# Background radiation and photons

1. May 14, 2007

### Theraven1982

1. The problem statement, all variables and given/known data
I have a fairly simple question, but for some reason I can't find the answer. It's about Planck's law, and cosmic background radiation.

We know Planck's law, and we know the temperature of the universe (approx 2.7 K). Furthermore, we know the mean number of photons in a field mode at T.

How much photons per litre does the background radiation contain?

2. Relevant equations
number of photons excited in a mode at T: 1/(e^{hw/kT}-1)

I've just started following a physics course, after a few years of inactivity...

Any kicks in the right direction are welcome ;).

2. May 14, 2007

### marcus

this should give the energy per unit volume

$$\frac{\pi}{15} \frac{k^4 T^4}{\hbar^3 c^3}$$

when I put this
(pi/15)*(k*2.75 kelvin)^4/(hbar*c)^3
in the google search box and press search,
what google gives back is:

((pi / 15) * ((k * (2.75 kelvin))^4)) / ((hbar * c)^3) = 1.37731725 × E-14 joules per cubic meter

that must be the energy density of the CMB
then you have only to divide by the average photon energy at that temperature

(I don't mean the Wien peak, I mean the average)

the average photon in Planck-law light at temperature T is 2.701 kT

Let's see what it would be for the CMB.

I get this from the google calculator, which always puts in extra decimal places where you dont want :-)

2.70100 * Boltzmann constant * (2.75 kelvin) = 1.02551253 × E-22 joules

So if the energy density of the CMB is 1.38 E-14 joules per cubic meter
and the average energy of a photon is 1.03 E-22 joules, and you divide you get around
130 million CMB photons per cubic meter. Is that right?

Last edited: May 14, 2007
3. May 15, 2007

### Theraven1982

You're a lot closer than me; the answer should be 5*10^5 photons per litre ... I'll look again at your calculation, maybe I can find the reason why there's a difference.

But a big thanks!

edit:
energy density is (according to my book): pi^2 *k^2*T^4/15c^3*hbar^3
There's an extra pi ;).

edit2:
energy density:
pi^2*k^4*2.7^4/(15*c^3*hbar^3) = 4.0207599 × 10-14 m-1 kg s-2 K-4

There must be a problem at the photon energy; if I divide by kT, i get 1.08*10^6 photons per litre. Only a factor 2 difference.

Last edited: May 15, 2007
4. May 20, 2007

### Chronos

I get about 400 million cmb photons/m^3/sec [just a pi off from marcus!]

5. May 20, 2007

### Theraven1982

Hmm ... that's the same answer I get now... I don't know why the book says 5x10^5 photons per litre :\. I tried everything; maybe I should go with this answer instead.

Thanks for all the help! Much appreciated! (You all do a great job, helping untangle the web of science ;) )

6. May 20, 2007

### marcus

right! I was going by memory and forgot a factor of pi. thanks for catching that!

7. Aug 31, 2007

### Theraven1982

I finally found the right solution (well, my teacher did.. when I got to the Riemann zeta function, I always thought: This can't be correct... but it was). But it appears the above mentioned answer was correct as well; or at least it gets very close to the result my teacher gave me. Maybe the book was wrong.
But at least there's a second way to find the answer:
You have to integrate <n> (number of photons excited in a mode) * rho (density of modes).