Can monochromatic light have units of per frequency?

Question

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?

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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?

Answer 1

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).

Answer 2

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.