<blockquote>
  <p>Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less? </p>
</blockquote>

<p>Yes, at least over human timescales. You could reasonably expect the fusion rate within the sun to be the same today as a few thousand years ago or into the future, give or take some small fraction.</p>

<blockquote>
  <p>Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.? </p>
</blockquote>

<p>The energy released by fusion is quickly distributed as thermal energy in the centre of the sun, and the temperature difference between surface (around 6000K) and centre (estimated 15 million K) drives an energy flow from hot to cold.</p>

<blockquote>
  <p>Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction? </p>
</blockquote>

<p>Fusion in the sun is not a runaway nuclear reaction (like a critical mass of uranium in a fission reaction).</p>

<p>It is possible in theory to have runaway fusion events, but the pressure and temperature for these to happen are not approached in the core of the sun. For stable stars like the sun, the forces and energy flows are in equilibrium - if the core grew slightly hotter, then the pressure would increase and the star expand slightly against the force of gravity to compensate. <a href="https://en.wikipedia.org/wiki/Thermal_runaway#Astrophysics" rel="noreferrer">Interesting things happen when stars fall out of equilibrium and runaway fusion ignition can happen in some scenarios</a>. </p>

<p>In addition, this equilibrium point moves during the lifetime of a star as its mix of elements changes due to fusion. This is predictable for many stars and forms the basis of the main sequence stars in the <a href="https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram" rel="noreferrer">Hertzsprung-Russell diagram</a></p>

<blockquote>
  <p>I have heard that the probability for fusion event to happen is only 1 in 10^12 for every collision</p>
</blockquote>

<p>I don't know the accuracy of that, but it seems reasonable. The definition of "collision" becomes somewhat arbitrary in such a hot dense environment. If you only include approaches close enough to make the strong nuclear force dominate the interaction, the ratio could be higher.</p>

<p>Another fact that I found interesting in the same area is that the power density from fusion - i.e. the Watts per cubic metre of substance - in the sun is <a href="https://en.wikipedia.org/wiki/Sun#Core" rel="noreferrer">roughly the same as that found in a typical compost heap</a>. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.</p>

CC BY-SA 4.0
2019-03-18T19:05:40.733Z
Re your last point, I find it fascinating that the sun's mighty power output tells us less about the power of fusion, and more about just how big the sun is! And it shows that the idea that a reactor is "recreating the power of the sun" is sort of unenlightening... the sun is doing it the easy way :-)
SusanW
CC BY-SA 4.0
2019-03-19T10:05:18.300Z
@SusanW You could also see the low power density as a demonstration of just how very very weak force gravity is. That small power output per volume is enough to stop the entire mass of sun from collapsing down to white dwarf matter. Stellar fusion is able to produce as much energy as is needed to stop the collapse (up to a point, ie. black hole density), demonstrated by how much faster the biggest stars can consume their hydrogen, and at the other end of the scale, how much longer the smallest can keep going, compared to our Sun.
hyde
CC BY-SA 4.0
2019-03-19T11:56:42.510Z
@SusanW The statistic is a bit disingenous, because what's producing all that power is just the core, which makes up less than 1% of the suns volume.
Cubic
CC BY-SA 4.0
2019-03-19T16:30:50.817Z
@Cubic: that is the power density *of the core* though, at least as far as I can tell. Total power ~ $4 \times 10^{26} W$, total volume of *core* ~ $2  \times 10^{25} m^3$. Power density ~ $20 Wm^{-3}$. Although I seem to of lost an order of magnitude somewhere compared to wikipedia . . . I am starting with their quoted total power of the sun though . . . :-)
Neil Slater
CC BY-SA 4.0
2019-03-19T16:33:33.450Z
@NeilSlater I might've misinterpreted the statement in that case
Cubic
CC BY-SA 4.0
2019-03-19T16:36:34.740Z
@Cubic: Possibly, although I've also just taken it on trust that Wikipedia has been fact-checked, and now scratching my head over whether there's some missing factor or a typo. Even taking conservative 20% radius as the core, I still cannot get power density over $40Wm^{-3}$ from the figures on that page
Neil Slater
CC BY-SA 4.0
2019-03-18T15:39:41.520Z
2019-03-18T22:06:18.123Z
2019-03-18T22:06:18.123Z
Neil Slater
<p>No, the fusion rate of the Sun is not absolutely constant in time. The Sun is gradually becoming more luminous and that luminosity is provided for almost exclusively by fusion in the core. However, the rate of increase is not large, of order 10% per billion years.</p>

<p>The fusion process is extremely slow (and inefficient in terms of energy release per unit volume) - the Sun releases only 250 W/m<span class="math-container">$^3$</span> in it's core. The reason for this is that fusion events are extremely unlikely, requiring two protons to overcome the Coulomb barrier between them <em>and</em> for one of the protons to inverse beta-decay into a neutron so forming a deuterium nucleus. </p>

<p>The average lifetime of a proton against this process in the core is <span class="math-container">$10^{10}$</span> years (the lifetime of the Sun), meaning the fusion rate per proton is about <span class="math-container">$3 \times 10^{-18}$</span> s<span class="math-container">$^{-1}$</span>. We can compare this to a collision rate between protons by assuming an average thermal speed of <span class="math-container">$v \simeq (3k_B T/m_p)^{1/2} = 600$</span> km/s for a core temperature of <span class="math-container">$15\times 10^{6}$</span> k, a proton number density of <span class="math-container">$n_p \sim 6 \times 10^{31}$</span> m<span class="math-container">$^{-3}$</span> in the core and a collisional cross-section of <span class="math-container">$\sigma \sim \pi (\hbar/mv)^2$</span>, where the term in brackets is the reduced de Broglie wavelength. Putting these things together, the collision rate is <span class="math-container">$n_p \sigma v \sim 10^{12}$</span> s<span class="math-container">$^{-1}$</span>.</p>

<p>Thus comparing the two rates, we can conclude that only about 1 in <span class="math-container">$3\times 10^{29}$</span> collisions ends up with fusion.</p>

<p>If the fusion rate of the Sun did increase rapidly then what would happen is that the Sun would expand, the core would become less dense and the fusion rate would fall. This basically acts as a thermostat, keeping the Sun at exactly the right temperature to support its own weight and supply the luminosity emerging from its surface.</p>

CC BY-SA 4.0
2019-03-19T10:13:49.650Z
I would replace the "fusion process is extremely inefficient" with some other way of saying it, because "inefficient" implies energy is wasted, and it's not. I mean, to me the fusion process is actually extremely *efficient*, allowing sun to stay stable for billions of years.
hyde
CC BY-SA 4.0
2019-03-18T18:32:06.420Z
2019-03-19T13:03:31.050Z
2019-03-19T13:03:31.050Z
ProfRob
<p>What's more, the usual explanation for why the fusion doesn't run away is incomplete. The simple story that can't be the full story is that if the fusion happens too fast, heat builds up and creates an overpressure.  That overpressure causes expansion, and expansion does work which lowers the temperature and dials back down the fusion until it matches the radiative escape rate.</p>

<p>The reason this is incomplete is that expansion work <em>doesn't</em> induce stability if it occurs only against a fixed external pressure, that amount of work is always insufficient to stabilize it (which leads to "shell flashes" later in the life of a star).  The only thing that is capable of stabilizing the fusion is the additional work against <em>gravity</em>, as you can easily see from how gravity gets included in any such analysis.  So it must be important that a local runaway has the net result of lifting gas away from the solar center, thereby doing gravitational work-- an important detail normally left out of the explanations.  Indeed, it would be more fair to say that solar fusion is stabilized by a <em>combination</em> of expansion work and gravitational lifting.</p>

CC BY-SA 4.0
2019-03-19T15:50:05.610Z
2019-03-19T16:31:21.803Z
2019-03-19T16:31:21.803Z
Ken G
<p>Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less?
Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.?
Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction?
I have heard that the probability for fusion event to happen is only 1 in 10<sup>12</sup> for every collision.</p>

CC BY-SA 4.0
2019-03-18T13:14:51.710Z
The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.
Martin Bonner supports Monica
CC BY-SA 4.0
2019-03-18T14:56:37.723Z
@MartinBonner I would agree with this - my question is that does every collision overcome the coulomb barrier or only a limited amount / percentage and why do you not have a run away reaction - ie more energy more diffusion
Kallie
CC BY-SA 4.0
2019-03-18T15:38:15.760Z
Fusion process is temperature sensitive. Becasue a star is typically staying in a hydrodynamic equilibrium, the temperature does not change, i.e., the fusion rate is constant.
Kornpob Bhirombhakdi
CC BY-SA 4.0
2019-03-19T17:03:45.007Z
If now were at a point in time where it were speeding up rapidly, we wouldn't be here to see it. ;-) So you can assume now is a time where the speed is fairly close to equilibrium
R.. GitHub STOP HELPING ICE
CC BY-SA 4.0
2019-03-21T07:30:31.110Z
One point is vaguely suggested in the question that is wrong, namely that there should be a direct relation between successive fusion events. A fusion event simply contributes to the general thermal energy, and it is thermal movement that cause fusion events; unlike in nuclear fission, a fusion event does not produce reaction products that _directly_ spark other fusion events. So the only question that remains is how _globally_ energy production and loss to the environment are balanced.
Marc van Leeuwen
<the-sun><stellar-astrophysics>
Why doesn't the fusion process of the sun speed up?
Kallie
# Why doesn't the fusion process of the sun speed up?

(Asked by: username_0 on 2019-03-18)

Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less?
Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.?
Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction?
I have heard that the probability for fusion event to happen is only 1 in 1012 for every collision.

**Tags:** the-sun, stellar-astrophysics

> **Comments:**
> **username_1**: The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.  
> **username_0**: @MartinBonner I would agree with this - my question is that does every collision overcome the coulomb barrier or only a limited amount / percentage and why do you not have a run away reaction - ie more energy more diffusion  
> **username_2**: Fusion process is temperature sensitive. Becasue a star is typically staying in a hydrodynamic equilibrium, the temperature does not change, i.e., the fusion rate is constant.  
> **username_3**: If now were at a point in time where it were speeding up rapidly, we wouldn't be here to see it. ;-) So you can assume now is a time where the speed is fairly close to equilibrium  
> **username_4**: One point is vaguely suggested in the question that is wrong, namely that there should be a direct relation between successive fusion events. A fusion event simply contributes to the general thermal energy, and it is thermal movement that cause fusion events; unlike in nuclear fission, a fusion event does not produce reaction products that _directly_ spark other fusion events. So the only question that remains is how _globally_ energy production and loss to the environment are balanced.  

---

## 3 answers

### Answer by username_5 on 2019-03-18. Score: 22

> Am I correct in saying that the fusion process of the sun is constant, i.e. X amount of fusion happens per day, more or less?

Yes, at least over human timescales. You could reasonably expect the fusion rate within the sun to be the same today as a few thousand years ago or into the future, give or take some small fraction.

> Why does this not speed up, i.e. one fusion event creates energy for two fusion events, etc.?

The energy released by fusion is quickly distributed as thermal energy in the centre of the sun, and the temperature difference between surface (around 6000K) and centre (estimated 15 million K) drives an energy flow from hot to cold.

> Does every collision of atom cause a fusion event, or is the probability small for a fusion event to happen thus it's not a runaway reaction?

Fusion in the sun is not a runaway nuclear reaction (like a critical mass of uranium in a fission reaction).

It is possible in theory to have runaway fusion events, but the pressure and temperature for these to happen are not approached in the core of the sun. For stable stars like the sun, the forces and energy flows are in equilibrium - if the core grew slightly hotter, then the pressure would increase and the star expand slightly against the force of gravity to compensate. [Interesting things happen when stars fall out of equilibrium and runaway fusion ignition can happen in some scenarios](https://en.wikipedia.org/wiki/Thermal_runaway#Astrophysics).

In addition, this equilibrium point moves during the lifetime of a star as its mix of elements changes due to fusion. This is predictable for many stars and forms the basis of the main sequence stars in the [Hertzsprung-Russell diagram](https://en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram)

> I have heard that the probability for fusion event to happen is only 1 in 10^12 for every collision

I don't know the accuracy of that, but it seems reasonable. The definition of "collision" becomes somewhat arbitrary in such a hot dense environment. If you only include approaches close enough to make the strong nuclear force dominate the interaction, the ratio could be higher.

Another fact that I found interesting in the same area is that the power density from fusion - i.e. the Watts per cubic metre of substance - in the sun is [roughly the same as that found in a typical compost heap](https://en.wikipedia.org/wiki/Sun#Core). It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.

> **Comments:**
> **username_6**: Re your last point, I find it fascinating that the sun's mighty power output tells us less about the power of fusion, and more about just how big the sun is! And it shows that the idea that a reactor is "recreating the power of the sun" is sort of unenlightening... the sun is doing it the easy way :-)  
> **username_7**: @username_6 You could also see the low power density as a demonstration of just how very very weak force gravity is. That small power output per volume is enough to stop the entire mass of sun from collapsing down to white dwarf matter. Stellar fusion is able to produce as much energy as is needed to stop the collapse (up to a point, ie. black hole density), demonstrated by how much faster the biggest stars can consume their hydrogen, and at the other end of the scale, how much longer the smallest can keep going, compared to our Sun.  
> **username_8**: @username_6 The statistic is a bit disingenous, because what's producing all that power is just the core, which makes up less than 1% of the suns volume.  
> **username_5**: @username_8: that is the power density *of the core* though, at least as far as I can tell. Total power ~ $4 \times 10^{26} W$, total volume of *core* ~ $2  \times 10^{25} m^3$. Power density ~ $20 Wm^{-3}$. Although I seem to of lost an order of magnitude somewhere compared to wikipedia . . . I am starting with their quoted total power of the sun though . . . :-)  
> **username_8**: @NeilSlater I might've misinterpreted the statement in that case  
> **username_5**: @username_8: Possibly, although I've also just taken it on trust that Wikipedia has been fact-checked, and now scratching my head over whether there's some missing factor or a typo. Even taking conservative 20% radius as the core, I still cannot get power density over $40Wm^{-3}$ from the figures on that page  

---

### Answer by username_9 on 2019-03-18. Score: 13

No, the fusion rate of the Sun is not absolutely constant in time. The Sun is gradually becoming more luminous and that luminosity is provided for almost exclusively by fusion in the core. However, the rate of increase is not large, of order 10% per billion years.

The fusion process is extremely slow (and inefficient in terms of energy release per unit volume) - the Sun releases only 250 W/m$^3$ in it's core. The reason for this is that fusion events are extremely unlikely, requiring two protons to overcome the Coulomb barrier between them *and* for one of the protons to inverse beta-decay into a neutron so forming a deuterium nucleus.

The average lifetime of a proton against this process in the core is $10^{10}$ years (the lifetime of the Sun), meaning the fusion rate per proton is about $3 \times 10^{-18}$ s$^{-1}$. We can compare this to a collision rate between protons by assuming an average thermal speed of $v \simeq (3k\_B T/m\_p)^{1/2} = 600$ km/s for a core temperature of $15\times 10^{6}$ k, a proton number density of $n\_p \sim 6 \times 10^{31}$ m$^{-3}$ in the core and a collisional cross-section of $\sigma \sim \pi (\hbar/mv)^2$, where the term in brackets is the reduced de Broglie wavelength. Putting these things together, the collision rate is $n\_p \sigma v \sim 10^{12}$ s$^{-1}$.

Thus comparing the two rates, we can conclude that only about 1 in $3\times 10^{29}$ collisions ends up with fusion.

If the fusion rate of the Sun did increase rapidly then what would happen is that the Sun would expand, the core would become less dense and the fusion rate would fall. This basically acts as a thermostat, keeping the Sun at exactly the right temperature to support its own weight and supply the luminosity emerging from its surface.

> **Comments:**
> **username_7**: I would replace the "fusion process is extremely inefficient" with some other way of saying it, because "inefficient" implies energy is wasted, and it's not. I mean, to me the fusion process is actually extremely *efficient*, allowing sun to stay stable for billions of years.  

---

### Answer by username_10 on 2019-03-19. Score: 1

What's more, the usual explanation for why the fusion doesn't run away is incomplete. The simple story that can't be the full story is that if the fusion happens too fast, heat builds up and creates an overpressure. That overpressure causes expansion, and expansion does work which lowers the temperature and dials back down the fusion until it matches the radiative escape rate.

The reason this is incomplete is that expansion work *doesn't* induce stability if it occurs only against a fixed external pressure, that amount of work is always insufficient to stabilize it (which leads to "shell flashes" later in the life of a star). The only thing that is capable of stabilizing the fusion is the additional work against *gravity*, as you can easily see from how gravity gets included in any such analysis. So it must be important that a local runaway has the net result of lifting gas away from the solar center, thereby doing gravitational work-- an important detail normally left out of the explanations. Indeed, it would be more fair to say that solar fusion is stabilized by a *combination* of expansion work and gravitational lifting.


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