| Topic: |
Science > Physics |
| User: |
"Professor Gauss" |
| Date: |
28 Aug 2004 09:13:16 PM |
| Object: |
Representing Bound Electrons as Standing Waves |
Your comments are welcome on the following, which is how I tend to think
about bound electrons. Please let me know if you agree or disagree with the
following explanation. Even though I think it is consistent with modern
physics, I want to find out if there are any major flaws before I share
these thoughts with my students:
Bound electrons in an atom were originally thought of as being in orbit
around the nucleus, but we soon realized that the continual acceleration
would radiate away energy, causing all of the electrons in an atom to
collapse into the nucleus within a fraction of a second, which obviously
contradicts our observation that matter tends to be much more stable than
that (!). So, instead, the Schroedinger wave equation treats electrons as
standing waves: there is a whole number of wavelengths in each "orbit,"
where the wavelength is consistent with the requirements of energy
conservation, quantum physics, etc. Now consider a standing wave on a
vibrating string stretched between two fixed supports. Each point on the
string sinusoidally oscillates transverse to the direction of wave
propagation, but the standing wave itself does not move along the string.
In fact, d'Alembert showed that a standing wave can be represented as the
superposition of two identical waves moving in opposite directions along the
string. Thus, the net motion along the direction of wave propagation is
zero. In the same sense, the standing wave representation of a bound
electron has zero net motion; hence, the bound electrons do not accelerate,
and stable atoms fit into our theoretical framework.
Thank you.
--
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand. -- Ancient Chinese proverb
(Remove the caps from my address if you are sending a reply.)
.
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| User: "Edward Green" |
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| Title: Re: Representing Bound Electrons as Standing Waves |
29 Aug 2004 07:42:06 AM |
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"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message news:<YtCdncvGt_NdpqzcRVn-rA@comcast.com>...
Your comments are welcome on the following, which is how I tend to think
about bound electrons. Please let me know if you agree or disagree with the
following explanation. Even though I think it is consistent with modern
physics, I want to find out if there are any major flaws before I share
these thoughts with my students:
Bound electrons in an atom were originally thought of as being in orbit
around the nucleus, but we soon realized that the continual acceleration
would radiate away energy, causing all of the electrons in an atom to
collapse into the nucleus within a fraction of a second, which obviously
contradicts our observation that matter tends to be much more stable than
that (!). So, instead, the Schroedinger wave equation treats electrons as
standing waves: there is a whole number of wavelengths in each "orbit,"
where the wavelength is consistent with the requirements of energy
conservation, quantum physics, etc.
It's all "quantum physics", man.
This whole number of wavelengths business is the Old Quantum Theory,
associated with the name of Bohr. The Schroedinger equation
essentially finds solutions the same way, but with more subtle
requirments for adjustment of the wave solution to the boundary
conditions, and the initial agreement of the simple argument in
finding some energy levels is regarded as largely coincidence or a
lucky guess which happened to work in a special case.
Now consider a standing wave on a
vibrating string stretched between two fixed supports. Each point on the
string sinusoidally oscillates transverse to the direction of wave
propagation, but the standing wave itself does not move along the string.
In fact, d'Alembert showed that a standing wave can be represented as the
superposition of two identical waves moving in opposite directions along the
string. Thus, the net motion along the direction of wave propagation is
zero. In the same sense, the standing wave representation of a bound
electron has zero net motion; hence, the bound electrons do not accelerate,
and stable atoms fit into our theoretical framework.
I think that's Ok.
One comment: solutions can be written in which the single electron
wavefunction _indeed_ shows traveling wave properties around the
center of the potential. If this were not so, there would be no
orbitals with net angular momentum.
.
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| User: "Professor Gauss" |
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| Title: Re: Representing Bound Electrons as Standing Waves |
29 Aug 2004 08:46:50 AM |
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"Edward Green" <spamspamspam3@netzero.com> wrote in message
news:eca320d0.0408290442.700f03de@posting.google.com...
"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message
news:<YtCdncvGt_NdpqzcRVn-rA@comcast.com>...
Your comments are welcome on the following, which is how I tend to think
about bound electrons. Please let me know if you agree or disagree with
the
following explanation. Even though I think it is consistent with modern
physics, I want to find out if there are any major flaws before I share
these thoughts with my students:
Bound electrons in an atom were originally thought of as being in orbit
around the nucleus, but we soon realized that the continual acceleration
would radiate away energy, causing all of the electrons in an atom to
collapse into the nucleus within a fraction of a second, which obviously
contradicts our observation that matter tends to be much more stable than
that (!). So, instead, the Schroedinger wave equation treats electrons
as
standing waves: there is a whole number of wavelengths in each "orbit,"
where the wavelength is consistent with the requirements of energy
conservation, quantum physics, etc.
It's all "quantum physics", man.
This whole number of wavelengths business is the Old Quantum Theory,
associated with the name of Bohr. The Schroedinger equation
essentially finds solutions the same way, but with more subtle
requirments for adjustment of the wave solution to the boundary
conditions, and the initial agreement of the simple argument in
finding some energy levels is regarded as largely coincidence or a
lucky guess which happened to work in a special case.
Now consider a standing wave on a
vibrating string stretched between two fixed supports. Each point on the
string sinusoidally oscillates transverse to the direction of wave
propagation, but the standing wave itself does not move along the string.
In fact, d'Alembert showed that a standing wave can be represented as the
superposition of two identical waves moving in opposite directions along
the
string. Thus, the net motion along the direction of wave propagation is
zero. In the same sense, the standing wave representation of a bound
electron has zero net motion; hence, the bound electrons do not
accelerate,
and stable atoms fit into our theoretical framework.
I think that's Ok.
One comment: solutions can be written in which the single electron
wavefunction _indeed_ shows traveling wave properties around the
center of the potential. If this were not so, there would be no
orbitals with net angular momentum.
OK, so now let's say that the old notion of an electron shell has been
replaced with a model that looks something like a solid hoop consisting of a
ring of standing waves centered on the nucleus. According to your statement
above, this hoop may or may not be rotating in its own plane, corresponding
to various amounts of angular momentum. Angular momentum (if present) would
generate a magnetic field perpendicular to the plane of the hoop.
Furthermore, the interaction of this magnetic field with other magnetic
fields might, under some circumstances (asymmetry), cause the hoop to
precess. This might correspond to the magnetic moment quantum number "m,"
while the radius and angular momentum might might correspond to the quanta
"n" and "l," respectively. To complete the picture, the spin quantum number
"s" would then correspond to whether the electron's intrinsic spin is
parallel or antiparallel to the direction of hoop rotation.
--
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand. -- Ancient Chinese proverb
(Remove caps from my address if replying)
.
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| User: "Edward Green" |
|
| Title: Re: Representing Bound Electrons as Standing Waves |
29 Aug 2004 03:56:36 PM |
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"Professor Gauss" <professor_gaussNO@SPAMcomcast.net> wrote in message news:<WfmdnaYBVK78Q6zcRVn-iA@comcast.com>...
OK, so now let's say that the old notion of an electron shell has been
replaced with a model that looks something like a solid hoop consisting of a
ring of standing waves centered on the nucleus.
"Electron shells" are not part of Old Quantum Theory: they are part of
the solution to the hydrogenic atom under the Schroedinger equation,
extended to approximation schemes for multiple electron atoms where
the electrons fill in "shells" corresponding to principal quantum
numbers.
And I definitely _don't_ want to think of a solid hoop, but, as
Gregory mentioned, something more like waves in a three-dimensional
rubber ball.
According to your statement
above, this hoop may or may not be rotating in its own plane, corresponding
to various amounts of angular momentum.
The wave patterns in the rubber ball may or may not look like
traveling waves when considered along circular paths ... your
statement is not too far off, except for the mention of "solid hoops".
The quintessential traveling wave in non-relativistic QM (the kind we
are considering here) is the momentum eigenstate -- a travelling plane
wave. I _think_ if we look at three dimensional solutions to the
hydrogenic along hoops perpendicular to the angular momentum (although
the solution extends through space) we we find something like the
plane wave momentum eigenstate going on, biting its own tail.
Angular momentum (if present) would
generate a magnetic field perpendicular to the plane of the hoop.
Hmm... here you exceed my limited expertise. I share with you what I
am tolerably sure of and readily admit uncertainty. But I think that
is correct: orbitals (solutions to the hydrogenic Schroedinger
equation corresponding to particular choices of quantum numbers)
having non-zero angular momentum have magnetic moments. Something
called "spin-orbit coupling" comes to mind, but do not press me for
details.
Furthermore, the interaction of this magnetic field with other magnetic
fields might, under some circumstances (asymmetry), cause the hoop to
precess.
Well, all of a sudden you are jumping from the most basic kinds of
questions about quantum mechanics to slightly less basic questions! I
believe what you say must be so, but I never went down that path, so
can't serve as your guide.
This might correspond to the magnetic moment quantum number "m,"
while the radius and angular momentum might might correspond to the quanta
"n" and "l," respectively.
Hmm... "n" corresponds to the energy (at least in the absence of
external magnetic field, or any other factor which would break the
degeneracy of all quantum states with given n), and has some relation
to the mean radius, "l" definitely corresponds to angular momentum,
while "m" I would have said corresponds to the component of angular
momentum along a fixed spatial direction. I think you are right, and
measurement of m would involve application of an external field, and
so some notion of precession, but you must appeal to wiser heads than
mine.
Me begins to think you are a kind of Platonic questioner, since you
knowledge suddenly expands, following the most innocent looking of
preliminaries.
To complete the picture, the spin quantum number
"s" would then correspond to whether the electron's intrinsic spin is
parallel or antiparallel to the direction of hoop rotation.
Again I suppose this is fundamentally correct -- given an external
field.
Until recently I would have agreed with your guess that solutions to
the wave function for the electron in a hydrogenic atom correspond to
standing waves. But I would have been mistaken: when we go from one
dimension to two or three -- or at least to the extend of considering
a one dimensional solution biting its tail (a loop) -- we have a
possibility not open in one dimension: a spatially bounded travelling
wave.
Good luck.
.
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| User: "Gregory L. Hansen" |
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| Title: Re: Representing Bound Electrons as Standing Waves |
29 Aug 2004 08:29:42 AM |
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In article <YtCdncvGt_NdpqzcRVn-rA@comcast.com>,
Professor Gauss <professor_gaussNO@SPAMcomcast.net> wrote:
Your comments are welcome on the following, which is how I tend to think
about bound electrons. Please let me know if you agree or disagree with the
following explanation. Even though I think it is consistent with modern
physics, I want to find out if there are any major flaws before I share
these thoughts with my students:
Bound electrons in an atom were originally thought of as being in orbit
around the nucleus, but we soon realized that the continual acceleration
would radiate away energy, causing all of the electrons in an atom to
collapse into the nucleus within a fraction of a second, which obviously
contradicts our observation that matter tends to be much more stable than
that (!). So, instead, the Schroedinger wave equation treats electrons as
standing waves: there is a whole number of wavelengths in each "orbit,"
where the wavelength is consistent with the requirements of energy
conservation, quantum physics, etc. Now consider a standing wave on a
vibrating string stretched between two fixed supports. Each point on the
string sinusoidally oscillates transverse to the direction of wave
propagation, but the standing wave itself does not move along the string.
In fact, d'Alembert showed that a standing wave can be represented as the
superposition of two identical waves moving in opposite directions along the
string. Thus, the net motion along the direction of wave propagation is
zero. In the same sense, the standing wave representation of a bound
electron has zero net motion; hence, the bound electrons do not accelerate,
and stable atoms fit into our theoretical framework.
And why should we not expect a vibrating string to radiate? Isn't it a
line of charge in motion? Pluck a guitar string!
Why does the electron not radiate away all its energy and fall into the
nucleus? The answer is that it has radiated away all its energy. It has
fallen into the nucleus! It has fallen as far as the uncertainty
principle and the Pauli exclusion principle will allow. Quantization is
still due to boundary conditions, to bound waves. But instead of thinking
of a vibrating string circling the nucleus, you should be thinking of
something more like three-dimensional waves in a rubber ball.
--
"Don't try to teach a pig how to sing. You'll waste your time and annoy
the pig."
.
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| User: "Professor Gauss" |
|
| Title: Re: Representing Bound Electrons as Standing Waves |
29 Aug 2004 09:01:51 AM |
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"Gregory L. Hansen" <glhansen@steel.ucs.indiana.edu> wrote in message
news:cgslo6$rs2$2@hood.uits.indiana.edu...
In article <YtCdncvGt_NdpqzcRVn-rA@comcast.com>,
Professor Gauss <professor_gaussNO@SPAMcomcast.net> wrote:
Your comments are welcome on the following, which is how I tend to think
about bound electrons. Please let me know if you agree or disagree with
the
following explanation. Even though I think it is consistent with modern
physics, I want to find out if there are any major flaws before I share
these thoughts with my students:
Bound electrons in an atom were originally thought of as being in orbit
around the nucleus, but we soon realized that the continual acceleration
would radiate away energy, causing all of the electrons in an atom to
collapse into the nucleus within a fraction of a second, which obviously
contradicts our observation that matter tends to be much more stable than
that (!). So, instead, the Schroedinger wave equation treats electrons as
standing waves: there is a whole number of wavelengths in each "orbit,"
where the wavelength is consistent with the requirements of energy
conservation, quantum physics, etc. Now consider a standing wave on a
vibrating string stretched between two fixed supports. Each point on the
string sinusoidally oscillates transverse to the direction of wave
propagation, but the standing wave itself does not move along the string.
In fact, d'Alembert showed that a standing wave can be represented as the
superposition of two identical waves moving in opposite directions along
the
string. Thus, the net motion along the direction of wave propagation is
zero. In the same sense, the standing wave representation of a bound
electron has zero net motion; hence, the bound electrons do not
accelerate,
and stable atoms fit into our theoretical framework.
And why should we not expect a vibrating string to radiate? Isn't it a
line of charge in motion? Pluck a guitar string!
No, a guitar string is a line of neutral molecules! In my analogy, the
waves of string motion correspond to the wavefunction (<psi>) of an
electron.
Why does the electron not radiate away all its energy and fall into the
nucleus? The answer is that it has radiated away all its energy. It has
fallen into the nucleus! It has fallen as far as the uncertainty
principle and the Pauli exclusion principle will allow. Quantization is
still due to boundary conditions, to bound waves. But instead of thinking
of a vibrating string circling the nucleus, you should be thinking of
something more like three-dimensional waves in a rubber ball.
--
"Don't try to teach a pig how to sing. You'll waste your time and annoy
the pig."
--
~~~~~~~~~~~
Professor Gauss
~~~~~~~~~~~
To hear is to forget,
To see is to remember,
To do is to understand. -- Ancient Chinese proverb
(Remove caps from my address if replying)
.
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| User: "Gregory L. Hansen" |
|
| Title: Re: Representing Bound Electrons as Standing Waves |
29 Aug 2004 06:37:16 PM |
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|
In article <H8OdnbvTp7JOfKzcRVn-vw@comcast.com>,
Professor Gauss <professor_gaussNO@SPAMcomcast.net> wrote:
"Gregory L. Hansen" <glhansen@steel.ucs.indiana.edu> wrote in message
news:cgslo6$rs2$2@hood.uits.indiana.edu...
In article <YtCdncvGt_NdpqzcRVn-rA@comcast.com>,
Professor Gauss <professor_gaussNO@SPAMcomcast.net> wrote:
Your comments are welcome on the following, which is how I tend to think
about bound electrons. Please let me know if you agree or disagree with
the
following explanation. Even though I think it is consistent with modern
physics, I want to find out if there are any major flaws before I share
these thoughts with my students:
Bound electrons in an atom were originally thought of as being in orbit
around the nucleus, but we soon realized that the continual acceleration
would radiate away energy, causing all of the electrons in an atom to
collapse into the nucleus within a fraction of a second, which obviously
contradicts our observation that matter tends to be much more stable than
that (!). So, instead, the Schroedinger wave equation treats electrons as
standing waves: there is a whole number of wavelengths in each "orbit,"
where the wavelength is consistent with the requirements of energy
conservation, quantum physics, etc. Now consider a standing wave on a
vibrating string stretched between two fixed supports. Each point on the
string sinusoidally oscillates transverse to the direction of wave
propagation, but the standing wave itself does not move along the string.
In fact, d'Alembert showed that a standing wave can be represented as the
superposition of two identical waves moving in opposite directions along
the
string. Thus, the net motion along the direction of wave propagation is
zero. In the same sense, the standing wave representation of a bound
electron has zero net motion; hence, the bound electrons do not
accelerate,
and stable atoms fit into our theoretical framework.
And why should we not expect a vibrating string to radiate? Isn't it a
line of charge in motion? Pluck a guitar string!
No, a guitar string is a line of neutral molecules! In my analogy, the
waves of string motion correspond to the wavefunction (<psi>) of an
electron.
So you have a dancing line of charge. That will radiate, unless you
invoke the aforementioned limitations of the uncertainty and exclusion
principles from a theory which doesn't do those lines of charge, or follow
Bohr's example in declaring certain states will radiate and others won't,
just because.
But there's a difference between the standing waves on a string, and a
wavefunction in quantum mechanics. A standing wave on a string makes the
string move. But what do you get for an electron density in an infinite
potential well? psi(x) ~ sin(pi*x/L) in the ground state. The time
dependence is given by operating with exp(-iHt/hbar), giving
psi(x,t) ~ exp(-iEt/hbar) sin(pi*x/L)
The electron density is
|psi(x,t)|^2 ~ sin^2 (pi*x/L)
What's moving? The phase. There is no physical movement, the particle
density is static. Same as a plane wave -- take the square magnitude and
you get a uniform and static density. Unlike classical field theory, we
don't take Re[exp(i(kx-wt))], the complex phase is taken seriously. You
might imagine the plane wave being composed of a field of little arrows
that spin around in the real and imaginary directions, if there was an
imaginary direction for the arrow to point.
--
"Don't try to teach a pig how to sing. You'll waste your time and annoy
the pig."
.
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