showing that the euler lagrange equations are coordinate independent

demonelite123

so i know for example that d/dt (∂L/∂x*_i) = ∂L/∂x_i for cartesian coordinates, where x_i is the ith coordinate in Rⁿ and x*_i is the derivative of the ith coordinate x_i with respect to time. L represents the lagrangian.

so using an arbitrary change of coordinates, q_i = q_i(x₁, x₂, ..., x_n)

i have that ∂L/∂x__i = ∂L/∂q_i * ∂q_i/∂x__i and that ∂L/∂x*_i = ∂L/∂q*_i * ∂q*_i/∂x*_i and substituting i get:

d/dt(∂L/∂q*_i * ∂q*_i/∂x*_i) = ∂L/∂q_i * ∂q_i/∂x_i and using the product rule i get:

d/dt(∂L/∂q*_i) * ∂q*_i/∂x*_i+ ∂L/∂q*_i * d/dt(∂q*_i/∂x*_i) = ∂L/∂q_i* ∂q_i/∂x_i

using the fact that ∂q*_i/∂x*_i = ∂q_i/∂x_i,

d/dt(∂L/∂q*_i) * ∂q_i/∂x_i + ∂L/∂q*_i * d/dt(∂q_i/∂x_i) = d/dt(∂L/∂q*_i) * ∂q_i/∂x_i + ∂L/∂q*_i * (∂q*_i/∂x_i) = d/dt(∂L/∂q*_i) * ∂q_i/∂x_i+ ∂L/∂x_i = ∂L/∂q_i * ∂q_i/∂x_i.

if i didn't have that ∂L/∂x_i on the left side then i could just cancel out the ∂q_i/∂x_i on both sides and i would be done. have i made a mistake somewhere or am i missing something? thanks.

vanhees71

Why don't you use the nice LaTeX functionality of this forum? It's really hard to read, what you have written. So I don't want to decipher it.

Here is my derivation of the form invariance of the EL equations under coordinate changes (diffeomorphisms). Let [itex]x^j(q)[/itex] denote the transformation with [itex]j \in \{1,\ldots,f \}[/itex]. Further [itex]q=(q^k)_{k \in \{1,\ldots, f\}}[/itex].

First of all we need an auxilliary equation for the time derivatives. From the chain rule we have

[tex]\dot{x}^j=\frac{\partial x^j}{\partial q^k} \dot{q}^k.[/tex]

Here and in the following I use Einstein's summation convention, according to which one has to sum over any pair of repeated indices. From this equation we see

[tex]\frac{\partial \dot{x}^j}{\partial \dot{q}^k}=\frac{\partial x^j}{\partial q^k}. \quad (1) [/tex]

Now we have

[tex]\frac{\partial L}{\partial q^k}=\frac{\partial L}{\partial x^j} \frac{\partial x_j}{\partial q_k} + \frac{\partial L}{\partial \dot{x}^j} \frac{\partial \dot{x}^j}{\partial q^j} \quad (2)[/tex]

and

[tex]\frac{\partial L}{\partial \dot{q}^k}=\frac{\partial L}{\partial \dot{x}^j} \frac{\partial \dot{x}^j}{\partial \dot{q}^k} \stackrel{(1)}{=}\frac{\partial L}{\partial \dot{x}^j} \frac{\partial x^j}{\partial q^k}.[/tex]

Taking the time derivative of this, we find

[tex]\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^k} = \frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{x}^j} \cdot \frac{\partial x^j}{\partial q^k} + \frac{\partial L}{\partial \dot{x}^j} \frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\partial x^j}{\partial q^k} \right). \quad (3)[/tex]

Now we obviously have

[tex]\frac{\mathrm{d}}{\mathrm{d} t} \left (\frac{\partial x^j}{\partial q^k} \right) = \frac{\partial \dot{x}^k}{\partial q^k}, \quad (4)[/tex]

and thus, if the EL equation are valid wrt. the coordinates, [itex]x^k[/itex] they are also valid wrt. to the coordinates, [itex]q^j[/itex], which can be seen by comparing (2) and (3) and making use of (4) and the EL equation wrt. [itex]x^k[/itex].

Petr Mugver

This is one of the mistakes. Instead, write:

[itex]\frac{\partial L}{\partial x_i}=\sum_j \frac{\partial L}{\partial q_j}\frac{\partial q_j}{\partial x_i}+\sum_j \frac{\partial L}{\partial \dot{q}_j}\frac{\partial \dot{q}_j}{\partial x_i}[/itex]

You fergot the second piece. Remember:

[itex]q_j=q_j(x_1,\dots,x_n)[/itex]
[itex]\dot{q}_j=\sum_k\frac{\partial q_j}{\partial x_k}\dot{x}_k[/itex]

so

[itex]\frac{\partial\dot{q}_j}{\partial x_i}=\frac{\partial^2 q_j}{\partial x_i\partial x_k}\dot{x}_k[/itex]