Applications of Special Relativity

Problem : What is the 4-vector inner product of ( E/c , (2E cosθ)/c , (2E sinθ)/c , 3E/c ) with the 4- momenta of a photon with energy G traveling in the x - z plane at an angle θ to the x -axis?

Solution for Problem 1 >>

A photon has no mass so the magnitude of its momentum is just p = = G/c . The photon is in the x - z plane so its 4-momentum is (G,(G cosθ)/c, 0,(G sinθ)/c) . Taking the inner product we have:

(E/c,(2E cosθ)/c,(2E sinθ)/c, 3E/c).(G/c,(G cosθ)/c, 0,(G sinθ)/c) = - - =

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Problem : The opposite to matter-antimatter annihilation is called 'pair creation.' This occurs when an photon 'decays' into one matter particle and one antimatter particle. Consider the simpler situation in which two photons, each with energy E , collide at an angle θ and create a particle of mass M . What is M in terms of E and θ ?

Solution for Problem 2 >>

As usual, write down the 4-momenta of the photons P _{γ 1} = (E/c, E/c, 0, 0) and P _{γ 2} = (E/c, E cosθ/c, E sinθ/c, 0) . The momentum is equal to the energy since the mass of the photons is zero and p = = 0 . The 4-momentum of the final particle, then, is P _M = (2E/c,(E + E cosθ)/c, E sinθ/c, 0) , since energy and momentum are conserved. Then taking the inner product:

P M 2 = (2E/c)2 - (E + E cosθ)2/c 2 - (E sinθ)2/c 2 = M 2 c 2
âá’M 2 =
âá’M = =

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Problem : A particle with mass m and energy G collides with a stationary particle of mass M . What is the speed of the frame in which the total momentum is zero?

Solution for Problem 3 >>

Say that m is moving in the positive x direction. The frame in which E is measured we will call the lab frame F . The frame of zero momentum call F' . The momentum of m is p = . We can employ the equivalent of the Lorentz transformations for energy and momentum which tell us that:

p' = γ(p - vE/c 2)

Let us transform the total energy and momentum. We know p' = 0 because the primed frame is the one with momentum zero. The only momentum comes from m but the total energy is E = G + Mc ² . Thus we have:

0 = γ( - v(G + Mc 2)/c 2)

Solving for v we have:

v =

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Problem : A particle with mass M _A decays into two particles with masses M _B and M _C respectively. What are the energies and momenta of M _B and M _C ?

Solution for Problem 4 >>

>From conservation of momentum we know that particles B and C must have equal and opposite momenta. Thus E _B ²/c ² + M _B ² c ² = p ² = E _C ²/c ² - M _C ² c ² . From conservation of energy we know: E _B + E _C = M _A c ² . Solving these equations for the energies gives:

E B =
E C =

Solving for p , which is the momentum of both B and C we have:

p =

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Problem : We can apply what we know about collisions to an analysis of Compton scattering. A photon of energy E = hν = hc/λ in incident on a particle of mass m . The photon scatters off at an angle θ with respect to its original direction. What is the new wavelength λ _f of the photon?

Solution for Problem 5 >>

Write down the 4-momenta for the photon and the particle: P _γ = (h/λ, h/λ, 0, 0) and P _m = (mc, 0, 0, 0) . The 4-momentum of the photon after the collision is P _{γ f} = (h/λ _f, h cosθ/λ _f, h sinθ/λ _f, 0) . It is not necessary to calculate the final 4- momentum of the particle in order to solve the problem. Conservation of energy and momentum tells us that P _γ + P _m = P _{γ f} + P _mf so P _γ + P _m - P _{γ f} = P _mf . Taking the inner product of both sides:

P γ 2 + P m 2 + P γ f 2 +2P m(P γ - P γ f ) - 2P γ P γ f = P mf 2
âá’0 + m 2 c 2 +0 + 2mc - -2 (1 - cosθ) = m 2 c 2

Multiplying through by and solving we find:

λ f = λ + (1 - cosθ)

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