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Showing that the euler lagrange equations are coordinate independent

  1. Jul 29, 2011 #1
    so i know for example that d/dt (∂L/∂x*i) = ∂L/∂xi for cartesian coordinates, where xi is the ith coordinate in Rn and x*i is the derivative of the ith coordinate xi with respect to time. L represents the lagrangian.

    so using an arbitrary change of coordinates, qi = qi(x1, x2, ..., xn)

    i have that ∂L/∂x_i = ∂L/∂qi * ∂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/∂qi * ∂qi/∂xi 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/∂qi* ∂qi/∂xi

    using the fact that ∂q*i/∂x*i = ∂qi/∂xi,

    d/dt(∂L/∂q*i) * ∂qi/∂xi + ∂L/∂q*i * d/dt(∂qi/∂xi) = d/dt(∂L/∂q*i) * ∂qi/∂xi + ∂L/∂q*i * (∂q*i/∂xi) = d/dt(∂L/∂q*i) * ∂qi/∂xi+ ∂L/∂xi = ∂L/∂qi * ∂qi/∂xi.

    if i didn't have that ∂L/∂xi on the left side then i could just cancel out the ∂qi/∂xi on both sides and i would be done. have i made a mistake somewhere or am i missing something? thanks.
     
    Last edited: Jul 29, 2011
  2. jcsd
  3. Jul 29, 2011 #2

    vanhees71

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    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].
     
  4. Jul 29, 2011 #3
    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]
     
    Last edited: Jul 29, 2011
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