Expanding Universe

Expansion of our universe can be considered as an adiabatic process for the photons of the microwave background radiation.Wavelength of photons grows linearly with the radius of the Universe (regarded a perfect sphere) .

Find the number of degrees of freedom of the photon gas (regarded ideal)


If you are looking for more such simple but twisted questions, twisted problems for jee aspirants is for you!
6 3 4 0 2 1 7 5

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Rohith M.Athreya
Jan 12, 2017

h c λ \displaystyle \large \frac{hc}{\lambda} is the energy of an individual photon.This energy is also proportional to T \displaystyle \large T ,the temperature(by kinetic theory)

T α 1 R \displaystyle \large T \alpha \frac{1}{R} R is the radius

t α 1 V 1 3 \displaystyle \large t \alpha \frac{1}{V^{\frac{1}{3}}}

This gives us γ \displaystyle \large \gamma is 4 3 \displaystyle \large \frac{4}{3}

and then i + 2 i \displaystyle \large \frac{i+2}{i} is γ \displaystyle \large \gamma

i = 6 \displaystyle \large i=6

Spandan Senapati
Feb 14, 2017

For this we can assume the photon gas inside a closed ideal box.Where the radiation pressure satisfies P V = U / 3 PV=U/3 .which means U = 3 n R T U=3nRT .this is equivalent to C v = 3 R Cv=3R .But then from equipartition of energy there must be 6 degrees of freedom each with ( k T 1 / 2 ) (kT*1/2) energy.

Weins law can be applied. Lambda * Temp.= constant.and lambda proportionals radius which is V^1/3.Now put T=PV/R.

Jatin Chauhan - 4 years, 3 months ago

Why does the radiation pressure satisfy that equation?

Gabriele Manganelli - 3 years, 11 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...