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The law of conservation of energy states that energy cannot be created or destroyed. Based on this principle, you can safely conclude that any effect resulting from a cause must somehow keep all energy within the system. This is clear for me.

Now, add in randomness. Suppose you hit the cue ball and it strikes the 8 ball into the corner. You have millions of atoms jumbling around and make an accurate prediction impossible, however statistically, you're going to end up with a very probable end result from the interactions of all those atoms contained in the 8 ball getting struck by all those atoms contained in the cue ball.

This behavior is so well-defined that it would be surprising if it did anything differently. However, from my understanding there is a clear distinction between improbable and impossible. It is improbable for say, as a result of all the atoms colliding, a single atom receives all the energy received and splits off from the 8 ball at a fraction of the speed of light while the rest of the 8-ball stays still? So long as the conservation of energy is not violated, we're talking about a very improbable but still possible scenario.

My question is this, at what point is such a scenario impossible and not just improbable? Could I come up with any scenario that doesn't require more or less energy than is possible and it would be improbable but not impossible?

This applies not only to my billiard example but in any reaction. In quantum mechanics, particles behave very strangely if we compare them to how objects typically behave on the macro scale (particles disappearing and reappearing, quantum entanglement, etc.). The only reason that doesn't happen on the macroscale is because it is extremely improbable that all atoms should behave strangely at the same time. Most interactions are the same and as such on the whole, you see a very average and predictable reaction from the cumulative of actions on the microscopic scale.

However, as improbable as it might be, could an object like a billiard ball simply vanish due to, by chance, all atoms contained in the billiard ball simultaneously having a quantum "strange" behavior? It would be so improbable that it has likely never happened in the existence of the universe almost certainly not even in some small measure.

Some might argue that extremely improbable and impossible are the same thing, though while it is improbable that it will rain 30 days in a row, I would very much consider it a different scenario to say, stumble across a four-sided triangle. However it is not clear where this line is. Since it is so improbable, my idea is that really quite a lot is possible and not probable, so long as the law of conservation of energy is respected. I would not, for instance, expect all the atoms of a unicorn to assemble together and form one, but that may even be possible. Though at that point, we enter discussions of boltzmann brains and what is more probable, I suppose.

TL;DR - What are the limits of what is possible in terms of what we know about the laws of conservation of energy? Can seemingly impossible things occur so long as these laws are respected?

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    $\begingroup$ If this is a "dumb" question and deserves to be downvoted, please explain to me at least why this is a dumb question. $\endgroup$
    – Neil
    Commented Apr 21, 2016 at 14:59
  • $\begingroup$ The situation you describe would be "impossible" because of momentum conservation. The forces between the atoms can't transfer that much momentum to a single atom, that would require a giant accelerator, far larger than anything we can do, today. When we talk about these kinds of scenarios in statistical mechanics we have specific interactions in mind which never include these extreme examples. The system "never" deviates significantly from its most likely state. $\endgroup$
    – CuriousOne
    Commented Apr 22, 2016 at 0:26
  • $\begingroup$ @CuriousOne If something is improbable enough, then it has never been observed. My question was what are the limits of what is possible? This explanation only tells me that we haven't ever observed atoms behave like that, but that doesn't answer my quesiton. $\endgroup$
    – Neil
    Commented Apr 22, 2016 at 6:47
  • $\begingroup$ I understand your question, but it's simply not relevant within the way physics operates. A cup of coffee doesn't become a particle accelerator that can put hundreds of J of thermal energy into a single particle. For that you need an astrophysical mechanics that possibly releases several solar masses of mass-energy within a few seconds, or so. If you want to discuss how your coffee can do the same, then YOU have to give an explicit mechanism and YOU have to experimentally demonstrate that it happens. If YOU can't, it's not science. Those are the rules, I am afraid. $\endgroup$
    – CuriousOne
    Commented Apr 22, 2016 at 6:52
  • $\begingroup$ @CuriousOne The types of interactions I'd expect to happen, however probable are likely the same interactions that can and did happen within the first second of the big bang. We can hardly observe a scenario quite like that, and so we are left to make conclusions based on what we observe in all its limitations. Using Hilmar's example, you say it is impossible for a perfect vaccuum to form on one side of a room and all other particles to settle on the other. What is the largest variance of pressure theoretically possible then and why can't it exceed that? $\endgroup$
    – Neil
    Commented Apr 22, 2016 at 7:07

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This probably more a philosophical than a physical discussion. Let's take a simple everyday example: The air molecules in the room where you are sitting are fairly evenly distributed through the room. Because the molecules are subject to random motion, it's perfectly POSSIBLE to have all molecules bunch up in one half of the room and that there is a perfect vacuum in the other half. However it's very unlikely. It is in fact so unlikely that the only pressure difference you will ever find throughout any room are exceedingly small (10-20 orders of magnitude below the average air pressure).

So in this case it makes not practical difference whether it's impossible or just very unlikely. There are no observable differences between the two interpretations and one could argue that hence the difference between the interpretations is meaningless.

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  • $\begingroup$ I think that's an excellent example. Maxwell's demon showed that such a thing is impossible without adding energy to the system, so perhaps in some convoluted way that we're not capable of understanding, that scenario is impossible though it merely seems improbable. It'd be like asking what the probability of a ball rolling uphill is which would clearly violate the law of conservation of energy, getting energy from nothing. There may be shades of gray with probability, but my hunch is that improbable and impossible are distinct, albeit hard to comprehend. $\endgroup$
    – Neil
    Commented Apr 22, 2016 at 6:58
  • $\begingroup$ @Hilmar But as you said is very unlikely. Does this mean that maybe in the future we will observe that? $\endgroup$ Commented Mar 27, 2020 at 12:18
  • $\begingroup$ No. The likelihood is so small that even when comparing it to the time & size scales of the entire universe, it's not going to happen. And as Neil said: you'd have to add energy. $\endgroup$
    – Hilmar
    Commented Mar 28, 2020 at 13:14
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There's nothing probabilistic when it comes to Conservation of Energy, and Quantum Mechanics. In fact, many equations from QM and Special Relativity are derived by first assuming Conservation of Energy. So, energy will be conserved, no matter what, by definition. The Schrodinger Equation itself is derived by saying "Here's a wavefunction, let's assume energy is conserved". And $E=mc^2$ is what what you get when you say "Let's measure Doppler shift, but assume energy is conserved".

However, in the QM case, energy is no longer the $\frac{1}{2}mv^2$ you may be used to, since $m$ and $v$ can be in superpositions. The energy of the system is encapsulated by what is called the "Hamiltonian Operator", and the statement of the Schrodinger Equation is that the Hamiltonian Operator of the Wavefunction remains conserved over time. You can read more here. The equation boils down to, if a particle can be in energy states 1, 2, and 3, then we know that P(1), P(2), and P(3), do not change over time (P(x) being the probability of energy state x). An Energy State can be described by $KE + PE$, where $KE$ is only dependent on your momentum, and $PE$ is dependent only on time and position. (More precisely, we have an "Energy Operator", which is what we call the Hamiltonian because that's what we presume is conserved. We could throw any conservation law in there, and get various results. Generally, $KE$ is set to the familiar $\frac{1}{2}mv^2$, and $PE$ is dependent on the force you want to model.)

So, in a purely Quantum environment, energy by this definition is neither created nor destroyed.

It's possible the universe started out in a superposition of energy states, and this remains consistent with current theories. But, we are a part of the universe, and trying to consider superpositions of conscious beings gets very weird from a philosophical perspective. So for our story let's assume for now that the big bang just happened and our macroscale bodies exist in a deterministic universe that started with a total energy E, with no Wavefunctions anywhere.

Let's say we decide to shoot two protons at each other with total energy $v$. Now, our deterministic part of the universe has energy $E - v$, and our protons have energy $v$ (With $v$ high enough to be interesting). Let's say that the protons are no longer a part of the deterministic side of the universe, so that they begin to experience quantum effects.

They collide, and split into many particles, through many different mechanisms (Feynman Diagrams), all simultaneously in superposition. However, from the Schrodinger Equation, we know that total energy of the Wavefunction after our experiment must equal $v$, with probability 1. After the protons collide, we no longer have one neat Wavefunction. We have a weighted sum of the Wavefunctions of each possible collision result.

In this case it's not so bad, since we know that each possible collision will also have energy $v$ (By Schrodinger). But, the collision resultant particles will have unknown energy levels (Since post-collision they are now their own closed system, we can consider each resultant particle to have its own wavefunction). Say our experimental results show that the proton split into one new particle with an energy of E1, and another particle with an energy of E2. I say "With an energy of", but what I really mean is that their wavefunction could be something like a normal distribution around E1 and E2, respectively (With the exact curve being a result of Schrodinger's Equation). What we know for a fact is that $E1+E2$ approaches $v$ as our measurements of E1 and E2 grow more and more accurate. It's important to realize here that energy conservation occurs via entanglement, since if a measurement of E1 is unexpectedly large, then E2 must be unexpectedly small, regardless of whether or not we measure it or care about its existence. (Ie, it will be small when it collides with air particles to distribute its energy, while our scientific devices observe E1 being large)

In our story, the results from our initial collision and all subsequent collisions will entangle, via the method described above, with the rest of our measurement apparatus and eventually the universe, so that everything at the quantum-scale becomes a mess of wavefunctions and superpositions of various energy levels. It does remain that the genesis wavefunction, and thus the global wavefunction over time, still has an explicit total energy $E$. The more precisely we try to measure the energy of everything, the closer we get to calculating that genesis $E$.

In reality, we did not get to measure the universe before the initial wavefunction. We can presume that the instant the big bang happened, the very first wavefunction happened. So we just don't know what $E$ is. We don't even know it exists, since there may be many $E_i$, as discussed earlier. For sanity, I like to presume there is indeed a genesis $E$ (Copenhagen does allow us to assume that macroscale objects do not exist in a state of superposition, and "The Entire Universe" is indeed macroscale). We can't calculate $E$ exactly, but we know that the sum of the energies of all things that we measure will never deviate from $E$ by any more than our own measuring instruments allow, no matter how lucky or unlucky we are.

P.S.

Maxwell's Demon, as mentioned previously, makes no claim to violate conservation of energy. What it attempts to violate is the 2nd law of thermodynamics. There's nothing that prevents pure luck from causing all the atoms to collect on one half of a box, and in-fact the 2nd law of thermodynamics is inherently probabilistic. The 2nd Law gives us many equations that are useful such as the ideal gas law, but these are macro-scale laws.

The 2nd Law of thermodynamics does not help us at the quantum level, and is in-fact entirely irrelevant. It's just not a law at all on the microscale, and you can get lucky and unlucky day and night on the quantum level (2nd Law is okay with being lucky on small scales). W Bosons with mass low enough to cause atomic decay are extremely, extremely, rare, but energy is conserved even when these events happen.

On the same token, the solution to Maxwell's Demon does not show that it is impossible to isolate gas temperatures without adding energy. It shows us that Maxwell's Demon can't do so without gaining entropy on average. If we assume infinite luck, and thus a functioning Maxwell's Demon, it may seem like "Oh, I can make perpetual motion out of this". But you can't. It would definitely give you a lot more energy than previously available, since you could spin a turbine with all of the luck you have in separating hot and cold particles. However, the velocity in the particles will be lost as the turbine spins faster, until eventually all of the particles in the chamber approach absolute zero.

QM exists such that, so long as you can't generate infinite kinetic energy, there's some valid function $PE(x, t)$ to describe potential energy. Then, we deduce the conservation of energy (Or rather, the equations were built to match such an assumption).

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Here is an impossible example. You have two balls of the same mass $m$, with one of them at rest. They elastically collide, and after the collision they both have the same speed in oppossite directions.

If energy is conserved, you get that $v_{final}=v_{initial}/\sqrt{2}$

But this is impossible because it violates conservation of momentum, you obtain:

$mv_{initial}=0$

which is a contradiction

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  • $\begingroup$ Granted, but assuming the law of conservation of energy is respected, what is considered an impossible scenario for a collision event? If I release a drop of water into a pool of water, typically a small amount rebounds and waves form, but respecting the laws of conservation of energy, conceivably that drop could rebound 100% off the water and leave no water wave, for instance. That would not be gaining or losing energy, correct? $\endgroup$
    – Neil
    Commented Jan 16, 2019 at 7:05
  • $\begingroup$ I do not know if the drop example is just impossible of improbable, due to its complexity. The example I gave you is simple, conservation of energy is respected, but it is impossible. or you have a different definition of impossible. $\endgroup$
    – user65081
    Commented Jan 16, 2019 at 11:31
  • $\begingroup$ There are many impossible examples. My question is regarding odd behavior that doesn't violate any laws. Technically a small subset of atoms is colliding with a small subset of atoms in the other ball. The rest of the ball moves because the crystalline structure means that force is efficiently distributed throughout the ball. However even if this is typical behavior, that doesn't necessarily make other behavior impossible, just unlikely. If by chance a single atom absorbed 50% of the energy by the collision, it would move very very quickly while the other atoms would absorb the other 50%. $\endgroup$
    – Neil
    Commented Jan 16, 2019 at 11:40

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