Plank outlines (page 209, equations 302-304) in his book that a monochromatic ray of frequency $v$ is has intensity of

\begin{equation} I_{\nu} = \frac{2 h \nu^{3} F \Omega}{c^{2}} \left( e^{\frac{h\nu}{k_{b}T}} - 1\right)^{-1} \end{equation}

Where $h$ is Planck's constant, $F$ is the area, $\Omega$ is the solid angle, $k_{b}$ is the Boltzmann constant, $c$ is the speed of light and $T$ is the temperature. Writing this as per unit area ($F = 1$) and per $4 \pi$ solid angle (i.e. multiplying by $4\pi$), the intensity on the monochromatic beam is then written as

\begin{equation} I_{\nu} = \frac{ 8 \pi h \nu^{3}}{c^{2}} \left( e^{\frac{h\nu}{k_{b}T}} - 1\right)^{-1} \end{equation}

We can write this as a photon flux by dividing by the energy of a photon ($h\nu$), i.e.

\begin{equation} B = \frac{ 8 \pi \nu^{2}}{c^{2}} \left( e^{\frac{h\nu}{k_{b}T}} - 1\right)^{-1} \end{equation}

This has units of $\frac{\text { photons }}{\mathrm{s} \ \mathrm{m}^{2}\ 4\pi\mathrm{~Sr} \ \mathrm{~Hz}}$, i.e. photons per unit area, per unit bandwidth, per unit time, and per $4 \pi$ solid angle. Others, for example, then use this expression to determine the entropy change associated with the loss of a photon from the incident ray as

$$ \Delta S=-K \log \left(1+\frac{8 \pi \nu^{2}}{c^{2} B}\right) $$

But, how can a monochromatic ray have units of per unit bandwidth? It doesn't have any dependence on bandwidth! But, surely, it must have units of $\textrm{Hz}^{-1}$ for the $\log$ in the above equation to have the correct dimensions?

Planck (page 17, after equation 7), does say

"The specific intensity $K$ of the whole energy radiated in a certain direction may be further divided into the intensities of the separate rays belonging to the different regions of the spectrum which travel independently of one another. Hence we consider the intensity of radiation within a certain range of frequencies, say from $\nu$ to $\nu'$. If the interval $\nu' - \nu$ be taken sufficiently small and be denoted by $d\nu$, the intensity of radiation within the interval is proportional to $d\nu$. Such radiation is called homogeneous or monochromatic"

And also (page 20, after equation 14), also states

"there exists in nature no absolutely homogeneous or monochromatic radiation of light or heat. A finite amount of radiation contains always a finite although possibly very narrow range of the spectrum. This implies a fundamental difference from the corresponding phenomena of acoustics, where a finite intensity of sound may correspond to a single definite frequency"

But now, if I have some bandwidth to the light - doesn't this means depending on my choice of $\delta \nu$ I can change the intensity of the monochromatic light? Doesn't that seem unphysical?

Edit: If I understand this correctly, the width of this infinitesimal band would ultimately affect the energy flux which is arguably the more physical and measured quantity.

So really there are two intrinsic physical quantities here — the purity of the beam which sets the bandwidth and the energy injected over this interval. The ratio of these will then set this somewhat misleading monochromatic intensity?

  • $\begingroup$ Do you mean spectral density at a given frequency: en.m.wikipedia.org/wiki/Planck%27s_law ? $\endgroup$
    – Roger V.
    Oct 18, 2021 at 18:26
  • $\begingroup$ No, not really - because I'm talking about a monochromatic beam. Not the spectral density at a specific wavelength I.e. As I understand in a monochromatic beam there shouldn't be any density? I imagine you'd need to include a dirac delta at some v0 somewhere - which would remove per unit Hz? $\endgroup$
    – Tomi
    Oct 19, 2021 at 9:02

2 Answers 2


It seems you misunderstand the use of the word 'monochromatic' here. It does not mean that the radiation has no finite bandwidth, but merely that he considers the specific spectral intensity at a given sharp frequency $\nu$. A band with zero width at this frequency would obviously have zero intensity, and therefore the intensity at frequency $\nu$ is defined by the intensity a band of 1 Hz would have if it had the same constant intensity within this band (1 Hz is quite a narrow frequency band, so there won't be much of an error by making this approximation).

  • $\begingroup$ Thanks. Just to clarify - what you are saying just by changing my definition of frequency band, I can change the intensity? My intensity is per unit bandwidth, so if I make the bandwidth infinitesimally small or biger - my intensity per unit frequency also changes? That seems wrong? Surely I should be able to get the same intensity for some monochomratic beam? I can't just pick some aribtrary bandwidth and the value changes? $\endgroup$
    – Tomi
    Oct 19, 2021 at 9:09
  • $\begingroup$ @Tomi There are no exactly monochromatic lines as their width would be zero and they would therefore have zero net intensity. If you measure a spectral intensity at frequency $\nu$, you always measure the intensity that comes in within a certain finite bandwidth $\delta\nu$ around $\nu$ which is determined by your instrument. In order to give this an objective meaning you normalize this to your unit bandwidth e.g. 1 Hz. So if you instrument had actually a bandwidth of 10 Hz, you have to divide the measured intensity by 10 in order to give the intensity per 1 Hz. $\endgroup$
    – Thomas
    Oct 23, 2021 at 16:34

But, how can a monochromatic ray have units of per unit bandwidth?

Due to Heisenberg uncertainty principle pulse bandwidth and duration is related like that :

$$ \Delta \omega \cdot \Delta t \geq 2 \pi $$

So if you'll compress laser pulse into short duration - then you'll get high frequency bandwidth. And vise-versa, if you want to achieve monochromatic-like pulse, then you you must generate very long pulses in time domain. Perfect "monochromatic" light is just an idealization, because for such light you would need to generate an infinite pulse, which is impossible, hence all light sources have a bandwidth.

  • $\begingroup$ Understood. So when Planck talks about monochromatic light, it must have some bandwidth. But the width of this infinitesimal band will affect the energy flux? How do I pick the width of this band? Isn't it strange just by changing this band I can change my energy flux? $\endgroup$
    – Tomi
    Oct 20, 2021 at 18:35
  • $\begingroup$ No, by choosing $\delta \nu$, you are not changing energy flux,- you just slide frequency window, where you compute/integrate total energy flux per frequency given interval. To change spectral energy flux distribution over frequencies- you need to change black body temperature, as per Wien law or something like that. $\endgroup$ Oct 21, 2021 at 19:47
  • $\begingroup$ Imagine you are standing in the river with different currents flow speed at various places, and by moving across river, you feel with your body that flow rate changes. Now by analogy you would say - that by moving in the different river places (or by combining flow effects from multiple people in different places) you are changing total water flow rates ! It's not like that. Hope this analogy has helped you to understand a flaw in your argument. $\endgroup$ Oct 21, 2021 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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