Solutions_HW6 - Theoretical Dynamics October 14, 2010...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

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

Unformatted text preview: Theoretical Dynamics October 14, 2010 Homework 6 Instructor: Dr. Thomas Cohen Submitted by: Vivek Saxena Problem 1 Part (a) The equation of motion is d d (( m + S ) u ) = S (1) which can be rewritten as d d ( mu ) = S- d d ( S u ) = S- d S d u + S du d = S- S u u - S du d (2) where we have used the fact that d S /d = u g S = u S . Part (b) Multiplying both sides of the above equation by u we get m du d u = u S- S u u u - S du d u = u S- S u - S du d u (as u u = 1) which implies ( m + S ) du d u = 0 (3) As S is a space-time dependent scalar field, it is not identically equal to- m and hence this implies du d u = 0 (4) 6- 1 Part (c) The action is S = Z d (- m + S ) (5) = Z dt p 1- v 2 (- m + S ) (6) = Z dtL (7) So the Lagrangian is L = p 1- v 2 (- m + S ) (8) For S << m and non-relativistic conditions, L = p 1- v 2 (- m + S ) - m- m 2 v 2 + S + O ( v 4 ) (9) 1 2 m ~x 2- S- m where ~x = v (10) So the Lagrangian equals 1 2 m ~x 2- S up to an irrelevant constant- m , which does not affect the equations of motion. The equations of motion in this regime are t L x i = L x i which, from equation (10), can be written as the vector equation, m ~x =- S (11) Problem 2 S = Z d (- m + A u ) (12) = Z dt p 1- v 2 (- m + A u ) as dt = d/ (13) = Z dtL (14) where L = p 1- v 2 (- m + A u ) (15) Now, L = p 1- v 2 (- m + A u ) =- m p 1- v 2 + 1 ( A g u ) =- m p 1- v 2 + 1 ( A t - A v ) =- m p 1- v 2 + A t- A j v j (16) 6- 2 The Euler-Lagrange equations are d dt L x i = L x i (17) = d dt mv i 1- v 2- A i = A t x i- A i x j v j = d dt mv i 1- v 2 = dA i dt + A t x i- A i x j v j = d dt ( mu i ) = A i t + A i x j v j + A t x i- A j x i v j = d d ( mu i ) = A i t + A i x j v j + A t x i - A j x i v j = d d ( mu i ) = A i t + A i x j v j + A t x i - A j x i v j = d d ( mu i ) = A i x u - A x i u (18) = d d ( mu i ) = ( A i- i A ) u (19) So the equation of motion is d d ( mu ) = A x - A x u (20) Problem 3 Part (a) For A to transform as a 4-vector, we must have A - A = L A (21) where L is the Lorentz transformation. Now, under a gauge transformation, A - A = A + G (22) So, under a gauge transformation, the field tensor F is invariant, and hence the fields (which are derivable from the field tensor) are also invariant: F = A - A (23) = ( A + G )-...
View Full Document

This note was uploaded on 12/29/2011 for the course PHYSICS 601 taught by Professor Hassam during the Fall '11 term at Maryland.

Page1 / 9

Solutions_HW6 - Theoretical Dynamics October 14, 2010...

This preview shows document pages 1 - 4. Sign up to view the full document.

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