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Consider a specific case of a hourglass. Here, because of gravity sand flows down and reaches the equilibrium state when all the sand is in the lower half. This way we can keep a measure of time. From this perspective reversal of time is a really absurd concept, is it not?

If time is reversed, do physical forces stop acting? So, isn't the nature of time inherently linked with the natural forces?

I know my analysis is very simplistic but is my logic or reasoning flawed at some point? Am I making some mistake?

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closed as unclear what you're asking by ACuriousMind, user36790, DilithiumMatrix, John Duffield, Gert Dec 4 '15 at 17:35

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  • $\begingroup$ It is abundantly clear that time reversal does not hold on the macroscopic scales (see the second law of thermodynamics). But the thing is that fundamental laws often exhibit this symmetry: you reverse time and the sign of some of the quantities (forces still have to act - though they might act in different directions) and you get valid physics. The problem with your "logic" is that you are not looking at the laws, you are looking at the world and make premises. However, that's not why people talk about time reversability. $\endgroup$ – Martin Dec 4 '15 at 13:22
  • $\begingroup$ Can you please elaborate on how I am not looking at the laws and instead looking at the laws and making premises? And "Why else do people talk about time reversability?". $\endgroup$ – sunsun Dec 4 '15 at 13:40
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As I read it, your argument is essentially the following: If I have a look around me, it's clear that I can't just reverse the arrow of time. So the concept of time-reversal must be nonsense.

I claim that physicists can see that just as well as you can. The trouble is that what you assume they are talking about is not what they are talking about, which means that there are two errors in your reasoning: a) your premises are false (reasoning error) and b) your argument then essentially boils down to an argument from ignorance (this is a logical error).

So what is time-reversal-symmetry really about?

Let us consider classical physics. You can model any classical system by a Hamiltonian function, which is somewhat related to the energy. Then the motion of the system is given by the equations of motion. Given a Hamiltonian function $\mathcal{H}$, the equations of motion are given by the Hamiltonian equations

$$\dot{p}=-\frac{\partial \mathcal{H}}{\partial x} \\ \dot{x}=\frac{\partial \mathcal{H}}{\partial p} $$

If you want a visualisation, you can consider a particle in one dimension attached to a spring. This means, we have two variables, the position in space $x$ and the momentum $p$. Both may vary in time, i.e. $p=p(t)$ and $x=x(t)$, then the Hamiltonian function becomes

$$\mathcal{H}=\frac{p^2}{2m}+\frac{1}{2}kx^2$$

where $k$ is the potential. If you write down the equations of motion given the above, you have $\dot{x}=p/m$ (just telling you that momentum is mass times velocity, as it should) and $\dot{p}=kx$ (since the change in momentum is exactly what we call "force", this is just Hooke's law, which it should be).

Now have a look at what happens when I reverse time: Position obviously doesn't change $x(-t)\mapsto x(-t)=x(t)$. Momentum is velocity times mass. If I just let the movie play backwards, the mass should be the same, but the velocity will be backwards. We can also see this:

$$p(-t)=m\dot{x}=m\frac{dx}{d(-t)}=-m\frac{dx}{dt}=-p(t)$$

Good, this is what we would expect. What happens to the equations of motion? You just plug in what we had above. For the left hand side:

$$\dot{p}(-t)=\frac{-dp(t)}{d(-t)}=\frac{dp(t)}{dt}=\dot{p(t)}$$

For the right hand side:

$$ -\frac{\partial \mathcal{H}}{\partial x(-t)}=-\frac{\partial \mathcal{H}}{\partial x(t)}$$

Since we know that $\dot{p}(t)=-\frac{\partial\mathcal{H}}{\partial x}$, we therefore also know:

$$\dot{p}(-t)=-\frac{\partial \mathcal{H}}{\partial x(-t)}$$

which has exactly the same form as before. This means that the equations of motion still hold in the same way. In our spring system for example, Hooke's law still holds. You can do this and see that the other equation also looks the same. This means that plugging in $t\mapsto -t$ leaves the equations of motion invariant. This is what is meant by "time-reversal invariance". It is true for every classical system, as we have just proven. And this is also what is meant by "looking at the laws of physics". The laws of physics here are the equations of motion given by the systems Hamiltonian and the Hamiltonian itself.

Since this also holds in quantum mechanics (basically it is enough to replace the Hamiltonian function by the Hamiltonian operator and the equations of motions by the Schrödinger equation), invariance under time reversal are inherent in these theories - and that is why people talk about it.

But that's not what you are talking about. You are just saying "We measure time by sand flowing down, reversal doesn't make sense, does it?" But that's not true: See, you can write down the equations of motion of the sand clock+earth and then reverse time. The physical picture would be that gravity acts just the same as always (downwards), but the earth has some momentum upwards. This momentum pushes the the sand, which is sort of "catapulted" upwards. Gravity will then result in the sand slowly losing speed until they reach the top of the sad clock. In principle, there is no problem here.

If you want to see that this is not ridiculous, instead of a sand clock, have a look at a what happened if the small whole through which the sand flows would be as wide as the whole vial. Now the sand would just plunk down. You can achieve the reversal picture by giving the clock a push from below: The sand will leap up. It's the same principle. With the sand clock, this is harder to achieve, because the push needs to be coordinated differently.

There are a few questions arising, but you can answer them by really having a look at the whole system - except for the problem with thermodynamics: You will see that you are going to need more energy to create the momentum (the push). This will need to come from heat, which is impossible in thermodynamics. But that's exactly the point I made in the comment: your system is already too complex for simple mechanics to apply, and once you have the second law of thermodynamics, you lose time-reversal invariance. Nobody really knows the details about all of that (this is known as the "arrow of time" problem) and this is one part of why people like to talk about time-reversal invariance.

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  • $\begingroup$ No problem! This is a common problem with physical and mathematical concepts (and I suspect most other disciplines): It seems clear what is meant by the concept to a layperson/beginner, but the experts actually mean something completely different... $\endgroup$ – Martin Dec 11 '15 at 16:30

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