Classical Mechanics | Lecture 3

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  • Опубліковано 21 вер 2024
  • (October 10, 2011) Leonard Susskind discusses lagrangian functions as they relate to coordinate systems and forces in a system.
    This course is the beginning of a six course sequence that explores the theoretical foundations of modern physics. Topics in the series include classical mechanics, quantum mechanics, theories of relativity, electromagnetism, cosmology, and black holes.
    Stanford University
    www.stanford.edu/
    Stanford Continuing Studies
    http:/continuingstudies.stanford.edu/
    Stanford University Channel on UA-cam:
    / stanford

КОМЕНТАРІ • 348

  • @zen-hx2hn
    @zen-hx2hn 6 місяців тому +12

    reading classical mechanics by the man himself and watching these videos really helps a lot, whatever i dont understand in the book i understand here and whatever i dont understand here i understand in the book, thank you stanford and Dr.Susskind.

  • @joabrosenberg2961
    @joabrosenberg2961 2 роки тому +52

    Law of least (extremum) action; Calculus of variations (minimal distance between points) 11:45; Light moving the shortest time between points 21:00; Motion on a line 24:00; Action definition, Lagrangian 31:00; Euler Lagrange equation 47:00; The Langrangian that produces Newton equations 50:40; Least action does not depend on the coordinate system (unlike the equations of motion) 1:01:00; Coriolis and centrifugal force 1:16:00; Polar coordinates 1:22:30; Conservation law (angular momentum) 1:31:00

  • @rodovre
    @rodovre 6 років тому +44

    I have never seen this topic explained with so much clarity. He is the greatest teacher in physics, and I admire his effort to go through all of physics for the benefit of beginning students. It is a great contribution to the field as a whole, and hopefully some of his listeners will become future physics stars thanks to this, just like the Feynman lectures.

    • @corgispotter
      @corgispotter 6 років тому +3

      As one would expect from a person who proved stephen hawking wrong

    • @MarcCastellsBallesta
      @MarcCastellsBallesta 4 роки тому +1

      @Joseph Winett I totally agree. I guess people thought the same about paperwork when Guttenberg invented the printer.

    • @gagadaddy8713
      @gagadaddy8713 2 роки тому +2

      @rodovre Yes! Can't agree more! ..... this is the most simply approach to explain(if not derive) the Eurler-Lagrange Equation. This video is really blow my mind .... though Prof. Susskind's drawing is a bit suck! 🤣 Will come back later for the rest of the topics!

  • @Rakeshkumar30
    @Rakeshkumar30 4 роки тому +17

    To me, this is the most important lecture of the series, the way Euler Lagrange equation was derived blew my mind.

  • @ozzyfromspace
    @ozzyfromspace 7 років тому +25

    1:00:20 "we have written down the law of...scones". Only at Stanford. Great lecture, Professor Susskind!

  • @ButtUglyParakeet
    @ButtUglyParakeet 11 років тому +64

    The textbook Classical Mechanics by John R. Taylor has many exercises that fit well with this course.

    • @waynelast1685
      @waynelast1685 4 роки тому

      it does seem like it has some good sections

    • @of8155
      @of8155 3 роки тому +3

      Goldstein

  • @TheGamingg33k
    @TheGamingg33k 4 роки тому +6

    Watch out at 1:06:00 guys. Its a rotation matrix basically. The equations are actually x = Xcos(wt) - Ysin(wt) and y= Xsin(wt) + Ycos(wt)

    • @ehabmalkawi194
      @ehabmalkawi194 4 роки тому +1

      He is a great physics teacher... I would also recommend a new channel for solving somewhat advanced problems in classical mechanics with thorough discussion... ua-cam.com/video/ofTlsMtdB98/v-deo.html

    • @praneetkumarpatra2661
      @praneetkumarpatra2661 Рік тому

      i think its inverse of rotation matrix.....i mean you are moving from X Y to x y , hence theta in that case is negative

    • @olivernorth7418
      @olivernorth7418 Рік тому

      Yup. Easy to draw the geometry to prove it.

  • @AlphaFoxDelta
    @AlphaFoxDelta 5 років тому +4

    His teaching makes me so happy, I couldn't ask for a better physics professor.

  • @willchan83
    @willchan83 9 років тому +37

    YES!! 0:40:00 to 0:44:00 is a clearer derivation of the Euler-Language equation than page 112-114 of the book, imho.

    • @FMasamune
      @FMasamune 6 років тому +2

      Thank you for your comment. Literally went to this video to see if he derived it differently here.

    • @paulnewton3556
      @paulnewton3556 6 років тому

      What is “the book”?

    • @drewmetra
      @drewmetra 6 років тому

      Probably The Theoretical Minimum

    • @aryamanmishra154
      @aryamanmishra154 5 років тому +1

      @@paulnewton3556 Theoretical minimum: classical mechanics or the 1st one which ever is in your country..

    • @kennywong7850
      @kennywong7850 4 роки тому

      The book’s derivation is driving me crazy, the delta t seems missing...

  • @jessstuart7495
    @jessstuart7495 6 років тому +84

    1:06:25 "That's correct, you can check this." I did, and it isn't correct.
    x = X*cos(ωt) - Y*sin(ωt)
    y= X*sin(ωt) + Y*cos(ωt)

    • @jesuesp1297
      @jesuesp1297 4 роки тому +3

      thank you, I thought I was crazy for a moment

    • @MarcCastellsBallesta
      @MarcCastellsBallesta 4 роки тому +2

      @stanford should pin this comment. Or add this to the description of the video.

    • @archiebrown1405
      @archiebrown1405 4 роки тому +9

      I think he must have gotten X and x and Y and y swapped by accident, as we want (x,y)→(X,Y). The correct equations are then X=xcos(θ)+ysin(θ) and Y=−xsin(θ)+ycos(θ)

    • @TheLevano22
      @TheLevano22 3 роки тому +2

      as a result the coriolis force written at 1:16:21 will actually be (mw/2) * ((dY/dt) * X - (dX/dt) * Y)

    • @maxmeier1015
      @maxmeier1015 3 роки тому +5

      @@TheLevano22 I think he also lost track of a factor of 2 when he expanded the squares before. So it really should be (mω) * ((dY/dt) * X - (dX/dt) * Y). This also causes his final result for the coriolis force at 1:20:17 to be off by exactly that factor. As far as I know it really should be 2mω dY/dt (or 2mω dX/dt for the y component of force).

  • @ermiasawoke192
    @ermiasawoke192 3 місяці тому +1

    The pursuit of physics with faith!

  • @andyjiao3114
    @andyjiao3114 5 років тому +1

    Professor, you are a proper theoretical physicist that I aim to be. You do not make over-arching assumptions, and promises the integrity of theory.

  • @andreamercuri9984
    @andreamercuri9984 3 роки тому +2

    We love you, Prof Susskind. Thank you so much for your free and marvellous lessons

  • @Euquila
    @Euquila 10 років тому +112

    why would you say that mathematical rigor is lacking? this is a physics lecture and he is trying to convey ideas. personally, i found this lecture to be very helpful. thank you very much.

    • @Deoxippus
      @Deoxippus 9 років тому +12

      +Euquila I second this. My professor is very fond of mathematical rigor and I feel it gets in the way sometimes.

    • @non-inertialobserver946
      @non-inertialobserver946 4 роки тому +7

      I find the lack of mathematical rigour quite refreshing

    • @orientaldagger6920
      @orientaldagger6920 3 роки тому +7

      You can always read Goldstein for something inscrutable.

    • @orientaldagger6920
      @orientaldagger6920 3 роки тому +7

      I don't get it either. There is no real high level math here. He is showing you the insights. You can read the derivation in all its detail from any book or internet site nowadays.

    • @joeboxter3635
      @joeboxter3635 3 роки тому

      Not just a question of mathematical rigor, unless you mean by rigor no basic math mistakes. His lectures are chalk full of them.

  • @VladTepesh409
    @VladTepesh409 5 років тому +1

    Looks like this is helpful if the actions are not only passing through a void or vacuum, but also when the actions transition between states of matter / elemental compounds from a to b. Like how the action of light passes through air, and transitions through water does not appear to be a straight line, but rather a path of least action such that it goes the distance from a to b in the least amount of time with respect to the transition between states of matter / elemental compounds.
    Very cool!

  • @samarthsai9530
    @samarthsai9530 6 років тому +25

    S for distance because of the latin word spatium which means space.

    • @sayanjitb
      @sayanjitb 4 роки тому

      Ahh really!

    • @joeboxter3635
      @joeboxter3635 3 роки тому

      Kewl ... so why "T" for kinetic energy? I know V is voltage and V is a potential energy.

    • @joeboxter3635
      @joeboxter3635 3 роки тому +2

      @@sayanjitb so here we are 2 years later.

    • @of8155
      @of8155 3 роки тому +1

      @@sayanjitb aha

  • @joelcurtis562
    @joelcurtis562 3 роки тому +1

    Very insightful derivation of Euler-Lagrange equations. Much in the style of EF Taylor. Much more intuitive than the typical textbook presentation that relies on integration by parts.

    • @gagadaddy8713
      @gagadaddy8713 2 роки тому +1

      Yes! The way to derive the Euler-Lagrange is what a gem to me. And this is the first time I really understand what the trick behind L = KE - PE (instead the common practice of summing up the kinetic and potential energy). Great presentation for the basic principle to "Calculus of Variation"! Love it!

  • @Rakeshkumar30
    @Rakeshkumar30 4 роки тому +7

    I keep coming to this lecture, this is such a gem.

  • @qbtc
    @qbtc 5 років тому +4

    When he was transforming between a rotating frame and a stationary frame around middle of the lecture, he meant to solve for X and Y, the rotating frame, but he unfortunately wrote x and y, the stationary frame. He would follow this using the rotating frame X and Y as an example which resulted in a Coriolis force. You would not get this in the stationary frame. That's how I know what he intended. He tends to get his notation mixed up at times and it's always good to use pencil and paper when following along to really understand and appreciate everything. Great lectures nonetheless.

  • @SM2005_
    @SM2005_ 3 роки тому

    I’d like to point out how relaxed he is just speaking. Just rolls off his tongue while multitasking while teaching strangers something complex.

  • @joeboxter3635
    @joeboxter3635 3 роки тому +1

    @32:30 Start Derivation of Lagrangian
    @48:44 Summary of Result
    There are other derivations that are closer to first principles. Some even have youtube videos. But in 10 min he shows energy is conserved is a constraint of motion and with this assumption alone leads to newton's equation which was simply accepted fact.
    So while there are presentations that are closer to first principles and thus more mathematical in nature, got to love this man's conveying the most physics with least effort and least time. (Took 10 min to convey. Starts around min 38 and done at min 48). Nature would be proud.
    But then could we expect any less from a world renown physicist and educator?

  • @LaureanoLuna
    @LaureanoLuna 7 років тому +1

    The problem with using infinitesimals is readily seen here, in the very beginning of the lecture: a mínimum in potential energy has these two inconsistent effects: one, if you change the input just a bit, energy increases (because it was at a minimum); two, if you change the input just a bit, nothing changes, because the derivative is 0.

  • @Gruemoth
    @Gruemoth 4 роки тому +4

    the camera(wo)man had one job to do and he/she did it magnificently well! thank you mr or mrs camera(wo)man

  • @jsh31425
    @jsh31425 4 роки тому +1

    One slightly confusing notational point: In his derivation of the Euler-Lagrange equations (around 44:00), he keeps writing del L / del v_i and del L / del v_{i+1}. But L itself is a function of only two variables, say x and v. He means to write the partial del L / del v, but *evaluated* at two different points.

  • @felicityc
    @felicityc 3 роки тому +1

    The comments about time in culture and writing is very interesting, since I truly think I was able to start thinking "backwards" in a sense, or rather, in alternate directions (since there is no 'right' way, per se, but relative to my upbringing), when I studied arabic script and Islamic languages. Being able to flip back and forth is quite incredible. I know it was a joke but it is an extremely poignant point to make.

  • @DrDress
    @DrDress 5 років тому +1

    12:00 calculus of variations
    45:00 E-L discrete derivation

  • @joandres1482
    @joandres1482 4 роки тому +11

    just to the little question after 15min: the history of using "s" as the variable for distance is that it comes from the German word "Strecke"

    • @ahmetkaraaslan8429
      @ahmetkaraaslan8429 4 роки тому +2

      are you sure? because I didn't look at historical articels but I learned from my physics classes that the integral symbol came from sum and we use the word "die Summe" still in German and it means sum. moreover, in latin the word Summa means sum maybe these words are coming from latin

    • @alalize
      @alalize 4 роки тому

      Latin Summa I believe.

  • @xhonshameti1749
    @xhonshameti1749 3 роки тому

    So Concise! He Knows and feels the fabric! Beautiful!

  • @benjamincordes207
    @benjamincordes207 10 років тому +4

    Great lecture, thanks so much for sharing this. I found this very helpful and well explained.

  • @John-lf3xf
    @John-lf3xf 4 роки тому +1

    The calculus of variations is concerned with finding functions which minimize certain quantities.

  • @rightwraith
    @rightwraith 9 років тому +43

    His rotated coordinate transformations at 1:06:10 are slightly wrong, he messed up the signs. The second term of the x transformation should be negative, and the first term of the y transformation should be positive.
    These are the proper transformations:
    x = Xcos(wt) - Ysin(wt)
    y = Xsin(wt) + Ycos(wt)

    • @MoronicAcid1
      @MoronicAcid1 9 років тому +7

      rightwraith For a second, I thought I was just retarded.

    • @NuclearCraftMod
      @NuclearCraftMod 9 років тому +3

      +rightwraith I think what he has is correct, as your equations are the transformations from x and y to X and Y, while he's doing X and Y to x and y.

    • @NuclearCraftMod
      @NuclearCraftMod 9 років тому +3

      +rightwraith I think what he has is correct, as your equations are the transformations from x and y to X and Y, while he's doing X and Y to x and y. I think what he has got wrong, though, is that he's got x and y the wrong way round.

    • @rightwraith
      @rightwraith 9 років тому +1

      +NuclearCraft Mod No, mine are correct for the transformation from the rotating (X, Y) frame to the unrotating (x. y) frame.

    • @NuclearCraftMod
      @NuclearCraftMod 9 років тому +3

      rightwraith Ah yes, my mistake ;)

  • @tshankomakech1875
    @tshankomakech1875 2 роки тому

    the concepts are well explained in his lectures for every one to understand, thanks for this lecture i appreciate it

  • @plutopulp
    @plutopulp 12 років тому +3

    it's just the chain rule seeing as v is a function of x: dL(x,v)/dx= dL/dx+ (dL/dv)*(dv/dx)
    The d's above should be partials though, and just wrote it with x and v to make it clearer (hopefully!)

  • @TebiByyte
    @TebiByyte 5 років тому +9

    Huh, what an interesting way to derive the Euler-Lagrange equation.

  • @kaushaltimilsina7727
    @kaushaltimilsina7727 4 роки тому

    The idea that I feel like is fundamental or rather synonymous to the stationary action , is that we are defining evolution of a system through evolution between instantaneous States of equilibrium; because by definition we are minimizing something. And I think evolution through "instantaneous eigenstates" is where this idea goes down to quantum mechanics.

  • @DoisKoh
    @DoisKoh 10 років тому +36

    It's Mike Ehrmantraut!

  • @vihaannair5165
    @vihaannair5165 2 роки тому +1

    I appreciate the effort by Susskind but the mathematics would not seem very easy to comprehend for beginners. The derivation by Susskind is different from most textbooks and I don’t think anyone can derive the Euler-Langrange equation who just learned calculus, Newtonian style. I would suggest that you should watch this lecture and read from the book only if you are well versed in the topic.

  • @waynelast1685
    @waynelast1685 4 роки тому

    Not to be picky, but at 1:15:11 the far right term in the Lagrangian does not have a 1/2 factor, so this works it way to the final result where the coriolis term has a factor of 2 , not 1 in it. So.... coriolis force in X = -2mwYdot and in Y = 2mwXdot . Doesn't change the explanation of the physics , but just numerical value.

  • @hasanshirazi9535
    @hasanshirazi9535 4 роки тому

    Stationary particle in a rotating frame does in fact experience force which keeps it stationary and hence it is accelerating. It will have potential energy but no kinetic energy. However, the same particle in a stationary frame will have kinetic energy.. So yes energy varies according to the Frame of Reference.

  • @JP-re3bc
    @JP-re3bc 6 місяців тому

    This guy is a great teacher.

  • @guyedwards22
    @guyedwards22 2 роки тому +2

    So I've always grappled with wondering why the hell the Lagrangian is KE - PE instead of the sum, but it finally clicked in the section where he was talking about that very weirdness.
    d/dxi of the negative Potential field are the components of the forces acting on a system(F=-grad(PE)), and d/dxi of Kinetic Energy is zero since it's independent of the system's position
    On the other hand, d/dx•i of classically defined KE are the components of linear momentum of the system, and d/dt of momentum is the sum of forces acting on a system (the more generalized form of Newton's second law), but d/dx•i of -PE is zero since PE depends purely on position, not the nature of motion
    The Euler-Lagrange equations set d/dt(d/x•i)(L) equal to d/dx(L), or, by the logic above, two ways of arriving at net force equal to each other using different derivatives of a single functional.
    That unique functional is the Lagrangian, and I see why it is what it is now!
    Edit: Also, holy cow, Coriolis forces seem super obvious now; if you have a velocity in the Y direction in the rotating frame, you're essentially "keeping up" with the rotation and counteracting the perceived centrifugal force. It scales with omega rather than omega squared because your Y velocity is independent of radial distance, so at greater x values, you'd have to compensate by having a greater Y velocity to cover that larger circular path

  • @flyingdutchman2649
    @flyingdutchman2649 7 років тому +1

    Finally find intermediate level mechanics lecture

  • @evanparsons123
    @evanparsons123 6 років тому +1

    (first 10 minutes) why wouldn't equilibrium be based on the point with respect to the axis not the actual tangent of the line at a point? I realize this is a very basic question but by his logic couldn't one argue all possible points are in equilibrium even if it is a derivative of V?

  • @NocturnalJin
    @NocturnalJin 10 років тому +3

    It's clear to me, looking at the shape of the letter S, that it is the only letter they could have chosen to represent the length of a curve.

    • @peterbonnema8913
      @peterbonnema8913 10 років тому +1

      even though it doesn't :P

    • @valentinjonas8995
      @valentinjonas8995 7 років тому

      I think it derives from the german word "Strecke" meaning distance in some situations.

    • @samarthsai9530
      @samarthsai9530 6 років тому +2

      It is derived from latin word spatium(=space).

  • @feiqu6693
    @feiqu6693 10 років тому +1

    I really appreciate that he just used a epsilon to prove the equation.

  • @waynelast1685
    @waynelast1685 4 роки тому

    Regarding his example at 1:00:00 ... he states in the beginning the particle is not moving in the x-y frame, and that the carousel is rotating, and that we want to know what the person on the carousel "sees". But this is a subtle point.... the particle is NOT fixed to the carousel nor held in any kind of rotational motion... it took me awhile to make that distinction when interpreting the results . If you assume the particle is held in a rotational motion, then there must exist a centripetal force in the x-y frame, thus a potential. In this case the Lagrangians in the x-y and X-Y frames look a little different than the case without a potential in the x-y frame, and thus give you slightly different equations of motion and solution interpretations. Mainly, if assuming a fixed particle in the rotating frame, there is no NET force on the particle (assuming it has no velocity relative to the rotating frame) , thus there is no X or Y doubledot ( acceleration in X-Y frame). This makes sense...if no net force, then no acceleration is observed. Too complicate a little, if the particle has a velocity relative to the rotating frame then there is another force ( coriolis force as a result of the coriolis acceleration). So why is "fixed" particle vs "not fixed" particle important? Because it helps to understand how the Potential function translates between the two frames, and to understand the different resulting dynamics . Not too necessary to explanation of Lagrangians, but helpful anyways in understanding the particle dynamics.
    So if the particle is NOT fixed to the carousel, it does not actually have a net force on it. It APPEARS to have a centrifugal force if you are riding on the carousel , making it move outwards, with corresponding acceleration. Likewise if the particle had already been moving when the carousel was rotating, with a velocity relative to the carousel, there APPEARS to be force ( coriolis) making it move perpendicular to the velocity . The person in the x-y frame observes NO motion at all ( thus no real forces). So this may appear obvious but it is subtle points in the rotation dynamics and understanding the potential functions in both frames. Sorry if I am o particular but I wanted to understand potentials in different reference frames, as it relates to using Lagrangians to do physics.

    • @Metallurgist47
      @Metallurgist47 2 роки тому

      I puzzled over this too .
      If the particle is stationary in the x-y frame , then presumably there is no friction between the carousel and particle -- otherwise , it would move in the x-y plane .
      That is , the surface of the carousel is a frictionless plane , rotating beneath the particle. But to the observer on the carousel, in his (X_Y)coordinates , the particle appears to be rotating in a circle, at a fixed radius.
      So to him , that would imply that there is a "centripetal" force keeping it at that fixed radius.
      Anyone ?

  • @robertfulton6397
    @robertfulton6397 3 роки тому +1

    I really enjoyed the questions at the end

  • @animimm
    @animimm 12 років тому +1

    At 1:04:40 he had to name the blue coordinates system as XY, and the red ones as xy, so that the transformation equations will be true!

  • @brendanward2991
    @brendanward2991 6 років тому +7

    1:06 - I think he's got the minus sign on the wrong term:
    x = Xcoswt - Ysinwt
    y = Xsinwt + Ycoswt

    • @florianwerne890
      @florianwerne890 5 років тому

      true

    • @qbtc
      @qbtc 5 років тому

      You are right. If you solve for the X and Y system, you just reverse the signs of the sinwt terms and that is what he has. He wrote the eqn's for the rotated system which is X and Y.

  • @masterineverything
    @masterineverything 11 років тому

    I think the difference is one of rotating direction, i.e the sign of w. Which would both plausible, but yours is preferred

  • @michaelgarcia812
    @michaelgarcia812 2 роки тому

    At 1:35 Dr. Suskind states that one term in the Lagrangian that has mass and velocity does not contribute to energy. Why doesn’t it? He didn’t elaborate…..

  • @devnull256
    @devnull256 7 років тому +4

    is that actually true? (very beginning)
    "... equillibrium is when object does't move... or accelerate?!... " then he says that net force must be zero. But surely that still impies that the object can move with constant velocity. Clearer Definition of equillibrium must be given.

    • @joelbiffin
      @joelbiffin 7 років тому +3

      The definition he gives is true. Provided that we are in an inertial frame of reference, equilibrium most commonly refers to when the net force is zero - i.e., acceleration is zero which implies the velocity is constant (remember that when velocity=0 is a special case of the velocity being constant)

    • @joelbiffin
      @joelbiffin 7 років тому +1

      The 'next stage' of the definition of equilibrium would be to say that there is no NET torque in the system. But we haven't covered rotational dynamics yet so this is not an issue.

    • @devnull256
      @devnull256 7 років тому

      Joel Biffin : that's true of course. I'm just having problem with first phrase that the "object does not move" at equilibrium.

    • @devnull256
      @devnull256 5 років тому +2

      @@ErreniumWell... in the object's own frame of reference the object is always "does not move" (it is it's stationary frame of reference.) , so it's pointless to say it does not move in it.

    • @devnull256
      @devnull256 5 років тому +1

      What was probably meant is: if the object does not move, it is at equilibrium, but not necessarily the other way around (if it is at equilibrium it does not move)

  • @Sans_K5
    @Sans_K5 Рік тому +1

    thanks sir for amazing lectures❤🙏

  • @mohabmetwally5749
    @mohabmetwally5749 4 роки тому

    why there is a negative sign between the kinetic Energy, and the potential, why Energy isn't used instead for example, T - V is very specific case, the lagrangian must include, the coordinate, and it's derivative, and T - V satisfy this, but why not E, or -E, or V - T?

  • @constantinlefter8235
    @constantinlefter8235 7 років тому

    The demo about coriolis force has many errors. 1. The last therm of the derivation is not omega * (x dot * y-y dot * x) . It is 2 * omega * (y dot * x - x dot * y). From this error it goes to the coriolis force formula where the 2 is lost and the signs are reverced. fx coriolis is 2 * m * omega * y dot. !!!!

  • @tylerboulware6510
    @tylerboulware6510 8 років тому +4

    Is the Coriolis force term correct? I think there's a factor of 2 that goes missing on the Coriolis term when he multiplies the Lagrangian out in the rotating reference frame. Can anyone confirm?

    • @clarity97
      @clarity97 8 років тому +4

      Yes there will be a 2 with the Coreolis term

  • @andrewtaylor9799
    @andrewtaylor9799 3 роки тому +1

    The form of L as L = KE - PE seems to be just pulled out a hat. Is there any intuition for why this should be right, other than it can be used to derive F = ma in Cartesian coordinates? I see that a coordinate-free description is extremely advantageous. But actually there was no proof that the L stated even gives this; the claim was just made. Is he appealing to the fact that this has been empirically verified, and that the Lagrangian formulation also an extremely convenient method of "bookkeeping"? Or is their some physical understanding that motivates this form of L?

    • @joeboxter3635
      @joeboxter3635 3 роки тому

      There are other derivations that are closer to first principles. Some even have youtube videos. But in the end it is a philosophical belief that energy moving between KE to PE and back will have no loss, ie energy is conserved.
      The two forms of E: KE is a function of time, where as PE is a function of space. Thus if there is a change in one into the other, KE must change in time in equality to PE change in space. Ie, energy is conserved. Nature will enforce this rule at every point in space and every moment in time and hence constrain the motion of partical accordingly. The constrained path is the same as newtons law.
      So while there are presentations that are closer to first principles and thus more mathematical in nature, got to love this man's conveying the most physics with least effort and least time. (Took 10 min to convey. Starts around min 38 and done at min 48). Nature would be proud. But then could we expect any less from a world renown physicist and educator?

    • @andrewtaylor9799
      @andrewtaylor9799 3 роки тому

      @@joeboxter3635 Conservation of energy is based on looking at KE + PE, not KE - PE. What is being posited here is not that total energy is conserved, but that the path followed is an extremum (or saddle point) of the integral of KE - PE. The video gives no logical explanation for this, and merely states it as if it was given by God...without even admitting to doing so. It's intellectually dishonest.

    • @joeboxter3635
      @joeboxter3635 3 роки тому

      @@andrewtaylor9799 No. That is only one way to look at it. KE + PE at start = KE + PE at finish is what you are saying. But rearrange with KE on one side and PE on other side and you see change between two forms occurs without loss follows. They mean the same thing and only seeing one way is to miss the point of this lecture.
      And who knows? Lagrange came up with this at 19 without any formal education in math or physics. So maybe God did give it to him. Lol.

  • @utility00
    @utility00 3 роки тому

    Having little to no exposure to variational calculus I find Euler-Lagrange equation explanation messy and unclear. I think Landau treats this issue with more clarity in his Classical Mechanics book. Probably will have to read a bit more on variational calculus because I've lost track of what prof. Susskind was trying to convey by the 47th minute.

  • @ozzyfromspace
    @ozzyfromspace 7 років тому

    So, the action integral minimizes the trajectory of a point-mass particle in generalized coordinates based on the lagrangian T(x,x')-V(x,x'), but the trajectory of light is minimized using a time lagrangian. Can someone explain to me why it seems that the principle of least action prefers to optimize trajectories based on different parameters depending on the system?
    For example, the brachistochrone problem in elementary variational calculus allows us to derive a path that minimizes the transit time of a particle between two points in a gravitational field, but the path is not the shortest path possible. That means that we can ENGINEER a system to transport a particle by minimizing the distance a particle covers between points A and B, or the time it takes said particle to go from A to B. If we have this level of choice about what action to minimize in the brachistochrone problem, how does nature decide when to optimize for energy and when to optimize for time.
    I believe this to be a problem of my understanding, and not the principle of least action itself. All help/insight is appreciated.
    Thanks, and great video series!

  • @arpitchoudhary6475
    @arpitchoudhary6475 Рік тому

    Excellent lecture content-wise but is it only I who is disturbed by the camera moving all the time?!

  • @charlesabernathy5842
    @charlesabernathy5842 3 роки тому

    Great discourse. I'm still hoping to learn that.

  • @shyamvijay8985
    @shyamvijay8985 4 роки тому

    But isn't it a circular argument (shortest distance is a straight line) since it uses dS at the short increment is the 'distance' and it's computed as dS = sqrt ( dx2 + dy2)

  • @maxfindus
    @maxfindus 12 років тому +1

    correction @ 1:13:26 the last term should be mw(X'Y-Y'X) (factor 2 error)

    • @willie5069
      @willie5069 6 років тому

      Checked this with mathematica and I believe you are correct. I also got wm [Y[t] Derivative[1][X][t]--X[t] Derivative[1][Y][t]. Only took me 5 years to answer.

  • @willie5069
    @willie5069 6 років тому

    I checked the result with mathematica and terms in the Langragain for the X,Y (upper case) are correct.

  • @waynelast1685
    @waynelast1685 4 роки тому

    He likes to to do derivations instead of just telling you the formulas and doing examples. Not all the examples here or in future lectures do not match the book by the way. But the example at 1:00:00 is in the book. I wanted to see the previous example (in the book) to have explanation why V(x) = V(X) in a translational frame where x= f(t) + X . Anybody know why please let me know. I guess I am missing something. Hope it is not so obvious.

  • @therealjordiano
    @therealjordiano 11 років тому +2

    that is some awesome concept imo.. the part with the angular momentum xd physicsgasm

  • @fjolsvit
    @fjolsvit 12 років тому

    OK. I now see what he was doing. I was thinking delta L_i = L_i -L_i-1. It was interesting to try my approach. It feels like it should work, but I don't get the difference @L_i+1/@v - @L_i/@v. Both partials end up negative.

  • @Asdun77
    @Asdun77 4 роки тому +1

    God bless you

  • @TM-et7wi
    @TM-et7wi 3 роки тому +1

    Funny how people that look the same also move the same and talk and sound the same. Mike Ehrmantraut thanks for this lecture

    • @Yan_Alkovic
      @Yan_Alkovic 3 роки тому

      That's exactly what I think every single time I watch one of his lectures
      Though TBH Mike has a rougher voice.

  • @abhinandanmehra7765
    @abhinandanmehra7765 3 роки тому

    Did anyone knew at which age he presented these lectures as he is 80 years old now in 2020 but didn't seem to be in this video.

  • @onebylandtwoifbysearunifby5475
    @onebylandtwoifbysearunifby5475 3 роки тому

    21:10 was hilarious.
    And hello from the future,
    where that problem is ongoing.

  • @martingreen436
    @martingreen436 6 років тому

    I didn't watch the entire lecture (yet). Does he cover using Lagrangians in systems with applied forces? Also I just realized these series of lectures are longer versions of the book "Theoretical Minimum (part1) "

  • @John-lf3xf
    @John-lf3xf 4 роки тому

    Shortest path between two points on a curved surface is called a geodesic.

    • @danielblumowski34
      @danielblumowski34 3 роки тому

      it actually isn't, you can imagine a plane with a hill, a geodesic from one side of the hill to the other side would go all the way up to the top but of course it isn't the shortest path (you can just go around the hill)

  • @allureee
    @allureee 12 років тому +1

    I don't understand how he gets the derivative of the second expression at 44:18..

  • @AdenKhalil
    @AdenKhalil 19 днів тому

    He does remind me of my teacher who who used to eat while teaching

  • @waatchit
    @waatchit 10 років тому +2

    Can someone please explain the differentiation that he does from 42nd minute to 47th minute to arrive at the Lagrange equation @ 47th minute.

    • @liberphilosophus7481
      @liberphilosophus7481 4 роки тому

      He's taking the multi-variable derivatives of L(X(i)), but he doesn't find the actual derivatives to avoid mathematical rigor.

  • @wafikiri_
    @wafikiri_ 10 років тому +2

    One of the few gorillas that were taught sign language and thus communicate with humans had this concept of time orientation: past, ahead; future, behind. For you can't see the future but can remember the past.

    • @alalize
      @alalize 10 років тому

      That's how aztecs used to see time.

  • @bigtimernow
    @bigtimernow 5 років тому

    He seems to be talking about perturbation theory and finding the steady states from 47:30 - 48:45.

  • @vishwanathkr22
    @vishwanathkr22 10 років тому +11

    The resolution of co-ords are wrong. x=X cos wt - Y sin wt; y=X sin wt + Y cos wt. The guy is really shaky with writing stuff.

    • @andydawson5678
      @andydawson5678 10 років тому

      Yes there is an error - however his equations are correct for a clockwise rotating frame - unfortuneatly he showed a anticlockwise rotation. By convention an increasing angle is taken to be a clockwise rotation.

    • @jadpole
      @jadpole 9 років тому +1

      Andy Dawson No, he did show a clockwise rotation! He expresses (x, y) as a clockwise rotation of (X, Y) by ωt.

    • @christineauyeung4983
      @christineauyeung4983 6 років тому +1

      He's correct and the rotating axis (X,Y) is rotating anticlockwise. He's expressing x in terms of X, that means you are observing the stationary axis (x,y) on the turntable (X,Y), which moves away from you clockwise.

  • @fjolsvit
    @fjolsvit 12 років тому

    @ 0:41:44 I believe he should be subtracting L_i-1 from L_i, not adding. This developement is intuitively very compelling, but mathematically a bit sloppy. Since the intuitive part is that which matters in this case, there's not much of a loss.

  • @jmath8988
    @jmath8988 5 років тому +2

    What's interesting is Lagrange was only 19 when he formulated some of this.

    • @seandafny
      @seandafny 4 роки тому

      He fudged the numbers of his age. Anyways much of this isnt that hard to do. Newton was better

  • @ares12265
    @ares12265 7 років тому

    The key question is, strength weighs something or not.

  • @12388696
    @12388696 10 років тому +1

    excellent as usual.

  • @nicholashalden601
    @nicholashalden601 8 років тому +8

    Love the fact that this guy is a fucking Genius and don't even know how to write: "for all i" in mathematical term ! XD

    • @bagoplayer7455
      @bagoplayer7455 8 років тому +2

      It doesn't matter if you know what you're talking about

    • @netrapture
      @netrapture 6 років тому +1

      @nicholas halden - you misquoted him. He said how to write it the way *mathematicians* write it.

  • @jenniferlaflora3293
    @jenniferlaflora3293 11 місяців тому +1

    LaFlora Sinish 41:43

  • @samerapornpan8423
    @samerapornpan8423 4 роки тому

    At 44:51, did he mess up indices? If the Lagrangian corresponding to the right piece is a function of x(i) and v(i), should not the second term of its derivative be minus one over epsilon times the partial derivative of L with respect to v(i) ? Using chain rule?

  • @potugadu5160
    @potugadu5160 9 років тому +2

    I have a question: I plan to watch all these lectures by Mr Susskind on classical mechanics, but will I get anything out of these lectures without an accompanying textbook?
    Thanks.

    • @lsbrother
      @lsbrother 9 років тому +1

      +potugadu I guess if you have any questions - which I would think is almost inevitable! - then a text book will be useful.
      Lecturers tend to occasionally make mistakes, mess up explanations, get lost and sometimes plain get it wrong and Lenny is no exception!

    • @BongboBongbong
      @BongboBongbong 8 років тому +3

      +potugadu I can recommend Goldstein's Classical Mechanics

    • @collegemathematics6698
      @collegemathematics6698 8 років тому +1

      the text book associated with these lecture is "theoratical minimume" by prof. susskind him self

    • @tehyonglip9203
      @tehyonglip9203 7 років тому +2

      What i recommend is, watch these lectures first, absorb the mistakes he make and forget all of them, you are here to learn physics not to argue with correct equations, his equations are explained at least 10 times before changing topic, so it should be very easy to learn physics here, you are here to learn the concepts, books are for the advanced level people where they keep track of everything and almost no explanation of the concept, I dont know... for me, books are too difficult for REMEMBER! CONCEPT IS WHAT MATTERS

    • @abhishekcherath2323
      @abhishekcherath2323 7 років тому +1

      I'm using goldstein's classical mechanics, these lectures make that book actually readable.

  • @tianmingguo8271
    @tianmingguo8271 4 роки тому

    Brilliant explanation.

  • @yingwang1053
    @yingwang1053 8 років тому

    1:26:48 Why =mr *(theta dot)^2- dV/dt? Is that the derivative of r?
    If is, why not r double dot instead?
    Thx

  • @alantew4355
    @alantew4355 8 років тому +4

    Must the action has exactly one stationary value?

    • @stressedowl795
      @stressedowl795 8 років тому +1

      By stationary values we mean to refer to the "shortest path"; The shortest path can only be one.

    • @alantew4355
      @alantew4355 8 років тому +2

      juhi singh I suppose your "shortest path" means the path that gives a global minimum value of action. While it's true that there is only one global minimum, it's not true that the action has to be minimum. The physically observed path may be one that makes the action a saddle point.

  • @filiepsamyn4450
    @filiepsamyn4450 2 роки тому

    The langrian is introduced as a known concept without explaining what the langrian is.

  • @merlinthegreat100
    @merlinthegreat100 7 років тому

    On the second problem, is there a potential for the left hand side?

  • @madhavpr
    @madhavpr 11 років тому

    Isn't the V_fictitious energy like the rotational kinetic energy of the particle= 1/2 * moment of inertia * r^2 ? Correct me if I'm wrong.

  • @peterbonnema8913
    @peterbonnema8913 10 років тому

    At 44:29, shouldn't the second term be 1/epsilon times the partial derivative of L with respect to v sub i-1? He writes down v sub i instead of v sub i-1 which I think is wrong since he is differentiating the L(x sub i-1, v sub i-1) term.

    • @tylerboulware6510
      @tylerboulware6510 8 років тому

      I think what you're saying is right, but in the end it doesn't matter since you're letting epsilon go to 0, so the only difference is you're moving in from the left or from the right and both give the same result.

  • @simontimothy7051
    @simontimothy7051 24 дні тому

    21:25 what did he mean by neutrinos travelling faster in Italy ? Can someone explain it?

    • @AdenKhalil
      @AdenKhalil 19 днів тому

      He was joking!

    • @troelsfrostholm
      @troelsfrostholm 14 днів тому

      en.wikipedia.org/wiki/2011_OPERA_faster-than-light_neutrino_anomaly
      The mistaken result made the rounds in the news the same year the lecture was given.

  • @orientaldagger6920
    @orientaldagger6920 3 роки тому

    Now that derivation for the Lagragian is cool ! Compared to Goldstein....

  • @DenizBoz
    @DenizBoz 8 років тому

    Regarding the first 20min of the lecture, why variation (for both cases of function and functional) is 0 when we're at the minimum? What does it mean "the change is 0 TO THE FIRST ORDER"? What does TO FIRST ORDER means? Does it have to do with Taylor expansion? If so, how? (I don't remember basic calculus stuff well, sorry) Thanks.

    • @bagoplayer7455
      @bagoplayer7455 8 років тому +1

      It just means the derivative of the function is 0 at that point

    • @LaureanoLuna
      @LaureanoLuna 7 років тому +1

      At a minimum of the function, the first derivative is 0 but the second is positive; this is why he says 'first order'. However, the use of infinitesimal quantities messes it all a bit up: if the function is at a mínimum, then any variation in its argument should increase it.

    • @DenizBoz
      @DenizBoz 7 років тому +1

      Laureano Luna Thanks

  • @danielstone8775
    @danielstone8775 7 років тому

    because it's a constant

  • @standardcoder1184
    @standardcoder1184 Рік тому

    1:16:22 Good stuff

  • @askhetan
    @askhetan 4 роки тому

    Isn't the frame with X and Y a non-inertial frame with respect to the frame with x and y? why does Lagrangian mechanics hold in that frame ?

    • @kharanshu2854
      @kharanshu2854 3 роки тому

      it isn't undergoing any acceleration wrt the inertial frame, so the XY frame is still an inertial RF