The Most Beautiful Result in Classical Mechanics

Рет қаралды 57,520

2 жыл бұрын

Noether's theorem says that a symmetry of a Lagrangian implies a conservation law. But to fully appreciate the connection we need to go to Hamiltonian mechanics and see how symmetries act on phase space! Get the notes for free here: courses.physicswithelliot.com...
The connection between symmetries and conservation laws is one of the deepest relationships in physics. Noether's theorem says that for every continuous symmetry of a Lagrangian, you'll find a corresponding conserved quantity. But to fully understand the connection between the two, we need to investigate their relationship in Hamiltonian mechanics. Any function on phase space generates a "flow," similar to how the Hamiltonian generates time evolution. Then the rate of change of any other function along the flow is given by its Poisson bracket with the generator. A quantity will be conserved if and only if the flow that it generates leaves the Hamiltonian invariant!
Get all the links here: www.physicswithelliot.com/ham...
Watch these first!:
- The principle of least action: • Explaining the Princip...
- Noether's theorem: • Symmetries & Conservat...
- Poisson brackets: • Before You Start On Qu...
Also check out:
A 20 minute intro to Lagrangian and Hamiltonian mechanics: • Lagrangian and Hamilto...
The relativistic action: • The Special Relativist...
Tutoring inquiries: www.physicswithelliot.com/tut...
If you find the content I’m creating valuable and would like to help make it possible for me to continue sharing more, please consider supporting me! You can make a recurring contribution at / physicswithelliot , or make a one time contribution at www.physicswithelliot.com/sup.... Thank you so much!
About physics mini lessons:
In these intermediate-level physics lessons, I'll try to give you a self-contained introduction to some fascinating physics topics. If you're just getting started on your physics journey, you might not understand every single detail in every video---that's totally fine! What I'm really hoping is that you'll be inspired to go off and keep learning more on your own.
About me:
I’m Dr. Elliot Schneider. I love physics, and I want to help others learn (and learn to love) physics, too. Whether you’re a beginner just starting out with your physics studies, a more advanced student, or a lifelong learner, I hope you’ll find resources here that enable you to deepen your understanding of the laws of nature. For more cool physics stuff, visit me at www.physicswithelliot.com.

Пікірлер: 108

@nerdsgalore5223 2 жыл бұрын

I've been waiting patiently for this video ever since I found your video on Poisson Brackets, great quality as always!

@PhysicswithElliot 2 жыл бұрын

Glad you liked it!

@eigenchris 2 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.

@narfwhals7843 2 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. "

@RizkyMaulanaNugraha 2 жыл бұрын

@eigenchris you both should do collab sometimes :D

@alphalunamare 2 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.

@narfwhals7843 2 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".

@alphalunamare 2 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.

@user-ws6ik1ch5c 2 жыл бұрын

We hail you as an exceptional physics instructor. Thank you so much for the amazing work!

@PhysicswithElliot 2 жыл бұрын

Glad you liked it!

@khandakerahmed7408 2 жыл бұрын

Please keep up and add more and more. Please do not stop.

@johnchessant3012 2 жыл бұрын

Great video! I've read about Noether's theorem before but this really made everything click for me

@bosonh7723 2 жыл бұрын

You are a legend mate, thank you so much for such a lovely simplified introduction to otherwise profoundly deep concepts

@PhysicswithElliot 2 жыл бұрын

Happy you liked it!

@edzielinski 3 ай бұрын

Fantastic. The explanation about the flow in the phase space was a revelation to me on how to think about this. Thanks!

@charlesgantz5865 2 жыл бұрын

Very nice start to the New Year.

@General12th 2 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. :)

@jms547 2 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!

@PhysicswithElliot 2 жыл бұрын

So glad you liked it Jonathan! Thanks for sharing it with your friends!

@jms547 2 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!

@rui-anchang139 Жыл бұрын

Very great video!!!! I have never seen such a clear explanation of the relation between symmetry and conservation law.

@dennylane2010 2 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!

@PhysicswithElliot 2 жыл бұрын

Glad to help!

@kquat7899 2 жыл бұрын

Fantastic series.

@eustacenjeru7225 10 ай бұрын

Nice and clear explanation

@bartpastoor1028 2 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!

@PhysicswithElliot 2 жыл бұрын

Thanks Bart!

@avnishhhhh 2 жыл бұрын

Your videos are really helpful, keep going 😍

@PhysicswithElliot 2 жыл бұрын

Thanks Avnish!

@alphalunamare 2 жыл бұрын

I understood every sentence but when I put them together my head fell off! This is fascinating stuff :-)

@sgurdmeal662 10 күн бұрын

Amazing video!

@canyadigit6274 2 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?

@PhysicswithElliot 2 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

@deepakjanardhanan7394 2 жыл бұрын

Great, really great. High quality content. Superb..... Expecting more and more.....

@PhysicswithElliot 2 жыл бұрын

Thanks Deepak!

@aadityapratap007 Жыл бұрын

Great video

@hishan.farfan Жыл бұрын

new favorite channel!! 😁

@kid0ftheOldblock 2 жыл бұрын

Great video, made me want to reopen my group theory in physics book i was reading

@devamjani8041 2 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.

@mplaw77 2 жыл бұрын

Well done, I learned something today!

@PhysicswithElliot 2 жыл бұрын

Thanks Michael!

@lazzatv 2 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.

@nice3294 2 жыл бұрын

This is an amazing vid

@mdabdulquader8995 2 жыл бұрын

Your videos are really outstanding. Please do more

@PhysicswithElliot 2 жыл бұрын

Thanks Abdul!

@mdabdulquader8995 2 жыл бұрын

@@PhysicswithElliot If possible, Please make a complete playlist of classical mechanics & classical electrodynamics.

@marcovillalobos5177 2 жыл бұрын

Wonderful video❤️

@PhysicswithElliot 2 жыл бұрын

Thanks Marco!

@ARAVINDKUMAR-ug7gt 2 жыл бұрын

I really love your video sir ❤️❤️❤️

@ChristAliveForevermore 2 жыл бұрын

Emmy Noether is an un-sung genius of a very high calibre.

@hundhund489 2 жыл бұрын

Holy shit, thanks!

@res0nanc320 2 жыл бұрын

I love the style of these videos. Would you mind sharing what software you use to produce them?

@PhysicswithElliot 2 жыл бұрын

Keynote and Final Cut Pro mainly!

@nassersaleh8327 2 ай бұрын

I'm very disappointed that you didn't turn on the subtitle function, so we non-English speakers lost important information

@koenth2359 2 жыл бұрын

Nice vid Eliot, very neat!

@PhysicswithElliot 2 жыл бұрын

Thanks Koen!

@DeepSeeker2809 2 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!

@alphalunamare 2 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.

@alphalunamare 2 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.

@PhysicswithElliot 2 жыл бұрын

Will hopefully talk more about it in the future!

@andrasgilicz8153 2 жыл бұрын

Great video :-) How you made the animation?

@c4knowledge562 23 күн бұрын

Can you do the symplectic basis regarding this topics like simplectic involution

@physicshuman9808 2 жыл бұрын

6:43 L is angular momentum and P is linear momentum L causes rotation and P causes space translation

@maurocruz1824 2 жыл бұрын

4:05

@rodrigoappendino 2 жыл бұрын

I read in my book the demonstration, involving variatioal calculus, which is a lot more complicated. This would be easier.

@General12th 2 жыл бұрын

I want to see videos about Lie algebras down the line.

@gowrissshanker9109 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

@dlrmfemilianolako8 2 жыл бұрын

Thank you so much for your videos . Please can you make videos about dark matter and energy ?

@PhysicswithElliot 2 жыл бұрын

Thanks Emiliano! Potentially!

@user-sl6gn1ss8p 2 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

@user-xt8to6yi8n 2 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.

@PhysicswithElliot 2 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: kzfaq.info/get/bejne/gbWVnpBzmLjanZs.html

@DeepLyricist 2 жыл бұрын

Not sure how you calculate dP/dLambda using "the chosen G" since real G's are supposed to be silent like lasagna.

@MrRyanroberson1 2 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

@mathadventuress 2 жыл бұрын

I’m not at this level yet but I want to be 😫

@r3lativ 2 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?

@PhysicswithElliot 2 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.

@meenalimbu2553 Жыл бұрын

ELLIOT WHY THE FUCK DIDNT YOU START A PHYSICS KZfaq CHANNEL BACK IN 2010 WHEN I STARTED MY PHYSICS DEGREE. 😭😭😭😭😭😭

@mimzim7141 2 жыл бұрын

Does one get conservation of charge and of other quantities with the same symmetry arguments?

@PhysicswithElliot 2 жыл бұрын

Yep!

@mimzim7141 2 жыл бұрын

@@PhysicswithElliot what are the corresponding symetries?

@sidkt7468 2 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.

@PhysicswithElliot 2 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.

@sidkt7468 2 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)

@PhysicswithElliot 2 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

@sidkt7468 2 жыл бұрын

@@PhysicswithElliot oh that makes sense now, I didn't quite understand the meaning of it being free at the beginning.

@samicalvo4560 2 жыл бұрын

Hello Elliot, I've already subscribed to your web to get the notes but I haven't received any email from you.

@PhysicswithElliot 2 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)

@chadgregory9037 2 жыл бұрын

are these "flows" basically just principle fibres from bundles

@chadgregory9037 2 жыл бұрын

I know this all ties into HJB equation too and stochastic optimal control

@PhysicswithElliot 2 жыл бұрын

They're the integral curves of vector fields on phase space

@ginaluzsalcedoalarcon4779 Жыл бұрын

No subtitles 😭

@drbonesshow1 2 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.