The Solar Temperature *

Estimate the solar temperature using

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

\begin{displaymath}\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\end{displaymath}

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

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

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 \bgroup\color{black}$E(\lambda,T)$\egroup peaks at 500 nm as one of the graphs shows. We need to transform \bgroup\color{black}$E(\nu,T)$\egroup. Remember \bgroup\color{black}$f(\nu)d\nu=g(\lambda)d\lambda$\egroup 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

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

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 \bgroup\color{black}$y=h\nu/kT$\egroup and solve the equation

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

I solved this iteratively starting at y=5 and got \bgroup\color{black}$y=4.97$\egroup implying

\begin{displaymath}\bgroup\color{black}h\nu=4.97kT\egroup\end{displaymath}

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

\begin{displaymath}\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\end{displaymath}

That's agrees well.

Jim Branson 2013-04-22