Tunneling That ClicksBonus ② / The real cold fusion ── muon catalysis

There is exactly one cold fusion that genuinely happens at room temperature ── it works, and still it does not pay

The real cold fusion ── muon catalysis Replace the electron with a muon and the molecule shrinks by a factor of 207.
The barrier thins and fusion occurs in picoseconds, not microseconds.
What stops it is not the barrier ── the muon sticks to the helium it just made.

Tools you'll need: the mass dependence of tunneling (Episode 1), the Gamow factor (Episode 2), bonus ① The heart of this episode: retirement after about 150 cycles, break-even needs about 300

Bonus ① said "if the electron were about 8 times heavier, cold fusion would happen." So what if you actually use a heavier particle? The answer has been known for a long time ── it happens. The muon is 207 times heavier than the electron; put one into a deuterium molecule and the molecule shrinks by a factor of 207, bringing the nuclei within 500 fm of each other, and they fuse within picoseconds. And this is not theory: Alvarez found it by accident in a bubble chamber in 1956. Fusion at the temperature of liquid hydrogen, i.e. 20 K ── cold fusion in the literal sense. And yet, seventy years on, it is not generating power. What stops it is not the barrier but something far more mundane. This episode is about that remaining factor of two.

01The muon is a heavy, short-lived electron

electronmuon
mass0.511 MeV105.7 MeV (206.8×)
lifetimestable2.197 microseconds
charge and spin−e, 1/2−e, 1/2 (same as the electron)

Since its charge and spin match the electron's, a muon can stand in for an electron inside an atom. But it is heavy. The Bohr radius is inversely proportional to mass, so ──

The molecule shrinks by a factor of 207
Moleculeinternuclear distance
ordinary D₂ (electrons)74,000 fm (0.74 Å)
ddμ (a muon)about 500 fm
the range of the nuclear forcea few fm

Still more than a hundred times the range of the nuclear force ── but for tunneling that is plenty. Episode 1's \(S_E=\int\sqrt{2m(V-E)}\,dx\) is an integral over distance, so cutting the distance by a factor of 150 shrinks the exponent dramatically.

02The result ── fusion in picoseconds

Fusion rates compared (per pair per second)
Systemfusion rateMeaning
a D₂ molecule (electrons)\(10^{-64}\)would not happen if you waited \(10^{47}\) times the age of the universe
ddμ\(\sim10^{9}\)nanoseconds
dtμ (deuterium + tritium)\(\sim10^{12}\)picoseconds. Effectively instantaneous

From \(10^{-64}\) to \(10^{12}\) ── a leap of 76 orders of magnitude. Bonus ① calculated that 53 orders were needed. The muon overshoots the requirement by 23 orders. Which is why it happens with room to spare.

Discovered by accident The prediction goes back to Frank in 1947 and Sakharov in 1948 (unpublished at the time), but the actual discovery was an accident. In 1956 Alvarez's group, watching cosmic-ray muons in a liquid-hydrogen bubble chamber, found tracks they could not explain. Following them up: a muon had shrunk a hydrogen molecule, caused fusion, and come back out.
Alvarez is said to have thought for a moment that he had found an unlimited source of energy. He realised almost immediately that the question was how many times one muon can work ── and that calculation is the rest of this episode.

03The catalytic cycle

It is called catalysis because the muon is not consumed and can be used repeatedly.

  1. A muon enters a deuterium/tritium mixture and is captured by an atom
  2. It pulls in the other nucleus to form a dtμ molecule (the rate-limiting step, about \(10^{-8}\) s)
  3. The nuclei are close, so they fuse immediately: \(d+t\to{}^4\mathrm{He}+n+17.6\) MeV (\(10^{-12}\) s)
  4. The muon is released and returns to step 1

One cycle takes about \(10^{-8}\) s, so within the muon's 2.2 μs lifetime it could in principle go round about 220 times.

04What stops it ── two walls of almost the same height

Two limits on how many times one muon can work
$$N=\frac{1}{\ \omega_s+\dfrac{1}{\lambda_c\tau_\mu}\ }$$

① α-sticking, \(\omega_s\) ── the helium (\(\alpha\)) produced by fusion has charge +2 and attracts the muon strongly. With a probability of about 0.45–0.6% the muon is captured by the α and carried away. Some are stripped back off in flight (reactivation); this figure is the net.
→ On its own that caps things at \(1/0.0045\approx220\) cycles.

② the muon lifetime \(\tau_\mu=2.2\) μs ── at a cycling rate \(\lambda_c\approx10^8\)/s, that is \(\lambda_c\tau_\mu\approx220\) cycles within the lifetime.

Remarkably, these two are almost the same size. Together they give \(N\approx110\). The best experimental value is about 150 cycles (with density and temperature optimised). By coincidence, nature has placed the two limits at the same height.

So how many are needed?

Making one muon costs, realistically, about 5 GeV in an accelerator (about 50 times the muon's rest mass of 106 MeV, because production is inefficient).
One d+t fusion yields 17.6 MeV. Therefore:

$$N_{\text{break-even}}=\frac{5000\ \mathrm{MeV}}{17.6\ \mathrm{MeV}}\approx 284\ \text{cycles}$$

And since converting heat back to electricity is only about 40% efficient, running an actual power plant in the black needs of order 700 cycles.
Achieved: 150. Needed: 300–700. Not orders of magnitude ── a factor of two to five. That distance has not closed in seventy years.

05Try it ── the last factor of two will not come

The figure is that balance sheet. Move the sticking probability, the cycling rate and the muon production cost and it shows how many times one muon works and how the books come out.

Look at the distance between the current state of the art (the blue point) and break-even (the green line). Whichever knob you turn, the line is hard to cross ── halving the sticking would do it, but that is a number fixed by nuclear physics and not ours to move.

Figure: the number of cycles one muon can achieve, N = 1/(ω_s + 1/λ_cτ_μ). Horizontal = α-sticking probability (log). Blue = the lifetime limit alone, red = the sticking limit alone, thick line = both. The green line is break-even. The distance to it is the "factor of two to five"
cycles actually achievable, N the sticking limit alone the lifetime limit alone break-even

06As science, it is a complete success

Failing to become a power source does not diminish the phenomenon. Muon-catalysed fusion is the demonstration that "thin the barrier and fusion happens even at room temperature."

Relation to bonus ① ── this is the baseline
Methodeffective massfusion rateVerdict
an ordinary electron×1\(10^{-64}\)does not happen
screening in a metal lattice×1.0–1.1of order \(10^{-60}\)not enough
(needed for 1 W)×8\(10^{-11}\)──
a muon×207\(10^{12}\)it happens (demonstrated)

Because muon catalysis exists, bonus ①'s argument can be stated in the far stronger form "not impossible in principle, merely insufficient in quantity." "Thin the barrier and it happens" is correct. The problem is that a lattice does not have anything like the muon's power.

And ironically ── the real one, having completely solved the barrier problem, is stopped for an entirely different reason (sticking and lifetime). Bonus ①'s lesson, "solve the entrance and you get stuck somewhere else," repeats here too.

◇ ◇ ◇
The honest line

Established: the muon's mass ratio of 206.77 and lifetime of 2.197 μs; that muon substitution shrinks the molecule by about a factor of 207; that the dtμ fusion rate is of order \(10^{12}\)/s; the predictions of Frank (1947) and Sakharov (1948) and the observation by Alvarez and colleagues (1956); the structure of the catalytic cycle and a net α-sticking probability of about 0.4–0.6%; that cycling rates of order \(10^8\)/s are reached at high density; that experiments observe of order 100–150 fusions per muon (the series of experiments by Jones and others); and the 17.6 MeV released by d+t. All established experimental facts.

Ranges and caveats: (1) The 5 GeV muon production cost is approximate. Depending on accelerator design it ranges from about 2 to 10 GeV, and that directly sets the break-even estimate; moving the slider shows the effect. (2) The sticking probability depends on temperature and density; the "effective" value used here includes reactivation by stripping in flight. (3) The break-even cycle count depends on assumptions about heat-to-electricity conversion and how the neutrons are used (breeding blankets and so on). The 284 in the body is the minimum line for recovering the muon production energy alone; an actual plant needs more margin. (4) The cycling rate and the sticking probability cannot in fact be varied independently (both depend on density and temperature); the figure simplifies them into independent knobs. (5) Concepts other than direct power generation exist (hybrid reactors triggered by muon catalysis to drive fission, for instance), but none has been made practical.

Exercises
  1. Why does replacing the electron with a muon make fusion happen?
    See the answer
    The Bohr radius is inversely proportional to mass, so the molecule shrinks by a factor of 207 (74,000 fm → about 500 fm). Episode 1's \(S_E=\int\sqrt{2m(V-E)}\,dx\) is an integral over distance, so the exponent shrinks dramatically and the fusion rate leaps from \(10^{-64}\) to \(10^{12}\)/s ── 76 orders.
  2. Why is it called a catalyst, and why is it not a perfect one?
    See the answer
    Because the muon is released after fusion and can be reused. But with a probability of about 0.45% it is captured and carried off by the helium (charge +2), so it is lost after about 200 cycles on average. A perfect catalyst would work forever; this one does not.
  3. Which matters more, sticking or the lifetime?
    See the answer
    They matter about equally. Sticking alone caps it at \(1/0.0045\approx220\) cycles; the lifetime alone at \(\lambda_c\tau_\mu\approx220\). Together, \(N\approx110\). Improve one and the other is waiting ── that is the essence of the difficulty.
  4. How does this episode reinforce bonus ①'s verdict?
    See the answer
    It demonstrates that thinning the barrier makes fusion happen even at room temperature. So bonus ①'s conclusion can be put as a quantitative claim ── not "impossible in principle" but "lattice screening is nowhere near the muon." Needed ×8, lattice ×1.1, muon ×207.

Bonus ② summaryIt works. And still the last factor of two to five will not come

The muon is 207 times heavier than the electron and shrinks the molecule by the same factor. The internuclear distance becomes 500 fm and the fusion rate leaps from \(10^{-64}\) to \(10^{12}\)/s ── 76 orders. Fusion in picoseconds. A real cold fusion, discovered by accident by Alvarez in a bubble chamber in 1956.

What stops it is not the barrier. ① the muon sticks to the helium it just made (about 0.45%, capping at 220 cycles) and ② the muon's 2.2 μs lifetime (also capping at about 220 given the cycling rate) ── two limits at almost the same height, giving 100–150 in practice. Meanwhile a muon costs 5 GeV to make, so break-even is about 284 cycles and a power plant needs of order 700. Not orders of magnitude but a factor of two to five. And it has not closed in seventy years.

As science it is a complete success, and it also provides the baseline for bonus ① ── "thin the barrier and it happens" is correct; a lattice simply does not have the muon's power. And the real one, too, is stuck somewhere else after solving the entrance. That repeated shape is the real lesson of this field.

This document is Bonus ② of the "Tunneling That Clicks" series, a reading piece for physics-loving high-schoolers and undergraduates. The muon's mass ratio and lifetime, the shrinking of the molecule under muon substitution, the dtμ fusion rate, the predictions of Frank (1947) and Sakharov (1948) and the observation by Alvarez and colleagues (1956), the catalytic cycle and net α-sticking of about 0.4–0.6%, the cycling rate at high density, the 100–150 fusions per muon observed experimentally, and the 17.6 MeV from d+t are all established experimental facts. That the 5 GeV production cost is an approximation depending on accelerator design, that the sticking probability is an effective value depending on temperature and density and including reactivation, that the break-even cycle count depends on conversion-efficiency and neutron-utilisation assumptions, that the figure simplifies the cycling rate and sticking into independent knobs, and that hybrid-reactor concepts have not been made practical ── all spelled out in the body's "honest line." ── To print, use your browser's "Print" and "Save as PDF" (in the print version the sliders and answers are frozen and hidden). Adjacent: Bonus ①, Does cold fusion happen? / Bonus ③, You can see a single atom because it is an exponential / Contents.

Print / make a PDF: ⌘+P (Ctrl+P on Windows). On screen, move the three knobs to see which of them, improved how far, would reach break-even. The sticking probability is fixed by nuclear physics and not ours to move. "See the answer" opens each solution.