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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?

Qmechanic
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SuperCiocia
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  • this is not the equation of motion, but a solution to some equation. – Phoenix87 Jan 06 '15 at 16:24
  • That's not the equation of motion but a solution to the equation of motion. The equation of motion would be $x''(t)=a$. – CuriousOne Jan 06 '15 at 16:25
  • Right, sorry, I'll change it – SuperCiocia Jan 06 '15 at 16:25
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    In normal time the ball speeds up. In reverse time it slows down. Of course the solution is not symmetric under time reversal. – John Rennie Jan 06 '15 at 16:27
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    But if $v_0=0$ then the equation is symmetric – SuperCiocia Jan 06 '15 at 16:28
  • imagine throwing the ball up. it will come down. now reverse the motion: the ball goes up, then it comes down (of course you might want to replace $t$ with $t_0 - t$ to actually see this). – Phoenix87 Jan 06 '15 at 16:29
  • And that's the key point. The motion isn't symmetric under $t \rightarrow -t$, but it is symmetric under $t \rightarrow -t + t_0$ for some suitably chosen value of $t_0$. – John Rennie Jan 06 '15 at 16:32
  • and what does $t_0$ signify? – SuperCiocia Jan 06 '15 at 16:35
  • @SuperCiocia - If you want to time-reverse reverse the initial conditions as well as the equation of motion, then since velocity is the first derivative with respect to time, $v = dx/dt$, changing $t$ to $-t$ should be accompanied by changing $v_0$ to $-v_0$ (likewise with any other quantities that are odd-numbered derivatives with respect to time). – Hypnosifl Jan 06 '15 at 16:41
  • For more on what needs to be reversed under a time-reversal, see the Microscopic phenomena: time reversal invariance section of the wikipedia article on T-symmetry. – Hypnosifl Jan 06 '15 at 16:44
  • I feel the comments are kind of misleading.... The first equation SuperCiocia wrote represents the motion of some object with constant acceleration, the second equation represents the time-reversed motion. It is exactly the same as the first equation but with the time variable flipped and it represents the motion of the object "as if the film was run in reverse". Surely when you run the film in reverse the velocity is automatically reversed, that is what anyone would expect to see, and what the equation says. Both trajectories are solutions of newton's equation with reversed initial conditions – Ignacio Dec 10 '16 at 06:14
  • Brethil's answer is also kinda weird. He seems to imply that the equation is "wrongly reversed" and that you should also reverse sign of the velocity, however, I don't think this is the case, the equation is correctly time-reversed as it stands, differentiating two times produces $-v_0$, which is the logical thing. I think that the question, as it stands, was correctly answered by John Rennie's first comment. Am I missing something? – Ignacio Dec 10 '16 at 06:19
  • Perhaps the confusion arises from the fact that newton's (or lagrange's) equations are symmetric under time reversal, but not the solutions to the equations; the solutions are symmetric under time reversal + reversal of the initial conditions (at least I think so)... – Ignacio Dec 10 '16 at 06:23

2 Answers2

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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

Kyle Kanos
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Brethil
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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 $$a \mapsto \frac{\mathrm d}{\mathrm d(-t)} \frac{\mathrm dx}{\mathrm d(-t)} = (-1)^{2}a = a$$

Zo the Relativist
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  • Sorry for commenting on such an old post, but when you write t->(-t), should you not also change x(t) to x(-t)? – GRrocks Dec 15 '18 at 07:52