Time reversal on potential $V(\frac{d^n {\vec x}( t)}{d t^n})$ with any odd or even power time-derivative on the position function
In this post, I tried to challenge what <PERSON> said in his answer in https://physics.stackexchange.com/a/633205/42982. I followed what he said, except I kept a common widely used notation $\overset{T}{\to}$ to imply the function assignment before and after the time reversal.
The symmetries of the action (or Lagrangian) are always symmetries of the associated Euler-Lagrange equations.
The symmetries of the Euler-Lagrange equations are not always symmetries of the associated action (or Lagrangian).
Question How do we exam the time-reversal symmetries? On the Lagrange that depends on not just the position function $\vec{x}(t)$ but also the explicit higher derivative $\frac{d^n {\vec x}( t)}{d t^n}$ and the explicit $t$ dependence: $$ L=\frac{1}{2} m \frac{d^2 \vec{x}(t) }{d t^2} -V ( \vec{x}(t), \frac{d^n {\vec x}( t)}{d t^n}, t) $$ The $V=V ( \vec{x}(t), \frac{d^n {\vec x}( t)}{d t^n}, t)$ is the potential term. The $n$ can be any positive integer $n=1,2,3,etc.$
1. We check whether $L$ is time reversal invariant. This means that we follow the rule (including the suggestion by <PERSON> that we should rewrite the old function $f(t)$ define a new function $f'(t)$ (Note: here $'$ is not a derivative but only a function labeling!!!) after the time reversal. 955
\begin{eqnarray} { t} &\overset{T}{\to}& { t},\nonumber\ \vec x(t) &\overset{T}{\to}& \vec x({-t}) \equiv \vec x'(t) ,\nonumber\ \frac{d^n {\vec x}( t)}{d t^n} &\overset{T}{\to}& (-1)^n\frac{d^n {\vec x}(\tilde t)}{d \tilde t^n} \bigg\vert_{ \tilde t = -t} = (-1)^n\frac{d^n {\vec x'}(-\tilde t)}{d \tilde t^n} \bigg\vert_{ \tilde t = -t} = \frac{d^n {\vec x'}(-\tilde t)}{d (-\tilde t)^n} \bigg\vert_{ -\tilde t = t} =\color{red}{ \frac{d^n {\vec x'}(t)}{d t^n} } , \nonumber\
V ({\vec x}(\tilde t), \frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}, \tilde t) \bigg\vert_{ \tilde t = t} &\overset{T}{\to}& V ({\vec x}(\tilde t), (-1)^n\frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}, \tilde t) \bigg\vert_{ \tilde t =- t}
=\color{red}{ V ({\vec x'}(t), \frac{d^n {\vec x'(t)}}{d t^n},
2. t) } \nonumber \end{eqnarray}
I follow <PERSON> advice to write down the higher time derivative on the position function. Tt seems leading to a weird result. For any odd or even power time derivative on the position function, we still get the same form under time reversal. Does it mean that it (like $n=1$ as the velocity) is really time reversal?
$$ \frac{d {\vec x}( t)}{d t} \overset{T}{\to} \color{red}{+ \frac{d {\vec x'}( t)}{d t}} .$$ <PERSON> advice gives a strange conclusion (! 804