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.