# Why does time-reversal mean negation of time i.e. $t\mapsto -t\;?$

Disclaimer : This is a follow-up to this question.

For long time, I've been pondering of this but couldn't come to a stern conclusion to the question:

Why is time-reversal the negation of time $t\;?$ Meant to say, how does negation of time mean the backward flow of time?

Dumb though the query maybe; but still I want to ask it.

For instance, this is forward flow of time: $0,10,20,30,40,50\;;$ when it is reversed, the time sequence would look like $50,40,30,20,10,0\;;$ is it the negative of the forward time flow $0,10,20,30,40,50\;?$ No. But still it is the backward flow of time, isn't it?

Why has time-reversal to do with the negation of time, for $50,40,30,20,10,0$ represents the backward flow of time $0,10,20,30,40,50$, although the former is not the negation of the later?

I am sure I'm missing something but couldn't point it out.

So, can anyone explain it to me why time-reversal means $t\mapsto -t\;?$

• I will explain myself better in an answer. – yuggib Jan 27 '16 at 10:28
• @yuggib: It would be great. Waiting: ) – user36790 Jan 27 '16 at 10:30
• It is usual to take a beginning for all space and time axis when one sets up a problem. The simplest origin is (0,0,0,0) . When one inverts the x direction( or the y or the z), one sets x ->-x and if the problem is setup with the symmetry origin different than (0,0,0,0) one does a transformation ( galilean for classical, Lorenz for special relativity) to define the symmetry point as the origin for simplicity of solutions. – anna v Jan 27 '16 at 10:52

The flow of time is not simply modelled by the real line. The real line with sum $(\mathbb{R},+)$ is an abelian group that is commonly used to model the time coordinate, and the fact that it is possible to "shift the origin" of time (choose which time should represent the starting point).

The flow of time could be modelled by a function $f:\mathbb{R}\to \mathbb{R}$ that works as follows. Its domain is a time variable, as it is its co-domain (both measured in seconds); and to a time $t\in\mathbb{R}$ we associate the flow $f(t)$ of $t$ seconds by $$f(t)=t_0 +t\; ,$$ where $t_0$ is a fixed reference time of origin, that is customary chosen to be zero. If $t_0=0$, the function $f$ reduces to the identity function.

Now what does it mean to take the reversed flow $f_\textrm{inv}$, i.e. "reverse time"? It is quite intuitive within the model above: $f_\textrm{inv}:\mathbb{R}\to\mathbb{R}$ such that $$f_\textrm{inv}(t)=t_0 - t\; .$$ In this way a time lapse of $t$ seconds has flown "backwards" (in the inverse direction wrt the group operation of the coordinates).

Now I would say that the time reversal is the mapping $TR:f\mapsto f_\textrm{inv}$ between the flow and the inverse flow, so it is a map between functions, and not between numbers. If $t_0=0$, it is easy to justify the abuse of notation $TR: t\mapsto -t$. However it remains a map of functions (time flows), and not simply of numbers.

# Edit:

To get the two sequences of time the OP mentioned using the flows, we should think as follows:

We start from $t_0$, and look at a sequence of 6 "forward" discrete time steps, each of 10 seconds. Using the flow $f$, we may describe them by $t_0+t$, for any $t\in \{0,10,20,30,40,50\}$. If $t_0=0$, we get the sequence of forward times $0,10,20,30,40,50$. Now we suppose that at the final counted time $t_0'=t_0+50$ we reverse time, and look at a sequence of 6 "reversed" discrete time steps of 10 seconds each. We have now to use the inverse flow $f_{\text{inv}}$, and we get $t'_0-t=t_0+50-t$, for any $t\in\{0,10,20,30,40,50\}$. With $t_0=0$, this gives $50,40,30,20,10,0$.

• Thanks for the answer; however could you tell where I can get a intuition of abelian group? I know nothing of it. – user36790 Jan 27 '16 at 11:41
• An abelian group is a commutative group. It is roughly speaking a mathematical construct consisting of a set $G$ together with an associative operation $\circ: G\times G\to G$ that satisfies certain properties. In particular: there exists an identity element $e$, such that for any $x\in G$, $x\circ e=e\circ x=1$; for any $x\in G$ there is an inverse element $x^{-1}\in G$ such that $x\circ x^{-1}=x^{-1}\circ x=e$; (for abelian groups only) for any $x,y\in G$, $x\circ y=y\circ x$. The reals (or the integers) with addition are an abelian group, with $e=1$ and $x^{-1}=-x$. – yuggib Jan 27 '16 at 11:49
• Anyways, the idea is that the time flow is quite naturally implemented by the group operation ($+$), and therefore is also quite natural that the inversion of the time flow should be implemented by taking the inverse (again wrt the group operation, i.e. taking $-t$ instead of $t$ in the flow). – yuggib Jan 27 '16 at 11:52
• Yes of course, sorry about that – yuggib Jan 27 '16 at 19:52
• Another thing is saying that you start from $t_0$, and look at a sequence of 6 "normal" time intervals of $10$ seconds each (described, for any $t\in\{0,10,20,30,40,50\}$ by $t_0+t$ time passed), and then at $t_0'=t_0+50$ you reverse the time and look at the 6 "reversed" time intervals of $10$ seconds each (described, for any $t\in\{0,10,20,30,40,50\}$ by $t_0'-t=(t_0+50)-t$). For the first sequence, letting $t_0=0$, you get $0,10,20,30,40,50$, and for the second you get $50,40,30,20,10,0$. – yuggib Jan 28 '16 at 11:15

The transformation $t \mapsto -t$ transforms an ascending function into a decreasing one. It thus inverts the flow of time. You have to think of it as a transformation on the variable $t$ which is an input in observables : each observable $O(t)$ will see time flowing backwards.

Assume you want to apply the time-inverting transformation at some time $t_0$. Because of time-translation invariance, you can choose the transformation to be $t \mapsto 2 t_0 - t$, it is equivalent to doing just $t \mapsto -t$. You thus get the sequence you described in your question : by the transformation $t \mapsto 100 - t$ starting at $t_0 = 50$, the sequence 50, 40, 30, 20, 10, 0 becomes 50, 40, 30, 20, 10, 0.

• What about the sequence: $50,40,30,20,10,0\;?$ Isn't it the backward flow of time? – user36790 Jan 27 '16 at 10:02
• I think it has to do with time-translation invariance. The sequence 50 - 40 - 30 - 20 - 10 - 0 is equivalent to 0, -10, -20, -30, -40, -50 by a shift of 50. – Dimitri Jan 27 '16 at 10:10
• Then do you want to say $50 \mapsto 0\;?$ That would mean $v(50)\mapsto -v(0)\;;$ really I'm not getting it : / – user36790 Jan 27 '16 at 10:14
• What I mean is that the transformation $t \mapsto -t$ is equivalent to doing $t \mapsto t_0 - t$ because of time-translation invariance. You can choose $t_0$ to be the time at which you do the transformation in the old system of time coordinates. But you're right something feels incomplete in my explanation, I'd like to see someone formuate this better than myself. – Dimitri Jan 27 '16 at 10:32
• I've not said you're wrong; this is all my fault that I'm not getting the fact yet : / – user36790 Jan 27 '16 at 10:35