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.
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. "
@@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.
@@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".
@@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.
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?
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
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!
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.
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.
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
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.
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: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-KOek-B3Rvmg.html
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.
@@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
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.
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. :)
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!
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!
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!
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?
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.
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!
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.
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.
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
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
