Feynman Diagrams and Perturbation Theory: Calculating in Particle Physics

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ZAP Physics

ZAP Physics

Күн бұрын

In this video, we talk about how physicists perform calculations in particle physics using perturbation theory and Feynman diagrams. We discuss what perturbation theory is, how it relates to Feynman diagrams, how to read Feynman diagrams and how one actually gets a mathematical result out of Feynman diagrams.

Пікірлер: 65
@timurpryadilin8830
@timurpryadilin8830 3 жыл бұрын
Finally, I saw a video about Feynman's diagrams, that actually showed the equations to which the graphical elements correspond. It is always frustrating to learn about them just as pictures and simply be told that "they simplify calculations" without doing any calculations with them. Not that I understood any of the formulae tho, but it is at least inspiring.
@zapphysics
@zapphysics 3 жыл бұрын
@Timur Pryadilin I'm glad you enjoyed this! I was a bit worried that the mathematical expressions would be a turn-off for a lot of people, but I didn't feel like I could do a discussion about Feynman diagrams justice without showing exactly how they are used in calculations!
@Arseniy_Afanasyev
@Arseniy_Afanasyev 2 жыл бұрын
@@zapphysics you're right!
@TheBigBangggggg
@TheBigBangggggg Жыл бұрын
I'm afraid that the inspiring effect will be over in a few years. Nobody here will do any useful calculation in his life!
@Littleprinceleon
@Littleprinceleon Жыл бұрын
@@TheBigBangggggg and so? Why do you think that most of us, wondering about the nature of quantum physics or the nature of "Nature" in general and the ways we humans use logic to describe regularities has to do "useful" calculations eventually? One can get inspiration from physics for some other field of science and use the knowledge to design better experiments or at least try to interpret the "old" data from new perspective(s). Einstein himself relied on other people with more advanced math, however if one doesn't even know how powerful methods are "hidden" in the depth of mathematics, then one has much lower chance of coming up with useful (whatever that means) ideas.
@TheBigBangggggg
@TheBigBangggggg Жыл бұрын
@@Littleprinceleon I find your reaction more inspiring than his video. But in my opinion, the mathematical bridge between what detectors "see" and what really happened becomes bigger and bigger. That can only mean that the outcomes of experiments will be more and more uncertain. "Think like a physicist" has one or two videos on 5-sigma. But she admits she can't explain what's behind the sigma (apart from the fact that it is a standard deviation).
@NovaWarrior77
@NovaWarrior77 3 жыл бұрын
The taylor series-feynman diagram-perturbation theory relationship blew my mind! I NEVER saw how people made calculations from feynman diagrams despite being told we know how to do that!
@PrettyMuchPhysics
@PrettyMuchPhysics 3 жыл бұрын
Great video! Can't believe how much info you stuffed into just one video :P
@zapphysics
@zapphysics 3 жыл бұрын
Thanks you! I appreciate it! I like my content like I like my black holes: with high information density
@PrettyMuchPhysics
@PrettyMuchPhysics 3 жыл бұрын
@@zapphysics 🤯
@prateeksharma5051
@prateeksharma5051 3 жыл бұрын
@@zapphysics 👍👍 which softwares you use to write? Writing on screen is fast and audio speed is relatively slow & that is interesting? How to make such videos? Please answer?
@zapphysics
@zapphysics 3 жыл бұрын
@@prateeksharma5051 I use a Microsoft program called Scrble Ink to write. I change the speeds by recording audio separately, screen capturing my writing using OBS and using Blender's video editor to edit and speed up the writing to sync up with the audio. I don't know if this is the most efficient way of doing things, but it works and is cost-effective (OBS and Blender are free programs and I think I paid like $7 for Scrble, though that also has a free, lite version). Blender can have a bit of a learning curve to it, but there are a lot of videos, etc. out there that can be very helpful.
@prateeksharma5051
@prateeksharma5051 3 жыл бұрын
@@zapphysics thanks for detailed answer, keep it up.
@JoeHynes284
@JoeHynes284 Ай бұрын
ive been working the last few years on teaching myself QFT, basically all of P and S book. I've never sat in a classroom and im 46, it easy for me to get lost in the weeds of the equations and you and Sean carrol really help me to focus on why the equations matter and what they are doing. thank you!
@jerboajelly
@jerboajelly 2 жыл бұрын
Wow this video is really amazing and helped greatly in understanding this topic. I am doing QFT in my post-grad and struggled to understand these concepts through books and lectures. But what you did here with the video is absolutely revolutionary. You made understanding of a difficult subject so natural and easy. ZAP Physics, you are a lifesaver and also a great teacher. Keep spreading the good word of physics through your great work! Thank you very much and cheers!
@tanvirfarhan5585
@tanvirfarhan5585 3 жыл бұрын
I have learned more from this channel than I have in several year's worth of schooling. I can’t believe this is free content. Thank you so much. great channel. I really appreciate your hard work
@MaeLSTRoM1997
@MaeLSTRoM1997 2 жыл бұрын
Hey, I'm a graduate student in physics and I really appreciate how your videos provide a nice context to all the calculations thrown at us in the class, reminding me why we're doing all this weird mental exercise. Thanks for making these.
@TheBigBangggggg
@TheBigBangggggg Жыл бұрын
That's very honest. I call it intellectual masturbation.
@MaeLSTRoM1997
@MaeLSTRoM1997 Жыл бұрын
@@TheBigBangggggg i agree, and i think people should do it sometimes. masturbation in moderation is healthy.
@TheBigBangggggg
@TheBigBangggggg Жыл бұрын
@@MaeLSTRoM1997 Of course, your own physical and intellectual needs don't always coincide with those of your partner. Fundamentally, we are individuals!
@61rmd1
@61rmd1 2 жыл бұрын
Well done...It's so useful. With the help of this clip I realized why in a bremsstrahlung against a Ze nucleus, we find a cross section depending on Z^2*e^3 (it's not obvious...) Thank you
@freyaday1167
@freyaday1167 Жыл бұрын
At long last, I finally understand what a virtual particle ‘is’! Thank you ^^
@jakublizon6375
@jakublizon6375 Жыл бұрын
Im trying to get more people to check out your channel. This is great because you actually show the math, as well as some historical narration that makes the content very digestible, yet satisfying to my curiosity.
@nimnim5149
@nimnim5149 2 жыл бұрын
hey , I love your videos ...there is only one request ; that you also link in some of the resources in the description for self study for the viewers .... that would be very much appreciated :) Thank you .
@ethanking123
@ethanking123 2 жыл бұрын
I wish I had found this channel sooner! Great content and the videos are very informative as well as enlightening. Keep up the good work, looking forward to seeing more videos down the road. I really appreciate the effort you put in.
@stevewhitt9109
@stevewhitt9109 Жыл бұрын
I like the fact that you do show the equations of QFT.
@robertschlesinger1342
@robertschlesinger1342 3 жыл бұрын
Excellent video. Very interesting, informative and worthwhile video.
@narfwhals7843
@narfwhals7843 3 жыл бұрын
This is probably the best science communication video on Feynman Diagrams I've seen. Putting the complicated math on screen for those who want it without going into detailed explanations is a good middle ground in my opinion. Maybe a silly question. In Bhabha scattering, isn't there an amplitude for the particles to just not interact? Or would that be the same as the first order one and the virtual photon just "doesn't do anything" so the incoming and outgoing momenta are all the same?
@zapphysics
@zapphysics 3 жыл бұрын
Hi Narf, I'm glad you liked the video! This is a great question. Mathematically, yes, this "non-interaction" term does show up in the calculation of S-matrix elements. You can sort of think of it as the "zeroth-order" term in the perturbative expansion, where there are no interaction vertices. However, we typically only care about the case where the particles have a measurable interaction, i.e. the case where their incoming and outgoing states are different. Since the states are all orthonormal, this non-interaction term essentially just corresponds to a total delta function between the two states. So, if the two states are different at all, this term will just always be zero, so in any reasonable particle physics experiment, this term is pretty much totally irrelevant and is typically subtracted off to calculate just the interacting S-matrix elements.
@maxmann6933
@maxmann6933 3 жыл бұрын
Thanks again. This is again a great video!
@williamworthy4091
@williamworthy4091 3 жыл бұрын
Great stuff
@felipemonteiro5877
@felipemonteiro5877 3 жыл бұрын
Awesome stuff, thank you very muvy
@youtubesucks1885
@youtubesucks1885 2 жыл бұрын
Can you do a video about string field theory and finding solutions?
@ffusk
@ffusk 2 жыл бұрын
Wow, thanks!
@dynapb
@dynapb 3 жыл бұрын
How do the Feynman diagrams take care of say e+e- trajectories coming together at different angles? Are they part of the momentum input? If so then would there be a different diagram for each set of trajectories?
@zapphysics
@zapphysics 3 жыл бұрын
@Peter Becher Great question. Yes, the information about angles is encoded in the external momentum states. One of the nice features of Feynman diagrams is that they typically allow us to do the calculations using arbitrary momenta, so we don't have re-do any calculations when we just change the angles of our external particles. However, when we do higher-order calculations, we find that not all of the momenta of the internal particles are fixed by momentum conservation. In principle, each of these corresponds to a new Feynman diagram (although there's no need to draw a new one every time!), and so we have to sum (really, integrate) over all possible un-fixed internal momentum states. On the surface, this can actually cause serious problems, since these integrals have an annoying habit of diverging. This problem is solved by the technique of renormalization. I'm hoping to make my next video addressing this exact topic, but I hope this answers your question in the mean time.
@dynapb
@dynapb 3 жыл бұрын
@@zapphysics Thanks for that good info. Look forward to seeing the next video!
@RazorBaze
@RazorBaze 3 жыл бұрын
Brilliant as always! One question - about VPs, virtual particles: so... if one VP, from any one Feynman diagram is *not* a representation of reality, but many VPs from many Feynman diagrams *kind of are* - does that mean that these many VPs are *also* a kind of representation of the quantum vacuum? And if so, are these VPs the same as the VPs in, say, the Casimir effect? I'm feeling silly asking these, but I'm confused with so many mathematically advanced physicists being so persistent in calling VPs "not real" - as if there were no (at least somewhat realistic) formulation of VPs at all! Beside actual evidence, like mentioned Casimir. Btw, I'm not familiar with a lattice QED a bit; but on a contrary I'm having an impression that String theory uses basically same Feynman diagrams, except that main "blocks" there are Worldsheets.
@zapphysics
@zapphysics 3 жыл бұрын
@Vladislav These are all good questions. I think that, at the end of the day, it is always dangerous to think of *any* Feynman diagrams as being a true representation of reality. They are really just a mathematical tool we use to do calculations, so I would try to refrain from thinking of virtual particles as any sort of explanation of nature. The Casimir effect is a bit different and doesn't have to rely on virtual particles at all. The idea is that a single quantum harmonic oscillator has some non-zero energy in its ground state. We can actually exactly calculate that the quantized free electromagnetic field (i.e. the photon field without any charged particles coupled to it) is really just an infinite number of decoupled momentum-space quantum harmonic oscillators (the fact that the oscillators decouple in momentum space is actually very important to being able to do this calculation analytically). But since each oscillator has a non-zero ground state energy, this means that each momentum mode has some amount of energy in it, even in the vacuum, and since we have an infinite number of momentum modes (a photon can have any momentum it wants), there is an uncountably infinite multiple of these ground state energies in the vacuum. However, when we put two large parallel plates near each other, not all of the momentum modes can exist between them. In fact, it turns out that only a *countably* infinite number can, which means that there is a mismatch between the number of ground-state momentum modes inside and outside the plates. Since each mode has energy to it, this means that there is more energy outside the plates than inside, which results in the plates being pushed together. We never have to talk about virtual particles or Feynman diagrams or even perturbation theory to get this result! I am perhaps not the best person to talk about string theory, since my knowledge of it is fairly limited, but yes, I do know that it is possible to use perturbation theory in string theory. However, there I think it is a bit different and I don't think Feynman diagrams are useful in the same way since string scattering is more a problem of topology than what we do in particle physics. The details are a bit hairy, and I am not confident enough in my own knowledge of the subject to say much more than that, unfortunately!
@RazorBaze
@RazorBaze 3 жыл бұрын
@@zapphysics I'm speechless :) thank you very much!
@brunovillalobos6944
@brunovillalobos6944 3 жыл бұрын
You're a genius
@zapphysics
@zapphysics 3 жыл бұрын
You're too kind! I am glad you enjoyed the video!
@sergeysmyshlyaev9716
@sergeysmyshlyaev9716 Жыл бұрын
Excellent video! I have a question: perturbation theory and Feynman diagrams were invented when most calculation were performed by hand, because computers were limited, especially for performing symbolic calculation. But now we have more powerful computers and more powerful mathematical engines. Shouldn't we invent a new technic, that simplifies the calculations with modern computers?
@jitteryjet7525
@jitteryjet7525 Жыл бұрын
I think they do. I seem to recall someone invented a simplification or "rule" a number of years ago which did exactly that, made it easier to integrate the series contributions by eliminating some calculation paths which were going to cancel out anyway.
@zapphysics
@zapphysics Жыл бұрын
@Sergey Smyshlayev this is a great question, and I apologize it took me so long to get to it. In short, the answer to your question is sort of yes, sort of no. I'll start with the sort of no side of things: the unfortunate thing about computers, no matter how advanced they are, they aren't particularly good at "thinking." What I mean by that is, it doesn't really matter how good your computer is if people don't know how to tell it how to solve something. And in the case of QFT, we don't know how to go about exactly solving the equations, so we can't really code a computer to come up with an exact solution. On the other hand, computers are great at doing a lot of computations orders of magnitude faster than any person can do them. So, if we have a different way of approximating the system that we know how to solve in principle, but is way too computationally difficult for any one person to do, we can code up a computer program to crunch these calculations. This is what is done for lattice QFT: you approximate the system as living on a discrete lattice of points instead of a continuous spacetime (you have to do this because, even if the computer is the best in the world, it can't do an infinite number of computations). This has the advantage that you don't have to rely on small numbers in the theory, so you can look at e.g. low-energy QCD where perturbation theory breaks down, but it has some downsides as well (it requires quite a lot of computing power, complex phases appear that computers don't really know how to handle, etc.) So unfortunately, if people don't know how to analytically solve the theory, a computer isn't going to know how to either. That said, computers are absolutely helpful in the modern era through both lattice as well as performing perturbative calculations much faster than any people can (assuming that the solution can be done algorithmically or numerically).
@rodrigoappendino
@rodrigoappendino Жыл бұрын
0:38 What are the others?
@andrewferris8169
@andrewferris8169 3 жыл бұрын
How do Electrons know when another Electron is approaching in order to fire a photon at it?
@franks.6547
@franks.6547 3 жыл бұрын
How do two cars in a car crash know how to find each other? All cars do their thing, and only sometimes we observe "car scattering" and call it that in hindsight. Only that electrons do all things they can possibly do at once, and one final state "emerges" in the not yet really understood process of "observation". But as the video implies, you can also regard "virtual particles" as mere names for mathematical terms in a calculation whose outcome often matches observation without actually imagining that any real photon is exchanged in electron scattering. It is just a scheme to organise a complex calculation - and only when the perturbation ansatz works in the first place.
@zapphysics
@zapphysics 3 жыл бұрын
@Frank S. I couldn't have said it better myself!
@jitteryjet7525
@jitteryjet7525 Жыл бұрын
Good video. I get annoyed at quatum mechanics videos which act as if virtual particles are "real" entities in some sense - clearly they are not.
@rodrigoappendino
@rodrigoappendino Жыл бұрын
9:52 But where is it written that special relativity is only for non virtual particles?
@user-zz3sn8ky7z
@user-zz3sn8ky7z Жыл бұрын
Special relativity can't apply to virtual particles because there is no such thing as a virtual particle in the real world, as the video said they are purely theoretical quirks of the approximate model we use and don't actually occur in nature. A good comparasion in my eyes are complex numbers, in the "real world" square root of a negative number makes no sense, but in math calculations they show up all the time
@Littleprinceleon
@Littleprinceleon Жыл бұрын
@@user-zz3sn8ky7z but the complex plane is "perfectly" good to describe the phase change of quantum waves, which at least gives nice intuitive visualization of a layer of abstraction
@wishamahmad2719
@wishamahmad2719 3 жыл бұрын
Mate, you are a very beautiful man for doing this, thank you!
@Littleprinceleon
@Littleprinceleon Жыл бұрын
I think "wonderful person" is a more appropriate phrase in English. Have a good time 😊
@hannah2843
@hannah2843 Жыл бұрын
why can we just say that p^2=m^2 for massive particles? where does that come from? i always thought it was p^2=m^2-E^2
@zapphysics
@zapphysics Жыл бұрын
Ah yes, this is an unfortunate bit of notation, I think. In this case, p is the magnitude of the *four-momentum*, whereas in the equation E^2=p^2+m^2, p is the magnitude of the *three-momentum*. The square of the four-momentum is a Lorentz invariant, meaning that it is the same in any inertial reference frame, so it is simplest to take the particle's rest frame, where it is fairly straightforward to show that the square of the four-momentum is just the rest-mass squared. Sorry for the confusion!
@MisterTutor2010
@MisterTutor2010 20 күн бұрын
@2:47 The solution to this equation is 42 :)
@prikarsartam
@prikarsartam Жыл бұрын
when you mentioned how such diagrams are not actually what's happening everywhere, I think you missed the idea that - the most-contributing case [1st order diagram(s)] even consists of internal photons, and given sufficiently long time, there indeed pops up virtual particle-antiparticle pair inside the internal photon line and that causes very small correction to predictions. So indeed each diagram symbolically represents what "happens", and they are separated into equivalence classes where each class has diagrams contributing equally in a process and are taken care of with symmetry factors. Nevertheless if the formalism of the theory significantly contains interactions where 'virtual' objects excite into existence and vanish in vacuum, of whose Feynman diagrams are symbolic representations of; then that is the case also in nature. But this matter becomes already quite ambiguous when asked in this way; rather we must look in term of 'phenomenon in nature' and 'representation of phenomenon in nature'; where only the later one, i.e. in our theories which are representation of phenomenon that happens in nature - we find symbolic forms like Feynman diagrams which represents events that (theoretically) contribute in a phenomenon. But saying that - since this is a perturbation theory it isn't exact but approximation, so it not the case in nature - is wrong since approximation captures the most significant contributions, which in turn are exactly those diagrams; i.e. the terms in perturbation series upto arbitrarily large order has 1-2-1 correspondence between diagrams. Then again the question arises about finitude of those internal processes, i.e. do they contribute infinitely? (then every transition amplitude in the theory will be infinite and won't capture any information so will be valueless) and renormalisation is there to take care of this. So the concern about "reality" of Feynman diagrams are to be understood as symbolic representation of the phenomenon that fundamentally takes place in nature, not literal processes.
@Littleprinceleon
@Littleprinceleon Жыл бұрын
What is 1-2-1 correspondence?
@nzuckman
@nzuckman 3 жыл бұрын
10/10, we stan \(°o°)/
@TheBigBangggggg
@TheBigBangggggg Жыл бұрын
ZAP = ZERO APPLICATIONS POINT
@frankkolmann4801
@frankkolmann4801 Жыл бұрын
Feynman diagrams are the craziest way to then claim that this results in the most accurate predictions we have of any physical theory. If your calculations are not accurate then make up some more duagrams until you get the answer you want. Further the equations of momentum and energy are low speed approximations of reality as they equations do not account for gamma. So starting from an approximation you then claim to get the most accurate calculation of physical reality. How is this in any way science?
@simonO712
@simonO712 Жыл бұрын
You don't simply make up new diagrams, they all correspond to specific terms in a perturbation series. They are easier ways to write complicated sums, that's all. You're essentially saying that perturbation theory is completely unscientific, which, if that's a position you actually hold, shows a lot of ignorance on the subject. If that isn't a position you hold you need to clarify how your issue with Feynmann diagrams can't be applied to perturbation theory in general. As for the relativistic part I've already adressed in a comment on the video on perturbation theory. At least I think it was your comment, if not then I can exand on it here.
@jitteryjet7525
@jitteryjet7525 Жыл бұрын
Wow! Do you subscribe to Anti-science Ideology by any chance (I have met some people that do). The claim of "accurate predictions" is justified because the predicitions agree with experiment very well. "How is this in any way science?" well rhetorical questions are always suspect. But if you can demonstrate your claim by writing a paper and getting it published in a credible, peer-reviewed scientific journal, I think your trip to Stockholm is in the bag.
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