I've been waiting patiently for this video ever since I found your video on Poisson Brackets, great quality as always!
@PhysicswithElliot2 жыл бұрын
Glad you liked it!
@eigenchris2 жыл бұрын
Super great video! It directly answers some of my questions from the previous video. I've never seen the "momentum is the generator of space translations" idea for classic physics explained so plainly. Do you know if there's a way to make Hamiltonian mechanics work together with special relativity? Seems in HM time and space play fundamentally different roles. I wonder if there is a way to "rescue" HM when moving to relativity.
@narfwhals78432 жыл бұрын
I had a similar question, though in regards to quantum mechanics a few weeks back. I hope nobody minds if I just copy/paste ZAP Physics' answer here. It seems like it will lead us inevitably to field theories. "One way to see that this can fit into special relativity is that if we just define H to be the zero-component of the 4-momentum operator, then we can see that the 4-momentum generates translations of the 4-dimensional spacetime vector with x^0 = t. However, the issue is that a lot is hidden in "H," and the form of H that we have been using is absolutely not suitable for a relativistic quantum theory. This can easily be seen since the energy of a free relativistic particle is Sqrt(p^2 c^2 + m^2 c^4) unlike our classical p^2/(2m). The square root makes things tricky since it isn't well-suited for the linear properties that we want when we upgrade the momentum to a momentum operator. There are sort of two ways around this: First, we can try to "square" both sides, in which case we end up with the Klein-Gordon equation. The problem with this is that it results in negative-norm states, so we can't interpret Psi^* Psi as a probability density and it is very tricky to figure out what this is actually telling us (also, it doesn't account for spin-1/2 particles) The other option is to use a Hamiltonian which is naturally relativistically invariant, even within Newtonian mechanics. This happens to be a property of many field equations, but the issue here is that we have to replace our position and momentum operators with corresponding field operators. This is what is known as canonical quantization. "
@RizkyMaulanaNugraha2 жыл бұрын
@eigenchris you both should do collab sometimes :D
@alphalunamare2 жыл бұрын
@@narfwhals7843 Doesn't 'Canonical' just mean 'going by the book'? ie 'follow the rules'. If it works then don't doubt it sort of thing? No actual explanation for why it works however.
@narfwhals78432 жыл бұрын
@@alphalunamare Canonical in this context refers to canonical coordinates. en.wikipedia.org/wiki/Canonical_coordinates I'm not sure why they're called that. Possibly because the canonical transformations leave the hamiltonian equations unchanged or "as written".
@alphalunamare2 жыл бұрын
@@narfwhals7843 I don't know why people are allowed to post such impenetrable gibberish on Wiki just because it is correct. Surely knowledge is about understanding? As such the referenced page totally fails. I could dig into it and take it apart but, to be honest, I can't be arsed. There is nothing in Mathematics that a child can not understand, that it is disguised so is a poor reflection on those professing to understand things in the first place. Wiki is a piss poor resource.
@ΝίκοςΓιαννόπουλος-λ5θ2 жыл бұрын
We hail you as an exceptional physics instructor. Thank you so much for the amazing work!
@PhysicswithElliot2 жыл бұрын
Glad you liked it!
@General12th2 жыл бұрын
I like it! One component of this is how we can take advantage of the fact that there's exactly one number N such that N = -N, and that just happens to be what these brackets ought to resolve to if there's a symmetry or a conserved quantity we can take advantage of. The N stands for Nifty. :)
@khandakerahmed74082 жыл бұрын
Please keep up and add more and more. Please do not stop.
@jms5472 жыл бұрын
This video is great, and landed in my feed just in time for me to bring it to a classical mechanics reading group that's starting this year! I'd love to see your explanation of Lie algebras, symmetry groups, and representations - that's exactly where my current understanding of mechanics dries up!
@PhysicswithElliot2 жыл бұрын
So glad you liked it Jonathan! Thanks for sharing it with your friends!
@jms5472 жыл бұрын
Not at all, thanks for making great videos. Incidentally, I've just discovered Lax's equation, which has opened up a whole new vista on the Poisson bracket/commutator structure of mechanics. I'd love to see how they all relate to each other within the Lie algebra/group context, if you ever make that video!
@dennylane20102 жыл бұрын
My professors have the talent of making simple things look so complicated. Yours is the reversed. You just summarized the missing connection that I am looking for (for years) between symmetry and conservation laws in classical mechanics. Thank you very much! Awesome explanation!
@PhysicswithElliot2 жыл бұрын
Glad to help!
@discopotato467310 күн бұрын
I know that this is two years old and you probably won't see this but I want you to know something. I absolutely LOVE Physics, and also want to be a professor in Physics and your way of teaching is absolutely inspiring. You've done a beautiful job with all your videos. 🫶
@bosonh77232 жыл бұрын
You are a legend mate, thank you so much for such a lovely simplified introduction to otherwise profoundly deep concepts
@PhysicswithElliot2 жыл бұрын
Happy you liked it!
@johnchessant30122 жыл бұрын
Great video! I've read about Noether's theorem before but this really made everything click for me
@edzielinski8 ай бұрын
Fantastic. The explanation about the flow in the phase space was a revelation to me on how to think about this. Thanks!
@alphalunamare2 жыл бұрын
I understood every sentence but when I put them together my head fell off! This is fascinating stuff :-)
@devamjani80412 жыл бұрын
In my opinion, Noether's theorem is the single most impactful and important result in all of physics. The entire standard model of particle physics builds upon it and also the entire mathematical physics. This theorems are what founded mathematical physics.
@bartpastoor10282 жыл бұрын
It’s late here now and I will will revisit this video to get a better grasp on me studying the Lie Group /Lie Algebra within Robinson “Standard Model and Particle Physics”. Keep up the good work!
@PhysicswithElliot2 жыл бұрын
Thanks Bart!
@c4knowledge5625 ай бұрын
Can you do the symplectic basis regarding this topics like simplectic involution
@rui-anchang139 Жыл бұрын
Very great video!!!! I have never seen such a clear explanation of the relation between symmetry and conservation law.
@canyadigit62742 жыл бұрын
Dude…this just opened my mind! I’ve studied generators and translations in the context of quantum mechanics, representation theory, classical mech, and quantum field theory-it had always been something like e^(theta)X where X “generated” the group (which made sense, since it could be expanded as a Taylor series and X more or less acts like the generator of group G where G is a cyclic group. But this new intuition on generators makes more physical sense! How can we tie these 2 seemingly different notions of generators? One notion because an exponent that generates a group (G= e^(theta)X), and the other being the one you defined in this video?
@PhysicswithElliot2 жыл бұрын
In the flow equation dx/\lambda = {x, G} = - {G, x}, the object {G, _} is a derivative operator D_G, and the solution to this equation can be written x(\lambda) = e^(-\lambda D_G ) x. For example {p, _} = -d/dx is minus the x derivative of whatever goes in the second slot. Then the solution of the flow equation dx/\lambda = 1 is x(\lambda) = e^(\lambda d/dx ) x = (1 + \lambda d/dx +1/2 \lambda^2 d^2/dx^2+...)x = x + \lambda. In quantum mechanics, {p, f} = -df/dx becomes [p, f] = -i\hbar df/dx
@deepakjanardhanan73942 жыл бұрын
Great, really great. High quality content. Superb..... Expecting more and more.....
@PhysicswithElliot2 жыл бұрын
Thanks Deepak!
@ChristAliveForevermore2 жыл бұрын
Emmy Noether is an un-sung genius of a very high calibre.
@jp-ratsratsrats2 жыл бұрын
Great video, made me want to reopen my group theory in physics book i was reading
@charlesgantz58652 жыл бұрын
Very nice start to the New Year.
@meenalimbu2553 Жыл бұрын
ELLIOT WHY THE FUCK DIDNT YOU START A PHYSICS KZbin CHANNEL BACK IN 2010 WHEN I STARTED MY PHYSICS DEGREE. 😭😭😭😭😭😭
@physicshuman98082 жыл бұрын
6:43 L is angular momentum and P is linear momentum L causes rotation and P causes space translation
@avnishhhhh2 жыл бұрын
Your videos are really helpful, keep going 😍
@PhysicswithElliot2 жыл бұрын
Thanks Avnish!
@DeepSeeker28092 жыл бұрын
You seem to be good at explaining the math of physics and its symmetries.. It would be great if you can squeeze in a series about group & representation theory of particles..!! Great video btw!
@alphalunamare2 жыл бұрын
Group theory is beautiful in of itself. One always worries about its usurpation by physicists. Not that I am being picky, I have just never seen a decent explanation for the ways in which they slam group structures together as if there is some underlaying miracle.
@alphalunamare2 жыл бұрын
Years ago' I asked Proff Weigold (Cardiff) what it was all about. He said that they were 'near' to understanding every possible group structure and I pondered why the effort. He just smiled at me ... he was a lovely Man.
@PhysicswithElliot2 жыл бұрын
Will hopefully talk more about it in the future!
@r3lativ2 жыл бұрын
Interesting video, as usual. But there are some issues: - Not everything that commutes with H is a conserved quantity. For instance, E^2 - p^2 is not a conserved quantity. He says that if dQ/dt = 0, then Q is conserved. But not necessarily. For instance, for a free particle, v is constant, dv/dt = 0, but v is not a conserved quantity, mv is. So, how exactly do we define the idea of a "conserved quantity"? - He says that every symmetry has a corresponding conserved quantity. Is this true? What's the conserved quantity corresponding to the Galileo or Lorentz transformations?
@PhysicswithElliot2 жыл бұрын
A conserved quantity Q(x(t),p(t),t) is a function that's constant in time, dQ/dt = 0. When Q(x(t),p(t)) doesn't depend explicitly on time, its rate of change is dQ/dt = {Q, H}. More generally, this becomes dQ/dt = {Q,H} + \partial Q/\partial t when Q does have explicit time dependence. The conserved quantity for a Galilean boost is K = p t - m x. It explicitly depends on time, so it doesn't commute with the Hamiltonian. Instead, for a free particle, dK/dt = {K, H} + \partial K/\partial t = -m p/m +p = 0.
@lazzatv2 жыл бұрын
Check Michael Penn video playlist on Differential Forms for a general mathematical formalism. Basically a Poisson bracket is a differential 2-form determinant of a quantity parametrized in (x,p) phase space which appears in the calculation of the integral of that quantity in phase space.
@mplaw772 жыл бұрын
Well done, I learned something today!
@PhysicswithElliot2 жыл бұрын
Thanks Michael!
@eustacenjeru7225 Жыл бұрын
Nice and clear explanation
@mdabdulquader89952 жыл бұрын
Your videos are really outstanding. Please do more
@PhysicswithElliot2 жыл бұрын
Thanks Abdul!
@mdabdulquader89952 жыл бұрын
@@PhysicswithElliot If possible, Please make a complete playlist of classical mechanics & classical electrodynamics.
@andrasgilicz81532 жыл бұрын
Great video :-) How you made the animation?
@DeepLyricist2 жыл бұрын
Not sure how you calculate dP/dLambda using "the chosen G" since real G's are supposed to be silent like lasagna.
@hishan.farfan2 жыл бұрын
new favorite channel!! 😁
@res0nanc3202 жыл бұрын
I love the style of these videos. Would you mind sharing what software you use to produce them?
@PhysicswithElliot2 жыл бұрын
Keynote and Final Cut Pro mainly!
@hundhund4892 жыл бұрын
Holy shit, thanks!
@kquat78992 жыл бұрын
Fantastic series.
@nassersaleh83277 ай бұрын
I'm very disappointed that you didn't turn on the subtitle function, so we non-English speakers lost important information
@mathadventuress2 жыл бұрын
I’m not at this level yet but I want to be 😫
@SanjayMishra-su7sz13 күн бұрын
Nice video and quality
@rodrigoappendino2 жыл бұрын
I read in my book the demonstration, involving variatioal calculus, which is a lot more complicated. This would be easier.
@dlrmfemilianolako82 жыл бұрын
Thank you so much for your videos . Please can you make videos about dark matter and energy ?
@PhysicswithElliot2 жыл бұрын
Thanks Emiliano! Potentially!
@nice32942 жыл бұрын
This is an amazing vid
@General12th2 жыл бұрын
I want to see videos about Lie algebras down the line.
@sgurdmeal6624 ай бұрын
Amazing video!
@MrRyanroberson12 жыл бұрын
Since continuous symmetries lead to conservation laws, would periodic yet discontinuous symmetries lead to periodic conservation laws? Such as a phase angle of something strangely dictating some other event
@ВасилийЛесничий-с9и2 жыл бұрын
Thanks for the video! Is anyone aware of the proof, that there are no more conservation laws (energy, momentum, angular momentum, parity, charge, center-of-momentum velocity)? There are a lot of different symmetries in various less general systems, of course. For example, there is a discrete translational symmetry in crystals, which under the condition of the incident particles' momentum conservation leads to many beautiful results for elastic scattering. And it is also not exactly the ordinary translational symmetry, of course.
@PhysicswithElliot2 жыл бұрын
The set of symmetries depends on the system you're looking at. The Hamiltonian for a particle in a 1/r gravitational potential for example has a very non-obvious symmetry that leads to the conservation of the Runge-Lenz vector: kzbin.info/www/bejne/gYDInGB4aLepo8k
@aadityapratap007 Жыл бұрын
Great video
@marcovillalobos51772 жыл бұрын
Wonderful video❤️
@PhysicswithElliot2 жыл бұрын
Thanks Marco!
@gowrissshanker91092 жыл бұрын
Hlo Elliot, How come the position and momentum are independent variables? Consider SHM Hamiltonian when we change position , Momentum changes right? Both are related to each other right? Thank you
@chadgregory90372 жыл бұрын
are these "flows" basically just principle fibres from bundles
@chadgregory90372 жыл бұрын
I know this all ties into HJB equation too and stochastic optimal control
@PhysicswithElliot2 жыл бұрын
They're the integral curves of vector fields on phase space
@ARAVINDKUMAR-ug7gt2 жыл бұрын
I really love your video sir ❤️❤️❤️
@user-sl6gn1ss8p2 жыл бұрын
I like how well this seems to generalize to the idea that you could define "conservation in space", for example, in the same way as the usual "conservation [in time]", with the same relationship to symmetries - the concept that symmetries and conservation are linked is deeper than just the sense we usually see
@koenth23592 жыл бұрын
Nice vid Eliot, very neat!
@PhysicswithElliot2 жыл бұрын
Thanks Koen!
@mimzim71412 жыл бұрын
Does one get conservation of charge and of other quantities with the same symmetry arguments?
@PhysicswithElliot2 жыл бұрын
Yep!
@mimzim71412 жыл бұрын
@@PhysicswithElliot what are the corresponding symetries?
@sidkt74682 жыл бұрын
So can you do this with any Q, i.e. say that if dx/dt = 0 then {x,H} = 0 so {H, x} = 0 (and by solving the equations for the flow you get that space is the generator for momentum translations). Then, "position is conserved if there is an invariance under momentum translations" because that doesn't seem correct.
@PhysicswithElliot2 жыл бұрын
{x, H} won't typically vanish. For a typical Hamiltonian H=p^2/2m + U(x) you'll get {x, H} = p/m.
@sidkt74682 жыл бұрын
@@PhysicswithElliot but isn't that also the case for {p, H} as it would yield -dU/dx for a typical Hamiltonian H = p^2/2m + U(x)
@PhysicswithElliot2 жыл бұрын
@@sidkt7468 That's right---that's why I mentioned that for a single particle the momentum would only be conserved if it's free, meaning U = 0 (or constant). But when you have multiple particles in an isolated system, the total momentum will be conserved, and the symmetry corresponds to picking up the whole system and sliding everything over
@sidkt74682 жыл бұрын
@@PhysicswithElliot oh that makes sense now, I didn't quite understand the meaning of it being free at the beginning.
@samicalvo45602 жыл бұрын
Hello Elliot, I've already subscribed to your web to get the notes but I haven't received any email from you.
@PhysicswithElliot2 жыл бұрын
Hi Sami, could it have gone in your spam folder? If you don't find it just send me an email (elliot@physicswithelliot.com)
@ginaluzsalcedoalarcon4779 Жыл бұрын
No subtitles 😭
@maurocruz18242 жыл бұрын
4:05
@drbonesshow12 жыл бұрын
The Most Beautiful Result in Classical Mechanics is better when explained to a pretty girl sitting across from you. At least, for Einstein and myself the physics professor.
@ventriloquistmagician47352 жыл бұрын
Change the subtitles language, it's trying to translate from Dutch.