The Cosmic Background Radiation — Question 6

Section 44-11 · Physics pages 7502–7503

Question 6 The background dipole can estimate an observer speed For small beta=v/c, use T(θ) = T0[1 + β cos θ]. Let T0=2.725 K and dipole amplitude 3.36 mK. Take c=3.00 × 105 km/s. cooler warmer Schematic; labels, not drawing lengths, specify the data. (a) Infer beta and speed. (b) Find the maximum-to-minimum temperature difference and distinguish it from the dipole amplitude.
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Question 6 - Solution The background dipole can estimate an observer speed cooler warmer (a) Method and calculation β = A/T0 = .00336/2.725 = .00123303 . v = cβ ≃369.9 km/s . (b) Interpretation and further calculation Tmax = T0 + A, Tmin = T0 −A. Tmax −Tmin = 2A = 6.72 mK . Physical check The exercise assumes a purely kinematic dipole after foreground removal. Amplitude is half the peak-to-peak difference.

Question 6

The background dipole can estimate an observer speed

For small beta=v/c, use T(θ)=T0[1+βcos⁡θ]T(\theta)=T_0[1+\beta\cos\theta]. Let T0=2.725 K and dipole amplitude 3.36 mK. Take c=3.00×1053.00\times10^5 km/s.

Diagram shown in the original SVG worksheet.

  1. Infer beta and speed.

  2. Find the maximum-to-minimum temperature difference and distinguish it from the dipole amplitude.

Question 6 - Solution

The background dipole can estimate an observer speed

Diagram shown in the original SVG worksheet.

(a) Method and calculation

β=A/T0=.00336/2.725=.00123303.\boxed{\beta=A/T_0=.00336/2.725=.00123303}.v=cβ≃369.9km/s.\boxed{v=c\beta\simeq369.9\ {\mathrm{km/s}}}.

(b) Interpretation and further calculation

Tmax=T0+A,Tmin=T0−A.T_{\max}=T_0+A,\quad T_{\min}=T_0-A.Tmax−Tmin=2A=6.72mK.\boxed{T_{\max}-T_{\min}=2A=6.72\ {\mathrm{mK}}}.

Physical check

The exercise assumes a purely kinematic dipole after foreground removal. Amplitude is half the peak-to-peak difference.