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The root mean square speed of an electron in the ground state of hydrogen is very close to the fine structure constant times the speed of light, approximately

$$ \frac{c}{137}. $$

This should also be true for the muon replacing the electron in muonic hydrogen. So, I'd expect the muon's half-life to be affected by time dilation. If I'm right, it should be about

$$ \frac{1}{\sqrt{1 - \frac{1}{137^2}}} \approx 1.0000266 $$

longer than usual. I have two questions, really: is this about right, and has anyone been able to measure this small effect?

When I say "about right", I'm suggesting that it's not exactly right, because I see no reason to compute the expected time dilation by using the root mean square speed: the square root of the expected value of the velocity squared. But it might still be a pretty good estimate.

By the way, here's how you compute the root mean square speed of an electron in its ground state in hydrogen. First you compute the expected value of its kinetic energy, using Schrödinger's equation. This takes real work, but it's done in the first answer here:

The answer is

$$ \langle K \rangle = \frac{\hbar^2}{2ma^2} $$

where $a$ is the Bohr radius. But there is a 'velocity squared' operator $v^2$ in quantum mechanics such that

$$ K = \frac{mv^2}{2}. $$

(It's defined by $v^2 = p^2/2m$ where $p$ is the usual momentum operator.) So, we get

$$ \langle v^2 \rangle = \frac{\hbar^2}{m^2 a^2}. $$

But the Bohr radius is

$$ a = \frac{\hbar}{m c \alpha} $$

where $\alpha$ is the fine structure constant, so a little calculation gives

$$ \langle v^2 \rangle = \alpha^2 c^2 $$

and the square root of this is $\alpha c$. The final result doesn't depend on the mass of the electron so it also applies to the muon.

Here I'm treating both hydrogen and muonic hydrogen using Schrödinger's equation, which is not exact but should be close to right.

David Bailey
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John Baez
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  • Yet the electron is not a little planet zooming around the nucleus. As a counter example, the binding energy of hydrogen is 13.6eV, the rest mass of an electron is 911keV, so I conclude that the electron in hydrogen is nowhere near relativistic. – Jon Custer Jan 18 '22 at 21:05
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    Indeed it's nowhere near relativistic! I showed you using quantum mechanics that according to Schrodinger's equation, the expected value of the square of the speed of the electron in the ground state of hydrogen is the fine structure constant times c - that is, about $c/137$. This has nothing to do with little planets zooming around. (Btw, the electron mass is 511 keV/$c^2$.) – John Baez Jan 18 '22 at 23:21
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    probably irrelevant but besides $\gamma(\langle v\rangle)$ maybe one could also try $\langle \gamma(v)\rangle$ – Dabed Jan 21 '22 at 15:12
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    I bet that's more accurate; they should be pretty close for energy eigenstates of hydrogen but not for heavy elements. I got into this because a crude calculation says "the innermost electrons of mercury move at 58% the speed of light". Then someone asked me about time dilation for muonic hydrogen and I realized I didn't know whether it's been studied. – John Baez Jan 21 '22 at 16:29
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    Muonic hydrogen is too light to see the effect, but time dilation has been observed for very heavy muonic atoms. The answer to Would superheavy muonic atoms be more stable than light muonic atoms such as muonic helium (hydrogen 4.1)? maybe be helpful. – David Bailey Oct 06 '22 at 04:47
  • This is a great question (I know I already upvoted but just to reiterate I'll say it here). And probably can only be answered by someone who works in bound state QED---and further someone who has a really deep conceptual understanding of this work and could calculate something that typically nobody ever calculates. Could we not put comments saying "but this effect is small! you could never measure it!" Says who? what's the precision limit on measuring the lifetime of the muon? It's lifetime is currently known to 7 digits. And the question may be conceptual, so the feasibility doesn't matter. – AXensen Jan 05 '24 at 23:28
  • @DavidBailey this is a useful comment but just to prevent this from getting closed as a duplicate... the answer there dodges the conceptual question and the conceptual question remains unanswered. – AXensen Jan 05 '24 at 23:31
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    Concretely, one can perform this calculation by evaluating the matrix element $\langle e^- \bar{\nu}e \nu\mu | L_{\mathrm{eff}} | \mu^- \rangle$ where $|\mu^- \rangle$ is the initial state, and $L_{\mathrm{eff}}$ is the four-fermion operator mediating the decay, and then integrating over final state momenta. Standard QFT tells us this yields a time dilated decay rate for a moving free muon, as expected. For a bound muon, the matrix element will involve an integral over its wavefunction. I'm 99% sure it'll produce the expected time dilation effect, but haven't checked since it seems clunky. – knzhou Jan 05 '24 at 23:44
  • Related: https://physics.stackexchange.com/questions/730311/would-superheavy-muonic-atoms-be-more-stable-than-light-muonic-atoms-such-as-muo?noredirect=1&lq=1 – hft Jan 06 '24 at 03:41

1 Answers1

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Your calculation for the time dilation effect is correct, but the muon in muonic hydrogen actually has a shorter lifetime than a free muon because the time dilation lifetime increase is much less than the reduction due to $\mu^- p \rightarrow n \nu_\mu$ captures. We are close to, but do not yet have, sufficient experimental and theoretical accuracy to observe the time dilation effect in muonic hydrogen. As noted in this answer, however, the time dilation effect has likely been observed in heavier muonic atoms.

Your calculation of the time dilation effect agrees with those of Überall ("Decay of μ− Mesons Bound in the K Shell of Light Nuclei", Eq. 44b) and Silverman ("Relativistic time dilation of bound muons and the Lorentz invariance of charge", Eq. 3).Überall actually has $1-\alpha_{\mathrm{QED}}^2/2$ after a more rigorous calculation instead of $\sqrt{1-\alpha_{\mathrm{QED}}^2}$, but they are equivalent to this order. The change in the muon decay rate due to time dilation in muonic hydrogen is: $$\Delta_{\mathrm{rel}}^{\mathrm{th}}=\frac{\sqrt{1-\alpha_{\mathrm{QED}}^2}-1}{\tau_\mu}=\frac{-2.663 \times 10^{-5}}{2.196981 \times 10^{-6}\,\mathrm{s}}=-12.1 \,\mathrm{s^{-1}}$$

I believe the best experimental measurement of the muonic hydrogen capture rate ($\Lambda_S$) is MuCap Collaboration's paper on "Measurement of Muon Capture on the Proton to 1% Precision and Determination of the Pseudoscalar Coupling $g_{P}$":

$$\Lambda_S^{\mathrm{exp}}=714.9 \pm 5.4\,(stat) \pm 5.1\,(syst) \,\mathrm{s^{-1}}$$

So the decreased decay rate ($\rightarrow$lifetime increase) due to time dilation is 60 times less than the increased decay rate ($\rightarrow$lifetime decrease) due to muon capture by the hydrogen nucleus. At the time of the MuCap measurement in 2013, the expected value for the capture rate, based on earlier experimental and theoretical values, was $$\Lambda_S^{\mathrm{th}}=712.7 \pm 3.0 \pm 3.0 \,\mathrm{s^{-1}}$$ in good agreement with experiment: $$\Lambda_S^{\mathrm{exp}}-\Lambda_S^{\mathrm{th}}=2.2 \pm 6.2\,(stat) \pm 6.0\,(syst) \,\mathrm{s^{-1}}$$

I can't immediately tell if time dilation is included in the theoretical calculation. (The authors are certainly familiar with the effect, since they have an earlier paper on "Muonium Decay" where the conclusion is that all other effects except time dilation cancel out.) It doesn't really matter either way, however, since the uncertainty is comparable to the effect, so we don't yet have sufficient accuracy to confirm or exclude the contribution from time dilation.

David Bailey
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