The Spallation Neutron Source (all details taken from this link) is described as firing a 1 GeV proton beam into a mercury target. As this makes the proton beam relativistic, a cyclotron cannot be used (not for all of it, anyway), and therefore a very large linac is used instead.

Could the setup be made smaller by accelerating mercury to 1 GeV and firing it into a hydrogen target? As mercury atoms have a mass of 186.85 GeV/c², I make the speed and γ at 1 GeV to be:

$$ \small {\begin{alignat}{7} && 1 \, \mathrm{GeV} &~=~ (\gamma -1) \cdot 186.85 \, \frac{\mathrm{GeV}}{c^2} c^2, \qquad γ = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\[2.5px] \therefore~~ && 1 \, \mathrm{GeV} &~=~ \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right) \cdot 186.85 \, \mathrm{GeV} \\[2.5px] \therefore~~ && \frac{1}{186.85} &~=~ \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1 \\[2.5px] \therefore~~ && 1 + \frac{1}{186.85} &~=~ \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \\[2.5px] \therefore~~ && \sqrt{1-\frac{v^2}{c^2}} &~=~ 1/(1 + \frac{1}{186.85}) \\[2.5px] \therefore~~ && 1-\frac{v^2}{c^2} &~=~ \frac{1}{\left(1 + \frac{1}{186.85}\right)^2} \\[2.5px] \therefore~~ && -\frac{v^2}{c^2} &~=~ \frac{1}{\left(1 + \frac{1}{186.85}\right)^2}-1 \\[2.5px] \therefore~~ && \frac{v^2}{c^2} &~=~ -\frac{1}{\left(1 + \frac{1}{186.85}\right)^2}+1 \\ && &~=~ 1-\frac{1}{\left(1 + \frac{1}{186.85}\right)^2} \\[2.5px] \therefore~~ && v^2 &~=~ c^2 \left(1-\frac{1}{\left(1 + \frac{1}{186.85}\right)^2}\right) \\[2.5px] \therefore~~ && v &~=~ \sqrt{c^2 \left(1-\frac{1}{\left(1 + \frac{1}{186.85}\right)^2}\right)} \\[2.5px] \therefore~~ && v &~≅~ 0.103 c, \qquad γ ≅ 1.0053471 \end{alignat}} $$

which is close to what I've been told is the relativistic limit for a cyclotron (I think on the lower side of the limit, but can't remember).

However, even if it's on the wrong side of that limit: a heavy element like mercury can be multiply-ionised, and even fully ionising mercury requires far less energy than accelerating it to that speed, so even a linac with that design could be 20 times shorter. If spallation works like that.

I'm mainly thinking of this in terms of a spacecraft powered by an accelerator-driven subcritical reactor, and reducing the linac length from 335 m to 16.75 m seems like a significant improvement for such a task.

[My physics level is {UK: AS-level equivalent, USA: probably highschool, I'm not sure}; my maths level is {UK: double-A2-level, USA: probably between freshman and somophore year at university, but I'm not sure}, please target answers at that sort of level]

  • 4
    $\begingroup$ You need to think in the center of mass frame, not the lab frame. The difference becomes clear when you do. $\endgroup$
    – Jon Custer
    Jun 15, 2018 at 14:55
  • $\begingroup$ @JonCuster center-of-mass-frame being the frame in which momentum is exactly zero? Or have I misunderstood? $\endgroup$
    – BenRW
    Jun 15, 2018 at 14:58
  • $\begingroup$ Related: physics.stackexchange.com/a/77999/44126 $\endgroup$
    – rob
    Jun 15, 2018 at 15:23
  • 2
    $\begingroup$ Yes, in the centre of mass frame the spatial components of the total momentum are zero. You can easily get the energy in the centre of mass frame by calculating the total momentum and using the fact that $p^\mu p_\mu = E^2 - p^2$ is invariant under Lorentz transformations. $\endgroup$ Jun 15, 2018 at 15:36

1 Answer 1


First, you have to analyze the collision in the center-of-momentum reference frame. Since the mercury nucleus is eighty times more massive than the proton, simply swapping the kinetic energies doesn't give you the same interaction. This is a famous introductory physics problem that I'll let you think about on your own.

Furthermore, in spallation on heavy-metal targets, the typical neutron yield is twenty or thirty neutrons for each proton entering the target. That suggests that, when the proton beam interacts with the metal target, each proton interacts with perhaps ten of the metal nuclei. If you were to accelerate the mercury nuclei, you would lose this advantage: you'd have to send each mercury nucleus into the hydrogen target with enough energy to drive the spallation reaction, and any secondary reaction would be between fast mercury fragments and protons at rest rather than between a still-fast proton and another intact mercury nucleus.

You might be interested to know that the spallation source at PSI, which like the SNS is a one-megawatt machine, is driven by a half-GeV cyclotron.

For beam-based space propulsion, you might read about ion drives.

  • $\begingroup$ I've just realised my phrasing was ambiguous; I meant an accelerator-driven subcritical reactor as a power source. This has definitely helped me think about the problem in a better way, but I am still confused by what you say about proton-beam-interacting-with-mercury being different from mercury-beam-interacting-with-proton, as if I correct the numbers for the correct frame of reference I would still expect them to be equivalent. $\endgroup$
    – BenRW
    Jun 15, 2018 at 16:09
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    $\begingroup$ Oh, a subcritical reactor is a different beast from a spallation source. At a spallation source the goal is to get the neutrons out; in an accelerator-driven reactor the goal is to produce a few extra neutrons to stimulate fission. In the reactor case you'd have to use a fissionable target, like uranium (as at the IPNS) or thorium, and the concerns about having the heavy nuclei near each other are even more important. You could never get a net energy gain by accelerating a beam of uranium nuclei onto a hydrogen target. $\endgroup$
    – rob
    Jun 15, 2018 at 16:22

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