lördag 2 april 2011

A Simple Radiation Model

Inspired by Claes Johnsson I will here make an attempt to formulate a simple radiation model aimed at illustrating how the temperature gradient changes with an increase in absorption/emission parameter A. The equation reads

(A - A^2)U(x) - 1/2*A^2U''(x) = 0.

U(x) stands for the temperature. In this model we assume that the radiation is proportional to T and not T^4. The amount of radiation going directly into space is propotional to A - A^2, since the amount A^2 is absorbed by the atmosphere. The remaining amount of radiation is instead transported as diffusion. The physical solution to this equation is

U(x) = exp(-ax)

where a^2 = 2(1 - A)/A, which is a strictly decreasing function of A in the interval (0,1). Hence, the gradient flattens as the optical activity A increases, contrary to the greenhouse hypothesis. The situation with A = 1 corresponds to a fully opaque atmosphere. The strength of this model is that the A^2 term can be replaced by any term less than A and the main conclusion stands still. All other forms of heat transfer are neglected here which is the cause of the singularity at A = 0. Comments are welcome.

PS

The formula follows from some very simple assumptions which is most clearly seen in a discrete form, create a discrete mesh with index n, U(n) is the temperature at position n. The equation is now

-(A - A^2)U(n) + 1/2*A^2(U(n-1) - U(n)) - 1/2*A^2(U(n) - U(n+1)) = 0

The first term is the heat lost by direct radiation to space. The second term is the heat gained from the warmer lower position. The third term is the heat lost to the cooler upper position. The equation can be rewritten

(A - A^2)U(n) - 1/2*A^2(U(n+1) - 2U(n) + U(n-1)) = 0

And we can now identify the discrete second derivative. Excersize: Spot the "back-radiation" terms ;)

1 kommentar:

  1. Hi Anders,

    I'm not sure where you got your physics training, but I hope you kept the receipt ;-)

    In constructing your "model", you assert that

    "The amount of radiation going directly into space is propotional to A - A^2, since the amount A^2 is absorbed by the atmosphere." And then you use the term

    (A - A^2)U(n)

    for "the heat lost by direct radiation to space.", where U(n) is the "temperature at position n".

    But this term takes no account of how much atmosphere the radiation has to pass through to escape into space. Even if "position n" somehow relates to a position in the atmosphere, this has no effect on the calculation.

    So your "model" assumes that a position at the planet surface or on the border of space loses exactly the same amount of energy "directly into space".

    In other words, it asserts something entirely unphysical and so will not return useful results.

    I'm afraid this kind of thing is not easy to do, and perhaps learning to walk before trying to run, on the maths front at least, might be something to consider.

    You also write:

    "The remaining amount of radiation is instead transported as diffusion."

    Before attempting to make models of this kind, it is advisable to familiarise oneself with the fundamental principles and properties involved. In this case, if you had troubled yourself to discover what the processes of radiation and diffusion physically entail, you would not have confused their properties so completely.

    SvaraRadera