{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW-571-1-solution

# HW-571-1-solution - K = u 2 K = 4 1-v 2/c 2 = 2 2 γ...

This preview shows page 1. Sign up to view the full content.

1 Phys 571 , Fall 2010 Solution to assignment # 1 1. Applying two successive boosts K K 0 K 00 : x 0 = γ ( x - vt ), t 0 = γ ( t - vx/c 2 ); and x 00 = γ 0 ( x 0 - v 0 t 00 ), t 00 = γ 0 ( t 0 - v 0 x 0 /c 2 ) we have x 00 = γ 0 [ γ ( x - vt ) - v 0 γ ( t - vx/c 2 )] = γγ 0 [ x (1 + v 0 v/c 2 ) - t ( v + v 0 )] On the other hand, the direct boost K K 00 : x 00 = γ 00 ( x - v 00 t ). Therefore, v 00 = v + v 0 1 + vv 0 /c 2 . 2. Let u μ = u μ 1 + u μ 2 be the sum of 4-velocities of two particles. u 2 = u μ u μ is boost-invariant, i.e. it must be the same in any inertial reference frame. In particular, it must be the same in the frame K in which the two particles move with the same velocities v towards each other and in the frame K 0 in which one of the particles is at rest. We have K : u μ K = { γ, γ~v/c } + { γ, - γ~v/c } K 0 : u μ K 0 = { 1 , ~ 0 } + { γ 0 , γ 0 ~v 0 /c } where v 0 is the velocity of a particle in the other particle rest frame. Thus, using the usual rules for 4-vector scalar products we have
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: K = u 2 K = 4 1-v 2 /c 2 = 2 + 2 γ yielding γ = 1+ v 2 /c 2 1-v 2 /c 2 . Since l = l /γ we ﬁnally obtain l = l 1-v 2 /c 2 1 + v 2 /c 2 3. Decompose r into two components: (a) r k = V r · V V 2 , r ⊥ = r-r k and similarly for r . Using Lorentz transformations r k = γ ( r k + V t ) , r ⊥ = r ⊥ we get after simple algebra r = γ ( r + V t ) + ( γ-1) ( r × V ) × V V 2 , t = γ ± t + r · V c 2 ² (b) Calculating d r and dt as functions of r and t and taking ratio yields v = v + V + ( γ-1) V V 2 [( v · V ) + V 2 ] γ ( 1 + v · V c 2 )...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern