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↦−t?
Answer
The flow of time is not simply modelled by the real line. The real line with sum (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:R→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∈R we associate the flow f(t) of t seconds by f(t)=t0+t,
Now what does it mean to take the reversed flow finv, i.e. "reverse time"? It is quite intuitive within the model above: finv:R→R such that finv(t)=t0−t.
Now I would say that the time reversal is the mapping TR:f↦finv between the flow and the inverse flow, so it is a map between functions, and not between numbers. If t0=0, it is easy to justify the abuse of notation TR:t↦−t. However it remains a map of functions (time flows), and not simply of numbers.
[ following the comments ].
To get the two sequences of time the OP mentioned using the flows, we should think as follows:
We start from t0, and look at a sequence of 6 "forward" discrete time steps, each of 10 seconds. Using the flow f, we may describe them by t0+t, for any t∈{0,10,20,30,40,50}. If t0=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=t0+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 finv, and we get t′0−t=t0+50−t, for any t∈{0,10,20,30,40,50}. With t0=0, this gives 50,40,30,20,10,0.
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