Schwarzschild's equation for heat transfer by radiation through the atmosphere is a two-stream upwelling-downwelling heat transfer model with the net heat transfer expressed as the difference between upwelling and downwelling streams. It is the basic model for radiative heat transfer used in climate models.
To get perspective on this two-stream model, let us see what a two-stream model for heat transfer by conduction instead of radiation would look like. We shall then compare with the standard one-stream model for vertical heat conduction through a horisontal layer, which is Fourier's Law:
- $q(x) = -\gamma\frac{T(x+dx)-T(x)}{dx}$ or $q(x) = -\gamma\frac{dT}{dx}$, (1)
where $q(x)$ is net heat transfer/flow, $\gamma$ is a heat conductivity coefficient, $T(x)$ is temperature, $x$ is a vertical coordinate and $dx$ a small increment.
A two-way version of Fourier's Law takes the form
- $q(x) =\gamma\frac{T(x)}{dx} - \gamma\frac{T(x+dx)}{dx}$ (2)
where the net heat transfer $q(x)$ is expressed as the difference of two gross streams $\gamma\frac{T(x)}{dx}$ and $\gamma\frac{T(x+dx)}{dx}$ in opposite directions. But this two-way version is unstable because you are dividing temperatures $T(x)$ and $T(x+dx)$ by the small quantity $dx$, and so it is both uncomputable (by dividing by small $dx$) and unphysical since an unstable system does not have permanence over time. Note that the derivative $\frac{dT}{dx}$ does not suffer from the same instability since $dT$ is small.
We conclude that a Schwarzschild two-stream model for heat transfer by conduction is unstable, uncomputable and unphysical. Can we expect that Schwarzschild's two-stream model for heat transfer by radiation does not suffer from the same deficiency?
It remains to formulate a stable one-stream model for heat transfer by radiation through an absorbing/emitting gas. An interesting such model is presented here, to which I will return.
We can take the argument one step further by expressing conservation of heat energy in the form
- $\frac{q(x)-q(x-dx)}{dx} = f(x)$ (3)
- $\frac{T(x)}{dx^2} - \frac{T(x+dx)}{dx^2} - \frac{T(x-dx)}{dx^2} +\frac{T(x)}{dx^2}=f(x)$ (4)
- $q_{up}(x)\equiv\frac{2T(x)}{dx^2} =\frac{f(x)}{2}$
- $q_{down}(x)\equiv\frac{T(x+dx)}{dx^2} + \frac{T(x-dx)}{dx^2}=-\frac{f(x)}{2}$
- $-\frac{d^2T}{dx^2} \approx -\frac{T(x+dx)-2T(x)+T(x-dx)}{dx^2}=f(x)$
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