<p>The mean density of the star is really only defined by the formula $\bar\rho=M/V=3M/4\pi R^3$.  The radius of a star is a generally a very complicated function of a star's other properties.  When we determine the radius in stellar models, it's only because we've solved equations that describe the structure of the whole star, and read off the value at what we define as the surface. So no simple formula in general.</p>

<p>That said, one can derive the approximate functional dependence for stars of various evolutionary states through the principle of <em>homology</em>. i.e. assuming that stars of a certain type are just rescaled versions of each other. Glancing at my old course notes, on the upper main sequence, where stars burn hydrogen principally through the CNO cycle and have radiative envelopes dominated by electron-scattering opacity, we derived $R\propto M^{15/19}$. The same principle (but with different assumptions about the star) is used to determine the location of the <a href="https://en.wikipedia.org/wiki/Hayashi_track">Hayashi track</a> for pre-main-sequence stars, along which $R\propto M^{-7}T^{49}$.  Particular formulae can be found for different types of star but the relationships between $M$ and $R$ vary wildly.</p>

<p>Neither the two stars you mentioned are typical main-sequence stars. R136a1 is a Wolf-Rayet star, which is basically a star that has blasted away most of its hydrogen envelope.  Mass-radius relations are usually strongly dependent on mean molecular weight, which is higher without hydrogen, so the relations break down (or, rather, would have to be derived separately). But usually higher mean molecular weight gives a more compact star. UY Scuti has probably finished burning hydrogen in its core and has moved off the main sequence. So again, it'll follow a different relation.</p>

CC BY-SA 3.0
2016-02-10T09:26:14.890Z
Maybe this should be a new question, but in the case of our sun, it's slowly turning hydrogen into helium which should make it more dense, but it's also, slowly growing hotter which should expand the material and make it larger.   Is our sun growing smaller or larger over time?   It's growing more luminous, but that doesn't necessarily mean larger, though I'd assumed that was the case, a very gradual increase in both size and temperature.
userLTK
CC BY-SA 3.0
2016-02-10T10:47:03.030Z
@userLTK: Yes, the Sun's radius and with it the luminosity increase over time. Wikipedia reference: http://arxiv.org/abs/0911.4872
AtmosphericPrisonEscape
CC BY-SA 3.0
2016-02-11T06:53:36.323Z
That's right. That falls into the part about how really you have to solve the full equations to know. The homology relations always rely on a particular set of assumptions (and approximations about things like opacity and nuclear reaction rates) that are valid only for some class of star. On the main sequence, they're usually for "zero age": i.e. when nuclear reactions begin and the star is still chemically homogeneous.
Warrick
CC BY-SA 3.0
2016-02-10T08:33:17.393Z
2016-02-10T08:33:17.393Z
Warrick
<p>If we take a look at stars more massive than the Sun, their densities vary a lot. UY Scuti is an extremely low-density star that's only <strong>8.5x</strong> more massive than the Sun, but is <strong>1000-2000x</strong> its size. r136a1, on the other hand, is <strong>256x</strong> more massive than the Sun, but is only <strong>30x</strong> its size. Neither of them are main-sequence. </p>

<p>The Sun has a lower mass than both of them, yet it's also smaller than both of them.</p>

<p>It doesn't seem like mass makes a big difference. Both reach hydrostatic equilibrium at completely different sizes. r136a1 has a much stronger gravitational pull than UY Scuti, but it should also have much more radiation pressure, right?</p>

<p>So how can we determine the density of a star? Is there a formula?</p>

CC BY-SA 3.0
2016-02-10T09:20:12.620Z
In a hugely oversimplified sense, during main sequence, more massive stars burn hotter and faster and grow significantly larger and as a result are much less dense, but that ignores red giant stages or white dwarf stages and a number of other factors.  The density of smaller red dwarf stars or brown dwarfs can be quite high.  Warrick's answer is great by the way, I'm just summarizing.
userLTK
<star><the-sun><mass><stellar-dynamics><density>
How do star densities work?
Sir Cumference
# How do star densities work?

(Asked by: username_0 on 2016-02-10)

If we take a look at stars more massive than the Sun, their densities vary a lot. UY Scuti is an extremely low-density star that's only **8.5x** more massive than the Sun, but is **1000-2000x** its size. r136a1, on the other hand, is **256x** more massive than the Sun, but is only **30x** its size. Neither of them are main-sequence.

The Sun has a lower mass than both of them, yet it's also smaller than both of them.

It doesn't seem like mass makes a big difference. Both reach hydrostatic equilibrium at completely different sizes. r136a1 has a much stronger gravitational pull than UY Scuti, but it should also have much more radiation pressure, right?

So how can we determine the density of a star? Is there a formula?

**Tags:** star, the-sun, mass, stellar-dynamics, density
> **Comments:**
> **username_1**: In a hugely oversimplified sense, during main sequence, more massive stars burn hotter and faster and grow significantly larger and as a result are much less dense, but that ignores red giant stages or white dwarf stages and a number of other factors.  The density of smaller red dwarf stars or brown dwarfs can be quite high.  Warrick's answer is great by the way, I'm just summarizing.  

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## 1 answers

### Answer by username_2 on 2016-02-10. Score: 5 | Accepted answer

The mean density of the star is really only defined by the formula $\bar\rho=M/V=3M/4\pi R^3$. The radius of a star is a generally a very complicated function of a star's other properties. When we determine the radius in stellar models, it's only because we've solved equations that describe the structure of the whole star, and read off the value at what we define as the surface. So no simple formula in general.

That said, one can derive the approximate functional dependence for stars of various evolutionary states through the principle of *homology*. i.e. assuming that stars of a certain type are just rescaled versions of each other. Glancing at my old course notes, on the upper main sequence, where stars burn hydrogen principally through the CNO cycle and have radiative envelopes dominated by electron-scattering opacity, we derived $R\propto M^{15/19}$. The same principle (but with different assumptions about the star) is used to determine the location of the [Hayashi track](https://en.wikipedia.org/wiki/Hayashi_track) for pre-main-sequence stars, along which $R\propto M^{-7}T^{49}$. Particular formulae can be found for different types of star but the relationships between $M$ and $R$ vary wildly.

Neither the two stars you mentioned are typical main-sequence stars. R136a1 is a Wolf-Rayet star, which is basically a star that has blasted away most of its hydrogen envelope. Mass-radius relations are usually strongly dependent on mean molecular weight, which is higher without hydrogen, so the relations break down (or, rather, would have to be derived separately). But usually higher mean molecular weight gives a more compact star. UY Scuti has probably finished burning hydrogen in its core and has moved off the main sequence. So again, it'll follow a different relation.

> **Comments:**
> **username_1**: Maybe this should be a new question, but in the case of our sun, it's slowly turning hydrogen into helium which should make it more dense, but it's also, slowly growing hotter which should expand the material and make it larger.   Is our sun growing smaller or larger over time?   It's growing more luminous, but that doesn't necessarily mean larger, though I'd assumed that was the case, a very gradual increase in both size and temperature.  
> **username_3**: @userLTK: Yes, the Sun's radius and with it the luminosity increase over time. Wikipedia reference: http://arxiv.org/abs/0911.4872  
> **username_2**: That's right. That falls into the part about how really you have to solve the full equations to know. The homology relations always rely on a particular set of assumptions (and approximations about things like opacity and nuclear reaction rates) that are valid only for some class of star. On the main sequence, they're usually for "zero age": i.e. when nuclear reactions begin and the star is still chemically homogeneous.  

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