How many photons can exist in a cubic box of unit volume, simultaneously?

Suppose that we have a cubic box of unit volume. Simultaneously, how many photons can exist in such box? Is there any limit?

• In the physical theory system, particles are represented by two separated but equally important groups: the fermions that have half-integer spin and obey the Pauli exclusion principle and the bosons that have integer spin and can exist in the same state. You should read their stories. DUN DUN
– Bort
Oct 19, 2016 at 7:14
• There is no limit if you ignore gravity. If you consider gravity, you can't have too much energy or you'll create a black hole: en.wikipedia.org/wiki/Kugelblitz_(astrophysics)
– Wood
Oct 19, 2016 at 7:15

Photons are bosons so unlike fermions such as electrons there is no restriction on multiple photons occupying the same energy state. Consequently there is no limit on the number of photons you can put in your box.

There is an upper limit to the total energy density in the box since if you make it too high the box willcollapse into a black hole. However this limit is absurdly high and for most purposes can be ignored.

Note that individual photons can have arbitrarily low energies, so for any given energy density you can have an arbitrarily high photon number density simply by using photons of a low enough frequency.

Later:

Wood raises an interesting point in a comment. Suppose we have a cubical box of side $d$ then the largest wavelength/lowest frequency photon that it is possible to fit in the box has a wavelength of $2d$, so the energy of that photon is $h\nu = hc/2d$.

Let's take a black hole with a Schwarzschild radius of $d/2$ (we'll use this approximation since black holes aren't cubes) in which case the mass is:

$$M = \frac{c^2r_s}{2G} \approx \frac{c^2d}{4G}$$

and the energy is just $E=Mc^2$ so:

$$E \approx \frac{c^4d}{4G}$$

The number of photons is just this nergy divided by the photon energy we calculated above of $hc/2d$ so the number of photons in the box is (approximately):

$$N \approx \frac{c^3d^2}{2Gh}$$

And there's your expression for the maximum number of photons you can get into your box before it turns into a black hole. For a $1$m box I make that about $3\times10^{68}$ photons.

• But if you know they're restricted to a cubic box with unit volume, they can't have precise frequencies, right?
– Wood
Oct 19, 2016 at 7:17
• en.wikipedia.org/wiki/Kugelblitz_(astrophysics)
– Wood
Oct 19, 2016 at 7:19
• @Wood why not? The frequencies would be quantized, but they can be precise as long as you allow infinite time for measurement. Oct 19, 2016 at 15:28
• @John Rennie So the maximum density decreases with $d$? This is very interesting. I looks like it gives $~8\times 10^{40}$ photons per cubic meter for the observable universe.
– Wood
Oct 19, 2016 at 22:31
• @Wood: oops, sorry, for some reason I read your comments as maximum density increases with $d$. Yes, you're quite correct, the maximum density decreases with $d$. The density of a black hole (defined as the mass divided by the voume inside the event horizon) decreases as the radius of the horizon increases. If you search this site you'll find several questions related to this. Oct 20, 2016 at 6:24

You have to keep in mind that if we are not at absolute zero the atoms in the walls of the container will have thermal energy and therefore they will vibrate and emit/absorb photons. In an equilibrium situation, the electromegnetic radiation in the box will be blackbody radiation (assuming we are dealing with an isolated system). The expected number of photons will be in this case $N \propto V T^3$. Therefore, since volume is fixed, the answer will depend on the temperature of the box.

• What if you make the assumption that the walls are made out of a "perfect mirror" that reflects 100% of all frequencies, absorbing nothing? Of course, such a material does not exist, but as this question is already highly theoretical... Oct 19, 2016 at 15:19
• @valerio92 But could the temperature of the box itself be a function of the number of photons (assuming we are dealing with an isolated system)? Oct 19, 2016 at 17:49
• @DevilApple227 Nothing would change. There are three parameters characterizing every material: transmittance $\tau$, absorptivity $\alpha$ and reflectance $\rho$. For an opaque body, $\tau=0$ and $\alpha+\rho=1$. A perfectly reflecting body (white body) is a special case of opaque body in which $alpha=0$ and $\rho=1$. In order for the radiation to be blackbody, you only need the walls to be opaque: the value of the reflectance does not matter. See here. Oct 20, 2016 at 10:13
• @J.Pak Well, I guess you could write $T=(N/V)^{1/3}$: nothing prohibits this. However, the fundamental thermodynamic quantity is $T$, not $N$. In the case of a photon gas, $N$ can be expressed only as an average value and not as a fixed constant like in a massive gas: the reason is that the photons are ceaselessly absorbed an emitted by the atoms in the walls. On the other hand, $T$ and $V$ are thermodynamic variables we can assume fixed. So $T$ and $V$ are the fundamental quantities, not $N$. Oct 20, 2016 at 10:18
• @J.Pak Errata corrige: I meant $T \propto (N/V)^{1/3}$, of course. Oct 20, 2016 at 10:32