Classical Mechanics | Lecture 4

[@joabrosenberg2961]

Susskind explains why he uses mathematics and not Scientific American general talk; Variation and deriving again Euler Lagrange equations 2:30; Conservation and symmetry 17:00; symmetry of free particle 32:00; Rotational Symmetry 42:10; General Statement on Symmetry and Conservation 55:00; Examples 1:07:00; Q&A from participants 1:12:00; Harmonic Oscillator 1:20:00; Symmetries without generator (mirroring) 1:28:00; continuation Q&A 1:30:30; A case where symmetry doesnt bring conserved quantity 1:43:00; an example of maximal action 1:49:00

[@AdenKhalil]

If you're good at math, you understand these things clearly! Blessed mathematics!

[@anmolraina3509]

just for the records, p is used because the word impetus formally in place of momentum comes from the Latin, petere, hence we get p.

[@DrDress]

20:00 I love how he doesn't just invoke the chain rule here. He gives an intuitive interpretation of it instead.

[@ibraheemkhurram6185]

can you explain more wym by that?

[@AdenKhalil]

Invoking chain rule is way better though!

[@kaitan8824]

In last lecture the conservation law was stated that when Lagrangian is independent of one coordinate (q_i), then dL/dv_i is conserved. In this lecture, the conserved quantity Q is derived given some symmetry. Both statements are equivalent, all depends on the choice of coordinate system. Take the rotational symmetry example, if we choose polar coordinate system, thus L will be independent of θ, i.e. dL/dθ will be zero, then we will get the same angular conservation but in polar coordinate expression.

[@camilodominguez4678]

All this lectures are very highly appreciate it, I think the subtitles on the students questions would make it even better, thanks in advance.

[@DrDress]

32:00 Symmetry 1:00:00 Noethers theorem

[@albertmichelson1615]

In lagrange function, if you change the sign of kinetic energy to eliminate the coefficient, it will cost the potential energy... Derivative of lagragian respect to velocity equals to momentum. Derivative of lagragian respect to the coordinates will give you the force. First order change of action equals to zero. That means we come cross the principle of least action. So, this lecture is pretty good.

[@cypts1]

I have an exam tomorrow morning on cell biology, why am i watching this?

[@mtark9988]

yo how did it go

[@ninadjadkar5062]

did you pass?

[@hawkevil_810]

@@ninadjadkar5062 8 years later dude probably has children now ahahahha

[@0xcc32sys_err4]

@@ninadjadkar5062 most likely not, I am not sure how classical mechanics could ever help you with Cell Biology.

[@orientaldagger6920]

@@hawkevil_810 Being a physics major is a pretty sure way to not get any. I can attest to that.

[@afifakimih8823]

How simply he explain everything but yet deep!and that is the characteristics of LS.💜❤❤

[@Delfigamer1]

1:23:30 > Anybody know what's the motion of p and q? I know, it's the tentacle orbit!

[@annamallett879]

Shoutout to the audience member who asked what a and b might represent. I got hung up on that for ages when reading the book and came to the lecture hoping for an answer.

[@zphuo]

I don't understand @39:40 why q prime = q + b*delta? when we shift a delta, then both q1 and q2 all move a delta at same direction, and that shouldn't be q prime = q + delta? For proving dV=0, let q1 and q2 move different distance, it that correct method??

[@sharpfang]

He neglected to add primes to every new (transformed) coordinate, and this got very confusing fast because you never knew when he'd take the original coordinates or the ones he transformed to. Generally, it shows he had a worse day that day, and it's a pity because this is a pretty difficult subject that should receive more TLC. I had to reach to other sources before I began understanding what he was doing. BTW, the question about 1:33:50, if we have a bunch of delta squared values, how do we know they won't add to anything significant if we do a bunch of steps, like combining all the small transformations. The answer is, that in the equation there's both a delta and a delta squared. Delta squared is as much smaller than delta, as delta is smaller from a normal "big" number. By the time you stop summing up all your deltas adding them up to a large transformation by a normal number, the delta-squares will only add up to a single delta.

[@bonononchev634]

I know that I am six years too late, but still... The question about 1:33:50 is actually about something else. To write it down more rigorously you need to assume delta is infinitesimal, and intuitively as he was explaining it they are strictly bigger than zero, but however you add them up a finite amount they never add up to any non-inifinitesimal quantity. However even if you add a countable infinite number of actual zeros, you still get zero. There is no ambiguity here. When you do your proof, you never sum up / integrate deltas, you are only allowed to infinitely sum up exact zeros - otherwise your operation is simply undefined. Another way to think about it is, arithmetic with infinitesimals is mathematically equivalent to ordinary calculus. How would that question look there? Well, if you have a function and compute its derivative, then you can integrate that derivative and get back your function. This is the fundamental theorem of calculus. Basically you define your deltas in the approximation, and with some mild regularity conditions you show that you can get your integral "discrepancy" from actual value to be smaller than any fixed nonzero epsilon by choosing a particular nonzero delta. This is the epsilon-delta definition of Cauchy. There you only sum up a finite number of finite values. I find infinitesimal arithmetic more intuitive in some ways, but as it is mathematically equivalent to calculus, the weird cases are as unintuitive as they come...

[@achillesmichael5705]

@@bonononchev634 hope he graduated by now :D

[@samlaf92]

1:01:28 He says that dL/dq' = p. Isn't that assuming that the potential is not dependent on velocity q'? We saw earlier that in rotating frames you get velocity-dependent potentials...

[@SpenserF]

at 19:00, why does the kinetic energy disappear when finding the momentum of the lagrangian? In other words, why is it assumed that d((q'_1 + q'_2)/2)/d(q_1) = 0? Is it because we are summing the velocities as two points to define the KE, and these velocities don't depend on their positions at that instant in time? I don't see how that could be since velocity is a function of position and time.

[@rosensofroniev1264]

In the Lagrangian equation you do the partial derivative with respect to q_1. This means that if q_1 doesn't stand explicitly in the kinetic Energy then del(kinE)/del(q_1) = 0. In this case only q_1' and q_2' stand explicitly there.

[@samlaf92]

16:00 He says Lagrangian is one line whereas writing the equations of motion would be 300 lines (for 300 particles). But if you can index the Lagrangian per particle i, then you can also do that with Newton's equations of motion...? And that's exactly what he has on the board. Both are just one liners. Am I missing something?

[@abhishekcherath2323]

Where can I get my hands on these notes he talks about?

[@BARUNIZ]

Because physics is awesome.

[@zoltar8x]

Could someone clear up his math from 40:40 - 41:48... I don't understand how he gets two terms with a*b. Is Professor Susskind arbitrarily shifting the potential's dependent terms by b*delta and a*delta unrelated to the shifts he drew before?

[@joplinsita]

If you go to 22:22 to 24:54, you will see that for this kind of potential the conservation quantity was obtained by multiplying d/dt P1 by b and d/dt P2 by a. The conservation quantity is actually bP1-aP2. Thus, If I am not wrong, I guess that's the reason why of the transformations for q1'and q2' you are asking for. Hope that helps!

[@vinitchauhan973]

Im not sure what you are referring to to but all he does is shift q1 by +b∆ to get instead of aq1 we get a(q1+b∆) which gives through distribution aq1+ab∆ and also shifting q2 by -a∆ to get instead of bq2 he gets b(q2-a∆) then distribute to get bq2-ba∆, since we where originally adding the 2 V(aq1+bq2) we just add our shifted quantities to get V(aq1+ab∆+bq2-ba∆) which through cancelation gives V(aq1+bq2) thus the transformation doesn't effect the potential energy function which is the only term we are concerned with in the lagrangian thus ∆L=0 symmetry

[@subashkc7674]

@@vinitchauhan973 Thank you so much❤️😊 I was also confiused on that same step but Now I understood with your nicely explained comment❤️

[@waynelast1685]

1:42:38 - I believe Emmy Noether's name is pronounced " Nootr " . Another video guy dug up research on it. (Physics Asylum)

[@ZeroXAlAttas]

listen to this man make me wanna go to this university!

[@ozzyfromspace]

Same. Maybe I'll apply. I'm one of those flipping-burgers type kids who just learns because I have a lot of questions on my mind. I gave up a prestigious engineering scholarship in my 2nd year after suffering anxiety attacks because I didn't know what I wanted to do with my life. Professor Susskind has me thinking maybe that something is physics, although, even now my purpose is not all that clear. Maybe Stanford's where I should be. I've been thinking this for a few months now. The irony is that it was my dream school for years when I was in high-school. I think I'm scared to say I'm getting things back on track again because, well, what happens if I burn out and fail AGAIN? I've lost friends and family; that is the price of my incompetence. I guess I was venting in a comment and didn't realize it (don't mind me, I'm sorry). I won't delete the comment though; I think it makes for a nice, little snippet of my history, I suppose.

[@HilbertXVI]

@@MrTroywoo >Mostly differential geometry and Lie groups You're ignoring the enormous prerequisites required for that, and it generally involves taking a course in a college. You can't really expect the average person to put in that much effort by themselves, especially when graduate courses like differential geometry are involved.

[@Pranav_Bhamidipati]

@@HilbertXVI I would like to respectfully correct that there are undergraduate offerings of differential geometry in most universities.

[@zphuo]

So @36:00 the dL=0 && translation symmetry are just a form of uniform speed motion( dq = d must be hold in every dt)??

[@noumena2655]

He is really an amazing teacher

[@muruhappanchidambaram9489]

While deriving the conservation law for a particular symmetry, he is factoring out del(the small number). SInce del tends to 0, won't we end up with 0/0 which has a value? If we look at it that way, dQ/dt will not be 0.

[@TheRamlaw]

emmy noether worked on this in 1908 to 1919

[@chandus2496]

Is there anyway we can get access to those notes he mentioned?

[@SARANSASIDHARAN1]

Chandu S www.amazon.in/Theoretical-Minimum-Start-Doing-Physics/dp/0465075681/ref=sr_1_1?ie=UTF8&qid=1435440268&sr=8-1&keywords=The+Theoretical+Minimum '

[@fjolsvit]

I sure don't want to sound ungrateful for having these wonderful lectures available at no cost, but I disagree with Dr. Susskind's explanation regarding the reason that the time derivative of the partial of the Lagrangian with respect to velocity shows up in the Euler-Lagrange equations. His heuristic derivation certainly supports that interpretation. But a more formal development shows his derivation to be only one of many possible scenarios. See ISBN: 0-691-02663-7 Section 2.2

[@netrapture]

@steven hatton - please listen to the statement he makes at 2:12. Considering "many possible scenarios" would not be consistent with the strategy of the presentation.

[@orbital1337]

Actually it is neither, really. This course is part of Stanford's continuing studies project and can as such be taken by any "adults in the Bay Area". Most of the people here are somewhat older (according to Susskind he's had people in their 80s in the course) and have graduated in technical or mathematical disciplines.

[@omaregb]

well things are starting to get confusing, anybody knows a good textbook to follow this course?

[@mearpanchatterjee]

As explained in this video, if we know the symmetry then we can find the conserved quantity associated with it. But can we do the reverse i.e can we find the symmetry if we know the conservation law?

[@postbodzapism]

did he address hamiltonian formalism in the course?

[@itsRAWRtime007]

He does in the later lectures I think. Google website called theoreticalminimum, that is the website of the course you can find the list of courses and lectures in each course there.

[@harshkumarjha5731]

does that book theorotical minimum contains the summary of these same lectures?

[@jordanedmunds4460]

Is there always only a single stationary path for the action for a given Lagrangian?

[@Pranav_Bhamidipati]

Good question. The answer is no. Since stationarity is with respect to small changes in the vicinity of a path, it is possible to have multiple stationary paths. Such a thing is very common when it comes to light/ electromagnetic radiation - when there is a pair of a receiver and a transmitter of electromagnetic waves near the interface of two media, there are multiple rays that travel between them since all those paths are stationary. For example, there can be a direct wave and a lateral wave.

[@martingreen436]

With regards to discrete symmetries not having a conserved quantity, I think he is saying he did not do, nor is there a derivation, to prove it? ( for classical mechanics)

[@waynelast1685]

I think some conservations exist for discrete symmetries but there are no mathematical associations

[@ibraheemmoosa]

1:33:22

[@DrDress]

1:41:15 It not the 30's, is it? I am quite sure Leonard is much closer.

[@DrDress]

Yeah, it was around 1915.

[@flixerstudios1862]

What quantity is actually conserved in Quantum Mechanics with the flipping of position q=-q in symmetric potentials?

[@SphereofTime]

15:44

[@AdiCherryson]

48:12 where did it come from?

[@non-inertialobserver946]

It's the Taylor series of cosx

[@AdiCherryson]

ViperDaniel yes, it is. Thank you! 🙂

[@randymartin5500]

At 46:00, here is an excellent derivation of the x' and y' co-ordinate in terms of x and y kzbin.info/www/bejne/hGe7pKqNptKYi6M

[@Sans_K5]

thanks sir for the amazing session❤🙏

[@ankannath9044]

While deriving the conservation law for continuous symmetry, sum over del L/ del qi * delta qi was written. Is this sum over all particles or all coordinates or both?

[@abhishekkolli7410]

sum over all the particles in the transformed coordinates after substituting f(q)

[@billyraybar]

Correct

[@angst_]

I'm pretty sure I mostly understand the differences when Δ, δ, and d are used. What confuses me is d/dt. I'm not used to d without something next to it. what is the d/dt called in "d/dt * δL/δẋ - δL/δx" and what does it do? I know δL/δẋ is the partial derivative, right? Thanks

[@bonononchev634]

δL/δẋ is actually a representation of the full derivative in arithmetic with infinitesimals. It is somewhat more intuitive and was introduced by physicists as the first way to introduce derivatives. Now mostly people use the delta-epsilon definition of Cauchy as it doesn't deal with infinitely small numbers and is easier to make rigorous statements in it. But when explaining, the infinitesimals can be more intuitive. ∂/∂t is the partial derivative. d/dt is the full derivative, that means that if F is represented as a function with 100 parameters, and each of those parameters is a function of t (can be constant, still a function), d/dt means that you vary t in all of them. So d F(x1,x2,...,x100) / dt = ∂F/∂x1 dx1/dt + ... ∂F/∂x100 dx1/dt.

[@valtih1978]

What does he demonstrate with viscous liquid equation? Does the equation demonstrate symmetry or conservation? I could not understand.

[@ozzyfromspace]

The viscous liquid equation is a good statistical equation that models the motion of fluids...it is not derived from fundamental physics. The viscous liquid equation is, then, just a really good model for the system. In particular, it says that a*x''=b*x', where a and b are the relevant constants of the equation. You will notice that the equation is translationally invariant (x'(t) = (x(t))' = (x(t) + d)') since d is a constant. However, the equation has no conservative quantity as he demonstrates in the video. Hope this helps (although I'm 3 years late lol). Also, sorry for any errors I may have made; I'm learning, just like you :)

[@SphereofTime]

13:22

[@Pidxr]

Is dL = 0 when qdot^2/2 the reason why photons experience no time?

[@Cosmalano]

Photons experience no time because of relativity.

[@Cosmalano]

Budiman Budimen 18, 16 when I posted that.

[@Cosmalano]

My original comment doesn't have any real meaning, sorry. Photons experience no time because if you put the speed of light into the Lorentz transformation you get out a result of 1/0, which makes no sense, but if you take the limit as the velocity approaches infinity and apply it to the photon's time then you see time slows infinitely, implying that time doesn't pass in a photon's reference frame. You can also apply the Lorentz transformation to the photon's displacement and you'll see that the space between it's emission and it's absorption contracts infinitely. Thinking about this, it would be hard for time to pass if you arrived at your destination instantly. Keep in mind this is only relative to the photon, and some physicists would argue that applying a Lorentz transformation to a photon doesn't make sense.

[@Pidxr]

It is the most mysterious thing to me. That and the idea of time slowing down as temperature approaches zero.

[@Cosmalano]

Piotr. Moes perhaps I'm not thinking about this properly, but what does temperature approaching absolute zero have to do with time slowing?

[@eliodecolli]

Are these the same lectures he does in an undergrad physics course or are these some kind of open lectures targeted to the general population?

[@赵悦-s5m]

I am now making the captions..

[@meowwwww6350]

Nice job

[@jamesdowns72]

How come he always explicitly writes out what KE equals (0.5mv^2), but always says PE is just some function U of x.

[@jamesdowns72]

@@Errenium thanks

[@DerrykSchieck]

why is the lagrangian so much 'better' than the hamiltonian? everyone talks about lagrangians.. like he said : 'if you get a lagrangian, you can derive everything from it' this should be the case for the hamiltonian as well right, it's the sum not the difference between kinetic and potential energy..

[@minecraftplayer560]

How I enjoy watching these lectures

[@seungsoolee1949]

Hi, I might just be deaf or dumb, but can someone better explain why dp_1/dt = -V'(q_1 - q_2) and dp_2/dt = V'(q_1 - q_2) please? Thank you

[@mlele7372]

He mixes up the signs as far as I can tell. dpi_1/dt = V’. Not -V’

[@sattarabd2201]

Exactly, I couldn't understand why he makes Pْ= -V, where did the minus sign come from?? because we know that the Pْ-V-0 and then Pْ=V. so V is V(q1-q2) when we differentiate it concerning q1 it becomes a plus V prime, not a minus.

[@waynelast1685]

I must be OCD. It bugs me when he doesnt erase the board completely.

[@SmilingCherryPie-oc4hl]

Can anyone tell me why space is measured in cubes. But the equation involves pie ?

[@natepolidoro4565]

46:11 Nope your signs are correct, i checked.

[@halilibrahimcetin9448]

not correct!

[@vladislavlukmanov4998]

You didn’t mention about the energy conservation law. It comes from time symmetry: the langrangian doesn’t change if you shift time a little bit

[@MrBobbyArcher]

He mentions there being notes for these lectures. Does anyone know if they're available anywhere online?

[@nurneues1662]

look up his 'Theoretical Minimum' books. They are based on these lectures.

[@sodacanman1]

does anyone know if this is a graduate or undergraduate course in classical mechanics??

[@Thenomadthought]

sodacanman1 graduate level

[@brothberg]

It's a general ed course for the public in the Stanford area. It covers the same topics as mechanics in upper level undergraduate physics, but with less hard calculation. As he says in the previous lecture, it's not how he would teach the material to people who want to become physicists.

[@physics110]

Leo never clears out the blackboard....he always leaves small drawings on it

[@MegaDespotic]

rotation transformation is wrong is not it?

[@artbess]

its right, if you take the x',y' as the base and rote the xy-coordinate system clockwise. You just apply the Point of view under the x'y'-reference frame for the counterclockwise rotation of x'y'

[@DerrykSchieck]

this is pre-doc level. well thats what he said in one of the vids

[@sandeepchandappillai9814]

Thank You !

[@flawns]

because science is fucking awesome! I do this all the time too, I think, we all do :)

[@FishLoveOcean]

My poor English...Could anyone be kind to give transcripts?

[@sundiego64]

i have two finals tomorrow... why am i seriously here

[@halilibrahimcetin9448]

when LS does physics, I say hell yeah babe, that's true :)

[@Seth-ri9hl]

I had to pause at 1hr28 and eat chocolate

[@РодионЧаускин]

Hall David Williams Patricia Gonzalez Mark

[@AbuSayed-er9vs]

Above all it's a better view.

[@haydenkwon7217]

legend.-

[@Chiavaccio]

👏👏👍

[@ДмитрийВербицкий-у7д]

Lee Charles Gonzalez Margaret Miller Kenneth

[@jackymarcel4108]

Davis Jennifer Martinez Anthony Lopez Amy

[@Ronnicus]

cuz u should be a physics major instead

[@lillianvaughner6491]

African American freedom by linwood B

[@MA-dh2bd]

Presentation is weak and scattered

[@SphereofTime]

35:42