Proving momentum equation for neutrino/nucleon scattering

  • #1
99
2

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 [itex]p'^{2}[/itex]=[itex]m_{1}E_{2}/2[/itex], where p' is the momentum of the neutrino or nucleon in the center of mass frame, [itex]m_{1}[/itex] is the mass of the nucleon, and [itex]E_{2}[/itex] 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'=[itex]\frac{1}{\sqrt{1-(v/c)^{2}}}[/itex][p-[itex]\frac{vE}{c^{2}}[/itex]]

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

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


[itex]E^{2}[/itex]=[itex]p^{2}[/itex][itex]c^{2}[/itex]+[itex]m^{2}_{0}[/itex][itex]c^{2}[/itex]

The Attempt at a Solution



First find the velocity of the center of mass frame:

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

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

[itex]p_{1}[/itex]-[itex]\frac{vE_{1}}{c^{2}}[/itex]+[itex]p_{2}[/itex]-[itex]\frac{vE_{2}}{c^{2}}[/itex]=0
[itex]0[/itex]-[itex]\frac{vE_{1}}{c^{2}}[/itex]+[itex]p_{2}[/itex]-[itex]\frac{vE_{2}}{c^{2}}[/itex]=0
[itex]p_{2}[/itex]=[itex]\frac{vE_{1}}{c^{2}}[/itex]+[itex]\frac{vE_{2}}{c^{2}}[/itex]
[itex]p_{2}[/itex]=[itex]\frac{vE_1+vE_2}{c^2}[/itex]
[itex]p_{2}[/itex]=[itex]\frac{v(E_1+E_2)}{c^2}[/itex]
[itex]\frac{p_{2}c^2}{E_1+E_2}[/itex]=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 [itex]p_{1}[/itex], and [itex]m_{1}c^{2}[/itex] for [itex]E_{1}[/itex]:

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

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

[itex]p'^{2}_{1}[/itex]=[itex]\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}}[/itex]
[itex]p'^{2}_{1}[/itex]=[itex]\frac{m^{2}_{1}c^{2}E^{2}_{2}}{m^{2}_{1}c^{4}+2m_{1}c^{2}E_{2}}[/itex]
[itex]p'^{2}_{1}[/itex]=[itex]\frac{m_{1}E^{2}_{2}}{m_{1}c^{2}+2E_{2}}[/itex]

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

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
13,262
3,543
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: Pμtot
where P0tot= Etot/c and P1tot = Ptot,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 Pμtot, final = Pμtot, 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.
 
Last edited:
  • #3
99
2
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.
 

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