So I know the basic gist is that fusion power's main issue is sustaining the fusion. I also know that there are two methods. The Torus method and the laser method. The torus magnetically contains plasma and heats it with radiation and accelerates the plasma around to make strong enough collisions that protons fuse. The laser method uses 192 lasers and focuses it on tiny frozen hydrogen pellets and aims to initiate fusion each time pellets are dropped.

The though struck me when we could sorta combine the two designs together. The torus doesn't have to worry about making fusion happen at a specific location but it has issues in that the plasma is unevenly heated and leaks. On the other hand, the laser design is extremely complicated in the level of precision needed and would have to repeat this for every pellet. This lead me to think to make something precise and contained at the same time.

I see that particle colliders are able to direct two beams of protons and have them collide at a specific spot with a very precise energy. Couldn't we tune the energy of the two beams of protons to the energy required for them to fuse? We have the ability to smash them into bits, surely we have the ability to have them fuse. (I'm thinking about the type of collider that circles two beams in opposite directions)

It would be at much lower energies than normal colliders and would be very precise and it would be possible to fuse at a specific location that has greater leeway because for protons that missed collision, they'd just circle around again! Thus protons would efficiently be used and very little would be wasted. There wouldn't be problems of plasma leakage because we are focusing them in a thin tight beam.

It seems that this idea has girth, or I feel this way at least, can someone back me up by offering some calculations on how to calculate the efficiency? How would I go about calculating the two circling beams of protons and at what specific velocity would be needed? etc.


4 Answers 4


A subtle problem you seem to overlook is that the proton-proton cross section is very small, about 0.07 barns (a barn is $10^{-28}$ square meters) at the LHC energies and not dramatically different at your lower "fusion energies". It means that at the LHC, much like at your dream machine, most of the protons simply don't hit their partners. It is not really possible to focus the proton beams arbitrarily accurately, for various reasons (the uncertainty principle is the truly unavoidable effect: you either localize the beams in the transverse direction, into a "thin pipe", or you specify that the velocity in this transverse direction is zero which is needed if you want to preserve the location "in the thin pipe" in the future, but you can't do both at the same moment). If it were possible, the LHC would be among the first ones that would use the method, to increase the luminosity.

So if you accelerate two beams of protons against each other, an overwhelming majority of them will simply continue in their original motion. (The protons in the LHC have to orbit for half an hour or so – tens of millions of revolutions – before one-half of them collides or disappears.) It costs some energy to accelerate the protons to these energies and you want this energy to be returned from fusion, with some bonus. But the fusion only returns you the energy from the protons that collided (some of them could create helium at your energies but there will always be nonzero probabilities of other final states; it's not a deterministic system that always produces the same final state for a given initial state; quantum mechanics says that the outcomes are random) which is a tiny portion of the protons. So you will be losing most of the energy you invested for the acceleration. Note that the LHC consumes as much energy as the households in Geneva combined and it just produces collisions of protons whose energy is smaller than a joule per pair.

To increase the fraction of the protons that hit their partners, you either need to send them to the collision course repeatedly, like at the LHC, but then you need to pump extra energy to the protons that they lose by the synchrotron radiation (which is always nonzero if the acceleration vector is nonzero, e.g. for all circular paths). Or you will need to dramatically increase the density of the beams.

But if there are many protons in the beam, they will electrically repel each other and you will become unable to focus them for collisions, too. So what you need to do is to electrically neutralize the high-density proton beam and then you have nothing else than the plasma and you face the usual tokamak problems how to stabilize it. Note that the electrons respond totally differently to the external electromagnetic fields than the protons. The LHC uses both electric and magnetic fields to accelerate the protons but to keep the plasma neutral, you must avoid electric fields.

Tokamaks only work with magnetic fields. Whether they will ever become fully working and feasible remains to be seen but the absence of the electric fields implies that they don't have much in common with particle accelerators.

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    $\begingroup$ Wow. This is a great answer! I was missing a lot of factors that go into it like you've pointed out. However, I didn't know that the collisions of the protons are smaller than a joule per pair. This seems confusing. $\endgroup$ Nov 2, 2012 at 19:37
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    $\begingroup$ Dear QEntanglement, nice to have helped you, perhaps. The total center-of-mass energy at the LHC is 2x 4 TeV which is 8 TeV right now. T means trillion, $10^{12}$, and eV means $1.602\times 10^{-19}$ joules. So the product is about $8 \times 10^{12}\times 1.602\times 10^{-19} = 10^{-6}$ joules per proton-proton pair. However, one must appreciate that the LHC has already organized over a quadrillion, $10^{15}$, proton-proton collisions, so it's $10^{9}$ joules just in the collisions. Still, most of the energy that went to the LHC was "lost" by the synchrotron radiation etc. $\endgroup$ Nov 2, 2012 at 19:41
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    $\begingroup$ BTW the LHC consumes about 1,000 GWh per year, see lhc-machine-outreach.web.cern.ch/lhc-machine-outreach/faq/… - that's $10^{12}$ times 3600 joules, i.e. $3.6\times 10^{15}$ joules per year. The total energy in the protons that have collided, see the previous comment, is roughly millions of times smaller: 99.9999% of the energy was "lost". This hapless efficiency can't be improved too much. $\endgroup$ Nov 2, 2012 at 19:56
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    $\begingroup$ This answer replies to the question asked, of course. the link @AlanSE provided in his now deleted answer offers an alternative to laser ignition by particle accelerator ignition, and is an interesting concept by itself , (the balance of energy in /energy out is different than in this answer) so I am giving the link for completion . fusionpowercorporation.com $\endgroup$
    – anna v
    Nov 3, 2012 at 5:38
  • $\begingroup$ So basically, they can't hit the broad side of a barn? $\endgroup$
    – Aloha
    Feb 4, 2016 at 15:16

Actually, this has been done, but it's not sustainable. Wikipedia has a brief explanation:

Accelerator-based light-ion fusion is a technique using particle accelerators to achieve particle kinetic energies sufficient to induce light-ion fusion reactions. Accelerating light ions is relatively easy, and can be done in an efficient manner—all it takes is a vacuum tube, a pair of electrodes, and a high-voltage transformer; fusion can be observed with as little as 10 kV between electrodes. The key problem with accelerator-based fusion (and with cold targets in general) is that fusion cross sections are many orders of magnitude lower than Coulomb interaction cross sections. Therefore the vast majority of ions end up expending their energy on bremsstrahlung and ionization of atoms of the target.

Basically, a particle accelerator provides many ways for the accelerating particles to lose energy that are unrelated to the actual collision.

Take the LHC as an example. It consumes over 200 MW of electricity, and generates up to 600 million collisions per second. If the particles being accelerated were deuterium and tritium nuclei, each fusion event would release about 17 MeV of energy. So if somehow every single one of those 600 million collisions per second was a successful D-T fusion event, the total power generated through fusion would be

$$\frac{17\text{ MeV}}{\text{collision}}\times \frac{6\times 10^8\text{ collisions}}{\mathrm{s}} = 1.6\text{ mW}$$

for an efficiency of 0.0000001%. That's not very good at all. A sustained nuclear fusion reactor needs to have an efficiency of greater than 100%.

Granted, most of the energy consumed by the LHC goes into giving the protons or ions kinetic energy, which isn't necessary for fusion. If you subtract off the 8 TeV of energy per proton, that could raise the efficiency by a factor of about a million (bearing in mind that there are still many reasons that what I'm talking about is impossible), but it'd still only be around 0.1%.

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    $\begingroup$ The LHC is an abhorrent example as it accelerates ions to energies dramatically greater than what a fusion reaction produces. Fusion cross-sections have an optimal energy, increasing it further can only hurt. These mechanisms by which the ions lose energy depends on the design and is unclear here. $\endgroup$ Nov 2, 2012 at 19:53
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    $\begingroup$ The LHC was identified in the question, plus it was the easiest accelerator to find information on. And it's only meant to demonstrate how the power produced would only be a small fraction of the required energy input, which is something that one could expect to hold for any accelerator. If someone asked a question about a specific lower-energy accelerator, then I suppose we could dig up some numbers for that case as well. $\endgroup$
    – David Z
    Nov 2, 2012 at 20:16
  • $\begingroup$ The question mentions colliding proton beams. The LHC is optimized to produce the Higgs particle, and maybe others, while this question is about a machine optimized to produce energy. I see some direct references to colliding ion beams including N. Rostoker (1998), but the applicability is still limited, as they suggest FRC to perform the act of recycling the ions, as opposed to the OP. The energy input of the LHC is not congruent to a lower energy accelerator, particularly when the loss is EM radiation! Surely an accurate president is out there, although I do not know of a good one. $\endgroup$ Nov 2, 2012 at 20:26
  • $\begingroup$ The question has the LHC tag. $\endgroup$
    – David Z
    Nov 2, 2012 at 20:28

The problem is that the power demsity of fusion is in the range of 10^20 watts per kilogram. To achieve this requires a magnetic field of > 12 tesala to compress the plasma and increase the fusion rate to a self sustaining reaction.

This is the goal of the ITER fusion project. To achieve this a new tube of power vacuum is being developed to drive the currents into the superconducting magnets.

The power density of the fusion I driver must be higher fusion. Many scheme fail be attempting to parallel multiple low power density devices. This was the issue with the NIF Laser system---the 'noise' adds up faster then the power.

The Sun uses gravity to compress the plasma in the Stars--of course a fission device is used to compress the fusion in an H-bomb; which has a power density much higher than fusion.

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    $\begingroup$ "The power density of the fusion I driver must be higher fusion." <- This sentence it is unclear a little bit. $\endgroup$
    – peterh
    Aug 14, 2016 at 10:14

A typical fusion reaction liberates about $10~$MeV of energy, therefore in order to obtain a moderate (thermal) power of $1~$GW we need about $6\times10^{20}$ reactions per second. The number of particles required in each of the two beams we be few order of magnitudes larger as we will manage to fuse only a fraction of them, say $1/100$. Assuming that these particles have a charge of $1~e$, we get an average beam current of $10~$kA.

We have no clue of how to provide that.

To give you an idea, to my knowledge, a proton injector never went above $100~$mA. We are several order of magnitudes away from even just a concept.


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