# Why isn’t CERN afraid of a fusion reaction in the LHC?

Given that they can reach terrifying energies and temperatures, why isn’t fusion of protons a concern? After all, they start with a plasma and ram protons into each other. At some point the strong force will overcome the proton-proton electric repulsion, no?

And as a corollary, can they re-purpose CERN to become a fusion reactor?

• Why would proton-proton fusion be a concern? May 13, 2020 at 11:21
• why would you consider the energies terrifying? overall i thought it was a matter of a few joules per particle. they have an absorption system they can dump it all into in case of emergency. May 15, 2020 at 21:11

Proton-proton fusion happens at energies around 15 keV. The LHC currently operates at an energy of 13 TeV, which is literally one billion times larger. Fusion is one of the lowest-energy processes that could occur at the LHC, and most of the interesting reactions that are studied there go way beyond nuclear fusion.

A proton is made up of fundamental, pointlike bits of matter called quarks, held together by force carriers called gluons. These quarks are bound together with a certain binding energy; the quarks are also relatively light, so this binding energy actually makes up most of the mass of the proton. In other words, the binding energy of the proton is roughly equal to the mass of the proton.

When you collide protons at an energy that's far below its mass, the proton acts as a single object. This is the regime in which nuclear physics typically occurs; all of the reactions that break nuclei apart, and the energy that holds them together, are typically small enough that it works to treat the proton as a single unit most of the time. This includes nuclear fusion; the mass of the proton is 938 MeV, and the amount of energy required for nuclear fusion (15 keV, as we said) is tens of thousands of times smaller than that.

If we instead decide to collide protons at energies that are higher than a few times the proton mass, then there is now enough energy in the reaction for the quarks inside the protons to interact with each other directly. We can no longer treat protons as single objects, since the energy involved in the reaction is enough to expose its inner components. As the collision energy becomes higher and higher, the binding energy of the proton becomes less and less relevant, and the picture of two dense clouds of quarks and gluons interacting becomes more and more accurate. These reactions have enough energy to produce all manner of exotic matter, particles that you could never find in a nucleus, nor could you create them by nuclear fusion. The current energy of the LHC is roughly ten thousand times greater than the mass of the proton; at those energies, the proton's mass and binding energy are completely negligible, and the interesting reactions are entirely driven by individual quarks and gluons interacting with each other. The proton is merely a vehicle for us to deliver bits of fundamental matter to the collision site. If there were a way to just collide free quarks and free gluons, without all the mess generated by the proton, many particle physicists would jump at the chance. (Unfortunately, this turns out to be impossible.)

There's certainly lots of extremely interesting physics to do at the nuclear-physics scale, and our understanding of the dynamics of nuclei is the subject of a great deal of active research in other experiments. But that's simply not what the proton-proton collisions at the LHC are meant to explore. The proton-proton collisions are actually better thought of as quark-quark collisions, or quark-gluon collisions, or gluon-gluon collisions. They're not meant to study protons, which is why the protons are often totally destroyed in the collision, transformed into exotic matter which then decays back into ordinary matter. Proton-proton collisions are meant to make conclusions about the interactions of fundamental particles, and that requires acceleration to extremely high energies, far above what you would need, or want, for nuclear fusion.

So, given all this, the LHC would make a terrible fusion reactor. Its energy is much too high to actually reliably trigger nuclear fusion, and the energy released in collisions is miniscule compared to the energy required to keep the beams running, so it would be incredibly inefficient.

• Yup. Fusion makes protons stick together. LHC and RHIC make protons melt. May 13, 2020 at 22:55
• Isn't 938 MeV/15 keV ~= 63k rather than a million?
– l0b0
May 13, 2020 at 23:22
• Well I just learnt a lot. If you ever decide to write a book called 'atomic physics for dummies' please send a link. May 14, 2020 at 4:08
• As an absolute layman when it comes to physics, this was by far the most intuitive explanation of what the LHC does and how quarks and gluons work. May 14, 2020 at 10:38
• @probably_someone Thank you. Dou you have a YouTube channel? May 14, 2020 at 14:25

As pointed out in the comments, the energy in the LHC is very high. High enough that the interactions are with partons inside the proton, where a parton is any one of the 3 valence quarks (u, u, d), or any of the divergent number of sea quarks, antiquarks, or gluons.

A fusion reaction such as:

$$p + p \rightarrow d + W^+ \rightarrow d + e^+ +\nu_e$$

is unlikely because the initial state involves 2 coherent protons, and the final state involves a coherent bound deuteron and 2 point leptons.

The energy dependence of such a reaction can be considered with constituent counting rules: basically, you count the number, $$n$$, of point particles participating in the reaction and get:

$$\sigma \propto \frac 1 {s^{n-2}}$$

where $$s$$ is the center of mass energy squared. With 6 quarks in both the initial and final states plus a W-boson in the final state, you get:

$$\sigma \propto \frac 1 {s^{11}} \propto \frac 1 {E^{22}}$$

This is basically dimensional scaling: the deuteron and each proton is a non-interacting collection of point particles, compressed into a flat disk by Lorentz contraction. The probability of each constituent being close goes as the wavelength squared, or $$E^2=s$$.

Put another way: the de Broglie wavelength at 7 TeV is 0.000029 fm (a proton is about 1.6 fm across), so you have to select the piece of the proton wave function that has all the quarks that close together, for each proton, and for the final state nucleons as well... and you need a W boson in that tiny disk as well. Anything else and the initial or final states are blown apart.

Given that the temperature in the Sun's core is 1.25 keV, and $$\sqrt s = 14 TeV$$ at the LHC, fusion should be suppressed by:

$$f = \big[\frac{1.25\,{\rm keV}}{7\,{\rm TeV}}\big]^{22} \approx 10^{-221}$$

Moreover, the half life of a solar core proton is around 5 billion years, in a "soup" with a density of 100 gram per cc. Comparing this with the LHC interaction point would need knowledge of the spot size and duty cycle... but suffice to say, the time scale for a single fusion reaction is on the order of the life expectancy of our Universe (vs. heat death), plus or minus $$10^{100}$$.

• I'd like to add a little more info about proton + proton fusion. In $p+p\to d+W^+$ a proton converts to a neutron via the weak interaction; that conversion has a very low probability. At the solar core temperature it's relatively easy for 2 protons to overcome their electrostatic repulsion to fuse into a diproton. But the diproton is unstable and it usually decays into 2 protons before the relatively slow weak interaction has a chance to occur. May 14, 2020 at 8:15
• The probability of a weak interaction transforming a diproton into a deuteron is on the order of $10^{-26}$. That is, you have to fuse around a gram of protons to produce a single deuteron. May 14, 2020 at 8:15
• I don't quite get your last sentence, JEB. You didn't specify the units of that $10^{100}$, but I guess it doesn't make much difference at that scale if it's seconds, years, or the current age of the universe. ;) May 14, 2020 at 8:52
• @PM2Ring Ah, you do get it. I was trying to do an estimate in my head, but I forgot the timescale of heat-death, so plus or minus a factor of a googol, and we're still safe.
– JEB
May 14, 2020 at 15:29

A simple experimental fact about the LHC is that all reactions of protons on protons happen in vacuum. As the other answers explain, the energy in the beams is too high to allow the two protons to fuse into deuterons , instead the energy "fuses" to generate the particles studied in the experiments. The trajectories of these particles go through detectors depositing energy so as to be able to study the main interaction.

And as a corollary, can they repurpose CERN to become a fusion reactor ?

A reactor means that more energy will come out then spent in generating the beams, and this is not possible with the LHC design, which functions with high energies. There are experiments for colliding beam fusion, and proton proton collisions are not their choice.

(Disclaimer: I am not a physicist. Any needful corrections will be duly incorporated.)

The other explanations are good, but for those who want something a bit more at the layman's level I offer this:

If you are concerned that somehow an experiment will result in a runaway reaction that winds up destroying a lot of stuff, this should help you understand why this is not a concern.

The fusion reactions that go on in the center of stars (such as our sun) require immense temperatures and pressures, and these pressures have to be sustained long enough that the reaction takes place and that the energy from it is contained so that it can cause the more reactants nearby to react as well.

In the center of a star (such as our sun), this is not a problem. The temperatures and pressures are present, and are contained by the weight of all the stellar material that surrounds it. The energy release by the reaction is also contained to a high degree (some of it, after some thousands of years, escapes and becomes the light that shines on us). Also, there is plenty of unreacted material (free protons and electrons), which take part in the continuing reaction. It should also be understood that in the center of a star there is really no runaway reaction; things proceed at a very steady pace that is surprisingly slow. It will take billions of years for the Sun to get to the point where this reaction ceases.

In a hydrogen bomb--which is a runaway reaction--the process starts with a smaller nuclear explosion (from a fission device), which creates the initial temperature and pressure required for fusion, and there is a large quantity of fusion material (hydrogen and lithium, if I understand these things correctly), which is heated and pressurized by the fission bomb to the point that the fusion reactants can fuse. This is contained only by the inertia of the fusion material, but that does the job for as long as it needs to. Boom.

In the CERN reactor, the initial energy is enough to raise only a small quantity of material to the necessary temperatures, so there is less original energy to start with.

On top of this, whatever fusible material there is in the target is present in far less quantity. This means that there is less total fusion possible, and also far less momentary containment from the inertia of the target materials.

There is no other source of containment that will preserve the conditions required for the reaction to continue. Instead of being held in place to help the reaction continue, a lot of the release energy escapes. In fact, the chief reason we do not have nuclear fusion as a feasible power source is because we have no such means of containment.

As a consequence, instead of a whole lot of fusible material suddenly getting hot enough to fuse, and releasing a huge amount of energy from this, a small amount of material heats up, undergoes whatever reaction is possible and then explodes in a very tiny burst, and that's the end of the reaction. Unless I am mistaken, the firecrackers you can buy at many roadsides are more powerful. Its effects do not escape that environment.

• That it is not a runaway reaction is the key point. You most likely do have some exothermic nuclear reactions triggered by the collisions, and probably 2nd, 3rd ... nth generation ones as well, but there is no chain reaction possible (in these pandemic times we would say that the reproduction rate R is much smaller than 1). A similar question would be why we don't use the tremendous concentrated energy of lasers to produce steam for a power plant. The answer is that you put much more energy in the laser than you get out of it, which is not true for a fire which is a runaway process. May 13, 2020 at 16:57
• Edited. Better? May 13, 2020 at 18:42
• I liked it before as well; it is the key point. May 13, 2020 at 19:15
• It might be a bit misleading to talk about temperature in the context of the LHC beams. They are very high energy, but actually low temperature, because all the articles are moving in the same direction at once. Similarly, paragraph 6 implies that the energies in the CERN reactor are small, but they are huge on a per-mass basis... In general I find this answer more complicated than it needs to be and raises more questions than it answers.
– craq
May 14, 2020 at 5:11
• @EvilSnack Thank you for another clear analogy. However, the proton beams that intersect at the LHC are a constant source of high energy particles colliding. That would seem to be an ideal construct for producing a controlled and constant fusion reaction. As you pointed out, any fusion that happens at a collision rapidly disperses its energy and the particles and photons produced scatter away without containment. But then new particles come to occupy the collision zone to produce another ‘little fusion’. Of course, probably_someone’s comments about the energy being wrong is relevant. May 14, 2020 at 14:30

And as a corollary, can they repurpose CERN to become a fusion reactor ?

In theory, yes. However, accelerator-driven fusion systems, more generally known as colliding beam fusion, have a number of purely theoretical problems that suggest it will never be a energy-positive system.

As a purely experimental system for studying fusion, CERN would work. But so would much smaller machines that cost orders of magnitude less :-)