I am interested in knowing how (Q1) the particle's masses are experimentally determined from accelerator observations.

What kind of particles? They must be as far as we know elementary and unstable (very short lifetime) and not subject to the strong interaction (for example, Higgs particle, Z boson, etc.) I'm not interested in neutrons (not elementary), electrons (stable) or quarks (hadron). I'm not particularly interested in neutrinos either, since I think that best constraints come from neutrino oscillations and cosmological observations.

Since the particles I'm asking about acquire their masses through the Higgs mechanism, I would like to know what is actually or more directly measured the mass or the Yukawa coupling. (Q2)

I also wonder what is actually measured the propagator's pole (this the magnitude reported as mass for stable leptons) or the running mass at certain energy scale (this is one of the magnitudes reported as mass for quarks). (Q3)

This question may be considered a follow-up of How Can We Measure The Mass Of Particle?

Thanks in advance.

Edit: In connection with the answers: From all your answer I deduce that the mass reported for the Higgs, W and Z is the mass (rest energy) that appears in the energy-momentum conservation law. I guest that this mass corresponds to the pole of the free propagator of the Higgs, W and Z, respectively (and not to the running mass). I also deduce that what is more directly measured is the mass of the Higgs and from that value one deduces the self-coupling of the Higgs (and not in the other way around). These were my Question 3 and 2. Do you agree with my conclusion?

  • $\begingroup$ I;ve stripped the [accelerator-physics] tag as that usual means the behavior of the beam and energy imparting mechanisms rather that physics done with the beam. And I see that there are many such misuses. Off to meta to poll the users. $\endgroup$ – dmckee Aug 14 '12 at 21:14

For sufficiently long-lived charged particles, one measures the helix-shaped track in an external magnetic field, and gets from this the 4-momentum (and hence the masss).

For very short-lived particles, one gets complex masses from resonance measurements.

Edit: Any mass of an unstable particle is complex and defined as the pole of a propagator. The mass of a particle like Higgs is determined quite indirectly, as it takes lots of scattering experiments to reliably determine the relevant cross sections. See http://arxiv.org/pdf/1207.1347.pdf for how to determine the Higgs mass from measurements. See also http://arxiv.org/pdf/1112.3007.pdf

  • $\begingroup$ The geometry of the helix really only gives you the momentum-to-charge ratio, then you get both the total energy and the charge by calorimetry along the track (the energy loss for "massive" particles is a function of $q\gamma\beta$) and so shows a experimentally useful pattern. $\endgroup$ – dmckee Aug 15 '12 at 14:57
  • $\begingroup$ @ArnoldNeumaier: I added a last paragraph to my answer, could you tell me if you agree? $\endgroup$ – Diego Mazón Aug 16 '12 at 0:52
  • $\begingroup$ @drake: I added a paragraph to my question. ;-) Swap question and answer in your and my comment ;-) $\endgroup$ – Arnold Neumaier Aug 16 '12 at 7:20

The simplest way is to try and produce a massive unstable particle. Let's say that we're trying to produce a muon from an electron and a photon via the process:

$$e + \gamma \rightarrow \mu^{-} + \nu_{e}+{\bar\nu}_{\mu}$$

Just to make our game easier, let's say that we have the electron kept initially with zero momentum, and we vary the intensity of the gamma particle. Then, the initial energy of the system is $m_{e}c^{2} + E_{\gamma}$, and the initial momentum of the system is $E_{\gamma}/c$.

Since the muon has a larger mass than the electron, though, the zero momentum state on the right hand side will have a higher energy than the zero momentum state on the right hand side (we can ignore the neutrino masses). Therefore, this process violates conservation of energy unless the photon is sufficiently energetic enough to make conservation of momentum possible. This happens precisely when $m_{e}c^{2}+E_{\gamma}=m_{\mu}c^{2}$. So, you can look at the exact point where muons start to get made, and voila! You have a measurement of the muon mass.

A more sophisticated approach would measure energies and momenta of the particles before and after the collision, and other parameters affected by mass. But I think this example is probably the clearest way to see this effect.

  • $\begingroup$ The fun part here is getting high energy photons with well known energies. The search term is "photon tagger" or "photon tagging". A very cool business in and of itself. Thresholds are generally not used for very heavy particles because the rate at threshold is vanishing. $\endgroup$ – dmckee Aug 14 '12 at 21:10
  • $\begingroup$ Can one extend that procedure to very unstable particles like the Higgs? $\endgroup$ – Diego Mazón Aug 14 '12 at 21:13
  • $\begingroup$ Yes the threshold approach can be extended to very unstable particles, but the production rate for the Higgs is too small by far. The Tevatron was chasing it for more than a decade, but didn't have the cross-section. $\endgroup$ – dmckee Aug 14 '12 at 21:18
  • $\begingroup$ By far the best way to get masses is to measure the decay products and use conservation laws as @dmckee says in his answer. $\endgroup$ – anna v Aug 15 '12 at 7:28
  • $\begingroup$ @annav: sure, but when teaching to beginners I like to introduce with sine qua non effects. $\endgroup$ – Jerry Schirmer Aug 15 '12 at 18:50

We measure the four-momenta of the decay products and reconstruct the four-momentum of the particle in question, then apply the usual relation between energy, three-momentum and mass: $$ E^2 = m^2 + \mathbf{p}^2 $$ (in $\hbar = c = 1$ units, of course). Such measurements suffer from both detector related uncertainties and the Heisenberg relation, but with many taken together we find a well defined resonance peak in the mass.

The observables from the detector are energy losses, and directions (which means changes in directions due to multiple scattering and magnetic fields). And a few odd-ball like Cerenkov detectors give you velocities (or at least above/below threshold). These can be used to ID particles and reconstruct both the energy and the three-momenta with some confidence once the detector is well understood. From that the mass associated with each track is clear (and this is used as a check of the particle ID mechanism).

With heavy particles many of the products may be unstable themselves, so we bootstrap the measurement from their decay products.

  • $\begingroup$ How is that exactly done with the Higgs and which mass parameter of the decaying particle enter in the equation? $\endgroup$ – Diego Mazón Aug 14 '12 at 21:16
  • $\begingroup$ Higgs has a lot of possible channels. I believe that only gamma--gamma and 2-weak boson channels were used for the recent announcement. In those cases there are some missing particles, so more tricks have to be played based on the assumed decay physics. The night of the announcement I could have interpreted it all for you, but it is not my subfield and I have forgotten much of what I heard. $\endgroup$ – dmckee Aug 15 '12 at 2:00
  • 2
    $\begingroup$ @drake For the reported Higgs have a look at the CMS paper which gets the mass from the invariant mass of the two gammas cms.web.cern.ch/news/… . To nail it as a higgs one needs good statistics in all channels where higgs can decay plus the angular distributions in order to confirm spin parity. $\endgroup$ – anna v Aug 15 '12 at 7:24
  • $\begingroup$ @annav Thanks! I added a last paragraph to my answer, could you tell me if you agree? $\endgroup$ – Diego Mazón Aug 16 '12 at 0:55
  • $\begingroup$ @dmckee: I added a last paragraph to my answer, could you tell me if you agree? $\endgroup$ – Diego Mazón Aug 16 '12 at 0:56

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