4 ms·
Not the original commenter, but also ex-HEP person: The invariant mass is the rest mass of the particle (i.e. it's "inherent" mass). You can calculate it by ta
by fnands 2y ago
Not the original commenter, but also ex-HEP person:
The invariant mass is the rest mass of the particle (i.e. it's "inherent" mass). You can calculate it by taking the final state decay products of the original particle (i.e. the particles that are actually observed by the detector) and summing up their four-vectors (squared).
You can plot the invariant mass calculated from any particular final state, and for a rare particle like the Higgs the majority of the contributions to your plot will be from background processes (i.e. not Higgs decays) that decay into the same final state.
If you have a lot of Higgs decays in your sample you should be able to see a clear peak in the distribution at the invariant mass of the Higgs boson, a clear sign that the Higgs (or something with the same mass) exists.
Often by the time the discovery has reached statistical significance, you might not really be able to see such a clear sign in the mass distribution. I.e. the calculations are telling you it's there but you can't see it that clearly.
I wouldn't really say this helps confirm the discovery in a scientific sense, just that it's reassuring that the signal is so strong that you can see it by eye.
- nick3443 2y agoLike this one? https://cds.cern.ch/record/1546765/files/figs_gamma_gamma_mass.png https://cds.cern.ch/record/1546765/files/figs_gamma_gamma_ma...
- exmadscientist 2y ago> just that it's reassuring that the signal is so strong that you can see it by eye It's really something when this happens. I worked on a big neutrino experiment searching for theta_13, where our goals were to (a) determine if theta_13 was dead zero or not (being truly zero would have a Seriously Major Effect in theories) and then (b) to measure its value if not. Our experiment was big, expensive, and finely tuned to search for very, very small values of theta_13. We turned the thing on and... right there there was a dip. Just... there. On the plot. All the data blinding schemes needed to guarantee our best resolution kind of went out the window when anyone looking at the most basic status plot could see the dip immediately! On the one hand, it was really great to know that everything worked, we'd recorded a major milestone in the field (along with our competition, all of whom were reading out at basically the same time), and the theorists would continue to have nothing to do with their lives because theta_13 was, in fact, nonzero. On the other hand... I wasted how many years of my life dialing this damned detector in for what now? (It wasn't wasted effort, not at all... but you get the feeling.)
- Filligree 2y agoSquared four-vectors? I'm only an amateur, but wouldn't that give different results depending on choice of units? I.e, I usually use C=1.
- dguest 2y agothe math is m^2 c^4 = E^2 - p^2 c^2 where m is mass, E is the total energy in the decay products and p is the 3-vector sum of the momentum. Those units should work out (they certainly do if you set c = 1).
- Filligree 2y agoAh, I see. I was assuming you meant the 4-momentum. Though I'm not sure this doesn't come out to the same thing.
- dguest 2y agowhat I showed is the squared 4 momentum when you use the Minkowski metric [1], assuming "squared" means "self-dot product". The formulation above is just another way to illustrate the Minkowski dot product. [1]: https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metric https://en.wikipedia.org/wiki/Minkowski_space#Minkowski_metr...
- sixo 2y agoyou use the same units on both sides of the equation, it's fine, it's like counting "meters squared"
- WalterBright 2y agoWhat about the loss of mass released as energy inherent to the decay process?
- cwillu 2y ago“Energy” is only released as the energy and momentum of the resulting particles.