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Is it possible to have an atmosphere that is essentially fog, but still be breathable by humans?

I'm basically looking to have an environment where visibility is very low (for suspense reasons) even in full daylight (which would obviously not be that bright in this case).

If such an atmosphere is technically possible, what effect would that have on plant/animal life? I assume that it would be like growing plants in constant misty rain, so only plants that thrive on lots of water would grow? But I am one of those not smart guys, so I could be way off.

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  • $\begingroup$ @MikeNichols, what would you consider "opaque"? I'm trying to use optical depth, and I was assuming you meant $\tau\approx1$, but I wasn't sure. $\endgroup$ – HDE 226868 Jun 7 '18 at 17:56
  • $\begingroup$ @HDE226868 Yes I believe an optical depth in the range of 1 to 10 or so should produce the desired visual effect. Low visibility, but not permanent darkness. $\endgroup$ – Mike Nichols Jun 7 '18 at 18:24
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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 $$\tau=n\sigma x,\quad\tau=\kappa\rho x$$ 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 $$I(x)=I_0e^{-\tau(x)}=I_0e^{-n\sigma x}=I_0e^{-\kappa\rho x}$$ 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 $\tau\geq 1$.


1 In reality, most mediums are not of uniform density, and so we would rewrite the first equation as$\tau(x)=\int_0^xN(x')\sigma dx'$, 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, $\lambda\sim10^{-7}$ meters. In air, the Rayleigh approximation holds well, but for fog, $d\sim10^{-6}$ 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'$3 and mean particle diameter $d$, we get $$\sigma_{\text{air}}=\frac{2\pi^5}{3}\frac{d_{\text{air}}^6}{\lambda^4}\left(\frac{n_{\text{air}}'^2-1}{n_{\text{air}}'^2+2}\right)^2$$ For air at standard temperature and pressure, $n_{\text{air}}'\approx1.000293$, and $d_{\text{air}}\sim2\times10^{-9}$ meters, generously. If we pick light midway through the visible spectrum - say, green light with $\lambda=550$ nm - then we find that $\sigma_{\text{air}}\sim5\times10^{-33}$ square meters.

For fog, let's bite the bullet and use the same approximation. We'll say $d_{\text{fog}}\sim5\times10^{-6}$ meters. I wasn't able to find great figures for the index of refraction, but this group indicates $n_{\text{fog}}'\approx1.5$ for some fogs. Using the same formula gives me $\sigma_{\text{fog}}\sim3\times10^{-6}$ square meters - much larger than $\sigma_{\text{air}}$, as one would expect.


3 I'm using $n'$ 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 $$n_{\text{fog}}=\frac{\tau}{\sigma_{\text{fog}}x}\approx3.32\times10^5\text{ particles/m}^3$$ which comes out to $1.7\times10^{-7}$ 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 $n_{\text{fog}}'=1.5$ (and an imaginary index of refraction of $-0.3$, as given in the linked paper), the calculator gave me $\sigma_{\text{fog}}\sim3\times10^{-8}$ square meters, lower than our Rayleigh result by a factor of 100. This, then, means a number density of $n_{\text{fog}}\approx3.32\times10^7$ 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.

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  • $\begingroup$ How could I generate a planet-wide situation just like the one you describe in the Great Banks? $\endgroup$ – Jano Moore Sep 23 '18 at 3:33
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Fog is just a suspension of minute droplets of water/liquid in the atmosphere. If the atmosphere was breathable before the fog, it stays breathable also with the fog.

There are cities in our world where fog is notoriously a major aspect of their life, still they flourish (think London or Frisco).

Water based fog can be highly beneficial for life: in the Namib desert the only supply of water comes from the fog forming on the dunes thanks to the close-by ocean.

Things can be different if the fog contains other chemicals: a fog of sulfuric acid would be highly aggressive and therefore not suitable for life forms. Such a case happens with smog (see Beijing or London), which is fine dispersion of fog and other pollutants.

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    $\begingroup$ Note that London Fog was really smog. $\endgroup$ – RonJohn May 16 '18 at 5:28
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    $\begingroup$ @RonJohn, the one in 50es yes, but London has been foggy as long as it exists $\endgroup$ – L.Dutch May 16 '18 at 5:53
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    $\begingroup$ There are lots of places with constant fog or nearly so. Check quora.com/Where-are-the-10-foggiest-places-on-earth-Why? Fog is fine. $\endgroup$ – Willk May 16 '18 at 21:10
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    $\begingroup$ "If the atmosphere was breathable before the fog, it stays breathable also with the fog." is not true. Water droplets and water vapor displace oxygen. If you have enough, there will not be enough oxygen to breathe and you will die. This is a cause of death in places like ships and power plants when steam lines explode. $\endgroup$ – user71659 Jun 8 '18 at 21:22
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    $\begingroup$ @Wompguinea, Fog requires warm water, cold air, and high humidity. A planet with a thin mantle (warm oceans via nearby magma) that was in the outer reaches of the habitable zone (distant sun, cold days) should produce world-wide fog. $\endgroup$ – JBH Jun 9 '18 at 15:02
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There's a difference between humid and foggy. Humid is what you get in swamps. We don't necessarily want that. Fog occurs for many reasons, but I believe the most popular is warm water and cold air. This is why there's a miniature bank of fog over the road where the river crosses under it early spring and late fall.

So, you need to heat the water and cool the air. How could you do this?

  • A thin mantle (warm magma is near the surface, think "hot springs" all over the planet, including the ocean), shallow oceans, and a planet near the outer extreme of the habitable zone (cool air). The closer the planet is to the sun, the thinner the mantle must be.

  • A heavily ruptured mantle (TONS of active volcanos) on a world with few if any continents (islands everywhere). Deep oceans are OK with this one and it can be better situated in the habitable zone.

Cheers!

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