Why is the branching fraction of strange $D$ mesons to tau leptons so high?

Question

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.

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.