Nuclear fusion - Core Fusion Physics

Introduction to Nuclear Fusion What is Nuclear Fusion? Nuclear fusion is a reaction in which two or more atomic nuclei combine to form a larger nucleus. This is fundamentally different from nuclear fission (which we won't cover here), where a heavy nucleus splits apart. Fusion releases enormous amounts of energy and is the process that powers the sun and all stars in the universe. The energy released in fusion comes from a simple but profound principle: mass can be converted into energy. According to Einstein's mass-energy equivalence equation: $$E = mc^2$$ When two nuclei fuse together, the mass of the resulting nucleus is slightly less than the sum of the original masses. This "missing mass" (called the mass defect) is converted directly into kinetic energy of the products and radiation. Even tiny amounts of mass convert into enormous amounts of energy because of the $c^2$ factor. Why Light Nuclei Matter Not all nuclei can usefully produce energy through fusion. The key insight is understanding the binding energy of nuclei—the energy that holds the nucleus together. The binding energy per nucleon (average binding energy per nucleon in the nucleus) increases as we go from light nuclei up to nickel-62, which is the most tightly bound nucleus. After nickel-62, binding energy per nucleon decreases for heavier nuclei. This has a crucial consequence: fusion of nuclei lighter than nickel-62 is exothermic (releases energy), while fusion of heavier nuclei is endothermic (requires energy). This is why fusion research focuses exclusively on light nuclei. When light nuclei fuse, they form products that are more tightly bound, and the energy difference is released. The most fusible nuclei—the ones that fuse most readily—are hydrogen's isotopes: deuterium (heavy hydrogen, with one proton and one neutron) and tritium (with one proton and two neutrons), plus helium-3 (with two protons and one neutron). Conditions Required for Fusion Initiating fusion is remarkably difficult. Here's why: atomic nuclei are all positively charged (they contain protons), so they repel each other strongly through the Coulomb force. Two nuclei cannot easily get close enough to fuse. To overcome this repulsion, fusion requires extreme conditions: very high temperature, high density, and the nuclei must be confined long enough for fusion to occur. These three requirements are captured in the Lawson criterion, which requires that the product of density ($n$), temperature ($T$), and confinement time ($\tau$) exceed a minimum threshold: $$n T \tau > \text{threshold}$$ This product is called the Lawson triple product, and it's the fundamental requirement for any fusion reactor to produce net energy gain. This criterion explains why fusion is so difficult: you need all three conditions simultaneously. The Physics of Fusion Reactions Forces at Work in the Nucleus To understand why fusion is possible at all despite the Coulomb barrier, you need to understand the competing forces inside a nucleus: The strong nuclear force is an extremely short-range force that attracts nucleons (protons and neutrons) to each other. It operates only at distances of about $10^{-15}$ meters. The Coulomb force is electromagnetic repulsion between positively charged protons. Unlike the strong force, it has a long range—it weakens gradually as distance increases. When nuclei approach each other, they initially feel only the long-range Coulomb repulsion. They must get close enough to feel the strong nuclear force to actually stick together. This is the fundamental barrier that fusion must overcome. The D-T Reaction: The Most Important Fusion Reaction The deuterium-tritium (D-T) reaction is the most practical fusion reaction and is the primary target for controlled fusion research: $$\text{D} + \text{T} \rightarrow {}^4\text{He} + n + 17.6 \text{ MeV}$$ In this reaction, a deuteron (deuterium nucleus) and a triton (tritium nucleus) fuse to produce helium-4 and a neutron. The 17.6 MeV of energy is distributed between the helium-4 nucleus (3.5 MeV of kinetic energy) and the neutron (14.1 MeV of kinetic energy). The D-T reaction is favored for practical fusion because: It has one of the highest reaction rates at moderate temperatures Deuterium is abundant in seawater The energy release per reaction is large The cross-section (probability of reaction) is well-understood Reaction Cross-Section and Reactivity The probability that two nuclei will fuse depends on their collision energy. The reaction cross-section $\sigma(E)$ is a measure of this probability as a function of the relative kinetic energy $E$ of the two nuclei. Think of it this way: if nuclei were macroscopic objects, the cross-section would be their geometric area. But for fusion, $\sigma$ is an effective area for the nuclear reaction, and it depends strongly on energy—particularly on whether the nuclei have enough energy to overcome the Coulomb barrier. In a hot plasma, ions have a range of energies following a Maxwell-Boltzmann distribution. The relevant quantity for calculating fusion rates is the reactivity $\langle\sigma v\rangle$, which is the velocity-averaged product of cross-section and relative velocity. It tells you, on average, how likely fusion is for particles at a given temperature. The reactivity increases with temperature (higher temperatures mean more energetic collisions), but at different rates for different reaction pairs. Notice that D-T has the highest reactivity at most temperatures, which is another reason it's the favored reaction. Understanding the Cross-Section: Classical vs. Quantum The Classical Problem In a purely classical picture, two nuclei are hard spheres that repel each other through the Coulomb force. They cannot fuse unless their collision energy is high enough to overcome the Coulomb barrier—the peak in potential energy encountered as they approach. This barrier is enormous: typically several MeV, far higher than the typical thermal energies in a plasma (fractions of a keV or a few keV). If only classical mechanics applied, fusion would never occur in any reasonable fusion reactor, because the barrier is far too high. Quantum Tunneling to the Rescue Quantum mechanics changes everything. Even when particles don't have enough energy to classically overcome the Coulomb barrier, they have a probability of tunneling through it—passing through the barrier despite not having sufficient energy. This is purely a quantum effect with no classical analog. The probability of tunneling is quantified by the Gamow factor: $$G = \exp\left(-\frac{2\pi Z1 Z2 e^2}{\hbar v}\right) = \exp(-2\pi\eta)$$ where: $Z1$ and $Z2$ are the nuclear charges (number of protons in each nucleus) $e$ is the elementary charge $\hbar$ is the reduced Planck constant $v$ is the relative velocity $\eta$ is the Sommerfeld parameter The exponential form shows that tunneling probability is exquisitely sensitive to the parameters. Small changes in the Coulomb barrier height or particle velocity lead to huge changes in tunneling probability. This sensitivity is why fusion rates vary so dramatically with temperature. Parameterizing the Fusion Cross-Section The full expression for the fusion cross-section incorporates both the quantum tunneling through the Coulomb barrier and the nuclear physics of actually forming the compound nucleus: $$\sigma(E) = \frac{S(E)}{E} \exp(-2\pi\eta)$$ This formula separates the physics into two parts: Tunneling: The exponential factor $\exp(-2\pi\eta)$ captures how difficult it is to tunnel through the Coulomb barrier. It depends on nuclear charge, velocity, and the Sommerfeld parameter. Nuclear resonances: The $S(E)/E$ factor is the astrophysical S-factor, which varies weakly with energy. It captures nuclear physics effects—in particular, whether the collision energy happens to match an excited state of the compound nucleus, which would dramatically enhance the cross-section. The beauty of this formula is that it separates the "easy" part (Coulomb barrier tunneling) from the "hard" part (nuclear structure effects). For most practical calculations, $S(E)$ varies slowly enough that it can be treated as approximately constant over a range of energies. Fusion Rates in Hot Plasmas Why Thermal Averaging Matters A fusion reactor operates with hot plasma—a gas of ions at temperatures ranging from millions to tens of millions of degrees Kelvin. Ions in this plasma don't all have the same velocity; instead, their velocities follow a Maxwell-Boltzmann distribution. Some ions are very fast, some are quite slow, and most are somewhere in between. To calculate the actual fusion rate in a plasma, you cannot simply use the cross-section at some "average" energy. Instead, you must average the product $\sigma(E) \cdot v$ (cross-section times relative velocity) over the entire distribution of ion energies. This gives the Maxwell-averaged reactivity $\langle\sigma v\rangle$. The physical meaning is clear: $\langle\sigma v\rangle$ tells you the average reaction rate per ion pair in the plasma. It's the fundamental quantity that determines how many fusion reactions occur per unit time. The Temperature Dependence of Reactivity For most fusion reactions, the Maxwell-averaged reactivity can be approximated by: $$\langle\sigma v\rangle \approx C \, T^{-2/3} \exp\left(-\frac{b}{T^{1/3}}\right)$$ where $C$ and $b$ are constants determined from experimental measurements and nuclear physics calculations, and $T$ is temperature in units of keV (kilo-electron-volts). This formula reveals something surprising: reactivity has two competing temperature dependences: $T^{-2/3}$ factor: This comes from the Maxwell-Boltzmann distribution—higher temperatures mean the distribution spreads out more, and this geometrical effect reduces the average product $\langle\sigma v\rangle$. Exponential factor: This comes from the Gamow tunneling factor—higher temperatures mean ions have higher velocities and can tunnel through the Coulomb barrier much more easily, dramatically increasing $\sigma(E)$. At low temperatures, the exponential factor dominates and reactivity increases rapidly with temperature. At high temperatures, the $T^{-2/3}$ factor becomes more important and reactivity increases more slowly. This creates a characteristic shape for $\langle\sigma v\rangle(T)$—it rises steeply at low temperatures, then flattens out or even decreases slightly at very high temperatures. Optimizing Fusion Reactor Conditions The Lawson criterion requires $nT\tau > \text{threshold}$, but which temperature is best? This is determined by asking: at what temperature is the Lawson triple product minimized for a given reaction? To see this, note that the fusion power is proportional to $n^2 \langle\sigma v\rangle T$. For a given power output and energy confinement time $\tau$, minimizing the required $nT$ product means maximizing $\langle\sigma v\rangle/T^2$. Since $\langle\sigma v\rangle$ varies with temperature in a complex way (the competing $T^{-2/3}$ and exponential factors), there is an optimal temperature that maximizes this ratio. For the D-T reaction, this optimal temperature is around 20-30 keV, which is why most magnetic confinement fusion reactors (like tokamaks) target temperatures in this range. Different reactions have different optimal temperatures. D-D fusion, for example, has a somewhat higher optimal temperature, which is why it's harder to achieve in practice but potentially more advantageous in some future reactor designs.