# What is the physics behind the cooling of the atmosphere during a total solar eclipse?

Consider the Great American Eclipse (The total solar eclipse over America in 2017). Observed atmospheric temperature drops were in the range of 3 C to 8 C in a matter of minutes. Surface temperatures dropped about the same amount as did the atmosphere, best I can tell. I'm lacking data on that though.

Other total solar eclipse observations universally (to my knowledge) document temperature drops.

In any event, such temperature drops represent a massive amount of energy presumably lost to space in a very short amount of time.

Edit:

How does this energy in the atmosphere radiate to space?

• The Energy must be radiated to space
• Greenhouse gases can absorb kinetic energy from O2 and N2
• Only greenhouse gases can radiate at atmospheric temperatures
• Without solar input, the atmosphere cools because greenhouse gases radiate more energy than the earth can supply

Is that correct?

• There must be a massive energy flux out to space, to balance the massive energy influx from the Sun. When the latter is suddenly removed, the balance shifts suddenly Commented May 22, 2023 at 16:15
• I’m not quite sure what the question is here. Are you asking whether the mechanism for temperature drops during an eclipse is different from the mechanism for making it get colder at night?
– rob
Commented May 22, 2023 at 16:58
• Energy transfer can only occur via conduction, convection, or radiation in the atmosphere. The only gasses capable of significant transfer of energy in the atmosphere are greenhouse gases. O2 and N2 radiation at atmospheric temperatures are negligible. In cool dry air, the only significant greenhouse gas that could be cooling this body of air that fast is CO2. Can CO2 actually cool the atmosphere? How does that square with CO2 trapping heat? Commented May 23, 2023 at 15:22

However, Earth's infrared optical depth due to all the Greenhouse gases is only $$\tau_{\rm IR}\approx2$$, so it is not surprising that a cooling effect can happen in minutes via radiation. The infrared radiation simply diffuses away once it is not resupplied anymore.
• @PaulSnow Without greenhouse gases, the atmosphere would be transparent to the outgoing infrared radiation, and hence cool maximally. It would reach radiative equilibrium with the surface only at $T=T_{\rm eq}$. When adding GHG, your cooling surface, which cools with the power $L = 4\pi r^2_{\tau=1} T^4_{\tau=1}$ from the $\tau=1$ transition, will move up the adiabatic gradient towards lower $T_{\tau=1}$ (while $r_{\tau=1}$ only changes very little), hence decreasing L, which means that radiation must be left behind. This is the greenhouse effect. Commented May 24, 2023 at 9:13
• @PaulSnow Not sure what you're talking about. To avoid confusion I specifically said "the infrared optical depth". Look up the definition if you're unsure. But essentially, when computing the integral over densities $\rho$, path length r and opacity of the gas mix $\kappa$ you end up with $\tau= \int \rho(r) \kappa(r,T,\rho) dr$ of approx. $\tau \approx 1.93$ for a saturated steam water atmosphere on Earth, and $\tau\approx 1.98$ when adding ~300ppm CO2 into the mix. This $\tau$ is enough to raise the temperature above $T_{eq}$ by a few tens of K, but also the diffusion time is not long. Commented May 24, 2023 at 9:18