I am trying to prove the classic problem to showcase Lagrangian's scalar invariant property.

Namely, that if you have $x_i = \{ x_1, ...., x_n; t \}$ , you can then represent $L(x_1,....,\dot{x_1},...,t)$ and find the Euler-Lagrangian equations of motion: \begin{equation} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x_i}}\right) - \frac{\partial L}{\partial x_i} = 0.\tag{1} \end{equation}

Now, if you introduce a new set of coordinates $q_j = \{ q_1, ...., q_n; t \}$, then you too can express the equations of motion via:

\begin{equation} \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q_j}}\right) - \frac{\partial L}{\partial q_j} = 0.\tag{2} \end{equation} I know I have to show that indeed this is the case, thus showing $L$ is scalar invariant under point transformations.

Now here's my problem.

I am going to use the simplest example I know: you only have $x$ and $q$ - only one degree of freedom.

I can define $x(q,t)$ and $q(x,t)$.

I can find the differential as follows:

\begin{equation} dx = \frac{\partial x}{\partial q}dq + \frac{\partial x}{\partial t}dt\tag{3} \end{equation}

And thus, I can find $\dot{x}$:

\begin{eqnarray} \frac{dx}{dt}= \frac{\partial x}{\partial q} \frac{dq}{dt} + \frac{\partial x}{\partial t} \tag{4}\\ \dot{x} = \frac{\partial x}{\partial q} \dot{q} + \frac{\partial x}{\partial t}.\tag{5} \end{eqnarray}

I have read that you can turn that last equation into: \begin{equation} \frac{\partial \dot{x}}{\partial \dot{q}}= \frac{\partial x}{\partial q} + \frac{\partial}{\partial q}\frac{\partial x}{\partial t} \tag{6} \end{equation}

which would yield: \begin{equation} \frac{\partial \dot{x}}{\partial \dot{q}}= \frac{\partial x}{\partial q}.\tag{7} \end{equation}

Here's my question:

I am not sure how to justify the change that yields this result. Because if I follow the equation closely, I would find this value: \begin{equation} \frac{\frac{dx}{dt}}{\frac{dq}{dt}} = \frac{\partial x}{\partial q} + \frac{d}{dq}\frac{\partial x}{\partial t}.\tag{8} \end{equation}

What am I missing or doing wrong that yields the intended result in partials rather than the result I just wrote above?

  • $\begingroup$ Read eqs. (6) and (8) where? $\endgroup$ – Qmechanic Sep 20 '19 at 17:58
  • $\begingroup$ If you look at eq (6), I am not sure how you justify taking $\frac{\dot{x}}{\dot{q}}$ into it's partial from that is shown on eq(7). $\endgroup$ – Enrique Segura Sep 20 '19 at 18:26
  • 1
    $\begingroup$ Eq. (7) is correct. $\endgroup$ – Qmechanic Sep 20 '19 at 18:28
  • $\begingroup$ Yes of course! I am not saying it's not. I want to understand how to get there. It does not seem clear to me in terms of how do you convert a $dq$ into a $\partial q$. $\endgroup$ – Enrique Segura Sep 20 '19 at 19:34
  • $\begingroup$ Eqs. (6) and (8) are not correct. $\endgroup$ – Qmechanic Sep 20 '19 at 19:37

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