<p>Very roughly.</p>
<p>The star is in hydrostatic equilibrium so
<span class="math-container">$$ \frac{dP}{dr} = -\rho g$$</span>.</p>
<p>Replacing <span class="math-container">$dP/dr$</span> by <span class="math-container">$P_c/R$</span>, where <span class="math-container">$P_c$</span> is the central pressure and <span class="math-container">$R$</span> the radius, and letting <span class="math-container">$\rho \propto M/R^3$</span> and <span class="math-container">$g \propto M/R^2$</span>, then we get
<span class="math-container">$$ P_c\propto \frac{M^2}{R^4}$$</span>
But the central pressure is proportional to <span class="math-container">$\rho T/\mu$</span>, where <span class="math-container">$T$</span> is the interior temperature and <span class="math-container">$\mu$</span> is the mean molecular mass. So
<span class="math-container">$$ \frac{\rho T}{\mu} \propto \frac{M^2}{R^4}$$</span>
<span class="math-container">$$ \frac{MT}{R^3 \mu} \propto \frac{M^2}{R^4}$$</span>
<span class="math-container">$$ R \propto \frac{\mu M}{T}$$</span></p>
<p>Now during the main sequence lifetime, the mass stays (roughly) constant and <span class="math-container">$T$</span> stays (roughly) constant because the temperature sensitivity of hydrogen burning is high.
However, <span class="math-container">$\mu$</span> increases because 4 hydrogens (4 protons + 4 electrons) are getting turned into 1 helium (1 nucleus + 2 electrons). So <span class="math-container">$\mu$</span> increases (in the central regions) from 0.5 to 4/3. Hence the radius increases.</p>
<p>The radius increase is indeed about a factor of 2 over the whole main sequence lifetime.</p>
<p>A different way to think about it, more appropriate for solar mass stars and above, is in terms of the radiative energy transport. The increase in <span class="math-container">$\mu$</span> is also accompanied by a <em>decrease</em> in the number of free electrons per mass unit. This reduces the opacity in the central regions and decreases the temperature gradient. This means the star is bigger for roughly the same central temperature.</p>

CC BY-SA 4.0
2021-03-16T17:58:17.977Z
2021-03-16T18:11:30.060Z
2021-03-16T18:11:30.060Z
ProfRob
<p>You might want to have a look into a stellar structure and evolution book for this, such as <a href="https://www.springer.com/de/book/9783642302558" rel="nofollow noreferrer">Kippenhahn</a>.</p>
<p>But in short, for a star such as the sun the following happens:</p>
<ol>
<li>Throughout the lifetime, H is slowly being converted to He in the core.</li>
<li>This changes the mean-molecular weight <span class="math-container">$\mu$</span> of the core with time.</li>
<li>The hydrostatic law demands for the stellar structure to fulfil <span class="math-container">$\frac{{\rm d}P}{{\rm d}r} = - \rho g(r)$</span> and <span class="math-container">$P = \frac{\rho k_B T}{\mu}$</span>. As <span class="math-container">$\mu$</span> increases with time in the core, this decreases the pressure support there, and demand it to shrink. This again increases the luminosity and expands the outer stellar shells, increasing the stars radius.</li>
</ol>

CC BY-SA 4.0
2021-03-16T08:27:54.913Z
2021-03-16T10:58:27.293Z
2021-03-16T10:58:27.293Z
AtmosphericPrisonEscape
<p>I know that main sequence stars increase in temperature and luminosity as they age. However, I do not know about how they change in size. Do they expand? If so, why do they expand? Do they grow very much? I’ve tried to search up the answer, but details were limited, so I would appreciate an answer here.</p>

<star>
How do stars change in size over their main sequence lifetime?
Jackson
# How do stars change in size over their main sequence lifetime?

(Asked by: username_0 on 2021-03-16)

I know that main sequence stars increase in temperature and luminosity as they age. However, I do not know about how they change in size. Do they expand? If so, why do they expand? Do they grow very much? I’ve tried to search up the answer, but details were limited, so I would appreciate an answer here.

**Tags:** star


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## 2 answers

### Answer by username_1 on 2021-03-16. Score: 6

Very roughly.

The star is in hydrostatic equilibrium so
$$ \frac{dP}{dr} = -\rho g$$.

Replacing $dP/dr$ by $P\_c/R$, where $P\_c$ is the central pressure and $R$ the radius, and letting $\rho \propto M/R^3$ and $g \propto M/R^2$, then we get
$$ P\_c\propto \frac{M^2}{R^4}$$
But the central pressure is proportional to $\rho T/\mu$, where $T$ is the interior temperature and $\mu$ is the mean molecular mass. So
$$ \frac{\rho T}{\mu} \propto \frac{M^2}{R^4}$$
$$ \frac{MT}{R^3 \mu} \propto \frac{M^2}{R^4}$$
$$ R \propto \frac{\mu M}{T}$$

Now during the main sequence lifetime, the mass stays (roughly) constant and $T$ stays (roughly) constant because the temperature sensitivity of hydrogen burning is high.
However, $\mu$ increases because 4 hydrogens (4 protons + 4 electrons) are getting turned into 1 helium (1 nucleus + 2 electrons). So $\mu$ increases (in the central regions) from 0.5 to 4/3. Hence the radius increases.

The radius increase is indeed about a factor of 2 over the whole main sequence lifetime.

A different way to think about it, more appropriate for solar mass stars and above, is in terms of the radiative energy transport. The increase in $\mu$ is also accompanied by a *decrease* in the number of free electrons per mass unit. This reduces the opacity in the central regions and decreases the temperature gradient. This means the star is bigger for roughly the same central temperature.


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### Answer by username_2 on 2021-03-16. Score: 3

You might want to have a look into a stellar structure and evolution book for this, such as [Kippenhahn](https://www.springer.com/de/book/9783642302558).

But in short, for a star such as the sun the following happens:

1. Throughout the lifetime, H is slowly being converted to He in the core.
2. This changes the mean-molecular weight $\mu$ of the core with time.
3. The hydrostatic law demands for the stellar structure to fulfil $\frac{{\rm d}P}{{\rm d}r} = - \rho g(r)$ and $P = \frac{\rho k\_B T}{\mu}$. As $\mu$ increases with time in the core, this decreases the pressure support there, and demand it to shrink. This again increases the luminosity and expands the outer stellar shells, increasing the stars radius.


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