Can the radiation balance value be estimated through an infinite geometric series?

Question

First a disclaimer: This is just an idea I had last night, so if my reasoning here is way off, please let me now.

Let's say that we have an initial thermodynamic system which is in complete balance. We assume, in order to have an easy number to work with, that the incoming radiation is 100 $W/m^2$ and this is perfectly balanced by an outgoing radiation of 100 $W/m^2$.

Now let us assume that a "lid" is placed over this system so that 50 % of the outgoing radiation is reflected back again through a greenhouse effect. The initial incoming radiation is, however, unaffacted. So in total, as 50 % is now reflected back, the total incoming radiation will now be 150 $W/m^2$.

As the system is now in imbalance, the temperature will rise, so that a new balance may be achieved. Eventually, the outgoing radiation will then also become 150 $W/m^2$.However, as 50 % will still be reflected back, the amount of radiation reflected back will now be 50 % of this. So now 75 $W/m^2$. will be reflected. If we add this to the original incoming radiation of 100 $W/m^2$, the total incoming radiation is now 175 $W/m^2$. And as the temperature rise again, and the outgoing radiation rises to this level, we now find that 50 % of this, that is 87.5 $W/m^2$. will be reflected back, and so forth.

The scenario may be easily summarized as a geometric series, where the total incoming and outgoing radiation, in $W/m^2$, becomes:

$100 + 50 + 25 + 12.5 + ...$

This we recognize as a convergent, infinite geometric series, and the sum may be easily calculated via the sum formula:

$s = \frac{a_1}{1 -k} = \frac{100}{1-0.5} = 200$

Through this reasoing, the system will therefore eventually stabilize at an incoming and outgoing radiation value of 200 $W/m^2$.

Now, obviously, this is a very simple example, and in a real, thermodynamic system, there would be a multitude of other factors which would influence the radiation balance. Furthermore, in reality, the process described here would be continuous, and not discrete. However, I still wondered if the general reasoing here may be applied if we have such a simple thermodynamic system, and if we assume that there are no other factors influencing the radiation balance.

Thanks in advance!

Answer 1

Given the assumptions you are making, the calculation method is correct - but I'm going to make a frame challenge.

As a question about radiation being reflected by a pot lid, this is fine. However, I note you say:

"Now let us assume that a "lid" is placed over this system so that 50% of the outgoing radiation is reflected back again through a greenhouse effect."

The greenhouse effect does not work by reflection. The pure radiative version works by absorption and re-emission of thermal black body radiation, with the layers at different temperatures. As the Stefan-Boltzmann law says the black body radiation emitted by a body at temperature $T$ is $\sigma T^4$, equal changes in temperature don't give rise to equal changes in radiation emitted, or vice versa. The temperature drops to about $-50^\circ$ C at 10 km altitude, so the difference is significant. It isn't likely to be a fixed proportion of the input like 50%.

We can take as a toy example a shallow pool of water - and for the purposes of demonstration we will ignore convection, conduction, and diffusion and assume heat is transferred only by radiation. Sunlight shines through and hits the bottom, warming it. That radiates infrared upwards, which immediately hits the water and is absorbed. 100% of the infrared can be absorbed by a millimetre thickness of water. That millimetre layer therefore warms, and it too starts radiating more, both upwards and downwards. And so on to the water's surface. If input power is $W$, then the top surface must radiate $W$ upwards (at equilibrium) and $W$ downwards (because radiation is isotropic), the layer below must radiate $2W$ upwards (so the net flow upwards will be $W$) and therefore $2W$ downwards. The next layer emits $3W$ up and down, and so on to the bottom of the pool. If the water is a metre deep, there would be 1000 layers, and the radiation emitted downwards by the bottom layers would be considerable! (In a solar pond, where convection is suppressed by dissolving salt in the lower layers of water, the bottom of the pool can reach $90^\circ$ C.)

In a convective fluid (like the Earth's atmosphere), the greenhouse effect works by a different mechanism entirely. Sunlight warms the ground, which warms the air in contact with it, which rises convectively to the top of the troposphere, cooling as the pressure drops, where it radiates to space. It then cycles back to the surface, warming as the pressure rises. Greenhouse gases raise the average altitude of infrared emission to space, which increases the pressure difference from there down to the surface, which increases the warming due to compression. (There are additional vertical thermal transfers due to water vapour evaporating and condensing, plus clouds, horizontal heat transfers and temperature gradients, and many other effects.)

Your approach is fine as an abstract physics problem about reflection of radiation, and the use of geometric series to sum repeated reflections, so long as readers are aware that it is entirely unrealistic as a model of the greenhouse effect.

Answer 2

Yes, actually this reasoning is quite similar with the one for calculating the reflection coefficient of a Fabry-Pérot interferometer. In you case, it is even more simple though.

Note that you can solve the problem directly. Writing $\phi_i$ the incident flux ($\phi_i=100 \,Wm^{-2}$ in you case), $\phi_e$ the emitted flux (the unknown) and $r$ the reflection coefficient of the lid ($1/2$ in your case), you have the energy balance equation: $$ \phi_e = r\phi_e+\phi_i $$ which you can directly solve: $$ \phi_e = \frac{\phi_i}{1-r} $$ Your method by geometric series is essentially a resolution of the fixed point equation by iteration: $$ \phi_e^{(n)} = r\phi_e^{(n-1)}+\phi_i $$ which by construction converges to the unique solution: $$ \phi_e^{(n)} \xrightarrow{n\to\infty} \phi_e $$ What is nicer about the iterative approach is that you expect you'd expect something like that to happen dynamically, i.e. you'd expect an exponential relaxation to the stationary state in time. It also explains what happens to the case $r=1$. Trying to solve the equation would tell you that there are no solutions, but the iterative method says that there is a linear blowup, so you'd need to typically take into account nonlinear effects.

Hope this helps.