# Background

While I was in graduate school, I put together some cartoon-like comparisons of multiple stars to show the order of magnitude differences in radii.

At the time, VY Canis Majoris was the largest known star by radius (it appears from my graphic that it was then thought to be ~1950 $R_{\odot}$, where it is now thought to have a radius of $1420 \pm 120 \ R_{\odot}$). I see now that UY Scuti has taken that title with a radius of $1708 \pm 192 \ R_{\odot}$. I recall that at the time the mass of VY Canis Majoris was not well known (as suggested by my cartoon image) but now I see that it is reported to be $17 \pm 8$ $M_{\odot}$. Even more interesting is that UY Scuti has an even smaller mass of ~7-10 $M_{\odot}$.

As a comparison, one of the more massive stars in our catalogues is Eta Carinae, which is a binary system where the primary, $\eta$ Car A, has $r \sim 60-800 \ R_{\odot}$ and $M \sim 100-200 \ M_{\odot}$.

A quick survey of Wikipedia shows me that there are over a dozen stars with $r \geq 1000 \ R_{\odot}$ and over a dozen different stars with $M \geq 100 \ M_{\odot}$.

# Questions

1. What causes a star like UY Scuti to have such a large "radius" but so little mass while the much more massive $\eta$ Car A is less than half the size?
• Is it their respective ages?
• Is it their composition (i.e., fuel source)?
• en.wikipedia.org/wiki/Red_supergiant – user83548 Jun 10 '16 at 12:28
• To be very honest with you, I don't think that drawing these objects like this is correct. These "stars" are losing a lot of mass due to their strong stellar winds and I don't think they are nearly as sharply defined as shown. – CuriousOne Jun 10 '16 at 12:36
• @CuriousOne - Yes, I agree. I only created this as a cartoon for some of my family members to try to help them understand the whole "orders of magnitude" thing. – honeste_vivere Jun 10 '16 at 12:57
• Can't blame you for that one... the bigger issue is that we probably don't have a good model for these objects, yet, and that we aren't quite capable of imaging their core, i.e. what one would call "the star" inside the surrounding nebula. – CuriousOne Jun 10 '16 at 13:00
• @Vinay5forPrime - The Chandrasekhar limit is for the separation of white dwarf and neutron stars, not pre-collapse stars. This does not really apply to my question. – honeste_vivere Jun 10 '16 at 15:05

The virial theorem is a way of expressing the concept of hydrostatic equilibrium in a star. In dimensional terms we can say that $$\Omega = -3\int P\ dV,$$ where $\Omega$ is the gravitational potential energy and $P$ is the pressure.

Assuming a perfect gas and a uniform sphere (OK for a dimensional analysis), we can rewrite this as $$-\frac{3GM^2}{5R} = -3\frac{M kT}{\mu m_u},$$ where $\mu$ is the number of atomic mass units per particle in the gas and $T$ is some characteristic interior temperature. From this, we get $$R \sim \frac{GM\mu m_u}{5kT}$$

Now, what this simple argument shows is that the radius of a star does not just depend on its mass. It depends on $\mu$, which is composition dependent, and it depends on the interior temperature (profile).

Thus two stars with a different interior composition or internal temperature can have quite different radii at the same mass.

The radius also crucially depends on where nuclear burning is taking place (in the core or in a shell). A general rule is that shell burning stars have much larger radii.

It is this latter point which is largely responsible for the large discrepancy you note. There are no easy handwaving ways to explain why this is, but most of the luminosity of stars like VY CMa will be coming from a H burning shell.

Shell burning begins when the temperatures at the core are insufficient to ignite the ash of the previous burning phase. A layer of fresh fuel outside the core is compressed and heated until it ignites, with a greater volume and higher burning rate than the original core. This means the luminosity of the star increases drastically. However, there is a maximum temperature gradient supportable by stellar material - the so-called adiabatic temperature gradient where the star becomes unstable to convection. This maximum to the temperature gradient means that in order to radiate away the increased luminosity at the photosphere (at a few thousand degrees where the atmosphere becomes optically thin), the star has to swell up, according to Stefan's law ($L= 4\pi R^2 \sigma T^4$), to a much larger size.

So that's the key, it's what the star is made of and where the nuclear burning is taking place inside the star.

• Thanks, this is very helpful. What is the term $m_{u}$ though? If it is the masses of the constituent particles, why add the $\mu$ term? Or is it the effective mass of the constituent particles? – honeste_vivere Jul 16 '16 at 20:09
• @honeste_vivere $m_u$ is an atomic mass unit. $\mu$ is the (average) number of atomic mass units per particle. Particles in the gas are not identically massive and includes electrons. eg For ionised He, $\mu = 4/3$. – Rob Jeffries Jul 16 '16 at 21:19

This will be a short answer, not going very deep into how stars work. Basically, a star is a ball of gas which is more or less in equilibrium between collapse due to gravitation and expansion due to heat.

The radius of the star is determined by this equilibrium. A star which is more massive can have a smaller radius due to a large gravitational pull inwards. The temperature of a star, and thus the expansive force due to heat (you can imagine this like for an ideal gas: if you heat it up, it expands), is determined by nuclear fusion in its core.

Red supergiants, like UY Scuti have used up all their hydrogen fuel, thus their core collapsed due to lack of outward force in the core and got extremely hot. Because of this heat, and the relatively low mass, the equilibrium is established at a large radius. Eta Carinae is not as hot in its core but has more mass, so its radius is smaller.

Also note that the color of a star is determined by its surface temperature, not its core temperature.

• I am confused. I thought the core temperature could be approximated by something like: $$T_{core}[K] \approx \left(\frac{ G \ m_{p} \ M_{\odot} }{ 3 \ k_{B} \ R_{\odot}/2 }\right) \frac{ \tilde{M} }{ \tilde{R} } \sim 1.54 \times 10^{7} \frac{ \tilde{M} }{ \tilde{R} }$$ where $\tilde{Q}$ is the parameter normalized to the solar equivalent. This would suggest that larger mass and lower radius stars have hotter cores, thus more radiation pressure to expand the star. I do not see why a lower mass star would necessarily expand to such extremes though... – honeste_vivere Jun 10 '16 at 18:32
• Please correct me if i am wrong, but this formula looks to me like it is only valid for hydrogen-burning stars. If i remember correctly, in its derivation one approximates the Star as an ideal gas consisting of hydrogen atoms. Red supergiants have no more hydrogen left in their core, so the formula might Not be applicable here. – SimonS Jun 10 '16 at 18:54
• Perhaps, but I am confused why the core of UY Scuti would have been hotter than Eta Carinae, thus causing UY Scuti to over expand so much... In fact, I am inclined to think that Eta Carinae would have a much hotter core than, say, our sun. Doesn't the star need much higher temperatures for the CNO cycle to be efficient/dominant? – honeste_vivere Jun 10 '16 at 19:32
• After a main sequence star has used up all its hydrogen fuel, its core collapses, because there is no longer an outward force generated by the heat from the fusion process. The collapse heats up the helium gas the star now consists of(again similar to an ideal gas) up to a point where the temperature is high enough to fuse helium into carbon. This process runs at a much higher temperature(100 Million Kelvin) than hydrogen burning(10 to 50 million Kelvin). – SimonS Jun 11 '16 at 11:42
• Yes I know, which is why I would think even higher temperatures are needed for the CNO cycle... – honeste_vivere Jun 11 '16 at 17:04

The mass of a star $M$ is given by the integral over its density distribution: $$M=\int_0^R 4 \pi r^2 \rho (r) \, dr$$ So only because the star is big (large radius $R$) does not necessarily mean that it is heavy. It depends on its density profile. This profile depends on central pressure, equation of state, temperature-/luminosity-profile and more. The mass/radius relation of a star is a non-trivial result of many parameters. So the reason for the different mass/radius relations is in general the different internal composition.

• This does not address my questions. – honeste_vivere Jun 10 '16 at 15:11
• The density profile $\rho(r)$ depends on the internal composition. So the mass depends on the internal composition. This was one of your questions. The internal composition of a star and how pressure, density, temperature, luminosity and evole is a different question. @SimonS gave a short review on that. I gave a very general answer on the mass of a sphere with variable density. – N0va Jun 10 '16 at 15:44
• Okay, I agree with the first part of your comment but it does not answer my question because it is a vague generality that is true across physics. Perhaps there is not currently a good answer and that is the problem? – honeste_vivere Jun 10 '16 at 16:07

The simplest reason why shell-burning stars have huge radii is that the self-regulation of the temperature of core-burning does not work for shell burning, so the shell has to regulate something else to achieve equilibrium. That "something else" is the pressure and amount of fusing matter in the shell.

The reason the temperature self-regulation works in the core but not the shell is that the core can adjust its temperature by adjusting its radius (via the virial theorem, described above-- but really only applies to the core because then the mass is self-gravitating). But the shell cannot adjust its radius, it is stuck with the radius of the inert core. Since the shell cannot self-regulate its temperature, its temperature tends to get very high, and fusion is steeply temperature sensitive, so the fusion rate in the shell goes through the roof.

Now, of course that situation cannot be sustained, something has to give. What gives is that the excess heat goes into lifting the envelope, which reduces its weight, which reduces the pressure and the amount of material in the shell, which turns back down the fusion rate. So that's the self-regulation in a giant and supergiant, and since it requires lifting a huge amount of weight off the shell, it requires expanding the star considerably.

Many places will claim that there is no simple explanation for the giant phenomenon. I hope I have convinced you that simply isn't true.

A main sequence star burns hydrogen in a shell around the core, and the shell migrates out as it burns hydrogen into helium. Eventually there is not enough material in the outer regions to maintain the pressure necessary to maintain fusion. The core then begins to collapse and helium burning starts, and it begins with a rapid "helium flash." This causes outer layers to expand outwards and a star that is only about 10 solar masses can balloon outwards to two orders of magnitude its former radius. This is the red giant phase. In the case of larger stars they form red supergiants.

These red giants have very low densities. This is in contrast to blue giants like the pistol star. This star is 100 times the mass of the star, though it is still not very dense. It is $10^{-5}$ times the density of the sun, but far more dense than a red giant that is $10^{-8}$ times the density of the sun.

• A main sequence star does not burn H in a shell. The expansion of a star to a red giant takes place before He ignition. He core burning stars are only a little bigger than main sequence stars. Asymptotic giant branch stars are burning H and He in shells. – Rob Jeffries Jun 10 '16 at 18:48
• @Rob can we temp you into (officially) posting am answer? The question sounds simple but it's of strong layman interest. – Emilio Pisanty Jul 14 '16 at 21:29
• @Rob fair enough. I'd be interested in your take at least in terms of the rough physics of what goes on or why it's difficult to know (but I must confess I'm pretty confused by your previous answer). – Emilio Pisanty Jul 14 '16 at 22:35