How does a volcanic eruption cool the planet's surface? For example, after the eruption of Mt. Pinatubo in 1991 according to this paper,

The introduction of large amounts of sulphuric acid aerosol into the stratosphere increases the planetary albedo (essentially the Earth's reflectivity of solar radiation) because these aerosol particles are efficient scatterers but only weak absorbers at solar wavelengths.

...

The changes in the Earth's albedo observed by ERBE resulted in a net cooling of approximately 8 W m-2 between 5°S and 5o N, with a net cooling of 4.3 W m-2 between 40° S and 40° N.

What I'm interested in is how exactly this cooling happens. Which is correct:

*

*the atmosphere is experiencing a deficit of input energy, or


*the surface is experiencing a deficit of input energy, which is then communicated to the atmosphere by a decreases longwave surface emission
I suspect (2) is mostly correct, since about 75% of input shortwave solar radiation to the earth is absorbed by the surface, while only 25% by the atmosphere itself, according to ref [1] (and perhaps even less so by volcanic aerosols).
If this is correct, then once the surface does communicate that energy deficit to the atmosphere, will it happen locally? That is, will a slab of the atmosphere nearest the surface be the first to experience a cooling rate?

[1] Petty, G.W., 2006. A first course in atmospheric radiation, 2nd ed. ed. Madison, Wis: Sundog Pub.
 A: The atmosphere is mostly opaque to infrared light, but mostly transparent to visible light. (You may have noticed we can see through it.) A layer of random scatterers means that some fraction of the incoming visible light is redirected upwards, where it can escape without being absorbed.
The immediate cooling effect is local. Think of being at the swimming pool on a not-too-hot summer day. If you are too warm, you can find some shade.  In a small shadow, like from a towel-sized tent flap, there’s no impediment to airflow, so the air temperature can’t be very different between the sun and the shade — the difference is the radiative heat transfer.  But if a big cloud blocks the sunshine for the entire pool, all of the little ones start to shiver and all of the moms start to pack up everyone’s things.
Your question seems to consider only radiative heat transfer between the ground and the atmosphere. But convective transport is important, too.  Consider that hot, humid days tend to create a low-density “inversion layer” near the ground, which is why summer days are more likely to get updraft-generated anvil thunderheads and tornados in the afternoons than in the mornings, even when there is not any weather front passing through.
