# Power in Sun's core

1. Nov 11, 2014

### stinsonbr

I'm curious about something I read comparing the power released in the core of the Sun being comparable to a compost pile (~276 W/m^3). This seems very low (Wikipedia confirms this value).

The value I find for power in the core is on the order of 10^26 W, and using the volume of the core of the sun the value I get is around 3.42x10^7 W/m^3.

2. Nov 11, 2014

### Bandersnatch

You'd need to show us your calculations. I get about the right value to the order of magnitude.

3. Nov 11, 2014

### SteamKing

Staff Emeritus
Like Bandersnatch said, it all comes down to how big your solar core is.

The radius of the core is estimated to be 20%-25% of the solar radius of approx. 696,000 km. Fusion ceases entirely in the sun once you are beyond 30% R.

4. Nov 12, 2014

### stinsonbr

Good thing I didn't post my original calculations. I went over it again, and using 3.86x10^26 W from the core, and assuming 25% solar radius is where almost all power is made (.25*695,800,000m = 173,950,000m) I get about 17.5 W/m^3. This value just seems very low, though I do understand the size of the core leads to such an enormous power output. If correct, this is less than what human metabolism produces!

5. Nov 12, 2014

### Staff: Mentor

Yes, the numbers are surprisingly low. As the power is not constant over the whole core, the value in the center is a bit higher.
This also shows a reason why fusion power plants are hard to build - in addition to the giant pressure we cannot reproduce, the power density has to be orders of magnitude higher to be useful.

6. Nov 14, 2014

### Ken G

Another thing to bear in mind is that the physics of fusion has almost nothing to do with that 276 W/m^3 number, that is essentially entirely controlled by the time it takes the light to diffuse out from deep in the Sun. So the power is low because it takes a long time to escape, and since fusion has almost nothing to do with this, it just kind of fools us that we expect fusion to be a process that releases a lot of heat very quickly. The fusion rate is self-regulated by the rate of escape, so it is self-regulated to occur very slowly in each m^3. Bear in mind, for example, that it takes ten billion years to burn up the hydrogen in each of those cubic meters-- how long does it take your compost heap to break down?

7. Nov 16, 2014

### Staff: Mentor

Yoy are right that stellar fusion processes don't match our expectation; we hear the word "fusion" and we think mushroom clouds, not compost heaps.

However, I cannot buy you claim that the physics of fusion has "almost nothing" to do with the surprisingly low number. Describe to me a cubic meter of material under the conditions of the solar core, ask me what happens in that material, and I'll respond that fusion process will generate 276 joules every second... And if that is not an exercise in the physics of fusion, I don't know would be.

8. Nov 16, 2014

### Staff: Mentor

I don't think Ken is saying that fusion isn't generating this energy. It has to, otherwise the core would slowly collapse since its only other source of energy generation is gravitational contraction. I think he's saying that whether the source of this energy is fusion, gravitational contraction, or another source, the power output is the same under the current conditions in the Sun. Increasing or decreasing the amount of time that it takes energy to escape the Sun would directly affect the temperature that the core would "stabilize" at, which would change the rate of power generation in the core, by whatever means that power is being generated.

A good example is Jupiter. Jupiter doesn't undergo fusion in its core, yet it generates more energy than it gets from the Sun. All of this energy is generated by the gravitational contraction of the planet, which is measured as a power output. If the time it takes the energy to escape from Jupiter is very long (it's "insulated" very well) then it would be take longer to collapse under gravity, which reduces the power output. If we remove the "insulation" and allow energy to escape at a faster rate, then the power generation through gravitational contraction is increased.

Taken to the extremes, a very well insulated Jupiter would be much larger than it is today since it will have converted less of its gravitational potential energy into heat, while a very poorly insulated Jupiter would be smaller and more compact, since more of its gravitational potential energy would have been converted to heat and lost by now.

The same goes for the Sun. If it takes longer for the energy to escape, then the temperature of the core is cooler and the rate of energy generation is less since it's harder for the core to collapse and initially heat up. If it takes a shorter amount of time for the energy to escape, then the core contracts faster and gets hotter, which would result in a higher fusion rate due to this increased temperature.

9. Nov 16, 2014

### Ken G

Yes, Drakkith has it exactly right. The words are a bit unclear, when I said that fusion does not "control" the 276 joules per second, I only meant we don't need to know if fusion is occurring, or any details about fusion, to get a number quite close to that. The number comes from the structure of the Sun, so if we turn off fusion, it still requires millions of years for that number to change significantly, which is the timescale for the Sun to change structure significantly. The reason I mention it is because there is a common misconception that the Sun's luminosity "turned on" when fusion began, or that higher mass stars are more luminous because there is something different about the fusion they are doing. It's the other way around-- there is something different about their fusion because they are more luminous, the fusion is always self-regulated, as Drakkith was also saying.

10. Nov 16, 2014

### Staff: Mentor

This doesn't make sense to me; it implies two things that seem highly questionable:

(1) It implies that the power density of any process is the same at the same temperature and density. I don't think that's even true for different fusion reactions, let alone for fusion vs. gravitational contraction.

(2) It implies that supernovas can't happen, because supernovas involve a very rapid (on the order of 10 seconds) and drastic change in the structure of a star when fusion reactions stop in its core.

11. Nov 16, 2014

### Ken G

Strictly speaking, you are right, but only if you are looking for exact rather than approximate results. Don't forget that the temperature in the core is self-regulated to produce the necessary 276 joules per cubic meter. If you turn off fusion in the core, as Drakkith mentioned, the stellar luminosity will still stay pretty much the same, all that happens is the source of that 276 joules per cubic meter shifts to gravitational contraction. There would need to be some small re-adjustment of the structure to make that crossover, but the 276 joules per cubic meter would not change much, because the luminosity would still be set by the rate that light diffuses out. I can't say exactly how much it would change, I wouldn't be suprised by maybe 10 or 20 percent changes, but basically it would stay the same even though you wouldn't have fusion-- so that's why the fusion can't control it in the general sense. It would affect it in the precise sense, it's all a question of how exactly you want to get that number right.
Well, supernovas can occur in some situations, but if fusion were permanently turned off in the Sun, it would never supernova. As I said, turning fusion off permanently would have rather little effect on the Sun for about a million years or so, though we would probably notice some difference in a few hundred thousand years due to the readjustments I clarified above. So again it's a question of how much difference you are looking for.

12. Nov 16, 2014

### Staff: Mentor

I think there may be some confusion over whether or not we need to know about fusion. To fully understand what goes on inside a star we of course need to know about fusion, the rate of fusion, etc. What I think Ken's getting at is that if we just want to know how much power is generated inside the star, then we can figure this out without even mentioning fusion as long as we know how fast energy is allowed to escape and the luminosity of the star. It's like wrapping a blanket around a heat source. If we know how fast energy escapes through the blanket and how much energy is lost from the surface of the blanket, then we know how much energy is being produced by the heat source, regardless of whether our heat source is an electric heater, a chemical reaction, etc. Is that right, Ken? (I hope my analogy is accurate. :p)

Edit: The reason that knowing how fast energy escapes from the star is important is that, unlike a heat source and a blanket, a star's energy production is variable and will depend on how fast energy is allowed to escape. Since energy production pushes back against gravity, preventing the star from collapsing, a "well insulated" star will produce less power since each unit of energy stays inside the star for a long time. If we take an extreme case and look at a star that has essentially no "insulation", we would find that because the energy escapes very quickly, the star collapses very fast and generates a huge amount of energy, raising the core temperature to much higher than a normal star. In essence, there's nothing holding the star up so it simply collapses in on itself, releasing huge amounts of gravitational potential energy in the process.

Oddly enough, this is very similar to a supernova. Much of the energy generated by the supernova is lost as neutrinos, which escape out of the star without interacting much. If neutrinos interacted much more strongly with matter, they wouldn't be able to escape as easily, the energy wouldn't be lost nearly as quickly, and the core collapse would take weeks/months/years instead of seconds.

Last edited: Nov 16, 2014
13. Nov 16, 2014

### Staff: Mentor

I would really like to see an actual model that shows how this happens. I don't see how gravitational contraction will automatically supply 276 joules per cubic meter just because that's what was being produced in the previous equilibrium with fusion reactions. Is there a good reference that describes mathematically how this works?

Agreed; it's not massive enough for that. But that's not the same as saying it will just stay at the same, or nearly the same, equilibrium state.

Again, I would really like to see an actual mathematical model that shows how this happens.

14. Nov 16, 2014

### Ken G

Yes, if we further stipulate that whatever is inside that blanket is self-regulated to produce heat at the rate it is escaping. Basically, when an object emits heat, there are two very different things that could be going on. It might be forced to emit heat at whatever rate the heat is being supplied, which is how a campfire might work. Or, it might be forced to supply heat at whatever rate the heat is escaping, as happens with a star. The latter is true whether the star is pre-main-sequence or on the main sequence, and indeed this is why nothing happens to the luminosity of the Sun as it enters the main sequence. Contrast that with what happens to a campfire when you light it!
Yes, exactly.

15. Nov 16, 2014

### Ken G

Yes, look at the early work done by Eddington, where he correctly inferred the internal structure of the Sun that would be needed to produce the luminosity it emits-- all before anyone knew that nuclear fusion even existed. Another important fact is that the luminosity of the Sun does not change as it enters the main sequence, so the initiation of fusion in the core does not have any significant impact on the structure of the Sun. What it mostly does is simply stall the contraction that used to supply that luminosity, before there was fusion.

Another important fact is that if you look at the main sequence in an H-R diagram, you get no clue whatsoever as to the place where CNO cycle fusion takes over from p-p fusion. Those processes obey very different physical laws, yet there is not even a hint in the structure of a main-sequence star where one process gives over to the other. The star really just doesn't care, what matters is the mass of the object, because that's what controls the rate of diffusion of light. A star is just a big leaky bucket of light with some kind of heat source that doesn't matter much, though you are certainly right that if you want exact models (more accurate than what Eddington was able to do, for example), then you do need the detailed physics to get it all correct.
There are many flavors of equilibrium in a star, based on the three key timescales. On a sound crossing time (hours), we get force balance, so that is very quick and does not care if there is nuclear fusion, it only cares how much energy is in there and how it is spread around. On a light diffusion time (a hundred thousand years or so), the energy can reapportion itself between gas and light, but that is still largely controlled by the temperature structure. Then there is the thermal timescale, millions of years, the timescale for the star to figure out if it is suffering a net loss of heat. That's the only timescale that really cares crucially what the heat source is.
The simplest description is called a "polytrope model", which can get the structure of the Sun to a reasonable approximation (even though it neglects convection and doesn't know anything about fusion). That's what Eddington used to understand the luminosity of the Sun, where he constrained the polytrope by assuming a constant ratio between gas pressure and radiation pressure. He had a rationale for doing it that way, though of course neglecting convection is a more serious quantitative problem than neglecting fusion. Eddington's main problem with not knowing about fusion was getting the evolutionary timescale completely wrong-- he thought the Sun should evolve on million-year timescales, which gave the geologists of his day fits.

16. Nov 17, 2014

### D H

Staff Emeritus
You just sent me on a wild goose chase, Drakkith! I've seen that statistic myself, many times over. This time I decided to check the veracity of that statement. This very widely cited statistic apparently is false. The following summarizes what I found:
• Websites and scientific articles vary widely and wildly on Jupiter's Bond albedo, it's effective temperature, and it's luminosity.
• Almost all websites that say something about Jupiter's albedo, effective temperature, and luminosity are internally inconsistent. This includes the wikipedia articles on Jupiter and Jupiter's atmosphere.
• Using an overly high albedo is one of the sources of this false statement. Some websites (and some college lectures; shame on them) claim an albedo of over 0.7, which is completely wrong. Using an overly high value reduces the amount of energy received from the Sun. better value is a Bond albedo of 0.343 based, on Pioneer and Voyager measurements.
• Almost all sources that make this claim rely on data from the 1960s. Those 1960s measurements were problematic for a number of reasons.
• Later measurements based on the Pioneer, Voyager, Galileo, and Cassini probes falsify this claim. Jupiter internally generates about 2/3 as much energy as it receives from the Sun. Close, but no cigar.
Even if Jupiter did generate more energy than it receives from the Sun, this still does not make Jupiter a good example. The problem is that Jupiter's Kelvin-Helmhotz timescale is in the hundreds of billions of years. For the Sun, it's about 30 million years.

This result led to a significant controversy between physicists, geologists, and biologists in the latter part of the 19th century. Based on this thermal timescale, Kelvin deduced that the Earth was no more than 10 million years old. This was completely wrong. Fusion, which Kelvin didn't know about, was the key to solving this controversy.

Ken G, you are ignoring this history, and you are distorting fact. To say that the fusion rate has nothing to do with fusion is beyond ludicrous.

Getting back to the original post, the reason that the Sun produces energy comparable to a warm compost pile is because fusion in the Sun is rather improbable. Fusion in a main sequence star such as the Sun eventually results in four protons combining to form a helium-4 nucleus. A simultaneous collision of four protons is so highly improbable that it essentially never happens. The pathway from hydrogen to helium instead involves a series of reactions. In our Sun, the production of deuterium represents a very significant bottleneck in this chain of events.

One way our Sun produces deuterium involves a near-simultaneous collision between two protons and an electron to produce a deuterium nucleus and a high energy neutrino. While not nearly as improbable as a simultaneous collision of four particles, a near-simultaneous collision of three particles is still extremely unlikely. The other way to produce deuterium is for two protons to collide, yielding a helium-2 nucleus. This happens a lot, but in most cases the helium-2 nucleus quickly decays into two protons. The net result: Nothing happens. Thanks to the weak interaction, a helium-2 nucleus occasionally decays into a deuterium nucleus, a positron, and a neutrino. While highly unlikely, this is much more likely than a three-way collision. Either way, it is still highly unlikely that protons will combine to form a deuterium nucleus.

Larger stars get around this bottleneck via the CNO cycle. Our Sun is too small to make use of this reaction. Fusion in our Sun relies primarily on the proton-proton (p-p) chain, and that is why our Sun produces energy comparable to a warm compost pile.

17. Nov 17, 2014

### Ken G

I know that, I referred to it above, in this very thread.
I'm not clear on how I could be ignoring history, when I first described that history. What's more, I showed why it supports my point. Do you have a more specific question about that connection?
Now ask the next question-- why is that? The answer is simple: because improbable fusion is the only kind necessary to replace the heat lost by the diffusion of light out of the interior of the Sun, the latter effect being what determines the luminosity of any star where radiative diffusion is the dominant energy transport mechanism over the body of the object. Let me again summarize the arguments that make this completely clear:
1) stars on their "Henyey track" (like all main-sequence stars with solar mass or more) have luminosities that do not sensitively depend on their state of evolution, including whether or not they are undergoing any type of fusion.
2) Eddington understood the luminosity of the Sun rather well, without knowing fusion existed
3) there is no trace on the main sequence where stars go from p-p fusion to CNO fusion, even though those two flavors of hydrogen fusion involve completely different physics, and completely different degrees of "how probable" fusion is.
The reason for all three of these points is simple: fusion is self-regulated to replace the heat lost by radiative diffusion, and it does just that, quite insensitively to the details of how fusion occurs. Indeed, if fusion is not occurring at all, it just means that gravitational contraction provides the heat instead of fusion. The luminosity is rather unaffected, that's the whole point of the "Henyey track," which applies both before, during, and in some cases even after, core fusion is going on.

I realize that most people don't know this, which is why I'm helping out. "Ignoring history"? No, I'm just ignoring common misconceptions about how stars work.
And that is a classic example of the kind of misconception I'm talking about. Many people think stars that undergo CNO fusion are more luminous because they undergo CNO fusion instead of p-p fusion. But that's quite wrong, even a main-sequence star with no trace of carbon in it would have virtually the same luminosity that it does have. Again, the reason is because fusion only replaces the heat lost by radiative diffusion-- it makes little difference what the fusion mechanism is, it will self-regulate either way. Without CNO fusion, a high-mass star would need a higher core temperature to be on the main sequence, but this will influence its luminosity very little-- indeed that is the whole point of a Henyey track.

18. Nov 17, 2014

### Staff: Mentor

It will self-regulate, but at a different level. Remove the carbon, the star shrinks a bit and the core gets hotter, which gives a larger temperature gradient and more overall power - or a larger star, that will depend on the details how the outer part of the star changes.

19. Nov 17, 2014

### Ken G

Yes, but differences in the luminosity will be fairly small, and it's not even clear if the luminosity would go up or down. Details like the presence of convection are probably more important, and convection is already treated heuristically in terms of a fit parameter called "the mixing length." So it all depends on how detailed you want to go, and how far away from basic physical understanding.

20. Nov 18, 2014

### Ken G

OK, so let's summarize the answer. We have the very interesting observation that the Sun only emits the same energy per gram as a compost heap. Why is this? It is because light takes a long time to diffuse out of the Sun, so the rate that energy needs to be generated to replace that lost heat is quite low. That answer has almost nothing to do with fusion, and it is the physically insightful answer both thousands of years before fusion initiates, during fusion, and thousands of years after it ceases. I'd call that an interesting fact, and little known at that.

21. Nov 18, 2014

### Staff: Mentor

This shows that whatever mechanism is generating heat inside the Sun must be consistent with the structure Eddington inferred. But it doesn't, by itself, show that gravitational contraction is such a mechanism.

Doesn't stalling the contraction count as a significant impact on the structure? A contracting star is not stationary; it's not in equilibrium. A star with fusion generating pressure that exactly balances gravity is in equilibrium.

Also, how quickly is fusion supposed to turn on inside the core? And how does that timescale compare with how fast fusion turns off at the other end of the star's lifetime?

22. Nov 19, 2014

### D H

Staff Emeritus
That light takes a long time to diffuse from the Sun is pretty much irrelevant in a small star such as our Sun. In our Sun, almost all of the pressure is attributable to electrons and nuclei. Photon pressure is indeed a factor in larger stars, but not so much in the Sun.

To once more get back to the original post, the reason fusion at the core of the Sun produces energy comparable to that of a warm compost pile (per unit volume) is because the temperature at the center of the Sun is a mere 15 million kelvin. Production of deuterium is rather low at such low temperatures. As an aside, the fusion reactions we humans create operate at much higher temperatures and bypass the production of deuterium, both in nuclear weapons and in fusion generators such as tokamaks. A fission-fusion-fission bomb that had pure hydrogen-1 as the fusion part of the bomb would be a dud of a nuclear weapon. The fission part would work, but the fusion part would not. Nuclear weapons and fusion reactors use a mix of deuterium and tritium. Our Sun has to generate deuterium before it can take advantage of the easy fusion that we use in bombs and fusion reactors.

The fusion rate in main sequence stars is self-regulating. A locally higher than nominal fusion rate will temporarily result in a locally-increased temperature and hence a locally-increased pressure. That higher pressure makes that region of the star expand, which reduces the temperature, which brings the fusion rate right back to nominal. Stars on the main sequence have a built-in thermostat that keeps pressure, temperature, energy production from fusion, and energy loss from outgoing radiation in the core in balance.

That built-in thermostat will fail in five to seven billion years when fusion at the Sun's core has converted almost all of the hydrogen at the core into helium. That's when our Sun will turn into a red giant. The Sun will still burn hydrogen at this point, but in a shell surrounding the dead helium core. It's only later in the Sun's life where it will start burning helium.

23. Nov 19, 2014

### Ken G

It isn't the mechanism in our Sun-- the mechanism in our Sun is fusion. Before fusion initiated, it was gravitational contraction. The point I'm making is the luminosity of the Sun did not change much when that crossover occurred, which tells us that we can understand that luminosity of the Sun without knowing whether fusion is occurring or not, all we need to track is the diffusion of light. This physical insight is the basis of what is known as a "Henyey track," a rather underappreciated concept that applies to stars that satisfy two criteria: the slow step in their energy transport is radiative diffusion (i.e., not fully convective), and the temperature of the star is regulated by ideal-gas physics (i.e., no degenerate core).
Yes, but that's a different question-- that's the question, what sets the radius of the Sun. The radius of the Sun is set by fusion physics, because that tells us when the contraction will stall. But the luminosity is not set by fusion physics, unless you want a really exact result with everything included (including convection). The way you can tell this is if you insert some special catalyst that doubles the fusion rate at given temperature and density-- naively you might think the Sun's luminosity would double, but in fact it would not change much at all, because the fusion would still self-regulate to replace the heat that is radiatively diffusing out. However, if you insert some special radiation-blocker that doubles the opacity of the Sun at the same temperature and density, then you will find that the luminosity, and fusion rate, will basically drop by a factor of 2 once the new equilibrium is achieved.
There are many types of equilibrium, at many different timescales. A contracting star is generally in force balance, so we can say it is in momentum-flux equilibrium. It is also usually in radiative equilibrium. The only sense to which it is not in equilibrium is on the Kelvin-Helmholtz timescale, which is so long as to be of little significance to anything but the evolutionary process. In particular, it is not relevant to the luminosity, so for that we may as well take the star in complete equilibrium.
It's gradual at both ends-- all structural changes occur on a Kelvin-Helmholtz timescale for stars like the Sun, and as D H mentioned, that's tens of millions of years. The radiative diffusion time is much shorter, perhaps 100,000 years, so we can assume the luminosity always figures out what it is supposed to be much faster than any structural changes are occurring.

24. Nov 19, 2014

### Staff: Mentor

But Eddington's analysis was done based on the Sun's current state, when it's in equilibrium, right? He didn't analyze a sun that was gradually contracting.

25. Nov 19, 2014

### Ken G

Yet what I am talking about has nothing to do with photon pressure, it has to do with how to understand the luminosity of the Sun. It is very clear that we can get a reasonable understanding of the luminosity of the Sun by noticing that the Sun's luminosity is determined well before fusion even begins in its core. This is easy to do, it's what Eddington did in fact. One simply uses what is called a "polytrope" model, where you peg the pressure to the density in a way that does not even refer to temperature, and get the stellar structure that way, and then it is a simple matter to calculate how fast light will diffuse out when you connect that pressure and density structure to the implied ideal-gas temperature. The rate of diffusion of light then tells you the luminosity. This is how the "Henyey track" is calculated, and the Sun is on the Henyey track. I'm not making this up, a lot of people don't understand the simplest ways to think about what stars are doing.
Yes, but why is that the temperature of the core of the Sun? This is what you are not answering, and it is the real answer here. You could also say "fusion occurs at the rate it does because that's the temperature and density there." Of course that statement is true everywhere in the universe, even in a nuclear bomb where the energy generation rate is way higher than a compost heap. So that answer really doesn't tell us anything.

Let's change the question a little to see what I mean here. Imagine inserting a special fusion catalyst into the core of the Sun that doubles the fusion rate at given temperature and pressure. By your way of thinking, you might expect the Sun's luminosity to end up doubling as a result of this addition, yes? But that isn't right, the Sun's self-regulated luminosity would not change much at all if you did that. However, if you add a special substance that doubles the opacity to light at every temperature and pressure, you will end up making the Sun's luminosity drop by pretty close to a factor of 2. I think these facts make the situation pretty clear as to whether fusion physics, or light diffusion physics, is responsible for determining the luminosity of the Sun.

Last edited: Nov 19, 2014