Text_Template

Example 7

Assuming the formula for black body radiation to the valid for the universe, calculate the number density of photons in the universe due to cosmic microwave background.

Solution

Taking the expression for energy density in the interval $\nu$ and $\nu+d\nu$

$\displaystyle N = \frac{P}{h\nu} = \frac{1400}{6.63\times 10^{-34}\times 5\times 10^{14}}= 4.21\times 10^{21}$

$\displaystyle u(\nu)d\nu = \frac{8\pi h\nu^3}{c^3}\frac{1}{e^{h\nu\beta}-1}d\nu$

the number density of photons with energy in this frequency interval is obtained by dividing the above expression by $h\nu$ . The total number density of photons is obtained by integrating the above expression over all frequencies

$\displaystyle n = \int_0^\infty \frac{8\pi \nu^2}{c^3}\frac{1}{e^{h\nu\beta}-1}d\nu$

Substitute $h\nu\beta = x$ ,

$\displaystyle n$	$\displaystyle =$	$\displaystyle \frac{8\pi}{(ch\beta)^3}\int_0^\infty \frac{x^2}{e^x-1}dx$
	$\displaystyle =$	$\displaystyle 1.65\times 10^8\int_0^\infty \frac{x^2}{e^x-1}$

The integral has to be done numerically, say by using Simpson's rule. The value of the integral is 2.4, which gives the number density of photons of the cosmic radiation to be $3.96\times 10^8$ per m

.