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Solutions_HW6

# Solutions_HW6 - Theoretical Dynamics Homework 6 Instructor...

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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 (( m + S ) u μ ) = μ S (1) which can be rewritten as d ( mu μ ) = μ S - d ( S u μ ) = μ S - d S u μ + S du μ = μ S - α S u α u μ - S du μ (2) where we have used the fact that d S /dτ = u μ g μν ν S = u α α S . Part (b) Multiplying both sides of the above equation by u μ we get m du μ u μ = u μ μ S - α S u α u μ u μ - S du μ u μ = u μ μ S - α S u α - S du μ u μ (as u μ u μ = 1) which implies ( m + S ) du μ u μ = 0 (3) As S is a space-time dependent scalar field, it is not identically equal to - m and hence this implies du μ u μ = 0 (4) 6 - 1

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Part (c) The action is S = Z ( - m + S ) (5) = Z dt p 1 - v 2 ( - m + S ) (6) = Z dt L (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 ( - m + A μ u μ ) (12) = Z dt p 1 - v 2 ( - m + A μ u μ ) as dt = dτ/γ (13) = Z dt L (14) where L = p 1 - v 2 ( - m + A μ u μ ) (15) Now, L = p 1 - v 2 ( - m + A μ u μ ) = - m p 1 - v 2 + 1 γ ( A μ g μν u ν ) = - m p 1 - v 2 + 1 γ ( A t γ - A · γ 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 ( mu i ) = ∂A i ∂t γ + ∂A i ∂x j γv j + ∂A t ∂x i γ - ∂A j ∂x i γv j = d ( mu i ) = ∂A i ∂t γ + ∂A i ∂x j γv j + ∂A t ∂x i γ - ∂A j ∂x i γv j = d ( mu i ) = ∂A i ∂x ν u ν - ∂A ν ∂x i u ν (18) = d ( mu i ) = ( ν A i - i A ν ) u ν (19) So the equation of motion is d ( mu μ ) = ∂A μ ∂x ν - ∂A ν ∂x μ u ν (20) Problem 3 Part (a) For A μ to transform as a 4-vector, we must have A μ -→ A 0 μ = L μ ν A ν (21) where L μ ν is the Lorentz transformation. Now, under a gauge transformation, A μ -→ A 0 μ = A μ + μ G (22) So, under a gauge transformation, the field tensor F μν is invariant, and hence the fields (which are derivable from the field tensor) are also invariant: F 0 μν = μ A 0 ν - ν A 0 μ (23) = μ ( A ν + ν G ) - ν ( A μ + μ G ) = μ A ν - ν A μ + ν μ G - ν μ G = μ A ν - ν A μ = F μν (24) 6 - 3

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So, the invariance of the field tensor under a gauge transformation holds for arbitrary scalar functions G which are (at least) twice differentiable in the space-time coordinates. Now, for
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Solutions_HW6 - Theoretical Dynamics Homework 6 Instructor...

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