Without a firm grasp on the basics it can be hard to choose between a good and bad explanation.
In The Hoover Incident I explained what would happen if the atmosphere didn’t absorb or emit radiation – i.e., if the radiatively active gases were “hoovered up”. Have a read of that post for a full explanation, but the essence of it is that with no atmospheric absorption or radiation the surface would be radiating around 390 W/m² into space while receiving only 240 W/m² from the sun. Therefore, the earth would cool down until it was only radiating 240 W/m² (it’s slightly more complicated) – leading to a surface temperature around -18°C (255K).
One of the statements I made was:
And no matter what happens to convection, lapse rates, and rainfall this cooling will continue. That’s because these aspects of the climate only distribute the heat.
Nothing can stop the radiation loss from the surface because the atmosphere is no longer absorbing radiation. They might enhance or reduce the cooling by changing the surface temperature in some way – because radiation emitted by the surface is a function of temperature (proportional to T4). But while energy out > energy in, the climate system would be cooling.
Recently, one commenter said in response to this (but in another article):
Convection etc distributes the heat in ways that affect radiative heat transport from the surface. While the intensity of radiation is a function of temperature, radiative heat transport is a function of a temperature difference. No heat may be exchanged between regions with the same temperature.
In an isothermal atmosphere there would be no temperature difference between it and the surface and therefore no heat loss from the surface. The accumulation of heat in a radiatively-constrained atmosphere by non-radiative means of heat transport from the surface would produce an isothermal atmosphere. Then, energy out = 0 and < energy in and the climate system would be heating.
The comment is confused, so I thought it was worth explaining in some detail.
Radiation and Temperature
This isn’t showing any heat transfer by conduction or convection, to keep the diagram simple.
The blue area – the troposphere, or lower atmosphere – is shown with a gap between it and the earth’s surface. This is just to make heat transfer values clearer – there isn’t really a gap. Notice that no radiation is emitted by the atmosphere (because this is a thought experiment where radiatively-active gases have been “hoovered” up).
Note as well that we are looking at averages in this diagram. The solar radiation absorbed in any one places is very rarely 240 W/m² – at night it is zero, and at midday in the tropics it is closer to 1000 W/m². If you want to understand why the average value of solar radiation absorbed is 240 W/m² take a look at Earth’s Energy Budget – Part One.
Rather than thinking of this as the average, if it helps, simply think of this as the heat transfer for one location where these are the actual values.
The equation for the emission of thermal radiation by the earth’s surface is only dependent on its temperature and emissivity. The equation is the well-known Stefan-Boltzmann law:
j = εσT4
where ε=emissivity, σ=5.67×10-8, T is temperature in K and j is energy per second per unit area (W/m²)
Emissivity is a value between 0 and 1, where 1 is a “blackbody” or perfect radiator. The surface of the earth has an emissivity very close to 1. See The Dull Case of Emissivity and Average Temperatures.
Now regardless of any heat transfer by conduction or convection with the atmosphere, the surface of the earth will continue to radiate in accordance with that equation. With an emissivity of 1, a surface of 15°C (288K) radiates 390 W/m².
Emission of thermal radiation is independent of any other heat transfer mechanisms and only depends on the temperature of the body and its emissivity.
Now the earth also absorbs solar energy by radiation. So for our initial conditions, the net heat transfer by radiation,
Hrad = 240 – 390 W/m² = -150 W/m² (i.e., a cooling of 150 W/m²)
The only heat transfer mechanism in a vacuum is radiation and therefore heat can only be transferred into and out of the total climate system by radiation. In our thought experiment the atmosphere is unable to absorb or emit radiation.
Therefore the solar energy absorbed at the surface minus the energy radiated from the surface of the earth gives the net heat transfer for the entire climate system.
Radiation, Sensible and Latent Heat
Now with the particular example above let’s add heat transfer between the surface and the atmosphere by conduction and convection. This is often termed sensible heat. Gases have a very low thermal conductivity, so most heat will be transferred by convection (bulk movement of air). This will also include latent heat, which is the heat used in evaporation of water from the surface of the earth.
There is no simple formula for convection because it depends on many factors including the speed of the air movement. The formula for latent heat removal is also complex. So to get started we will use the average value derived by Kiehl and Trenberth in their well-known 1997 paper. Note that their calculation of latent heat was derived from the amount of rainfall (what comes down, must have been evaporated up in the first place).
Here is the updated diagram, still showing the initial conditions, just after the Hoover Incident has taken place:
Note that the conduction and convection from the atmosphere into space = 0 W/m².
And with conduction and convection it is conventional to show the net flow of heat – which is why there is no arrow with heat from the atmosphere to the surface. (With radiation, because heat is exchanged across distances it is more usual to show radiation emitted from each body).
What happens now?
To calculate dynamic processes is more difficult, especially if we wanted to do it for all points on the earth.
What everyone should be able to see is that the surface of the earth is losing heat.
If we use the value from K&T for sensible and latent heat removal, we can see that net heat transfer from the surface of the earth at time = 0 is now 252 W/m². That is, a cooling of 252 W/m².
Let’s consider the atmosphere. It is gaining heat from the surface of the earth, and not radiating it into space (or back to the surface), because in this post-Hoover world we have an atmosphere with no ability to emit radiation.
Therefore, within a relatively short space of time, the heat transfer (averaged around the globe) between the surface of the earth and the atmosphere will drop to almost zero. If the atmosphere heats up and the earth’s surface cools down – the result has to be that this heat transfer reduces.
But whatever happens to the temperature difference and heat transfer between the atmosphere and the earth’s surface – the radiation from the earth’s surface into space will still follow the Stefan-Boltzmann law and be proportional to T4. The only way this can change is if fundamental physics turns out to be wrong..
How fast will the earth’s surface cool down?
This is a more challenging question. It involves calculating the heat flow out from the rocks, soil, sand, vegetation and most importantly, from the oceans. For each of these materials we would need to know the thermal diffusivity, which is the ratio of the thermal conductivity (how well heat travels through a material) to the heat capacity (how much heat is stored in a material per K of temperature change). As the earth’s surface cools down the rate of heat loss from radiation will reduce. This is because, using our earlier equation with the term for radiation stated explicitly:
Hrad = 240 – εσT4 W/m² ( = solar radiation absorbed – radiation emitted from the surface)
That is, the solar radiation absorbed stays constant while the radiation emitted reduces as the temperature decreases. It’s not so easy to visualize if you haven’t seen this kind of function before. Here is a very simple model of how the temperature (and net radiation) might change with time:
Click for a larger view
This graph is calculated by assuming that the climate system’s heat is stored in an ocean 4km deep and a very high thermal conductivity of water (that is, the heat can flow from the depths of the ocean to the surface with almost no resistance).
A more complete treatment takes account of the thermal conductivity of water. This value varies greatly depending on whether the water is still or well-mixed.
Below, the graph on the left shows the surface temperature against time. The graph on the right is more interesting and shows the temperature profile against depth of ocean for a few different times:
Click for a larger view
In this right hand graph the lower curves are later times. The initial temperature profile against depth is a straight line from 288K at the surface to 273K at 4000m – this is my assumption, my initial conditions.
What you can see from this graph is that the surface is much better at radiating heat away than the ocean is at conducting heat from its depths to the surface. That’s why the temperature stays higher for longer lower down in the ocean.
In this more thorough treatment the surface cools more quickly initially but will take much longer to reach the equilibrium of 255K.
And of course, alert readers will have noticed that it all changes when the surface freezes as the heat conductivity through ice will be different, and the albedo of the earth will change..
In fact, the problem can be made more and more complex, but that doesn’t change the essential elements.
Heat transfer by radiation is conceptually simple.
A surface emits radiation with a well-known formula which depends on temperature of that surface (and its emissivity).
The net heat transfer by radiation depends on how much radiation is incident on that surface from other bodies – whether near or far – and what proportion is absorbed.
In the case of a planet with an atmosphere – if the atmosphere cannot absorb or emit radiation then the equilibrium condition for that climate system will be where the radiation emitted by the planetary surface equals the radiation absorbed by the planetary surface.