A tutorial dialogue about nuclear fusion

 

Nigel:
So, in this post I'd like you to remind me, in simple terms, how nuclear fusion works, and how we get more energy out than we put in. I already understand that nuclear fusion is about getting hydrogen nuclei - deuterium and tritium - close enough to overcome their repulsive forces and fuse. When they do, they form helium and release a lot of energy, right?

ChatGPT:
Absolutely, that’s the gist of it. Fusion brings hydrogen nuclei - often deuterium and tritium - close enough to overcome their repulsion. When they fuse, they form helium, releasing energy, extra neutrons, and high-energy photons, typically gamma rays. You’ve got it right: fusion transforms mass into radiant energy.

Nigel:
And the preferred mixture in reactors is deuterium (one proton and one neutron) and tritium (one proton and two neutrons)?

ChatGPT:
Exactly. Deuterium–tritium fusion is favoured because it happens at lower temperatures and with higher probability (a larger cross-section). If you’re aiming for practical fusion, that’s the go-to isotope combination.

Nigel:
But two single protons - without neutrons - can’t just fuse into something stable, right, because the helium nucleus is two protons and two neutrons?

ChatGPT:
Correct. A pair of bare protons can’t directly fuse into a stable nucleus. In stars, the proton–proton chain is a multistep process involving beta decays that eventually produce helium, but a simple two-proton fusion doesn’t yield a stable product. So your intuition is right.

Nigel:
So the binding comes from the strong nuclear force - but we’re not really talking about quarks and gluons here, are we?

ChatGPT:
Yes, the binding force is the strong nuclear force, more precisely its residual form that holds protons and neutrons together. Quantum chromodynamics (QCD) describes the underlying quark–gluon interactions, but fusion discussions usually stay at the nucleon level. You’re right: we don’t go down to quarks and gluons for practical fusion physics.

Nigel:
Could we think of it as like an asteroid falling into Earth’s gravity well: energy released as it falls in and impacts?

ChatGPT:
That’s a good analogy. As nuclei “fall” into the potential well of the strong force, they move to a lower-energy, more tightly bound state. The difference in potential energy is released as radiation, just like the energy liberated when an asteroid falls into a gravity well.

Nigel:
So, classically, that image of little billiard balls crashing together still helps?

ChatGPT:
Yes, it’s a useful simplification. Imagine the nuclei as billiard balls ramming together with enough kinetic energy to overcome electrostatic repulsion. Once they’re close enough, the strong force pulls them into a bound state, releasing energy. It’s not quantum-accurate, but it gives the right intuition.

Nigel:
And in Newtonian terms: how fast are these nuclei moving in a fusion-ready plasma?

ChatGPT:
In a deuterium–tritium plasma at tens of millions of kelvin, nuclei typically move at a few thousand to about ten thousand kilometres per second. That’s only a few per cent of the speed of light; fast, but still comfortably non-relativistic.

Nigel:
So they slow down slightly as they push through the Coulomb barrier, then snap together?

ChatGPT:
Exactly. They approach fast, decelerate under repulsion, then once they breach the barrier, the strong force pulls them sharply together. You can picture it as a quick fall into the potential well, releasing a burst of binding energy, again on the order of a few percent of c for their speeds.

Nigel:
And the observer just sees that extra kinetic energy turning into gamma radiation and fast neutrons?

ChatGPT:
Yes. From the outside, it looks as if kinetic energy has been transformed into radiation and high-energy neutrons. Fundamentally, the released energy comes from the difference in nuclear binding energies - that helium nucleus is more tightly bound than its parts - but your description is a good macroscopic picture.

Nigel:
So the bookkeeping really comes down to binding energy differences.

ChatGPT:
Exactly. The input kinetic energy only gets the nuclei close enough. The net gain comes from the drop in total nuclear energy. The more stable helium 'product' nucleus has lower mass–energy, and that difference is emitted as usable fusion energy. That’s the accounting.

Nigel:
So if we stay within a classical Newtonian picture, the approaching positively charged hydrogen nucleus has to climb that Coulomb-repulsion hill high enough to get close enough to the other nucleus for the strong force to take over — overcoming the electrostatic push and slamming the two together so they fuse and form helium, releasing all that energy. It just shows, doesn’t it, how powerful that residual strong nuclear force is?

ChatGPT:
It really does. But in reality, quantum mechanics changes the picture. The hydrogen nucleus doesn’t always need enough energy to climb over the Coulomb barrier — instead, it has a finite probability of tunnelling through it. Quantum theory allows it to appear within the range of the strong force even when, classically, it shouldn’t be able to. That’s the quantum “magic” that makes fusion in stars — and in reactors — possible at achievable temperatures.

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Energy Conservation in Nuclear Fusion

Let the reactants have rest masses m₁, m₂ and an initial (centre-of-mass) kinetic energy Kᵢ.

Let the products be j = 1…N with rest masses mⱼ, total kinetic energy K_f, and total photon (gamma) energy Eγ.

Energy conservation:

m₁c² + m₂c² + Kᵢ = (Σ mⱼc²) + K_f + Eγ

Define the Q-value as the mass-energy difference between reactants and products:

Q = (m₁ + m₂ − Σ mⱼ)c²

Then the energy-balance equation becomes:

K_f + Eγ = Kᵢ + Q

If Q > 0 (an exothermic reaction), the final kinetic energy plus radiated gamma energy exceeds the initial kinetic energy — the surplus Q comes from the mass defect, i.e. the greater nuclear binding of the products.

If Q < 0 (an endothermic reaction), at least |Q| of extra kinetic energy must be supplied.

The Coulomb “hill” and the short-range strong attraction simply redistribute kinetic energy during approach; they do not change the conserved total. Quantum tunnelling allows fusion even when Kᵢ is below the classical barrier height. Momentum conservation then fixes how K_f is shared among the reaction products.

Example: Deuterium–Tritium Fusion

D + T → ⁴He + n  Q ≈ 17.6 MeV

In the centre-of-mass frame (for small Kᵢ), nearly all 17.6 MeV appears as kinetic energy of the products: about 14.1 MeV for the neutron and 3.5 MeV for the alpha particle. Prompt gamma emission is negligible in the main branch.

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