c^{2}}]+\gamma[p_{2}-\frac{vE_{2}}{c^{2}}]=0
p_{1}-\frac{vE_{1}}{c^{2}}+p_{2}-\frac{vE_{2}}{c^{2}}=0

Since the nucleon is at rest in the laboratory frame, p_{1} is 0:

p_{1}-\frac{vE_{1}}{c^{2}}+p_{2}-\frac{vE_{2}}{c^{2}}=0
0-\frac{vE_{1}}{c^{2}}+p_{2}-\frac{vE_{2}}{c^{2}}=0
p_{2}=\frac{vE_{1}}{c^{2}}+\frac{vE_{2}}{c^{2}}
p_{2}=\frac{vE_1+vE_2}{c^2}
p_{2}=\frac{v(E_1+E_2)}{c^2}
\frac{p_{2}c^2}{E_1+E_2}=v

Then get an expression for the squared momentum of the nucleon in the center of mass frame, using the above for the velocity, 0 for p_{1}, and m_{1}c^{2} for E_{1}:

p'_{1}=\frac{1}{\sqrt{1-(v/c)^{2}}}[p_{1}-\frac{vE_{1}}{c^{2}}]
p'_{1}=\frac{1}{\sqrt{1-(v/c)^{2}}}[0-\frac{vm_{1}c^{2}}{c^{2}}]
p'_{1}=\frac{1}{\sqrt{1-(v/c)^{2}}}[{-vm_{1}}]
p'^{2}_{1}=\frac{m^{2}_{1}v^{2}}{1-(v/c)^{2}}
p'^{2}_{1}=\frac{m^{2}_{1}}{1-(\frac{p_{2}c}{E_1+E_2})^{2}}(\frac{p_{2}c^{2}}{E_1+E_2})^{2}
p'^{2}_{1}=\frac{m^{2}_{1}(E_1+E_2)^{2}}{(E_1+E_2)^{2}-p^{2}_{2}c^2}\frac{p^{2}_{2}c^{4}}{(E_1+E_2)^{2}}
p'^{2}_{1}=\frac{(m^{2}_{1}c^2)(p^{2}_{2}c^2)}{(E_1+E_2)^{2}-p^{2}_{2}c^2}
p'^{2}_{1}=\frac{(m^{2}_{1}c^2)(p^{2}_{2}c^2)}{(m_{1}c^2+E_2)^{2}-p^{2}_{2}c^2}
p'^{2}_{1}=\frac{(m^{2}_{1}c^2)(p^{2}_{2}c^2)}{m^{2}_{1}c^{4}+2m_{1}c^{2}E_{2}+E^{2}_{2}-p^{2}_{2}c^2}

Since the book was written before it was discovered that neutrinos have mass, the p_{2}c was assumed to be equal to just E_{2}:

p'^{2}_{1}=\frac{(m^{2}_{1}c^2)(E^{2}_{2})}{m^{2}_{1}c^{4}+2m_{1}c^{2}E_{2}+E^{2}_{2}-E^{2}_{2}}
p'^{2}_{1}=\frac{m^{2}_{1}c^{2}E^{2}_{2}}{m^{2}_{1}c^{4}+2m_{1}c^{2}E_{2}}
p'^{2}_{1}=\frac{m_{1}E^{2}_{2}}{m_{1}c^{2}+2E_{2}}

And then the only way to get the answer from there would be to assume that E_{2} was much greater than m_{1}c^2, but I'm not sure if I'm allowed to assume that. Or did I screw up the calculations somewhere?

 

[TSny]

Hello, gildomar.

I haven't checked all your steps, but I think your answer is correct. As you say, it appears that you need to assume E₂ >> m₁c² and make an approximation to get the result given in the statement of the problem.

You can avoid a lot of algebra if you work with the concept that the "length" of a 4-vector is the same in all frames.

Total energy and total momentum can be combined to make a 4-vector: P^μ_tot
where P⁰_tot= E_tot/c and P¹_tot = P_tot,x. You can forget the y and z components for this problem.

The square of the length of the 4-vector is (P⁰)²-(P¹)²

Conservation of energy and momentum implies P^μ_{tot, final} = P^μ_{tot, initial} for each component μ.

Therefore the length of the final 4-vector must equal the length of the initial 4-vector. And you can evaluate the length in any frame since the length is invariant.

See what you get if you set the square of the length of the initial 4-vector in the lab frame equal to the square of the length of the final 4-vector in the CM frame.

 

[gildomar]

I'll try that, but the way that it's worded in the book implies that I need to use the momentum/energy transformation equation that I showed at the start.