# number density of photons

My lecturer put up as a solution to the number density of photons to be $$n_{o\gamma}=\tfrac{8\pi}{c^3}\int^{\infty}_0(\tfrac{kT}{h})^3\tfrac{x^2dx}{e^x-1}$$. The integral on the right hand side being the Riemann zeta function with value which is approximately 1.2. But then he said this gives the value of the number density as $$n_{0,\gamma}=400cm^{-3}$$ for T=2.7. But if we input all the value in the first equation I gave this doesn't give anywhere close to that value.

I think perhaps what he is done is made a small blunder and the first equation gives the total number of photons in the universe and then we must divide by the volume of the universe, but even then $$\tfrac{8\pi}{c^3}\int^{\infty}_0(\tfrac{kT}{h})^3\tfrac{x^2dx}{e^x-1}=1.98\times10^8$$ and the volume of the universe is about $$4\times10^{80}$$ but dividing the first of the by the second returns a value of order $$10^{-73}$$which is still way off from what he said it should be.

Could anyone please explain this to me ?

• Why did you post essentially the same question twice? physics.stackexchange.com/questions/453118/… – G. Smith Jan 9 at 17:03
• @G.Smith Because they are not the same question one they involve using two entirely different methods – exodius Jan 9 at 17:30

Putting this exact formula into Wolfram Alpha actually gives me $$400\text{ cm}^{-3}$$, assuming that your evaluation of the integral is correct. We can check to see that the units here make sense by doing some dimensional analysis. The equation is $$\frac{8\pi}{c^3}\left(\frac{kT}{h}\right)^3\int^{\infty}_0\frac{x^2dx}{e^x-1}$$ Let's break down the units here: $$\frac{8\pi}{c^3}\text{ is in }\frac{\text{seconds}^3}{\text{meters}^3}$$ $$\left(\frac{kT}{h}\right)^3\text{ is in }\left(\frac{\text{Joules}\cdot\text{Kelvin}^{-1}\cdot{\text{Kelvin}}}{\text{Joules}\cdot\text{seconds}}\right)^3=\text{seconds}^{-3}$$The product of these two quantities is in $$\text{meters}^{-3}$$ - the units of number density. Since the integral as I've written it is dimensionless, the result of this equation is indeed a number density, with a value of $$4\times10^8{\text{ m}}^{-3}=400\text{ cm}^{-3}$$ in cgs units - closer to the value your professor gave you. The difference in scale is simply that your professor wrote their answer in $$\text{cm}^{-3}$$ rather than $$\text{m}^{-3}$$, as you calculated it.
The remaining discrepancy with your result of a factor of $$2$$ is an incorrect evaluation on your part of the integral. The Riemann zeta functions is $$\zeta(s)=\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{x^{s-1}}{e^{x}-1}$$ You want to calculate $$\zeta(3)\Gamma(3)\approx2.4$$.
Note, finally, that experiments tell us that the number density of photons is indeed about $$410\text{ cm}^{-3}$$.