HW-571-3-solution

HW-571-3-solution - 2 2 + 2 m 1 m 2 1-v 2 . V = p = m 1 v m...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
1 Phys 571 , Fall 2010 Solution to Assignment #3. 1. In the proper frame element of the solid angle is d Ω 0 = 2 π sin θ 0 0 . Therefore, the fraction of photons moving inside d Ω 0 is dN = 1 4 π d Ω 0 = 1 2 | d cos θ 0 | Now, we derived before that cos θ 0 = cos θ 0 - β 1 - β cos θ Substituting we get d Ω 0 = 1 - β 2 (1 - β cos θ ) 2 d Ω Thus, in the lab frame dN = 1 4 π 1 - β 2 (1 - β cos θ ) 2 d Ω Of course R dN = 1. 2. Integrating dN over 0 d cos θ 1 and - 1 d cos θ 0 and taking the ratio we derive f = 1 + β 1 - β This can be inverted as v ( f ) and then ε = mc 2 f + 1 2 f 3. m 2 1 = m 2 + m 2 2 - 2[ q ( p 2 + m 2 )( p 2 2 + m 2 2 ) - pp 2 cos θ 2 ] , c = 1 4. m 2 = ε 2 - p 2 = m 2 1 + m
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 2 2 + 2 m 1 m 2 1-v 2 . V = p = m 1 v m 1 + m 2 1-v 2 , c = 1 2 5. Equation of motion: p x =-| eE | can be integrated as p x =-| eE | t + p Particle energy E = p m 2 + p 2 (we set c = 1). In particular, E = p m 2 + p 2 . Using v = x x = p / E we get l = Z p / | eE | p- | eE | t E dt = Z p dp x | eE | p x p m 2 + p 2 x = E-m | eE |...
View Full Document

Page1 / 2

HW-571-3-solution - 2 2 + 2 m 1 m 2 1-v 2 . V = p = m 1 v m...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online