Relativistic elastic scattering in CM and Breit frames

[yelneerg]

Homework Statement

Take the case of elastic scattering (A+B -> A+B); if particle A carries energy E_A', and scatters at an angle θ, in the CM (center of mass/momentum frame), what is its energy in the Breit* frame? Find the velocity of the Breit frame (magnitude and direction) relative to the CM.

*The Breit frame (or 'brick wall' frame) is the system in which A recoils with its momentum reversed (p_after = -p_before), as though it had bounced off a brick wall.

Homework Equations

(Note: bold letters are 3-vectors, letters with a subscript x, y or z are vector components, and letters with a super or subscript μ are contravariant and covariant 4-vectors respectively.)

p^μ = (E/c,p_x,p_y,p_z)

Lorentz transformations for momentum and Energy
E/c = γ(E'/c - β*p_x')
p_x = γ(p_x' - β*E'/c)
p_y = p_y'
p_z = p_z'

(Where β = v/c and γ² = 1/(1-β²)

Lorentz Invariant
p_μ*p^μ = E²/c² - p² = m²c²
or in the more common form:
E² = p²c² + m²c⁴

Maybe more, I'm not sure.

The Attempt at a Solution

So in the CM frame the scattering angle is related to the relative masses/energies of the particles and since we "know" the energy of particle A in the CM we can say that the scattering angle, θ, is related to the total energy/mass of the system. This means I can write a set of 4 equations of the following form:

|p_Af'|*cos(θ) = p_Afx'

Which might be useful...

In the CM I can say that the total momentum 4-vector

p^μ_TOT' = ((E_A' + E_B')/c,0)

and in the Breit frame

p^μ_TOT = ((E_A + E_B)/c,p_A)

I think if I were to take the inner product of each of the total momentum 4-vectors with itself I could set the results equal to each other... Maybe mass comes in handy since B initially only has rest mass in the Breit frame?

Somehow I need to get γ so I can do the second part of the question...

I feel like I'm close to some pivotal revelation, just not seeing it. There are so many unknowns I'm not sure how to start reducing the number of unknowns I'm dealing with.

 

[yelneerg]

Alright, so after spending several more hours on this I think the key is to take the inner product of the initial and final momentum 4-vectors in each frame and set them equal:

Let p and q represent the initial and final momentum of particle A (with primes to indicate CM frame and non-prime as breit frame).

p_μq^μ = p'_μq'^μ

Now this comes out to:

E² - p*q = E'² - p'*q'

(Where E is the energy of particle A and * is the dot product)

From here it all works out (using the definition of dot product and some trig identities)

!HOWEVER!

This is ASSUMING that the initial and final energies of particle A are equal. CAN I ASSUME THIS?? It is a relativistic elastic collision so mass of each particle is conserved, total energy is conserved, KE is conserved, and momentum is conserved. I feel like there is a reason I can say that energy of A initial = energy of A final, but I can't quite come up with a reason that convinces myself. If someone can give me a good reason that I can say that, then the rest of this problem I can do.

 

[Dick]

Well, m^2=E^2-p^2. As you said m doesn't change, and in the Breit frame |p| doesn't change. So E can't change. In general, a "brick wall" can absorb momentum, not energy. How's that?