Question

What is the momentum four-vector for the photons in the lab frame, higgs decay?

Answer 1

ANSWER }
 If two photons with energies E1E1 and E2E2 collide, with their trajectories making an angle θθ. Is the following right about individual initial 4-momentums and total initial 4-momentum? and how do we find the four-momentum in the center of momentum frame(COM)?

Attempt at Answer:(assuming without loss of generality that particle 1 is moving in + x direction and the whole motion is in x-y plane)
- The initial momentum for each particle is:

pμ1=(E1,E1,0,0)pμ2=(E2,−E2cosθ,E2sinθ,0)p1μ=(E1,E1,0,0)p2μ=(E2,−E2cosθ,E2sinθ,0)


hence the total initial momentum is:

pμT=(E1+E2,E1−E2cosθ,E2sinθ,0)pTμ=(E1+E2,E1−E2cosθ,E2sinθ,0)


- squared length of 4-momentum in lab frame (LBF):

(pTμpμT)LBF=(E1+E2)2−(E1−E2cosθ)2−(E2sinθ)2=2E1E2(1+cosθ)(pTμpTμ)LBF=(E1+E2)2−(E1−E2cosθ)2−(E2sinθ)2=2E1E2(1+cosθ)


- square 4-momentum in COM = square length of 4-momentum in LBF:

(pTμpμT)LBF=(pTμpμT)COM=2E1E2(1+cosθ)(pTμpTμ)LBF=(pTμpTμ)COM=2E1E2(1+cosθ)


Now to find the actual momentum in COM I use the above fact that square of the total four momentum is invariant under frame transformation and combining this with the fact that the square of 4-momentum of photon is zero we get:

pμ1(COM)=(E′,E′x,E′y,0)pμ2(COM)=(E′,−E′x,−E′y,0)p1(COM)μ=(E′,Ex′,Ey′,0)p2(COM)μ=(E′,−Ex′,−Ey′,0)


-So the total 4-momentum in COM is:

pμT(COM)=(2E′,0,0,0)pT(COM)μ=(2E′,0,0,0)


squaring it and equating to sqr of total 4-momentum in COM its gives:

E′2=E1E22(1+cosθ)=E1E2cos2θ2E′2=E1E22(1+cosθ)=E1E2cos2θ2


As I have the magnitude and I know that

E′2x+E′2y=E1E2cos2θ2Ex′2+Ey′2=E1E2cos2θ2

its evident that the two components will have the following form:

E′x=E1E2−−−−−√cosθ2cosθ′E′y=E1E2−−−−−√cosθ2sinθ′Ex′=E1E2cos⁡θ2cos⁡θ′Ey′=E1E2cos⁡θ2sin⁡θ′

Question: The way to find the angle θ′θ′ is using Lorentz boost matrix such that the spatial components of four momentum go to zero.