2
$\begingroup$

I was reading over some assignment solutions that my lecturer put up and one of them said that instead of calculating the number of photons in the universe with the familiar method of $$\int^{\infty}_0(\tfrac{k_bT}{h})^3\tfrac{x^2dx}{e^x-1}\times \tfrac{8\pi}{c^3}.$$ That we can can take the mean energy of a photon to be $kT$ and use that as an estimate instead. He didn't go any further into explaining how to do so and I'm not sure I understand how to do what he meant.

Would it be something like $$n_{o,\gamma}=\tfrac{\text{Total energy density of universe}}{Energy one photon}?$$ If so then how does one calculate the total energy of the universe in this expression?

$\endgroup$
  • $\begingroup$ I think your numerator should be total energy "density" of the universe. $\endgroup$ – Lewis Miller Jan 9 at 16:40
  • $\begingroup$ I'll Change it now but how does one calculate the total energy density of the universe ? $\endgroup$ – exodius Jan 9 at 17:32
  • $\begingroup$ @LewisMiller is it just $\epsilon_r=\tfrac{4 \sigma T^4}{c^3}$ ? $\endgroup$ – exodius Jan 9 at 17:34
  • $\begingroup$ @LewisMiller wait it would be $\epsilon_{tot}=\epsilon_r+\epsilon_m+\epsilon_{vac}$ wouldn't it ? I'm not sure how to calculate the vacuum energy density though... can we consider it as negligible compared to the other two ? $\endgroup$ – exodius Jan 9 at 19:34
  • $\begingroup$ Sorry, can't help there. $\endgroup$ – Lewis Miller Jan 9 at 19:36
1
$\begingroup$

The photon number per $m^3$ can be calculated from, $$n_{\gamma}=\frac {\text{Energy density of the photon}} {\text{Energy per photon}}$$

and If we want to calculate the total number of photons in the universe, we can write $$N=n_{\gamma}\,V_{universe}$$ Where $V_{universe}=4\,10^{80}m^3$

Lets calculate it

We can take the mean energy of the photon as $$E_{mean}=2.70\,kT$$ where $k=1.380\,10^{-23}JK^{-1}$ and $T=2.725K$ (Tempature of the CMB)

From here we get $$E_{mean}=1.015\,10^{-22}J$$

Energy density of the photons can be calculated from,

$$\epsilon_{\gamma}=aT^4=\frac {4\sigma} {c}T^4$$ where $a= 7.56×10^{-16} Jm^{−3} K^{−4}$

From here we have $$\epsilon_{\gamma}=4.171\,10^{-14}Jm^{−3} $$

Hence, $$n_{\gamma}=\frac {\epsilon_{\gamma}} {E_{mean}}\frac {4.171\,10^{-14}Jm^{−3}} {1.015\,10^{-22}J}=4.109\,10^{8}m^{−3}$$

Finally, $$N=4.109\,10^{8}m^{−3}\,4\,10^{80}m^3=1.6436\,10^{89} photon$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.