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Does the time reversal flip the differential measure $dt$ or integral measure $dt$?

Does the time reversal flip the differential measure $dt$ or $\Delta t$?

  1. Suppose we look at the derivative on a classical position $x(t)$, naively if we think the differential measure as $\Delta t$ then we do NOT flip it under time reversal. So $$ \frac{d {x}( t)}{d t } \vert_{ t = t_1} \equiv \lim_{\Delta t \to 0} \frac{{x}( t_1+\Delta t ) -{x}( t_1)}{\Delta t }$$ $$\overset{T}{\to} \lim_{\Delta t \to 0} \frac{{x}( -t_1-\Delta t ) -{x}( -t_1)}{\Delta t }= \color{red} {(-1)} \frac{d {x}( t)}{d t } \vert_{ t = -t_1}. $$ Here we flip $x(t) \overset{T}{\to} x(-t)$ for any $t$ on the time axis, so we flip ${x}( t_1+\Delta t ) \overset{T}{\to} {x}( -t_1-\Delta t )$ and flip ${x}( t_1 ) \overset{T}{\to} {x}( -t_1 )$.

  2. Again on a classical position $x(t)$, naively if we think the differential measure as $\Delta t = t_{j+1}-t_j$ then we seem to have to flip it under time reversal. So $$\Delta t = t_{j+1}-t_j\overset{T}{\to} t_{-(j+1)}-t_{-j} =-(t_{-j} - t_{-(j+1)})=- \Delta t (!!!)$$ $$ \color{red}{dt \overset{T}{\to} - dt (!!!)}$$ Here we label $t_{j+1} > t_j > 0 > t_{-j}=-t_j >t_{-{j+1}}=-t_{j+1}$ in an easily understandable way from the large $t>0$ value to the $t<0$ value along the time axis.

Then we derive a totally different opposite sign of the time reversal on the time derivative: $$ \frac{d {x}( t)}{d t } \vert_{ t = t_1} \equiv \lim_{\Delta t =t_2-t_1\to 0} \frac{{x}( t_2 ) -{x}( t_1)}{t_2-t_1 }\overset{T}{\to} \lim_{t_2-t_1 \to 0} \frac{{x}( t_{-2} ) -{x}( t_{-1})}{ t_{-2}-t_{-1} }$$ $$= \lim_{t_2-t_1 \to 0} \frac{{x}( -t_2 ) -{x}( -t_1)}{ -t_2+t_1 }= \lim_{t_2-t_1 \to 0} \frac{-({x}( -t_1)-x(-t_1-(t_2-t_1 ) ))}{ -(t_2-t_1) }= \color{red}{(+1)} \frac{d {x}( t)}{d t } \vert_{ t = -t_1}. $$ Here we label $t_2 > t_1 > 0 > t_{-1}=-t_1 >t_{-2}=-t_2$ in an easily understandable way from the large $t>0$ value to the $t<0$ value along the time axis.

All the discussion above applies to the classical state, position $$x(t)$$ and also to a quantum state (state vector in a Hilbert space) $$| \psi(t)\rangle$$ So you can answer from a quantum theory perspective too. Thanks in advance.

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  • $\begingroup$ I highly recommend that you stop using the "$\rightarrow$" notation to specify transformations, as I described to you in my answer to another of your questions. While this notation can be computationally efficient, it only promotes confusions like you seem to be having. Instead, define the new function which is to be the transformed version of the old function explicitly. $\endgroup$ May 2, 2021 at 1:12
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    $\begingroup$ Not only do I completely agree with @RichardMyers, I've also written a rant about why the $\to$ notation is very confusing. $\endgroup$
    – knzhou
    May 2, 2021 at 1:26
  • $\begingroup$ I see. thanks but in any case, I will still appreciate an answer. This confusion is another issue, not answered by your previous answer! $\endgroup$ May 2, 2021 at 2:11

1 Answer 1

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No, the time reversal does NOT flip the differential measure $dt$ or integral measure $dt$, if we choose the active transformation on changing the state function $x(t)$ only.

The reason is that we need to first fix either choose

  • the active transformation (the associated field or state function changes but the coordinates stay the same) $$ x(t) \to x'(t)\equiv x(-t) $$ $$ t \to t $$
  • the passive transformation (the coordinate changes but the associated field or state function stays the same) $$ x(t) \to x'(t)\equiv x(t) $$ $$ t \to t'\equiv -t $$

We should not do both active transformation and passive transformation altogether, then it will lead to totally null unchanged transformations.

  1. So my original post in the part 1 uses the active transformation only, which leads to the correct result. The time coordinate should stays the same $$\Delta t = t_{j+1}-t_j\overset{T}{\to} t_{(j+1)}-t_{j} = \Delta t $$ $$ \color{blue}{dt \overset{T}{\to} dt }$$

  2. So my original post in the part 2 uses both the active and passive transformation, which leads to the wrong result. This is not a time reversal transformation, but a null transformation. In that case, I did: $$ \color{red}{x(t) \to x'(t)\equiv x(-t)}. $$ $$ \color{red}{t \to t'\equiv -t}. $$ Which is \color{red}{wrong for a time reversal transformation}.

So, no, the time reversal does NOT flip the differential measure $dt$ or integral measure $dt$, if we choose the active transformation only.

The time reversal can flip the differential measure $dt$ or integral measure $dt$, if we choose the passive transformation only. But in that case, we do not perform active transformation on the state $x(t)$.

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