# What elements would be created in a star composed entirely of gold?

I'm working with the idea of creating a sun made out of pure gold. Of course, this would be completely man-made. Why would anyone want to do this? Because I want a cool concept like that in my story! :)

First, you start small, combining thousands of pounds of gold. And you continue to add more and more gold. And you end up with a small gold planet. How marvelous.

But we're not done! We continue to add more and more mass by continuing to add more and more gold.

Eventually, we'll reach a point where the mass is so great that the pure gold atoms begin to undergo nuclear fusion within the core of this massive gold planet. And thus, a star is born!

But, I have no idea what elements would be created from this.

After doing some basic research, I've learned that the only known stable isotope of gold is 197Au, so that should be a good starting point for people answering this question.

# What elements are created during the initial nuclear fusion state of a gold star?

I also looked into the nuclear fusion process, but I have learned that it is a very complex process that I simply can't learn in the amount of time I have.

• Why do you expect fusion to happen without energy input? Gold is well past the peak of the nuclear binding energy curve – Patricia Shanahan Oct 16 '19 at 14:15
• Your gold star would probably collapse into a black hole before you were able to add enough for the gold atoms to fuse. – Futoque Oct 16 '19 at 14:20
• I did some quick math. The Chandrasekhar limit is hit by a sphere of gold 200,000 times larger than the Schwarzschild radius – MongoTheGeek Oct 16 '19 at 22:02
• For those who want to know what the Chandresekhar Limit has to do with anything: It's the amount of mass needed to go from a white dwarf (which a solid mass of gold would be at those scales) to a neutron star. At that point, it would no longer be gold, it would be "degenerate matter." @MongoTheGeek's calculations mean that it would not go straight from a solid gold white dwarf to a black hole as some are saying. (And I think that this would make an excellent answer from Mongo.) – Ghedipunk Oct 16 '19 at 22:55
• @MongoTheGeek That's almost certainly incorrect. You're likely using the density of non-degenerate gold to calculate the radius, whereas a white dwarf made of gold would be degenerate, and therefore have a massively higher density. Degenerate iron is thousands of times denser than metallic iron, so a similar factor should be expected for degenerate gold. – March Ho Oct 16 '19 at 23:19

Unfortunately, no matter how much pure gold you add to your mass, you will never end up with a star. The reason for this is that fusing gold is an endothermic process, meaning that it requires energy, rather than releasing it. In fact, all elements with an atomic mass greater than or equal to that of iron consume energy upon fusing, rather than releasing it, as all atoms which are smaller than iron do. The reasons for this are complex and have to do with the binding energy of atoms.

The binding energy of an atom is always positive (as if it were not, the hypothetical atom in question would spontaneously fly apart as soon as it formed) and increases with the size of the atom. Up to a certain point, the binding energy of an atom created by fusing two smaller atoms is greater than the sum of the binding energy of the component atoms. For example, when combining two atoms of Hydrogen into one atom of Helium, the binding energy of the Helium is greater than the total binding energy of the two Hydrogens. The net potential energy of the system has now decreased (as it would take more energy to separate the new atom into its component parts than it would have for the previous atoms), and thus energy is released (generally as heat). However, as the size of atoms increases, the binding energy begins to increase by smaller and smaller steps, until one reaches the tipping point. When one fuses, say, an Iron atom and a Helium atom, the result is a single atom with a binding energy lower than the sum of the binding energy of the two original atoms. Thus, we have increased potential energy and actually had to consume energy from outside the system, making the reaction endothermic.

As Gold is a much, much larger atom than Iron, it is incapable of fusing exothermically, and thus your star will never get off the ground. If you add sufficient quantities of gold at a quick enough rate, it may release energy via contraction, but it will be massively dimmer and shorter-lived than an equivalently massed star. In the end, if you just continuously add gold to your mass, you'll end up with a black hole before you ever get a star.

• I was going to follow up by asking a brand new question regarding whether a black hole can be formed from pure gold. So you not only answered the question, you answered my next question as well. – overlord Oct 16 '19 at 18:02
• @overlord, You can form a black hole by sticking enough of any material in a small enough space. Gold would be an... inventive choice, but pretty much anything works, up to and including light. – Gryphon Oct 16 '19 at 18:04
• Would a pure gold dwarf “star” really turn into a black hole, or would it turn into a neutron star first? – nick012000 Oct 17 '19 at 6:20
• @nick012000 IANAA (I am not an astrophysicist), so I'm not sure. However, even if it does form a neutron star, if you keep funneling gold into your neutron star, you will eventually end up with a black hole. So regardless of whether a neutron star is an intermediary stage, you will eventually end up with a black hole. – Gryphon Oct 17 '19 at 14:45
• @Gryphon To be fair, the process of converting a bunch of gold into a neutron star or a white dwarf would release a lot of energy. In a sense, a star is just a gravitational collapse slowed down by the potential energy of fusion; if there was a phase change for gold it might work. I doubt it tho. – Yakk Oct 17 '19 at 17:46

This star would not fuse gold.

Fusion reactions producing elements beyond zinc-60 are not energetically favorable; they are endothermic, and so consume energy. Several elements heavier than iron are formed through this fusion chain and subsequent decay (cobalt, nickel, copper and zinc), but these are unstable and decay back to iron, meaning that iron is essentially the heaviest stable element that can be formed in stars or involved in significant fusion.

Two exotic processes - the r-process and the s-process - can fuse heavier elements (see Burbidge et al. 1957, Clayton et al. 1961). These involve a neutron being captured by a so-called seed nucleus; repeated neutron capture produces heavier and heavier nuclei, and it is not unrealistic to think that gold could be involved.

However, these processes need neutron sources; even the slower s-process requires neutron densities of $$10^{13}$$ neutrons per cubic centimeter (Lugaro et al. 2016); the r-process may require neutron densities on the order of $$10^{24}$$ neutrons per cubic centimeter (see Burbidge et al.). As the star is made purely of gold, there is no existing neutron source (e.g. the fusion of carbon or neon nuclei with alpha particles), and therefore neither process can proceed.

• @val A neutron star is an extremely heavy element on its own :) – Hagen von Eitzen Oct 17 '19 at 10:07
• Wouldn't this create a fission-based star instead, until all the gold is turned to iron? I can't imagine it would just do nothing. – kloddant Oct 17 '19 at 13:10
• I think it is going to be one of two possibilities - either the degenerate matter in star's core starts slowly fusing atoms into neutrons, in which case it might be active for some time. Or it fuses them quickly, the star explodes and we're left with a neutron star. – TStancek Oct 17 '19 at 14:46
• @Gloweye I did consider mentioning them. Thing is, supernovae aren't particularly stable, and the matter inside them won't exist in those states for long. So I think it might not be a great analog of an (ideally stable) star. – HDE 226868 Oct 17 '19 at 18:12
• @TStancek I think an interesting starting point there would be trying to investigate the substances gold would form under those pressures. I did some searching but have been unable to figure out just what would happen (it's, er, apparently not a huge scientific priority). I might end up asking a question about that, come to think of it. – HDE 226868 Oct 17 '19 at 18:13

Not really a star, but it could still shine.

As others have said, no fusion would happen. However, depending up how you add the gold, it would fall until it hit the current mass.

For example, meteors impact the earth at speeds greater than escape velocity (at least when they impact the upper atmosphere) Earth escape velocity is about 11.2 km/s. Solar escape velocity is about 617.9 km/s.

Considering the well known 0.5 M*V^2 relations for kinetic energy, solar impacts are going to be 3000 times as energetic as earth impact (which are already white hot)

If you consider solar asteroid bombardment, they are going to be well over white hot after impact. Given the mass of this "sun", it is going to glow for a very long time.

Historically, gravitational collapse of the sun was considered as a possibility for the source of the sun's heat. Even if you carefully set the newly arrived gold on the surface, it is still going to compress under gravity and release enough energy to glow for a long time.

As others have said, you can't have self-sustained fusion with gold. But is there a chance you might make fission work?

If you look at all the atoms, the light ones can give off energy by fusion. Like hydrogen combining to make helium (sometimes a little bigger atoms, but still on the small side, getting bigger. This is what occurs in the sun or in hydrogen bombs.

In contrast, the heavy elements can give off energy by FISSION. Heavy atoms splitting into smaller ones. The best example here is uranium or plutonium in fission bombs or reactors.

In the middle, you have Fe. It's basically dead. Can't get energy from fission or fusion of iron.

See graph:

https://en.wikipedia.org/wiki/Nuclear_binding_energy#/media/File:Binding_energy_curve_-_common_isotopes.svg

Now gold is to the heavier side of iron. So it's not going to do fusion. It might do FISSION though. I totally haven't thought this through...but on a worldbuilding site with sufficient handwaving (with a bias to making the idea work), perhaps you could imagine some massively huge/dense bunch of gold, where neutrons can build up and be moderated/absorbed/cause fission. Given the huge size, leakage would be low, so even a very crappy reaction might become self-sustaining.

Perhaps some trace isotopes that emit neutrons (or perhaps seeding with a minor amount of uranium or other neutron sources), then perhaps you could get a self-sustaining fission reaction where enough neutrons are produced from gold fission to sustain the reaction.

Gravity would tend to keep the stuff from flying apart (same as fusion). And you end up with some sort of equilibrium of gravity/nuclear reaction in terms of the stuff hanging out as a star (not blowing up, not shutting down).

If this were possible, you'd have a shining fission star, not a fusion star. Maybe not quite as bright as a fusion star. But still a ball of very hellish energy.

In terms of the atoms that would result, you get a distribution of smaller elements. With uranium, this tends to be bimodal. Centered at half of the nuclear mass, but with peaks to either side. It's called the Mae West curve. See here:

https://idahospudsblog.blogspot.com/2013/10/some-odd-quirks-that-nuclear-reactors.html

As you can see, the two peaks are actually about 20 mass units below/above the half-weight of U-235.

I'm not sure about gold, but it seems reasonable to expect some similar bimodal distribution centered around half the nuclear mass. If not, then a normal distribution centered around the middle. I sort of think bimodal is more likely though. Has to do with the jellium* model of how fissioning atoms split.

If we assume something similar for Au-197 (stable isotope of Au), then we could assume about 100 for the midpoint. So something around 80 and 120 for the two peaks of the Mae West curve. Might be a little tighter like 85 and 115, given gold is smaller than uranium. Stable nuclide atoms in that weight are, respectively rubidium and indium.

Of course, you are going to get a soup of atoms and nuclides of atoms. Some of these may react a little further in various fashion to move up/down slightly in the atomic number. But the big picture: lots of rubidium and stuff close to it. And lots of indium and stuff close to it.

*Not joking that's what it's called...nukes love their little names...look up "barns" for instance.

• No, this is not possible with real physics. First, gold is observationally stable (not radioactive at all) as well as theoretically stable. So, no natural fission or other decay either. For a fission reactor, you must use fissile (atoms split spontaneously, e.g. U-235) or fertile (atoms absorb a neutron and become fissile, U-238 becomes Pu-239 eventually). I did some calcs on a Th-232 "star" once, though not fissile, energy is released during the normal decay chain - enough to make the whole white hot when equal in mass to our sun. But gold, nope, no natural decay. – Gary Walker Nov 5 '19 at 18:05

Nothing will happen until you get enough gold to overcome electron degeneracy pressure.

Then your gold fuses into one ginormous nucleus and the protons capture the electrons.

• This is called a neutron star. – cmaster - reinstate monica Oct 17 '19 at 6:59
• When the electrons are forced out of their shells (electron degeneracy, aka white dwarf) it will cease to be gold colored. – Loren Pechtel Oct 19 '19 at 4:09

As noted, Iron is at the end of the curve of binding energy, and when the fusion of heavier elements ends with iron, you cease the fusion reaction. The core "goes out" and the massive gravitational energy of the star draws everything back towards the center (up to now, the energy of the nuclear fusion reactions have been "pushing back" against the gravitational collapse of the star).

The rapid infalling of matter onto the iron core creates the implosive energy that results in a Type II Supernova and incidentally is where all the elements heavier than iron are produced in the universe, including gold.

As noted, simply dumping more and more of any element into one place will simply create a black hole (possibly you might stop at an intermediate step like a neutron star if you carefully control the amount of matter you add and stop at the correct time). Since our understanding of Neutron stars is incomplete, there is a possibility that the layer of atoms on the surface of the neutron star might be transmuted into something besides gold or neutronium, but this is only a guess.

The only "real" (for some versions of real) way to do something like this would be to create a gold sphere the size of the core of a star and teleport it in to replace the existing core. The stellar collapse will create a Type II supernova, but given the extreme energies involved, the actual material of the core really makes no difference at that point, the implosion will tear everything apart and fuse it into all the heavier elements. If there is any way to determine if there is a gold core rather than an iron one, it would probably take a careful analysis of the ratios of various elements created in the supernova explosion (although I have no idea of how you would mathematically calculate the elemental ratios of a non iron supernova core explosion).

As a story element, this could be the rather subtle indicator of the presence of super science or an advanced alien race, who are looking to mine the stellar debris for elements and might want a larger proportion of transuranic elements (they need to move fast before they decay).

You would use all the gold as gravitational mass to produce fusion of hydrogen in the center.

Stars burn by fusing light elements; hydrogen at first, then helium and so on. As has been pointed out in other answers, once you get to iron you do not get energy back from fusing elements. Heavy ones like gold are only created in supernovas that have loads of excess energy which is drunk up by the fusion of heavy elements.

A regular fusion star has so much stuff that it compresses the stuff in the middle until fusion starts. Most of that stuff is hydrogen and light elements.

But you happen to have a lot of gold. In the center of your gold star, you leave some hydrogen or helium-4 or muons or other things that like to do fusion. The mass of the gold compresses the center until fusion starts. Because it is gold you don't need as much of it to achieve the needed central compression so your gold star is nice and small; smaller than a star for sure. The heat of the fusion melts the gold, of course, but molten gold is also awesome. The gold star is heated by the fusion to white hot and so glows with the same frequency as a star.

You might want to build in a way to refill the hydrogen center.

@Muuski in the comments! OK. Here is a chart of hydrogen phases

https://www.pnas.org/content/107/29/12743

If there is enough pressure the hydrogen remains a liquid metal. If the pressure comes down it changes to a plasma. In the center of the gold star, the pressure is high so you have metallic hydrogen. The adjacent gold would probably be solid too; it is also under pressure. But gold is an excellent thermal conductor, and so at some distance from the core, the gold would go through its various phases until the exterior would be gold plasma.

We are for the moment handwaving away objections about how the hydrogen core stays put in the center.

• If it's star-temperature, the gold isn't going to be molten, it's going to be a gas or a plasma. Which is also awesome, so it's not much of an issue. – Gryphon Oct 16 '19 at 18:02
• Maybe temper the fusion until it is just white hot but still solid. It does not need to be star temp to shine like a star. – Willk Oct 16 '19 at 20:08
• Shouldn't the heavier elements "sink to the bottom" and the lighter ones "bubble up"? – Muuski Oct 16 '19 at 20:47
• Metallic hydrogen! That would be denser than gold plasma, surely. A metallic hydrogen core can retain its righteous glowing place in the center of the glowing gold ball. – Willk Oct 17 '19 at 1:04
• @Willk I really am not trying to be a pain, but, hydrogen which is hot enough to turn gold into plasma, but cool enough to not be plasma itself? I don't know the math, but I'm doubtful. – Muuski Oct 17 '19 at 22:00

# A gold star may be possible, with a fission-fusion reaction.

The other answers are correct, gold is heavier than iron, so fusing those atoms will consume more energy than it releases. The extra energy gets encapsulated in the binding energy holding the nucleus together.

As far as I can tell, it's not known what the product of an AU + AU fusion is. But, like all elements heavier than Lawrencium (and not in the Island of Stability), it will probably produce a very unstable element with a very short (less than 1 second) half-life. And this is the key point here.

Let's call this element Doublegoldium. This means that the Doublegoldium atoms will actually release a lot of the energy it sucked up in the fusion process. It will release this energy as radioactive decay, and these decay products may trigger fission in each other via a nuclear reaction. Or, Doublegoldium may even just directly undergo "spontaneous" fission.

In fact, it is possible that the fission will release more energy than the fusion absorbed.

The question is whether this process can complete a chain reaction of

heat -> fusion -> unstable element -> fission -> heat


where the heat generated is greater than the heat absorbed.

How much extra heat will determine how hot the star is.

A fission-fusion star would probably be unique in the universe.

All in all, nobody truly knows what would happen. Nobody has ever tried to fuse gold atoms, and decay of unstable elements can be very complex.

But it is possible.