Corpuscular Rays in St. Peter's Basilica (the Vatican)

A few days ago, I stayed in the Vatican (more about this one day symposium in another post). During the stay, I naturally visited St. Peter's Basilica.

Near Bernini's altar (made from copper stolen from the Pantheon which of course has nothing to do with our story...) I saw corpuscular rays. It may seem like some godly thing (quite appropriate for the location I guess), but from a physicists point of view, it was merely scattering by dust particles. So, I wondered, what could one say about that holy dust?

Here is the sight I saw: 

The first thing to note is the rather obvious - the dust is colorless! This means that the scattering dust particles are much larger than the wavelength of visible light. Namely, $r \gtrsim 1 \mu m$.

If the particles were to have radii much smaller than the visible wavelength, Rayleigh scattering would have been the dominant scattering process, and it would have given bluish light which scatters more effectively (as is the case with the sky). If the particles radii would have been comparable to the wavelength, we would have witnessed complex behavior, and certainly get some color behavior at some angles (e.g., as is the case with the scattering by small water droplets, which gives rise to colored haloes, arcs, glows and so forth, especially near the forward and backward scattering directions). But again, since we don't see any color, the size must be relatively large.

Next, because the dust is airborne, it cannot be too large. In fact, it cannot be falling at a speed much larger than the typical air motions in the basilica. (say,  a few cm/sec).

Because the sizes are so small, we are obviously dealing with laminar flows (i.e., non-turbulent flows around the dust... if you don't beleive me, you can check the Reynolds number for yourself). Thus, the terminal speed of the dust due to Earth's gravity is obtained by equating the Stokes drag to the gravitational acceleration:

$$f_{drag} = 6 \pi \mu r v ~~~ =~~~ f_{grav} = m g = {4\pi r^3 \rho g\over 3},$$

where $\mu$ is the visocity of air. Thus, the terminal speed is:

$$v_{max} = {2 r^2 \rho g \over 9 \mu}.$$

The radius must therefore be smaller than:

$$r \lesssim \sqrt{{9 \mu v_{max} \over 2 r^2 \rho_{dust} g }} \sim 10 \mu m$$

In other words, we know that the holy dust should typically be between 1 and 10 μm, otherwise there would have been at least some color, or the dust would have settled down too quickly.

We can also say something about the amount of dust, but this requires a few rough approximations.

We know that the scattered light (e.g., at 30° in the picture) has roughly the same brightness (give or take a factor of a few...) as the ambient light within the basilica. From photography, we know that outdoors we would typically need a shutter which is perhaps 100 times shorter than in the basilica (say 0.01 sec instead of 1 sec). This means that over a distance of about 10 meters (which is the length that my line of sight passes inside the "corpuscular beam"), the dust scatters light which is perhaps 100 times brighter than the ambient, to the ambient level.

This implies that over one meter, the light beam attenuates by 0.1%. Hence, the total projected area of the dust in a cubic meter of air should be of order 0.001 of a m². If the typical radius is 3μm, we then need about 30 million particles, which add up to about 4 mg per cubic meter... give or take an order of magnitude.

We could discuss the angular dependence of the light scattering, and reach the conclusion for example that the dust particles don't have a large diffusive component, at least not one which is semi-isotropic since otherwise the backscattering would have been much brighter... but given the late hour I'll quit here.