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In this post, I tried to challenge what @Richard Myers 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 @Richard Myers 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. \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}, - t) } \nonumber \end{eqnarray}

I follow @Richard Myers 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?
$$ \frac{d {\vec x}( t)}{d t} \overset{T}{\to} \color{red}{+ \frac{d {\vec x'}( t)}{d t}} .$$ @Richard Myers advice gives a strange conclusion (!?). Similarly, is the velocity dependent potential $V$ (or any odd $n$ power $ \frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}$ dependent $V$) really time reversal? $$ V ({\vec x}(\tilde t), \frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}, \tilde t) \bigg\vert_{ \tilde t = t} \overset{T}{\to}\color{red}{ V ({\vec x'}(t), \frac{d^n {\vec x'(t)}}{d t^n}, - t) } $$ It seems only that the odd $t$ function of $V(t)$ matter for breaking time reversal, but not the velocity dependence? It is weird!

  1. We also check whether the Equation Of Motion is time reversal invariant. But I think the conclusion above still holds -- it leads to the result that any even or odd $n$ power $ \frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}$ dependent $V$ does not break time reversal. This seems to contradict with the common sense that the velocity-dependent potential leads to violate time reversal.

Please point out any logical flaw if there is any.

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As I and other have pointed out to you elsewhere, the arrow notation is the source of your confusions. For completeness, the "advice" I have given in my other answers is to stop using this notation entirely. It may be exceedingly common notation within physics, but so is confusion, so that is hardly an argument in favor of using it.

To begin, let me point out that the conclusion you have reached about odd-order time derivatives being even under time reversal clearly cannot be correct. This can be seen in the example $x(t)=t$, the derivative of which is $1$, while the time reversal of this function is $x^\prime(t)=-t$, which clearly has first derivative $-1$.

As I described in my answer here, we must be clear about what we are doing. I define the time reversal of a function $x(t)$ to be the function $x^\prime(t)$ defined by $x^\prime(t)=x(-t)$. No arrows to be seen or needed. This is an entirely unambiguous statement defining one function in terms of another.

If we would now like, we may compute the time derivative of this new function, $x^\prime$, and use its definition to rewrite it in terms of the derivative of the old function $x$: $$ \frac{d x^\prime(t)}{dt}=\frac{d x(-t)}{dt}=-\frac{d x(t^\prime)}{d t^\prime}\Bigg|_{t^\prime=-t}. $$ That is it. Nothing more and nothing less. Note that the final evaluation at $t^\prime=-t$ is necessary. For example, take $x(t)=t^2$ which has derivative $2t$. This function is reversal symmetric, so $x^\prime(t)=t^2$ as well. Using the expression found above with this particular function, we would find $\frac{d}{dt}t^2=-(2t^\prime)|_{t^\prime=-t}=2t$, so this extra evaluation is needed to make sure things work out in the end.

With this in mind, assume once more that $$ \frac{d^2x(t)}{dt^2}=\partial_xV\left(x(t),\frac{dx(t)}{dt},t\right). $$ As in my previous answer, we now ask what differential equation is satisfied by $x^\prime(t)$: $$ \frac{d^2 x^\prime(t)}{dt^2}=\frac{d^2x(-t)}{dt^2}=\frac{d^2x(t^\prime)}{dt^{\prime\,2}}\Bigg|_{t^\prime=-t}=\left[\partial_xV\left(x(t^\prime),\frac{d x(t^\prime)}{dt^\prime},t^\prime\right)\right]\Bigg|_{t^\prime=-t}. $$ The question we now need to ask is: what is $\frac{d x(t^\prime)}{dt^\prime}\Big|_{t^\prime=-t}$ in terms of $x^\prime(t)$?

In fact, this is precisely the right hand side of the identity found above, which we may rewrite as $$ \frac{dx(t^\prime)}{dt^\prime}\Bigg|_{t^\prime=-t}=-\frac{d x^\prime(t)}{dt}. $$ So we find that the function $x^\prime(t)$ obeys the differential equation $$ \frac{d^2 x^\prime(t)}{dt^2}=\partial_xV\left(x^\prime(t),-\frac{dx^\prime(t)}{dt^\prime},-t\right). $$

So, in fact, the method I described in my previous answer, if actually applied, produces precisely the result one would expect from time reversal, which may also be verified in specific examples as I noted.

The issue you are having, I will reiterate, is only enabled by this arrow notation you insist upon using. If you abandoned the arrow notation, as I recommend, you would have noticed immediately that your calculation showing that odd-order time derivatives are symmetric under time reversal is actually the tautology $$ \frac{d^n x^\prime(t)}{dt^n}=\frac{d^n x^\prime(t)}{dt^n}. $$ This fact is entirely obscured by the arrow notation.

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  • $\begingroup$ thanks so much really +1 - I will accept if I agree, and abandon the bad notation in the future $\endgroup$ Commented May 3, 2021 at 22:09
  • $\begingroup$ btw. Should we also think $V$ after time reversal as $V'$ assigned to the time coordinate $t$, based on the active transformation view? At least we have $V'(t) = V(-t)$ if the $V$ is only time-dependent without other dependence. Agree? Thanks! $\endgroup$ Commented May 3, 2021 at 22:12
  • $\begingroup$ Also how about the possible improvement on $∂_𝑥𝑉$ part in the final answer $\frac{d^2 x^\prime(t)}{dt^2}=\partial_xV\left(x^\prime(t),-\frac{dx^\prime(t)}{dt^\prime},-t\right). $, should it also has the $∂_{𝑥'}$ on $V'$ thus it is $∂_{𝑥'}V'$? $\endgroup$ Commented May 3, 2021 at 22:25
  • $\begingroup$ @anniemarieheart that derivative is being taking before evaluation, so it is entirely irrelevant to the point I am making here. $\endgroup$ Commented May 3, 2021 at 22:28
  • $\begingroup$ OK, I agree what you said. But I think what I did is logically mostly correct except I used the notation $\overset{T}{\to}$. What I did possibly wrong is the crucial step which I cannot yet debug: $\endgroup$ Commented May 3, 2021 at 23:03
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My earlier question made a mistake here: $$ V({\vec x}(\tilde t), \frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}, \tilde t) \bigg\vert_{ \tilde t = t} \overset{T}{\to} V ({\vec x}(\tilde t), (-1)^n\frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}, \tilde t) \bigg\vert_{ \tilde t =- t} =\color{red}{ V ({\vec x'}(t), \frac{d^n {\vec x'(t)}}{d t^n}, - t) \quad(\text{wrong})}$$ It should be instead $$ V({\vec x}(\tilde t), \frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}, \tilde t) \bigg\vert_{ \tilde t = t} \overset{T}{\to} V ({\vec x}(\tilde t), \frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}, \tilde t) \bigg\vert_{ \tilde t =- t} =\color{blue}{ V ({\vec x'}(t), (-1)^n\frac{d^n {\vec x'(t)}}{d t^n}, - t) \quad(\text{correct})}$$ This is because that the $V(\frac{d^n {\vec x}(\tilde t)}{d \tilde t^n})$ is an explicit function depends on $\frac{d^n {\vec x}(\tilde t)}{d \tilde t^n}$. We do the time reversal on the $V$ and we obtain the new $V'(t)=V(-t)$.

Although this step is still true: $$ \frac{d^n {\vec x}( t)}{d t^n} \overset{T}{\to} (-1)^n\frac{d^n {\vec x}(\tilde t)}{d \tilde t^n} \bigg\vert_{ \tilde t = -t} = (-1)^n\frac{d^n {\vec x'}(-\tilde t)}{d \tilde t^n} \bigg\vert_{ \tilde t = -t} = \frac{d^n {\vec x'}(-\tilde t)}{d (-\tilde t)^n} \bigg\vert_{ -\tilde t = t} =\color{blue}{ \frac{d^n {\vec x'}(t)}{d t^n} \quad(\text{correct})}$$ This is because that the $x(t)$ is an explicit function depends on $t$, and we do the time reversal on the $x(t)$ as $x'(t)=x(-t)$ and do the derivative respect to the $x'(t)$.

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