Why doesn't the fusion process of the sun speed up? The Next CEO of Stack OverflowWhy doesn't Earth's axis change during the year?Why there is no smoke around the Sun?Why doesn't the sun pull the moon away from earth?What would the Sun be like if nuclear reactions could not proceed via quantum tunneling?Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the SunWhy are main sequence stars more massive than the Sun less dense? e.g. Vega, Spica etcWhy doesn't the Sun explode?Why do stars twinkle but the Sun doesn't (I'm asking this because the Sun is also a star)Why isn't the Sun hollow?Why we define Stellar motions with respect to sun?
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Why doesn't the fusion process of the sun speed up?
The Next CEO of Stack OverflowWhy doesn't Earth's axis change during the year?Why there is no smoke around the Sun?Why doesn't the sun pull the moon away from earth?What would the Sun be like if nuclear reactions could not proceed via quantum tunneling?Is the Sun slightly blue in the center? - Wavelength-dependent limb darkening of the SunWhy are main sequence stars more massive than the Sun less dense? e.g. Vega, Spica etcWhy doesn't the Sun explode?Why do stars twinkle but the Sun doesn't (I'm asking this because the Sun is also a star)Why isn't the Sun hollow?Why we define Stellar motions with respect to sun?
$begingroup$
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.
the-sun stellar-astrophysics
edited Mar 18 at 16:27
Glorfindel
1,9712925
asked Mar 18 at 12:31
KallieKallie
5613
$endgroup$
add a comment |
$begingroup$
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.
the-sun stellar-astrophysics
edited Mar 18 at 16:27
Glorfindel
1,9712925
asked Mar 18 at 12:31
KallieKallie
5613
$endgroup$
1
$begingroup$
The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.
$endgroup$
– Martin Bonner
Mar 18 at 13:14
$begingroup$
@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
$endgroup$
– Kallie
Mar 18 at 14:56
2
$begingroup$
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.
$endgroup$
– Kornpob Bhirombhakdi
Mar 18 at 15:38
1
$begingroup$
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
$endgroup$
– R..
Mar 19 at 17:03
$begingroup$
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.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:30
add a comment |
$begingroup$
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.
the-sun stellar-astrophysics
edited Mar 18 at 16:27
Glorfindel
1,9712925
asked Mar 18 at 12:31
KallieKallie
5613
$endgroup$
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.
the-sun stellar-astrophysics
the-sun stellar-astrophysics
edited Mar 18 at 16:27
Glorfindel
1,9712925
asked Mar 18 at 12:31
KallieKallie
5613
edited Mar 18 at 16:27
Glorfindel
1,9712925
asked Mar 18 at 12:31
KallieKallie
5613
edited Mar 18 at 16:27
Glorfindel
1,9712925
edited Mar 18 at 16:27
Glorfindel
1,9712925
edited Mar 18 at 16:27
Glorfindel
1,9712925
1,9712925
asked Mar 18 at 12:31
KallieKallie
5613
asked Mar 18 at 12:31
KallieKallie
5613
asked Mar 18 at 12:31
KallieKallie
5613
5613
1
$begingroup$
The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.
$endgroup$
– Martin Bonner
Mar 18 at 13:14
$begingroup$
@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
$endgroup$
– Kallie
Mar 18 at 14:56
2
$begingroup$
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.
$endgroup$
– Kornpob Bhirombhakdi
Mar 18 at 15:38
1
$begingroup$
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
$endgroup$
– R..
Mar 19 at 17:03
$begingroup$
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.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:30
add a comment |
1
$begingroup$
The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.
$endgroup$
– Martin Bonner
Mar 18 at 13:14
$begingroup$
@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
$endgroup$
– Kallie
Mar 18 at 14:56
2
$begingroup$
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.
$endgroup$
– Kornpob Bhirombhakdi
Mar 18 at 15:38
1
$begingroup$
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
$endgroup$
– R..
Mar 19 at 17:03
$begingroup$
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.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:30
1
1
$begingroup$
The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.
$endgroup$
– Martin Bonner
Mar 18 at 13:14
$begingroup$
The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.
$endgroup$
– Martin Bonner
Mar 18 at 13:14
$begingroup$
@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
$endgroup$
– Kallie
Mar 18 at 14:56
$begingroup$
@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
$endgroup$
– Kallie
Mar 18 at 14:56
2
2
$begingroup$
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.
$endgroup$
– Kornpob Bhirombhakdi
Mar 18 at 15:38
$begingroup$
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.
$endgroup$
– Kornpob Bhirombhakdi
Mar 18 at 15:38
1
1
$begingroup$
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
$endgroup$
– R..
Mar 19 at 17:03
$begingroup$
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
$endgroup$
– R..
Mar 19 at 17:03
$begingroup$
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.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:30
$begingroup$
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.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:30
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
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.
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
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. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
$endgroup$
15
$begingroup$
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 :-)
$endgroup$
– SusanW
Mar 18 at 19:05
1
$begingroup$
@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.
$endgroup$
– hyde
Mar 19 at 10:05
2
$begingroup$
@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.
$endgroup$
– Cubic
Mar 19 at 11:56
$begingroup$
@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 . . . :-)
$endgroup$
– Neil Slater
Mar 19 at 16:30
$begingroup$
@NeilSlater I might've misinterpreted the statement in that case
$endgroup$
– Cubic
Mar 19 at 16:33
|
show 1 more comment
$begingroup$
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 $15times 10^6$ 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 $3times 10^29$ 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.
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
$endgroup$
$begingroup$
Thank you for the explanation
$endgroup$
– Kallie
Mar 19 at 5:51
1
$begingroup$
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.
$endgroup$
– hyde
Mar 19 at 10:13
add a comment |
$begingroup$
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.
answered Mar 19 at 15:50
Ken GKen G
3,834412
$endgroup$
add a comment |
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3 Answers
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3 Answers
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oldest
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$begingroup$
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.
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
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. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
$endgroup$
15
$begingroup$
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 :-)
$endgroup$
– SusanW
Mar 18 at 19:05
1
$begingroup$
@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.
$endgroup$
– hyde
Mar 19 at 10:05
2
$begingroup$
@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.
$endgroup$
– Cubic
Mar 19 at 11:56
$begingroup$
@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 . . . :-)
$endgroup$
– Neil Slater
Mar 19 at 16:30
$begingroup$
@NeilSlater I might've misinterpreted the statement in that case
$endgroup$
– Cubic
Mar 19 at 16:33
|
show 1 more comment
$begingroup$
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.
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
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. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
$endgroup$
15
$begingroup$
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 :-)
$endgroup$
– SusanW
Mar 18 at 19:05
1
$begingroup$
@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.
$endgroup$
– hyde
Mar 19 at 10:05
2
$begingroup$
@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.
$endgroup$
– Cubic
Mar 19 at 11:56
$begingroup$
@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 . . . :-)
$endgroup$
– Neil Slater
Mar 19 at 16:30
$begingroup$
@NeilSlater I might've misinterpreted the statement in that case
$endgroup$
– Cubic
Mar 19 at 16:33
|
show 1 more comment
$begingroup$
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.
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
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. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
$endgroup$
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.
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
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. It is a very different environment to the inside of a fusion reactor experiment or a fusion bomb, which have much higher power densities.
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
edited Mar 18 at 22:06
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
answered Mar 18 at 15:39
Neil SlaterNeil Slater
32115
32115
15
$begingroup$
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 :-)
$endgroup$
– SusanW
Mar 18 at 19:05
1
$begingroup$
@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.
$endgroup$
– hyde
Mar 19 at 10:05
2
$begingroup$
@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.
$endgroup$
– Cubic
Mar 19 at 11:56
$begingroup$
@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 . . . :-)
$endgroup$
– Neil Slater
Mar 19 at 16:30
$begingroup$
@NeilSlater I might've misinterpreted the statement in that case
$endgroup$
– Cubic
Mar 19 at 16:33
|
show 1 more comment
15
$begingroup$
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 :-)
$endgroup$
– SusanW
Mar 18 at 19:05
1
$begingroup$
@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.
$endgroup$
– hyde
Mar 19 at 10:05
2
$begingroup$
@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.
$endgroup$
– Cubic
Mar 19 at 11:56
$begingroup$
@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 . . . :-)
$endgroup$
– Neil Slater
Mar 19 at 16:30
$begingroup$
@NeilSlater I might've misinterpreted the statement in that case
$endgroup$
– Cubic
Mar 19 at 16:33
15
15
$begingroup$
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 :-)
$endgroup$
– SusanW
Mar 18 at 19:05
$begingroup$
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 :-)
$endgroup$
– SusanW
Mar 18 at 19:05
1
1
$begingroup$
@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.
$endgroup$
– hyde
Mar 19 at 10:05
$begingroup$
@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.
$endgroup$
– hyde
Mar 19 at 10:05
2
2
$begingroup$
@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.
$endgroup$
– Cubic
Mar 19 at 11:56
$begingroup$
@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.
$endgroup$
– Cubic
Mar 19 at 11:56
$begingroup$
@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 . . . :-)
$endgroup$
– Neil Slater
Mar 19 at 16:30
$begingroup$
@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 . . . :-)
$endgroup$
– Neil Slater
Mar 19 at 16:30
$begingroup$
@NeilSlater I might've misinterpreted the statement in that case
$endgroup$
– Cubic
Mar 19 at 16:33
$begingroup$
@NeilSlater I might've misinterpreted the statement in that case
$endgroup$
– Cubic
Mar 19 at 16:33
|
show 1 more comment
$begingroup$
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 $15times 10^6$ 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 $3times 10^29$ 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.
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
$endgroup$
$begingroup$
Thank you for the explanation
$endgroup$
– Kallie
Mar 19 at 5:51
1
$begingroup$
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.
$endgroup$
– hyde
Mar 19 at 10:13
add a comment |
$begingroup$
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 $15times 10^6$ 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 $3times 10^29$ 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.
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
$endgroup$
$begingroup$
Thank you for the explanation
$endgroup$
– Kallie
Mar 19 at 5:51
1
$begingroup$
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.
$endgroup$
– hyde
Mar 19 at 10:13
add a comment |
$begingroup$
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 $15times 10^6$ 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 $3times 10^29$ 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.
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
$endgroup$
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 $15times 10^6$ 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 $3times 10^29$ 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.
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
edited Mar 19 at 13:03
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
answered Mar 18 at 18:32
Rob JeffriesRob Jeffries
53.7k4110170
53.7k4110170
$begingroup$
Thank you for the explanation
$endgroup$
– Kallie
Mar 19 at 5:51
1
$begingroup$
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.
$endgroup$
– hyde
Mar 19 at 10:13
add a comment |
$begingroup$
Thank you for the explanation
$endgroup$
– Kallie
Mar 19 at 5:51
1
$begingroup$
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.
$endgroup$
– hyde
Mar 19 at 10:13
$begingroup$
Thank you for the explanation
$endgroup$
– Kallie
Mar 19 at 5:51
$begingroup$
Thank you for the explanation
$endgroup$
– Kallie
Mar 19 at 5:51
1
1
$begingroup$
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.
$endgroup$
– hyde
Mar 19 at 10:13
$begingroup$
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.
$endgroup$
– hyde
Mar 19 at 10:13
add a comment |
$begingroup$
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.
answered Mar 19 at 15:50
Ken GKen G
3,834412
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 19 at 15:50
Ken GKen G
3,834412
$endgroup$
add a comment |
$begingroup$
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.
answered Mar 19 at 15:50
Ken GKen G
3,834412
$endgroup$
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.
answered Mar 19 at 15:50
Ken GKen G
3,834412
edited Mar 19 at 16:31
answered Mar 19 at 15:50
Ken GKen G
3,834412
answered Mar 19 at 15:50
Ken GKen G
3,834412
answered Mar 19 at 15:50
Ken GKen G
3,834412
3,834412
add a comment |
add a comment |
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$begingroup$
The energy produces by fusion in the core is almost exactly balanced by energy lost by diffusion of radiation from the core.
$endgroup$
– Martin Bonner
Mar 18 at 13:14
$begingroup$
@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
$endgroup$
– Kallie
Mar 18 at 14:56
2
$begingroup$
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.
$endgroup$
– Kornpob Bhirombhakdi
Mar 18 at 15:38
1
$begingroup$
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
$endgroup$
– R..
Mar 19 at 17:03
$begingroup$
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.
$endgroup$
– Marc van Leeuwen
Mar 21 at 7:30