Laplace's extinction theorem

Toward the end of his monumental work on Celestial Mechanics, Laplace has a long chapter dealing with atmospheric refraction. In the course of it, he derives a remarkable result, which seems to have been forgotten by most people today :the extinction of the atmosphere is related to its refraction. This concluson is called Laplace's extinction theorem. Exactly it sais that the logarithm of the intensity of light, of any star, is proportional to its refraction, divided by the cosine of its apparent altitude. We must immediately notice that the proportionality constant is negative, because the stars become dimmer as they approach the horizon, and their refractions increase. Furthermore, one should strictly say the relation is a linear one, rather than a simple proportionality. So we have the equation :

$$M\propto\frac{r}{\sin z}$$ (1.4)

where $M$ is the air mass $r$ is the refraction and $z$ is the zenith angle. Let's see how this equation is derived.

The differential of airmass is $dM$ it is just proportional to the differential of path length $ds$ in figure 1.2, multiplied by the local density, $\rho$

Figure 1.2: Calculation of Laplace's theorem

Let's regard the differential of refraction from a different point of view: $dr$ must be equal to the ray curvature multiplied by the element of path length $ds$. The curvature in turn, is proportional to the component of the atmospheric density gradient $d \rho/dh$ normal to the path of the ray. As the atmospheric model is horizontally stratified, its density gradient is a vector directed vertically i.e., toward the zenith. The component of this gradient normal to the ray is just proportional to $\sin z$. So we have :

$$dr\propto \frac{d\rho}{dh}\sin zds$$ (1.5)

Now we need to assume that the density gradient is proportional to the density itself. Very near the horizon, the diminution of light depends, like the refraction, upon the constitution of the atmosphere. We may, without fear of any sensible error, use the hypothesis of a uniform temperature. So we assume an exponential atmosphere, in which $d\rho /dh \propto \rho$, and have :

$$dr\propto \rho\sin z ds$$ (1.6)

But also we have $dM \propto \rho ds$. If we eliminate the product $\rho ds$ on the right side of the last proportionality for $dr$ by replacing it with $dM$, we see that :

$$dM \propto \frac{dr}{\sin z}$$ (1.7)

Then integration will give Laplace's result, $M \propto r / \sin z$, provided that we can factor some mean value of $\sin z$ out of the integrand when we integrate this differential relation. In fact, this is not a bad approximation, because $\sin z$ is nearly constant through the atmosphere near the zenith (where $z$ is nearly constant along the line of sight) and $\sin z$ is itself nearly constant near the horizon, because the sine function flattens out near unity when $z$ is near $90^{o}$.

Notice that Laplace's result is trivially true for the flat-Earth approximation, where $M =\sec z$ and $r \propto \tan z$. Also, in a long duct, where the ray simply follows the curve of the Earth, and $\sin z$ is unity, the refraction in the duct is obviously proportional to the path length there.