The necessary requirements for a “Greenhouse effect”

First, as so many have so many times said, the so called “Greenhouse effect” is completely different in physics to a real greenhouse. Instead as I outlined in my last post, the effect stems from the lapse rate: that the earth AS SEEN FROM SPACE must emit radiation equivalent to a black body at 255C (the temperature at which incoming and outgoing radiation balances at our distance from the sun).

The reason for the higher temperature at the surface, is because there is a drop in temperature with height due to the lapse rate, and this means the planet’s surface will be warmer. Yes, there is radiation trapping, yes greenhouse gases have a marginal effect on the average height (and therefore temperature) of the earth as seen from space (or to reverse, the temperature of the surface must change to bring the temperature as seen from space back to the stable blackbody temperature). However, the radiation trapping is important, only in that it drives the lapse rate. However, I jump the gun. So what are the key requirements for the Greenhouse effect:

An Atmosphere with Pressure

The first requirement is that the planet must have an atmosphere. The reason for this, is that there is no temperature profile through the atmosphere without an atmosphere.

Greenhouse gases OR IR emitters like Clouds

The next requirement is that the atmosphere, which will have a thermal gradient due to the lapse rate, must emit radiation from within the atmosphere from either IR interactive molecules like water vapour, CO2, etc. or from liquid droplets in clouds.

That there must be heating to drive convection

Whilst the lapse rate explain the thermal gradient, it is not possible for heat to be emitted from the atmosphere without simply cooling the atmosphere, unless  unless there is a heat flow from lower down in the atmosphere. In other words, there must be an excess of heat in the lower atmosphere, that drives convective currents to take that heat to higher up where it will be emitted.

This is where the “Mickey mouse” explanation that we hear so often about global warming is partly right. The scale of the greenhouse effect is almost entirely due to the lapse rate. In that regard, the “heat trapping” concept is completely unhelpful. However, unless there were heat trapping: radiant energy is absorbed at the surface and cannot escape via IR, then there would be no energy source to drive the convective currents which stabilise the temperature gradient in the atmosphere DESPITE massive cooling as IR is emitted from out of the top of the atmosphere.

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9 Responses to The necessary requirements for a “Greenhouse effect”

  1. Pablo says:

    You might be interested in:

    ..which discusses the reduction of the gravitational lapse rate via intermolecular radiation which “tends to bring the air of different levels to the same temperature.”

    • Scottish-Sceptic says:

      Thanks – yes an interesting article. I have of course glossed over the issue of what happens if there is not a lapse rate to create the temperature gradient from the surface to the “topmost” molecules, which is the process of creating the “greenhouse” temperature. It’s obvious on earth that there’s plenty of spare energy to drive that gradient most of the time, but on planets like Venus and Jupiter with a thick atmosphere that simplification no longer holds unless there is an internal source of energy.

  2. oldbrew says:

    Radiation works both ways – up and down, or all ways to be more exact. What you get with an atmosphere is pressure.

    Temperature and atmospheric pressure are closely related, as per the ideal gas law. In the lower troposphere the temp and pressure gradients are almost identical (see linked graphs), which is the basis of the Standard Atmosphere used by the aviation industry and others.

    • Scottish-Sceptic says:

      Very well put, within the atmosphere where radiation is effectively trapped, the whole greenhouse effect is determined by pressure (as seen by lapse rate gradient).

      However, that relationship no longer works when you get to the edge of the atmosphere where radiation is no longer simply transferring energy within the atmosphere, but is losing out to space. Then you get a discontinuity in the relationship. This is where Ned has made such an enormous booper. He’s got it right within the atmosphere, but he’s misunderstood where the “top” of the atmosphere is. When calculating the greenhouse temperature, the “top” is the point where radiation leaves it – not the physical “top” beyond which there are no molecules.

  3. Pablo says:

    An atmosphere without water vapour or other radiating gases will still have the lapse rate of 9.8C/km due to the acceleration of gravity at 9.8m/sec/sec.
    The presence of water vapour enables surplus energy at the surface to be radiated to its compatriot molecules aloft to cool the surface air and warm the higher levels thus shrinking the thermal gradient.
    This means that stable air is usually 3.3C/km warmer than it should be for its height (an increase of potential temperature of 3.3C/km) and has a lapse rate of 6.5C. So at a 10km high tropopause the air would be 16.5C warmer and the surface air 16.5C cooler by intermolecular radiation. ( NASA’s 33ºC warming?)
    Mixing by turbulence and wind in the boundary layer during day time always corrects that discrepancy back to the gravitational lapse rate so usually up to the cloud base there is no potential temperature increase with height (ie. the 9.8ºC/km lapse rate).

    • Scottish-Sceptic says:

      I couldn’t have put that better (I mean literally couldn’t).

      I don’t pretend to understand the lapse rate in all its detail, because what is important is that the atmosphere has a relatively stable temperature profile that doesn’t (massively) change with “greenhouse” gas concentration. This means the temperature gradient of the atmosphere depends on the lapse rate and not greenhouse gas concentration. But the height from which IR escapes = topmost molecules/surfaces is dependent on the concentration (rather like the distance we can see into a wood depends on the density of trees).

      Thus the exact height (and so temperature) of the topmost emitting surface is dependent mostly on the lapse rate, but also marginally on the density of greenhouse gas (CO2 +H2O + water droplets, dust, etc.)

  4. I would rephrase this:

    “The first requirement is that the planet must have an atmosphere. The reason for this, is that there is no temperature profile through the atmosphere without an atmosphere.”


    A planet has an atmosphere. The reason is that gravity does work which we call pressure.

    Heat expands and that is why matter is excited into an atmospheric low density shell. Gravity contracts it and create pressure. Whithout gravity it would be blown into space.

    Gravity acting in a point along a meter in a second: 9.8Nm worth of force. (surface acceleration, 9.78m/s,

    Gravity acting on a surface area of 1m^2: 9.78×9.78=95.7N/m^2
    N/m^2 is equal to W/m^2.
    TSI is 1360.8W/m^2, average/m^2=340.2W/m^2

    First law:

    dU(340.2)-W(95.7)=Q(244.5W/m^2)=256.3K effective temperature.

    When making units comparable, force and added radiation fits emission perfectly.

    I also noticed lots of other peculiar correlations. Like how gravity and TSI is equal in the equation shown here:

    TSI/(4/3pi*r^3)= (4/3pi*r^3)*8g^2

    And also according to the inverse square law: 4g^2(4*95.7)=383W/m^2=287K
    Also: g^2*4/3=average tropopause temp

    Surface temp from TSI: double shell sphere irradiated on one side:
    (TSI*pi*r2)/(4/3pi*r^3))/(2pi*r^2), or 1/2((TSI/(4/3))

    Works for Mars as well. It is a bit trickier on Venus, but the relationship is definately there. For average moon temperature, just use inverse square law, volume of a sphere, and TSI.

    More weird stuff:

    There is only heat and gravity in the universe, and only heat and work can raise temperature.

    • Sorry, this:
      (TSI*pi*r2)/(4/3pi*r^3))/(2pi*r^2), or 1/2((TSI/(4/3))

      should be:
      (TSI*pi*r2)/(4/3pi*r^3)^2)/(2pi*r^2), or 1/2((TSI/(4/3)^2)

      • Scottish-Sceptic says:

        I would change that from “and only heat and work can raise temperature.” … in the absence of energy loss or gain.

        That is very true in much of the atmosphere, but it breaks down at the top where all the heat flow is emitted.

        The difference between what I have and Ned, is that Ned assumes no heat loss from the atmosphere (but that implies all heat is lost at the very topmost molecules). Whereas I take a more realistic approximation, which is that heat is lost from the topmost IR active molecule. The difference is often relatively small, which is why Ned’s approximation is not too bad at modelling planetary atmospheres.

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