If I understand correctly the term "spectral flux density" describes radiant flux for a given wavelength, right?

Like the given wavelength is the $\nu$ in:

$$F(\mathbf{x},t;\nu)=\oint_\Omega I(\mathbf{x},t;\hat{\mathbf{n}},\nu)\hat{\mathbf{n}}\,d\omega(\hat{\mathbf{n}})$$

For example, the EM wavelength of green light is about $500\,\mathrm{nm}$.

So can we talk about the spectral flux density $D$ of green light at a given point on a surface? Or have I got it wrong?

If I've got it right - what I don't understand is why is it per unit meter wavelength? Do we need to divide through by $500 \times 10^9$ to get the value of D in SI units?

Let's take a concrete example: say there are $1000\,\mathrm{lx}$ (ie $1000\,\mathrm{lm/m^2}$) of uniformly white light hitting a surface. What is the spectral flux density of green light in the SI unit of $\mathrm{Wm^{-3}}$?

  • $\begingroup$ What has the equation (written in terms of frequency) got to do with the question? $\endgroup$
    – ProfRob
    Jan 22 at 9:05
  • $\begingroup$ I'm thinking spectral flux density would usually be watts per square meter (of incident area) per Hertz (of optical bandwidth). $\endgroup$
    – Roger Wood
    Jan 22 at 9:08
  • $\begingroup$ "White light" is a very ambiguous term for a physicist. A lot of different spectra look white due to metamerism. And, to add to the confusion, white color is also not unique due to different color temperatures of the white point one may choose. What exactly do you mean by "uniformly white light"? $\endgroup$
    – Ruslan
    Jan 22 at 9:12
  • $\begingroup$ @ProfRob: See here: en.wikipedia.org/wiki/…' $\endgroup$ Jan 22 at 9:17
  • $\begingroup$ @RogerWood: I believe the SI unit is formally Wm^-3. I assumed the extra per meter is for wavelength. See above link. $\endgroup$ Jan 22 at 9:19

2 Answers 2


Spectral flux density is required because the number of photons (or amount of energy) at some precise wavelength is zero. You have to integrate over a finite range of wavelengths to get a finite flux. The thing that is integrated over a range of wavelengths must therefore be expressed per metre (or more conveniently per whatever unit of wavelength you are using). i.e. To work out the power received in total you have to integrate the flux density over a wavelength bandwidth.

It is difficult to answer your second question which mixes up units of perceived brightness with physical units of power. It would depend on the conversion factor from lumens to Watts and what the actual spectrum of a perceived white light source was. But let's assume your white light is a scaled solar spectrum. Direct sunlight has about 100 lumens/W and a power per unit area of 1000 W/m$^2$, so your light source is about 1% as bright as direct sunlight.

The spectral flux density of direct sunlight peaks in the green at about 1.3 W/m$^2$ per nm, so your source would be 100 times fainter/smaller than that. If you then really want to express that as W/m$^3$, multiply by $10^9$ to get $1.3\times 10^7$ W/m$^2$ per m (or W/m$^3$).

  • $\begingroup$ Yeah I think I get it now. You are taking a derivative of flux over the domain of an infantesimaly small range of wavelengths surrounding the given wavelength - so the per unit wavelength in the SI unit is to reflect the denominator of that derivative. (You don't need to divide through by the given absolute wavelength). $\endgroup$ Jan 22 at 9:49

The flux of green light over a "point" could be measured with a filter green light and a photoreceptor beside. As ProfRob comments, the probability of finding a photon with an exact frequency is zero, therefore we need to integrate over frequency or, what is equivalent, wavelength.

Mathematically speaking we can represent a photon either in the space-time domain or in the energy-momentum domain, a "pure" photon" of a single frequency would be represented in the momentum space as a delta and as a harmonic wave overall the space, however in the real world the photons do not behave that way, they move through specific regions of space with specific frequencies, but we can argue that neither the position nor the frequency of the real photons is a pure delta, that is the reason why a small detector filtering around a small range of green light frecuencies can detect photons.


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.