2
$\begingroup$

Shorter version: I am wondering if non-elastic collisions preserve time-symmetery; i.e., given a set of objects with positions and velocities known at a given time, we can calculate forward in time and predict when they will collide, but is it possible to calculate backward in time and see which objects were perhaps formed through inelastic collisions of two smaller objects?

Longer version: Given an isolated (isolated meaning there is nothing around for it to collide with) classical object and knowing its current position and velocity one can determine where it will be for all time and where it has been for all time. If the object were to be placed in some sort of force field we can still determine where it will be for all time and where it has been for all time. The reason we can know both where it will be and where it has been, I believe, is due to the time-reversal symmetry of physics (which, if I remember right, leads to the conservation of energy).

Moving to non-isolated objects, if we know the velocity and position of every object in a system such as the solar system we can similarly determine where all these objects will be and where they have been. Moving forward in time, some objects will collide and will either fracture or stick together. However, even though we can calculate collisions when moving forward in time, moving backwards in time it is a lot more difficult to determine when collisions occurred, especially if the result was the sticking together of two bodies.

In the specific example of the solar system, I could run a full simulation forward and know what the solar system will look like years from now (barring any unmodeled perturbations from passing stars) but I can't run the same simulation backwards and know when objects have experienced collisions or, running the simulation back in time very far, what the primordial solar system looked like before anything resembling the planets was formed.

There seems to be an asymmetry here: I can calculate collisions moving forward in time, but I can't always calculate collisions when moving back in time. Why is this? Is it due to these collisions quite often being inelastic collisions?

Improve this question
$\endgroup$
4
  • $\begingroup$ Physics is not exactly time-reversible, there exist time-asymmetrical processes, one of the is the nuclear decay. $\endgroup$
    – Sofia
    Commented Feb 18, 2015 at 23:33
  • $\begingroup$ Which collisions remain for the future if you exclude influence from "passing stars"? $\endgroup$
    – Sofia
    Commented Feb 18, 2015 at 23:43
  • $\begingroup$ @Sofia Earth collides with meteors every day and occasionally larger ones (like the Chelyabinsk meteor two years ago). $\endgroup$
    – NeutronStar
    Commented Feb 19, 2015 at 2:09
  • $\begingroup$ Isn't this about the arrow of time? "Non-reversible time" has more to do with entropy and probability - the unlikeliness that a collection of scattered shards will spontaneously absorb the requisite energy to self-assemble into a vase than any prohibition in the laws of physics. $\endgroup$
    – Anthony X
    Commented Feb 22, 2015 at 22:48

3 Answers 3

Reset to default
2
+50
$\begingroup$

You're confusing many different things together, and I'll try to clarify and separate them to make the problem clear. The problem you're facing isn't a time reversal issue. It's lack of information. So let's go through this one by one.

  1. Classical physics is 100% symmetric in time. This is the concept of determinism. There's no doubt in that. The non-determinstic nature of Quantum Mechanics is what drove many scientists (like Einstein) crazy. This doesn't exist in classical mechanics.

  2. Quantum field theory isn't 100% symmetric in time. If CPT theorem holds (Charge conjugation with Space mirroring (parity) and Time reversal) in our universe, then time isn't reversible with the same physics. There's slight irreversibility in it. This was first discovered when Kaon decays have shown that they're CP violating (and they got the Nobel prize in 1980 for that).

  3. You have to understand that physics is modeling of the universe. It's not a full description of the universe (we hope we'll reach that level with a Grand Unified Theory). This means that we created the classical physics model (with newton's laws) to be 100% reversible in time. That's why we assign an accuracy with each model. Classical physics is only accurate to about $10^{-6}$ relative accuracy. I don't know any classical mechanics experiment that could produce better accuracy than that.

  4. The reason why you can't reverse the collision is that you lost information during the collision. If you take into consideration the heat dissipated in the system and the forces that hold the constituents of your bodies and their motion, then you should be able to reverse everything with no problem at all. But what you're thinking about is ignore the information about energy and heat and deformation, and still be able to reverse the dynamics of the system. No doubt this is not realistic and doesn't work. Your model is simply incomplete.

I hope this makes the picture clear.

Improve this answer
$\endgroup$
6
  • $\begingroup$ The fact that the collisions are inelastic means (from its definition) that energy is not preserved which means (from Noether's theorem) you can't describe the system with time-translation symmetry. Now, time-reflections are not the same as time-translations, but without time-translation symmetry it actually matters what instant you reflect about. So that's probably sufficient to violate anything that you are interested in. Fortunately, you can often model an inelastic collision of things as aggregates of constituent particles elastically colliding. $\endgroup$
    – CR Drost
    Commented Feb 25, 2015 at 21:00
  • $\begingroup$ @chris sorry, but you can't be more wrong. Energy is never not conserved in classical physics. Energy is always conserved. However, in inelastic collisions there is heat dissipation and deformation of objects which is usually difficult to measure. Your last statement is right, and is what I mentioned when I talked about the incompleteness of the model the asker is wondering about. If the model included the constituents and their forces and the relevant energy transfer, then modeling inelastic collisions would be possible in both time directions with no problems. $\endgroup$
    – The Quantum Physicist
    Commented Feb 25, 2015 at 21:26
  • $\begingroup$ Your words are a little murky. Do you mean "classical physics is fundamentally incapable of modeling a system where energy is not conserved?" That is false; see gravitational slingshots, friction forces, modeling gravity when $G = G(t)$, etc. Or, do you just mean "we happen to think (in both classical and quantum physics) that energy is always conserved if you use a sophisticated enough model"? That is true, but it's not what I mean when I say that the particle collisions don't preserve the energy of the particles -- there I am clearly talking about the kinetic energy of the particles. $\endgroup$
    – CR Drost
    Commented Feb 25, 2015 at 21:41
  • $\begingroup$ @ChrisDrost Building systems that "waste" energy is a math trick (like the Rayleigh dissipation function). You force adding a term that will waste energy with time. But in reality, with this trick, you're explicitly saying that you don't care about this wasted energy, thus arriving at the result of energy non-conservation. But can you start from a Hamiltonian or Lagrangian and derive the equation of motion of an energy non-conserving system? Of course not! And that's my point! Manipulating equations doesn't mean that we used classical mechanics, but just a manipulated version of it. $\endgroup$
    – The Quantum Physicist
    Commented Feb 26, 2015 at 10:21
  • $\begingroup$ @ChrisDrost This is for sure incomplete! And for that you can't do any time reversibility because you chose to ignore parts of your system. $\endgroup$
    – The Quantum Physicist
    Commented Feb 26, 2015 at 10:22
0
$\begingroup$

The laws of classical physics are strictly symmetric under time reversal.

So given the backwards motion of every single point particle (which allows you to know convectional/conduction flow of heat, as well as longitudinal sound waves), as well as every single light ray (radiational flow of heat), you must be able to run a simulation backwards in time and see for example the Earth fusing into the moon, and then all the heat and light converging into the combined system to then split into the two original planets they were. Much the same as if you put a cube of ice in a cup of water enclosed in an ideally isolated box, and watch it dissolve; if you were able to account for every single air, glass and water molecule, then if you run the simulation backwards you should be able to see the miraculous heat flowing out of the cup and a cube of ice magically forming out of nothing.

Both of these processes seem miraculous, in a sense that you can tell using your intuition that this definitely looks like going back in time! The way you can tell is through the only quantity in physics that distinguishes between past and future: $Entropy$. As the second law of thermodynamics has it, in every process the total entropy will increase (and in idealized situations might remain unchanged). When entropy which measures the amount of disorder increases, the process is called irreversible in thermodynamics..

Inelastic collisions inherently increase entropy, because they involve the release of heat (loss of kinetic energy), which means they increase the disorder in the considered system (imagine all these new waves and photons flying around). Whereas in elastic collisions entropy change is zero, and it is in thermodynamic lingo a reversible process, because the amount of disorder in the system did not increase, and the reverse of such processes are not deemed miraculous according to our intuition of the passage of time.

read http://en.wikipedia.org/wiki/Entropy_(arrow_of_time)

The statement that given the entirety of information about a system in a certain instant, you must be able to predict its past is a central tenet in physics, whose consistency in the most extreme realm of quantum gravity sparked a historic debate regarding the loss of information in black holes through Hawking evaporation.. It seems that even in that extreme realm, conservation of information is still valid (that is the predictability of the past). Notice that even in quantum mechanics information is never lost, it is just that quantum information has a different character than the classical counterpart (determinism etc..)

read http://en.wikipedia.org/wiki/Black_hole_information_paradox

Improve this answer
$\endgroup$
-1
$\begingroup$

Inelastic collisions are so that kinetic energy is not conserved, for example, collision of macroscopic objects in real life. Part of kinetic energy is transformed in heat (atomic motion in participants). In order to write time-reversible, symmetrical equations, you have to include losses of kinetic energy.

Briefly, "a set of objects with positions and velocities known at a given time" is not enough to describe a system undergoing inelastic collision, equations will not be consistent.

However, if you include heat transfered, just looking on the equations, you will see nothing wrong when $t$ is substituted for $-t$. That is exactly what time-reversible means. But it should be noted that reversed process will not happen in a million years. For entertainment see Feynman's "The Character of Physical Law - 5 -The Distinction of Past and Future"

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.