Lecture 4 | Modern Physics: Classical Mechanics (Stanford)

[ArabyGUC]

I felt so bad about myself when I knew this about the Hamiltonian! It's like when you're playing those games like age of empires and you explore a new territory and the field LIGHTS UP. This lecture made a lot of hidden connections in my mind light up!!!!

[grunder20]

Give me more time to absorb his discussion! great work!

[SalsaTiger83]

I think I get the "how" but I can't figure out the motivation behind all this yet... What is this displacement bussiness useful for?

[csmcmillion]

"I am a software engineer by profession... Am actually planning to quit my job and do a Phd...incidentally in Physics"

= UNEMPLOYED

[SalsaTiger83]

@csmcmillion computer engineer + physics phd = on the best way to become a highly paid quant specialist on wall street

[csmcmillion]

@SalsaTiger83 These lecture series pretty much show me I'm not physics PhD material. I mean, these aren't even full-blown grad classes and I'm barely able to hang on.

[SalsaTiger83]

@csmcmillion I have the same problem. But I think that is just the feeling of overwhelm, which indicates my brain working to overcompensate for the demands put on it. The problem is that in such lectures it is not always possible to make sense of where the stuff is going. Here you need to know that the goal is to find either the complete functions for the coordinates of the particles/etc or enough information to solve it numerically. Also try to write down the equations yourself.

[csmcmillion]

@SalsaTiger83 I wasn't talking so much about Classical Mechanics - this isn't bad. I've been through some of the others, such as Quantum Entanglements, QM, SR/Relativistic Field Theory and GR that are way tougher. Those are the ones that I'm barely hanging onto.

[SalsaTiger83]

@csmcmillion ok, then I am even stupider ;-)

[savvvvvvvvvvy]

I was just wondering whether somebody could help me out regarding what's said in the lecture. When Susskind talks about the function of many variables [at about 23 mins], f (alpha i), he says at the minimum, delta f = 0. I don't really understand what he means when he says that if a small change in delta alpha i occurs, the function doesn't change to first order. I'd be really grateful for any help :)

[PLecN]

@savvvvvvvvvvy

In one variable ... suppose we know value of f(a), then we can approximate f(a + dx) where dx is small by f(a + dx) approx f(a) + f'(a)dx

Now, if f'(a) = 0 then this simplifies to f(a + dx) approx f(a)

So if f'(a) = 0 then changing 'a' by a small amount doesn't change f hardly at all. Exactly the same can be said about fn's of many vars but now have to consider small changes in more than direction. Works out very similar.

[Joelbesse1]

Great lectures, even if the vocable "least action" should not be used, but "stationary action" instead!

[joeglimmix]

isn't it funny that with 0 dislikes youtube still displays a short bar

[thyorison]

He's not writing dA/epsilon. He's writing just dA as a result of epsilon, which equals dA/epsilon times epsilon - that's where the epsilon factor comes in. The differential at the beginning of the integral, which he initially wrote as dq, was later correctly changed to dt. Integration over dt makes sense (as opposed to integration over epsilon, which I proposed in my previous comment).

[thyorison]

I'm still confused on derivation of conservation of energy. Now I think 0 = not dA/dq nor dA/dt but rather dA/epsilon. Epsilon is an infinitesimal. The dq at the beginning of the integral should be epsilon. The 1st part of the integral, dL/epsilon, = dL/dq times dq/dt-shifted times dt-shifted/epsilon. dq/dt-shifted = q-dot. t-shifted = t - epsilon, so dt-shifted/epsilon = -1. All those factors combine to equal dL/dq times (minus q-dot times epsilon). Still wrong - integral over epsilon?

[thyorison]

At 1:17:43, he writes down explicitly "the change in the action from the black trajectory to the red trajectory" - the change in the action due to a shift in time, which is zero because of the time symmetry of physical laws. He writes down an integral starting with dq. That should be dT - it's actually an integral over time of the time derivative of the Lagrangian of the original path plus the time-shifted path. At 1:20:19, he corrects the dq to dt.

[thyorison]

Later in the lecture, Professor Susskind was, as he said, getting tired, and he made a slight error. At 1:18:47, when writing down the second part of the time derivative of the Lagrangian, he correctly said "dL/dq-dot", but he omitted the dot when he wrote it down. He repeated that error a few minutes later when he wrote it again, but the third time he wrote it, he included the dot. I'm not trying to nitpick, but that confused me the first time I watched this.

[thyorison]

When deriving conservation of energy, at 1:18:20, Professor Susskind writes the first term of the time derivative of the Lagangian as dL/dt times delta-q. By the chain rule, that should be dL/dt times dq-shifted/dt, but since L is a function of q-shifted = q - epsilon * t,

d(q - epsilon * t)/dt = dq/dt time minus epsilon (by the chain rule),

but Professor Susskind had previously shown that was equal to delta-q.

[thyorison]

My previous comment was incorrect. The d/dt was brought inside the integral sign right at the beginning: dA/dt = d/dt ( integral of L) = intregral of dL/dt.

The reason for dropping the integral sign on the end segments is that (as Susskind says at 1:26:00), "the integral over a sufficiently small interval, the integral is just equal to the integrand times the time interval".

segment to integrate (add up) so dropping the integral sign doesn't change the value.

[thyorison]

My previous comment was incorrect. The d/dt was brought inside the integral sign right at the beginning: dA/dt = d/dt ( integral of L) = intregral of dL/dt.

The reason for dropping the integral sign on the end segments is that (as I guessed before) the end segments are infinitesimal, so there is only one infinitesimal segment to integrate (add up) so dropping the integral sign doesn't change the value.

[thyorison]

Maybe a better explanation is simply that the derivative of the definite integral of a function is just the difference of the values of the function at the end-points. In this case, the time derivative of the integral is zero (because it's a symmetric displacement of a function that obeys the Action Principle). Professor Susskind should have kept the integral sign until he took the derivative.

- Ray Eston Smith Jr

[thyorison]

Please read my comments from the bottom up. They wouldn't fit into a single comment & YouTube lists them in reverse order.

The reason the difference is the same for any two end-points is because the part in-between is zero because the trajectory obeys an action principle, and the whole integral is zero because it's a symmetry displacement, therefore the other part of the integral (the difference between the end-points) must also be zero.

- Ray Eston Smith Jr

[thyorison]

In that one infinitesimal term, the dq was still present - the term was not integrated because it was just a single infinitesimal. dq = epsilon * f(q), but he had explained that the epsilon could be dropped because it wouldn't change the final result of zero.

[thyorison]

That's the justification for taking the term out of the integral. In essence, he picks two adjacent end-points (because the difference is the same, zero, for any end-points), then changes the integral from a sum of infinite infinitesimal terms to a sum of just one infinitesimal term, so the integral sign isn't needed to sum just one term.

[thyorison]

A minute later, he explained a little (49:50): "What we've discovered is that, as a consequence of the symmetry under this [pointing to f(q)] particular displacement, a quantity is conserved, and that quantity is (dL/dq-dot) * f(q). In other words, the time derivative of this is equal to zero. We derived it in a slightly different form. We derived it by saying it's the same at the two end-points, but if it's the same for every pair of end-points, it means it just doesn't change with time."

[thyorison]

When applying integration by parts, there is a term contributed by the end-points, because the symmetry transation, f(q), is non-zero at the end-points. That end-points term is (dL/dq-dot) * f(q). What confused me is that Professor Susskind took that term out of the integral without immediately explaining why (49:07): "So let's write it down now. The conserved quantity is...").

[thyorison]

This lecture is the only proof of Noether's Theorem that I've found that I can come even close to understanding. So it was worth watching it a few times to understand it as well as I could. The one part that was confusing me is when the conserved quantity was taken out of the integral without an immmediate explanation of why that was justified. In case somebody else is confused by the same thing, what follows is as close as I could come to an explanation.

[rahulilrplac]

I am a software engineer by profession but just fell in Love with physics after 'understanding' it through these lectures.

Am actually planning to quit my job and do a Phd...incidentally in Physics.

Thank you Stanford. Thank you Suskind.

You are doing a great service to the humanity. Trust me on this one :)

[ibreakkidslegs]

its a long and lonely road to physics phd

[rahulilrplac]

I know. But sadly, (or interestingly), I have nothing better to do :-)

[ThrashAbaddon]

@rahulilrplac you do that,physics is amazing way :)

[guardianhyuga]

@rahulilrplac I'm also a software engineer albeit with a love of physics. Though I was already thinking of pursuing a degree in Physics before I saw these lectures, they've really helped me get ready and get excited about physics again. I hope that Ph.D. works out for you!

[paulojunior201]

@rahulilrplac But don't you need to do undergraduate in physics before phd ?

[TheOvermaster]

@paulojunior201 As long as you get the right supervisor - you can do a PhD in anything if you have a degree, it's just that you would have to upgrade your knowledge significantly if you did a PhD in a field other than the one you received your degree in.

[opendevs]

Thank you for putting up these great lectures Stanford.

[jonnylazer]

do you even know who lenny susskind is ?

[sikory]

actually i agree that susskind isn't the best lecturer, he's good, but not the best. Yes, I do know that he is pretty much the founder of string theory and he's a very smart guy. Teaching however doesn't require knowledge as much as teaching skills. Susskind does have some teaching skills and there are way worse teachers. I might even say I've learnt the most about physics by his lectures.

[bilb0oooo]

super ignorant spammer !