Optical depth

Let's start by looking at the idea of optical depth. Optical depth is a quantity that describes how light is attenuated as it travels through a medium. There are two commonly used equations for optical depth, $\tau$. In a homogeneous¹ medium, they are $$\tau =n\sigma x,\tau =\kappa \rho x$$ Here, $n$ and $\rho$ are the number and mass densities of the medium, $\sigma$ is the cross-sectional area for absorption², $\kappa$ is a quantity called the opacity of the medium, and $x$ is the distance traveled by light. As an example, if light travels through a uniform slab of optical depth $\tau (x)$, the intensity of the light after traveling a distance $x$ will be $$I(x)=I_0{e}^{-\tau (x)}=I_0{e}^{-n\sigma x}=I_0{e}^{-\kappa \rho x}$$ In stellar astrophysics, the surface of the Sun - the edge of its photosphere - is defined as the radius $r$ at which $\tau =2/3$. In general, we can say that an object is opaque if $\tau \ge 1$.

¹ In reality, most mediums are not of uniform density, and so we would rewrite the first equation as$\tau (x)={\int }_0^{x}N(x^{\prime })\sigma d{x}^{\prime }$, where $N(x)$ is the column density. However, on your planet, I'd guess that over short distances, $N$ is approximately uniform.
² $\sigma$ is usually wavelength-dependent, and should really be written as ${\sigma }_{\lambda }$. The same goes for the opacity.

Scattering cross-sections

When talking about scattering in gases, there are two main regimes. When the scattering of light of wavelength $\lambda$ is due to particles of diameter $d$, the theory of Rayleigh scattering holds when $d\ll \lambda$ and the more general Mie scattering holds when $d\simeq \lambda$. Now, for visible light, $\lambda \sim {10}^{-7}$ meters. In air, the Rayleigh approximation holds well, but for fog, $d\sim {10}^{-6}$ meters, and Mie theory would be more appropriate.

Regrettably, all solutions for the cross-sections in Mie scattering are numerical, not analytical. I'm going to therefore try to use Rayleigh scattering as an example. Let's calculate some cross-sections. Assuming that air has an index of refraction $n^{\prime }$³ and mean particle diameter $d$, we get $${\sigma }_{\text{air}}=\frac{2{\pi }^5}{3}\frac{d_{\text{air}}^6}{{\lambda }^4}{\left(\frac{n_{\text{air}}^{\prime 2}-1}{n_{\text{air}}^{\prime 2}+2}\right)}^2$$ For air at standard temperature and pressure, $n_{\text{air}}^{\prime }\approx 1.000293$, and $d_{\text{air}}\sim 2\times {10}^{-9}$ meters, generously. If we pick light midway through the visible spectrum - say, green light with $\lambda =550$ nm - then we find that ${\sigma }_{\text{air}}\sim 5\times {10}^{-33}$ square meters.

For fog, let's bite the bullet and use the same approximation. We'll say $d_{\text{fog}}\sim 5\times {10}^{-6}$ meters. I wasn't able to find great figures for the index of refraction, but this group indicates $n_{\text{fog}}^{\prime }\approx 1.5$ for some fogs. Using the same formula gives me ${\sigma }_{\text{fog}}\sim 3\times {10}^{-6}$ square meters - much larger than ${\sigma }_{\text{air}}$, as one would expect.

³ I'm using $n^{\prime }$ and not the normal $n$ here so as to not confuse it with number density, $n$.

Putting it together

It should now be clear that the dominant source of extinction is due to fog. The number densities of air and fog should be relatively similar, and so $n_{\text{fog}}{\sigma }_{\text{fog}}\gg n_{\text{air}}{\sigma }_{\text{air}}$. We can therefore say that the optical depth is largely thanks to fog. Now, say we define "opaque" as meaning that $\tau =1$ at a distance of $x=1$ meters. We then would need to have a number density of $$n_{\text{fog}}=\frac{\tau }{{\sigma }_{\text{fog}}x}\approx 3.32\times {10}^5{\text{ particles/m}}^3$$ which comes out to $1.7\times {10}^{-7}$ kg/m${}^3$ . . . which is much less dense than water vapor, by a factor of about ${10}^6$. Clearly, the Rayleigh approximation fails.

Applying Mie theory

So, fortunately, smart people out there have built tools that let the rest of us calculate some important factors. I used this Mie scattering calculator. Plugging in a radius of $d=5$ microns, a light wavelength of $\lambda =550$ nm, and an index of refraction of $n_{\text{fog}}^{\prime }=1.5$ (and an imaginary index of refraction of $-0.3$, as given in the linked paper), the calculator gave me ${\sigma }_{\text{fog}}\sim 3\times {10}^{-8}$ square meters, lower than our Rayleigh result by a factor of 100. This, then, means a number density of $n_{\text{fog}}\approx 3.32\times {10}^7$ particles per cubic meter, and a mass density off of water vapor by a factor of ${10}^4$, rather than ${10}^6$.

Now, you wouldn't expect the atmosphere to have the density of water vapor. On Earth, fog constitutes a much smaller fraction of air, which is why we don't suffocate on a foggy day. The factor of ${10}^4$, then, seems somewhat reasonable, though possibly off by an order of magnitude or so. At any rate, Mie theory does, as expected, give a much better result, and it seems like the sort of atmosphere you require would not be unreasonable.

What could cause this?

The Grand Banks are arguably the foggiest place on Earth, where the warm Gulf Stream mixes with the cold Labrador current. However, that sort of mixing won't occur planet-wide - this is the only place on Earth it occurs on such a large scale - as currents of different temperatures will only meet like this in certain regions.

The Swiss Plateau might be a better example. Fully foggy days occur from November to January, and slightly less foggier days occur between October and February. This happens because of a specific kind of temperature inversion, thanks to a wind current called a bise, which interacts with the mountains. Turbulence underneath the inversion layer leads to low-level stratus clouds and, eventually, fog. Again, though, we have the problem that this sort of current won't occur everywhere in the world.

San Francisco provides another interesting case. The Bay Area has great conditions for fog: moisture from the Pacific, a large temperature gradient between ocean currents and the land, and mountains to further trap clouds and fog. This sort of coastal region isn't foggy year-round, but when fog develops, it becomes extremely thick.

Essentially, the ingredients you want for really thick fog are

• Some sort of temperature gradient, whether that be colliding air/water currents or temperature differences between land and sea.
• A source of moisture, such as an ocean.
• Mountains or valleys to trap the fog and low-level clouds and prevent them from dissipating.

Combine elements from these three regions, hand-wave the currents a bit, and you have the potential for some very foggy regions. I'm thinking plenty of coastlines, plenty of mountains and valleys, and lots and lots of water.

posted almost 7 years ago

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