# Mean optical depth in plane parallel atmospheres

I have the second edition of Houghton's "The Physics of Atmospheres". In section 2.2 he says one can do simple radiative transfer calculations in a plane parallel atmosphere by assuming that there are two fluxes, one going vertically up and one vertically down. In reality the radiation is going in all directions, but the two streams represent the amount integrated over the two respective hemispheres.

Now in reality different rays will see a given layer of atmosphere as having different optical depths, depending on the angle a given ray makes with the normal. Houghton says that detailed calculations show that one can take this into account by replacing the true thickness of a given layer dz with 5/3 dz.

This is my question-- Where does that 5/3 factor come from? When I do the calculation for an optically thin absorbing layer (as one would expect for something of infinitesmal thickness dz) I find that an initially isotropic stream (or rather, that half which is going either down or up) will have twice as much energy absorbed as would happen if all the radiation were travelling along a line normal to the layer. So I would have guessed you'd replace dz with 2 dz rather than 5/3 dz, but Houghton is the expert.

Houghton uses this 5/3 factor several times in the book.

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Perhaps he is assuming the atmosphere is optically thick? Radiative transfer in a thick atmosphere probably works differently. An atmosphere like our planets is probably too far from either asymptote for simple 2beam approximations to be very accurate. – Omega Centauri Jan 18 '11 at 4:57
Maybe he is assuming optical thickness for the whole atmosphere, but when he says "replace dz with 5/3 dz", that suggests it applies to optically thin slices. As for the accuracy of the approximation when applied to real atmospheres, I couldn't comment. – Donald Jan 18 '11 at 16:12

I imagine you did some sort of integral like $\frac{\int \sin\theta \cos\theta \left(\frac{dz}{\cos\theta}\right) d\theta}{\int \sin\theta \cos\theta d\theta} = 2$

The problem is, even for optically thin atmospheres, the photons emitted at oblique angles ($\theta \approx \pi/2$), there is a chance for them to scatter into small angles and then escape, so the $dz/\cos\theta$ factor must be replaced with a higher-order approximation for the optical depth as a function of initial emission angle. I did this problem once, but remember it being rather messy, and I think I just wrote a mathematica program to get the answer, and it was close to 5/3.

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Thanks Jeremy. That sounds plausible. I was assuming the photons were either absorbed or not and didn't think to consider what might happen if they were scattered before absorption. Offhand I wouldn't even know how to set that up. This would also explain why Houghton says "detailed calculation"--my back of the envelope calculation (more or less the integral you wrote) wasn't exactly detailed. – Donald Jan 18 '11 at 22:17
I suspect it is simply assuming an optically thick atmosphere. Then you assume the intensity emitted (or scattered) is a linear function of Z. Then you calculate the net vertical flux of radiation. This should define the optical "resistance" to radiation transport. Several decades back I had this in astrophysics, but I don't recall the details. Once the optical thickness is not high, then such a simple theory like this breaks down (i.e. you can use it for the stellar interior, but not for the atmosphere/photosphere). – Omega Centauri Feb 17 '11 at 22:35