For beta decay of mesons, the heavy quark approximation and surely other techniques can be used to show that the decay width scales as the quintic power of mass. This can be checked from pdg files very easily with this gnuplot script:

  set logscale xy
  set key noautotitle
  plot "<cat Downloads/mass_width_2020.mcd|grep -v ^* | cut -c 32-" using "%20lf%*8lf%*8lf%16lf"
  replot 2.4952*(x/91.1876)**3
  replot 2.99591E-19*(x/0.105658)**5

The plot includes all the particles having mass and decay width listed in the particle data group datafile: mesons and baryons, some of their excited states, and also the Z0 and the W and the two unstable leptons, muon and tau. And the top quark. Higgs boson does not appear because its decay width is not yet in the table, proton does not appear because it is stable AFAWK, and well, the neutron does not appear because I have zoomed it out)

Units in the table are GeV (mass in the horizontal, decay width in the vertical). And you can appreciate the aforementioned quintic scaling as the blue line here.


But surprise, there is also an alignment for neutral, purely electromagnetic, decays. It is a cubic scaling. Is it an artifact, or can it be shown analytically?

To put more detail: the particles on the green line are $\pi^0, \eta, \Sigma^0, J/\Psi, \Psi(2S)$ and, rather peculiarly, the gauge bosons.

Particle Mass (Gev) Total Width (GeV) "Reduced Width" $MeV^{-2}$
$\pi^0$ 0.135 7.73E-9 3.1
$\eta$ 0.548 1.31E-6   8.0
$\Sigma^0$ 1.19 8.90E-6  5.3 
$D^*$ 2.01 8.34E-5  10.3 
$J/\Psi$ 3.10 9.29E-5   3.1
$\Psi(2S)$ 3.69 2.94E-4  5.9 
$B(s2)^*$ 5.84  1.49E-3 7.5 
$Z^0$ 91.2 2.50  3.3 

The particles above the green line are all the mesons and baryons that have strong decay. For some of these, when the partial EM decay width is known, usually to $\gamma\gamma$, it also aligns on the green line. I have checked this for $\omega, \phi$ and $\eta'$.

It would seem that the hadronic scale, extracted via the "reduced width" $\Gamma/M^3$, has some general bound, the only exception being the three first states in the bottom quark sector: $\Upsilon(1S,2S,3S)$ happen to have some extra of stability allowing them to lie under the cubic line. But note $\Upsilon(4S)$ is already above it.

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    $\begingroup$ This question would be easier to understand if the axes and points on your graph were labeled. I used to know how to get gnuplot to position a text label at a data point, so you wouldn’t need a huge connect-the-dots legend, but nowadays I don’t use gnuplot enough to need that skill any more. $\endgroup$ – rob Feb 10 at 17:26
  • $\begingroup$ Hmm problem was to extract the labels too. But I am going to try some explanation with more detail. $\endgroup$ – arivero Feb 10 at 17:36
  • $\begingroup$ Even a small number of handwritten labels would be helpful. (If your computer has a tool that lets you draw red circles around important things in pictures, you could use that.) Also some helpful autocorrect has written “decay with” where you meant “decay width.” $\endgroup$ – rob Feb 10 at 17:53
  • $\begingroup$ It is amusing that $Z^0$ is not just in the ballpark, but only a 5% correction up the line (in fact, $\Gamma_\pi = \Gamma_Z (M_\pi/M_Z)^3$ is only two sigmas away, given the huge error in the measurement of pion decay). We could bring it towards the line asking not for the total decay but only the part from the axial couplings, couldn't we? $\endgroup$ – arivero Feb 11 at 17:21

You might get better answers if you actually identified the particles involved and their decays.

I'll just remind you of the dimensional analysis aspect of all of them, which you drill in an introductory HEP course:

For weak decays involving one dominant scale, Γ with units of energy must go as the amplitude-squared, involving an exchange of a virtual W in the amplitude, so $1/M_W^2$, so $$ \Gamma\propto \frac {1}{M_W^4} ; $$
for dimensional consistency, for a single scale problem, the mass of the decaying particle, must involve the 5th power of its mass, as you first learned in muon decay, $$ \Gamma \sim m^5/M_W^4. $$ This might hold for mesons, dominated by heavy quarks, etc...

By contrast, for e.g. em decays of the pion, $$ \Gamma_{\pi^0} \sim \alpha^2 \frac{m_\pi^3}{f_\pi^2}, $$ by dint of the PCAC nature of the pion, which resolves to a triangle of quarks inversely as $f_\pi$ in the amplitude. This might be at the root of your observation, but I can't tell which meson decays you are talking about.

  • Note added after clarification OK, for the pseudoscalar mesons, all decay constants are a low energy hadronic scale not too different from $f_\pi$. For the J/Ψ, the EM decays spring to prominence because of the OZI suppression of the strong modes, and, as in the case of the $\Sigma^0$ hadronic decay, the mass constant in the denominator of the amplitude is some low hadronic scale quantifying how a meson couples to a quark-antiquark pair current coupling to a photon, so not that dissimilar to neutral pion decay. You'd get the square of that scale in the denominator, and so the cube of the dominant mass, that of the decaying particle, in the numerator. There is such a plethora of circumstances, around, however, wavefucntions at the origin, etc... that I am not sure what general systematics could be seen to prevail.
  • $\begingroup$ Hmm I will add some detail on the particles which are involved more closely, I had skipped this detail because it is basically all of them. All the EM neutral meson decays seen to be aligned by scaling the formula from pion. $\endgroup$ – arivero Feb 10 at 17:48
  • $\begingroup$ My main doubt is that while the quintic scaling is general enough as it depends of W mass, or Fermi constant if you wish, the cubic formula for pion depends on parameters coming from QCD, not from electroweak theory, so I can not see why it should be a general bound extended for all the possible masses. $\endgroup$ – arivero Feb 10 at 18:13
  • $\begingroup$ Comment after the Note after clarification: I guess that the general systematics could be argued to be corrections of order one to a main "low hadronic scale" then, or at least bounded by this scale. That would explain mostly, and of course the $Z^0$ just coincidence. $\endgroup$ – arivero Feb 11 at 16:24
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    $\begingroup$ Well, 70% of the Z decays are hadronic, and something has to hadronize the quarks it couples to, and the massive gauge boson mass is irrelevant/washed out by its own mass, so... $\endgroup$ – Cosmas Zachos Feb 11 at 16:35

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