1
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ What lecture? Who was giving it? $\endgroup$ – James K Feb 3 at 23:31
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.