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?

  • $\begingroup$ What lecture? Who was giving it? $\endgroup$
    – James K
    Feb 3, 2021 at 23:31

1 Answer 1


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.

This ratio does not give the solar luminosity.


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