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I want to understand the following:

The coordinates in our original coordination system are:

$\vec x = (x_1,..., x_{3N})^T,\dot{\vec x} = (\dot x_1,..., \dot x_{3N})^T$

We want to use generalized coordinates: $\vec q = (q_1,..., q_{3N})^T,\dot{\vec q} = (\dot q_1,..., \dot q_{3N})^T$

The Lagrangian-function is given by $$L=L(\vec x,\dot{\vec x} , t)=L(\vec x(\vec q),\dot{\vec x}(\vec q, \dot{\vec q},t) , t)$$

So $L: \mathbb{R}^{3N}\times \mathbb{R}^{3N}\times \mathbb{R}^+ \rightarrow \mathbb{R}$

I haven't had any multidimensional analysis yet, so it's difficult for me to understand the following partial derivative:

$$\frac{\partial L}{\partial \dot q_j}=\sum_{i=1}^{3N}(\frac{\partial L}{\partial x_i}\frac{\partial x_i}{\partial \dot q_j} + \frac{\partial L}{\partial \dot x_i}\frac{\partial \dot x_i}{\partial \dot q_j}) $$

Can somebody explain to me, what has been done here?

It looked a bit like the chain-rule, so I read this section: https://en.wikipedia.org/wiki/Chain_rule#General_rule

What I find confusing, is that I can't identify an $f$ and $g$, if you interpret $\vec x:\mathbb{R}^{3N}\rightarrow \mathbb{R}^{3N}, \vec q\mapsto \vec x(\vec q)$ as a function and $L: \mathbb{R}^{3N}\rightarrow \mathbb{R}, \vec q\mapsto L(\vec x(\vec q))$

$$D(L\circ x)=D_L\cdot D_x\\ =\begin{pmatrix} \frac{\partial L}{\partial x_1}L(\vec x) & ... & \frac{\partial L}{\partial x_{3N} }L(\vec x)\end{pmatrix}\cdot \begin{pmatrix} \frac{\partial x_1}{\partial \dot q_1} & ... & \frac{\partial x_1}{\partial \dot q_{3N} } \\ \vdots &... & \vdots \\ \frac{\partial x_{3N}}{\partial \dot q_1} & ... & \frac{\partial x_{3N}}{\partial \dot q_{3N} }\end{pmatrix}\\ =(\sum_{i=1}^{3N} \frac{\partial L}{\partial x_i}L(\vec x)\cdot \frac{\partial x_{i}}{\partial \dot q_1},...,\sum_{i=1}^{3N} \frac{\partial L}{\partial x_i}L(\vec x)\cdot \frac{\partial x_{i}}{\partial \dot q_{3N}}) $$

Analogue for $\dot x$: $$D(L\circ \dot x)=D_L\cdot D_{\dot x}\\= \begin{pmatrix} \frac{\partial L}{\partial \dot x_1}L(\dot{\vec x}) & ... & \frac{\partial L}{\partial \dot x_{3N} }L(\dot{\vec x})\end{pmatrix}\cdot \begin{pmatrix} \frac{\partial \dot x_1}{\partial \dot q_1} & ... & \frac{\partial \dot x_1}{\partial \dot q_{3N} }\\ \vdots &... & \vdots \\ \frac{\partial \dot x_{3N}}{\partial \dot q_1} & ... & \frac{\partial \dot x_{3N}}{\partial \dot q_{3N} }\end{pmatrix}\\ =(\sum_{i=1}^{3N} \frac{\partial L}{\partial \dot x_i}L(\dot{\vec x})\cdot \frac{\partial \dot x_{i}}{\partial \dot q_1},...,\sum_{i=1}^{3N} \frac{\partial L}{\partial \dot x_i}L(\dot{\vec x} )\cdot \frac{\partial \dot x_{i}}{\partial \dot q_{3N}}) $$

So if I add my results, I get ALMOST the given form when comparing the components, the differences are the factors $L(\dot{\vec x}), L(\vec x)$:

$$D(L\circ x)+D(L\circ \dot x)=(\sum_{i=1}^{3N} \frac{\partial L}{\partial x_i}L(\vec x)\cdot \frac{\partial x_{i}}{\partial \dot q_1},...,\sum_{i=1}^{3N} \frac{\partial L}{\partial x_i}L(\vec x)\cdot \frac{\partial x_{i}}{\partial \dot q_{3N}})+(\sum_{i=1}^{3N}\frac{\partial L}{\partial \dot x_i}L(\dot{\vec x})\cdot \frac{\partial \dot x_{i}}{\partial \dot q_1},...,\sum_{i=1}^{3N} \frac{\partial L}{\partial \dot x_i}L(\dot{\vec x} )\cdot \frac{\partial \dot x_{i}}{\partial \dot q_{3N}}) $$

AND I can't explain why I should add $D(L\circ x)$and $D(L\circ \dot x)$?

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  • $\begingroup$ Are you sure that $\vec x$ is only a function of $\vec q$ and does not depend on $\dot{\vec q}$? In that case $\frac{\partial x_i}{\partial \dot q_j} = 0$. What is the source of the formula? $\endgroup$
    – Paul Frost
    Oct 2, 2022 at 14:31
  • $\begingroup$ Yes, there is actually a question/answer about this dependency and I can't justify it better: math.stackexchange.com/questions/4131929/… $\endgroup$ Oct 2, 2022 at 18:33
  • $\begingroup$ I understand, why $\frac{\partial x_i}{\partial \dot q_j} = 0$ has to be true, if there is no dependency of $\vec x$ from $\dot{\vec{q}}$. The source of the formula is a scipt about analytical mechanics, more specifc: It is a chapter about the invariance of the Euler-Lagrange-Equations if you change the coordinates from cartesian to generalized coordinates. My question relates to the proof of this fact. $\endgroup$ Oct 2, 2022 at 18:42

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Let us write more generally $\vec x = \vec x(\vec q, \dot{\vec q},t)$. I do not know whether it makes sense physically that we assume a dependency of $\vec x$ from $\dot{\vec q}$ and $t$, but formally it is certainly okay. If $\vec x$ only depends on $\vec q$, then we get $\dfrac{\partial x_i}{\partial \dot{\vec q_j}} = 0$ and $\dfrac{\partial x_i}{\partial t} = 0$ which leads to the formula $$\frac{\partial L}{\partial \dot q_j}=\sum_{i=1}^{3N}\frac{\partial L}{\partial \dot x_i}\frac{\partial \dot x_i}{\partial \dot q_j} .$$ So let us consider the function $$\tilde L(\vec q, \dot{\vec q},t) = L(\vec x(\vec q, \dot{\vec q},t),\dot{\vec x}(\vec q, \dot{\vec q},t), t)$$ defined on $\mathbb R^{3N} \times \mathbb R^{3N} \times \mathbb R^+$. Note that I wrote $\tilde L$ to distinguish it from $L$ which is a function of $\vec x, \dot{\vec x},t$. We have $$\tilde L = L \circ X$$ where $X : \mathbb R^{3N} \times \mathbb R^{3N} \times \mathbb R^+ \to \mathbb R^{3N} \times \mathbb R^{3N} \times \mathbb R^+$ is given by $X(\vec q, \dot{\vec q},t) = (\vec x(\vec q, \dot{\vec q},t),\dot{\vec x}(\vec q, \dot{\vec q},t), t)$.

Since we are only interested in the partial derivatives with respect to the $\dot{\vec q_j}$, we simplify notation by ignoring the variables $\vec q$ and $t$. That is, we consider the functions $$\vec x = \vec x(\dot{\vec q}) ,$$ $$\dot{\vec x} = \dot{\vec x}(\dot{\vec q}) ,$$ $$L(\vec x,\dot{\vec x}) ,$$ $$\tilde L(\dot{\vec q}) = L(\vec x(\dot{\vec q}),\dot{\vec x}(\dot{\vec q})) ,$$ $$X(\dot{\vec q}) = (\vec x(\dot{\vec q}),\dot{\vec x}(\dot{\vec q})) .$$

The chain rule tells us that the Jacobian matrix of $\tilde L$ at a point $\dot{\vec q} \in \mathbb R^{3N}$ $$J \tilde L (\dot{\vec q}) = \begin{pmatrix} \dfrac{\partial \tilde L}{\partial \dot q_1}(\dot{\vec q}) & \dots & \dfrac{\partial \tilde L}{\partial \dot q_{3N}}(\dot{\vec q}) \end{pmatrix}$$ is the product $JL(X(\dot{\vec q})) \circ JX(\dot{\vec q})$ of the Jacobians of $L$ and $X$. Suppressing the argument $\dot{\vec q}$ on the RHS we get $$JL(X(\dot{\vec q})) = \begin{pmatrix} \dfrac{\partial L}{\partial x_1} & \dots & \dfrac{\partial L}{\partial x_{3N}} & \dfrac{\partial L}{\partial \dot x_1} & \dots & \dfrac{\partial L}{\partial \dot x_{3N}} \end{pmatrix} $$ $$JX(\dot{\vec q}) = \begin{pmatrix} \dfrac{\partial x_1}{\partial \dot q_1} & \dots & \dfrac{\partial x_1}{\partial \dot q_{3N}} \\ \dots & \dots & \dots \\ \dfrac{\partial x_{3N}}{\partial \dot q_1} & \dots & \dfrac{\partial x_{3N}}{\partial \dot q_{3N}} \\ \dfrac{\partial \dot x_1}{\partial \dot q_1} & \dots & \dfrac{\partial \dot x_1}{\partial \dot q_{3N}} \\ \dots & \dots & \dots \\ \dfrac{\partial \dot x_{3N}}{\partial \dot q_1} & \dots & \dfrac{\partial \dot x_{3N}}{\partial \dot q_{3N}} \end{pmatrix}$$ A simple matrix multiplication shows then that $$\frac{\partial \tilde L}{\partial \dot q_j}=\sum_{i=1}^{3N}(\frac{\partial L}{\partial x_i}\frac{\partial x_i}{\partial \dot q_j} + \frac{\partial L}{\partial \dot x_i}\frac{\partial \dot x_i}{\partial \dot q_j}) .$$

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