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Newton's law are form invariant under the coordinate substitutions:

$$ \tilde{x^{i}}=x^{i}+a^{i} $$

This means that Newtons' equation of motion,

$$ F^{i}=m \frac{d^{2} x^{i}}{d t^{2}} $$

(where $i=1,2,3)$ is the same in terms of the $\tilde{x}^{i}$ Is this the Gallielian Invariance ? How would I prove this ?

  1. If we define new set of coordinates as

$$ y^{i}=y^{i}\left(x^{k}\right) . $$

How would I find the equations of motion in these new cordinates assuming that $x^{k}$ are coordinates in an inertial frame? Which textbook should I be looking at ?

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  • $\begingroup$ Is i equal k ? In your question $\endgroup$
    – Eli
    Sep 5, 2023 at 7:47
  • $\begingroup$ No it doesn’t seem. There are two parts to the question. $\endgroup$
    – HRTninja
    Sep 5, 2023 at 22:44

1 Answer 1

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Newton's equation \begin{align*} &m\,\ddot{\mathbf{x}}=\b F(\b x~,\dot{\mathbf{x}})\tag 1 \end{align*} with the new set of coordinates (where $~i=k~$) \begin{align*} &\b y=\b y(\b x)\quad \Rightarrow \dot{\b{y}}=\underbrace{\frac{\partial \b y}{\partial \b x}}_{\b A(\b x)}\dot{\b x}\quad\Rightarrow\\ \end{align*} assume that $~\det(A)\ne 0~$ \begin{align*} &\dot{\b x}=\b{A}^{-1}\ \dot{\b{y}}=\b J(\b x)\,\dot{\b{y}}\tag 2 \end{align*} with \begin{align*} \ddot{\b{x}}=\b J\,\ddot{\b{y}}+\frac{\partial \dot{\b{x}}}{\partial\b y}\,\dot{\b{y}} \end{align*} and equation (1)

\begin{align*} &m\,\left(\b J\,\ddot{\b{y}}+\frac{\partial \dot{\b{x}}}{\partial\b y}\,\dot{\b{y}}\right)=\b F(\b x~,\dot{\mathbf{x}})\quad,\text{or} \end{align*}

\begin{align*} &m\,\b J^T\,\b J\ddot{\b{y}}=\b J^T\,\b F(\b x~,\dot{\b{x})}-m\,\b J^T\,\frac{\partial \dot{\b{x}}}{\partial\b y}\,\dot{\b{y}}\tag 3 \end{align*}

thus the Newton's equation (1) transfered to equation (2) and (3)

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