# What causes the dimensions of a star increase when its hydrogen fuel is exhausted?

What causes the dimensions of a star increase when its hydrogen fuel is exhausted? For example, the Sun is expected to increase its radius 250 times. What causes this if its temperature is expected to fall? How gas can expand if the temperature falls?

• Note that a cooling of the surface does not generally imply a cooling of the core, and that in order to start burning a new fuel, the power generation region must get hotter (if that fuel would have burned at the previous temperature this process would already have been underway). – dmckee --- ex-moderator kitten Dec 29 '13 at 17:29
• @dmckee: I understand what you are saying, but it bugs me when I hear talk of "fuel" and "burning" related to nuclear reactions. – Olin Lathrop Apr 19 '14 at 21:36
• Another resource on stellar evolution: ads.harvard.edu/books/1989fsa..book Chapter 5 has a section which directly addresses red giant expansion and helium core temperatures. Lot's of mid-level physics and math here. – Bill N Jan 26 '15 at 22:15

I had hopes that someone who knew this subject well would answer as it's been about 20 years since I had the relevant course, but I guess I'll give it a try.

What follows may be out of date in some ways, as I am going on my memory of a course I took in 1993 and on the basis of the text we used, Schwarzchild's 1958 Structure and Evolution of the Stars. I understand that there has been much progress in the last 20 years, especially in how the magnetic field behaves and transports energy.

There is a lot of background knowledge to be reviewed.

## Stars live in equilibrium

A star is a ball of plasma held together by self-gravitation and always in local fluid equilibrium, so that it is held "up" by the kinetic pressure of the plasma. That means that the pressure at any given place is the areal weight of the plasma further out and the density and temperature are related to the pressure by the equation of state. It is hotter in the middle and heat moves outward by either radiative or convective processes depending on the local density and temperature gradients.

In a small, low mass star like our sun large portions of the star are non-convective and that means that heat takes many thousands of years to make it from the core to the surface.

The energy to maintain the high core temperature comes from two sources

1. From fusion of light elements
2. From the conversion of potential energy to thermal kinetic energy as the mass contracts (Virial heating)

## Composition of stars

At least in the early days stars were classified in terms of the fractions of hydrogen, helium and everything else (called "metals") they contained. Due to nucleosynthesis in the big bang most stars are roughly 75% hydrogen, 25 percent helium and a pinch of metals. In population I stars like our sun this skews a little bit toward higher metalicity, but that goes beyond my discussion.

## The structure of a young, low-mass star

I'm simply going to take our Sun as the exemplar. The core and the outer layers are non-convective and there is a convection region in the middle. The convective layer is efficiently mixed, but the core is nearly static. Mixing occurs only by diffusion and the mean free paths are very short so the core is taken to be almost unmixed.

## The fusion cycles

• proton--proton chain (hydrogen burning) $$\mathrm{H} + \mathrm{H} \to \mathrm{D} + e^+ + \nu_e + 1.44\,\mathrm{MeV}$$ $$\mathrm{D} + \mathrm{H} \to {}^3\mathrm{He} + \gamma + 5.46\,\mathrm{MeV}$$ $${}^3\mathrm{He} + {}^3\mathrm{He} \to {}^4\mathrm{He} + \mathrm{H} + \mathrm{H} + 12.85\,\mathrm{MeV}$$ The time scale of this reaction in the sun is controlled by the first step which has a halflife for any particular proton of about 14 Billion years. (Energy gains exclude that carried away by the neutrinos.)

• carbon cycle (a carbon-catalyzed hydrogen burning process) $${}^{12}\mathrm{C} + \mathrm{H} \to {}^{13}\mathrm{N} + \gamma + 1.95\,\mathrm{MeV}$$ $${}^{13}\mathrm{N} \to {}^{13}\mathrm{C} + e^+ + \nu_e + 2.22\,\mathrm{MeV}$$ $${}^{13}\mathrm{C} + \mathrm{H} \to {}^{14}\mathrm{N} + \gamma + 7.54\,\mathrm{MeV}$$ $${}^{14}\mathrm{N} + \mathrm{H} \to {}^{15}\mathrm{O} + \gamma + 7.35\,\mathrm{MeV}$$ $${}^{15}\mathrm{O} \to {}^{15}\mathrm{N} + e^+ + \nu_e + 2.71\,\mathrm{MeV}$$ $${}^{15}\mathrm{N} + \mathrm{H} \to {}^{12}\mathrm{C} + ^4\mathrm{H}e + 4.96\,\mathrm{MeV}$$ When carbon is available this proceeds with a halflife of only 30 million years, so it is very fast. However, at the start of it's life even a population I star has very little carbon to work with, so energy production in the sun is still dominated by the proton-proton reaction.

• triple alpha process (helium burning) At much higher temperature it becomes possible to see $$^4\mathrm{He} + ^4\mathrm{He} \to ^8\mathrm{Be} - 95\,\mathrm{keV}$$ and $$^8\mathrm{Be} + ^4\mathrm{He} \to ^{12}\mathrm{C} + 7.4\,\mathrm{MeV}$$ to occur. However, the Beryllium-8 nucleus is unstable with a very short half-life, so this requires not only high temperature but also high Helium concentration to proceed at any speed.

## The evolution of a small, cool star

A star like the sun starts it's life slowly burning hydrogen by the proton-proton cycle at the very center of a non-convective core. Over substantial periods of time the core becomes increasingly enriched in helium and depleted of hydrogen.

If the star was a static object this would cause the fusion rate to drop and the star to cool, but instead the inner-layers of the star contract slightly increasing core densities and temperatures resulting in a slow increase in the net power output. As a side effect the outer layers expand slightly, so that the whole star grows fractionally.

This represents the behavior of the sun up until now and for another few billion years.

As the core warms, the proton cycle can proceed at slightly higher radii than before, so this process builds an increasingly-large, hydrogen-depleted, hot dense core. Because more fusion is occurring on the boundary of that core than inside the core is nearly isothermal. This process proceeds until the core is nearly pure helium and all hydrogen burning occurs in a shell around the fully-isothermal, helium core.

Two things are happening at this point, first hydrogen burning is moving outward to regions that have not previously seen any burning to speak of and are therefore still at the initial hydrogen fraction, and secondly the volume over which this burning is occurring is steadily increasing. The result is, over a fairly short period of time a vast increase in the total power of the star.

As that heat migrates outward all overlying plasma expands and much of it can be treated roughly with the ideal gas law, so the expansion is considerable. Of course the temperature of the surface is a function of both total power and the radius of the star like $$P \propto R^2 T^4 \,.$$

The results is a vastly increased radius but a cooler temperature, i.e. a red giant.

• +1. Schwarzschild's book is still considered by many to be the best summary of basic stellar physics out there, even if its opacities and nuclear reaction rates are out of date. – user10851 Jan 4 '14 at 4:32
• All the sources say the temperature of the Sun is expected to decline over its lifetime to about 60% of today's. So your answer does not tell how the gas can expand if its temperature is declining. – Anixx Jan 4 '14 at 17:01
• Also you postulate that the core will compress because the temperature will drop, and then you assume the temperature will rise. If it will rise, why the core not to return to its previous state then? – Anixx Jan 4 '14 at 17:03
• (1) The temperature of the core and that of the surface are not tightly coupled, but I don't know what source you are reading that say that. Schwarzchild has the surface temperature remaining near steady as the star grows slightly while it remains on the main sequence. (2) The core compression happens to prevent a temperature drop and that brings new fuel into a state suitable for hydrogen burning, and new equilibrium is reached. The equilibrium business is important because it is responsible for almost all the structure of the star. – dmckee --- ex-moderator kitten Jan 4 '14 at 17:34
• The thing here is that this was a 10 week, mixed graduate/upper-division course at a pretty good physics department, and it only made me a reasonably well informed beginner, ready to start learning. It is a big subject. That is why I was hoping for a real expert. – dmckee --- ex-moderator kitten Jan 4 '14 at 17:36

There are two questions here. The second,

What causes this if its temperature is expected to fall? How gas can expand if the temperature falls?

is, IMO, answered by dmckee's answer. (You should probably also read it just for some background on the next bit.) The first question,

What causes the dimensions of a star increase when its hydrogen fuel is exhausted?

is different and, actually, the answer is not clearly known. We have some good ideas but there is no consensus. We don't really have predictive power to look at a star and know if it will evolve into a giant or not, except that we've now run so many models that we've seen what happens and can say with hindsight. What's a bit more useful is that if a star is a giant, we can usually identify the region that causes the expansion. (This is thanks to a result by Eggleton & Cannon (1991).) But sometimes you can find a region where similar things are true, but the star isn't a giant. Given that the basic, modern picture of stellar structure and evolution has pretty much been known since the mid-20th century, this may come as a surprise, so I'll try to explain. Ask away if you aren't satisfied though: it's a tricky issue!

There are two key features that distinguish a main-sequence star (like the Sun) from a red giant, and both are related to the separation of a distinct, inactive helium core, and an extensive envelope. They're separated by a layer in which nuclear reactions (specifically, hydrogen burning) are taking place. So between the core and envelope there is (a) a small region in which nearly all the luminosity is generated (and the temperature gradient is thus nearly isothermal), and (b) a steep gradient in the mean molecular weight. i.e. the average mass per free particle*. When we calculate stellar models, we find that, if the star becomes a giant, there must be a region where at least one of these is true (or something with a similar structural effect is happening).

The temptation is to then say that these are clearly the cause. But! There are three types of star that cause problems. First, there are stars formed in mergers that have a three-layered appearance, because the core of one of the stars is smeared around the core of the other. Sometimes these stars have all the right features of a giant (mean molecular weight gradient, strong shell burning) but don't become giants. Second, we have stars that lose their envelopes, either because they have strong winds that drive them off, or they interact with a binary companion. Sometimes, these stars have nearly no envelope left, but they still manage to become giants, at least in the models. Third, we can just construct exotic, unrealistic models (like pure helium stars) that also show the "causes" of gianthood, but don't actually expand. For a question like this, these are still useful, because the cause is somewhere in the equations, so as long as you're using the same equations, then the output is telling you about the answer.

Thus, we want to say shell-burning + mean molecular weight jump = giant, but that's just not true. It must be the right track, but the counterexamples muddy the water. And finally, just to really mess things up, there's another issue. What if a star starts to become a giant but the process is interrupted? We then have to disentangle two effects: why the star started become a giant, and why did the process stop. (And how do we even decide if it "started"?)

For some more details, I'll shamelessly plug Sections 6.1 and 7.2 of my PhD thesis and the references therein. I did some research into a very theoretical result that appears to be related and thus had to summarize previous work on the problem.

*To understand the mean molecular weight, consider first ionized hydrogen. There are two free particles: the proton and the electron. Their total mass is $m_p+m_e$, so the mean molecular weight is $(m_p+m_e)/2$. Now, consider ionized helium. There are three free particles: the helium nucleus and two free electrons. Their total mass is roughly $4m_p+2m_e$ so the mean molecular weight is $(4m_p+2m_e)/3$. So going from pure ionized hydrogen to pure ionized helium, there is a factor of about $8/3$ change.