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Consider the solution to the equation of motion for a particle with a constant acceleration: $$ x(t) = x_0 + v_0t + \frac{1}{2}at^2.$$

If I let $t \rightarrow -t$, then the equation becomes: $$ x(-t) = x_0 - v_0t + \frac{1}{2}at^2,$$

which is different. Does this mean that this equation is not symmetric under time-reversal? What is the physical meaning of this? What if this represented a ball falling under gravity being recorded on tape: surely we should see the same thing if we run the film in reverse?

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  • $\begingroup$ this is not the equation of motion, but a solution to some equation. $\endgroup$
    – Phoenix87
    Jan 6, 2015 at 16:24
  • $\begingroup$ That's not the equation of motion but a solution to the equation of motion. The equation of motion would be $x''(t)=a$. $\endgroup$
    – CuriousOne
    Jan 6, 2015 at 16:25
  • $\begingroup$ Right, sorry, I'll change it $\endgroup$
    – SuperCiocia
    Jan 6, 2015 at 16:25
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    $\begingroup$ In normal time the ball speeds up. In reverse time it slows down. Of course the solution is not symmetric under time reversal. $\endgroup$ Jan 6, 2015 at 16:27
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    $\begingroup$ But if $v_0=0$ then the equation is symmetric $\endgroup$
    – SuperCiocia
    Jan 6, 2015 at 16:28

2 Answers 2

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If you substitute $t\to-t$, the sign of the velocity also changes, thus the equation maintains the same functional form

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Everyone else here is right, but I just want to add a little something about why you would reverse the time, rather than the implication that it is just some arbitrary rule.

you are correct that all of this time-reversal business starts with the mapping $t \mapsto -t$. But when you do this, you have to time-reverse EVERYTHING.

Since $v = \frac{\mathrm dx}{\mathrm dt}$, we need to reverse the time in the denominator, which gives us $v \mapsto \frac{\mathrm dx}{\mathrm d(-t)} = -v$

The acceleration is unchanged, since $$a \mapsto \frac{\mathrm d}{\mathrm d(-t)} \frac{\mathrm dx}{\mathrm d(-t)} = (-1)^{2}a = a$$

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  • $\begingroup$ Sorry for commenting on such an old post, but when you write t->(-t), should you not also change x(t) to x(-t)? $\endgroup$
    – GRrocks
    Dec 15, 2018 at 7:52

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