Is this how it works?

1. Jan 25, 2012

questionpost

Do atoms themselves violate thermodynamics and relativity?

If I have an atom in a completely isolated system where there's no energy coming in or going out, let's say it's already at it's ground state since the second law of thermal dynamics implies all systems will eventually lose all possible energy to entropy. It can't lose any more energy so at this point it doesn't attempt to violate thermodynamics, and then if we measure those electrons, we would find that at that energy level, an electron still has a probability of existing at a larger distance away form the nucleus, but is the electron considered to have a greater energy at a greater distance, or is that just part of it's probability wave?
Because if it's just part of it's probability wave to have a chance of existing at a larger distance away without increasing in energy, why is an mean location of even further away considered to be at a higher energy level?
And then, couldn't that still technically be used for electrons to by chance happen to exist at a a greater distance form the nucleus and so form a compounds that could normally only be formed by adding more energy?

Also, someone told me that nothing can reach absolute 0 and that if we did it means time has stopped, but another person made a good point by saying if we pick up a block and hold it in our hand, from our point of reference it is not moving and therefore has 0K from our view, and the clock will still continue to count time. So...can absolute 0 actually be measured just by a mere frame-of-reference, or is technically anything that is not moving having 0K relative to you?

And to add on to that, because of the uncertainty within particles, we should be able to measure something not moving at all, however given the proper instruments we would measure that all particles within any object are in one location and at a later time in another or traveling distance over time, suggesting the presence of kinetic energy or at least motion.
So, how does that not violate relativity if we can always measure that an atom is moving even though motion depends on the frame-of-reference and therefore no motion can be measured from a frame-of-reference? Because I have heard form people that relativity isn't broken at that point.
And when I say broken, I don't mean all of it is flat out wrong, I mean for that particular instance it fails to accurately describe the situation.

2. Jan 26, 2012

Simon Bridge

No.
It doesn't but go on...
It is total energy that is conserved - the electron detected at a great distance from the center of the atomic potential has greater kinetic energy but less potential energy.

No - it has zero kinetic energy - the K would be a unit of temperature.
0K is just a point on a graph.
The uncertainty principle means that there are limits to the precision with which we can simultaneously know the position and momentum of a particle. In principle it is possible to for a particle to atually be stationary with respect to us but we only measure it's speed to be 0±Δv (some value).

Think of it another way - the particle knows for a fact, to infinite precision, that it is stationary - the uncertainty principle means that it cannot know what it's speed is - and it doesn't. From it's POV, everything else is moving :)

Both these examples result from missing out half of the situation.
With the energy of a bound electron you forgot about potential energy and with uncertainty in relativity you forgot that it is a relationship.

3. Jan 26, 2012

questionpost

Doesn't the second law state that nothing can be 100% efficient because in a process there will always be some energy that is irreversibly lost in the form of entropy? So technically shouldn't the universe over much much time come to a lowest possible energy state?
http://en.wikipedia.org/wiki/Second_law_of_thermodynamics

So I guess because K is the temperature, it has 0 temperature, but it still has kinetic energy itself?

That's what I use to think, but an electron can't move in a classical sense because it would radiate it's energy away. It also doesn't have a defined location when it isn't a measurement, so how can it have a defined frame-of-reference?

4. Jan 26, 2012

Simon Bridge

That's not what you said though. You said:
... and that is not what the Wikipedia article is saying. The second law of thermodynamics says that the total entropy of a closed system cannot decrease - it can, however, stay the same. You don't "lose energy to entropy", you lose energy to other energy. The energy does not go away.

Lets say you set up your box with an atom inside, which is in an excited state.
There is a chance the atom will decay by emitting a photon (say) so at some time this happens - now you have an atom (gs) as well as a photon, in the box. If the walls are perfectly reflective (no energy can get out remember) then there is a chance the atom will encounter the photon again and absorb it - returning to an excited state.

But make the box very big compared with the atom, and the chance of reabsorbtion gets smaller. For an infinite box - so you effectively have a Universe with one atom and one photon in it, that chance is zero...

Of course - it can have bulk kinetic energy - you can still throw a snowball can't you?

You mean that if it had a classical orbit it would accelerate and therefore radiate it's energy away? Electrons can, indeed, move as a classical particle.
That is of of the many problems that a relativistic model of quantum mechanics has to solve and why the standard model tends to be thought of as a particle rather than a wave model.

But one way to think about this is that when you say "does not have a position" (it does have a "defined" position - it's the expectation of it's position operator on it's wave-packet) you mean it does not have one with respect to you as the observer. However, with respect to itself, it always knows where it is (0,0,0) and it is always stationary and not accelerating at all so it does not radiate any energy away so no harm no foul.

Also remember that an electron has to come from some measurement event and go to some measurement event so it always has some defined position with some uncertainty. I think Roger Penrose would say that the bits of an atom are constantly measuring each other when they exchange photons.

Finally, you don't need absolute positions and speeds to do relativistic mechanics - it works quite well with uncertain measurements too. After all, none of out actual measurements in the lab are anything like Heisenberg level for most things... my ruler is only good to a quarter millimeter for eg.

5. Jan 26, 2012

questionpost

Well I mean particles can move physical distance, but it's not written in their wave mechanics that they oscillate by traveling physical distance.

Ok, but so if it doesn't have a defined position, wouldn't we always measure that an electron is "jumping around" no matter what frame of reference your in and therefore always measure it has energy, other that I suppose traveling at light or above? Although technically, you can still have photons hit your retinas and send information if your traveling at light speed.
Also, I don't know how proven it is that an electron has such a clear frame of reference of itself as you are saying, it doesn't make sense to me that it would. In the classical would it would make sense, but, how is the electron measuring itself? How does that measurement not collapse its wave function?

Hmm, I don't understand how that could be true based on my current understanding, unless you can clarify it more, because if an atom is always determined or certain, that implies we are getting infinitely continuous measurement data or a completely unbroken stream of photons, which isn't happening. We get little tiny bursts of measurements every second and by the time a photon has left the atom, the atom has to then be existing in an undefined state since since at that point it's not measured until it hits your retina. And with virtual virtual photons they don't do exactly the same thing or we would never be able to observe entanglement experiments because the virtual photons of Earth's magnetic field and of it's gravity are still hitting the entangled particle during an experiment.

Hmm, what about with liquid helium?

6. Jan 27, 2012

Simon Bridge

I got that backwards ... the farther from the center, the less KE and more PE. The takeaway lesson remains the same though - it is total energy that is conserved.

Where did you get that information from?

Your current understanding is incorrect - I thought that had been established.
Didn't say that though. If you are going to read-into what I say I cannot help you.

It is not proven at all - it is a way of looking at things which makes the math come out right. Have a look at "self energy of an electron".

But it is certainly consistent that a particle has pretty good knowledge of itself - that is really just saying that it is stationary in it's own reference frame. Which is what relativity says of everything.

This "jumping around" between successive measurements happens classically too ... you take a meter-rule out and measure the length of the carpark: do it right now. Right down the measurement and it's uncertainty. Tomorrow, go out and make the same measurement. The again the next day. Bet they are different each time.

The measurement of the carpark jumps about. Randomly.
You understand why this is. This is also why positions of QM objects jump around ... only, in this case, the measurement is all we have so we've gotten used to treating the measurement as the thing. But what is the difference anyway?

You know, I can summarize your objections by saying that non-relativistic wave-mechanics does not sit easily with relativity ... and you are right, of course. But why would you expect it to? Why would you even expect it to be intuitive?

I'm sorry you don't seem to like what I'm telling you but thems the rules ... it is very difficult to describe them in the limitations of the English language, I'm sorry for that too. How's your math?

7. Jan 27, 2012

questionpost

I don't think your particularly right or wrong it's just a matter of the evidence presented, and "my understanding"= everything I learned before this topic and including this topic, so my understanding includes what you have mentioned as well.
Although I don't think what your saying is entirely accurate based on what other physicists have told me and the fact that there isn't a general consensus on implications of different aspects of QM in the real world, I'm still now convinced that an atom itself doesn't violate thermo-dynamics or relativity based on our current knowledge so ty for clearing it up.

The main reason though that I would try to look into rules being broken though, is because if rules are being broken it means progress is being made.

8. Jan 27, 2012

fellupahill

Kinda seems like question is 2 different people. One person in the first posts, someone who thinks they are more knowledgeable (politely not making any assumptions either way) in the later posts.

9. Jan 27, 2012

questionpost

I'm simply stating why I think what I think or stating why something should work the way I state it works given my current understanding. If I thought I was more knowledgeable I wouldn't post questions on here.