Converting between $F_{\nu}$ and $F_{\lambda}$ spectral density

Question

In papers, spectral energy distributions are given either in $F_\nu$, $\nu F_\nu$, $F_\lambda$ or $\lambda F_\lambda$. $F_\nu$ has units of Janskys, for example.

Is there a clear explanation online I can point my undergrad students to for how to convert between these and why it works this way? We are trying to plot archival spectra and the units are not always clearly documented.

What I have found online so far is the explanation on the Planck's law wikipedia page, which I paraphrase here. Planck's law integrated over the sphere, or any other spectral density, can be given as either

• $F_\nu$, which is the spectral density along the frequency axes, or
• $F_\lambda$, which is the spectral density along the wavelength axis.

Then the two have to be equal when integrating between two wavelength/frequencies:

$$ \int _{\lambda _{1}}^{\lambda _{2}}F_{\lambda }(\lambda)d\lambda = \int _{\nu (\lambda _{2})}^{\nu (\lambda _{1})}F_\nu (\nu)d\nu = \int _{\lambda _{2}}^{\lambda _{1}}F_\nu(\nu){\frac {d\nu }{d\lambda }}d\lambda = \int _{\lambda _{1}}^{\lambda _{2}}-F_\nu(\nu){\frac {d\nu }{d\lambda }}d\lambda $$

where $\nu=c/\lambda$ is used, which implies

$$\frac {d\nu }{d\lambda }=-c/\lambda^2$$

This means the exchangable integrand is $$F_{\lambda }(\lambda)=F_\nu \frac{c}{\lambda^2}$$

In words, because the integration axes are different and inversely related, converting between $F_\nu$ and $F_\lambda$ involves multiplication by $1/\lambda^2$.

It is a convenient convention to plot in

$$\lambda F_{\lambda }(\lambda)=\nu F_\nu(\nu)$$

because then the curves look the same regardless of whether wavelength or frequency is plotted, and both axes can be placed on the same plot. (nu-Fnu, lambda-Flambda)

Regarding the units: On the left side, if $\lambda$ is in nm, and $F_{\lambda }(\lambda)$ is in $W/nm/m^2$ (or erg/s/nm/cm^2), then the combined unit is $W/m^2$ (or erg/s/cm^2), which is consistent with the right side, if $\nu$ is in Hz, and $F_\nu$ is in $Jy$.

Luminosities: If luminosity densities are considered, multiplying by the area ($m^2$), for example $4\pi\times D_L$ with $D_L$ the (luminosity) distance, removes $m^2$ from the units of $F_{\lambda }(\lambda)$, and $Jy$ become $J$ for $F_{\nu}(\nu)$ .

Answer 1

Good news. You didn't make any mistakes. So, I'm going to use this as an excuse to elaborate wildly. More generally, pretty much any quantity that is a density in physics actually lives inside of an integral of some sort, and you can change the integration variable to change what the density is defined with respect to. Any valid integration change of variables is allowed. Basically, here are the rules:

1. the boundaries to the integration region before and after the change need to correspond, and
2. everywhere within the region of integration the coordinate change needs to be invertible (with the exception of isolated points, like the center of polar coordinates).

Now, let's have some fun with densities. As you've noted, $F_\nu$ has units similar to Janskies, defined as $10^{-26}\,\mathrm{W}\,\mathrm{m}^{-2}\,\mathrm{Hz}^{-1}$. I often find that writing the integration sign and limits is not actually needed, it's sufficient to write $F_\nu\, \mathrm{d}\nu$ (this notation is very similar to differential forms and is used, for example, in these articles by David W. Hogg et al: Distance Measures in Cosmology and The K-Correction). As you've noted, $F_\nu\, \mathrm{d}\nu = F_\lambda\,\mathrm{d}\lambda$ (remember: a negative sign corresponds to flipping the order of limits, so it isn't meaningful that $\mathrm{d}\nu / \mathrm{d}\lambda < 0$). You can go further, though. From basic calculus, $\mathrm{d}\ln\nu = \nu^{-1}\,\mathrm{d}\nu$. The unit for natural logarithms is e-folds (i.e. factors of e). Thus, $F_\nu\,\mathrm{d}\nu = \nu F_\nu \mathrm{d}\ln\nu$ meaning that $\nu F_\nu$ is still a spectral flux density it just has a pseudo-unit in the denominator (kind of like how $\omega = 2\pi f$ is in radians per second). I've often written $\nu F_\nu$ as having units $\mathrm{Jy\,GHz\,e}$-$\mathrm{fold}^{-1}$. Most astronomers aren't familiar with the idea of e-folds, so you can convert to dex (decades = factors of 10) by: $\ln(10) \nu F_\nu \,\mathrm{d}\log_{10}\nu$. That gives a quantity with units like $\mathrm{Jy\,GHz\,dex^{-1}}$.

We're just getting warmed up, though!

Recall that $E = h\nu$, so frequencies are equivalent to energies. So, if we divide $F_\nu$ by $h$, the energies in the numerator and denominator should cancel. But, just like with $\nu F_\nu$ they leave behind a pseudo-unit, because the energy in the numerator is the energy of individual photons, and in the denominator its the energy spanned by an interval. In fact, the units for $F_\nu / h$ are equivalent to $\mathrm{photons\, sec^{-1}\, m^{-2} \,e}$-$\mathrm{fold^{-1}}$.

We can go a step further, though. First, let's add all of the differentials in: $F_\nu \,\mathrm{d}\nu\,\mathrm{d}t\,\mathrm{d}A$. Notice you have seconds and meters both in the denominator? Well, you can exchange time for distance by noting that light travels at $c$. In other words, we exchange the wait by $\mathrm{d}t$ with the length of the box of photons that will travel through our $\mathrm{d}A$ in $\mathrm{d}t$ to get $F_\nu c^{-1}\, \mathrm{d}\nu\, \mathrm{d}V$. Thus our new quantity is the spectral energy density (in space and over frequency) of photons.

Combining both of the above is really more fruitful if we change from just considering flux to thinking about spectral radiance $I_\nu$. This quantity has units $\mathrm{W\,m^{-2}\,sr^{-1}\,Hz^{-1}}$. We can convert it into a quantity that is much much more interesting, and easier to conceptualize. First, divide by $c h\nu$ to get a density with units like $\mathrm{photons\,m^{-3}\,sr^{-1}\,Hz^{-1}}$ that looks like $$\frac{I_\nu}{h\nu c} \,\mathrm{d}V\,\mathrm{d}\Omega\,\mathrm{d}\nu.$$ The presence of that solid angle looks funny, to be honest. Where does solid angle come from? It comes from changing to polar coordinates and looking at volume there. $$\mathrm{d}V = \mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z = r^2\sin\theta \,\mathrm{d}r\,\mathrm{d}\phi\,\mathrm{d}\theta \equiv r^2\,\mathrm{d}r\,\mathrm{d}\Omega.$$

$\nu$ is a scalar, though, not the length of a vector. It's equivalent to the wave-number, though $kc = \nu$, so we can change our density to: $$\frac{I_\nu}{h\nu} \,\mathrm{d}V\,\mathrm{d}\Omega\,\mathrm{d}k$$ which suggests that we can go to $$\frac{I_\nu}{k^2h\nu} \,\mathrm{d}V\,\mathrm{d}\Omega\,k^2\mathrm{d}k = \frac{I_\nu}{hc k^3} \,\mathrm{d}V\,\mathrm{d}^3k.$$

What's the use of doing these transformations? Well, if you recall the $p = hck$ we can do one more transformation to get: $$\frac{I_\nu}{hc k^3} \,\mathrm{d}V\,\mathrm{d}^3k = \frac{I_\nu}{hc p^3} \,\mathrm{d}^3x\,\mathrm{d}^3p,$$ which is a long winded way of saying that $\frac{I_\nu}{hc p^3}$ is the phase-space density of photons. What's the use of this quantity? Well, I don't know about you, but I have an easier time thinking about the density of something in a nice Euclidean space than worrying about seemingly unrelated and abstract solid angles and energy density over a spectrum, even if the dimensionality of that Euclidean space is high. Even nicer, the phase space density of photons is invariant under the expansion of space-time (the scale factors for the spatial volume exactly cancel the ones for the three factors of wave-number/momentum).

Even more fun than that, in terms of phase space density of photons the black-body spectrum is supremely simple. It's $$\frac{B_\nu}{hcp^3} = \frac{2}{h^3 (e^{h\nu / k_B T} - 1)},$$ where the $h^3$ is there to get the units right (it's not even there if you use wave-number instead of momentum), and the 2 is the number of photon polarizations.

I used phase space photon density to derive Equation 4 in my paper on the contribution of galaxies to the background light at $3.4\,\mu$m. It is a useful tool for calculating the background at some redshift given the background at a previous redshift and the intervening luminsity density.