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Let me say I consider this a mathematics post, not a physics one -- correct me if I'm wrong.

In Lagrangian Mechanics we consider the Lagrangian function, which is a function of the position, the velocity and time $L(x, \dot x,t)$. In arbitrary coordinates it has the form $L= L (q, \dot q,t) $ (where $q$ is a vector if we have a problem in $2$ or $3$ dimensions). Suppose we change the coordinate system to $q'=f(q,t)$ so that $\dot {q'}={\partial f \over \partial q} \dot q + {\partial f \over \partial t}$.

Then we get a new Lagrangian by

$$L'\left(q', \dot q',t\right)= L'\left(f(q,t),{\partial f \over \partial q} \dot q + {\partial f \over \partial t},t\right) = L(q,\dot q,t).$$

We know these two Lagrangians describe the same system, when the Euler-Lagrange equations are implemented on them (a bit a grammar issue here maybe- sorry).

These two functions are different, but for a given time $t$, their values coincide.

I want to show that $$ {d \over dt} \left({ \partial L \over \partial{\dot q}}\right) - {\partial L \over \partial q} = {\partial f \over \partial q} \left[ {d \over dt} \left( { \partial L' \over \partial{ \dot q'}}\right) -{ \partial L' \over \partial q'}\right] \space \space (1)$$ So my question is how do I approach this? The functions $L$, $L'$ are not equal (as functions), I only have a relation between the values of them in time $t$. More specifically how can I use the chain rule for a function like $L'$, what are the partial derivatives and what's its time derivative? I'm not so much interested in showing $(1)$ as I'm interesting in knowing how to find these derivatives.

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  • $\begingroup$ Silly question probably, but don't we have : ${d \over dt} \left({ \partial L \over \partial{\dot q}}\right) - {\partial L \over \partial q} =0$ on both sides of $(1)$? $\endgroup$ Commented Apr 21, 2017 at 17:59
  • $\begingroup$ @RutgerMoody What you said holds for the solution of the Euler-Lagrange equations, not for every $q, \dot q, t$ $\endgroup$
    – Arbiter
    Commented Apr 21, 2017 at 18:05
  • $\begingroup$ But then your claim is that $(1)$ is true in general for every function $L$? What makes you think it should be? $\endgroup$ Commented Apr 21, 2017 at 18:11
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    $\begingroup$ This is usually known as coordinate covariance. The key point here is that $f(q,t)$ is an invertible, differentiable transformation that does not depend on velocity. Here's an example derivation: physics.usu.edu/Wheeler/ClassicalMechanics/… $\endgroup$
    – Alex R.
    Commented Apr 21, 2017 at 18:13
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    $\begingroup$ @RutgerMoody I found it in a mechanics book as a problem left for the reader. We know $L$ as a function of $q, \dot q,t$. What we want to show is that for any transformation $q'=f(q,t)$ (1) holds, where $L'$ is the "same" Lagrangian expressed in different coordinates. $\endgroup$
    – Arbiter
    Commented Apr 21, 2017 at 18:14

1 Answer 1

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Ok first we use the chain rule:

\begin{align*} \frac{\partial L}{\partial \dot q} &= \frac{\partial L'}{\partial q'} \frac{\partial q'}{\partial \dot{q}} + \frac{\partial L'}{\partial \dot{q}'} \frac{\partial \dot{q}'}{\partial \dot q} + \frac{\partial L'}{\partial t} \frac{\partial t}{\partial \dot q}\\ &= \frac{\partial L'}{\partial q'} 0 + \frac{\partial L'}{\partial \dot{q}'} \frac{\partial \dot{q}'}{\partial \dot q} + \frac{\partial L'}{\partial t} 0\\ &= \frac{\partial L'}{\partial \dot{q}'} \frac{\partial f}{\partial q}, \end{align*} since $\dot q ' = \frac{\partial f}{\partial q} \dot q + \frac{\partial f}{\partial t}$.

Using the same line of argument we can show that: \begin{align*} \frac{\partial L}{\partial q} &= \frac{\partial L'}{\partial q'} \frac{\partial f}{\partial q} + \frac{\partial L'}{\partial \dot q'} \frac{\partial \dot q'}{\partial q}\\ &= \frac{\partial L'}{\partial q'} \frac{\partial f}{\partial q} + \frac{\partial L'}{\partial \dot q'} \left(\frac{\partial^2 f}{\partial q^2}\dot q + \frac{\partial^2 f}{\partial q \partial t} \right) \end{align*} Now let calculate $\frac{d}{dt} \frac{\partial L}{\partial \dot q} $: \begin{align*} \frac{d}{dt} \frac{\partial L}{\partial \dot q} &= \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q}\right) \frac{\partial f}{\partial q}+ \frac{\partial L'}{\partial \dot q'} \left(\frac{\partial^2 f}{\partial q \partial t} + \frac{\partial^2 f}{\partial q^2} \dot q\right) \end{align*} Finally equating $\frac{d}{dt} \frac{\partial L}{\partial \dot q}$ with $\frac{\partial L}{\partial q}$ we get that: \begin{align*} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot q}\right) \frac{\partial f}{\partial q} = \frac{\partial L'}{\partial q'} \frac{\partial f}{\partial q} , \end{align*} as we wanted.

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