# Estimating Sun luminosity from first principles

In a lecture, it was claimed that luminosity can be estimated by dividing how much energy there is in the Sun with the typical time it takes for a photon to get out from the centre of the Sun to its surface. I don't see how that argument holds. Why would that give me luminosity?

• What lecture? Who was giving it? – James K Feb 3 at 23:31

The thermal energy of the Sun is something like $$E \sim \left(\frac{3k_BT}{2}\right)(N_i + N_e)\ ,$$ where $$N_i$$ is the number of ions and $$N_e$$ is the number of electrons and the interior temperature $$T \sim 10^7$$ K.
To first order we can consider the Sun to be made of hydrogen, so $$N_e = N_i \sim M/m_p$$, where $$m_p$$ is the mass of a proton.
Putting in the numbers gives $$E \sim 5\times 10^{41}$$ J.
The typical mean free path of a photon in the solar interior is $$l\sim 10^{-3}$$ m. The solar radius is $$R\sim 7\times 10^{8}$$ m. Since the photons get out by a "random-walk" diffusion process, then it takes about $$(R/l)^2$$ steps, each of which takes a time $$l/c$$. Thus the diffusion timescale $$\tau \sim \left(\frac{R}{l}\right)^2 \left(\frac{l}{c}\right) = \frac{R^2}{lc} = 1.6\times 10^{12}\ {\rm s}$$.
The ratio of these two numbers gives $$3\times 10^{29}\ {\rm W}$$, so about 3 orders of magnitude larger than the solar luminosity.