The Solar Temperature *

Estimate the solar temperature using

• the solar radiation intensity on the earth of 1.4 kilowatts per square meter. ( $r_{sun}=7\times 10^8$ m, $d_{sun}=1.5\times 10^{11}$ m)
• and the solar spectrum which peaks at about 500 nm.

First we compute the power radiated per unit area on the solar surface.

$$\bgroup\color{black}R=(1400\mathrm{W/m^2})(4\pi d_{sun}^2)/(4\pi r_{sun}^2)=6.4\times 10^7 \mathrm{W/m^2}\egroup$$

We compare this to the expectation as a function of temperature.

$$\bgroup\color{black}R(T)=(5.67\times 10^{-8} \mathrm{W/m^2/\;^\circ K^4}) T^4\egroup$$

and get

$$\begin{eqnarray*} T^4&=&{6.4\times 10^7\over 5.67 \times 10^{-8}}\\ T&=&5800\;^\circ K\\ \end{eqnarray*}$$

Lets assume that $E(\lambda,T)$ peaks at 500 nm as one of the graphs shows. We need to transform $E(\nu,T)$. Remember $f(\nu)d\nu=g(\lambda)d\lambda$ for distribution functions.

$$\begin{eqnarray*} E(\nu,T)&=&{2\pi\nu^2\over c^2}{h\nu\over e^{h\nu/kT}-1}\\ ... ...^{h\nu/kT}-1}={2\pi\nu^4 \over c^3}{h\nu\over e^{h\nu/kT}-1}\\ \end{eqnarray*}$$

This peaks when

$$\bgroup\color{black}{\nu^5\over e^{h\nu/kT}-1}\egroup$$

is maximum.

$$\begin{eqnarray*} {5\nu^4\over e^{h\nu/kT}-1}-{\nu^5 (h/kT)e^{h\nu/kT}\over (e^... ...}-1)\over e^{h\nu/kT}}=h\nu/kT\\ 5(1-e^{-h\nu/kT})=h\nu/kT\\ \end{eqnarray*}$$

Lets set $y=h\nu/kT$ and solve the equation

$$\bgroup\color{black}y=5(1-e^{-y})\egroup$$

I solved this iteratively starting at y=5 and got $y=4.97$ implying

$$\bgroup\color{black}h\nu=4.97kT\egroup$$

If we take $\lambda=500$ nm, them $\nu=6\times 10^{14}$.

$$\bgroup\color{black}T={h\nu\over 5 k}={(6.6\times 10^{-34})(6... ...10^{14})\over (5)(1.4\times 10^{-23})}=6\times 10^3=5700\egroup$$

That's agrees well.