Suppose a ball falls from a certain height and reaches the ground. Later on, somehow we managed to reverse time. Now on reversing time, will the ball move upward to reach the same point from where it was fallen (as literally, reversing time means moving backward in time to reach the same point from where the events were originated), or will it stay still on the ground obeying the laws of gravitation? This appeared quite puzzling to me because either way one has to abandon the result arising due to time reversibility or the fundamental laws of nature.
$\begingroup$
$\endgroup$
0
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
0
Good question. Let's first consider the ball falling immediately before it hits the table. Neglect friction with the air for simplicity. The ball has a velocity in the downward direction. If we reverse time, the ball is in the same position right above the table, but now it has an upward velocity. A ball with an upward velocity will rise with a negative acceleration due to the force of gravity. Therefore, when we reverse time the ball WILL obey the laws of gravitation and Newton's laws. Together these physical laws predict that the ball will rise due to the velocity in the upward direction. In response to @CuriousOne's comment, here the equations of motion ARE time reversal invariant as long as we aren't considering the collision with the ground.
Now let's pick a different point in time to consider. Consider several seconds after the ball hits the ground. Let's call this time $t_0$. In the forward direction, the ball started in the air, hit the ground, and then released kinetic energy to the ground in the form of heat which causes the ball to stop and the atoms in the ground and the ball to vibrate. This brings us to time $t_0$. Now consider time in the reverse direction. From a microscopic mechanical viewpoint, it is totally possible that the atoms are vibrating in the ground in such a way that they transfer kinetic energy and momentum to the ball so that the ball is "kicked" up into the air. This "kick" and the following rise of the ball would be totally consistent with the equations of motion, regardless of the fact that time is reversed.
So why don't we see balls at rest on the ground spontaneously rising up in the air? It has nothing to do with the time reversibility of the equations of motion (Newton's second law) since this event is totally possible according to the equations of motion. The reason balls don't rise in the air is that such an event is statistically extremely improbable. It is highly unlikely that the atoms in the ground organize their motion in such a way that they kick the ball into the air. You would have to have extremely precise conditions for the atoms in the ground for them to kick the ball into the air from rest on the ground. This is in direct accordance with the Second Law of Thermodynamics which comments on the nature of entropy in the universe. In this case, due to the statistical improbability of the ball being kicked into the air, our macroscopic equations of motion ignore the possibility of the ball being kicked into the air. Thus, our simplified macroscopic equations of motion don't account for the full microscopic physics of the atoms in the ground, so we can't expect these macroscopic equations to be time reversible. See @CuriousOne's comment above.
Hopefully this example illustrates the contrast between the time reversibility of a microscopic theory like classical mechanics or quantum mechanics, and the time irreversibility of a macroscopic theory like thermodynamics.
$\begingroup$
$\endgroup$
0
Is time reversible?
Look at the stroboscopic photograph. Is the ball "falling up" or falling down? The answer is surely we don't know!
A motion picture of this sort of sequence of the event could be run backward & would inevitably be impossible for the viewer to detect any violation of Newton's laws. A time-reversal changes both $t,v \to -t,-v$ leaving the acceleration unchanged . Thus any conclusion about forces that we reach as a result of watching a dynamical process in reverse sequence are identical with what we would conclude from the process itself. After all, attractions don't turn into repulsions or any of that sort.
However, when we see an ordinary picture in reverse, it quickly becomes apparent from the behaviour of inanimate objects-leaving aside the ludiucrous effects of reversing human actions, which appear strange for quite different reasons-that most physical actions have a well-defined direction. Imagine, for example, a sequence in which a glass falls from a table & shatters into small fragments on the floor.If we saw a motion picture in which the fragments gathered themselves together into a whole glass, which then jumped up onto the table, this would clearly be impossible - nature doesn't act like that. Yet a "micromovie"of the individual atomic encounters at every stage of the process ought to be perfectly time-reversible.
Thus, we are faced with a puzzle: Newton's law implies that the fundamental dynamical behaviour of an individual particle is reversible in time, but when one takes a system of very large numbers of particles, apparently the behaviour ceases to be time-reversible.
Remarked by A.P.French.
$\begingroup$
$\endgroup$
0
To expand a bit on Ian's answer, let's consider the case of a ball that has dropped from a given height and has bounced repeatedly against the ground, losing energy through partially inelastic collision with the ground until all its energy is dissipated as heat. Now if at this point time was reversed, what we might expect to see is that the flow of entropy would reverse, and the seemingly random motions of individual particles would spontaneously converge on the spot under the ball and impart a slight, almost imperceptible upward motion. But going further, it would not be a single upward impulse, but rather a periodic sequence of convergences on the spot under the ball, at the exact moment the ball was in contact with the ground each time. Furthermore, after the final imparting of momentum from the ground to the ball, the Earth would suddenly achieve a stable point and the oscillations would cease, and the ball would rise to its highest point where it would engage with whatever mechanism released the ball in the first place.
However, nobody can say for certain that this will really happen. First, because we have no means to reverse time, and so we cannot perform experiments or gather data on this scenario, and from that it follows that our models of time are just that - models - which may or may not behave as expected outside the range of experience they were developed under. Everything in the first paragraph is a thought experiment which assumes that whatever mechanism reverses time must necessarily cause everything that happened to be deterministically be reversed, but an alternate explanation would be that the constraint of deterministic behavior be relaxed, in which case the backwards flow of time might revert to different initial conditions. If memory serves, some of Hawking's recent works on the present selecting initial conditions of the past may be relevant here.
$\begingroup$
$\endgroup$
0
To understand my answer, you have to imagine space as a 2-D plane and time as a vector perpendicular to it. "Normal" time is defined as a vector pointing to the "right."
If time is now instantaneously reversed (vector now pointing to the left), nothing would change (laws of physics, etc.), because we could just as easily define "normal" time as pointing to the "left." In other words, our frame of reference is arbitrary!
If time is reversed in the "middle" of a plate falling, it will continue to fall and shatter, just as if time had not been reversed!
