These are amazing lectures. Susskind's presentation is perfectly clear and perfectly paced, with the relaxed energy and humor of a big personality, but without the arrogance and implied intimidation that so often accompany it.
@markkennedy97675 жыл бұрын
I completely agree. He's at the top of his profession and he makes you feel he's just figuring things out himself as he goes. Making the student want to think about it. A great quality in a teacher.
@lelomambueliane49152 жыл бұрын
the best video ever seen on classic mechanics kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@lucasmcguire15549 ай бұрын
Perfectly put. He has confidence without arrogance, very good but somewhat rare combination
@joabrosenberg29612 жыл бұрын
An example of a wedge; Second example of double pendulum 27:00; Hamiltonian and Harmonic Oscillator in Phase space 53:30; Hamilton Equations 1:13:30; Q&A 1:29:00
@halilibrahimcetin94484 жыл бұрын
Thanks to Leonard Susskind , I feel more confident to learn more advanced topics.
@ozzyfromspace7 жыл бұрын
These lectures get better and better every class!
@seandafny9 жыл бұрын
this man is in love with harmonic oscillators.
@tobywhite110011 жыл бұрын
This is fun, sort of. Susskind has a great sense of just how fast he can push without getting me totally confused and giving up. I feel like a greyhound chasing a mechanical rabbit. I never quite catch up, but I can get close enough to stay in the race if I keep running as fast as I can. (There's probably some Lagrangian to describe that kind of motion, but I don't even want to know it)
@TheMaximumGForce12 жыл бұрын
Loving this series! cant wait to get to quantum mechanics!
@lelomambueliane49152 жыл бұрын
the best video ever seen on classic mechanics kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@camilodominguez46784 жыл бұрын
" If you know the rules of Lagrangian Mechanics, it's a mechanical exercise, a completely mechanical exercise you can be dumb as hell and still solve the problem"- Leonard Susskind.
@EdSmiley5 жыл бұрын
Anybody who doesn't have an intuitive feeling for how complexly odd the double pendulum's equations of motion laid out by Dr. Suskind are in practice should look at a KZbin video of a double pendulum. You'll be glad you did!
@abhishekcherath23237 жыл бұрын
I'm not getting the same expression for the kinetic energy for the double pendulum Nvm got it I'm amazed how well I'm understanding this. Prof is so good at explaining this stuff
@habblebabble7011 жыл бұрын
Canonical momentum is essentially a definition of momentum given by the LaGarange equation. It is, by definition, the partial of the LaGarange with respect to the time derivative of the space coordinate (most recognizable as velocity). I recommend wrestling with the LeGendre transformation when you get the chance. Since the LaGarange, in general, is not always in a conservative field, the units tend to change when the cononical momentum is taken.
@ivancozza77025 жыл бұрын
LaGrange = Lagrange
@ArabyGUC5 жыл бұрын
1:20:03 Here one has to smile with satisfaction at the simplicity and elegance of the derivation :)
@roberthumphreys55945 жыл бұрын
Check out his "Theoretical Minimum" series of books, available on Amazon and other booksellers! They're great, especially for people approaching the subject for the first time, or coming back having forgotten calculus!
@hasanshirazi95354 жыл бұрын
This is a best explanation of Lagrangian and Hamiltonian.
@seandafny9 жыл бұрын
Very nice and smooth. I hope the rest go like this. Hamiltonians seem so much less complicated then them lagrangians.
@seandafny9 жыл бұрын
Sean Dafny Though Lagrangians seem more useful.
@aeroscience98347 жыл бұрын
Sean Dafny really? I find the lagrangian to be simpler
@jessstuart74956 жыл бұрын
I think there is a sign error at 38:00. The cos(alpha) term should have a (theta_dot + alpha_dot) factor, not (theta_dot - alpha_dot).
@aarshchotalia53165 жыл бұрын
Right
@jamesdowns725 жыл бұрын
His explanation for how beautiful the Hamiltonian is is how you can see these phase portraits and gain an understanding of the cycles and non convergent behavior, etc... but I keep thinking to myself, "but you could take Newton's equations and just plot x vs xdot and see that too..." Also, I am slightly confused how he initially explains that it was vitally important in coming up with the Lagrangian that it must be T-U, not T+U, in order to obtain the equations of motion after applying the Euler Lagrange equation. Then Hamilton comes along and decided, "no, it's better to make it T+U" after all? Is his formulation kind of like an add-on or adjustment to the Lagrangian so that it can still give the equations of motion using the Euler Lagrange operations but also does not change with time? What does it mean, intuitively, that the action that the Lagrangian minimizes is T-U? Is it that as kinetic and potential energy trade back and forth through the course of a system in motion, that it is done so in such a way that their difference is kept as minimal as possible? Since there is no constraint on holding total energy constant with time in the Euler Lagrange, is this achieved just accidentally?
@karolispetruskevicius81789 жыл бұрын
Hi all. Hope you have all enjoyed these lectures as much as I did. I was just wondering if anyone knows where I could find original lecture notes for this series? Much appreciated!!
@cubbtom9 жыл бұрын
Karolis Petruskevicius www.lecture-notes.co.uk/susskind/
@seandafny9 жыл бұрын
Thomas Cubbins THANK YOU SO MUCH I LOVE THE WORLD ! U ARE AWESOME !
@Sans_K5 Жыл бұрын
thanks sir for these amazing lectures❤🙏
@brainstormingsharing13093 жыл бұрын
Absolutely well done and definitely keep it up!!! 👍👍👍👍👍
@joeboxter36353 жыл бұрын
+mg was right and he changed it to -mg at the urging of the same student who tried to convince him derivative sin is -cos. If you have d/dt of momentum = -mg then you are saying the F (derivative of momentum) is acting on a falling object is not in the same direction as the acceleration. Both have to be in same direction and hence same sign. The F is down and mg is down. So the signs on both sides have to be same. If sign is reversed then resulting Force would be in opposite direction of the mass time acceleration: g. When you do L = T - V, V = mgy. As y increases V increases. L = T - mgy. He had this right. Then take partials bringing the contribution from d/dt (dT/dq-dot) = mg. Lastly, let's say he was right: F=-mg. Then this means F + mg = 0. But the gravitational force is suppose to be conservative. The plus on left breaks this. On other hand, F = mg, then F - mg = 0 is valid as we would expect for a conservative force. So it has to be F = mg. This man absolutely knows his physics. But he lets his students questions and comments fluster him into making mistakes. No wonder he admitted in lecture 3 that he makes algebraic mistakes. It's cause he is not in a quiet place to think but has to perform in the presence of confusion.
@lelomambueliane49152 жыл бұрын
Can it be denied that this guy solves the most difficult problems? kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@thebestofthebest94948 жыл бұрын
Bon appetit, Leonarodo!
@jubilaeumskagen7 жыл бұрын
Scone energy transfers into Susskind energy. Use the lagrangian to calculate the laws of chewing noises.
@joeboxter36353 жыл бұрын
+mg was right and he changed it to -mg at the urging of the same student who tried to convince him derivative sin is -cos. If you have d/dt of momentum = -mg then you are saying the F (derivative of momentum) is acting on a falling object is not in the same direction as the acceleration. Both have to be in same direction and hence same sign. The F is down and mg is down. So the signs on both sides have to be same. If sign is reversed then resulting Force would be in opposite direction of the mass time acceleration: g.
@Yan_Alkovic3 жыл бұрын
His q at 1:36:00 actually has units of square root of action, surprisingly
@lelomambueliane49152 жыл бұрын
Can it be denied that this guy solves the most difficult problems? kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@matschreiner2 жыл бұрын
1:29:30 - 'Quantum Mechanics would go to hell in a handbag if you tried to take some of these other cases' Haha,
@leonig10010 жыл бұрын
At 1.30 there was a discussion on dimensions. q initially is the definition of a coordinate system and in my view the dimensions of this must be position as one of the students said. How does the dimension demonstrated by th professor relate to this?. Is the coordinate system selected in a special way to make this true? Perhaps another way to put it is how does one express this in terms of the coordinates geometrically.
@toshiro0o10 жыл бұрын
To answer your final question: straight lines. The coordinates he chose were proportional to those of space but by a factor that has units of its own.
@mrfrankincense10 жыл бұрын
Is the first problem supposed to be that it is kept on the plane by a normal force?
@toshiro0o10 жыл бұрын
You can think about it as being kept on a horizontal frictionless surface if there is gravity, but the way he explained it was that it was in absence of gravity or any potential field, so if initially you give it motion that keeps both pendulums in the same plane that's how it will stay. He just didn't mention the initial conditions or he was just thinking of 2D space.
@orientaldagger69203 жыл бұрын
Yes if you do the balance of force thing, but not really relevant here.
@willie50696 жыл бұрын
I am desperate to find out how he came up with the omega parameter at 1:03:40. I can not seen to get form the standard fornm wuth K and m. Any helpl would be greatly appreciated.
@y09t9b36 жыл бұрын
At 1:32:00 he explains how he derives omega from k and m via a change in coordinates.
@TheLevano223 жыл бұрын
There is a known formula for springs: T = 2pi * sqrt(m/k). f = 1/T, so we have to take both sides to the power of -1, which will result in: f = (1/2pi)* sqrt(k/m) or w = sqrt(k/m).
@rfmo838510 жыл бұрын
do you have to use the rectangular (x,y) components of velocity when you write down the kinetic energy?
@toshiro0o10 жыл бұрын
No, those just happen to be the easiest components to figure out a lot of the time. For example, you can also use polar coordinates (for 2D space, as that is what he was using) where the kinetic energy can be written with the sum of the radial and angular velocities squared. As long as what you write for velocity is actually the length of the velocity vector.
@bautibunge7376 жыл бұрын
you can use any coordinate, but for most of them it may be difficult to find out how is the expression of the kinetic energy. Most of the time you just write your coordinate system as a function of cartisian or polar and then work out the kinetic energy from there
@angst_3 жыл бұрын
When he works out the equations of motion, he's doing a "partial lagrangian"? Can anyone tell me what that's called so I can learn more about it? Thanks! I probably haven't taken the math needed for this course, but that hasn't stopped me yet. It's a very interesting topic.
@davidv29863 жыл бұрын
He takes the partial derivative of the Lagrangian w.r.t. a given variable, be it q or q dot. It's where you differentiate a multi-variable function w.r.t just one of the variables and assume the other variables are kept constant.
@angst_3 жыл бұрын
@@davidv2986 Thank You
@welovfree11 жыл бұрын
is there a textbook and exams to follow this lectures?
@chamberlainandrew267211 жыл бұрын
Is P_theta really conserved? I think the Lagrangian involves theta and P_theta is not a conserved quantity since V involves theta. Prof. Susskind just mistake T as T-V. Someone agree with me?
@ZraveX10 жыл бұрын
He is discussing an example where there is no gravitational force. So V = 0, and P_theta is conserved. Otherwise, you'd be right, and he mentions that at 40:30
@chamberlainandrew267210 жыл бұрын
Woo, seems right. But I really can't remember what's it about since it's too long time ago.Anyway, thank you~
@sathyanarayanansubramaniam51252 жыл бұрын
He has assumed V=0
@michaelgarcia8122 жыл бұрын
It appears that the way he solved the inclined plane problem, by assuming the point mass remains on the inclined plane, in no way reflects reality if the inclined plane is accelerated in a direction away from the point mass. In reality, the inclined plane would leave the point mass since it is not exerting a force on the point mass. This could be addressed by utilizing Newton’s laws of motion. The problem is posed as a reversible problem by requiring the point mass to remain on the inclined plane, while in reality it is not. Caveat emptor.
@welovfree11 жыл бұрын
pdfs would be great yes you can send them
@davidv29863 жыл бұрын
Fantastic lecture! I don't understand why the change in p q dot at 1:17:10 can be re-expressed by p (delta q dot) + q dot (delta p); it seems to me that it should equal p (delta q dot) + the transformed q dot value (delta p) or alternatively q dot (delta p) + the transformed p value (delta q dot). I'm probably doing something wrong, any help would be greatly appreciated. :)
@lelomambueliane49152 жыл бұрын
the best video ever seen on classic mechanics kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@lelomambueliane49152 жыл бұрын
Can it be denied that this type solves the most difficult problems? kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@savanadoll10033 жыл бұрын
hey where can i get your lectures on electromagnetism...😁😁😄😄😄....
@lelomambueliane49152 жыл бұрын
Can it be denied that this type solves the most difficult problems? kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@hemantjoshi83888 жыл бұрын
why did he change coordinates from x to q?
@martingreen4366 жыл бұрын
He just uses different labels for the same thing depending on the context.
@y09t9b36 жыл бұрын
With that change in coordinates it is easier to visualize the orbits in phase space.
@physicspoint33563 жыл бұрын
May God bless you sir
@lelomambueliane49152 жыл бұрын
the best video ever seen on classic mechanics kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@lelomambueliane49152 жыл бұрын
Can it be denied that this type solves the most difficult problems? kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@조원준-e6b8 жыл бұрын
at about the last 2 min, why people laugh at "why neutrino can't go faster than the speed of light" ? Is there a western cultural background that I don't know? ( I'm Asian )
@Afewwilliams7 жыл бұрын
there was an experiment that came out around the time of these lectures where the experimenters thought they had observed neutrinos travelling faster than the speed of light. The discrepancy was subsequently explained.
@조원준-e6b7 жыл бұрын
Oh thanks, I knew that experiment but didn't exactly know it was neutrino :3
@amirhosseinkzm92782 жыл бұрын
Thank you very much.
@lelomambueliane49152 жыл бұрын
the best video ever seen on classic mechanics kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@lelomambueliane49152 жыл бұрын
the best video ever seen on classic mechanics kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@06KingDave11 жыл бұрын
A great textbook to learn classical mechanics from is (imaginatively titled) "Classical Mechanics" by Herbert Goldstein which includes great explanations and problems. If you want I can send you p.d.f 's of the book itself and solutions to the problems.
@of81553 жыл бұрын
Yes
@manoranjansahu71614 жыл бұрын
Is anyone aware of the book on General Relativity by Prof. Suskind
@orientaldagger69203 жыл бұрын
Prof. Suskind? Who is that?
@meowwwww63503 жыл бұрын
What?? Please tell me the name of the book please
@manoranjansahu71613 жыл бұрын
@@meowwwww6350 Well on page no 392 of "Special Relativity and Classical Field Theory", Prof. Susskind mentioned this. "See you in General Realtivity"
@meowwwww63503 жыл бұрын
@@manoranjansahu7161 oh!! But it'll be fun if he writes a theoretical minimum Series on general relativity!!
@manoranjansahu71613 жыл бұрын
@@meowwwww6350 Let's hope it becomes reality
@achillesmichael57052 жыл бұрын
Cameraman was sleeping during this one
@cubedude7611 жыл бұрын
excuse my possible ignorance but what exactly is canonical momentum? From the way he is using that term it sounds like momentum in a particular direction (like a component of momentum). and why wouldn't it always have the same units?
@aeroscience98347 жыл бұрын
cubedude76 canonical momentum is the partial derivative of the lagrangian with respect to the time derivative of a coordinate. Therefore, it's unit depends on the unit of the coordinate in question. See the earlier lectures in this series for more info on canonical momentum
@seandafny9 жыл бұрын
lhh dis nigga said Herman 😂😂
@tomaszdzieduszynski8 жыл бұрын
+Sean Dafny It's a reference to the older version of those lectures from 2008, where the 2nd pendulum bob was actually called Herman. :D
@seandafny8 жыл бұрын
Tomasz Dzieduszynski hahaha. These lectures were entertIning for reasons I would otherwise not have guessed.
@SphereofTime Жыл бұрын
6:00
@prakashchaudhary455912 жыл бұрын
NOT
@Chiavaccio2 жыл бұрын
👍
@lelomambueliane49152 жыл бұрын
the best video ever seen on classic mechanics kzbin.info/www/bejne/ppzaamWVf9WpZ6c
@utku19037 жыл бұрын
Herman lol
@06KingDave11 жыл бұрын
Hi, sorry it took me so long to reply youtube inbox is messed up... If you pm me your e-mail I'll send it as soon as i can. Cheers ; )