# Euler-Lagrange question about strange differentiation

I'm watching Susskind's Classical Mech. YouTube lecture series and am really confused about something he's doing where otherwise I've followed everything up until this point without a problem. In Lecture 3 he's dealing with the Euler-Lagrange equation applied to minimizing the distance between two points, and I understand his work up until here, where he starts taking the partial derivatives of the lagrangian with respect to v_i and v_i-1 rather than x_i. Why does he do this rather than continuing to take the derivatives w.r.t. x_i? He flatly says, "We're differentiating with respect to x-sub-i here" and then proceeds to take a partial derivative w.r.t. v_i and v_i-1 instead, I don't get it.

BvU
Homework Helper
The second argument of ##\mathcal L## is ##v_i\quad ## So for the second ##\mathcal L\ ## : $${\partial \mathcal L \over \partial x_i} = {\partial \mathcal L \over \partial v_i} \; {\partial v_i \over \partial x_i} = {1\over \epsilon} {\partial \mathcal L \over \partial v_i}$$

Zacarias Nason
In Lecture 3 he's dealing with the Euler-Lagrange equation applied to minimizing the distance between two points, and I understand his work up until here, where he starts taking the partial derivatives of the lagrangian with respect to v_i and v_i-1 rather than x_i.

where he starts taking the partial derivatives of the lagrangian with respect to v_i and v_i-1 rather than x_i. Why does he do this rather than continuing to take the derivatives w.r.t. x_i? He flatly says, "We're differentiating with respect to x-sub-i here"

if he is doing calculation for finding out the minimum distance between two points his functional must be functions v's rather than x's thats why he is interested in the partial derivative w.r.t. v's

vanhees71
Gold Member
The second argument of ##\mathcal L## is ##v_i\quad ## So for the second ##\mathcal L\ ## : $${\partial \mathcal L \over \partial x_i} = {\partial \mathcal L \over \partial v_i} \; {\partial v_i \over \partial x_i} = {1\over \epsilon} {\partial \mathcal L \over \partial v_i}$$
This is misleading since by assumption the ##x_i## and ##v_i## are independent variables, concerning the partial derivatives of the Lagrangian. What's behind this is of course the action principle, which is about variations of the action functional
$$S[x_i]=\int_{t_1}^{t_2} \mathrm{d} t L(x_i,\dot{x}_i).$$
The variation of the trajectories ##x_i(t)## is taken at fixed boundaries ##\delta x_i(t_1)=\delta x_i(t_2)=0## and time is not varied. The latter implies that
$$\delta \dot{x}_i=\frac{\mathrm{d}}{\mathrm{d} t} \delta x_i$$
and thus
$$\delta S[x_i]= \int_{t_1}^{t_2} \left [\delta x_i \frac{\partial L}{\partial x_i} + \frac{\mathrm{d} \delta x_i}{\mathrm{d} t} \frac{\partial L}{\partial \dot{x}_i} \right ] = \int_{t_1}^{t_2} \delta x_i \left [\frac{\partial L}{\partial x_i} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{x}_i} \right ] \stackrel{!}{=}0.$$
In the last step, I've integrated the 2nd term by parts. Since this equation must hold for all ##\delta x_i##, you get to the Euler-Lagrange equations,
$$\frac{\delta S}{\delta x}=\frac{\partial L}{\partial x_i} - \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{x}_i} \stackrel{!}{=}0,$$
which are the equations of motion for the trajectories ##x_i(t)##.

Zacarias Nason
BvU