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jgourlay
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Why don't the electrons slam into the protons?
Yet another stumper from my 4 year old (the insightful one).
"Daddy, if the negative electron and positive proton are attracted, how come the electron acts like a planet around the sun? How come they don't bam
into each other?"
I assume the right answer is not, "shut up and eat your carrots!"


watson.fawkes
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Quote: Originally posted by jgourlay  "Daddy, if the negative electron and positive proton are attracted, how come the electron acts like a planet around the sun? How come they don't bam
into each other?"  "They don't because of quantum mechanics." Really, this is the right answer. When you get
to electrons in atoms, naive classical physics analogies break down and no longer provide true insight. The thing to teach is that models are
approximations to reality, not generators of reality, and that every model has some nonuniversal realm of applicability.
The atomic problem is, indeed, the first major triumph of QM, not for hydrogen (for which the semiclassical model sufficed), but for helium, for
which the Schrodinger equation correctly predicted the spectral lines.


jgourlay
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Yeesh....
I get it. However, it's difficult to communicate that to a youngster. They conceptualize so literally. Their little brains have difficulty with
constructs like "model".


turd
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"Daddy, if the earth and the sun are attracted, how come the earth acts like a planet around the sun? How come they don't bam into each other?"
Just don't tell him that charged particles going around corners are supposed to lose energy and you won't have to explain him QT.


Sobrero
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Hmm the answer "Because Daddy says so." is quite unscientific too ^^.
Perhaps you can tell this too:
Electrons are negatively charged and protons positively, but that doesn't imply that an electron orbiting around a proton will collapse on the proton,
because the centrifugal force opposites the Coulomb force.
But from electrodynamics we know that an accelerating charged particle radiates energy in the form of electromagnetic radiation. Thus the kinetic
energy of the electron decreases, and because the radius of the orbit is related to the kinetic energy of the electron, the radius decreases too and
so after a very short while they collapse.
Bohr's semiclassical model resolved this problem by bluntly postulating that:
 the angular momentum of the electron is quantised (and so also the radius and energy of the electron is quantised).
 the electron in its lowest energy state doesn't radiate energy.
The quantum mechanical model postulates that a system's wave function obeys the dynamic SchrÃ¶dinger equation. When solving that equation for a system
comprising a proton and an electron, it turns out that energy is quantised, and that the lowest energy state is nonzero, so the system is stable.
Well at least that's what I recall from my course of Molecular Structure .
"There exists a world. In terms of probability, this borders on the impossible." (Jostein Gaarder)


watson.fawkes
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Quote: Originally posted by jgourlay  I get it. However, it's difficult to communicate that to a youngster. They conceptualize so literally. Their little brains have difficulty with
constructs like "model".  Well, you don't have to explain QM with wave functions and eighenvalues, any more
than you've explained Newtonian mechanics with the EulerLagrange equation.
How about this: "An electron is a little cloud of sparkle. It looks like a single spark from far away, but up close you can see it's a cloud. You
can't squish the cloud down to a little point, so it spreads out a little but mostly surrounds the proton."


jgourlay
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Watson, that's perfect!


watson.fawkes
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Thanks.
The language "when it's up close" vs. "when it's far away" is the implicit cue to switch models. Whenever you need to switch models when explaining
science, the key is to find a distinction in context, then to put language to the distinction. This provides the opportunity to tell a different
story.
In the present case it's a scale difference that's physically relevant. So the key is to find sensory intuition that is homologous to a scale
difference. Large objects start to look like points from far away, so that's the grounding of the alternate context.


len1
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I dont think these are the answers. Newtons second law explains why the moon doesnt fall into the earth, but Schrodinger's equation does not explain
'why' electrons and protons dont slam, into each other. In fact according to it they do, there is a small probability for the electron to be located
over the nucleus  in common parlance 'hitting' the nucleus. Nor is semiclassical theory adequate for the hydrogen atom but not for others, I dont
know where that comes from.
One must remember that the electron 'wanders' over an average ball of radius 10^10 meters. The nucleus is constrained to 10^15. So 10^15 of the
electron is located over the nucleus. For a child  the electrin swirls around the house while the proton the size of the tip of a needle sits in the
centre. This reduces the probability of their interaction. What really protects the atom is Baryon number conservation  nucleons are Baryons and
any known reaction conserves their number. The electron and proton can not annihilate. But I guess we are just running around in circles 
explaining one empirical fact, by another of more general applicability, at the expense of greater abstraction.
The ordinary Schrondinger equation also does not explain why the atom is stable against radiation  it just uses Coulombs law  so it knows nothing of
radiation. For that you need quantised Maxwells equations (called the Dirac equation).
[Edited on 1252009 by len1]


turd
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Yes, len1, you gave the only correct answer so far. IIRC, the probability density function for s electrons takes its highest value at the centre(!).
Therefore your estimate of 10^15 is probably pessimistic (multiply the probability density functions of electron an protons and integrate?). The
correct answer would therefore be: "they constantly smash into each other" or "they do overlap". But there's are reason 4 year olds aren't quantum
physicists and therefore the really correct answer might actually be "shut up and eat your carrots". (I kid...)
[Edited on 1352009 by turd]


watson.fawkes
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Quote: Originally posted by len1  Nor is semiclassical theory adequate for the hydrogen atom but not for others, I dont know where that comes from.  The semiclassical model adequately predicts the Lyman, Balmer etc. series of emission lines—all those from the Rydberg
equation. The reason is that there's only a single electron. The helium atom has electronelectron interactions that were not accounted for adequately
in the semiclassical model. When Schrodinger's equation first came out, it was immediate to derive Rydberg's equation from it. That wasn't persuasive
in itself, though, because there was existing theory that accounted for that equation. It wasn't until the helium spectrum was derived that proof was
definitive.


DJF90
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turd: The probability density function for the selectron does not take its highest value at the nucleus!! There is however, a nonzero probablility
of finding an selectron there though, because the selectron is deeply penetrating. The actual probability of finding the electron at the nucleus is
small.
The bohr condition leads to a quantisation on the radius of the electron orbit (the electron can only occupy an orbit that possesses an angular
momentum, L, that is an integer multiple of the reduced plank's constant (h/2pi), i.e. L=(nh)/2pi. This is known as the Bohr condition, and leads to
the conclusion that energy is quantised.
By combining De Broglie's equation with the Bohr condition shows that the circumference of the orbit of radius r must be an integer multiple of the
electrons De Broglie wavelength.
This quatisation of energy must somehow link into the fact that the electron does not collide with the nucleus. Perhaps the electron does head for the
nucleus but experiences the "catapult effect" and so never actually collises with nucleus?


watson.fawkes
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Quote: Originally posted by DJF90  turd: The probability density function for the selectron does not take its highest value at the nucleus!!  Actually, it does for 1s electrons, but only for those. All other electrons have a density zero at the origin. The central potential
solutions of Schrodinger's equation are a spherical Bessel function times a spherical harmonic. The spherical Bessel functions are approximately a
monomial in the power of the principal quantum number 'n' (total energy). Only when n=0 is this density not zero at the origin.


turd
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Of course, because the nucleus is small. The probability of finding an electron in a volume element V is given by the integral of the density function
over V. Therefore (since our density function is real*) the probability of finding an electron at (x,y,z)=(0,0,0) must be 0. But the probability of
finding an selectron in an infinitesimally small volume element around (x,y,z)=(0,0,0) is higher than finding it in _any_other_ volume element of the
same size. Maybe you were thinking about the radial probability density function, which gives the probability of finding an electron at a given
distance r from (0,0,0). This one is 0 for r=0 also for selectrons (the esteemed reader may work out why this is the case. ).
PS: I like the notion of "finding an electron".
* In theoretical applications density functions often are composed of Dirac delta functions. These are not real anymore and then you can indeed obtain
an integral over a zero volume which is bigger than 0.


DJF90
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The radial distribution function for any sorbital electron is not zero (quite close though) at the nucleus, but neither is the maxima at the nucleus.
http://winter.group.shef.ac.uk/orbitron/AOs/1s/radialdist.h...


watson.fawkes
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Oops. The wave function for a
1s electron has a maximum, but the probability density doesn't. Sorry.


turd
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Note that I didn't say "at the nucleus", I said at r=0. Now read the link you posted: For sorbitals, the radial distribution function is given by
multiplying the electron density by 4*pi*r^2. The _radial_ distribution function _must_ be 0 at r=0. Just think about it for a minute.
Quote:  The radial distribution function for any sorbital electron is not zero [...]
Oops. The wave function for a 1s electron has a maximum, but the probability density doesn't. Sorry. 
Oh, now I'm dissapointed, watson. You were right before.
Think about how the wave function and the probability density are related. Of course the probability density function has a maximum at
(x,y,z)=(0,0,0). That's the most probable place you'll find an electron at.
The _radial_ probability function is something completely different: it tells you at which _distance_ from the core an electron is most likely to be
found. This must be 0 for r=0. Think about the ratio of the surface of a sphere with r=0 and with r>0 then this should be immediately obvious.
Check this out: http://en.wikipedia.org/wiki/Conditional_probability


watson.fawkes
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yeah. I was a little hasty. For 1s electron, the spatial probability density has a maximum and the origin. The radial density does
not (it's a local minimum).


len1
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There are many aspects observed then in atomic spectra that the semiclassical theory could not account for: fine structure due to relativistic
effects, effects due to spinorbit coupling, hyperfine splitting due to the spin of the nucleus, and the electron interactions. QM is able to account
for all of these, although the last was very difficult to do accurately then because the three body problem in QM is not analytically solvable. You
separate the last as having been the only aspect the known  have you any reference to this?
This argument about prob density function is a waste of time, the prob density for l=0 is ~ dV e^(r/bohr radius) where if you assume spherical
symmetry dV = 4 pi r^2 dr. The prob of finding the electron in a small volume is highest at the nucleus as turd says. To say its zero is to put dV=0
in above formula and so meaningless. I could do the integral r^2 e() dr, or look it up in a book to get exact number, but its just a small
multiplicative correction so why bother here.
While one cant test the atom for stability against radiation using just a QM treatment of Coulombs law, which is the ordinary SE, the exact solution
for a two EM particle relativistic quantum system is a many body problem and so hard to do. One can assume stability through the fact that the
orbital problem, being a time varying one in classical theory, is a static problem in quantum theory where the only information is a prob density
dependent on potential. Since nothing is time varying there is no charge current and one doesnt expect to get radiation.


Magpie
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Quote: 
Since nothing is time varying there is no charge current and one doesnt expect to get radiation.

Len, does this imply that the electron is actually not a particle? And is certainly not a moving particle? If not, just what is an electron? What
the QM math says it is?


watson.fawkes
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Quote: Originally posted by len1  There are many aspects observed then in atomic spectra that the semiclassical theory could not account for: fine structure due to relativistic
effects, effects due to spinorbit coupling, hyperfine splitting due to the spin of the nucleus, and the electron interactions. QM is able to account
for all of these, although the last was very difficult to do accurately then because the three body problem in QM is not analytically solvable. You
separate the last as having been the only aspect the known  have you any reference to this?  I had hoped
that, when I repeated what I had said, elaborating on it, by emphasizing that it was a historical account, that I had addressed the point sufficiently
for you to understand it. Apparently not.


turd
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Actually, I think it's useful! For the mathematically naive person it may seem contradictorily that on one hand the highest probability region of an
electron is the origin, while on the other hand r=0 is the most unlikely distance from the origin you'll find an electron at. And this thread is proof
that it really does confuse people. Once you understood that you learned a deal about probability and infinitesimal calculus.
Quote:  The radial density does not (it's a local minimum). 
True, but you can formulate that in an even stronger way: the radial probability density function has a _global_ minimum at r=0. Because it has to be
0 for any real* probability distribution and that's the lowest density you can have. So for selectrons r=0 is the most unlikely distance, but defines
the most likely place.
* I mentioned Dirac deltas before.


len1
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Quote: Originally posted by Magpie  Quote: 
Since nothing is time varying there is no charge current and one doesnt expect to get radiation.

Len, does this imply that the electron is actually not a particle? And is certainly not a moving particle? If not, just what is an electron? What
the QM math says it is? 
Hi Magpie, nice to hear from you again. Quantum mechanics says the electron is a particle but all its particles are also waves. The most standard
approach in physics is 'feel your way as far as you can, then just do the maths because it works.' Schrodinger himself never accepted the way
physicists use his equation, neither did Einsten. The former chanced up on an equation for a function whose meaning he struggled to glean, a short
time later Bohm gave it its present meaning, on which all its subsequent success was based, and the former two spent most of their remaining lives
protesting against it  in vain, they could propose no alternative, while quantum mechanics  Schrodinger's, as interpreted by Bohm, worked.
In classical mechanics its relatively easy
F = ma
which reduces to a simple second order differential equation for the motion of an electron (or the moon or the earth) about the centre of mass
K/x^2 = m d^2x/dt^2
You get all your planetary motion from there.
In the discussion on lenses in another thread I mentioned diffraction. If you shorten your apperature, your resolving power deteriorates. From the
formula I gave there you get
aperature * angular resolution > constant
For light this is a mathematical consequence of its wave nature which can be got from Parsevals inequality  a purely mathematical result  the width
of a distribution multiplied by the width of its Fourier transform is bound from below.
There was a lot of evidence in the early 20s that all problems with atomic spectra, stability of the atom, etc could be explained if this held for
particles also. That is if you use electrons say instead of light in your camera (as in an electron microscope) you still have the same basic
constraint.
After a lot of torturous trial and error maths Heisenberg arrived at
spread of momentum * spread of position > Planks constant
As a fundamental law of nature! More fundamental than F=ma  which then becomes untrue. This is because you can no longer talk of trajectories.
When you say an electron is stationary or it is moving you mean there is a function, position as a function x(t). This would then give velocity as
its derivative, and you would know the velocity and position exactly, in contradiction to the basic tenet of the world. So this answers your second
question  its senseless to talk of the electron moving instanteneously since it has no trajectory, as for its mean location  quantum mechanics does
not deny the existance of this, then we know the answer  its centre of mass is moving nowhere. So thats what quantum mechanics says  theres no
motion, the problem is a static one, and so no radiation.
To calculate something Schrdonger used Hamiltons formulation of mechanics. This is an energy formulation. For the problem above you know
electrostatics conserves energy so one can write the kinetic energy of the electron plus its potetntial energy in the electrostatic field is a
constant
1/2 p^2/2m + k/r = constant
Which is just an integral of the second order equation and so in classical mechanics gives the same results. But quantum mechanics treats this as
its basic equation  it is not derived from a law of motion, which according to quantum mechanics does not exist.
To get the particles position X(x) and momentum P(p) functions to satisfy the Heisenberg relation as a mathematical consequence, in the same way as
for light, they must be Fourier transforms of each other. It follows from equating probabilities in both mometum and position that the momentum is a
first order differential with respect to x, p ~ d/dx. So you get
h^2 d^2/dx^2 X(x) + kX(x)/x = constant X(x)
which is Schrodingers original equation.
You prob wont get the last bit, I thought I could find an easier way when I started writing this. Once I got this far it seemes a pity to throw it
away, so I post it.
@fawkes, dont know whats theres to get upset about. I asked for a reference to what you were saying. Schrodinger only mentioned the hydrogen atom in
his paper, effects such as Zeeman, Stark, fine splitting etc. If you dont have a reference that the Helium spectrum was what was considered most
convincing, fine.
[Edited on 1452009 by len1]


Magpie
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Thank you Len for that concise explanation. The lack of radiation loss of energy of "an electron" was always a stumbling block for me. I certainly
don't understand it all but this explanation does advance my primitive grasp of the subject.


jgourlay
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Wow....okay. Serves me right for asking impertinent questions!


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