Classical Electrodynamics · Problem 11.12

Radiating Systems, Multipole Fields and Radiation — John David Jackson

Physics — E&M · thomas-precession-algebraic

Problem

Apply the result of Problem 11.11 to a purely algebraic deviation of (11.116) on Thomas precession. (a) With L=−β⋅K(tanh⁡−1β)βL = -\frac{\boldsymbol{\beta} \cdot \mathbf{K}(\tanh^{-1}\beta)}{\beta} L+δL=−(β+δβ∥+δβ⊥)⋅K(tanh⁡−1β′)β′L + \delta L = -\frac{(\boldsymbol{\beta} + \delta\boldsymbol{\beta}_{\|} + \delta\boldsymbol{\beta}_{\perp}) \cdot \mathbf{K}(\tanh^{-1}\beta')}{\beta'} where β′=(β+δβ∥)2+(δβ⊥)2\beta' = \sqrt{(\boldsymbol{\beta} + \delta\boldsymbol{\beta}_{\|})^2 + (\delta\boldsymbol{\beta}_{\perp})^2}, show that δL=−γ2δβ∥⋅K−δβ⊥⋅K(tanh⁡−1β)β\delta L = -\gamma^2 \delta\boldsymbol{\beta}_{\|} \cdot \mathbf{K} - \frac{\delta\boldsymbol{\beta}_{\perp} \cdot \mathbf{K}(\tanh^{-1}\beta)}{\beta} (b) Using the commutation relations for K\mathbf{K} and S\mathbf{S}, show that C1=[L,δL]=−(tanh⁡−1ββ)2(β×δβ⊥)⋅SC2=[L,C1]=(tanh⁡−1β)2δL⊥C3=[L,C2]=(tanh⁡−1β)2C1C4=[L,C3]=(tanh⁡−1β)4δL⊥\begin{aligned} C_1 &= [L, \delta L] = -\left(\frac{\tanh^{-1}\beta}{\beta}\right)^2 (\boldsymbol{\beta} \times \delta\boldsymbol{\beta}_{\perp}) \cdot \mathbf{S}\\ C_2 &= [L, C_1] = (\tanh^{-1}\beta)^2 \delta L_{\perp}\\ C_3 &= [L, C_2] = (\tanh^{-1}\beta)^2 C_1\\ C_4 &= [L, C_3] = (\tanh^{-1}\beta)^4 \delta L_{\perp} \end{aligned} where δL⊥\delta L_{\perp} is the term in δL\delta L involving δβ⊥\delta\boldsymbol{\beta}_{\perp}. (c) Sum the series of terms for AT=A2A1−1A_T = A_2 A_1^{-1} to obtain AT=I−(γ2δβ∥+γδβ⊥)⋅K−γ2γ+1(β×δβ⊥)⋅SA_T = I - (\gamma^2 \delta\boldsymbol{\beta}_{\|} + \gamma \delta\boldsymbol{\beta}_{\perp}) \cdot \mathbf{K} - \frac{\gamma^2}{\gamma + 1}(\boldsymbol{\beta} \times \delta\boldsymbol{\beta}_{\perp}) \cdot \mathbf{S} correct to first order in δβ\delta\boldsymbol{\beta}. [See D. Shelupsky, Am. J. Phys. 35, 650 (1967).]

Related theory

Browse the problem library