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?

  • $\begingroup$ this is not the equation of motion, but a solution to some equation. $\endgroup$ – Phoenix87 Jan 6 '15 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 '15 at 16:25
  • $\begingroup$ Right, sorry, I'll change it $\endgroup$ – SuperCiocia Jan 6 '15 at 16:25
  • 1
    $\begingroup$ In normal time the ball speeds up. In reverse time it slows down. Of course the solution is not symmetric under time reversal. $\endgroup$ – John Rennie Jan 6 '15 at 16:27
  • 1
    $\begingroup$ But if $v_0=0$ then the equation is symmetric $\endgroup$ – SuperCiocia Jan 6 '15 at 16:28

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


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

  • $\begingroup$ 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 '18 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.