Explicitly show covariance of Euler Lagrange equations

I know that the Euler Lagrange equation (here only in 1D)

$$\left(\frac{d}{dt}\frac{\partial}{\partial\dot{x}}-\frac{\partial}{\partial x}\right)L\left(x,\dot{x},t\right)=0$$

is invariant under (invertible) coordinate transformations of the kind $q=q\left(x,t\right)$. Most simply because the derivation using the principle of least action can be performed in any coordinate system. Suppose however that I wish to explicit show that if EL is satisfied for $x$ that it will also be satisfied for $q$ by actually changing variables in the equation.

I begin by rewriting $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial\dot{x}}$ as

\begin{eqnarray*} \frac{\partial}{\partial x} & = & \left(\frac{\partial q}{\partial x}\right)\frac{\partial}{\partial q}+\left(\frac{\partial\dot{q}}{\partial x}\right)\frac{\partial}{\partial\dot{q}}\\ \frac{\partial}{\partial\dot{x}} & = & \left(\frac{\partial q}{\partial\dot{x}}\right)\frac{\partial}{\partial q}+\left(\frac{\partial\dot{q}}{\partial\dot{x}}\right)\frac{\partial}{\partial\dot{q}} \end{eqnarray*} so that my EL now reads

$$\left(\frac{d}{dt}\left[\left(\frac{\partial q}{\partial\dot{x}}\right)\frac{\partial}{\partial q}+\left(\frac{\partial\dot{q}}{\partial\dot{x}}\right)\frac{\partial}{\partial\dot{q}}\right]-\left[\left(\frac{\partial q}{\partial x}\right)\frac{\partial}{\partial q}+\left(\frac{\partial\dot{q}}{\partial x}\right)\frac{\partial}{\partial\dot{q}}\right]\right)L\left(q,\dot{q},t\right)=0$$

I then let $\frac{d}{dt}$ act from the right and collect terms, at some point i should possibly use that $\dot{q}\left(x,\dot{x},t\right)=\frac{\partial q}{\partial t}+\frac{\partial q}{\partial x}\dot{x}$ and $\dot{x}\left(q,\dot{q},t\right)=\frac{\partial x}{\partial t}+\frac{\partial x}{\partial q}\dot{q}$ to in the end obtain

$$\left(\mbox{some function}\right)\left(\frac{d}{dt}\frac{\partial}{\partial\dot{q}}-\frac{\partial}{\partial q}\right)L\left(q,\dot{q},t\right)=0$$

However expanding the action of $\frac{d}{dt}$ gives a horrible mess that i will not reproduce for you.

The question then is: Is the setup I'm trying above correct (albeit ugly), or is there some neater way, without making use of the principle of least action?

Iv'e found some questions related to this, such as Euler Lagrange equation in different frames but i'm not sure how to make use of them.

• – Qmechanic Sep 14 '16 at 16:56

$$\left(\frac{\mathrm d}{\mathrm dt}\left[\left(\frac{\partial q}{\partial\dot{x}}\right)\frac{\partial}{\partial q}+\left(\frac{\partial\dot{q}}{\partial\dot{x}}\right)\frac{\partial}{\partial\dot{q}}\right]-\left[\left(\frac{\partial q}{\partial x}\right)\frac{\partial}{\partial q}+\left(\frac{\partial\dot{q}}{\partial x}\right)\frac{\partial}{\partial\dot{q}}\right]\right)L\left(q,\dot{q},t\right)=0$$

Dear Mikael, if you had carried your derivation a little bit further, you would have gotten the correct answer!

First notice that since you wrote: $q = q(x,t)$, $q$ is not explicitly dependent on $\dot x$. So: $$\frac{\partial q}{\partial \dot x} = 0$$ In addition, as you wrote: $$\dot q = \frac{\partial q}{\partial t} + \frac{\partial q}{\partial x}\dot x$$ Which means that: $$\frac{\partial \dot q}{\partial \dot x} = \frac{\partial q}{\partial x} \quad \text{simply reading from the expression of \dot q}$$ So we can simplify your original expression down to: $$\left( \frac{\mathrm d}{\mathrm dt}\left(\frac{\partial q}{\partial x}\frac{\partial}{\partial \dot q}\right) - \frac{\partial q}{\partial x}\frac{\partial}{\partial q} - \frac{\partial \dot q}{\partial x}\frac{\partial}{\partial \dot q} \right) L(q,\dot q,t) = 0$$ In order to make this resemble EL equations more, we apply chain rule and do some rearrangements of terms: $$\left( \frac{\mathrm d}{\mathrm dt}\frac{\partial q}{\partial x} \cdot\frac{\partial}{\partial \dot q} - \frac{\partial \dot q}{\partial x}\frac{\partial}{\partial \dot q}+ \frac{\partial q}{\partial x} \cdot \frac{\mathrm d}{\mathrm dt}\frac{\partial}{\partial \dot q} - \frac{\partial q}{\partial x}\frac{\partial}{\partial q} \right) L(q,\dot q,t) = 0$$

Notice that the first two terms actually cancel because: $$\frac{\partial \dot q}{\partial x} = \frac{\partial}{\partial x}\frac{\mathrm d}{\mathrm dt} q = \frac{\mathrm d}{\mathrm dt} \frac{\partial}{\partial x} q$$

So now we are only left with: $$\left( \frac{\partial q}{\partial x} \cdot \frac{\mathrm d}{\mathrm dt}\frac{\partial}{\partial \dot q} - \frac{\partial q}{\partial x}\frac{\partial}{\partial q} \right) L(q,\dot q,t) = 0$$

But that just means: $$\frac{\partial q}{\partial x} \cdot \left( \frac{\mathrm d}{\mathrm dt}\frac{\partial}{\partial \dot q} - \frac{\partial}{\partial q} \right) L(q,\dot q,t) = 0$$ Since the coordinate transformation is not singular, $\frac{\partial q}{\partial x} \neq 0$, which implies that: $$\left( \frac{\mathrm d}{\mathrm dt}\frac{\partial}{\partial \dot q} - \frac{\partial}{\partial q} \right) L(q,\dot q,t) = 0$$

P.S. The same derivation would fail if $q$ is dependent upon $\dot x$. See Qmechanic 's answer in the question you quoted.