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It is clear that there is time reversal symmetry for Newton's second law if we have time-independent potential $V(x)$.

However, it seems that we still have time reversal symmetry for Newton's second law if we have time-dependent potential $V(x,t)$. My thinking is the following, under time reversal transformation $T$, we have sent $$ t' \overset{T}{\to} -t'. $$ $$ x(t') \overset{T}{\to} x(-t'). $$ $$ \frac{d {x}( t)}{d t } \vert_{ t = t'}\overset{T}{\to} (-1) \frac{d {x}( t)}{d t } \vert_{ t = -t'}. $$ $$ \frac{d^2 {x}( t)}{d t^2} \vert_{ t = t'}\overset{T}{\to} \frac{d^2 {x}( t)}{d t^2} \vert_{ t = -t'}. $$ Then we can check the Newton's second law with time-dependent potential $V(x,t)$, $$ m \frac{d^2 {x}( t)}{d t^2} \vert_{ t = t'} = - {\nabla}_{ x} V(x,t) \vert_{ t = t'} $$ $$ \overset{T}{\to} $$ $$ m \frac{d^2 {x}( t)}{d t^2} \vert_{ t =- t'} = - {\nabla}_{ x} V(x,t) \vert_{ t = -t'}. $$ Namely, we just rewrite the Newton's second law in terms of $V(x, t') \overset{T}{\to}V(x, -t') $. So:

  1. Do you agree with my result that time reversal symmetry remains for Newton's second law with time-dependent potential $V(x, t)$? This means that even if the energy $E$ is not conserved (there could be external forces), we can still have time reversal symmetric force law???

  2. In which case, can time reversal symmetry be broken for Newton's second law? (One famous example is that the right hand side force involves the linear velocity term $\frac{d {x}( t)}{d t } $.)

Note: Time reversal symmetry only requires the equation of motion to be invariant under 𝑡′→−𝑡′, in my viewpoint.

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    $\begingroup$ What you have found is not a symmetry but the statement that ordinary differential equations determine the solution both forward and backward in time from the initial condition. This is equivalent to the statement that if we were approximating with an Euler method, we could step backwards in time just as easily as forward in time. $\endgroup$ Apr 30, 2021 at 23:28
  • $\begingroup$ Can you point out the mistakes in my argument precisely and write an answer about it? Thank you!! $\endgroup$ Apr 30, 2021 at 23:30
  • $\begingroup$ There's not a mistake I can see in your work, it's just a misidentification of what this result means that I'm pointing out. $\endgroup$ Apr 30, 2021 at 23:31
  • $\begingroup$ But I mean the equations (the force law) look the same on both sides except we replace $𝑡′→ −𝑡′.$ This looks like a time reversal symmetry. (you can check time reversal symmetry for classical mechanics is exactly this) $\endgroup$ Apr 30, 2021 at 23:33
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    $\begingroup$ I am guessing that 𝑉(𝑥,𝑡′) = 𝑉(𝑥,-𝑡′) may be required to be time reversal symmetric, but I doubt this is needed. Time reversal symmetry only requires the equation of motion to be invariant under $ 𝑡′→−𝑡′. $ $\endgroup$ Apr 30, 2021 at 23:34

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Let me try and write and answer here which I hope OP will find sufficiently satisfying. Though it is exceedingly common in physics to use arrows to specify transformations, this is exceedingly bad notation and can often lead to confusions about what it is we are really trying to do.

So to start with, we suppose that we have some function $x(t)$ which satisfies the differential equation $$ \frac{d^2 x(t)}{dt^2}=-\partial_x V(x(t),t) $$ with some initial condition and where $V(x,t)$ is some known function.

So, what is time reversal symmetry actually supposed to mean to us? Typically when we refer to symmetries, we mean them only in the sense of preserving the action of the system. This is the correct technical requirement for (continuous) symmetries to give us conserved Noether charges. Though discrete symmetries, of which time reversal is one, do not lead to conserved charges, the same language is usually applied. I'll also point out that this is not just being pedantic as while symmetries of the action are always symmetries of the associated Euler-Lagrange equations, the converse need not hold.

With all that out of the way, let's be specific here about what we mean by a symmetry of the equations of motion. A symmetry is some function of the solution $x(t)$ which returns to us some other function $x^\prime(t)$ which also satisfies the same differential equation. That is, symmetries of the equations of motion map solutions to solutions. So let's write down the proposed map for time reversal symmetry. Given a solution $x(t)$, we want to check whether $x^\prime(t)=x(-t)$ is also a solution.

So let's see what differential equation $x^\prime(t)$ satisfies: $$ \frac{d^2 x^\prime(t)}{dt^2}=\frac{d^2 x(-t)}{dt^2}=\left(\frac{d^2 x(t^\prime)}{dt^{\prime\,2}}\right)\Bigg|_{t^\prime=-t}=\left(-\partial_x V(x(t^\prime),t^\prime)\right)\Bigg|_{t^\prime=-t}=-\partial_x V(x^\prime(t),-t) $$ where in the last step I have used $x(-t)=x^\prime(t)$.

What we have found is that this $x^\prime(t)$ does not satisfy the same differential equation as $x(t)$ unless $\partial_x V(x,t)=\partial_x V(x,-t)$. This is perhaps not unexpected: it's the statement that we have time reversal symmetry so long as our potential is time-reversal symmetric. We could have reached the same conclusion much faster and with less hassle by looking at the Lagrangian and noting that the kinetic term is time-reversal symmetric, so the only real way for the action to remain invariant is if the time dependence of $V$ is reversal symmetric (even function in $t$).

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Chris
    May 2, 2021 at 4:24
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Time reversal symmetry means both the left hand side and right hand side of the equation have the same symmetry under time reversal. Consider the Lorentz Force Law:

$$ \vec F = q(\vec E + \vec v \times \vec B) $$

Under time reversal:

$$ \vec F \rightarrow +\vec F$$ $$ \vec E \rightarrow +\vec E$$ $$ q \rightarrow +q $$ $$ \vec v \rightarrow -\vec v$$ $$ \vec B \rightarrow -\vec B$$

so the law becomes:

$$ +\vec F = +q(+\vec E + -\vec v \times -\vec B) $$

which is the same (even).

The other option is odd, such as the Biot-Savart Law:

$$\vec B(\vec r) = \frac{\mu_0}{4\pi}\int_C\frac{Id\vec l \times \vec r'}{|\vec r'|^3}$$

where: $$ \vec B \rightarrow -\vec B$$ $$ \vec r \rightarrow +\vec r$$ $$ d\vec l \rightarrow +d\vec l$$ $$ I \rightarrow -I $$

Both sides change sign.

When you specify $V(x, t)$, it doesn't have any time symmetry, it's arbitrary. You can break-it up into even and odd parts:

$$ V_+(x, t) = \frac 1 2 (V(x,t) + V(x,-t)) $$ $$ V_-(x, t) = \frac 1 2 (V(x,t) - V(x,-t)) $$

and try to make progress from there, but as is, it is meaningless. The reflection is relative to $t=0$, which is totally arbitrary.

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  • $\begingroup$ Maybe you can comment/answer on this too physics.stackexchange.com/q/633445/42982 ? $\endgroup$ May 2, 2021 at 0:38

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