Where does the radiation released by nuclear fusion ultimately come from?

Factual Questions
1

I am trying to ‘understand’ where the radiation released by nuclear fusion ultimately comes from.

In particular, if the ultimate source of the energy released by fusion originates from the strong nuclear force, how is a ‘strong phenomenon’ channeled into EM radiation (and is it only transformed into EM radiation)?

Basically, I think I’m asking how the strong nuclear force winds up involving leptons.

ETA: I understand that antimatter/matter annihilation occurs and that produces photons. Is that the sole source of the EM energy?

2

If the radiation that you are talking about is the electromagnetic radiation in the form of gamma rays, then it comes from the binding energy of the strong force.

If you accelerate a charged particle, it will give off a photon. The rather violent shifting around of charges involved in a fusion event releases a fair amount of energy. In the case of 4 hydrogen into 1 helium, that converts about 4% of the rest mass into energy, which emerges as high energy photons.

It’s “like” falling down a well. You are converting the potential energy of the 4 separate H atoms into kinetic energy and a lower energy He atom.

The other radiation given off is neutrons, as we don’t do 4H into He fusion, but rather something like Deuterium Tritium, which tosses off an odd neutron. You are also going to get electrons flying about from secondary decays, and technically a Helium nucleus is an alpha particle, even if we usually think of them as decay products.

There are also neutrinos involved, but they contribute extremely negligibly to any recoverable energy.

3

Your question talks about both EM radiation and leptons. Fusion can involve one, the other, both, or neither.

Leptons
To start with leptons, the relevant fusion reactions here are not purely strong processes. As you correctly note, a purely strong process can’t (intrinsically) involve leptons.

Consider the beta decay of a neutron. This decay is mediated by the weak interaction. In the decay, a neutron turns into a proton, electron, and antineutrino. Since the neutron has more mass than that collection of daughter particles, this decay is energetically allowed.

However (as you know), if the neutron is embedded in a nucleus, the strong interaction plays a role in allowing or forbidding the neutron’s decay. A neutron in a nucleus is in a bound state with the other protons and neutrons thanks to the strong force, and altering that bound state by turning a neutron into a proton (plus the leptons that leave the decay) may cost net energy, even if the “bare” neutron decay releases energy. In cases with a net energy cost for neutron decay, the nucleus is stable to that decay path.

Second, consider the “beta decay” of a proton. If a naked proton could do this, it would look like: proton --> neutron + positron + neutrino. The weak interaction fundamentally sees this in exactly the same light as neutron decay. They are both perfectly fine physical processes. When energy conservation is taken into account, we see that the proton is lighter than the neutron (nevermind the sum of all three daughter particles), so proton beta decay doesn’t happen in vacuum.

However, if the proton is embedded in a nucleus, proton beta decay may be energetically favorable in light of the energetics of the nuclear bound state. In those cases, proton beta decay does happen (e.g., carbon-11, sodium-22).

In both cases above, we are seeing the weak decay of a nucleon (proton or neutron), with the energetics of the process – including whether it is even allowed or not – being influenced by the details of bound state the nucleon is a part of. The bound state is itself governed largely by the strong force.

The simplest fusion process is proton + proton --> proton-neutron (bound) + positron + neutrino. The bound state is just deuterium, which is energetically favorable enough due to the strong force to be stable. In the case of fusion, this favorable bound state also allows proton beta decay to occur if you can get two protons into suitable initial conditions.

So, the weak force and the strong force are both critical pieces of the p-p fusion story. And as you mention, positron annihilation leads to gamma rays as a side effect.

Gamma rays directly
If the number of neutrons and protons is conserved throughout the fusion reaction, then the weak interaction is not needed. There are a number of favorable reactions, though, where you would want to end up with a single object in the end. For example, you can squish a proton into carbon-12 to make nitrogen-13, and that’s energetically favorable. However, it is impossible to conserve overall energy and momentum unless you have two bodies in the final state. The only candidate for that extra product is a gamma ray. So, the electromagnetic force is part of the story in these reactions (e.g., 12C + p --> 13N + gamma) just like the weak force was in p-p fusion.

Neither leptons or gammas
The process 2H + 2H --> 3H + 1H is both energetically allowed and has a two-body final state. Processes like these need neither the weak force or the electromagnetic force to proceed.

Gamma rays indirectly
In all situations where a nucleus is being created or altered, the result may not be in the lowest energy state (the “ground state”). Just like an atomic system (nucleus + electrons) can be in an excited state that will eventually decay by emitting photons, so too can a nuclear system be in an excited state. And a violently created nucleus has a good chance of not being in the ground state at birth. To shed energy and rearrange the nucleons into a happier lower-energy configuration, something has to be emitted, and a photon is usually the only viable candidate.

4

Thank you. Extremely helpful.

I don’t understand why sometimes gamma radiation is the “only” candidate but for other situations it is just “usually” (i.e. suggesting there may be different possibilities).

That leads me to ask a naive question while I’m at it. In those situations where the gamma photon is the only option, why is the EM field the one that balances the energy requirement? Could other fields not participate (or is their range is too small or there effect to minimal to factor in)?

5

My freshman physics understanding is that it’s a mass to energy conversion. The two protons and two neutrons of a helium atom have less mass than the four protons that make up four hydrogen atoms, with the difference being turned into energy in the form of radiation. As I understand it, the iron nucleus is the most stable / least energetic. Any nuclear reactions that shift in the direction towards iron will release energy.

6

It is both symmetry and energy conservation.

You can’t just create most particles. If an electron comes into being, a positron has to be created as well. Same with quarks, which make up the protons and neutrons. If they are created, they will really want to annihilate as quickly as possible.

When a photon is created, it actually is created with its antiparticle as well, but a photon is its own antiparticle, so that’s okay.

At the end of the day, you have to have the same charges, electric, weak, and strong, as you started with, and you also have to have the same overall spin. The photon is by far the easiest to satisfy these conservations.

Then you have energy requirements, an electron is about half a Mega-electron volt(MeV), so to create the positron electron pair requires a bit more than one MeV. Quarks and the particles made of them are even higher.

A photon can be at any arbitrarily low energy, though, so unless you have enough energy to create a particle, what you are going to get is a photon.

The energy released in DT fusion is high enough to theoretically make some lepton pairs, but not many, (a fair amount of energy is going to go into that spare neutron as well) and they would immediately annihilate and release gamma rays anyway.

If you smack space hard enough, you can make any particles you want. That’s what particle accelerators do. CERN runs at about 15 Tera-electron volts, for comparison.

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Another point, DT fusion does release more radiation than just photons. It also releases a neutron, which will be decaying into an electron and neutrino if it doesn’t combine with some other atom.
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I just had another thought. If your question is “Why gamma, rather than X-ray or microwaves?” Then that’s just because we classify the electromagnetic spectrum by its energy. There is no difference between a gamma photon, a microwave photon, or a visible light photon, other than the energy that they carry.

As you can see, fusion releases a whole lot of energy, so the light that is emitted is in the gamma part of the spectrum.

7

It’s just that a photon is sort of a freebie if you need to shed energy or momentum. It carries away no charge (or other conserved quantum numbers); it is massless; and it doesn’t “cost” very much in terms of interaction rate since it is electromagnetically produced (in comparison to something that requires the weak interaction, which would be heavily suppressed). The photon does carry away angular momentum, but that is a mere detail in the discussion at the moment.

It would be even better if the nucleus could shed something using the strong force rather than the electromagnetic force – the rate could in principle be higher and out-compete the EM process – but the lightest candidate particle to emit is the neutral pion, which weighs 135 MeV/c2, which is way more than the typical energy scale of nuclear rearrangements (a few MeV). In other words, there’s not enough energy being freed up in the rearrangement to emit something as heavy as a pion, so we must fall back an electromagnetic product.

You astutely noticed that I said “only” in one place and “usually” in the other. In the first, I was talking about a specific example (12C + p → 13N + gamma), where the gamma is indeed the only thing that works. In fact, this is true for all reasonable examples. My hedge in the second case (“usually”) is because it is possible to set up violent enough processes where there is enough excess energy in the system to emit a pion. So that rare corner case led to my distracting (in exchange for pedantry) use of “usually”. In fact, we would have to take a tangent into excited baryons for that corner case to be fleshed out. It’s not relevant at all in routine radioactive processes.

It’s just that if a particle needs to be emitted for kinematic reasons (energy/momentum conservation), the photon brings all the aforementioned advantages: neutral, lightweight (massless, even), and readily producible via electrodynamics since there are charged particles involved. Nothing else fits that bill.

8

Thank you both! This is really helpful and so very clear.
Much obliged.

9

I guess this a natural time to ask if the total energy of the universe is the sum of that in all its various fields. Or is there field-free energy?

10

A few minor comments on your post, k9bfriender.

You can also create an electron alongside an (anti)neutrino, as in beta decay. It doesn’t have to be a positron on the other side.

There is no need for photons to be created in pairs generally. You can make single photons without a problem.

You have to have the same overall angular momentum, but you don’t have to have the same overall spin. Orbital angular momentum can also be present, so you can end up with different spin in the system before and after. This is an important aspect of the classification of nuclear transitions, as transitions that involve changes in spin have typically longer lifetimes.

11

The majority of the total energy of the universe appears to be something we are unable to yet explain (dark energy), so whatever it is, it’s not what you said. :slight_smile:

12

But, presumably, there’s a dark energy field? Or could it exist separately from what we define as a field?

Thanks.

13

It’s worse than all that. Step 1 in this line of thinking is to calculate the total vacuum energy density due to the fields we know about and compare that to the observed amount of dark energy. The result: dark energy is some one hundred orders of magnitude smaller than that calculation suggests. So it’s not even that we just need to add some other field. The whole thing is busted from the start.

14

Maybe this is what you reap from the sin of renormalization.

15

E.g. with lithium-6 in designs that include a closed tritium cycle.

Anything know is there’s an equivalent of the [sup] tag here?

16

Use X<sup>2</sup> to get X2.
Use X<sub>2</sub> to get X2.

Why Discourse doesn’t like square brackets [ & ] is a mystery to me.

17

This is correct, but I was talking in the case of creating an electron just out of the energy released in fusion. Both charge and lepton number need to be conserved, in your case, charge is conserved as the neutron ends up with the positive charge and becomes a proton, and lepton number is conserved by the antineutrino.

Yeah, I messed that up in editing. There was a tangent about pair production that I took out as it’s pretty insignificant in DT fusion in anything less than super heavy stars. I actually intended that line as a contrast to photon creation by acceleration of charged particles. In the case of pair production and annihilation, you would end up with at least 2 photons. I apparently removed the wrong part. (No facepalm emoji?)

Though it did elude me as to how many photons would be created in a typical fusion event? Would it just be one, or would it be many? Seems like it should be more than one, if only just to keep things uncertain. Also seems like if should be quite a few, as the charges move around, finding their new equilibrium.

Right, the entire reason that we call that property of an electron “spin” and “angular momentum”, even though we are pretty sure that the electrons are not actually little balls spinning around, is because that spin can be transferred to and from the angular momentum of the orbitals.

I was talking in the free space case, where they are not coupled to an atom that a particle can either dump or borrow spin from. But yes, that is definitely important to keep in mind, as it is less often that particles are having interactions in free space than they are interacting around and in atoms.

18

Thanks. I was going for 6Li and realized the old tags didn’t work.

19

All correct, but this does happen in fusion reactions, too. p-p fusion would be the prototypical example.

Makes sense!

It’s a quantum state transition, from state A to state B. The number of photons isn’t a random variable unless you change the initial and final states (making them not states A and B anymore). For the 12C + p → 13N + gamma example, that’s the reaction, full stop. If you try to emit two photons, the angular momentum and energy accounting all changes, and you don’t have the same (and typically not even a viable) transition anymore.

There are cases where the nucleus that results from some decay or other reaction is not in its ground state. En route to the ground state it may transition to several intermediate states, and each transition is a distinct reaction with a distinct (and often very short) lifetime. That can lead to a cascade of gamma emissions, but it’s still one per step.