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A lagrangian $L(q,\dot q, t)$ is invariant under the point transformation $$q_i=q_i(s_1,...,s_n,t)$$

To prove this I show that

$$\frac{d}{dt} \frac{\partial L}{\partial \dot s_i} - \frac{\partial L}{\partial s_i} = 0$$

where

$$\frac{\partial L}{\partial s_i} = \frac{\partial L}{\partial q_j}\frac{\partial q_j}{\partial s_i}$$

$$\frac{\partial L}{\partial \dot s_i} = \frac{\partial L}{\partial \dot q_j}\frac{\partial \dot q_j}{\partial \dot s_i}$$

$$\frac{\partial \dot q_j}{\partial \dot s_i}= \frac{\partial q_j}{\partial s_i}$$

which gives $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot q_j}\frac{\partial q_j}{\partial s_i} \right) - \frac{\partial L}{\partial q_j}\frac{\partial q_j}{\partial s_i} = 0$$

but in order to get into the form $$\left(\frac{d}{dt} \frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j}\right)\frac{\partial q_j}{\partial s_i} = 0$$

I need to prove that $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot q_j}\frac{\partial q_j}{\partial s_i} \right) = \frac{d}{dt} \left(\frac{\partial L}{\partial \dot q_j}\right)\frac{\partial q_j}{\partial s_i} $$

What is the logic behind this? Any help would be greatly appreciated!

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    $\begingroup$ This equation is not correct: $$\frac{\partial L}{\partial s_i} = \frac{\partial L}{\partial q_j}\frac{\partial q_j}{\partial s_i}$$ It should be $$ \frac{\partial L}{\partial s_i} = \frac{\partial L}{\partial q_j}\frac{\partial q_j}{\partial s_i} + \frac{\partial L}{\partial \dot q_j} \frac{\partial \dot q_j}{\partial s_i} $$ $\endgroup$
    – md2perpe
    Jul 23, 2018 at 20:03

1 Answer 1

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In case anyone wants the whole proof.

Invariance of Lagrange's Equations under Point Transformations

Let $q_1,q_2,\cdots,q_n$ be a set of independent generalised coordinates for a system on $n$ degrees of freedom, with Lagrangian $L(q,\dot{q},t)$. Suppose we transform to another set of independent coordinates $s=s_1,s_2,\cdots,s_n$ be means of transformation equations: \begin{equation} q_i=q_i(s,t) \qquad i=1,2,\cdots,n \qquad (1) \end{equation} and \begin{align} \dot{q}_i=\frac{dq_i}{dt}=\sum_j \frac{\partial q_i}{\partial s_j}\dot{s}_j+\frac{\partial q_i}{\partial t} \nonumber \\ \dot{q}_i=\dot{q_i}(s,\dot{s},t)\qquad i=1,\cdots,n \qquad (2) \end{align} Such a transformation is called a point transformation. Then: \begin{align} \frac{\partial L}{\partial s_j}=\sum_i\left[\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial s_j} +\frac{\partial L}{\partial \dot{q}_i}\frac{\partial \dot{q}_i}{\partial s_j}\right]\qquad (3) \end{align}

\begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)&=\frac{d}{dt}\left[\sum_i\left(\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial \dot{s}_j}+\frac{\partial L}{\partial \dot{q}_i}\frac{\partial \dot{q}_i}{\partial \dot{s}_j}\right) \right] \qquad (4) \end{align} but we also know that $\partial q_i/\partial \dot{s}_j=0$ and also: \begin{align} dq_i=\sum_j\frac{\partial q_i}{\partial s_j}ds_j+\frac{\partial q_i}{\partial t}dt \nonumber \\ \frac{dq_i}{dt}=\sum_j\frac{\partial q_i}{\partial s_j}\frac{ds_j}{dt}+\frac{\partial q_i}{\partial t} \nonumber \\ \dot{q}_i=\sum_j\frac{\partial q_i}{\partial s_j}\dot{s}_j+\frac{\partial q_i}{\partial t} \nonumber \end{align} \begin{equation} \frac{\partial \dot{q}_i}{\partial \dot{s}_j}=\frac{\partial q_i}{\partial s_j} \qquad (5) \end{equation} Making these changes to (4) we get: \begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)&=\frac{d}{dt}\left[\sum_i \frac{\partial L}{\partial \dot{q}_i}\frac{\partial q_i}{\partial s_j}\right] =\nonumber \\ &=\sum_i \left[\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)\frac{\partial q_i}{\partial s_j}+\frac{\partial L}{\partial \dot{q}_i}\frac{d}{dt}\left(\frac{\partial q_i}{\partial s_j}\right)\right] \qquad (6) \end{align} But we know that $L$ satisfies the Lagrangian equations for the $q$ set of coordinates. Hence: \begin{equation} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)=\frac{\partial L}{\partial q_i} \qquad (7) \end{equation} as well as the property: \begin{align} \frac{d}{dt}\left(\frac{\partial q_i}{\partial s_j}\right)&=\sum_k\frac{\partial}{\partial q_k}\left(\frac{\partial q_i}{\partial s_j}\right)\dot{q_k}+\frac{\partial^2q_i}{\partial t\partial s_j}=\nonumber \\ &=\frac{\partial}{\partial s_j}\left[\sum_k\frac{\partial q_i}{\partial q_k}\dot{q}_k+\frac{\partial q_i}{\partial t}\right]\nonumber \\ &=\frac{\partial}{\partial s_j}\left(\frac{d q_i}{dt}\right)=\nonumber \\ &=\frac{\partial \dot{q}_i}{\partial s_j}\qquad (8) \end{align} and (6) becomes \begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)&=\sum_i\left[\frac{\partial L}{\partial q_i}\frac{\partial q_i}{\partial s_j}+\frac{\partial L}{\partial \dot{q}_i}\frac{\partial \dot{q}_i}{\partial s_j}\right] \qquad(9) \end{align}

Now making the substitutions (3) and (9), we get the important result that: \begin{align} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{s}_j}\right)-\frac{\partial L}{\partial s_j}=0 \qquad(10) \end{align}

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  • $\begingroup$ Hi, nice write up! I have a question about equations (3) and (4). Why we don't have to include the terms involving $\frac{\partial L}{\partial t}$, while in the equation just below (4), you have the term $\frac{\partial q_i}{\partial t} dt$? $\endgroup$
    – hb12ah
    Apr 2, 2020 at 7:23
  • $\begingroup$ I think I've asked a bad question. Now I see the reason. $q$ is a function of $s$ but $t$ is not. $\endgroup$
    – hb12ah
    Apr 2, 2020 at 9:26
  • $\begingroup$ Nice answer, just a typo in the third line of (8), should be $\frac{dq_i}{dt}$ instead of $\frac{dq}{dt}$ $\endgroup$
    – ydhhat
    Oct 24, 2020 at 16:22

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