Doppler Effect for Light — Question 8

Section 37-9 · Physics pages 6206–6207

Question 8 The classical longitudinal approximation misses a second-order correction For recession, the exact frequency factor is D = √ (1 −β)/(1 + β). A low-speed estimate uses Dlin = 1 −β. β D Schematic; labels, not drawing lengths, specify the data. (a) Expand D through second order in β. (b) At β = 0.10, find the exact factor and the fractional error of the linear estimate relative to the exact factor.
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Question 8 - Solution The classical longitudinal approximation misses a second-order correction β D (a) Method and calculation D = (1 −β)(1 −β2)−1/2. D = 1 −β + 1 2β2 + O(β3) . (b) Interpretation and further calculation D = √ 9/11 = 0.904534 , Dlin = 0.900000. D −Dlin D = 1 − √ 1 −β2 = 0.00501256 . Physical check The linear estimate is too low for positive recession speed. Its relative error starts at one half beta squared.

Question 8

The classical longitudinal approximation misses a second-order correction

For recession, the exact frequency factor is D=(1−β)/(1+β)D=\sqrt{(1-\beta)/(1+\beta)}. A low-speed estimate uses Dlin=1−βD_{\mathrm{lin}}=1-\beta.

Diagram shown in the original SVG worksheet.

  1. Expand D through second order in β\beta.

  2. At β=0.10\beta=0.10, find the exact factor and the fractional error of the linear estimate relative to the exact factor.

Question 8 - Solution

The classical longitudinal approximation misses a second-order correction

Diagram shown in the original SVG worksheet.

(a) Method and calculation

D=(1−β)(1−β2)−1/2.D=(1-\beta)(1-\beta^2)^{-1/2}.D=1−β+12β2+O(β3).\boxed{D=1-\beta+\tfrac12\beta^2+O(\beta^3)}.

(b) Interpretation and further calculation

D=9/11=0.904534,Dlin=0.900000.\boxed{D=\sqrt{9/11}=0.904534},\qquad D_{\mathrm{lin}}=0.900000.D−DlinD=1−1−β2=0.00501256.\boxed{\frac{D-D_{\mathrm{lin}}}D=1-\sqrt{1-\beta^2}=0.00501256}.

Physical check

The linear estimate is too low for positive recession speed. Its relative error starts at one half beta squared.