If you were inside of a gas that was producing its own light (for example, air at 900°C), what would it look like? Would it look foggy, or would objects that aren't the same colour as the glow still be visible?

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    $\begingroup$ Is this gas covering an entire planet (i.e. is it dominating the atmosphere?) or is it just a localized pocket? $\endgroup$ – HDE 226868 Apr 23 '20 at 15:12
  • $\begingroup$ Can you give us a general light level? is it closer to glowstick dim light or more like spotlight in fog? I would imagine that the closer the light reaches the point of the ambient light of the area the less you would be able to see. $\endgroup$ – IT Alex Apr 23 '20 at 15:16
  • $\begingroup$ The gas is a localized pocket, and the light would be quite dim $\endgroup$ – Ichthys King Apr 23 '20 at 15:19
  • $\begingroup$ Can you give us any more information about the gas itself - molecular composition, opacity, etc.? $\endgroup$ – HDE 226868 Apr 23 '20 at 15:39
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    $\begingroup$ Have you ever seen fire? It is glowing air. Can you see through the fire? $\endgroup$ – AlexP Apr 23 '20 at 20:05

I agree largely with Matthew's answer; this is intended to put everything on a more quantitative footing.

The answer to your question primarily depends on three things: the mass of molecules in the gas, the density of the gas and the opacity. These will play the key roles in determining how the gas absorbs light traveling through it and how it emits light of its own.

Part 1: Absorption

What might be helpful here is a quantity called the optical depth, $\tau$. If you have an object a distance $z$ away, the optical depth is $\tau=nm\kappa z$, where $n$ is the gas number density, $m$ is the mass of a particle in the gas, and $\kappa$ is something called the opacity. A higher optical depth means more absorption and therefore less transmission of light.

If we assume that the gas is an ideal gas, we can calculate the number density through the ideal gas law. Let's assume that it's in equilibrium with the surrounding atmosphere. Then we get $$n=\frac{P_{\text{atm}}}{k_BT}=6.26\times10^{-24}\text{ m}^{-3}$$ which is less than the number density of air, because the gas is hotter.

How about the opacity? That's harder to calculate, because we don't know much about the gas, and it's also wavelength-dependent. At visible wavelengths, Earth's atmosphere has, under decent conditions, an opacity of roughly $\kappa\approx10^{-8}\text{ m}^2\text{ g}^{-1}$. We can't really calculate the opacity for your gas because we don't know anything about it.

The final piece of information we'd need is $m$, the mean mass per particle. We also don't know this. One of the heaviest molecules in Earth's atmosphere is carbon dioxide, with a molecular mass of $m_{\text{CO}_2}=7.31\times10^{-23}\text{ g}$. We can assume that our gas will have $m=m_{\text{CO}_2}$ in a worst-case scenario. The optical depth for our source a distance $z$ away is $$\tau(z)=(6.26\times10^{-24}\text{ m}^{-3})(7.31\times10^{-23}\text{ g})(10^{-8}\text{ m}^2\text{ g}^{-1})z$$ which is actually better than on Earth by a factor of 3, as the gas is hot and therefore not very dense.

Part 2: Emission

As Matthew wrote, we also have to consider that the gas itself is emitting radiation. By Wien's law, it has peak emission at about $2.49\;\mu$m, firmly in the infrared. This does make me doubtful that we'll see substantial emission at visible wavelengths. We can, however, calculate the volume emissivity by $$j_{\lambda}=\kappa B_{\lambda}$$ where $B_{\lambda}$ is a form of Planck's law, and therefore the whole thing is wavelength-dependent. You can play around with this at various wavelengths, keeping in mind that $T=1173.2\text{ K}$.

Part 3: Putting it together.

As there is both absorption and emission, we can use the equation of radiative transfer to determine what a source a distance $z$ away looks like. The specific intensity ends up being $$I(z)=I(0)e^{-\tau(z)}+B_{\lambda}\int_0^{\tau(z)}e^{-\tau(z)-\tau'}d\tau'$$ the integral can be simplified and we do get $$I(z)=I(0)e^{-\tau(z)}+B_{\lambda}\left(1-e^{-\tau(z)}\right)$$ That will allow you to determine whether the emission or the source dominates. As $\tau$ is likely small, we can approximate $e^{-\tau}\approx1-\tau$, so we then get $$I(z)\approx I(0)(1-\tau(z))+B_{\lambda}\tau(z)=I(0)+(B_{\lambda}-I(0))\tau(z)$$ which is maybe easier to calculate.


It would be foggy.

In fact, it would probably look somewhat like fog illuminated by a car's headlights.

I assume the reason this isn't obvious is because you're thinking that this glowing air is not as opaque as fog... and that's not an unreasonable line of thought. However, because it is also glowing, you are still getting a bunch of "stray" light from the gas that is going to tend to overpower the light from any object in the space, in much the manner that fog or haze occludes objects.

It will probably be a little different because most of the light of the object is still reaching you (probably¹), and what you're seeing is a bunch of extraneous light added, rather than light being blocked. However, I would expect that you will have the same effect of objects further away being harder to see.

For a real life example, consider the sun.

(¹ Depending on how uniform the temperature is inside your gas cloud, you may get a bunch of heat shimmer as well. This doesn't occlude light, exactly, but it will twist it around so that what you see is blurry/wavy due to refraction.)


If the gas is 100% transparent, there would be no blackbody radiation(i.e. something looking 'red hot' because of its temperature), and you would be able to see though it normally.

From Kirchoff's laws of radiation, Emissivity(=absorbtion) + transmissivity + reflectivity = 1. If the gas if fully transparent (transmissivity = 1), the emissivity (how much it emits blackbody radiation) = 0, i.e. no radiation is emitted.

If the emissivity was greater than 0, you would effectively be flooded by radiation from all sides if you were in it assuming the emissivity was high enough and the gas pocket large enough, so even if light from other sources wasnt absorbed by the gas it would be drowned out by the gas' emissions. (This would also be the case if the gas emitted at a wavelength that it was transparent to, like scintillators).

  • $\begingroup$ What if you were seeing it through a camera that did not pick up the colour of the glow, would the glow still cause an effect? $\endgroup$ – Ichthys King Apr 23 '20 at 15:44
  • $\begingroup$ If you were looking through a camera that did not pick up the glow (the cameras spectral range was outside the range of wavelengths the gas emitted), the glow would have no effect on camera. However, the gas might have an effect on what the camera sees. If the gas was fully transparent in the cameras spectral range, to the camera it would look like the gas didn't exist. If the gas was fairly absorbing in those wavelengths, it would be like trying to look through smoke/fog $\endgroup$ – Jack Apr 23 '20 at 15:49
  • $\begingroup$ In addition, you can still have heat shimmer... Also, black body radiation is very broad-spectrum. $\endgroup$ – Matthew Apr 23 '20 at 16:03
  • $\begingroup$ @Matthew in principle, you could have a wavelength range where the gas had a very low emissivity, and use a narrowband filter with your detector so that you dont see any of the emission. But that is kind of pushing it. Also, yep heat shimmer would probably be a big issue. $\endgroup$ – Jack Apr 23 '20 at 16:15

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