<p>The thermal energy of the Sun is something like
<span class="math-container">$$E \sim  \left(\frac{3k_BT}{2}\right)(N_i + N_e)\ , $$</span>
where <span class="math-container">$N_i$</span> is the number of ions and <span class="math-container">$N_e$</span> is the number of electrons and the interior temperature <span class="math-container">$T \sim 10^7$</span> K.</p>
<p>To first order we can consider the Sun to be made of hydrogen, so <span class="math-container">$N_e = N_i \sim M/m_p$</span>, where <span class="math-container">$m_p$</span> is the mass of a proton.</p>
<p>Putting in the numbers gives <span class="math-container">$E \sim 5\times 10^{41}$</span> J.</p>
<p>The typical mean free path of a photon in the solar interior is <span class="math-container">$l\sim 10^{-3}$</span> m. The solar radius is <span class="math-container">$R\sim 7\times 10^{8}$</span> m. Since the photons get out by a &quot;random-walk&quot; diffusion process, then it takes about <span class="math-container">$(R/l)^2$</span> steps, each of which takes a time <span class="math-container">$l/c$</span>.
Thus the diffusion timescale
<span class="math-container">$$ \tau \sim \left(\frac{R}{l}\right)^2 \left(\frac{l}{c}\right) = \frac{R^2}{lc} = 1.6\times 10^{12}\ {\rm s}$$</span>.</p>
<p>The ratio of these two numbers gives <span class="math-container">$3\times 10^{29}\ {\rm W}$</span>, so about 3 orders of magnitude larger than the solar luminosity.</p>
<p>This ratio does not give the solar luminosity.</p>

CC BY-SA 4.0
2021-02-04T01:22:36.313Z
2021-02-04T01:22:36.313Z
ProfRob
<p>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?</p>

CC BY-SA 4.0
2021-02-03T23:31:36.457Z
What lecture? Who was giving it?
James K
<the-sun><luminosity>
Estimating Sun luminosity from first principles
zabop
# Estimating Sun luminosity from first principles

(Asked by: username_0 on 2021-02-03)

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?

**Tags:** the-sun, luminosity

> **Comments:**
> **username_1**: What lecture? Who was giving it?  

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## 1 answers

### Answer by username_2 on 2021-02-04. Score: 2

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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