# Why do practical fusion reactions need more than one product?

The Wikipedia article about nuclear fusion says that

To be a useful energy source, a fusion reaction must ... have two or more products: This allows simultaneous conservation of energy and momentum without relying on the electromagnetic force.

I'm not sure I understand what this means. Why would a reaction need the electromagnetic force to conserve both momentum and energy?

• It is poorly worded, perhaps. The electromagnetic force (i.e. a photon) can carry away energy, but very little momentum compared with the initial momentum of one particle smashing into another. So, you need a second particle (or more) to make energy and momentum balance. – Jon Custer Jul 17 '15 at 21:05
• @JonCuster So does that mean in a fusion reaction that emits a photon, the photon will receive most of the energy? – Dan Jul 17 '15 at 21:12
• No, it means that the 2 (or more) fusion products will go flying apart, and a photon or two might be emitted along the way. The (momentum,energy) of a photon just does not compare with with what a proton, neutron, or even an electron can take away. The problem is mainly on the momentum side. – Jon Custer Jul 17 '15 at 21:36

# What if there's literally only one resulting particle?

Suppose two relativistic particles with rest masses $m_1, m_2$ and velocities (in the lab frame) $\vec v_1, \vec v_2$ collide and stick together. Their energy-momentum 4-vectors are $\gamma_i m_i [c, ~\vec v_i] = [\bar m_i ~ c,~ \vec p_i]$ where $\bar m_i = \gamma_i m_i$ is the effective mass and $\vec p_i = \gamma_i m_i \vec v_i$ is the momentum seen in the lab coordinates.

There is then a center-of-mass frame given by the boost $\vec\beta$, which is calculated in the lab frame as: $$c~\vec \beta = \frac{\vec p_1 + \vec p_2}{\bar m_1 + \bar m_2}.$$Now, the two particles in this frame come together with some new velocities $\vec u_i$, new $\hat \gamma_i = 1/\sqrt{1 - (u_i/c)^2}$, and new relativistic masses $\hat m_i = \hat \gamma_i m_i$, with $\hat m_1 \vec u_1 + \hat m_2 \vec u_2 = 0$.

The resultant particle must necessarily have rest mass $m = \hat m_1 + \hat m_2 > m_1 + m_2$. So this reaction simply cannot release any energy and must instead absorb kinetic energy. Moreover, generally $m, m_1, m_2$ are fixed numbers. The above equation for $m$ therefore constrains $u$, so this reaction could only take place at a very particular family of impact velocities. So even when the reaction absorbs energy, it is pretty much impossible due to the specificity of the velocities.

In general there will always be two or more particles coming out of any non-contrived process. Yes, you can probably fire an electron and a photon and an antineutrino at a proton in just the right way to turn it into a neutron and nothing else, but that's going to be difficult to engineer and in a big complicated reactor this effect would be swamped by other interactions between them.

## So if there are always two particles out, isn't the statement vacuous?

Not quite. Very often those other "particles" are neutrinos or photons. The statement in context is only regarding a nucleus as a "product". So if there's only one "product" then the other "particles" are photons, needed to make the conservation laws balance out for a reaction which emits, rather than absorbs, energy.

## Why avoid photons?

Part of the problem is that photons and neutrons are uncharged particles and cannot be confined. Since cold fusion seems impossible, we're stuck with hot fusion, and hot fusion melts Earth materials: you want to limit your thermal contact with the fusing material as much as possible. So if a substantial proportion of the nuclear energy is carried off in gamma rays, you're hemorrhaging energy from the hot plasma where you need it. Photons and neutrons in these regimes can also be carcinogenic and may erode the materials that you're using to house the reactor, which is also not good. Photons from strong nuclear interactions are typically going to be gamma rays, which are pretty nasty biologically, but even if you avoid these your plasma will typically be hot-enough and moving-enough that you create X-rays via braking-radiation (Bremsstrahlung). Those losses can also be bad.