[1+ m2 c 4 4EE γ. The equations of conservation of energy and momentum are. E + E γ p p γ

Size: px

Start display at page:

Download "[1+ m2 c 4 4EE γ. The equations of conservation of energy and momentum are. E + E γ p p γ"

Alannah Day
5 years ago
Views:

1 Physics 403: Relativity Homework Assignment 2 Due 12 February Inverse Compton scattering occurs whenever a photon scatters off a particle moving with a speed very nearly equal to that of light. Suppose that a particle of rest mass m and total energy E collides head on with a photon of energy E γ. Show that the scattered photon has energy E [1+ m2 c 4 4EE γ Ultra-high energy cosmic rays have energies up to ev. How much energy can a cosmic ray proton transfer to a microwave background photon? The collinear collision is pictured below: ] 1 Proton: energy E γ Photon: energy E Before Collision ///////// Photon: energy E γ Proton: energy E //////// After Collision The equations of conservation of energy and momentum are E + E γ p p γ = E + E γ = p + p γ Since the initial and final photon are massless, we have E γ = cp γ and E γ = cp γ. Let us solve for the final energy and momentum of the proton:

2 E cp = E + E γ E γ = cp E γ E γ Thus E 2 (cp ) 2 = (E + E γ E γ )2 (cp E γ E γ )2 = E 2 (cp) 2 + 2E(E γ E γ)+2cp(e γ + E γ) 4E γ E γ Let us make use of the energy-momentum relations E 2 (cp ) 2 = m 2 c 4 = E 2 (cp) 2 to obtain E γ [E cp+2e γ ] = E γ [E + cp] In the extreme relativistic limit, we make the replacement cp E in this expression to obtain E γ = EE γ E γ + m 2 c 4 /(4E) As a crude estimate, that mc 2 = 10 9 ev and E γ = 10 4 ev, corresponding to microwave black body radiation at a temperature of a few Kelvins. For an incident cosmic ray proton of energy ev, we obtain E γ ev. Consequently, the cosmic ray loses 4% of its energy in such a collision. It is felt that this effect is responsible for the decrease in the number of high energy cosmic rays at about this energy. 2. A ray of light is reflected from a plane mirror, which moves in the direction of its normal with velocity v. Prove that the angles of incidence and reflection, θ and φ, respectively, are related by the formula sinθ cosθ ± v/c = sinφ cosφ v/c The incident and reflected rays are shown in the rest frame of the mirror.

3 Reflected Ray ρ Mirror ρ Incident Ray Rest Frame of Mirror Their four-vector momenta, (k i ) µ and (k f ) µ, are given in that frame as follows: (k i ) µ = (k,k cosρ,k sinρ,0) (k f ) µ = (k, k cosρ,k sinρ,0) In the moving frame, these rays are shown below: Reflected Ray φ θ Mirror Incident Ray Moving Observer Their four-vector momenta, (k i )µ and (k f )µ, are given in that frame as follows: (k i) µ = (k i,k i cosθ,k i sinθ,0) (k f )µ = (k f, k f cosρ,k f sinφ,0)

4 The respective four-momenta in the two frames are related by the Lorentz transrform. It is convenient to apply the inverse transforms. We do so for the x and y components of the incident four-momentum: (k i ) 1 = γ ((k i )1 + β(k i )0 ) k cosρ = γk i (cosθ+β) (k i ) 2 = (k i )2 k sinρ = k i sinθ Taking the ratio of these equations, we obtain tanρ = sinθ γ (cosθ+β) Now we consider the x and y components of the reflected beam: (k f ) 1 = γ ((k f )1 + β(k f )0 ) k cosρ = γk f ( cosφ+β) (k f ) 2 = (k f )2 k sinρ = k f sinφ Taking the ratio of these equations, we obtain tanρ = sinφ γ(cosφ β) Since the two expressions for tanρ must be identical, we obtain The factors of γ cancel out, so that sinθ γ (cosθ+β) = sinφ γ (cosφ β)

5 The result is thus established. sinθ cosθ+β = sinφ cosφ β 3. A particle of mass m and total energy E decays into two massless particles that travel symmetrically with respect to the direction of motion of the original particle. Determine the angle θ of the decay products with respect to the original direction. As particular cases, consider the non-relativistic case and the extreme relativistic case. Here is a picture of the situation before and after the decay process: θ θ Before After Decay of Massive Particle into Two Photons Let the incident particle have initial energy E and momentum p to the right. The final photons have momenta p ± and energies E± = cp ± : E p = E + + E = p + + p It follows from symmetry that E + = E, and that the momenta p ± are of equal magnitude. Thus E = 2E + p = 2p + cosθ As a consequence, with v the speed of the incident particle, we obtain v c = cp E = cp + cosθ = cosθ E +

6 Note that, when the speed of the incident particle is small compared with the velocity of light, the opening angle between the two photons (2θ) is close to 90. As the speed of the incident particle increases, that opening angle becomes smaller, approaching zero in the extreme relativistic limit. 4. Calculate the area of a circle of radius r (distance from center to circumference) in the two-dimensional geometry that is a sphere of radius a. Show that this reduces to πr 2 when r a. Choose a spherical polar coordinate system with the polar axis (North pole) at the center of the circle. The radius of the sphere is r = a θ 0, where θ 0 is the polar angle subtended by the perimeter of the circle. The surface area inside the (curved) circle is A = a 2 Z θ0 0 From the relation Z 2π dθ sinθ dφ = 2πa 2 (1 cosθ 0 ) = 2πa 2 (1 cos r 0 a ) we obtain cosθ 0 = 1 θ θ ( θ A = 2πa 2 2 ) ( ) ) 0 2 θ4 0 = πr 2 1 θ2 0 = πr (1 2 r a 2 When r = 100 m, the area is approximately π 10 4 m 2 = π hectares, differing from that value by about When = 100 km, the deviation is approximately Consider a particle with four-momentum p and an observer with four-velocity u = (γc,γ v). Show that, if the particle goes through the observer s laboratory, the magnitude of the three-momentum measured is ( p u ) 2 p 2 = p p c Suppose that the observer is moving in the x-direction with speed v, so that u µ = γ(c,v,0,0) = γc(1,β,0,0). The four-momentum of the particle in the initial frame is p µ = (p 0, p), so that

7 p p = p µ p µ = (p 0 ) 2 p p = p 2 0 p2 x p2 y p2 z p u = p µ u µ = p 0 u 0 p u = p 0 u 0 p x u x = γc(p 0 βp x ) We thus obtain Consequently ( p u ) 2 = γ 2 (p 0 βp x ) 2 c ( p u ) 2 p p = γ 2 (p 0 βp x ) 2 ((p 0 ) 2 p 2 x p 2 y p 2 c z) = (γ 2 1)(p 0 ) 2 2γ 2 βp 0 p x +(1+β 2 γ 2 )p 2 x + p2 y + p2 z = γ 2 (p 2 x 2βp x p 0 + β 2 (p 0 ) 2 )+ p 2 y + p 2 z = γ 2 (p x βp 0 ) 2 + p 2 y + p 2 z The components of spatial momentum in the frame of the moving observer are p x = γ(p x βp 0 ) p y p z = p y = p z so that The result is thus established. p 2 = γ 2 (p x βp 0 ) 2 + p 2 y + p 2 z

Physics 111 Homework Solutions Week #9 - Thursday

Physics 111 Homework Solutions Week #9 - Thursday Monday, March 1, 2010 Chapter 24 241 Based on special relativity we know that as a particle with mass travels near the speed of light its mass increases