- #1

hokhani

- 483

- 8

- TL;DR Summary
- I don't understand how does time-reversal applies to the free-fall problem.

I try to justify time-reversal symmetry in a very simple classical problem; Free Fall. The position, ##x##, and the velocity, ##v## are obtained versus time from the equation ##-g=\ddot x##. So, if we consider the primary conditions as ##t_0,x_0,v_0## it is clear that, ##x(t)=\frac{1}{2}(-g)(t-t_0)^2+v_0(t-t_0)+x_0## and ##v(t)=-g(t-t_0)+v_0## give teh position and velocity at any time. Plesase see the attached figure. AT the point (1) we have$$x(t_1)=x_1=\frac{1}{2}(-g)(t_1-t_0)^2+v_0(t_1-t_0)+x_0$$ and $$v(t_1)=v_1=-g(t_1-t_0)+v_0.$$ We expect by applying time reversal to obtain ##x(-t_1)=x(t_1)## and ##v(-t_1)=-v(t_1)##. However, inserting ##-t_1## instead of ##t_1## doesn't neccesarily fulfill this requirement:

$$x(-t_1)=\frac{1}{2}(-g)(-t_1-t_0)^2+v_0(-t_1-t_0)+x_0$$ and $$v(-t_1)=-g(-t_1-t_0)+v_0.$$ I would like to know what is my mistake?

$$x(-t_1)=\frac{1}{2}(-g)(-t_1-t_0)^2+v_0(-t_1-t_0)+x_0$$ and $$v(-t_1)=-g(-t_1-t_0)+v_0.$$ I would like to know what is my mistake?