<p>Suppose you collect 11.0114 grams of carbon-11, come back 27.11 hours (80 half lives) later, and see what your sample has become. The most likely outcome: You'll find that you have 11.0093 grams of boron-11 and absolutely no carbon 11. You might find an atom or two of carbon-11 amongst those 11.0093 grams of boron-11.</p>

<p>What about other results, for example, what is the probability that not a single one of those Avogadro's number of carbon-11 atoms will decay into boron-11 over a 27.11 hour interval?  </p>

<p>It is "technically possible" that none of those Avogadro's number of carbon-11 atoms will decay in 27.11 hours. The probability that a single atom of carbon-11 won't have decayed in that time interval is $2^{-80}$. That's a tiny number. The probability that not a single one of those Avogadro's number of carbon-11 atom won't have decayed is $\left(2^{-80}\right)^{6\times 10^{23}}$, or less than one out of one in $10^{10^{25}}$. That's not just a tiny number. The difference between "technically possible" and "can't happen" is essentially zero.</p>

<p>How is this applicable? This result is, for example, applicable to a white dwarf stealing mass from a binary pair and turning into a neutron star rather than becoming a type 1A supernova as the white dwarf's mass approaches the Chandrasekhar limit. All it takes is a tiny number (maybe just one) carbon-carbon fusion reactions. The odds this won't happen as the mass gets ever closer to the Chandrasekhar limit is not just small. It is smaller than smaller than small.</p>

<p>But it is "technically possible."</p>

CC BY-SA 3.0
2015-01-06T03:33:16.177Z
2015-01-06T03:33:16.177Z
David Hammen
<p>Stellar fusion and supernovae are governed by quantum particle interactions (on massive scales).  In general there are many possible ways for particles to interact, decay, etc. at various probabilities.  To understand the physics of a system correctly you must take account of all possibilities (up to an error tolerance in actual practice).  </p>

<p>One often sees phrases of the form "situation X favors reaction Y".  This means the conditions (X) make the the probability of Y greater than any other possible reaction, but it doesn't equate to Y having probability 1.  Some may prefer to use this phrasing for when Y's probability is, in some sense, much larger than any other event.</p>

<p>The <a href="http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction" rel="noreferrer">proton-proton chain</a> that governs the main sequence stage of a star is a prime example of why you cannot discount the reactions which aren't favored.  If you did, you would conclude that our own sun is incapable of fusion.  The 'favored' reaction for two interacting protons in a star like our sun is to simply bounce off of each other thanks to <a href="http://en.wikipedia.org/wiki/Coulomb%27s_law" rel="noreferrer">Coulomb repulsion</a>.  The protons simply don't have enough energy to overcome this repulsion in order to undergo fusion.  Technically this statement is also not probability 1, but rather very close to it.  But an improbable <a href="http://en.wikipedia.org/wiki/Quantum_tunnelling" rel="noreferrer">quantum tunneling</a> event is possible to bypass this potential barrier, allowing the interacting protons to fuse when they collide just right and the dice favor them.  But a diproton is unstable.  The heavily favored reaction in this state is for the diproton to immediately <a href="http://en.wikipedia.org/wiki/Beta_decay" rel="noreferrer">decay</a> into two protons.  But there is a small chance that one of the protons will undergo a <a href="http://en.wikipedia.org/wiki/Beta_decay" rel="noreferrer">beta decay</a> into a neutron, instead, at which point you have a stable helium nucleus.  That these things are so unlikely is a major factor in why our sun needs billions of years to exhaust enough of its hydrogen (aka protons) in order to initiate helium fusion and exit the main stage.  Even very large stars still need millions of years because of this.</p>

<p>So while the proton-proton reaction is rather unlikely, stars are simply so immense that it is not only a virtual certainty that some proton-proton reaction will yield helium, but in fact that it is a virtual certainty that the reactions will occur in large numbers and continuously until the available reactants dwindle.</p>

<p>At the other extreme, we have <a href="http://en.wikipedia.org/wiki/Supernova" rel="noreferrer">supernovae</a>.  There are quite a lot of varieties of supernovae, but the commonly understood types--the outer layers collapse and then violently rebound off the core--feature a sudden loss of radiation pressure due to certain endothermic reactions becoming favorable.  Iron production from fusion is the culprit here: it takes more energy to fuse it than is released.  The final split second of this sort of star's life is when conditions favor the production of iron via fusion.  Ridiculously favors it, as I really do mean it is the final split second.  There is an inconceivably tiny chance that this iron production would fail to occur for some chosen period of time, and if it happens to stall long enough then the star would likely undergo some other event.  See also David's answer for another wildly improbable, but technically possible, example of random chance altering a star's death.</p>

CC BY-SA 3.0
2015-01-06T16:21:39.007Z
Suggested correction, re *The final split second of this sort of star's life is when conditions favor the production of iron via fusion.* This isn't what triggers a type II supernova. A star needs to have been producing iron for about five days before the supernova happens. The trigger is the iron in the core reaching the Chandrasekhar mass limit. Adding mass to a white dwarf (the iron core is essentially a white dwarf) makes the white dwarf shrink. For a non-relativistic white dwarf, radius is inversibly proportional to the cube root of mass. ...
David Hammen
CC BY-SA 3.0
2015-01-06T16:30:10.040Z
... Weird things happen when the electrons move so fast that relativity comes into play. The radius drops precipitously toward zero at the Chandrasekhar limit. This zero radius can't happen; the iron core instead starts collapsing and turning into a neutron star. This core collapse is the trigger that starts the supernova, and events proceed extremely quickly once the collapse starts.
David Hammen
CC BY-SA 3.0
2015-01-06T03:51:07.710Z
2015-01-06T03:51:07.710Z
zibadawa timmy
<p>This comes from <a href="https://astronomy.stackexchange.com/questions/8468/is-there-an-element-of-chance-chaos-in-stellar-evolution#comment10132_8468">a comment</a> on <a href="https://astronomy.stackexchange.com/q/8468/1909">this question</a>, to quote:</p>

<blockquote>
  <p>The death of (large mass) stars is also based on quantum events with probabilities technically not 1 (and is very fast), so it is technically possible for a star not to die in a way it really ought to by sheer chance.</p>
</blockquote>

<p>The question is what quantum events are involved in the death of a star (likely and unlikely)? If many of these very unlikely quantum events happened at the same time all over the star, how could the death of a star differ form the typical nova or simply cooling down?</p>

CC BY-SA 3.0
2015-01-06T00:21:51.690Z
I think it would be nice if zibadawatimmy (the original commentor) explained it. That would be an interesting answer.
HDE 226868
<star><supernova>
Quantum death of stars
CDspace
# Quantum death of stars

(Asked by: username_0 on 2015-01-06)

This comes from [a comment](https://astronomy.stackexchange.com/questions/8468/is-there-an-element-of-chance-chaos-in-stellar-evolution#comment10132_8468) on [this question](https://astronomy.stackexchange.com/q/8468/1909), to quote:

> The death of (large mass) stars is also based on quantum events with probabilities technically not 1 (and is very fast), so it is technically possible for a star not to die in a way it really ought to by sheer chance.

The question is what quantum events are involved in the death of a star (likely and unlikely)? If many of these very unlikely quantum events happened at the same time all over the star, how could the death of a star differ form the typical nova or simply cooling down?

**Tags:** star, supernova

> **Comments:**
> **username_1**: I think it would be nice if zibadawatimmy (the original commentor) explained it. That would be an interesting answer.  

---

## 2 answers

### Answer by username_3 on 2015-01-06. Score: 5 | Accepted answer

Stellar fusion and supernovae are governed by quantum particle interactions (on massive scales). In general there are many possible ways for particles to interact, decay, etc. at various probabilities. To understand the physics of a system correctly you must take account of all possibilities (up to an error tolerance in actual practice).

One often sees phrases of the form "situation X favors reaction Y". This means the conditions (X) make the the probability of Y greater than any other possible reaction, but it doesn't equate to Y having probability 1. Some may prefer to use this phrasing for when Y's probability is, in some sense, much larger than any other event.

The [proton-proton chain](http://en.wikipedia.org/wiki/Proton%E2%80%93proton_chain_reaction) that governs the main sequence stage of a star is a prime example of why you cannot discount the reactions which aren't favored. If you did, you would conclude that our own sun is incapable of fusion. The 'favored' reaction for two interacting protons in a star like our sun is to simply bounce off of each other thanks to [Coulomb repulsion](http://en.wikipedia.org/wiki/Coulomb%27s_law). The protons simply don't have enough energy to overcome this repulsion in order to undergo fusion. Technically this statement is also not probability 1, but rather very close to it. But an improbable [quantum tunneling](http://en.wikipedia.org/wiki/Quantum_tunnelling) event is possible to bypass this potential barrier, allowing the interacting protons to fuse when they collide just right and the dice favor them. But a diproton is unstable. The heavily favored reaction in this state is for the diproton to immediately [decay](http://en.wikipedia.org/wiki/Beta_decay) into two protons. But there is a small chance that one of the protons will undergo a [beta decay](http://en.wikipedia.org/wiki/Beta_decay) into a neutron, instead, at which point you have a stable helium nucleus. That these things are so unlikely is a major factor in why our sun needs billions of years to exhaust enough of its hydrogen (aka protons) in order to initiate helium fusion and exit the main stage. Even very large stars still need millions of years because of this.

So while the proton-proton reaction is rather unlikely, stars are simply so immense that it is not only a virtual certainty that some proton-proton reaction will yield helium, but in fact that it is a virtual certainty that the reactions will occur in large numbers and continuously until the available reactants dwindle.

At the other extreme, we have [supernovae](http://en.wikipedia.org/wiki/Supernova). There are quite a lot of varieties of supernovae, but the commonly understood types--the outer layers collapse and then violently rebound off the core--feature a sudden loss of radiation pressure due to certain endothermic reactions becoming favorable. Iron production from fusion is the culprit here: it takes more energy to fuse it than is released. The final split second of this sort of star's life is when conditions favor the production of iron via fusion. Ridiculously favors it, as I really do mean it is the final split second. There is an inconceivably tiny chance that this iron production would fail to occur for some chosen period of time, and if it happens to stall long enough then the star would likely undergo some other event. See also David's answer for another wildly improbable, but technically possible, example of random chance altering a star's death.

> **Comments:**
> **username_2**: Suggested correction, re *The final split second of this sort of star's life is when conditions favor the production of iron via fusion.* This isn't what triggers a type II supernova. A star needs to have been producing iron for about five days before the supernova happens. The trigger is the iron in the core reaching the Chandrasekhar mass limit. Adding mass to a white dwarf (the iron core is essentially a white dwarf) makes the white dwarf shrink. For a non-relativistic white dwarf, radius is inversibly proportional to the cube root of mass. ...  
> **username_2**: ... Weird things happen when the electrons move so fast that relativity comes into play. The radius drops precipitously toward zero at the Chandrasekhar limit. This zero radius can't happen; the iron core instead starts collapsing and turning into a neutron star. This core collapse is the trigger that starts the supernova, and events proceed extremely quickly once the collapse starts.  

---

### Answer by username_2 on 2015-01-06. Score: 4

Suppose you collect 11.0114 grams of carbon-11, come back 27.11 hours (80 half lives) later, and see what your sample has become. The most likely outcome: You'll find that you have 11.0093 grams of boron-11 and absolutely no carbon 11. You might find an atom or two of carbon-11 amongst those 11.0093 grams of boron-11.

What about other results, for example, what is the probability that not a single one of those Avogadro's number of carbon-11 atoms will decay into boron-11 over a 27.11 hour interval?

It is "technically possible" that none of those Avogadro's number of carbon-11 atoms will decay in 27.11 hours. The probability that a single atom of carbon-11 won't have decayed in that time interval is $2^{-80}$. That's a tiny number. The probability that not a single one of those Avogadro's number of carbon-11 atom won't have decayed is $\left(2^{-80}\right)^{6\times 10^{23}}$, or less than one out of one in $10^{10^{25}}$. That's not just a tiny number. The difference between "technically possible" and "can't happen" is essentially zero.

How is this applicable? This result is, for example, applicable to a white dwarf stealing mass from a binary pair and turning into a neutron star rather than becoming a type 1A supernova as the white dwarf's mass approaches the Chandrasekhar limit. All it takes is a tiny number (maybe just one) carbon-carbon fusion reactions. The odds this won't happen as the mass gets ever closer to the Chandrasekhar limit is not just small. It is smaller than smaller than small.

But it is "technically possible."


---

