Leonard Susskind : Classical Mechanics

dx · Jun19-08, 08:35 PM

http://www.youtube.com/view_play_lis...9C0DCE90CB6D81

**lolgarithms** · Oct10-09, 09:56 PM

isn't the derivative of the first derivative squared:
d/dt (x')^2 = 2x'x''? why does susskind claim it is 2x'', in his classical lecture 3?

**zetafunction** · Oct11-09, 06:26 AM

he assumes X(t) so 'x' is a function of time not only of x , he is using the chain rule

**lolgarithms** · Oct11-09, 09:50 AM

Ithink he is not using the chain rule properly. if x is a function of time only, d/dt (dx/dt)^2 = 2dx/dt * d^2x/dt^2

**cristo** · Oct11-09, 10:17 AM

Originally Posted by lolgarithms

Ithink he is not using the chain rule properly. if x is a function of time only, d/dt (dx/dt)^2 = 2dx/dt * d^2x/dt^2

You've already asked this question, and had it answered in another thread. I don't know where abouts in the video you've seen this, but I'm guessing it has to do with the Euler-Lagrange equations:

$LaTeX Code: <BR>\\frac{d}{dt}\\Big(\\frac{\\partial\\mathcal{L}}{\\parti al\\dot{x}}\\Big)=\\frac{\\partial\\mathcal{L}}{\\partia l x}$

So, the LHS is not taking the time derivative of the Lagrangian, but is instead the time derivative of the derivative of the Lagrangian with respect to the coordinate velocity. It is important to treat the coordinate velocity as a variable; that is $LaTeX Code: \\mathcal{L}\\equiv\\mathcal{L}(x, \\dot{x})$ .

If this doesn't clear things up, let me know the exact time in the video that you're confused with, and I'll try and look at it.

**lolgarithms** · Oct11-09, 02:47 PM

Originally Posted by cristo

You've already asked this question, and had it answered in another thread.

I had the thread deleted because i decided I wanted to post it here.