Time reversal $t\rightarrow -t$: $0,10,20,30,40,50$ to $-50,-40,-30,-20,-10,0$ or $0,-10,-20,-30,-40,-50\;?$

Question

Time reversal symmetry is defined as $$T:t\mapsto-t\;.$$

Suppose, a stroboscopic film of a ll falling from a certain height to the ground is run forward with time-instances given as: $0,10,20,30,40,50\;.$

When the film is run backward, the ball would move up having the same velocity at each place during its descend but now in the opposite upward direction.

But, what would be the time-reversal sequence?

Would it be $-50,-40,-30,-20,-10,0$ or $0,-10,-20,-30,-40,-50\;?$

For me, it would be the former as the reversal definition says $t\rightarrow -t$ which implies, IMO, $50\rightarrow -50; \; 40\rightarrow -40$ etcetera.

That is $$\text{Forward}\\ \begin{array}{|c|c|} t& v \\ \hline 0& v_1\\ 10& v_2\\ 20& v_3 \\ 30& v_4\\ 40& v_5\\ 50& v_6\end{array}$$

Which is correct:

$$\text{Backward}\\ \begin{array}{|c|c|} t& v \\ \hline -50& -v_6\\ -40& -v_5\\ -30& -v_4 \\ -20& -v_3\\ -10& -v_2\\ 0& -v_1\end{array}$$ or $$\text{Backward}\\ \begin{array}{|c|c|} t& v \\ \hline 0& -v_6\\ -10& -v_5\\ -20& -v_4 \\ -30& -v_3\\ -40& -v_2\\ -50& -v_1\end{array}$$?

If it is former, then the velocity function must be odd that is $v(-50)= -v(50)\;$ now since, time-reversal negates velocity, does that mean the velocity function must always have to be an odd function? This would be ridiculous to think there would only be odd velocity function; neither even nor otherwise. That's why I'm in doubt whether time-reversal sequence would be the first one.

So, can anyone please explain this to me which would be the time-reversal sequence?

Answer 1

1. Your talk of the "sequence" of times does not make formal sense, but time reversal is indeed the map $t\mapsto -t$, so $50\mapsto -50$, etc.

2. The velocity is not constrained to be an odd or even function of time. That time reversal "negates" the velocity does not mean that $v(-t) = -v(t)$. It means that $v(t)\mapsto -v(t)$ under the time reversal map, but the point is that the image of $v(t)$ under $t\mapsto -t$ is not $v(-t)$. Look at the definition of velocity, which is $ v(t_0) = \left.\frac{\mathrm{d}x}{\mathrm{d}t}\right\rvert_{t=t_0}$, so we have $$v(t_0) = \left.\frac{\mathrm{d}x}{\mathrm{d}t}\right\rvert_{t=t_0}\mapsto \left.\frac{\mathrm{d}x}{\mathrm{d}(-t)}\right\rvert_{-t=-t_0} = \left.-\frac{\mathrm{d}x}{\mathrm{d}t}\right\rvert_{-t=-t_0} = -v(t_0)$$ given that you already know that $x$ doesn't change under time reversal. Again, that $x$ doesn't change doesn't mean that $x(t) = x(-t)$, it means $x(t_0)\mapsto x(t_0)$ under time reversal.

3. I can see how this is confusing. If you are familiar with a bit of geometry, the following explanation might help: We imagine here a "time manifold", which is just $\mathbb{R}$. On it, we have scalars, and vectors, but since the vectors are one-dimensional, we cannot really distinguish them from scalars if we have fixed a parametrization of time $t$. But: We are free to choose any parametrization of time that we like, in particular, we can choose to use $t' = -t$ instead of $t$. The scalars (such as $x$) don't care, but anything that's a vector - i.e. contains a single time derivative - will change its sign in the natural basis for vectors that comes with this parametrization. (Usually, you will have that the change of basis of the tangent vectors is the Jacobian. In one dimension, the Jacobian is simply the number $\frac{\mathrm{d}t'}{\mathrm{d}t} = -1$.)