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

Question

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.

Answer 1

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 $x(t)$.