The Cosmic Background Radiation — Question 5

Question 5

A small temperature anisotropy produces a larger fractional energy-density contrast

A patch has temperature 50 microkelvin above a 2.725 K mean. For bolometric radiation u is proportional to T to the fourth power.

1. Find the fractional temperature contrast and first-order energy-density contrast.

2. Estimate the second-order correction and judge whether the linear approximation is adequate.

Question 5 - Solution

A small temperature anisotropy produces a larger fractional energy-density contrast

(a) Method and calculation

$$\boxed{\epsilon=\Delta T/T=50\times10^{-6}/2.725=1.83486\times10^{-5}}.$$$$\boxed{\Delta u/u\simeq4\epsilon=7.33945\times10^{-5}}.$$

(b) Interpretation and further calculation

$$(1+\epsilon)^4-1=4\epsilon+6\epsilon^2+4\epsilon^3+\epsilon^4.$$$$\boxed{6\epsilon^2\simeq2.02\times10^{-9}\ll4\epsilon}.$$

Physical check

This is a bolometric contrast; the intensity response in a narrow frequency band has a different temperature derivative.