<p>The destiny of a star basically depends upon its mass.
All its activities variety depends upon its mass.
If a star's core has a mass that is below the Chandraseckhar limit ($M\sim1.4M_{sun}$), then is destined to die as a white dwarf (or, actually, as a black dwarf in the end).
The <em>composition</em> of the white dwarf, also depends upon the original mass of the star. Different masses will lead to different compositions.
More precisely, the more massive is the star, the heavier are the elements composing the final object.
This is because more mass means more <a href="http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/gravity.html" rel="nofollow">gravitational potential energy</a></p>

<p>$dU = - \frac{GM(r)dm}{r}$</p>

<p>that in turns can be converted into heat.</p>

<p>The hydrogen nuclear fusion starts, for the <a href="https://en.wikipedia.org/wiki/Proton-proton_chain" rel="nofollow">proton-proton reaction</a>(that is the dominant process for Sun-like stars) at around $10^7 K$. This is the value that allows the particles to overcome their coulombian barrier (i.e., to <em>fuse</em>).
After the hydrogen fusion, when the most of the core is composed by helium, then of course the hydrogen fusion can't happen anymore. The core starts to collapse, and heats itself. For a Sun-like star, there is enough mass to compress up to a level that heats the core enough to start the He burning. But that is all. When also the Helium is converted into Carbon, the star has not enough mass to compress again up to a level that starts another nuclear fusion reaction. <em>This is why the core nuclear reactions stop</em>. For the shell burning question, it is important to understand two things: $(1)$ the shell structure of a star is only an approximation, and $(2)$ there is a gradient of temperature within Sun-like stars, that means that (besides the corona) the temperature increases when you go from the outside to the core. Now, if the core is compressed and became so hot to burn helium, the shell "outside" the core (that in a onion-like schema was within the radius of the previous hydrogen burning core), is still hot enough to burn hydrogen. The <em>size</em> of the helium-burning core is smaller than the hydrogen-burning core (this is compression <em>by definition</em>). The shell has still enough hydrogen, and contemporary is deep enough inside the star (that means high temperature), to allow nuclear fusion of hydrogen.
If the star was more massive, more things could happen, like heavier elements core fusion, and more and more burning shells.</p>

<p>Take a look at these:
<a href="http://abyss.uoregon.edu/~js/ast122/lectures/lec16.html" rel="nofollow">Ref 1</a>, <a href="http://www.astro.virginia.edu/class/whittle/astr1220/17_Star_Evolution/star_evolution.pdf" rel="nofollow">Ref 2</a>.</p>

<p><a href="http://www.astro.princeton.edu/~burrows/classes/403/shell.burning.pdf" rel="nofollow">Ref 3</a> for some numbers too.</p>

CC BY-SA 3.0
2014-03-14T07:32:13.250Z
not brown dwarf in the end, after white dwarf it becomes black dwarf (but the universe is too young to actually have those). Brown dwarf is an object that is too low mass to fuse hydrogen. White dwarfs are carbon/oxygen cores as a remainder of a stellar life.
usethedeathstar
CC BY-SA 3.0
2014-03-14T07:58:28.170Z
Yes right. I will correct it in the answer.
Py-ser
CC BY-SA 3.0
2014-03-14T09:41:54.577Z
could you edit to avoid the word "burning"?
Jeremy
CC BY-SA 3.0
2014-03-14T11:11:21.587Z
@Jeremy, please feel free :)
Py-ser
CC BY-SA 3.0
2014-03-15T23:25:50.043Z
For Sun-like stars it's the Bethe-Weizsäcker-cycle (https://en.wikipedia.org/wiki/CNO_cycle), not proton-proton.
Gerald
CC BY-SA 3.0
2014-03-16T00:01:30.967Z
Sorry, forget my previous comment. It seems to have been reassessed since the last time I studied Sun's nucleosynthesis in detail.
Gerald
CC BY-SA 3.0
2015-09-17T16:03:21.283Z
Single white dwarfs are produced with a *maximum* mass of about $1.1M_{\odot}$. A core that is larger than this will become hot enough to fuse carbon and the star will end in some kind of supernova. Subtle point; but as written your first paragraph is incorrect. Also can you point me to a paper that discusses the composition of white dwarfs as a function of mass? Or are you just referring to the three types - He, C/O, and Ne/Mg?
ProfRob
CC BY-SA 3.0
2015-09-19T16:58:35.660Z
@RobJeffries, thank you I will edit the answer. Could you say if the $1.1M_{sun}$ limit comes from recent estimations? And yes, I was referring to those types, I will try to clarify that as well.
Py-ser
CC BY-SA 3.0
2014-03-14T07:29:38.417Z
2014-03-14T07:59:37.267Z
2014-03-14T07:59:37.267Z
Py-ser
<p>For a more fundamental understanding, it is helpful to realize the difficulties of fusing He-4 into C-12. This is called the Triple-Alpha process. </p>

<p>When two He-4 nuclei (alpha particles) have sufficient energy to overcome to the Coulomb barrier and have their cross sections align, it produces Be-8. The Be-8 nucleus is so unstable (due to it being energetically favourable for the subject nucleons to be arranged in two alpha particles) that it has a half-life of around 10^-17 seconds which is amazingly brief. Therefore, to produce C-12 three alpha particles have to come together nearly instantaneously, two produce Be-8 and in that half-life threshold, a third interacts. </p>

<p>Take a moment to think about how extreme the conditions of the core must be to allow the probability of three alpha particles to come together and successfully interact nearly instantaneously and for it to happen enough times to produce the energy needed to bring the core out of degeneracy. Helium fusion takes about 100 million K to start as opposed to the 15 million K of the sun's core (undergoing proton-proton chain for about 99% of reactions) at present time. This temperature is provided both by the incredible pressure of the degenerate core and by additional energy supplied by the shell. </p>

<p>Shell fusion starts before the triple-alpha process because as the core contracts and becomes degenerate, there is so much energy being radiated from the core that it heats the immediate surrounding layers to the point where it can start to fuse H-to-He, in fact it is so hot that shell fusion is by the CNO cycle.  </p>

<p>The outer layers of the star expand rapidly as there is an enormous amount of energy being radiated away from this shell, which is fusing at a temperature much hotter than the core is today. </p>

CC BY-SA 4.0
2019-06-06T18:06:50.757Z
2019-06-06T18:06:50.757Z
ShroomZed
<p>I think you're like me and need more of a layman's answer. If you want a good, easy to understand explanation of what happens look at "Formation and Evolution of the Solar System" in Wikipedia then click on 5.3 (The Sun and planetary environments). The sun will actually expand twice: Once when the core gets so hot from accelerated hydrogen fusion (as the sun's core gets hotter the hydrogen burns faster) that the hydrogen in the shell around the core starts to fuse (this hydrogen fusion in the shell is what pushes the outer layers out to about 1AU). Then after as long as about 2 billion years. The core reaches a critical density/temperature (due to the increased amount of helium) that the helium starts to fuse into carbon. At this point, there is a helium "flash" and the sun shrinks back down to about 11 times its original size. The helium in the core fuses into carbon for about 100 million years until the same sort of thing happens (except this time hydrogen and helium in the shell around the core start to fuse causing outer layers to expand again. It's after the helium starts getting used up (or "polluted" with carbon enough to stop the fusion process) and there is not enough mass to start carbon fusion that a planetary nebula is ejected and the star starts to "die".</p>

CC BY-SA 3.0
2015-09-15T03:26:16.350Z
2015-09-15T03:45:15.447Z
2015-09-15T03:45:15.447Z
Jack R. Woods
<blockquote>
  <p>My question is this: Why does fusion cease in the core?! It seems to
  me that as gravity crushes the star, stellar fusion would reignite in
  the core itself—not in a sphere around the core. Why does the stellar
  core remain “dead” while its “shell” begins fusion???</p>
</blockquote>

<p>Our sun is about halfway through its "main sequence" or the hydrogen fusing stage. Fusion in a star's core is part of its <em>dynamic equilibrium</em>.</p>

<ul>
<li><p>The gravitational field of the star (produced by its mass) tends to compress its mass toward the core.  The more compressed the matter, the hotter it becomes.</p></li>
<li><p>The release of energy produced by fusion of elements at the core tends to disperse matter away from the core.  Dispersion of matter from the core tends to reduce its temperature.</p></li>
</ul>

<p>The size of a star is due then, at least in part, by the dynamic equilibrium formed at which the gravitational compressive forces are equal to the fusion-produced expansive forces.  This is called a star's <em>hydrostatic equilibrium</em>.</p>

<p>The amount of energy that is released on a per-mass basis declines as heavier elements are fused.  The most energy is released for fusing hydrogen, less is released by fusing helium, and so on.  Eventually, a point is reached (fusion of iron) at which the amount of energy needed to fuse the elements is greater than the energy released by the fusion reaction.  The iron core of such stars is thought to be "non-fusing" because if the core were heated to a temperature to enable iron fusion, insufficient energy would be released from the reaction to maintain the temperature.</p>

<p>At this point, the star becomes increasingly unable to maintain its hydrostatic equilibrium, even as its mass condenses.  What happens next depends on how massive the star is and whether its gravitational field is strong enough to exceed the electron degeneracy pressure of its mass.</p>

CC BY-SA 3.0
2015-09-20T08:09:46.603Z
Sun-like stars never reach iron. They form degenerate helium cores, which then fuse in a sudden "flash", fusing helium to carbon in a few seconds. Cores in stars of the sun's size never reach the temperatures to fuse carbon.
James K
CC BY-SA 3.0
2015-09-15T16:01:09.450Z
2015-09-15T16:07:37.847Z
2015-09-15T16:07:37.847Z
scottb
<p>I suggest you to read <a href="http://www.space.com/22471-red-giant-stars.html" rel="nofollow noreferrer">this</a> article on <a href="http://www.space.com/" rel="nofollow noreferrer">http://www.space.com/</a>.</p>
<p>Quoting from it:</p>
<blockquote>
<p>Most of the stars in the universe are main sequence stars — those converting hydrogen into helium via nuclear fusion. A main sequence star may have a mass between a third to eight times that of the sun and eventually burn through the hydrogen in its core. Over its life, the outward pressure of fusion has balanced against the inward pressure of gravity. Once the fusion stops, gravity takes the lead and compresses the star smaller and tighter.</p>
<p>Temperatures increase with the contraction, eventually reaching levels where helium is able to fuse into carbon. Depending on the mass of the star, the helium burning might be gradual or might begin with an explosive flash. The energy produced by the helium fusion causes the star to expand outward to many times its original size.</p>
</blockquote>
<p><strong>EDIT:</strong>
Wikipedia provides some more insight :</p>
<blockquote>
<p>When the star exhausts the hydrogen fuel in its core, nuclear reactions can no longer continue and so the core begins to contract due to its own gravity. This brings additional hydrogen into a zone where the temperature and pressure are adequate to cause fusion to resume in a shell around the core. The higher temperatures lead to increasing reaction rates, enough to increase the star's luminosity by a factor of 1,000–10,000. The outer layers of the star then expand greatly, thus beginning the red-giant phase of the star's life.</p>
</blockquote>

CC BY-SA 3.0
2015-09-15T16:54:49.940Z
2015-09-16T04:44:30.080Z
2020-06-17T09:47:40.653Z
Yashbhatt
<p>(This is somewhat simplified but I hope it gets the idea across.)</p>

<p>The reactions stop in the core because it runs out of fuel. During the main sequence, the star is supported by the fusion of hydrogen into helium. Eventually, the hydrogen runs out at the centre, so hydrogen fusion is no longer possible there.</p>

<p>Why doesn't it start fusing helium into carbon right away? That's because the core isn't hot or dense enough yet. Different reactions broadly rely on the presence of different resonant states in the nuclei and, in the case of helium, such a state cannot be reached often enough until the core temperature is about $10^8$ kelvin.</p>

<p>In order to get that hot, the core has to contract and heat up. It eventually does (if the star is massive enough) but it doesn't happen instantaneously. Remember that the gas is still hot and at high pressure, which it exerts on itself and its surroundings.</p>

<p>Meanwhile, at the edge of the core, the star (partly as a result of said contraction) <em>is</em> hot enough to turn hydrogen into helium, so it does so. This is exactly the nuclear-burning shell that distinguishes the internal structure of a red giant.</p>

<p>So maybe think of it this way. Imagine a star at the end of the main sequence. Where is it hot enough to fuse hydrogen into helium? Everywhere up to the edge of the core! Does it fuse in the core? No, because it's out of fuel. So where does it fuse? At the edge of the core, which we recognize as the shell.</p>

CC BY-SA 3.0
2015-09-15T19:46:07.313Z
2015-09-15T19:46:07.313Z
Warrick
<p><strong>Disclaimer:</strong> I’m not a career astronomer.  I don’t own a telescope.  I have no professional credentials.  But I do find this stuff fascinating, and I consume all the astronomy documentaries I can.  </p>

<hr>

<p>So, I’ve watched lots of documentaries describing stellar evolution.  I understand that below a certain threshold, stellar death does not involve supernovae.  I understand that above that threshold, supernovae may create neutron stars, magnetars, or (if the supernova qualifies as a hypernova) black holes.</p>

<p>However, for a long time, I was curious about <strong>why</strong> stars below the supernova threshold—like our own Sun—become Red Giants.</p>

<hr>

<p>From documentaries, I have been instructed that (for stars below the supernova threshold), when the star’s core’s fusion cannot continue…fusion ceases, and the star begins to collapse under gravity.  </p>

<p>As gravity crushes the star, I understand that the star heats up as gravity crushes it.  As a result, although the stellar core remains “dead” (no fusion occurs), a “shell” of gas around the stellar core becomes hot enough to begin fusing helium.  Since the fusion occurs as a “shell” around the stellar core, the outward-push from the fusion is what pushes the star’s outer layers further.  The result is that the star grows into a Red Giant.</p>

<hr>

<p><strong>My question is this:</strong>  Why does fusion cease in the core?!  It seems to me that as gravity crushes the star, stellar fusion would reignite in the core itself—not in a sphere around the core.  Why does the stellar core remain “dead” while its “shell” begins fusion???</p>

<the-sun><red-giant><stellar-evolution>
Why do stars become red giants?
Zearin
# Why do stars become red giants?

(Asked by: username_0 on 2014-03-14)

**Disclaimer:** I’m not a career astronomer. I don’t own a telescope. I have no professional credentials. But I do find this stuff fascinating, and I consume all the astronomy documentaries I can.

---

So, I’ve watched lots of documentaries describing stellar evolution. I understand that below a certain threshold, stellar death does not involve supernovae. I understand that above that threshold, supernovae may create neutron stars, magnetars, or (if the supernova qualifies as a hypernova) black holes.

However, for a long time, I was curious about **why** stars below the supernova threshold—like our own Sun—become Red Giants.

---

From documentaries, I have been instructed that (for stars below the supernova threshold), when the star’s core’s fusion cannot continue…fusion ceases, and the star begins to collapse under gravity.

As gravity crushes the star, I understand that the star heats up as gravity crushes it. As a result, although the stellar core remains “dead” (no fusion occurs), a “shell” of gas around the stellar core becomes hot enough to begin fusing helium. Since the fusion occurs as a “shell” around the stellar core, the outward-push from the fusion is what pushes the star’s outer layers further. The result is that the star grows into a Red Giant.

---

**My question is this:** Why does fusion cease in the core?! It seems to me that as gravity crushes the star, stellar fusion would reignite in the core itself—not in a sphere around the core. Why does the stellar core remain “dead” while its “shell” begins fusion???

**Tags:** the-sun, red-giant, stellar-evolution

---

## 6 answers

### Answer by username_11 on 2015-09-15. Score: 15 | Accepted answer

(This is somewhat simplified but I hope it gets the idea across.)

The reactions stop in the core because it runs out of fuel. During the main sequence, the star is supported by the fusion of hydrogen into helium. Eventually, the hydrogen runs out at the centre, so hydrogen fusion is no longer possible there.

Why doesn't it start fusing helium into carbon right away? That's because the core isn't hot or dense enough yet. Different reactions broadly rely on the presence of different resonant states in the nuclei and, in the case of helium, such a state cannot be reached often enough until the core temperature is about $10^8$ kelvin.

In order to get that hot, the core has to contract and heat up. It eventually does (if the star is massive enough) but it doesn't happen instantaneously. Remember that the gas is still hot and at high pressure, which it exerts on itself and its surroundings.

Meanwhile, at the edge of the core, the star (partly as a result of said contraction) *is* hot enough to turn hydrogen into helium, so it does so. This is exactly the nuclear-burning shell that distinguishes the internal structure of a red giant.

So maybe think of it this way. Imagine a star at the end of the main sequence. Where is it hot enough to fuse hydrogen into helium? Everywhere up to the edge of the core! Does it fuse in the core? No, because it's out of fuel. So where does it fuse? At the edge of the core, which we recognize as the shell.


---

### Answer by username_1 on 2014-03-14. Score: 3

The destiny of a star basically depends upon its mass.
All its activities variety depends upon its mass.
If a star's core has a mass that is below the Chandraseckhar limit ($M\sim1.4M\_{sun}$), then is destined to die as a white dwarf (or, actually, as a black dwarf in the end).
The *composition* of the white dwarf, also depends upon the original mass of the star. Different masses will lead to different compositions.
More precisely, the more massive is the star, the heavier are the elements composing the final object.
This is because more mass means more [gravitational potential energy](http://burro.astr.cwru.edu/Academics/Astr221/StarPhys/gravity.html)

$dU = - \frac{GM(r)dm}{r}$

that in turns can be converted into heat.

The hydrogen nuclear fusion starts, for the [proton-proton reaction](https://en.wikipedia.org/wiki/Proton-proton_chain)(that is the dominant process for Sun-like stars) at around $10^7 K$. This is the value that allows the particles to overcome their coulombian barrier (i.e., to *fuse*).
After the hydrogen fusion, when the most of the core is composed by helium, then of course the hydrogen fusion can't happen anymore. The core starts to collapse, and heats itself. For a Sun-like star, there is enough mass to compress up to a level that heats the core enough to start the He burning. But that is all. When also the Helium is converted into Carbon, the star has not enough mass to compress again up to a level that starts another nuclear fusion reaction. *This is why the core nuclear reactions stop*. For the shell burning question, it is important to understand two things: $(1)$ the shell structure of a star is only an approximation, and $(2)$ there is a gradient of temperature within Sun-like stars, that means that (besides the corona) the temperature increases when you go from the outside to the core. Now, if the core is compressed and became so hot to burn helium, the shell "outside" the core (that in a onion-like schema was within the radius of the previous hydrogen burning core), is still hot enough to burn hydrogen. The *size* of the helium-burning core is smaller than the hydrogen-burning core (this is compression *by definition*). The shell has still enough hydrogen, and contemporary is deep enough inside the star (that means high temperature), to allow nuclear fusion of hydrogen.
If the star was more massive, more things could happen, like heavier elements core fusion, and more and more burning shells.

Take a look at these:
[Ref 1](http://abyss.uoregon.edu/~js/ast122/lectures/lec16.html), [Ref 2](http://www.astro.virginia.edu/class/whittle/astr1220/17_Star_Evolution/star_evolution.pdf).

[Ref 3](http://www.astro.princeton.edu/~burrows/classes/403/shell.burning.pdf) for some numbers too.

> **Comments:**
> **username_2**: not brown dwarf in the end, after white dwarf it becomes black dwarf (but the universe is too young to actually have those). Brown dwarf is an object that is too low mass to fuse hydrogen. White dwarfs are carbon/oxygen cores as a remainder of a stellar life.  
> **username_1**: Yes right. I will correct it in the answer.  
> **username_3**: could you edit to avoid the word "burning"?  
> **username_1**: @Jeremy, please feel free :)  
> **username_4**: For Sun-like stars it's the Bethe-Weizsäcker-cycle (https://en.wikipedia.org/wiki/CNO_cycle), not proton-proton.  
> **username_4**: Sorry, forget my previous comment. It seems to have been reassessed since the last time I studied Sun's nucleosynthesis in detail.  
> **username_5**: Single white dwarfs are produced with a *maximum* mass of about $1.1M_{\odot}$. A core that is larger than this will become hot enough to fuse carbon and the star will end in some kind of supernova. Subtle point; but as written your first paragraph is incorrect. Also can you point me to a paper that discusses the composition of white dwarfs as a function of mass? Or are you just referring to the three types - He, C/O, and Ne/Mg?  
> **username_1**: @RobJeffries, thank you I will edit the answer. Could you say if the $1.1M_{sun}$ limit comes from recent estimations? And yes, I was referring to those types, I will try to clarify that as well.  

---

### Answer by username_6 on 2019-06-06. Score: 2

For a more fundamental understanding, it is helpful to realize the difficulties of fusing He-4 into C-12. This is called the Triple-Alpha process.

When two He-4 nuclei (alpha particles) have sufficient energy to overcome to the Coulomb barrier and have their cross sections align, it produces Be-8. The Be-8 nucleus is so unstable (due to it being energetically favourable for the subject nucleons to be arranged in two alpha particles) that it has a half-life of around 10^-17 seconds which is amazingly brief. Therefore, to produce C-12 three alpha particles have to come together nearly instantaneously, two produce Be-8 and in that half-life threshold, a third interacts.

Take a moment to think about how extreme the conditions of the core must be to allow the probability of three alpha particles to come together and successfully interact nearly instantaneously and for it to happen enough times to produce the energy needed to bring the core out of degeneracy. Helium fusion takes about 100 million K to start as opposed to the 15 million K of the sun's core (undergoing proton-proton chain for about 99% of reactions) at present time. This temperature is provided both by the incredible pressure of the degenerate core and by additional energy supplied by the shell.

Shell fusion starts before the triple-alpha process because as the core contracts and becomes degenerate, there is so much energy being radiated from the core that it heats the immediate surrounding layers to the point where it can start to fuse H-to-He, in fact it is so hot that shell fusion is by the CNO cycle.

The outer layers of the star expand rapidly as there is an enormous amount of energy being radiated away from this shell, which is fusing at a temperature much hotter than the core is today.


---

### Answer by username_7 on 2015-09-15. Score: 1

I think you're like me and need more of a layman's answer. If you want a good, easy to understand explanation of what happens look at "Formation and Evolution of the Solar System" in Wikipedia then click on 5.3 (The Sun and planetary environments). The sun will actually expand twice: Once when the core gets so hot from accelerated hydrogen fusion (as the sun's core gets hotter the hydrogen burns faster) that the hydrogen in the shell around the core starts to fuse (this hydrogen fusion in the shell is what pushes the outer layers out to about 1AU). Then after as long as about 2 billion years. The core reaches a critical density/temperature (due to the increased amount of helium) that the helium starts to fuse into carbon. At this point, there is a helium "flash" and the sun shrinks back down to about 11 times its original size. The helium in the core fuses into carbon for about 100 million years until the same sort of thing happens (except this time hydrogen and helium in the shell around the core start to fuse causing outer layers to expand again. It's after the helium starts getting used up (or "polluted" with carbon enough to stop the fusion process) and there is not enough mass to start carbon fusion that a planetary nebula is ejected and the star starts to "die".


---

### Answer by username_8 on 2015-09-15. Score: 0

> My question is this: Why does fusion cease in the core?! It seems to
> me that as gravity crushes the star, stellar fusion would reignite in
> the core itself—not in a sphere around the core. Why does the stellar
> core remain “dead” while its “shell” begins fusion???

Our sun is about halfway through its "main sequence" or the hydrogen fusing stage. Fusion in a star's core is part of its *dynamic equilibrium*.

* The gravitational field of the star (produced by its mass) tends to compress its mass toward the core. The more compressed the matter, the hotter it becomes.
* The release of energy produced by fusion of elements at the core tends to disperse matter away from the core. Dispersion of matter from the core tends to reduce its temperature.

The size of a star is due then, at least in part, by the dynamic equilibrium formed at which the gravitational compressive forces are equal to the fusion-produced expansive forces. This is called a star's *hydrostatic equilibrium*.

The amount of energy that is released on a per-mass basis declines as heavier elements are fused. The most energy is released for fusing hydrogen, less is released by fusing helium, and so on. Eventually, a point is reached (fusion of iron) at which the amount of energy needed to fuse the elements is greater than the energy released by the fusion reaction. The iron core of such stars is thought to be "non-fusing" because if the core were heated to a temperature to enable iron fusion, insufficient energy would be released from the reaction to maintain the temperature.

At this point, the star becomes increasingly unable to maintain its hydrostatic equilibrium, even as its mass condenses. What happens next depends on how massive the star is and whether its gravitational field is strong enough to exceed the electron degeneracy pressure of its mass.

> **Comments:**
> **username_9**: Sun-like stars never reach iron. They form degenerate helium cores, which then fuse in a sudden "flash", fusing helium to carbon in a few seconds. Cores in stars of the sun's size never reach the temperatures to fuse carbon.  

---

### Answer by username_10 on 2015-09-15. Score: 0

I suggest you to read [this](http://www.space.com/22471-red-giant-stars.html) article on <http://www.space.com/>.

Quoting from it:

> Most of the stars in the universe are main sequence stars — those converting hydrogen into helium via nuclear fusion. A main sequence star may have a mass between a third to eight times that of the sun and eventually burn through the hydrogen in its core. Over its life, the outward pressure of fusion has balanced against the inward pressure of gravity. Once the fusion stops, gravity takes the lead and compresses the star smaller and tighter.
>
> Temperatures increase with the contraction, eventually reaching levels where helium is able to fuse into carbon. Depending on the mass of the star, the helium burning might be gradual or might begin with an explosive flash. The energy produced by the helium fusion causes the star to expand outward to many times its original size.

**EDIT:**
Wikipedia provides some more insight :

> When the star exhausts the hydrogen fuel in its core, nuclear reactions can no longer continue and so the core begins to contract due to its own gravity. This brings additional hydrogen into a zone where the temperature and pressure are adequate to cause fusion to resume in a shell around the core. The higher temperatures lead to increasing reaction rates, enough to increase the star's luminosity by a factor of 1,000–10,000. The outer layers of the star then expand greatly, thus beginning the red-giant phase of the star's life.


---

