# I Reversibility of physical laws

1. Mar 28, 2016

### Happiness

Must an initial state always be mapped onto exactly one final state and vice versa?

Consider a slope that is shaped like a curvy "w". Releasing a ball from either end will result in the same final state where the ball is stationary at the highest point in the middle. So does it mean that physical laws need not be reversible?

2. Mar 29, 2016

### Simon Bridge

Classical laws are time-reversible: but you may need more information than just the final position and velocity to determine the initial position and velocity (and vice versa): i.e. you need more than the two numbers to describe a state.
http://www.lecture-notes.co.uk/susskind/classical-mechanics/lecture-1/continuous-physics/
... also suggest Susskind's lectures on classical physics on youtube: lecture 1 answers your question in fine detail.

Of course, there is also physics that is not classical.

Last edited: Mar 29, 2016
3. Mar 29, 2016

### Happiness

The axes of a phase space are position and momentum. So these 2 quantities should be all that is required to specify a state.

4. Mar 29, 2016

### Simon Bridge

Those are only the axes of one particular phase space.
A classical state is, by definition, the smallest list of numbers you need to be able to uniquely determine the next state, given the rules for evolution of the states.
In monoatomic gas thermodynamics that is (P,V,T) - 3 numbers. Atomic state usually needs 4 numbers (n,l,m,s), and so on.

Your example specifies a 3-position system - the rules specified are:
1->2, 3->2, 2->2 ... this is not a closed system because how did you get to 1 or 2? So there is information missing.
It is not a classical system either - see lecture 1 above.

You really need at least 2 numbers - yes: position and momentum are OK here, propose maybe: $x\in\{1,2,3\}$ and $p\in\{-1,0, 1\}$ ... and you see you have not specified the rules for the entire phase space, only the points $(x,p)\in\{(1,1),(2,0),(3,-1)\}$ These rules being: $(1,1)\to (2,0), (3,-1)\to (2,0), (2,0)\to (2,0)$

We can fix that ... say: $(1,-1)\to (3,-1), (3,1)\to (1,1), (2,1)\to (3,1), (2,-1)\to (1,-1), (1,0)\to(1,0), (3,0)\to(3,0)$
Those would be "periodic boundary conditions" - so going off the left and reappear on the right.
You could choose an "infinite barrier" instead, so, for eg. $(1,-1)\to (1,1)$ etc. a ball travelling too far to the right or left rebounds.
... you should plot the complete rules out in the proposed phase space: actually draw the diagram.

You should see there are three singular states: all the momentum states of x=2. It seems the peak there is particularly "sticky".
Notice that momentum is not a conserved quantity, and there is no way to get to states (2,±1) from within the proposed phase space: so the system is not closed... the model, although quite predictive, is incomplete. (the equation of motion is not finished) - and are there any physical laws here that exist in Nature?

But that is for classical physics: remember, there is always the possibility of a non-classical law of physics.

5. Mar 29, 2016

### Staff: Mentor

If you want the ball to stop (at the top of the mid point), energy will have to be dissipated.

6. Mar 29, 2016

### Simon Bridge

I imagine @Happiness is thinking: if the centre of the W is a bit lower than the arms, then you can start the ball on a slope at just the right height that it comes to rest exactly on top of the centre, without the need to dissipate energy. (We also need the bottoms of the W to be more rounded...)
But that condition is unstable - and physically not realisable in Nature. In an example of how an idealization leads to a non-physical law.
(There is a similar situation in the phase diagram for a simple rigid rotator... where the rotator could have exactly the energy needed to balance on it's end.)

A simpler, and possibly clearer, example would be to put the object in a U shape, with a dimple in the bottom to catch the object.
If it starts anywhere, but stationary, it ends up in the hole. Is there a classical (i.e. reversible) model for this?

7. Mar 29, 2016

### Simon Bridge

Note: Susskind's lectures answer the question.
Lecture 1, at about 22mins in, deals with allowed laws in classical mechanics and fields questions about valid laws which don't follow the rules.
Lecture 2. at about 10mins, deals with an example very similar to the above in the continuous case as part of a discussion of Aristotle's physics.

8. Mar 29, 2016

### Happiness

The state in the middle is unstable, but I don't get why it is not realisable.

But when the ball falls into the hole, it loses energy.

9. Mar 29, 2016

### Happiness

Consider the following potential-energy functions (typical of the intermolecular potentials considered in many kinetic-theory problems)

where $s_3>s_2>s_1>s$

According to the above paragraph, a particle with parameters $E_1$ and $s_1$ would spiral inwards until $r_1$ and circle around the centre of force forever. We may launch the particle from the left or the right or from infinitely many different directions and they would all end up in the same final state of circular orbit of radius $r_1$.

So we have an example where multiple initial states all give rise to the same final state.

Last edited: Mar 29, 2016
10. Mar 29, 2016

### DrStupid

It would require infinite precision. In classical mechanics this is possible in theory but not in practice.

11. Mar 29, 2016

### Andy Resnick

If there is no dissipation, then yes. If you include dissipation (viscosity, friction, diffusion), then then answer is no. You can't unscramble an egg, for example.

12. Mar 29, 2016

### DrStupid

In theory there are special cases where the final state is not determined (e.g. if the central maximum in the setup of the OP is shaped like Norton's dome and the ball is replaced with a point mass).

13. Mar 29, 2016

### Andy Resnick

True, I neglected to mention unstable equilibrium states and bifurcating systems.

14. Apr 1, 2016

### Simon Bridge

Not realisable in Nature ... same reason you cannot balance a pin on it's point even though the geometric pointed cylinder can balance on it's point.

... it also loses energy when it rolls down the slope: that doesn't seem to be a problem.
Have you watched the lectures yet?

Always bear in mind that I am talking about classical mechanics ... which will be where you have heard that physical laws must be time reversible.
Do watch the lectures, Susskind answers your questions at the about times indicated but the total lecture will help you understand better what you have been told.

15. Apr 1, 2016

### Svein

Well, you have to take entropy in consideration. Putting an ice cube in a hot drink will result in a tepid drink and no ice. The reverse is wildly improbable...

16. Apr 2, 2016

### Simon Bridge

You also have to take account of the context of the statement made: re Susskind's lectures on classical mechanics (linked to), where these sorts of issues are addressed.
i.e. entropy is an emergent property of underlying processes which are time reversible.
I think he argues that entropic systems, like statistical ones, are not classical mechanics.

That si why the short answer to the question posed in post #1, initially suggested in post #2 and repeated several times, is "no".

17. Apr 2, 2016

### Happiness

Your "not realizable in nature" should be phrased as "theoretically realizable but practically not realizable in nature". That clears the confusion right away, avoiding the need for long sentences.

We consider the ideal case where there is no loss of mechanical energy as a point particle moves (no rolling) down the slope, but if it falls into the hole and stays there, there is such a loss.

Yes, I've watched the lecture. That's how I realized Susskind failed to mention these special cases where classical laws are not reversible.

It is unclear how these special cases should be handled to avoid violating a commonly assumed tenet of science—that in principle complete information about a physical system at one point in time should determine its state at any other time.

18. Apr 2, 2016

### Happiness

That's incorrect. Statistical mechanics is still classical, or classical statistical mechanics is still classical.

In principle, in a classical world, a scrambled egg could become unscrambled. The process is reversible in principle. It's just unlikely in practice.

19. Apr 2, 2016

### Simon Bridge

It is not incorrect: the law of entropy is not time reversible, but statistical mechanics is.
Susskind is defining what is meant by "classical mechanics" in lecture one.
It is important to agree on the meaning of words in order to communicate.

You have misunderstood - ... it is not just impractical, it is not possible because the model we are using to suggest that it is does not describe reality. It is wrong. If we did manage to get a needle to stand on it's point it will not be due to anything in the model involving balancing on a tip.
This does not mean the model is not useful - we just need to understand the limitations.

Total energy is always conserved though ... when the ball gets stuck in a hole, it's energy had to go someplace just like when it rolls down a slope it's potential energy is converted to kinetic. To model the system, classically, and completely, you need to account for that exchange.
Basically I'm trying to get you to think about a simpler system than the one you set up...

... during the lecture a student asks him about these systems, he responds to the question and I have given you the timestamps above (post #7).
It's a brief responce and he says he'll revisit it later on in the course material.

Just to be clear, your original question was:
My answer was and still is "no - only in classical mechanics". See posts 2,4,7, and 16.
If the model is not time reversible, then it is not a classical physical law... it may still be a physical law, just not a classical one.
Take care, sometimes the word "classical" is used to refer to physics before Einstein ... i.e. in the sense of being old or "classic".
However, Einstein's theories are classical theories in the technical sense which Susskind is expounding in his lectures.

Perhaps I was giving a too-complete answer.
Where did you hear that physical laws had to be time reversible?

20. Apr 2, 2016

### Happiness

A scramble egg could become unscrambled in principle, but it's highly unlikely. I'm pretty sure Susskind said something similar.

Classical models allow a needle to be balanced on its tip. Again, we mean this in principle.

The laws of classical mechanics are time-reversible except for some special cases. About how to handle these cases, no one says a word.

Last edited: Apr 2, 2016
21. Apr 2, 2016

### Simon Bridge

Classical models do not say the real life needle balances on it's tip - they say that a particular mathematical idealisation may, in principle, be balanced on it's tip, given assumptions like no thermal motion ... do not confuse the map with the territory.

The cases, and there are many, where a law is not time reversible is well known and there are many ways to handle them that are discussed copiously.
This is the subject of many books and papers - many of them text books for physics courses. eg. any work on thermodynamics will show you how to handle systems with entropy. Lossy systems are also well studied as in the damped pendulum and any system with friction.... and the entire body of quantum mechanics which comprises the foundation for the standard model of particle physics. The fields of biology and chemistry themselves are mostly concerned with such systems. So "about how to handle these cases, no one says a word" is a bit off.

They are just not laws of classical mechanics.
Classical mechanics does not deal with them except indirectly as in statistical mechanics.
This is the same reason that gravity theory does not deal with electromagnetism.

Though, you are free to reject Susskind's definition of "classical mechanics" if you want.
Whatever, I believe your actual question is answered.
If you have any more questions, feel free to start another thread.

22. Apr 4, 2016

### Andy Resnick

This is a high-school level of understanding: you also have a non-zero probability of spontaneously tunneling through your floor.

The second law of thermodynamics does not require a statistical interpretation.

23. Apr 4, 2016

### Happiness

Could you enlighten us?

24. Apr 4, 2016

### Andy Resnick

25. Apr 4, 2016

### Happiness

I mean enlighten us with the main idea of the alternative interpretation in a simple and concise manner.

Well, someone else may counter argue that no non-statistical interpretation exist and refer you to some textbooks too.