I'm trying to prove that the Euler-Lagrange equation $$\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_i})-\frac{\partial L}{ \partial q_i}=0$$ is invariant under an arbitrary change of coordinates $$q_i \rightarrow \bar{q}_i (q_1,...,q_n,t), \space \space\space i=1,...,n.$$

I performed a change of variables for the Euler-Lagrange equation and I ended up with the following equation (omitting intermediate steps): $$ \sum_k\{ \frac{d}{dt}(\frac{\partial L}{\partial \dot{\bar{q}}}_k )-\frac{\partial L}{\partial \bar{q}_k} \} \frac{\partial \bar{q}_k}{\partial q_i} =0.$$

What further argument can I make to say that $$ \frac{d}{dt}(\frac{\partial L}{\partial \dot{\bar{q}}}_k )-\frac{\partial L}{\partial \bar{q}_k} = 0?$$

I'm thinking along the lines of saying that $$\frac{\partial \bar{q}_k}{\partial q_i}$$ are independent functions and hence their coefficients have to be zero. But then why do they have to be independent?

Also, for the arbitrary change of coordinates, is it required for the $\bar{q}_i$ to be independent coordinates?


1 Answer 1


The quantity $\bar{q}_{k}$ is an arbitrary function of any number of the old coordinates $q_{i}$ and not independent of them. You can already see it from your second equation. Now, from the third equation you found, you can see that for it to hold for the sum of all $k$, there are two possibilities:

  • $\frac{\partial \bar{q}_{k}}{\partial q_{i}} = 0 \, , \, \forall \, i, k$

  • $\frac{d}{dt} \left (\frac{\partial L}{\partial \dot{\bar{q} _{k}}} \right )- \frac{\partial L}{\partial \bar{q} _{k}} = 0 \, , \, \forall \, k$

The first case is simply the trivial case of your new set of coordinates being constants and thus independent of any of the old coordinates. There is no dynamics and hence you can dismiss it. Thus the only non-trivial case is the latter i.e. the Euler-Lagrange equation for your new set of coordinates.

  • $\begingroup$ This is the right idea, but it's not strictly accurate that the third equation leads to one of the two equations you've written. If we define $$M_{ik} \equiv \frac{\partial \bar{q}_{k}}{\partial q_{i}} = 0, \qquad v_k \equiv \frac{d}{dt} \left (\frac{\partial L}{\partial \dot{\bar{q} _{k}}} \right )- \frac{\partial L}{\partial \bar{q} _{k}}$$then the OP's third equation is $M_{ik} v_k = 0$. This means that $v_k$ lies in the null space of the matrix $M_{ik}$, but it is certainly possible to have a non-zero vector $v_k$ in the null space of a non-zero matrix $M_{ik}$. ... $\endgroup$ Jun 14, 2022 at 14:11
  • $\begingroup$ ... Instead, if we need $M_{ik} v_k = 0$ to imply $v_k = 0$, it follows that $M_{ik}$ must be an invertible matrix; this is the usual condition on coordinate transformations (not just that the transformation is non-constant.) $\endgroup$ Jun 14, 2022 at 14:14
  • $\begingroup$ All true, albeit I took the invertibility for granted since for all reasonable physical systems the mapping is going to be a diffeomorphism. $\endgroup$
    – rhomaios
    Jun 14, 2022 at 14:51

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