Ignoring the Sun, what is the most common wavelength of a photon that is received by us on average? (I am not asking what is the average wavelength, but rather what is the most common wavelength received on average).

In other words (and please do correct me if this does not make any sense), what wavelength of a photon has the largest number density within our observable universe?

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    – DanielSank
    Aug 10, 2015 at 22:30
  • 3
    $\begingroup$ Which do you want, the average wavelength or the most likely wavelength? And is that per unit wavelength or per unit frequency? I also presume you mean when the Sun is on the other side of the Earth, otherwise it is round about 550 nm. $\endgroup$
    – ProfRob
    Aug 10, 2015 at 22:40
  • $\begingroup$ @RobJeffries thanks for that Rob, my answer might be wrong, but leaving out the Sun is really wrong. $\endgroup$
    – user81619
    Aug 10, 2015 at 22:45
  • $\begingroup$ Can you elaborate in layman's terms $\endgroup$ Aug 10, 2015 at 23:19
  • $\begingroup$ I think the OP is asking what photon wavelength has the highest number density, and have edited accordingly. @NeilGraham please check that this is what you want to ask. $\endgroup$
    – Kyle Oman
    Aug 10, 2015 at 23:41

2 Answers 2


I'm going to interpret your question a bit liberally - you ask for the case where we ignore the Sun; I'm going to go a little bit further and ignore the entire galaxy (and in fact other nearby galaxies) and talk about the cosmic background radiation. The cosmic microwave background gets a lot of attention, but in fact there are cosmic backgrounds at a very broad range of wavelengths. I'm not sure if the following is the most up-to-date compilation (as it's 5 years old I'm betting it's not), but it will give an idea. The figure and caption are from this paper.

enter image description here

Labelled are the cosmic radio background (CRB), microwave (CMB), infrared (CIB), ultraviolet and optical (CUVOB), x-ray (CXB) and gamma-ray (CGB). The shaded blocks are theoretically excluded regions, and the points, lines and arrows are measurements/upper limits. The axes are labelled in both frequency and energy (sorry, no wavelength in this paper$^1$). That left hand axis is frequency times specific intensity, which gives you the power deposited on a detector per unit area per unit solid angle, while the right hand axis looks like an energy density, the energy per unit volume of the photons out in space. To get something like a number density of photons you'd have to divide by the photon energy. Attempting to do this in my head/by eye, I think the peak of the CMB comes out ahead in terms of total number of photons. This agrees with what I know colloquially: the vast majority of photons in the Universe are CMB photons. This is true for the Universe on average; if you go sit at a particular location, photons from a particular source may outnumber photons from the CMB, at least locally.

$^1$ Note that asking what frequency is most common is a bit ambiguous because frequency is discrete; you're really asking where some distribution as a function of frequency is maximum, and depending what you want to maximize, the maximum as a function of frequency need not necessarily correspond to the analogous maximum as a function of wavelength as you'd expect from $\nu\lambda=c$.

  • $\begingroup$ Just two quick yes /no questions for my own benefit, to me as a newbie, it's the area that is as important as anything else in the question as regards intensity. So globally, although it's physically impossible to measure, the CMB is the most intense source, and locally, on the smallest practically measureable area, it's cosmic rays, is that correct I wonder? No offense to the OP, but it was originally unclear. Just to nail it for myself, thanks Kyle. If it's hassle, please ignore this. $\endgroup$
    – user81619
    Aug 10, 2015 at 23:52
  • $\begingroup$ @AcidJazz the area refers to the area of a (hypothetical) detector, so a detector with double the surface area will absorb double the energy per unit time at all wavelengths. The CMB is the most numerous component of photons globally. For a given location, it really depends on what's nearby. Not sure where you got cosmic rays? $\endgroup$
    – Kyle Oman
    Aug 11, 2015 at 5:35

EDIT: Now that the question has been reformulated, this isn't the actual answer (it gives the value of the average detected wavelength)

Two assumptions:

  • We're looking for some average values in the cosmos, not the specific value we get here on Earth (since the sun messes the whole calculation up)
  • We're looking for the average wavelength, not the most probable one (the one given by Wien's law)

We're using the third part of the answer to this question (the averaging by total mass) and Planck's distribution of black-body spectral energy density in terms of wavelength:

${u}_{\lambda }=\frac{8\pi hc}{{\lambda }^{5}}\cdot \frac{1}{\mathrm{exp}\left(\beta hc∕\lambda \right)-1}$, where $\beta =\frac{1}{{k}_{B}T}$, ${k}_{B}$ being the Boltzmann constant.

Note that $U=\underset{0}{\overset{+\infty }{\int }}{u}_{\lambda }d\lambda $, we're not talking about radiance here (the formula in the Wikipedia article on Planck's law only shows the flux of radiation per steradian). Another nota bene, by taking the average by total mass we're including all photons, not just those from CMB. Wien's wavelength comes out as ${\lambda }_{max}\approx 0.29 nm$, so we should expect the mean to be a bit bigger than that.

Since Planck's law may be viewed as a distribution, we may use the standard technique to find the mean value of $\lambda$. After integrating I got $⟨\lambda ⟩=\frac{30hc\zeta \left(3\right)}{T{\pi }^{4}{k}_{B}}$ as the end result, the numerical value being $⟨\lambda ⟩\approx 0.53nm$, which seems fine compared to the above prediction by Wien's law.

EDIT: Of course, we could have got the same result by using the black body radiance instead of the spectral energy density distribution. I merely made the remark because of the constants applied in the expression.

Another EDIT: if we negate the mass inside the universe (let's say that we're away from everything possible and all but the CMB radiation is concealed), then by plugging in the average temperature T=2.73K we get $⟨\lambda ⟩\approx 1.95mm$.


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