As I understand it, photons have no mass and thus have no "dimensions". So I would expect there is no limit at how many photons can hit, say, a square meter of a given surface. But I am not sure, so -

Is there any theoretical limit on how many photons can hit a given unit of surface?

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    $\begingroup$ photons have no mass and thus have no "dimensions" Mass and size are independent. Electrons do have mass but have no size. $\endgroup$
    – Ghoster
    Dec 25, 2022 at 0:54
  • $\begingroup$ @Ghoster They don't have size? I've heard that it is only a theoretical thing. $\endgroup$ Dec 25, 2022 at 14:43
  • 3
    $\begingroup$ All elementary particles are point particles, according to the Standard Model of particle physics. No high-energy scattering experiment has ever found a finite size for an electron. It’s possible that at higher energies than we can currently produce we’ll someday discover that they have structure. $\endgroup$
    – Ghoster
    Dec 25, 2022 at 17:39
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    $\begingroup$ Our intuitive concept of "size" is a macroscopic phenomenon resulting from electrostatic repulsion between valence electrons that means that each atom has their own region of space that other atoms can't intrude into. Any microscopic concept of "size" is at best an analogy capturing some, but not all, of this intuitive understanding, and at worst a misleading application of terminology. $\endgroup$ Dec 26, 2022 at 5:27
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    $\begingroup$ @Ghoster an electron or photon has no internal structure (techinally, they are represented by irreducible representations of the Lorentz group). That isn't really a statement about "size". When we shoot particles at each other, the frequency of reactions is proportional to a quantity called the "cross-section" which, as the name suggests, is an area. While reaction-dependent, these non-zero cross-sections are probably closer to our classical intuition of size than anything else you could define for a particle. $\endgroup$
    – tobi_s
    Dec 26, 2022 at 8:39

3 Answers 3


There is generally no hard limit to the number of photons per unit area in the sense that photons can overlap without issue (i.e., don't think of them as hard billiard balls). This ultimately comes about because photons are bosons, so the Pauli exclusion principle does not place restrictions on them like it does for fermions.

However, there can be many practical limits to the number of photons per unit area. These will depend on the frequency of the photons, as it may be easier to have lots of low-frequency RF photons compared to lots of gamma ray photons. One such limit is the Schwinger limit, where photons start to interact with each other and thinking of them as independent particles does not work anymore. This limit arises when the total electric field of the photons is large enough to create electron-positron pairs out of the vacuum.

If you have enough photons to reach the Schwinger limit, then you reach a non-linear regime, where some of those photons will (nearly) instantly disappear and turn into electron-positron pairs (and eventually back to photons), and the remaining photons will not behave at all like the light you are familiar with, so they should really be called something else - perhaps a "photon liquid"? Maxwell's equations don't apply here as in linear optics, so the physics is truly different.

The Schwinger limit is when the electric field reaches $$E_s = \frac{(m_e c^2)^2}{e \hbar c}$$

The electric field from a source with $N$ photons per second with energy $\hbar \omega$ per photon over an area $A$ is given by

$$E=\sqrt{\frac{2 N \hbar \omega}{c \epsilon_0 A}}$$

Equating the two and solving for $N/A$ gives

$$N/A=\mathrm{Photons/Area}=\frac{1}{\hbar \omega} (\frac{(m_e c^2)^2}{e \hbar c})^2$$ $$\mathrm{Photons/cm^2}_{\mathrm{Schwinger}} = \lambda [nm] \times 2.33\cdot10^{45} \frac{\mathrm{Photons}}{\mathrm{Second\cdot cm^2}}$$

where $\lambda$ is the photon wavelength in nanometers.

This limit is extraordinarily high, and has not been achieved on earth with any man-made light source yet.

Another practical limit is the number of photons needed to destroy any material via ionization. Here you could perhaps use the Rydberg constant ($13.6$ Volts) and Bohr radius ($\sim 50$ picometers) to construct an estimate for an electric field that would rip apart electrons from any atom. This estimate gives

$$\mathrm{Photons/cm^2}_{\mathrm{ionization}} = \lambda [nm] \times 10^{32} \frac{\mathrm{Photons}}{\mathrm{Second\cdot cm^2}}$$

  • $\begingroup$ There's also the fact that "photon hitting a surface" is generally understood to mean that first there's an atom/molecule in some steady state, then it interacts with a photon for a period of time that can, relatively speaking, be considered practically instantaneously for most purposes, and the the atom/photon arrives at another steady state. For instance, in reflection, an atom's outer electron is kicked into a excited state, then it goes back to the ground state, emitting a photon. As the number of photons increases, the validity of that model breaks down. $\endgroup$ Dec 26, 2022 at 5:39
  • $\begingroup$ Photons aren't interacting with a steady state; the electrons never go back to their ground state, they just keep bouncing around different excited states, and we reach a point where simplifications such as assuming that electrons exist in a eigenstate of the energy operator don't work. $\endgroup$ Dec 26, 2022 at 5:40
  • $\begingroup$ Yes, one could also construct a limit in terms of runaway heating. That gets a bit involved because it would depend on the lights frequency, absorption spectra, and thermal properties of the surface. $\endgroup$
    – KF Gauss
    Dec 26, 2022 at 13:09
  • $\begingroup$ Useful limit, but just one note - it's not a theoretical limit, because it doesn't mean you can't increase photon density anymore as it reaches non-linear regime. Besides electron-positron pairs will annihilate over time returning energy to photon. Hence total energy density of $e^- + e^+ + \gamma$ stays the same, which you still can increase. $\endgroup$ Dec 29, 2022 at 21:20
  • $\begingroup$ True, but it is a limit where the usual notion of light no longer applies due to the nonlinearity. It doesn't make sense to talk about independent photons anymore, they are not the right quantum numbers $\endgroup$
    – KF Gauss
    Dec 30, 2022 at 4:56

I think you're asking how close photons can get to each other. i.e. if they were like 1 cm marbles then only one marble per sq cm could fit on a surface. They are not truly dimensionless, but they are waves, and if you look up Bose–Einstein statistics https://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics you'll learn that they do not repel each other like electrons etc tend to do, so they can be put very close together. Laser light hitting a surface is a good example of many photons in a small area.

One factor is probably the density of the surface material. When photons impact a surface they are typically being absorbed by electrons, and there could theoretically be too few electrons to absorb a very dense stream of photons. Just as likely, imagine a semi-porous material where some of the photons just go right through. To maximize absorption you also need a dense and thick material. Other factors could affect the answer you're looking for, like the reflectivity of the surface, for instance.


My answer would be rather based on absolute theoretical limit. Schwinger limit is nice, but given that in case of non-linear effects some of the photons will be converted into electron-positron pairs,- one can still try to increase photon amount per unit area, conserving total energy per unit area in the form of $~e^-~+~e^+ ~+~ \gamma$.

Besides at least some electron-positron pairs will annihilate back into photons, so we will have concurrent processes. And even more,- vacuum has limited energy density. Which means that you can't generate as many electron-positron pairs as you wish.

Anyway, based on Planck units absolute theoretical energy density limit is : $$ \Phi(J/cm^2) = m_{_P} c^2 ~\ell_{_P}^{−2} \tag 1,$$

which is about $\approx 10^{74}~\text{J/cm}^2$.

Converting (1) as number of photons :

$$ \Phi(n/cm^2) = \frac {m_{_P} c^2}{h~\nu} ~\ell_{_P}^{−2} \tag 2,$$

Taking as a reference visible light spectrum, for example when $\nu = 600~ \text{THz}$, from (2) we get that for visible photons this absolute energy density limit is $\approx 10^{93}~\text{photons/cm}^2$, or else vacuum break-down doesn't make sense.


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