If time reversal symmetry was preserved would past and future notions be defined

Question

If time reversal symmetry was preserved would someone be able to define future as a time interval from a given time moment as measured by a clock (assuming they would still function as normal)?

Answer 1

So the "arrow of time" (https://en.wikipedia.org/wiki/Arrow_of_time) is about thermodynamics (or statistical physics, which miraculously give the same laws), and of course entropy and the 2nd Law of Thermodynamics.

Now books have been written about entropy, and it has many formations, so far to brief summary is: In thermodynamics, it's a measure of the how much energy is not available to do work, while in statical physics it's just a count of the number of microscopic quantum states that have the observed macroscopic parameters.

Anyway, it usually increases, making (non-adiabatic) macroscopic process irreversible, in spite (most) of the laws of physics being time-reversal symmetric.

The laws of physics being time reversal symmetric means the equations are still valid if you reverse time. For example, Newton's Law:

$$ \vec F = m\vec a = m\frac{d^2\vec x}{dt^2} $$

Under time reversal:

$$ \vec F \rightarrow + \vec F $$ $$ \vec x \rightarrow + \vec x $$ $$ m \rightarrow + m $$ $$ dt \rightarrow -dt $$

so Newton's law is:

$$ +\vec F = m\frac{d^2(+\vec x)}{(-1)^2dt^2} = (+\vec m)(+\vec a)$$

is the same equation. Likewise, the Lorentz force law:

$$ \vec F = q(\vec E + \vec v \times \vec B) $$

transforms to:

$$ +\vec F = (+q)\big(+\vec E + (-\vec v) \times (-\vec B)\big) $$

which is equivalent ($\vec B$ depends on current, which switches direction un $T$).

You can do the same with the full set of Maxwell's eqs and see they are the same under $T$, though sometimes each side flips sign, e.g.:

$$\vec\nabla \times \vec E = \mu_0\big( \vec J + \epsilon_0 \frac{\partial E}{\partial t} \big) $$

$$+\vec\nabla \times (-\vec B) = \mu_0\big( -\vec J + \epsilon_0 \frac{\partial (+\vec E)}{\partial (-t)} \big) $$

and that equation is said to be "time odd".

So that is what time reversal invariance is all about.

Since all the kinematic and electromagnetic laws are time reversal "invariant" (really: covariant, but we say "invariant"), macroscopic processes are time reversal symmetric, at the microscopic level.

Note the other discrete symmetric are "Parity", or coordinate inversion. Normal vectors flip sign in the mirror, and so-call axial vectors (they have a cross product, e.g. $\vec B=\vec \nabla \times \vec A)$ do not flip sign.

...and "Charge Conjugation": replace electrons with positrons, or $q \rightarrow -q$.

It is a good exercise to apply this to Newton (linear and angular momentum) and Maxwell to see how it all works out.

Originally, it was thought that $T$, $P$ and $C$ were all good symmetries, but the weak interaction messed that up. In beta decay (https://en.wikipedia.org/wiki/Wu_experiment) it was observed that

$$ \vec p_{e^-} \propto \vec S_{^{60}Co} $$

which is even (odd) on the LHS (RHS), and transforms to:

$$ -\vec p_{e^-} \propto +\vec S_{^{60}Co} $$

which maximally violated parity.

Then the weak interaction violated the product $CP$, which then must violate $T$, which is understood, but how that led to the matter/antimatter asymmetry is an open problem.

Anyway, the two meanings of "time reversal" are quite different.

In your question, which seems to be, "if macroscopic processes were time symmetric...", can be removed from the hypothetical to be: "In adiabatic (aka: isentropic) processes, can the future be given by a clock ticking?", which seems to be "Yes".