# Spallation Neutron Source design

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]

• You need to think in the center of mass frame, not the lab frame. The difference becomes clear when you do. – Jon Custer Jun 15 '18 at 14:55
• @JonCuster center-of-mass-frame being the frame in which momentum is exactly zero? Or have I misunderstood? – BenRW Jun 15 '18 at 14:58
• – rob Jun 15 '18 at 15:23
• 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. – Sebastian Riese Jun 15 '18 at 15:36