Page 191 - Physical Principles of Sedimentary Basin Analysis

6.18 Periodic heating of the surface 173

6.18 Periodic heating of the surface
The surface temperature of the Earth changes regularly on several different time scales. On
a short time scale it changes with night and day, and on the time scale of a year it changes
with the seasons. Finally, on a time scale of several hundred years to millions of years the
surface temperature is also controlled by climatic changes. The impact of periodic changes
on the surface temperature is found by solving the temperature equation

2
∂T ∂ T
− κ = 0 (6.266)
∂t ∂z 2
with a periodically changing surface temperature
T (z = 0, t) = T cos(ωt) (6.267)

as a boundary condition on the surface z = 0. The amplitude of the temperature change
is T and the circular frequency is ω. The second boundary condition is an unchanged
temperature at great depth
T (z =−∞, t) = 0. (6.268)
We see that the time it takes to complete one cycle, or one period, is

2π
t p = . (6.269)
ω
The inverse of circular frequency, t 0 = 1/ω, is a natural choice for a characteristic time,
and a characteristic length then follows from the characteristic time combined with the
√ √
thermal diffusivity, l 0 = κt 0 = κ/ω. We have already seen several examples where a
characteristic length is used in combination with the thermal diffusivity to obtain a charac-
teristic time. The characteristic length is “large” for a “small” frequency, and “small” for a
“large” frequency, which reflects the fact that slow oscillations in the surface temperature
penetrate deeper than fast oscillations. We will soon see how this follows from the tem-
perature solution. The temperature equation (6.266) can then be written in dimensionless
form as
∂T ˆ ∂ T
2 ˆ
− = 0 (6.270)
∂ ˆ t ∂ ˆz 2
ˆ
where T = T/ T , ˆ t = t/t 0 and ˆz = z/l 0 . The dimensionless boundary conditions become
T (ˆz = 0, ˆ t) = cos(ˆ t) and T (ˆz =∞, ˆ t) = 0. (6.271)
ˆ
ˆ
The scaled temperature equation (6.270) with boundary conditions (6.271) has no (explicit)
parameters, and the solution is

ˆ z 1

ˆ
T (ˆz, ˆ t) = exp −√ cos ˆ t − √ ˆz , (6.272)
2 2
as shown in Note 6.10. A first thing to notice about the temperature solution is that
the temperature decays exponentially with depth, because it is proportional to the factor