Purcell Problem 5.6                    Viva Horowitz

Mathematica did some algebra for me...

The speed of the frame:

v = c (1 - 1/γ^2)^(1/2) ;

u prime is the speed of the antiproton in frame S'.

up = (2v)/(1 + (v/c)^2) ;

γ prime is the γ for the antiproton in frame S'.

γp = 1/(1 - (up/c)^2)^(1/2)   // FullSimplify

1/1/(1 - 2 γ^2)^2^(1/2)

Choosing the positive square root of the above expression, we find that γ prime is:

γp = 2 γ^2 - 1 ;

γp /. γ→ 100

19999

β prime:

βp = ( 1 - 1/γp^2)^(1/2)

βp = βp /. γ→ 100

1 - βp //N

(1 - 1/(-1 + 2 γ^2)^2)^(1/2)

(600 1111^(1/2))/19999

1.25013*10^-9

Here's my numerical solution to equation (*).

θsol = NSolve[(Sin[θ]^2 (1 - βp^2))/(1 - βp^2 Sin[θ]^2)^(3/2) == 1/21/(1 - βp^2)^(1/2), θ, 20]

{{θ→ -1.570758003831914837}, {θ→1.570758003831914837}}

θp = θ /. θsol[[2]] ;

π/2 - θp  // ScientificForm

3.8322962981782×10^-5

c = 29979245800 cm/s ; γ = 100 ; d = 10^(-8) cm ; t = (2 d)/(βp c Tan[θp])

2.5566328989138*10^-23 s

Our main error in this approach arises from γ=100 being approximate.

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