From the PDG (https://pdg.lbl.gov/2008/listings/s031.pdf & https://pdg.lbl.gov/2010/listings/rpp2010-list-Ds-plus-minus.pdf) the branching rate for strange D mesons to tau leptons is about 3 orders of magnitude greater than that for D mesons. Why is this? I've thought of a few things, but I'm not completely confident.

The Ds mass is slightly larger than the D mass, and this means that it can produce a pion as well as a tau. The lighter D meson can't do this, so I assume what must happen is that the c quark turns into a d quark through the emission of the W boson, and this d quark annihilates with the remaining antidown quark in the meson to give photons. Because this is an electromagnetic process, it is slower than the strong process in the Ds. It also involves c to d and is Cabibbo suppressed.

My second thought is related. The Ds meson could also decay in this manner, except now the c quark could turn into an s quark, which can then annihilate with the anti-s. This is Cabibbo favoured.

Are these valid reasons, and is there more to be said?

References if link dies:

Nakamura, N.; et al. (Particle Data Group) (2010). " D± s" (PDF). Particle listings. Lawrence Berkeley Laboratory.

Amsler, C.; et al. (Particle Data Group) (2008). " D± " (PDF). Decay modes. Lawrence Berkeley Laboratory.

  • $\begingroup$ I would toss the π semileptonic mode as it is so close to threshold to be unobservable, no? Dividing the BRs by the lifetimes to compare widths, we are down by 2, not 3 orders of magnitude. The square of the Cabbibo cotangent for the $D_s$ gives you a factor of 18 in the ratio. The extra factor of 5 is easily accounted for by the free momentum in the phase space, 182 vs 90 MeV. So there really is no mystery there... $\endgroup$ Commented Jun 4, 2023 at 20:40

1 Answer 1


It is a combination of Cabbibo suppression and phase space.

Because the $\tau$ lepton is so heavy, semileptonic ($D\rightarrow \tau \nu + \mathrm{hadrons}$) are absolutely forbidden for $D^\pm$ mesons and so heavily suppressed for $D^\pm_s$ that they have never been observed accorded to the current PDG Listings. So the only decays involving the $\tau$ are purely leptonic decays, their decay widths can be calculated in the Standard Model, giving branching fractions

$$ B(D^\pm_s\rightarrow l\nu) = \frac{G^2_F}{8\pi}\tau_{D^\pm_s} f^2_{D^\pm_s}\left|V_{cs}\right|^2 m_{D^\pm_s} m_l^2 \left(1-\frac{m^2_l}{m^2_{D^\pm_s}}\right)^2 $$

$$ B(D^\pm\rightarrow l\nu) = \frac{G^2_F}{8\pi}\tau_{D^\pm} f^2_{D^\pm}\left|V_{cd}\right|^2 m_{D^\pm} m_l^2 \left(1-\frac{m^2_l}{m^2_{D^\pm}}\right)^2 $$

The decay rates are proportional to the mass of the charged leption ($m_l$) because the left-handed nature of the Weak interaction means that purely leptonic pseudoscalar meson decays into relativistic charged leptons of pseudoscalar mesons such as the $D^\pm$ and $D^\pm_s$ are heavily suppressed as explained in this answer about pion decay.

If we make a quark-model approximation the decay constants are equal, i.e. assume $f_{D^\pm}=f_{D^\pm_s}$, then using PDG values we expect the ratio of branching ratios to be

$$ R_{th}\equiv\frac{B(D^\pm_s\rightarrow l\nu)}{B(D^\pm\rightarrow l\nu)} = \frac {\tau_{D^\pm_s} \left|V_{cs}\right|^2 m_{D^\pm_s} \left(1-\frac{m^2_l}{m^2_{D^\pm_s}}\right)^2} {\tau_{D^\pm} \left|V_{cd}\right|^2 m_{D^\pm} \left(1-\frac{m^2_l}{m^2_{D^\pm}}\right)^2}\approx 37 $$

The experimental branching ratios are $B(D^\pm_s\rightarrow l\nu)=(5.32\pm0.11)$% and $B(D^\pm\rightarrow l\nu)=(0.120\pm0.027)$%, giving $R_{exp}=44\pm10$, in good agreement with theory.

  • $\begingroup$ You are comparing partial widths, not BRs, right? $\endgroup$ Commented Jun 4, 2023 at 22:22
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    $\begingroup$ @Cosmas_Zachos No, unless I have made some mistake (always a possibility), I am comparing BRs, which is why the lifetimes are in the formulas. The partial width of the $D^\pm_s$ is about 77 times that of the $D^\pm$, but the total width is also about a factor of 2 larger, so the ratio of branching ratios is half that. $\endgroup$ Commented Jun 4, 2023 at 23:32
  • $\begingroup$ Sorry, I missed the lifetimes. Why do you bother? The widths are more direct… $\endgroup$ Commented Jun 5, 2023 at 1:26
  • $\begingroup$ ... but the width of the non strange D is 1.2 x$10^{-3}$ which originally impressed the OP, and the ratio of the phase space factors is ~ 4.5 ..., as per my comment to him... $\endgroup$ Commented Jun 5, 2023 at 13:24
  • $\begingroup$ Thanks @Cosma_Zachos. I think my answer just spells out some details of what is stated in your comments, especially that there is no mystery about why the BR are so different (although not "3 orders-of-magnitude"). The OP asked about BR, so the answer is for that, but I agree partial widths are more direct since BR require the total width which is not so easy to estimate precisely from theory (e.g. because of effects such as non-leptonic enhancement) and so is best taken from experiment. $\endgroup$ Commented Jun 5, 2023 at 14:40

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