Question

5. A proton of mass at rest is bombarded from an accelerator beam by an antiproton of equal mass, the two particles annihilating each other into a pair of photons with momenta p, P Show that the angle between directions of flight of the photons is given by -me(A+ p.) PiP cos θ = 1 Show further that the angles 0,, 0, the scattered photons make with the incident beam obey cos = CP.P where Po, Eo are the momentum and energy of the incident antiproton. Verify that the 3-dimensional electric and magnetic fields measured by an observer with timelike 4-velocity u are given by 6.

Answer 1

5.Conservation of 4-momentum

in units of c=1.

p2, p2 COS内.

$E_0+m=p_1+p_2$Eq.1

  Eq.2

1 CO

$p_1\sin\theta_1=p_2\sin\theta_2$   Eq.3

Also,

$E_0^2=p_0^2+m^2$Eq.4

Using the equations 1,2 and 4, as follows

$(p_1+p_2-m)^2-(p_1\cos\theta_1+p_2\cos\theta_2)^2=m^2$

along with Eq.3 gives

$\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2=\cos(\theta_1+\theta_2)=\cos\theta$

to be

cos θ-1-rn (P1 + P2) p1p2

Using only equations 2 and 3 to eliminate $\theta_2$, we get

2pipo

Eq. 1 and 4 will give

$\cos\theta_1=\frac{(p_1-m)(E_0+m)}{p_1p_0}$

A similar expression can be obtained for $\theta_2$

6. Just use the definitions of F and u in one inertial frame of reference. Express in invariant forms and hence convert it into tensor equation. The most convenient frame is the stationary frame. So let the components of F be,

$F_{\mu\nu}=\begin{bmatrix} 0 & E_1 & E_2 & E_3\\ -E_1 & 0 & -B_3 & B_2\\ -E_2 & B_3 & 0 & -B_1\\ -E_3 & -B_2 & B_1 & 0 \end{bmatrix}$

and for stationary observer

u(1,0,0,0)

The signs may be different because different texts use different signs for the F tensor. The calculation from here is straight forward.