The Fountain of Discordia
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Subatomic violation of c?`
Speaking of electrons here...
As I understand it, an oribital is where the electron of Atom X has a 95% probability of being at at any given time, however the probability that it
will appear in an arbitrary location outside the orbital is small, but nonzero. Given that the location is arbitrary, it is posible for one of the
electrons on my face to suddenly appear on the other side of the galaxy.
How is this possible without violating c?
[Edited on 27-5-2009 by The Fountain of Discordia]
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chemrox
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That it will appear at a location outside the orbital shape does not mean it has to be a long way from 'home' or that the possibility that it could
appear on the other side of the galaxy is finite. Also you are prohibited from knowing both it's location and velocity at the same time.
"When you let the dumbasses vote you end up with populism followed by autocracy and getting back is a bitch." Plato (sort of)
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The Fountain of Discordia
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Yes, but if location compared to the previous location was further away then a photon could travel in the same length of time, the velocity must
generally be higher than c. Also, note that the galaxy it simply an example. The actual distance could be much less, or much more.
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Sauron
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The Heisenberg Uncertainty Principle is what you are fumbling with.
Sic gorgeamus a los subjectatus nunc.
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ScienceSquirrel
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Although the electron has a 95% chance of being inside the orbital volume, the chance of it being outside it falls away with distance at a very high
rate. It should also be remembered that electrons are not macroscopic particles.
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The Fountain of Discordia
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Yes, I realize this, however, due to uncertainty and the infinitely regressive nature of floating-point probabilities, it is nonzero and thus possible
for an electron to travel a distance that exceeds the equivalent distance a photon could travel in the same time period. I relize how enormously
unlikely this is, but I'm not asking about it in a real sense, but a hypothetical one. We know it can happen, so let's assume it just did.
How did the electron get there? It it still bounf to the nucleus of it's originator atom? If so, then will it reappear in the orbital after the next
interval, or will it seek the nerest + ionic nuleus it can bond with?
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Twospoons
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You may find, due to the uncertainty principle, that the orbital probability only applies the first time you locate the electron. At that point in
time I suspect that the orbital probability size becomes zero ( it must, as you know exactly where the electron is), from then on, the orbital
probability volume would increase with time - doubtless in such a way that c is not violated, and at a rate dependent on the accuracy of your initial
position(as velocity uncertainty goes up as position uncertainty goes down). It may even be that you cannot know position to an accuracy such that
the magnitude of the velocity uncertainty exceeds c.
I've never liked the Uncertainty Principle. When it was explained at university I always had the nagging feeling that the reasoning behind it was
circular - though I could never put my finger on why. Maybe I just didn't understand the lecturer properly.
[Edited on 4-6-2009 by Twospoons]
Helicopter: "helico" -> spiral, "pter" -> with wings
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not_important
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I think that the problem lies hee
Quote: Originally posted by The Fountain of Discordia  | ...
As I understand it, an oribital is where the electron of Atom X has a 95% probability of being at at any given time, however the probability that it
will appear in an arbitrary location outside the orbital is small, but nonzero. ... |
You are thinking of the electron as an object, a point; however at this scale the electron is a wave, the probability density function is not the
chance of finding an electron somewhere so much as a value describing in effect how much of the electron is there.
Think of atoms of helium, for instance. If the electrons were points, it would be fairly probably that both electrons would be in the same hemisphere
of the He atom, leaving a naked nucleus exposed in the other hemisphere. These localised charges would interact with similar localisation charges,
resulting in a fair amount of bond-like interactions and possibly even electrical conductivity. The heavier noble gases would exhibit less of such
behaviour as they have more electrons and thus less of a chance of all of them being in the same 'sector' at the same time.
Now with wave-like electrons, they appears as a fog around the nucleus, thinker at places and thinner in other regions. Each electron is always
surrounding the nucleus, not dashing from place to place.
Note that the shape of p, d, and f orbitals would require a point-like electron to pass through the nucleus, or to tunnel to the other side.
It's not that simple, as heavier elements exhibit properties (the inert pair effect, colour of gold, low melting point of mercury) best explained as
the effect of the relativistic speed of the s and d electrons. It's not easy to see how a wave function has a speed. Take a couple of years of
university level courses and it will become clearer.
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12AX7
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The probability of the event you propose is vanishingly small. In fact, so small that it will "never" occur, for any electron, during the entire life
of the universe.
There's no problem at all exceeding the speed of light, as long as it's "paid back" with an amount of energy corresponding to the event's amount of
"cheating". For example, a particle-antiparticle pair can spawn into and out of existence transiently without any apparent energy having been
delivered to produce it. But if no one sees the pair (it doesn't interact with anything) in the couple Planck times it exists, who's to say it even
happened? So what does it matter? It doesn't. The universe keeps going on the way it always has, pair or not.
But what if the pair does interact with something? Let's say the pair spawns on the event horizon of a black hole. One falls in, the other safely
escapes into free space (more or less). Down goes an entire particle, into that from which it cannot escape, while this other particle mysteriously
materialized from nothing. Well, no, that's not entirely true, because the other particle gave up its energy to the black hole -- actually, the black
hole gave *it* up, so in effect, what has happened is, the black hole radiated a particle.
Another view: the matter at the singularity has to quantum tunnel past the event horizon to escape. It cannot escape, relativistically speaking,
because that would require it to exceed the speed of light. But on the quantum level, barriers are just barriers, so it's just another question of
potential to overcome. There is no "speed of particle", just probability of particle. It's no big deal.
'Ya know, in light of the foggy  but correct explanation given by not_important,
it's quite astonishing that I can get away with such a particulate explanation here. To explain that, one must always keep in mind that "particles"
are always entities which have a foggy wave function which wraps around potentials and such, but a wave function with a definite overall shape. On
the scale of picometers, electrons get lost in the mist; on the scale of micrometers, electrons are point structures (even for rather cold
electrons!). It works the same way in everyday life: an automobile in orbit is just a hunk of metal, with a well-defined mass, inertia, center of
gravity and so on. But when you step into a car, you need to know a lot more about its fine structure; where the doors are, and so on. Details that,
on the large scale, you can safely ignore.
I guess the most bizarre behavior still comes from those wave/particle experiments where electrons obviously diffract and interfere with a fairly wide
wavelength, then produce teensy points on impact with the photographic film. In this type of experiment, both scales are being mixed (that is,
point-like and wave-like behavior are purposely overlapping), which has produced no end of confusion to us simple-minded humans. Take it on faith,
then, that neither the wave model nor the particle model is wholly sufficient to describe an electron, which is in fact a waveparticle (or sometimes a
particlewave).
Tim
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len1
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| Quote: | | Take a couple of years of university level courses and it will become clearer. |
I lectured quantum mechanics both ordinary and relativistic (which is called quantum field theory) at university quite a few years I dont think it
answers that question - as I alluded elsewhere scientists who invented modern physics such as Einstein and even Schrodinger himself where not happy
with what transpired.
You can only talk in probabilities of something being located somewhere, but once you make a measurement of position thats where it is, with whatever
accuracy the measurement was made. The electron is not treated as being partly located where you measure it - its wholly located there - that is what
is meant by the collapse of the eigenfunction - something regarded as unnatural by many scientist at the time when QM was developed and still subject
of resarch (decoherence theory has been the most importnant development there).
An electron in the lowest energy state around a proton (hydrogen) has probability
e^-r/a
of being located at radius r from proton. So the prob that you find it on the other side of the universe is non zero! These 'violations' of
relativity are not real in that they can not be used to transmit information - which is actually what relativity theory says! So no violation as we
have no trajectory. (Einsten protested about this very aspect in his famous EPR paradox).
Another point - measuring position accurately means hitting the electron with high energy particles - this measurement does not commute with the
hamiltonian - so the particle does not have to remain in its intial eigenstate (as it would if the measurement operator commuted with H)
[Edited on 4-6-2009 by len1]
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watson.fawkes
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| Quote: Originally posted by not_important | | You are thinking of the electron as an object, a point; however at this scale the electron is a wave, the probability density function is not the
chance of finding an electron somewhere so much as a value describing in effect how much of the electron is there. |
I feel the best perspective is Unruh's: "a particle is what a particle detector detects". Just because you measure an electron as a
particle doesn't mean the electron is, in any sort of fundamental way, a point particle. From a macroscopic point of view, you can treat it as such to
some approximation. The uncertainty principle is the limit to that approximation. Push the analogy too hard <cough>string
theory</cough> and you get nonsense. What a particle detector does is to treat the electron as a particle, to measure the macroscopic
parameters of that electron. Those parameters, though, measure the interaction of the detector, which is designed to see particles, and the electron.
To see an electron as necessarily a particle is to confuse (1) the interaction of an electron and a detector with (2) an electron outside the presence
of a detector.
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not_important
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But note that the question was more of a thought experiment without any actual measurements, just an all-seeing Observer noting where the electron is,
getting by collapsing the state vector.
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