On the physical origin of the semiannual component of surface air temperature over oceans

Abstract

With the understanding that seasonal cycle of the temperature is forced principally by the annually evolving solar irradiance, many previous studies have defined seasonal cycle of surface air temperature (SAT) as the sum of yearly-period sinusoidal component and its harmonics, especially semiannual component. This perception of directly forced response is confirmed for tropical SAT in this study. However, in mid-latitude and subpolar oceans, the ratio between the semiannual and annual components of solar irradiance is negligibly small but that of the SAT over oceans is not, which remains to be understood. To solve this puzzle, a simple energy budget model including main energy sources and sinks of oceanic mixed layer is designed. It is revealed that, when the oceanic mixed layer is prescribed as a layer of constant depth, the phase and amplitude of the modeled SAT are not consistent with these of observations. However, when the annually changing heat capacity of oceanic mixed layer is included, both the amplitude and phase of the modeled SAT share these of the observed SAT, proving that the semiannual component of SAT over mid-latitude and subpolar oceans is a result of the heat capacity-varying oceanic mixed layer in response to annually evolving solar irradiance.

Appendix: Analytical solutions for seasonal heat capacity

With

$$\begin{aligned} S \approx s_0 + s_1 \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _{s1}\right) , D = d_0 + d_1 \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _d\right) , \end{aligned}$$

and F being assumed as a constant, Eq. (2) can be written as

$$\begin{aligned}&C_p \left[ d_0 + d_1 \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _d\right) \right] \frac{d T}{d t} \nonumber \\&\quad = (1-\alpha ) \gamma \left[ s_0 + s_1 \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _{s1}\right) \right] \nonumber \\&\qquad - (\epsilon -\beta ) \sigma (4 T_0^3T - 3T_0^4) - F. \end{aligned}$$

(A1)

By multiplying $1 - \frac{d_1}{d_0} \mathrm {sin} (\frac{2 \pi }{\tau _a} t + \phi _d)$ to both sides and taking advantage of

$$\begin{aligned} 1 - \frac{d_1^2}{d_0^2} \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _d\right) \approx 1 \end{aligned}$$

with assumption of $d_1 \ll d_0$, Eq. (A1) becomes

$$\begin{aligned} C_p d_0 \frac{d T}{d t}&= \left[ 1 - \frac{d_1}{d_0} \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _d\right) \right] \nonumber \\&\quad \left\{ (1-\alpha ) \gamma \left[ s_0 + s_1 \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _{s1}\right) \right] \right. \nonumber \\&\qquad \left. -(\epsilon -\beta ) \sigma (4 T_0^3T - 3T_0^4) - F\right\} . \end{aligned}$$

(A2)

After reorganization, we obtain

$$\begin{aligned} \frac{d T}{d t}+B_4 T&= B_1 + B_{21} \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _{s1}\right) + B_{22} \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _d\right) \nonumber \\&\quad + B_3 \mathrm{cos} \left( \frac{4 \pi }{\tau _a} t + \phi _{s1} + \phi _d\right) , \end{aligned}$$

(A3)

where

$$\begin{aligned} \left. \begin{array}{r} B_1 = \frac{1}{C_p d_0} [(1-\alpha ) \gamma s_0 + 3 (\epsilon -\beta ) \sigma T_0^4 \\ - F - \frac{1}{2} (1-\alpha ) \gamma s_1 \frac{d_1}{d_0} \mathrm{cos} (\phi _{s1} - \phi _d)] \\ B_{21} = \frac{1}{C_p d_0} (1-\alpha ) \gamma s_1 \\ B_{22} = - \frac{d_1}{C_p d_0^2} [(1-\alpha ) \gamma s_0 + 3 (\epsilon -\beta ) \sigma T_0^4 - F] \\ B_3 = \frac{d_1}{2 C_p d_0^2} (1-\alpha ) \gamma s_1 \\ B_4 = \frac{4}{C_p d_0} (\epsilon -\beta ) \sigma T_0^3 \left[ 1 - \frac{d_1}{d_0} \mathrm{sin} \left( \frac{2 \pi }{\tau _a} t + \phi _d\right) \right] \end{array}\right\} . \end{aligned}$$

Note that the semiannual part in Eq. (A3) (the fourth term on the right-hand-side) does not come directly from solar irradiance.

The solution to this equation is hard to express explicitly. To obtain an approximate solution, we ignore the parameter, $\frac{d_1}{d_0} \mathrm{sin} (\frac{2 \pi }{\tau _a} t + \phi _d)$, in $B_4$, which results in the nonlinearity in thermal radiation term. Thus, in a stable solution, T can be expressed in a similar way of Eq. (5):

$$\begin{aligned} T = a_0' + a_1' \mathrm{sin} \left( \frac{2 \pi }{\tau _a}t+\phi _1'\right) + a_2' \mathrm{sin} \left( \frac{4 \pi }{\tau _a} t + \phi _2'\right) , \end{aligned}$$

(A4)

where

$$\begin{aligned} \left. \begin{array}{r} B_3 = \frac{d_1}{2 C_p d_0^2} (1-\alpha ) \gamma s_1, B_4' = \frac{4}{C_p d_0} (\epsilon -\beta ) \sigma T_0^3 \\ a_2' = \frac{B_3}{\sqrt{B_4'^2 + \frac{16 \pi ^2}{\tau _a^2}}}, \phi _2' = \phi _{s1} + \phi _d + \mathrm{arctan} \frac{\tau _a B_4'}{4 \pi } \end{array}\right\} . \end{aligned}$$