Euler-Lagrange equation: derivation and application

Рет қаралды 25,415

Күн бұрын

Classical Mechanics and Relativity: Lecture 3
0:00 Introduction
0:51 Principle of Least Action and the Lagrangian
6:01 Generalized Coordinates
11:38 Derivation of the Euler-Lagrange equation in generalized coordinates
25:29 Generalized momentum and generalized force
27:55 Polar coordinates
38:33 Example: pendulum
46:18 Newtonian vs Lagrangian mechanics
49:47 Global vs Local approach
57:47 The Hamiltonian
1:03:18 Legendre transformation
1:07:26 Hamilton's Equations
1:08:44 The Hamiltonian and Energy
1:12:27 Conservation of Energy
Theoretical physicist Dr Andrew Mitchell presents an undergraduate lecture course on Classical Mechanics and Relativity at University College Dublin. This is a complete and self-contained course in which everything is derived from scratch.
In this lecture I use the Principle of Least Action to derive the Euler-Lagrange Equation of Motion in generalized coordinates and perform the Legendre transformation to obtain Hamilton's equations. We will explore the connection between the classical Hamiltonian and the energy, and show that it is conserved. The concepts are illustrated with simple examples.
Full lecture course playlist: • Classical Mechanics an...
Course textbooks:
"Classical Mechanics" by Goldstein, Safko, and Poole
"Classical Mechanics" by Morin
"Relativity" by Rindler

Пікірлер: 30

@mtb095 Күн бұрын

I’ve been loving your channel since I came across it recently. Keep up the good work. One note: The armchair physicists in the comments of your videos are almost as entertaining as the videos. They are theoretical physicists, because their physics degrees are theoretical 😂

@jolez_4869 3 жыл бұрын

Great lectures! Sad to see that there are few views as soon as the physics get really interesting.

@Snowmaners 2 жыл бұрын

Dr. Mitchell's presentations are clear and fresh, giving a view that really enforces ones understanding if you have a earlier introduction.

@drmitchellsphysicschannel2955 2 жыл бұрын

Many thanks! I am very glad the lectures are helping people with their studies!

@shashanks.k855 3 жыл бұрын

This lecture was such a pleasure to follow. Thanks a ton, Dr Andrew Mitchell, ur the best.

@oded2304 Жыл бұрын

A couple of cirrections: 1. on Min. 37:13 - the derivative od the potential should be w.r.t. the angle \theta. 2. on min. 41.40 - z should probably be y, for consistency.

@christophertamina8569 3 жыл бұрын

Simply a great teacher 👏🏿👏🏿👏🏿 I take my heart off t you Sir

@jamestseng4014 2 жыл бұрын

37:14 EOM in theta: Left side should be - partial derivative of V w.r.t theta instead of w.r.t. r. Great job, thank you.

@suguruk1817 2 ай бұрын

14:10 Principle of Least Action

@aafeer2227 7 ай бұрын

Brilliant. Thank you.

@andredavis4657 2 жыл бұрын

Brilliant.

@BarakaGagiri-xb3zz 5 ай бұрын

Such good presentation

@mariogalindoq Жыл бұрын

Good video, congratulations.

@JP-re3bc 3 ай бұрын

This lecture would improve a lot IMHO if the speaker focused on the concept of "action" before stating the Lagrangian. Explain what is going on with all the trajectories, what is the problem mathematically. As it is action remains something arbitrary and mystical, while the Lagrangian pops up "deos ex machina" of sorts.

@andrewwrobel2255 8 ай бұрын

Good lecture, but the application of EL EoMs to the pendulum around 43:30 is somewhat faulty: the constraint r(t)=l should be taken into account from the very start, but here it is used only after writing out the EoMs. This does not matter for the EoM in theta, but if the EoM in r is written out as on the preceding slide, then it will falsely say that mr-dbldot=centrifugal force + radial component of gravity, instead of r-dbldot=0 in reality (the radial gravity term comes from differentiating -V = mgrcos(theta) w.r.t. r, with V=0 at the pivot). This says that the bob flies off the rod. The mistake comes form leaving out the reaction force of the constraining rod, which in reality balances both the radial component of gravity and the centrifugal force (or, from the inertial frame PoV, provides the centripetal force). In other words, the EL EoMs are applied to the problem with two coordinates (r, theta) as though there were no constraints except at the endpoints. The simplest remedy is, of course, to eliminate r from the start and consider the problem with just a single coordinate, theta. (Or, if one insists on keeping both coordinates (r, theta) then one has to apply the version of EL EoMs with additional holonomic constraints (which here would be r(t)=l for each t).)

@zeroUnknown117 Жыл бұрын

Could you do a video explaining Euler Lagrange using Cosines? Also within an algebraic setting.

@camac7988 2 жыл бұрын

WOW.

@gibbogle 8 ай бұрын

Nice clear derivation. But at 24:20 did you justify that the sum of the integrals equal to 0 implies that each individual integral = 0 and that each integrated term = 0? This is the hard part for me.

@drmitchellsphysicschannel2955 8 ай бұрын

Good question: it's a subtle point. The idea is that if we start out with the path with least action then the action will not change (to first order) if we change the path a bit. But there are many ways we can change the path: we can vary each of the several generalized coordinates q_i at any time between the start and end points. The change in the action, dS, is zero for *any* change we make to the path, changing any of the coordinates at any time, dq_i(t). But how can dS always be equal to zero for any dq_i(t)? It can only happen if the function itself being integrated is zero. Then multiplying by the arbitrary function dq_i(t) for each i won't affect the integral -- it's always just zero!

@wei-chihchen8647 3 жыл бұрын

20:22 time derivative of (q) --> time derivative of (dq)

@drmitchellsphysicschannel2955 3 жыл бұрын

Thank you for the correction, well spotted!

@wei-chihchen8647 3 жыл бұрын

@@drmitchellsphysicschannel2955 Thank you for uploading your lectures. They are fantastic!

@khnahid5807 5 ай бұрын

Why aren't we considering ∂L/∂t*dt in the differential of L at 17:26 ?

@josuelima5033 4 ай бұрын

I believe that is because dS = 0 only when the Taylor expansion is truncated at first-order terms. For example, consider dS = S[r(t) + ŋ(t)] - S[r(t)], where ŋ(t) represents an infinitesimal deviation from the stationary action path. Therefore, S[r(t) + ŋ(t)] can be expanded as S[r(t)] + ∇S[r(t)]·ŋ(t) + high-order terms. Here, ∇S[r(t)] = 0 because it lies on the path of least action. Thus, when ignoring high-order terms, dS = 0. If we add (∂L/∂t)*dt to dL, we will have a term with the integral [ ∫(∂L/∂t)dt ]*dt, resulting in a high-order term. I'm not sure if this is correct, but it occurred to me. I would appreciate it if someone could confirm if this makes sense.

@bejitasansensei Жыл бұрын

41:40 z is l*costheta, it's not dz. But anyway, with derivation it does not matter in the end.

@danielkonstantinovsky108 2 жыл бұрын

since you're setting delta_A to be 0, is that enough to say you are deriving the principle of least action (as in there's a global minimum at the Euler-Lagrange equation). Isn't it more like the principle of stationary action? I've heard that phrase somewhere but I'm not sure.

@danielkonstantinovsky108 2 жыл бұрын

i guess if you're finding the only situation in which the delta(S) is zero that must be either a global maximum or a global minimum. is it possible to take the second derivative of the functional w.r.t. the path to prove that it is in fact a minimum?