3 ms·
can some physicist explain why the calculations are so hard... something something Taylor series?
by sjkoelle 3y ago
can some physicist explain why the calculations are so hard... something something Taylor series?
- l33tman 3y agoIn essence, you need to enumerate and sum up all possible ways everything can happen in a very little piece of the universe that corresponds to the "reaction center" of the experiment. All possible ways you can pop out particles from the vacuum that interacts with your experimental particle and pop back into the vacuum etc. It's a totally impossible calculation to do in full, so a large chunk of theoretical high energy physics is about finding shortcuts to do that. All these ways contribute to the reaction probabilities (which are the ones you get a number on at the end of the day). The muon g-2 experiment is so interesting because there is a vast number of pretty exotic particles and particle pathways that have to be taken into account, and if our understanding of any of those is wrong, or there are (even more interestingly) particles that can pop out and contribute that we don't even know about, it will show up in the muon g-2 number as compared to the theoretical calc.
- frutiger 3y ago> All possible ways you can pop out particles from the vacuum that interacts with your experimental particle and pop back into the vacuum etc. This is a neat statement, but really those diagrams are notation for elements in a perturbation expansion series. Nothing in the theory says that these particles actually pop out and actually interact with the incoming/outgoing momenta. In fact they travel off-shell so we know that they cannot be physical.
- l33tman 3y agoIf you're referring to Feynman diagrams, sure in the modern treatment they don't mean what people usually think they mean, but it's also not accurate to state that particles don't pop in/out. It's a simplification for sure, but you really do need to enumerate all the field configurations and there is really no difference from a first principles point of view between your "physical" particles and the "off-shell" excitations. You can argue about the reality of the contributing terms but it sure looks like they're needed or you won't get the right results.
- ketralnis 3y agoNot a physicist but basically yes. Most quantum mechanics calculations involve the path integral[0], a sort of sum of all possible Feynman diagrams[1] that an interaction could involve and the more of these you involve the more accurate you hope to be. Further the field theory that they're calculating within is generally an effective field theory[2] which is known to be an incomplete approximation of the system it's modelling but "close enough" for the energies involved. That approximation costs accuracy as you get closer to the energy cutoff you included but you included it because you don't know what happens above it so it's not like you can just not do that. Depending on how renormalization[2] is done in that effective theory the constants involved can have their values affected. In this particular case you're also looking "closer" at an interaction than you can model with the particle intuition. For simple electron/photon scattering you can pretend that things are particles and use that intuition to guide you but when you start looking really close at things you can't really do that anymore and have to work with fields and waves in a way that belies normal physical intuition. Part of the way that manifests is "virtual particles" popping in and out of existence (really field interactions that aren't really particle-like but with the virtual particle trick can be modelled almost as if they were). These interactions do affect the real world and the way they affect it depends on the fields involved. If there's some unknown 4th generation of lepton for instance a virtual one of those could interact with the experiment, but since you didn't know about it you didn't include it in your calculation. Also depending on how you're calculating things you aren't just saying "1+2=3", you're saying "1±theory error + 2±theory error = 3±theory error" so you end up with a range of calculations. That might manifest as "well if there's a graviton then I need to calculate it this way but if there isn't then I need to calculate it this way, and we don't have experimental evidence to guide me one way or the other". Lastly, QFT is quantum mechanics + special relativity and I think it goes without saying that both of these are just detailed, difficult fields with difficult maths involved. So you're taking an already difficult field and since you're trying to be accurate out to a lot of decimal places while interacting with messy real-world experiments you're forgoing all of the nice physics-class "massless particle in a spherical box" approximations that normally let you skip over all of that trouble. See https://indico.fnal.gov/event/7309/contributions/101335/attachments/66130/79359/notes-3.pdf https://indico.fnal.gov/event/7309/contributions/101335/atta... for some lecture notes on the base problem. 0: https://en.wikipedia.org/wiki/Path_integral_formulation https://en.wikipedia.org/wiki/Path_integral_formulation 1: https://en.wikipedia.org/wiki/Feynman_diagram https://en.wikipedia.org/wiki/Feynman_diagram 2: https://en.wikipedia.org/wiki/Effective_field_theory https://en.wikipedia.org/wiki/Effective_field_theory 3: https://en.wikipedia.org/wiki/Renormalization https://en.wikipedia.org/wiki/Renormalization