Why do electrons spin?
Peter Jordan
>>Empty space ... a chain link fence is also mostly made up of empty space.
>>Can you have a field without matter ?
>>
>Yes.
Show me one.
>>what is matter if it is not structure,
>>can you represent information without some substance (matter).
>Does a transmission of electromagnetic energy in the emptyness of
>space (a pure vacuum) convey information? Yes.
Can you show me information that does not have mass ?
Electromagnetic energy is substance.
It has mass. Something like E=mc^2
Remember I said F=ma and W=F.d
so Joules, the unit of energy is equivalent to kg.m^2/s^2
If you look at E=mc^2 then you get the same units .....
c is the velocity of the speed of light, m is mass so we get (kg)(m/s)(m/s)
= kgm^2/s^2
The transmission of electromagnetic energy is equivalent to photons
having a mass speeding though space.
Light can be described in terms of electromagnetic wave propagation or
in terms of particles with mass.
And as far as spinning electrons go, it is not meaningful to talk about
the radius of an electron for that would require you to pinpoint an elctrons
position in space, which completely changes an electrons momentum.
Electrons exist amongst nuclei in certain integer multiple stable
energy levels. That is, their is a reasonable probability of finding
an electron inside a nucleus, or near it, but at a certain energy level
(orbital height) the probability of finding that electron will be zero
yet you can find the same electron at a higher energy level with
a reasonable probability.
If it can't have the intermediate energy levels, how does it get
from one to the other by orbiting ?
It doesn't.
Hmmm, if this ain't made up, what happens to the mass of an elctron when
it emits or absorbs a photon ?
Did I miss my own point ?
PeterJordan
Lloyd R. Parker
: ca...@winternet.com (caMel) writes:
: >>Empty space ... a chain link fence is also mostly made up of empty space.
: >>Can you have a field without matter ?
: >>
: >Yes.
: Show me one.
: >>what is matter if it is not structure,
: >>can you represent information without some substance (matter).
: >Does a transmission of electromagnetic energy in the emptyness of
: >space (a pure vacuum) convey information? Yes.
: Can you show me information that does not have mass ?
: Electromagnetic energy is substance.
: It has mass. Something like E=mc^2
No, it has a rest mass equivalent in energy, but a photon does not have mass.
: Remember I said F=ma and W=F.d
: so Joules, the unit of energy is equivalent to kg.m^2/s^2
: If you look at E=mc^2 then you get the same units .....
: c is the velocity of the speed of light, m is mass so we get (kg)(m/s)(m/s)
: = kgm^2/s^2
: The transmission of electromagnetic energy is equivalent to photons
: having a mass speeding though space.
: Light can be described in terms of electromagnetic wave propagation or
: in terms of particles with mass.
Particles yes; mass, no.
: And as far as spinning electrons go, it is not meaningful to talk about
: the radius of an electron for that would require you to pinpoint an elctrons
: position in space, which completely changes an electrons momentum.
: Electrons exist amongst nuclei in certain integer multiple stable
: energy levels. That is, their is a reasonable probability of finding
: an electron inside a nucleus,
The probability of finding the electron inside a nucleus is zero.
: or near it, but at a certain energy level
: (orbital height) the probability of finding that electron will be zero
: yet you can find the same electron at a higher energy level with
: a reasonable probability.
: If it can't have the intermediate energy levels, how does it get
: from one to the other by orbiting ?
: It doesn't.
: Hmmm, if this ain't made up, what happens to the mass of an elctron when
: it emits or absorbs a photon ?
It gains mass when it absorbs one and loses mass when it emits one. But
the mass is of the electron is around 100,000,000 (10^8) times the mass
change due to the photon energy. Essentially negligible.
Martin Ystenes
>Can you show me information that does not have mass ?
Any energy corresponds to mass, so we must find information that
does not contribute to energy.
How about the the figures you make when putting your hands in front
of a light projector? OK, they may have a negative mass, beeing an
area of reduces energy. Can you make a negative mass swan?
Any kind of modulation can in principle be made without any
net contribution to the energy. Hence you can make e.g. modulation
of a radio wave and thereby generate information that has no mass.
Martin Ystenes
Peter Jordan
>Martin Ystenes
>
That information still has to come from somewhere.
Modulations in amplitude must represent changes in transmitted energy
with time . . . since we can observe contrasts ...
this post contains modulated bytes of memory which required considerable
glucose expenditure to include 'meaningful' information ...as opposed
to a huge string of zeroes ....so ..
Negative swans sure, but 'is information conserved ?'
PeterJordan
(hey my accounts still active)
--
http://ugweb.cs.ualberta.ca/~pjordan/
Peter Jordan
>: Can you show me information that does not have mass ?
>: Electromagnetic energy is substance.
>: It has mass. Something like E=mc^2
>No, it has a rest mass equivalent in energy, but a photon does not have mass.
Ok, I dipped into the text-books and saw the
_________
E\/1-v^2/c^2 = mc^2
used to explain that since E must be finite for the photon and v squared
over c squared in a vacuum is equal to 1, mass must be zero.
Now I'm lost.
What about solar wind and space sails, is that due to electromagnetic
radiation or to particles ... and how do you know which is which when ..
So now I suppose I'm supposed to go read that stuff about diffraction and
etcetera ...
why do we bother ?
Is it because the economics of nuclear bombs means there is a research
grant to feed someone somewhere ?
(huh)
>Particles yes; mass, no.
>The probability of finding the electron inside a nucleus is zero.
Is that because the inside of the nucleus can't be seen ?
>: Hmmm, if this ain't made up, what happens to the mass of an elctron when
>: it emits or absorbs a photon ?
>It gains mass when it absorbs one and loses mass when it emits one. But
>the mass is of the electron is around 100,000,000 (10^8) times the mass
>change due to the photon energy. Essentially negligible.
This book has a chart inside the cover ...
"electron rest mass 9.11EE-31 kg = 0.511 MeV/c^2 (whatever that means)"
Are electrons supposed to become infinite mass with speeds approaching
speed of light ?
Did I just differentiate between a particle and a photon (quantum packet?)?
Anybody recomend any books .. hard math is ok provided I can appreciate
the application.
PeterJordan
--
http://ugweb.cs.ualberta.ca/~pjordan/
DMurphy3
Jordan's comment on em radiation and mass:
: Electromagnetic energy is substance.
: It has mass. Something like E=mc^2
>No, it has a rest mass equivalent in energy, but a photon does not have
>mass.
I think you have this backwards. A photon has no rest mass, but does have
relativistic mass. This mass, slight though it may be, may be calculated
*if* the momentum is known.
: Light can be described in terms of electromagnetic wave propagation or
: in terms of particles with mass.
>Particles yes; mass, no.
Same argument. Besides, how can you describe a particle without reference
to mass? And how can a particle with no mass impart momentum to a particle
(such as an electron) that has mass? In fact, if the photon is of
sufficient energy, it may completely knock an electron from its orbital
(ionizing radiation) imparting the electron with kinetic energy. This very
definitely sounds as though mass were involved.
>The probability of finding the electron inside a nucleus is zero.
If there is a node preventing the probability, yes, the probability is
zero. For an electron in the first quantum shell, which is essentially the
spherically symmetric 1s orbital, there is a very definite probability of
finding the electron at the nucleus.
: Hmmm, if this ain't made up, what happens to the mass of an elctron when
: it emits or absorbs a photon ?
>It gains mass when it absorbs one and loses mass when it emits one. But
>the mass is of the electron is around 100,000,000 (10^8) times the mass
>change due to the photon energy. Essentially negligible.
Back to your previous argument that a photon has no mass: From where do
you gain said mass? Sure, mass and energy are interconvertible, but that
wasn't your original premise.
Tonek Jansen
> (orbital height) the probability of finding that electron will be zero
> yet you can find the same electron at a higher energy level with
> a reasonable probability.
> If it can't have the intermediate energy levels, how does it get
> from one to the other by orbiting ?
> It doesn't.
Electrons do not have to be in well-defined energy levels. A hydrogen atom, for
example, can be in a state with an energy (=expectation value) which can be
anything greater than or equal to the ground state energy. However, if you
measure the energy of the hydrogen atom you will find only bound states with
discrete energy levels. The states with energies in between are superpositions of
states corresponding to these discrete levels.
Tonek
tgt...@chem.tue.nl
Lloyd R. Parker
: Lloyd Parker in article <47vv2r$l...@larry.cc.emory.edu> writes in reply to
: Jordan's comment on em radiation and mass:
: : Electromagnetic energy is substance.
: : It has mass. Something like E=mc^2
: >No, it has a rest mass equivalent in energy, but a photon does not have
: >mass.
: I think you have this backwards. A photon has no rest mass, but does have
: relativistic mass. This mass, slight though it may be, may be calculated
: *if* the momentum is known.
: : Light can be described in terms of electromagnetic wave propagation or
: : in terms of particles with mass.
: >Particles yes; mass, no.
: Same argument. Besides, how can you describe a particle without reference
: to mass? And how can a particle with no mass impart momentum to a particle
: (such as an electron) that has mass?
There's nothing that says a particle must have mass. Or that you must
have mass to have momentum.
: In fact, if the photon is of
: sufficient energy, it may completely knock an electron from its orbital
: (ionizing radiation) imparting the electron with kinetic energy. This very
: definitely sounds as though mass were involved.
No, just energy and momentum.
: >The probability of finding the electron inside a nucleus is zero.
: If there is a node preventing the probability, yes, the probability is
: zero. For an electron in the first quantum shell, which is essentially the
: spherically symmetric 1s orbital, there is a very definite probability of
: finding the electron at the nucleus.
No, because the volume there is zero. You must plot psi^2 times r
(distance from nucleus).
: : Hmmm, if this ain't made up, what happens to the mass of an elctron when
: : it emits or absorbs a photon ?
: >It gains mass when it absorbs one and loses mass when it emits one. But
: >the mass is of the electron is around 100,000,000 (10^8) times the mass
: >change due to the photon energy. Essentially negligible.
: Back to your previous argument that a photon has no mass: From where do
: you gain said mass? Sure, mass and energy are interconvertible, but that
: wasn't your original premise.
When a particle emits energy, the energy comes from some of the mass of
the particle being converted to energy. Except in a nuclear reaction,
the decrease in mass is insignificant. Similarly, when a particle
absorbs energy, the energy is converted into extra mass of the particle.
The photon had no mass at any time.
Lloyd R. Parker
Organization: Emory University
Distribution:
Peter Jordan (pjo...@gpu4.srv.ualberta.ca) wrote:
: lpa...@larry.cc.emory.edu (Lloyd R. Parker) writes:
: >: Can you show me information that does not have mass ?
: >: Electromagnetic energy is substance.
: >: It has mass. Something like E=mc^2
: >No, it has a rest mass equivalent in energy, but a photon does not have mass.
: Ok, I dipped into the text-books and saw the
: _________
: E\/1-v^2/c^2 = mc^2
: used to explain that since E must be finite for the photon and v squared
: over c squared in a vacuum is equal to 1, mass must be zero.
: Now I'm lost.
: What about solar wind and space sails, is that due to electromagnetic
: radiation or to particles ... and how do you know which is which when ..
: So now I suppose I'm supposed to go read that stuff about diffraction and
: etcetera ...
Photons DO have momentum; they just don't have mass.
: >Particles yes; mass, no.
: >The probability of finding the electron inside a nucleus is zero.
: Is that because the inside of the nucleus can't be seen ?
No, it's because there is zero volume for electrons there.
: >: Hmmm, if this ain't made up, what happens to the mass of an elctron when
: >: it emits or absorbs a photon ?
: >It gains mass when it absorbs one and loses mass when it emits one. But
: >the mass is of the electron is around 100,000,000 (10^8) times the mass
: >change due to the photon energy. Essentially negligible.
: This book has a chart inside the cover ...
: "electron rest mass 9.11EE-31 kg = 0.511 MeV/c^2 (whatever that means)"
That's the energy equivalent to the mass, from e = mc^2.
: Are electrons supposed to become infinite mass with speeds approaching
: speed of light ?
Yes, everything does.
John Milligan
dmur...@aol.com (DMurphy3) wrote:
>>The probability of finding the electron inside a nucleus is zero.
>
>If there is a node preventing the probability, yes, the probability is
>zero. For an electron in the first quantum shell, which is essentially
the
>spherically symmetric 1s orbital, there is a very definite probability
of
>finding the electron at the nucleus.
This is incorrect. The probability of finding the electron at a
particular distance, r, from the nucleus is given by r^2*Psi*2, where
Psi is the wavefunction. Since at the nucleus, r=0 the probablity of
finding the electron at the nucleus is zero. You should stop looking
things up in your general chemistry textbook, most of which are wrong on
this point.
Jeff E. Janes
: In article <484fnj$i...@newsbf02.news.aol.com>,
: dmur...@aol.com (DMurphy3) wrote:
:
:>
:>If there is a node preventing the probability, yes, the probability is
:>zero. For an electron in the first quantum shell, which is essentially
:the
:>spherically symmetric 1s orbital, there is a very definite probability
:of
:>finding the electron at the nucleus.
:
: This is incorrect. The probability of finding the electron at a
: particular distance, r, from the nucleus is given by r^2*Psi*2, where
: Psi is the wavefunction. Since at the nucleus, r=0 the probablity of
: finding the electron at the nucleus is zero. You should stop looking
: things up in your general chemistry textbook, most of which are wrong on
: this point.
OK, I will look it up in my Physical chemistry book, Joe Noggle's
_Physical Chemistry_ 2nd edition.
page 740:
"The proton, like the electron, has an intrinsic angular momentum (spin)
with a quantum number I=1/2. An electron in an s state has a *finite
probability of being at the site of the nucleus*; this causes the Fermi
contact interaction." *emphasis* mine. This Fermi contact interaction
is also known as hyperfine coupling. This is the same coupling used
to define the second.
Do you have a more impressive reference than Noggle? Perhaps you should
post your CV so we can judge who you are to scorn our general chemistry
textbooks.
The equation you use gives the probability limit of the electron being
in a spherical shell of radius r and thickness dr, divided by dr, as dr
approaches zero. If you instead look at the probability limit on a per
volume basis for the s orbitals, you will notice the electron has a
greater probability of being at the nucleus (i.e. 0 + dr as dr
approaches zero) than at any other region whose volume is similarly
vanishing.
--
Jeff Janes jej...@mtu.edu at Michigan Technological University
Second Law of Thermodynamics:
The enormousfees of the university is ever increasing.
Andy Felton
Parker) wrote:
>There's nothing that says a particle must have mass. Or that you must
>have mass to have momentum.
>
>: In fact, if the photon is of
>: sufficient energy, it may completely knock an electron from its orbital
>: (ionizing radiation) imparting the electron with kinetic energy. This very
>: definitely sounds as though mass were involved.
>
>No, just energy and momentum.
>
>: Back to your previous argument that a photon has no mass: From where do
>: you gain said mass? Sure, mass and energy are interconvertible, but that
>: wasn't your original premise.
>
Hold on a minute... is there experimential proof that a photon is
massless? Are you _sure_ it doesn't have mass, or we just can't
measure it with current equipment?
Weird L. Marcowitz
Felton) wrote:
[a lot erased]
> Hold on a minute... is there experimential proof that a photon is
> massless? Are you _sure_ it doesn't have mass, or we just can't
> measure it with current equipment?
Just read Einstein's theory on relativity. Maybe this doesn't provide any
experimental proof, but photons DO have a mass. Unfortunately they travel
at light speed, which means that one can not tell the difference between
energy and mass. Anyway, I believe there are some books on this topic.
Marco
--
"Real knowledge is to know the extent of one's ignorance"
Confucius
Lloyd R. Parker
: page 740:
Huh? My Daniels and Alberty shows the same probability distributions
I've always seen -- 0 at the nucleus. The maximum is reached a short
distance from the nucleus. And 0+dr is not AT the nucleus -- 0 is.
Lloyd R. Parker
: On 13 Nov 1995 15:10:03 -0500, lpa...@larry.cc.emory.edu (Lloyd R.
: Parker) wrote:
: >There's nothing that says a particle must have mass. Or that you must
: >have mass to have momentum.
: >
: >: In fact, if the photon is of
: >: sufficient energy, it may completely knock an electron from its orbital
: >: (ionizing radiation) imparting the electron with kinetic energy. This very
: >: definitely sounds as though mass were involved.
: >
: >No, just energy and momentum.
: >
: >: Back to your previous argument that a photon has no mass: From where do
: >: you gain said mass? Sure, mass and energy are interconvertible, but that
: >: wasn't your original premise.
: >
: Hold on a minute... is there experimential proof that a photon is
: massless? Are you _sure_ it doesn't have mass, or we just can't
: measure it with current equipment?
Only luminons can move at c. Luminons don't have mass; only leptons and
hadrons do. If luminons had mass, the mass would have to be infinite
since they're at the speed of light.
Lloyd R. Parker
: In article <48blre$m...@ixnews4.ix.netcom.com>, afe...@ix.netcom.com (Andy
: Felton) wrote:
: [a lot erased]
: > Hold on a minute... is there experimential proof that a photon is
: > massless? Are you _sure_ it doesn't have mass, or we just can't
: > measure it with current equipment?
: Just read Einstein's theory on relativity. Maybe this doesn't provide any
: experimental proof, but photons DO have a mass. Unfortunately they travel
: at light speed, which means that one can not tell the difference between
: energy and mass. Anyway, I believe there are some books on this topic.
: Marco
: --
: "Real knowledge is to know the extent of one's ignorance"
: Confucius
They do NOT have a mass. You can calculate the mass they would have if
they were at rest from e=mc^2. But a particle with mass cannot travel
at c, and that does come from relativity.
Robert Parson
John Milligan <alch...@pacificnet.net> wrote:
>This is incorrect. The probability of finding the electron at a
>particular distance, r, from the nucleus is given by r^2*Psi*2, where
>Psi is the wavefunction. Since at the nucleus, r=0 the probablity of
>finding the electron at the nucleus is zero. You should stop looking
>things up in your general chemistry textbook, most of which are wrong on
>this point.
The radius of a nucleus, while small (~10^-5 the radius of an atom), is not
zero. Finite nuclear volume effects must be taken into account in
high-precision atomic physics. See Cohen-Tannoudji et al, _Quantum
Mechanics_, Volume II, pp, 1141-1144.
In muonic atoms (where a negative muon replaces an electron) nuclear
volume effects can be quite large, since the Bohr radius of a muon
is (because of its mass) much smaller than that of an electron.
The radius of a Pb nucleus is 8.5x10^-15 m, the Bohr radius of a
muon bound to a Pb+ is 3x10^-15 m.
Hyperfine structure in EPR spectra is, in part, due to the "contact"
interaction between the electron and the nucleus. For simplicity the
nucleus is usually treated as a point in this calculation (this then
requires that the strength of the interaction be infinite, i.e. the
interaction is treated as a delta-function potential.) More realistic
models use a finite nucleus; there's a paper on this by S. M. Blinder
in _Phys. Rev. A_, late 1970's I think.
------
Robert
J. Spashett
'"The proton, like the electron, has an intrinsic angular momentum (spin)
'probability of being at the site of the nucleus*; this causes the Fermi
'contact interaction." *emphasis* mine. This Fermi contact interaction
'is also known as hyperfine coupling. This is the same coupling used
'to define the second.
I can second this. it seems counter intuitive but as I
understand it this is how hyperfine splitting arises. I'm no expert in
the field however. If you can explain to me hyperfine splitting another
way I'd love to here it cos I hate being taught something wrong.
-John
Simon Hogg
>Jeff E. Janes (jej...@mtu.edu) wrote:
>: John Milligan (alch...@pacificnet.net) wrote:
>: : In article <484fnj$i...@newsbf02.news.aol.com>,
>: : dmur...@aol.com (DMurphy3) wrote:
>: :
>: :>
>: :>If there is a node preventing the probability, yes, the probability is
>: :>zero. For an electron in the first quantum shell, which is essentially
>: :the
>: :of
>: :>finding the electron at the nucleus.
>: :
>: : This is incorrect. The probability of finding the electron at a
>: : Psi is the wavefunction. Since at the nucleus, r=0 the probablity of
>: : finding the electron at the nucleus is zero. You should stop looking
>: : things up in your general chemistry textbook, most of which are wrong on
>: : this point.
>
>: _Physical Chemistry_ 2nd edition.
>
>: page 740:
>
>: probability of being at the site of the nucleus*; this causes the Fermi
>: contact interaction." *emphasis* mine. This Fermi contact interaction
>: is also known as hyperfine coupling. This is the same coupling used
>: to define the second.
>
>: post your CV so we can judge who you are to scorn our general chemistry
>: textbooks.
>
>: The equation you use gives the probability limit of the electron being
>: in a spherical shell of radius r and thickness dr, divided by dr, as dr
>: approaches zero. If you instead look at the probability limit on a per
>: volume basis for the s orbitals, you will notice the electron has a
>: greater probability of being at the nucleus (i.e. 0 + dr as dr
>: approaches zero) than at any other region whose volume is similarly
>: vanishing.
>
>: --
>: Jeff Janes jej...@mtu.edu at Michigan Technological University
>: Second Law of Thermodynamics:
>: The enormousfees of the university is ever increasing.
>
>Huh? My Daniels and Alberty shows the same probability distributions
>I've always seen -- 0 at the nucleus. The maximum is reached a short
>distance from the nucleus. And 0+dr is not AT the nucleus -- 0 is.
Yes but the nucleus is NOT an infinitessimal point, it has volume, so th
elctron may occur within the nuclues volume (which is diffuse anyway)
--
Simon
Simon Hogg
>: On 13 Nov 1995 15:10:03 -0500, lpa...@larry.cc.emory.edu (Lloyd R.
>: Parker) wrote:
>
>: >There's nothing that says a particle must have mass. Or that you must
>: >have mass to have momentum.
>: >
>: >: In fact, if the photon is of
>: >: sufficient energy, it may completely knock an electron from its orbital
>: >: (ionizing radiation) imparting the electron with kinetic energy. This very
>: >: definitely sounds as though mass were involved.
>: >
>: >No, just energy and momentum.
>: >
>: >: Back to your previous argument that a photon has no mass: From where do
>: >: you gain said mass? Sure, mass and energy are interconvertible, but that
>: >: wasn't your original premise.
>: >
>: massless? Are you _sure_ it doesn't have mass, or we just can't
>: measure it with current equipment?
>
>hadrons do. If luminons had mass, the mass would have to be infinite
>since they're at the speed of light.
Sorry, me again.
I thought that photons had mass, but they are 'born' travelling at c, they do
not accelerate to get to c. Relativity only prohibits acceleration to c, but
it does allow spontaneous creation travelling at c
-- Simon
Simon Hogg
>Andy Felton (afe...@ix.netcom.com) wrote:
>: On 13 Nov 1995 15:10:03 -0500, lpa...@larry.cc.emory.edu (Lloyd R.
>: Parker) wrote:
>
>: >There's nothing that says a particle must have mass. Or that you must
>: >have mass to have momentum.
>: >
>: >: In fact, if the photon is of
>: >: sufficient energy, it may completely knock an electron from its orbital
>: >: (ionizing radiation) imparting the electron with kinetic energy. This very
>: >: definitely sounds as though mass were involved.
>: >
>: >No, just energy and momentum.
>: >
>: >: Back to your previous argument that a photon has no mass: From where do
>: >: you gain said mass? Sure, mass and energy are interconvertible, but that
>: >: wasn't your original premise.
>: >
>: Hold on a minute... is there experimential proof that a photon is
>: massless? Are you _sure_ it doesn't have mass, or we just can't
>: measure it with current equipment?
>
>Only luminons can move at c. Luminons don't have mass; only leptons and
>hadrons do. If luminons had mass, the mass would have to be infinite
>since they're at the speed of light.
Sorry, me again,
Jeff E. Janes
:
: : page 740:
:
: : "The proton, like the electron, has an intrinsic angular momentum (spin)
: : with a quantum number I=1/2. An electron in an s state has a *finite
: : probability of being at the site of the nucleus*; this causes the Fermi
: : contact interaction." *emphasis* mine. This Fermi contact interaction
: : is also known as hyperfine coupling. This is the same coupling used
: : to define the second.
:
: : in a spherical shell of radius r and thickness dr, divided by dr, as dr
: : approaches zero. If you instead look at the probability limit on a per
: : volume basis for the s orbitals, you will notice the electron has a
: : greater probability of being at the nucleus (i.e. 0 + dr as dr
: : approaches zero) than at any other region whose volume is similarly
: : vanishing.
:
: distance from the nucleus. And 0+dr is not AT the nucleus -- 0 is.
As other people have pointed out, the nucleus does have a radius, no matter
how small, lets call it Rn. Now, by the magic of calculus, I can make dr
smaller than Rn. In fact, for any finite Rn you give me, I can make dr a
million times smaller, or 10^(10^10) times smaller, or any other amount.
What Daniels and Alberty did you look in? I found the 1966 edition, and
on page 451 (fig 12.9) they give the wavefunction of electrons in various
orbitals. If you square it, you will find the probability (per volume) of
an electron being at a certain radius, and the highest probability is at
zero. If you turn the page, you will find radial distribution, which
gives the probability on a per linear basis, to be what you are used to.
The likelyhood of finding an electron in a s orbital in a sphere of space
surrounding 0, is greater than finding it in any other convex region of
equal volume. This is true no matter how small the volume is.
Since you seem more familiar with Daniels and Alberty than I, would you
find out what it says about hyperfine coupling and Fermi contact interactions?
Lloyd R. Parker
: What Daniels and Alberty did you look in? I found the 1966 edition, and
: on page 451 (fig 12.9) they give the wavefunction of electrons in various
: orbitals. If you square it, you will find the probability (per volume) of
: an electron being at a certain radius, and the highest probability is at
: zero. If you turn the page, you will find radial distribution, which
: gives the probability on a per linear basis, to be what you are used to.
: The likelyhood of finding an electron in a s orbital in a sphere of space
: surrounding 0, is greater than finding it in any other convex region of
: equal volume. This is true no matter how small the volume is.
I was just reading in another q.m. book and it makes this argument:
If an electron were to be found in the nucleus, its wave length would have
to be approximately the same as that of the proton and nucleus confined to
the same small volume. By de Broglie's equation, this means the momentum
of the electron would have to be the same. Since the mass of the electron
is much less than that of the proton and neutron, the energy of the
electron must be much greater to have the same momentum. Calculation
reveals this energy value to be greater than electrons emited in nuclear
decay; it is doubtful an electron with this much energy could remain in an
atom, much less in the nucleus.
G. Lane
>On 13 Nov 1995 15:10:03 -0500, lpa...@larry.cc.emory.edu (Lloyd R.
>Parker) wrote:
>>There's nothing that says a particle must have mass. Or that you must
>>have mass to have momentum.
>>
>>: In fact, if the photon is of
>>: sufficient energy, it may completely knock an electron from its orbital
>>: (ionizing radiation) imparting the electron with kinetic energy. This very
>>: definitely sounds as though mass were involved.
>>
>>No, just energy and momentum.
>>
>>: Back to your previous argument that a photon has no mass: From where do
>>: you gain said mass? Sure, mass and energy are interconvertible, but that
>>: wasn't your original premise.
>>
>Hold on a minute... is there experimential proof that a photon is
>massless? Are you _sure_ it doesn't have mass, or we just can't
>measure it with current equipment?
Theoretically, no particle with mass can go the speed of light. Light,
depending on how it is observed, is either electromagnetic radiation, or a
stream of photons. Both go the same absolute speed in a "pure" vacuum.
If there experimental "proof" then the mass would be very, very, very, very
small, on the order of ten to the neg. fifty, or less, though theoretically
speaking, (relitivity and all, a huge paradigm developed this century by
Einstein) the absolute speed is light, and nothing with mass can reach or
exceed that speed.
george
Lloyd R. Parker
: lpa...@curly.cc.emory.edu (Lloyd R. Parker) wrote:
: >Jeff E. Janes (jej...@mtu.edu) wrote:
: >: John Milligan (alch...@pacificnet.net) wrote:
: >: : In article <484fnj$i...@newsbf02.news.aol.com>,
: >: : dmur...@aol.com (DMurphy3) wrote:
: >: :
: >: :>
: >: :>If there is a node preventing the probability, yes, the probability is
: >: :>zero. For an electron in the first quantum shell, which is essentially
: >: :the
: >: :>spherically symmetric 1s orbital, there is a very definite probability
: >: :of
: >: :>finding the electron at the nucleus.
: >: :
: >: : This is incorrect. The probability of finding the electron at a
: >: : particular distance, r, from the nucleus is given by r^2*Psi*2, where
: >: : Psi is the wavefunction. Since at the nucleus, r=0 the probablity of
: >: : finding the electron at the nucleus is zero. You should stop looking
: >: : things up in your general chemistry textbook, most of which are wrong on
: >: : this point.
: >
: >: _Physical Chemistry_ 2nd edition.
: >
: >: page 740:
: >
: >: "The proton, like the electron, has an intrinsic angular momentum (spin)
: >: with a quantum number I=1/2. An electron in an s state has a *finite
: >: probability of being at the site of the nucleus*; this causes the Fermi
: >: contact interaction." *emphasis* mine. This Fermi contact interaction
: >: is also known as hyperfine coupling. This is the same coupling used
: >: to define the second.
: >
: >: post your CV so we can judge who you are to scorn our general chemistry
: >: textbooks.
: >
: >: in a spherical shell of radius r and thickness dr, divided by dr, as dr
: >: approaches zero. If you instead look at the probability limit on a per
: >: volume basis for the s orbitals, you will notice the electron has a
: >: greater probability of being at the nucleus (i.e. 0 + dr as dr
: >: approaches zero) than at any other region whose volume is similarly
: >: vanishing.
: >
: >: Jeff Janes jej...@mtu.edu at Michigan Technological University
: >: Second Law of Thermodynamics:
: >: The enormousfees of the university is ever increasing.
: >Huh? My Daniels and Alberty shows the same probability distributions
: >I've always seen -- 0 at the nucleus. The maximum is reached a short
: >distance from the nucleus. And 0+dr is not AT the nucleus -- 0 is.
: Yes but the nucleus is NOT an infinitessimal point, it has volume, so th
: elctron may occur within the nuclues volume (which is diffuse anyway)
That's called electron capture, a form of nuclear decay that leads to a
new element.
: --
: Simon
Simon Hogg
then squaring psi to find the probability ditribution. Since the solution is
spread over all space, we find that any body exists in some probability in any
area of space. So the electron has a probability (non-zero) of being found
(within Heisenberg's uncertainty principle) within the area occupied by the
nucleus. It also has a non-zero probability of being found 200 light years
away from the nucleus. Just as there is a non-zero probability of me being
found sitting next to you as you read this
n'est ce pas?
-- Simon
Jeffrey N Woodford
: page 740:
According to my Quantum Chemistry textbook (Lowe):
|psi|^2 (= psi times psi*) is the measure of the probability per unit
volume for the electron being at various distances from the nucleus
So you would have to plot 4*pi*r^2*|psi|^2 (one volume element times
probability per unit volume) vs. r to get the volume-weighted
probability density, and from this you can calculate the most probable
radius. This plot shows zero probability at r=0. If you just plot
|psi|^2 vs. r, you indeed get a finite (and large) probability at r=0.
But this probability is not volume-weighted. You are comparing apples
and oranges.
I have no CV, I'm just a grad student.
-Jeff
--
Jeffrey N. Woodford || Email: jwoo...@unlgrad1.unl.edu || Physical Chemistry
Homepage: http://wildcat.dementia.org/jeffw/index.html || Graduate Student
"The devils of truth steal the souls of the free" --NIN || (2nd Year) at UN-L
Ernst U. Wallenborn
>
>According to my Quantum Chemistry textbook (Lowe):
>
>|psi|^2 (= psi times psi*) is the measure of the probability per unit
>volume for the electron being at various distances from the nucleus
>
>So you would have to plot 4*pi*r^2*|psi|^2 (one volume element times
>probability per unit volume) vs. r to get the volume-weighted
>probability density, and from this you can calculate the most probable
>radius. This plot shows zero probability at r=0. If you just plot
>|psi|^2 vs. r, you indeed get a finite (and large) probability at r=0.
>But this probability is not volume-weighted. You are comparing apples
>and oranges.
>
>I have no CV, I'm just a grad student.
>
>-Jeff
>--
>Jeffrey N. Woodford || Email: jwoo...@unlgrad1.unl.edu || Physical Chemistry
>Homepage: http://wildcat.dementia.org/jeffw/index.html || Graduate Student
>"The devils of truth steal the souls of the free" --NIN || (2nd Year) at UN-L
Folks, you're mixing things up.
First, "the probability of finding an electron at a point x0" is not
an observable. It's an interpretation. The observable is "electron density"
and its operator is
rho(x0) = delta(x-x0)
delta being the usual dirac delta. So the expectation value of rho is in
Dirac's formalism
<rho(x0)>=<psi(x)|delta(x-x0)|psi(x)> = psi(x0)^2
by the most trivial property of the delta function. Hence, if the wavefunction
has a local maximum at the point x0, the density has a local maximum, too.
Now if you really want an answer in terms of "probability of finding"
you have to integrate over a small volume element
dV = dx dy dz = r^2 sin(theta) dphi dtheta
so that
Integral[psi[x,y,z],{x,xa,xb},{y,ya,yb},{z,za,zb}]/Integral[psi[x,y,z],
{x,-Infty,Infty},{y,-Infty,Infty},{z,-Infty,Infty}]
(the denominator being one in the case of normalized wavefunctions) gives
you a number that you can interpret as "the probability of finding the
electron in the box {[xa,xb],[ya,yb],[za,zb]}".
It is, however, clear that "the probability of finding an electron at
the point x0" (and not in a small but finite volume element around it)
is zero. This is trivially the case for all points (not only local maxima
of the wavefunction), the reason for it simply being that any set containing
just one triplet of real numbers is a subset of RxRxR with Lebesgue measure
zero. Read any book on Masstheorie or probability calculus if you don't
believe me.
But if you want "the probability of finding the electron at a radius r"
you have to do something else. Your volume element in this case is not
dxdydz. "at radius r" should mean "at a radius between ra and rb", and that
means "in a shell around the nucleus with ra<=r<=rb". So
you have to transform psi(x,y,z) to polar coordinates and you have to
integrate over the angular coordinates and over [ra,rb]. That leaves you
with
Integral[psi[r,phi,theta] r^2 Sin[theta],{phi,0,2 Pi},{theta,0,Pi},{r,ra,rb}]
the r^2 comes from the coordinate transformation. Now you immediately see
two things:
- "The probability of finding an electron at radius r from tha nucleus"
is again zero. A sphere is a subset of R^3 with Lebesgue measure zero.
- "The probability of finding an electron between ra and rb goes to zero
as ra goes to zero. The reason for this is just the r^2 above. This
is the famous "radial distribution function".
So we can conclude:
1. The probability of finding a s-electron in a small volume element has
a maximum at the nucleus.
2. The probability of finding a s-electron in a r+dr shell goes to zero for r=0
3. The probability of finding the electron "at a point" is zero everywhere
Hope that clears up the confusion.
---
-ernst wallenborn.
i'm not a bug.
i'm an undocumented feature.
Tonek Jansen
> I was just reading in another q.m. book and it makes this argument:
>
> If an electron were to be found in the nucleus, its wave length would have
> to be approximately the same as that of the proton and nucleus confined to
> the same small volume. By de Broglie's equation, this means the momentum
> of the electron would have to be the same. Since the mass of the electron
> is much less than that of the proton and neutron, the energy of the
> electron must be much greater to have the same momentum. Calculation
> reveals this energy value to be greater than electrons emited in nuclear
> decay; it is doubtful an electron with this much energy could remain in an
> atom, much less in the nucleus.
If you try to confine an electron in the nucleus, it will indeed get a
very large energy. However, if you talk about the probability that
an electron is in the nucleus, you're not confining it. It has also a
probability to be elsewhere. The energy will therefore remain small.
This changes when you want to measure the position of the electron.
When your measurement tells you that the electron is in some small
volume element, then there is a collapse of the wavefunction,
the electron is confined to that volume element, and the energy becomes
large. It doesn't matter if that volume element is the nucleus or
somewhere else.
Tonek
tgt...@chem.tue.nl
Tonek Jansen
> Huh? My Daniels and Alberty shows the same probability distributions
> I've always seen -- 0 at the nucleus. The maximum is reached a short
> distance from the nucleus. And 0+dr is not AT the nucleus -- 0 is.
Are you looking at the probability of finding the electron at distance r
from the nucleus, or at the probability of finding the electron at some
position as a function of r? In the former case your looking at r^2*|psi(r)|^2,
in the latter at |psi(r)|^2. The former is zero at the nucleus, the latter
need not be.
Tonek
tgt...@chem.tue.nl
Lloyd R. Parker
: What we're talking about here is the soln. to the Schrodinger eqn. for a body,
: n'est ce pas?
: -- Simon
Does the term "node" mean anything to you?
Simon Hogg
had a quick glance through and couldn't see it.
Also is there a Tompkins description of the Schrodingers cat experiment (and
the logic behind it?) Alternatively is there a Tompkin-esque discussion of
this anywhere else. I don't just want a 'popular science' book.
-- Simon
Robert Parson
Lloyd R. Parker <lpa...@curly.cc.emory.edu> wrote:
>I was just reading in another q.m. book and it makes this argument:
What book?
>If an electron were to be found in the nucleus, its wave length would have
>to be approximately the same as that of the proton and nucleus confined to
>the same small volume. By de Broglie's equation, this means the momentum
>of the electron would have to be the same.
This is merely an argument against the likelihood of _confining_ an
electron to a nucleus (or to any volume of similar size) and is
irrelevant to the point at issue, which is that an electron has a
small but nonzero probability of being found in a nucleus, and that
this overlap is responsible for the Fermi Contact term in the
hyperfine interaction. I refer you once again to Cohen-Tannoudji et.
al.'s Quantum Mechanics textbook, Volume II, pp. 1141-1148 and 122-57.
------
Robert
K. Murray
>Why do electrons spin?
>
>They don't spin in the physical sense we are used to. It's a math construct
>designed to help us understand their behaviour.
Yeah, but that's what gets me: they don't spin in the physical sense I'm
used to, yet they still have angular momentum. Seems like those electrons
are getting away with something here.
The question remains, why _do_ electrons do whatever it is that they are doing
when they are doing that thing that we call spin? Mr. Tompkins?
----------
Kermit K. Murray
mailto:kmu...@emory.edu
http://tswww.cc.emory.edu/~kmurray/
Per Rosqvist
LRP> : >: definitely sounds as though mass were involved.
LRP> : >
LRP> : >No, just energy and momentum.
LRP> : >
LRP> : >: Back to your previous argument that a photon has no mass: From
LRP> where do
LRP> : >: you gain said mass? Sure, mass and energy are interconvertible, but
LRP> that
LRP> : >: wasn't your original premise.
LRP> : >
LRP> : Hold on a minute... is there experimential proof that a photon is
LRP> : massless? Are you _sure_ it doesn't have mass, or we just can't
LRP> : measure it with current equipment?
LRP> Only luminons can move at c. Luminons don't have mass; only leptons and
LRP>
LRP> hadrons do. If luminons had mass, the mass would have to be infinite
LRP> since they're at the speed of light.
The photon has a mass. When light from the sun comes down to earth during a
solar eclipse it bends around the mass of the moon. A clear sign that
light (and therefore the photon) has a mass..
-Per
Joshua B. Halpern
: Jeff E. Janes (jej...@mtu.edu) wrote:
: : John Milligan (alch...@pacificnet.net) wrote:
: : : In article <484fnj$i...@newsbf02.news.aol.com>,
: : : dmur...@aol.com (DMurphy3) wrote:
Argument about finding the electron at r=0 or in the nucleus
appended below.
Hi,
Well, you're all right, but you are answering different questions.
First, the probability of finding the electron _exactly_ at r=0 is, as
the Jeffs say _exactly_ 0. Second, the probability of finding the
electron inside the nucleus is as John says _finite_ (but real small),
because the radius of the nucleus is _finite_ (a proton's radius is
0.322 Barns - a Barn is 1E-28 m^2)
By the way, if you really want to break your head, why does the electron
have no measurable physical extent? It behaves as a perfect point charge.
Regards
Josh Halpern
: : :>If there is a node preventing the probability, yes, the probability is
: : :>zero. For an electron in the first quantum shell, which is essentially
: : :the
: : :>spherically symmetric 1s orbital, there is a very definite probability
: : :of
: : :>finding the electron at the nucleus.
: : :
: : : This is incorrect. The probability of finding the electron at a
: : : particular distance, r, from the nucleus is given by r^2*Psi*2, where
: : : Psi is the wavefunction. Since at the nucleus, r=0 the probablity of
: : : finding the electron at the nucleus is zero. You should stop looking
: : : things up in your general chemistry textbook, most of which are wrong on
: : : this point.
: : OK, I will look it up in my Physical chemistry book, Joe Noggle's
: : _Physical Chemistry_ 2nd edition.
: : page 740:
: : "The proton, like the electron, has an intrinsic angular momentum (spin)
: : with a quantum number I=1/2. An electron in an s state has a *finite
: : probability of being at the site of the nucleus*; this causes the Fermi
: : contact interaction." *emphasis* mine. This Fermi contact interaction
: : is also known as hyperfine coupling. This is the same coupling used
: : to define the second.
: : Do you have a more impressive reference than Noggle? Perhaps you should
: : post your CV so we can judge who you are to scorn our general chemistry
: : textbooks.
: According to my Quantum Chemistry textbook (Lowe):
John Milligan
>
>OK, I will look it up in my Physical chemistry book, Joe Noggle's
>_Physical Chemistry_ 2nd edition.
>
>page 740:
>
>"The proton, like the electron, has an intrinsic angular momentum
(spin)
>with a quantum number I=1/2. An electron in an s state has a *finite
>probability of being at the site of the nucleus*; this causes the Fermi
>contact interaction." *emphasis* mine. This Fermi contact interaction
>is also known as hyperfine coupling. This is the same coupling used
>to define the second.
>
>Do you have a more impressive reference than Noggle? Perhaps you
should
>post your CV so we can judge who you are to scorn our general chemistry
>textbooks.
I don't need to prove my credentials to you or anyone else, but since
you asked, I have a Ph.D. in physical chemistry from UCLA. My
dissertation was on quantum and classical mechanics of small systems
(such as the hydrogen atom).
>
>The equation you use gives the probability limit of the electron being
>in a spherical shell of radius r and thickness dr, divided by dr, as dr
>approaches zero. If you instead look at the probability limit on a per
>volume basis for the s orbitals, you will notice the electron has a
>greater probability of being at the nucleus (i.e. 0 + dr as dr
>approaches zero) than at any other region whose volume is similarly
>vanishing.
>
Most general chem texts give the MAXIMUM probability of finding the
electron at the nucleus. This is incorrect. The MAXIMUM probability is
found at the Bohr radius (approx. 0.5 Angstroms). The way in which they
get the probability is by squaring the wavefunction. Again this is
incorrect. The probability is found, as you stated, for a thin shell by
squaring the probability multiplying by r squared and integrating. A
plot of the function r^2*(Psi)^2 will show a decreasing probability as
you approach the nucleus from the Bohr radius. At r=0 the probability
is zero. There is always a node at the nucleus.
________________________________________________
Dr. John Milligan "Spoon !!!!!!"
LAVC -The Tick
CSULA
UCLA Extension
John Milligan
comparing apples and oranges. Psi^2 is the probability DENSITY,
r^2*psi^2 is the probability of finding the electron in a thin shell
around the nucleus. These are two different concepts. Unfortunately,
most of the current general chemistry texts also get these two concepts
confused. I can't remember seeing a general chem text which does not
state that Psi^2 is the probability of finding the electron within a
given volume. I have seen one study guide that somewhat correctly
states that Psi^2 gives the probability of finding the electron in the
energy level given by Psi. But that is another barrel of monkeys
altogether.
Robert Parson
Ernst U. Wallenborn <wa...@phys.chem.ethz.ch> wrote:
>So we can conclude:
>
>1. The probability of finding a s-electron in a small volume element has
> a maximum at the nucleus.
>
>2. The probability of finding a s-electron in a r+dr shell goes to zero for r=0
>
>3. The probability of finding the electron "at a point" is zero everywhere
>
>Hope that clears up the confusion.
There's an excellent figure in Atkins' _Physical Chemistry_ book that
illustrates this. (Figs. 13.9 and 13.10, p. 361 of the 4th edition.)
Anybody who is still confused should take a look at it.
Simon Hogg
--Simon
Simon Hogg
>solar eclipse it bends around the mass of the moon. A clear sign that
>light (and therefore the photon) has a mass..
>
>-Per
>
>
--Simon
Robert Parson
John Milligan <alch...@pacificnet.net> wrote:
>This is part of the problem. As someone has pointed out this would be
>comparing apples and oranges. Psi^2 is the probability DENSITY,
>r^2*psi^2 is the probability of finding the electron in a thin shell
>around the nucleus. These are two different concepts.
At the risk of being pedantic, *both* are densities. Psi^2 is a
three-dimensional density, r^2*Psi^2 is a one-dimensional density.
Psi^2 dxdydz is the probability of finding the electron in a little
box with dimensions dx,dy,dz and for a 1s orbital this probability is
a maximum when you put the box around the origin.
4 Pi Psi^2 r^2 dr is the probability of finding the electron in a
spherical shell of radius r and thickness dr. For a 1s orbital this
probability is a maximum at the Bohr radius.
Figures 13.9-13.10 of Atkins (4th edition) illustrates this nicely.
------
Robert
Tonek Jansen
> Most general chem texts give the MAXIMUM probability of finding the
> electron at the nucleus. This is incorrect. The MAXIMUM probability is
> found at the Bohr radius (approx. 0.5 Angstroms). The way in which they
> get the probability is by squaring the wavefunction. Again this is
> incorrect. The probability is found, as you stated, for a thin shell by
> squaring the probability multiplying by r squared and integrating. A
> plot of the function r^2*(Psi)^2 will show a decreasing probability as
> you approach the nucleus from the Bohr radius. At r=0 the probability
> is zero. There is always a node at the nucleus.
It depends on what the question is. You're answering the question
- What is the probability distribution of finding the electron as a function
of distance from the nucleus?
This is indeed r^2*psi^2 integrated over theta and phi. But if you ask
- What is the probability distribution of finding the electron at some
position?
then the answer is psi^2.
If you have an electron in a 1s orbital the answer to the first question
has a maximum at the Bohr radius, but the answer to second at the nucleus.
There is not always a node at the nucleus. A node is a surface defined by
psi=0, and s-orbitals are not equal to zero at the nucleus.
Do you really have a Ph.D. in physical chemistry?
Tonek
tgt...@chem.tue.nl
Peter Jordan
>-- Simon
Is this the author of the book aimed at children which describes
thermodynamics using examples such as all the air in the room suddenly
being found under the coffee table, and a ship crossing the ocean using
a the heat energy in a teaspoon of water ?
Loved that book.
PeterJordan
Robert Parson
John Milligan <alch...@pacificnet.net> wrote:
>>
>Atkins (5th Ed.) also states that Psi^2 is a probability DENISTY. This
>is not a 3-D probability. It is a function which indicates what the
>probability will be when integrated over the volume element.
>BTW, you would not use dxdydz for a hydrogen atom since the wavefunction
>is in spherical coordinates.
Why not? Spherical coordinates are merely a convenient system for
solving the schrodinger equation. Once I have the wavefunction I
can write it in any coordinate system I chose. I write it in
rectangular coordinates, and then multiply by dxdydz. Let the
volume element be small and I have:
Probability of finding the electron in volume dxdydx centered at
a point (x1,y1,z1) is psi^2 dxdydz
> When you integrate over the volume element
>drd(theta)d(phi) you must also multiply by 4 Pi r^2, remember your
>calculus.
Which makes the volume element *increase with radius*. That's why it
is less confusing to use rectangular coordinates.
>Atkins states the the radial distribution function (ie. the
>probability as a function of radius) is given by the function I have
>stated above.
From which you can correctly conclude that the probability of finding
the electron's *radius* in the range (r1, r1+dr) is largest when
r1 is the Bohr radius.
Nevertheless the probability of finding the electron's *position* in
the neighborhood of the point x,y,z is largest when x,y,z is at
the origin. Not when x,y,z sits out at the Bohr radius.
John Milligan
rpa...@spot.Colorado.EDU (Robert Parson) wrote:
>In article <48nrjq$8...@zippy.cais.net>,
>John Milligan <alch...@pacificnet.net> wrote:
>
>>This is part of the problem. As someone has pointed out this would be
>>comparing apples and oranges. Psi^2 is the probability DENSITY,
>>r^2*psi^2 is the probability of finding the electron in a thin shell
>>around the nucleus. These are two different concepts.
>
> At the risk of being pedantic, *both* are densities. Psi^2 is a
>three-dimensional density, r^2*Psi^2 is a one-dimensional density.
>
>Psi^2 dxdydz is the probability of finding the electron in a little
>box with dimensions dx,dy,dz and for a 1s orbital this probability is
>a maximum when you put the box around the origin.
>
>4 Pi Psi^2 r^2 dr is the probability of finding the electron in a
>spherical shell of radius r and thickness dr. For a 1s orbital this
>probability is a maximum at the Bohr radius.
>
>Figures 13.9-13.10 of Atkins (4th edition) illustrates this nicely.
>
is not a 3-D probability. It is a function which indicates what the
probability will be when integrated over the volume element.
BTW, you would not use dxdydz for a hydrogen atom since the wavefunction
origin. Atkins states the the radial distribution function (ie. the
stated above.
________________________________________________
Tonek Jansen
> Hyperfine structure comes from the interaction of the electron's
> magnetic moment with the magnetic field of the nucleus.
>
> ------
> Robert
Is it really a magnetic interaction, or is the equation describing the
interaction similar to one describing a magnetic interaction?
I'm wondering if this is like the Heisenberg-exchange interaction between
the atoms in a ferromagnet. It is often said that it is a magnetic
interaction between the magnetic moments of the atoms, but it's really
an effect due to the antisymmetrization of the electronic wavefunction
of two atoms; i.e., the Pauli-principle.
There is a true magnetic interaction between two atoms in a ferromagnet,
but that is a much smaller interaction. Besides, the magnetic interactions
would result in atoms with anti-parallel moments. The magnetic interactions
do have an effect, in that they cause the formation of domains. That is
because the magnetic interactions have a longer range than the Heisenberg-
exchange interaction.
Tonek
tgt...@chem.tue.nl
Robert Parson
>John Milligan <alch...@pacificnet.net> wrote:
>Atkins (5th Ed.) also states that Psi^2 is a probability DENISTY. This
>is not a 3-D probability. It is a function which indicates what the
>probability will be when integrated over the volume element.
>BTW, you would not use dxdydz for a hydrogen atom since the wavefunction
>is in spherical coordinates.
>
solving the schrodinger equation. Once I have the wavefunction I
can write it in any coordinate system I chose. I write it in
rectangular coordinates, and then multiply by dxdydz. Let the
volume element be small and I have:
Probability of finding the electron in volume dxdydz centered at
a point (x1,y1,z1) = |Psi(x1,y1,z1)|^2 dxdydz
This probability is maximal when x1,y1,z1 is at the nucleus.
> When you integrate over the volume element
>drd(theta)d(phi) you must also multiply by 4 Pi r^2, remember your
>calculus.
Which makes the volume element *increase with radius*. So the
probability of finding the electron in a volume element
drd(theta)d(phi) centered at the point (r1,theta1,phi1) increases
as one moves away from the origin - for the trivial reason that
the size of the volume element is increasing.
That's why it is less confusing to use rectangular coordinates.
>Atkins states the the radial distribution function (ie. the
>probability as a function of radius) is given by the function I have
>stated above.
From which you can correctly conclude that the probability of finding
the electron's *radius* in the range (r1, r1+dr) is largest when
r1 is the Bohr radius.
Nevertheless the probability of finding the electron's *position* in
the neighborhood of the point x,y,z is largest when x,y,z is at
the origin. Not when x,y,z sits out at the Bohr radius.
------
Robert
Robert Parson
Tonek Jansen <t...@fyslab.hut.fi> wrote:
>>In article <48g8hv$b...@peabody.Colorado.EDU> rpa...@spot.Colorado.EDU (Robert Parson) writes:
>> Hyperfine structure comes from the interaction of the electron's
>> magnetic moment with the magnetic field of the nucleus.
>>
>> ------
>> Robert
>
>Is it really a magnetic interaction, or is the equation describing the
>interaction similar to one describing a magnetic interaction?
It's a real magnetic interaction. At least, in a nonrelativistic theory
it takes that form.
Per Rosqvist
No!.. Are you sure?
-Per
Lloyd R. Parker
: Im Artikel <ftn_2.200.116.13_4b...@p13.cw.ct.se>,
: per.ro...@p13.cw.ct.se (Per Rosqvist) schreibt:
: >The photon has a mass. When light from the sun comes down to earth during
: a
: >solar eclipse it bends around the mass of the moon. A clear sign that
: >light (and therefore the photon) has a mass..
: >
: >
: It shure has:
: hf
: E=--------
: c^2
: Another proof is, that one can observe a redshift when a photon travells
: upwards in a gravitaion field. The gain of potential energy is equivalent
: to the energy
: loss by the shift of frequency
: Greetings
: Pete
Neither the original post nor yours have anything to do with mass. Which
the photon does not have.
Kurt Kramer
LRP> DMurphy3 (dmur...@aol.com) wrote:
LRP> : Lloyd Parker in article <47vv2r$l...@larry.cc.emory.edu> writes in
LRP> reply to
LRP> : Jordan's comment on em radiation and mass:
LRP> : : Electromagnetic energy is substance.
LRP> : : It has mass. Something like E=mc^2
LRP> : >No, it has a rest mass equivalent in energy, but a photon does not
LRP> have mass.
LRP> : I think you have this backwards. A photon has no rest mass, but does
LRP> have
LRP> : relativistic mass. This mass, slight though it may be, may be
LRP> calculated *if* the momentum is known.
Very interesting discussion - may be somebody is able to answer my
question in relation to a mass/energy-problem in the context of the
quark-theory. Maybe it's a rather silly question, but I, myself
cannot found the solution and til now, nobody could help me to find
the issue out of my black hole of ignorance...
Protons and Neutrons consist of up and down quarks with a mass of
about 5 resp. 10 MeV. The sum of the mass of this three quarks has
to be diminished by the attractive power between the quarks, so it
should be less then 20 resp. 25 MeV.
But the real mass of protons and neutrons is nearly 1000 MeV.
Are you (or is anybody else capable to explain this discrepancy
for an interested non specialist? Thank you in advance!
CU, Kurt kkr...@ghost.aare.chnet.ch
Ernst U. Wallenborn
[snipped my utterings]
>Look at the math again. For finding the probability of the electron in
>"a small volume element" you must integrate over that volume element.
>Since the wavefunction is given (usually) in spherical coordinates you
>at some point must multiply by r^2 which will not give a maximum at the
>nucleus but at the Bohr radius for a hydrogen atom. Atkins (3rd ed.)
>states that the probability is proportional to Psi^2 d(tau) which again
>in spherical coords means you must multiply by r^2 dr sin(theta)
>d(theta) d(phi)
What does that mean? Of course you must integrate over a volume
element. That's exactly what i did. Wether you do this in cartesian
or spherical coordinates doesn't matter, as long as you get your
coordinate transformation right. And as long as you compare equally
sized volume elements, the integral of rho over dV is maximal where
the wavefunction is maximal, which for a s-electron is at the nucleus.
Is that so hard to understand?
[snip]
>If that were true then the probabilty of finding the electron in all of
>space would also be zero. How exactly does a function which is "zero
>everywhere" have a total probability of 1 when integrated over all
>space?
that's undergrad math. A point in R^3 is a subset of Lebesgue measure 0.
Hence the integral of any non-diverging function over this point is 0.
The following is 3rd semester physics here at ETH:
Integral[psi^2 dV over V] = sup(of all subsets of V){sum(i over all subsets)
[inf(of the ith subset) psi^2] * measure(ith subset)}
In case of a point, that is V=(x,y,z) you don't have a decomposition
except {(x,y,z)} itself, so the sup vanishes and so does the sum.
Psi^2 is uniquely defined at (x,y,z) and the inf goes away, too.
Integral[psi^2 dV over (x,y,z)] = psi^2 * measure((x,y,z))
which is zero, because psi^2 is finite and the measure of a point is 0.
You can check this in any elementary book on integration, e.g.
Soo Bong Chae, Lebesgue Integration, Springer, New York, 1995, ISBN 0-387-94357-9
Everything clear?
John Milligan
rpa...@spot.Colorado.EDU (Robert Parson) wrote:
>In article <48i1hk$t...@elna.ethz.ch>,
>Ernst U. Wallenborn <wa...@phys.chem.ethz.ch> wrote:
>
>>So we can conclude:
>>
has
>> a maximum at the nucleus.
>>
Look at the math again. For finding the probability of the electron in
"a small volume element" you must integrate over that volume element.
Since the wavefunction is given (usually) in spherical coordinates you
at some point must multiply by r^2 which will not give a maximum at the
nucleus but at the Bohr radius for a hydrogen atom. Atkins (3rd ed.)
states that the probability is proportional to Psi^2 d(tau) which again
in spherical coords means you must multiply by r^2 dr sin(theta)
d(theta) d(phi)
>>2. The probability of finding a s-electron in a r+dr shell goes to
zero for r=0
>>
Correct. No problems here.
>>3. The probability of finding the electron "at a point" is zero
everywhere
>>
If that were true then the probabilty of finding the electron in all of
space would also be zero. How exactly does a function which is "zero
everywhere" have a total probability of 1 when integrated over all
space?
>>Hope that clears up the confusion.
>
> There's an excellent figure in Atkins' _Physical Chemistry_ book that
> illustrates this. (Figs. 13.9 and 13.10, p. 361 of the 4th edition.)
> Anybody who is still confused should take a look at it.
>
The diagrams I believe you refer to are
Dann Corbit
>
>>The photon has a mass. When light from the sun comes down to earth during a
>>solar eclipse it bends around the mass of the moon. A clear sign that
>>light (and therefore the photon) has a mass..
>>
>>
>>
>Isn't that due to space distortion of massive objects?
>
>
What about diffraction? Considering the small mass of the moon,
the bending of light by its mass should be pretty small.
I believe that a ray from a distant star has been observed to have
been bent by a massive object between that star and earth.
If the ray path was not so close as to be diffracted, then
that would be physical evidence that light has mass (but not
necessarily arrest mass).
--
The opinions expressed in this message are my own personal views
and do not reflect the official views of Microsoft Corporation.
Jeff E. Janes
: In article <48o0k1$n...@peabody.Colorado.EDU>,
: rpa...@spot.Colorado.EDU (Robert Parson) wrote:
: >In article <48i1hk$t...@elna.ethz.ch>,
: >Ernst U. Wallenborn <wa...@phys.chem.ethz.ch> wrote:
: >
: >>So we can conclude:
: >>
: >>1. The probability of finding a s-electron in a small volume element
: has
: >> a maximum at the nucleus.
: >>
:
: Look at the math again. For finding the probability of the electron in
: "a small volume element" you must integrate over that volume element.
: Since the wavefunction is given (usually) in spherical coordinates you
: at some point must multiply by r^2 which will not give a maximum at the
: nucleus but at the Bohr radius for a hydrogen atom. Atkins (3rd ed.)
: states that the probability is proportional to Psi^2 d(tau) which again
: in spherical coords means you must multiply by r^2 dr sin(theta)
: d(theta) d(phi)
In order to have equal volumes, segements with larger r's must have smaller
regions of theta and phi which are integrated over, which cancels the r2
term. What you want to do is multiply by r2 again, which then no longer
gives probability as volume, but as shells.
:
:
: >>3. The probability of finding the electron "at a point" is zero
: everywhere
: >>
: If that were true then the probabilty of finding the electron in all of
: space would also be zero. How exactly does a function which is "zero
: everywhere" have a total probability of 1 when integrated over all
: space?
Draw a square with corners at (0,0) , (1,0), (1,1), and (0,1). What is
the area of this square? One, isn't it? What is the area of the
portion of the square laying on the line x=0.5? x=0.25?
x=0.76532987? It is zero for all of them, correct? Then how exactly
does this integrate to give an area of 1?
--
Jeff Janes jej...@mtu.edu at Michigan Technological University
Second Law of Thermodynamics:
The enormousfees of the university is ever increasing.
Martin Ystenes
If you had so much problems being understood, you would spin, too.
Martin Ystenes
K. Murray
>It is because of sheer frustration.
>
>If you had so much problems being understood, you would spin, too.
How about this: What would Mr. Tompkins see if electrons had spin
= 0?
G.K. Chan
Spin is just a name for a property electrons have.
Spin has NO CLASSICAL ANALOGUE ie it is not observable on the macroscopic
scale, which is why perhaps it is difficult to visualise what spin is.
If one considers the expression for angular momentum
2 2
L = l(l+1) h(bar)
then one can see that although the quantum of angular momentum is very
small, (h-bar), it is possible to have a body in a very high quantum state
ie l is very large. The transition to the classical system is by letting h(
bar) tend to 0; however if we then let the quantum number l tend to
infinity (which is what happens in classical systems, by the Bohr
correspondence principle) then one can see (rather unrigorously ) that one
could end up with a finite angular momentum. Thus angular momentum does
still exist in a classical world, in classical systems.
However let us now consider spin
2 2
S =s(s+1)h(bar)
again the quantum of spin is the same as that of angular momentum. More on
that later. However if we let h(bar) tend to zero - we cannot then let s
tend to infinity because s is an intrinsic property of the system: particles
must have fixed spin, thus the spin "angular momentum" tends to zero as one
approaches the classical system. There is thus no classical analogue of spin.
One cannot think of it in turns of "spinning tops" in three space. It exists
formally in an orthogonal dimension, spin space.
Why is spin classified as an angular momentum? Well, this can be answered
several ways. As you can see the unit of spin is the same as angular
momentum. This is normally enough to convince people that it is the same
thing . . . after all in classical mechanics, if we get something that is
of the same units eg L^2/2I is the same units as force * distance, then we
tend to say they belong to the same physical quantity, to within some
constant(which is not a physical quantity).
Why do they have the same units? We have to go one step deeper into the
quantum argument. The expression of spin is fully derivable from the
commutation relations:
[S^2,sz] = 0, [sx,sy] = ih(bar) sz
I don't know how much quantum mechanics is taught in America, but I hope
that is familiar. The angular momentum obey the same commutation relations;
and thus possess the same properties as spin. In fact we could call anything
which has the same commutators as angular momentum, angular momentum eg
electric charge, but that really would be quite strange.
Thus this concept of spin obeys all the properties of angular momentum, so
we don't have much choice but to call it that.
But why does spin exist?
Spin is an intrisically relativistic effect. It is normally just grafted on
to the non-relativistic Schrodinger equation by including the Pauli
principle, but still it's origins are in Dirac's equation.
Dirac's equation is not particularly difficult to derive, though it is
something that chemists, even theoretical ones, seldom use.
One takes the classical Schrodinger equation:
ih(bar)d(psi)/dt = H psi
and then substitutes the relativistic expression for the Hamiltonian.
(This is just contains a correction to the energy: just look up the
expression for the energy of a free particle in special relativity)
By factorising the equation and using constants which obey the quarternion
group, one obtains a very pretty equation, the Dirac equation.
One finds that one no longer has one solution to the equation(as in the
Schrodinger equation) , but 4 solutions. Two of these correspond to negative
energy states (these are antiparticles) and the others correspond to your
everyday electron, in two different states, which are degenerate at low
energies. As they are two different states we must characterise them by a
new quantum number, spin.
Thus you may have wondered how spin arose: as all the other quantum numbers
n,l, m arise from solving the Schrodinger equation, well spin arises from
solving the Dirac equation.
I have related spin to angular momentum above, but there is probably another
argument of more direct chemical significance.
If one solves the Dirac equation for what physicist call a central Coulomb
potential, but what all chemists love as the hydrogen atom, one obtains
expressions for the matrix containing the four wavefunctions (known as the
spinor) . . . if one then tries to see whether angular momentum is conserved
in the system, it is not. This is clearly bad news for physics as angular
momentum is conserved classically in a system of a body rotating around a
central potential. Something must be wrong . . . but luckily another
quantity is conserved, which is the "vector sum" of the angular momentum and
some quantity which is related to the spin quantum number. Thus this
combined quantity J is conserved, and the quantity that contains the spin
quantum number in it is known as the spin angular momentum (see above
formula). So spin does seem to play the role of the angular momentum of the
electron, even if we cannot assign a spinning frame of reference in normal
space. Such coupling between orbital angular momentum and spin angular
momentum is known as spin-orbit coupling: it's magnitude depends on the
gradient of the nuclear potential with respect to radius and is thus
important in heavy atoms with high Z eg the later transition elements. Much
of the chemistry is related to spin-orbit coupling effects.
I hope this makes things clear.
Perhaps just one last word . . . someone has already posted a lucid
explanation of whether the "probability" of finding the s-electron at the
nucleus is zero or not . . . but I guess I will just re-emphasize his point.
1. the probability density at the nucleus is non-zero
2. the probability of finding the electron at any point is zero (points are
just too small!)
3. the probability of finding an electron within any non-zero volume element
is not zero.
4. the radial distribution function is zero at r=0.
There just seems to have been some confusion at what one should call the
probability.
1. One cannot observe the "probability density".
2. One observes the radial distribution function.
As far as I can see the rdf is more relevant in atoms anyway as they are
spherically symmetric and rotating, so one only sees the electrons "smeared
out" in a shell. However the psi(squared) is still an observable.
Well that's the end.
Garnet Chan
third year chemist
Cambridge (England!)
ps _do_ stop throwing the names of Physical Chemistry books around! It is a
bit silly, and even more so as they are hardly authoritative when it comes
to theoretical chemistry.
G.K. Chan
>There just seems to have been some confusion at what one should call the
>probability.
>1. One cannot observe the "probability density".
>2. One observes the radial distribution function.
Just a correction . . .
As far as I can see the rdf is more relevant in atoms anyway as they are
spherically symmetric and rotating, so one detects spherical volume
elements. However the psi(squared)dxdydz is still an observable.
Garnet
YourName Here
>
>In article <48ntao$l...@oban.cc.ic.ac.uk>, s.h...@ic.ac.uk says...
>>
>>>The photon has a mass. When light from the sun comes down to earth during a
>>>solar eclipse it bends around the mass of the moon. A clear sign that
>>>light (and therefore the photon) has a mass..
>>>
>>>-Per
>>>
>>>
>>Isn't that due to space distortion of massive objects?
>>
>>--Simon
>
Photons do not have mass. Photons suffer a lensing effect around massive objects because
the space around a massive object is bent. Think of a ball in a rubber sheet.
Think on this, the laws of reletivity say that as an object with mass
travels closer to the speed of light it mass increases. The object can never reach
the speed of light because it would have infinate mass. Since light travels
at the speed of light, it either has infinate mass or does not have any
mass. If light has infinate mass it should be easy to weigh it.
Sorry about the spelling errors, but its getting late
K. Murray
>"K. Murray" <kmu...@emory.edu> wrote:
>>Martin Ystenes <yst...@kjemi.unit.no> wrote:
>>>It is because of sheer frustration.
>>>
>>>If you had so much problems being understood, you would spin, too.
>>
>>How about this: What would Mr. Tompkins see if electrons had spin
> >= 0?
>>
>
>
>Martin Ystenes
>
Only if he can get home through all the Bose condensation.
Kermit Murray
DMurphy3
>Neither the original post nor yours have anything to do with mass. Which
>the photon does not have.
The original post did indeed deal with photon mass. And if, as you still
assert, photons have no mass, how do you account for the fact that photons
have momentum? By the very definition of momentum, a photon must have mass
in order to possess momentum. Rest mass, no. Relativistic mass, yes. The
concept of wave-particle duality insists on this. Or do you know a
definition for momentum of which I am unaware?
Tonek Jansen
> Photons do not have mass. Photons suffer a lensing effect around
> massive objects because the space around a massive object is bent. Think
> of a ball in a rubber sheet. Think on this, the laws of reletivity say
> that as an object with mass travels closer to the speed of light it mass
> increases. The object can never reach the speed of light because it would
> have infinate mass. Since light travels at the speed of light, it either
> has infinate mass or does not have any mass. If light has infinate mass
> it should be easy to weigh it.
It seems to me that many people confuse rest mass and mass at some
velocity. The relation between the two, if I remember correctly, is
m=m0/sqrt(1-v^2/c^2), where m0 is the rest mass and m is the mass at
velocity v. As v=c for a photon, the denominator is 0. This means that
m0=0, because otherwise m is infinite. So the rest mass is zero. The
mass at v=c is not defined by this relation, because denominator and
nominator are both zero.
We can get the mass from the momentum of the photon. We know that it
has momentum from the Compton-effect, and from the de Broglie relation
p=h/lambda. This relation and p=mc gives you the mass.
Tonek
tgt...@chem.tue.nl
Lloyd R. Parker
: Lloyd Parker in article <48vsbd$b...@curly.cc.emory.edu> asserts:
Since photons have no mass, momentum does have a more general definition
than mass times velocity. From de Broglie's eq, lambda = h/(m*v), you
can get lambda = h/p (where p = momentum), and therefore p = h/lambda.
So the momentum of a photon should be Planck's constant divided by its
wavelength. The units do work out to kg-m.
Martin Ystenes
Robert Parson
>Martin Ystenes <yst...@kjemi.unit.no> wrote:
>>It is because of sheer frustration.
>>
>>If you had so much problems being understood, you would spin, too.
>
>How about this: What would Mr. Tompkins see if electrons had spin
>= 0?
I once asked a similar question on a P.Chem. exam: "What would the
periodic table look like if electrons had spin zero?"
PeteKonig
>The photon has a mass. When light from the sun comes down to earth during
a
>solar eclipse it bends around the mass of the moon. A clear sign that
>light (and therefore the photon) has a mass..
>
>
It shure has:
Robert Parson
G.K. Chan <gkc...@hermes.cam.ac.uk> wrote:
>
[ *Very* nice discussion of angular momentum in Q.M. deleted.]
>As far as I can see the rdf is more relevant in atoms anyway as they are
>spherically symmetric and rotating, so one only sees the electrons "smeared
>out" in a shell. However the psi(squared) is still an observable.
Ah, but it is only the *Hamiltonian* that is spherically symmetric; the
L>0 probability densities are not. So the atom need not be spherically
symmetric; indeed, people study collisions between "aligned atoms."
In the example I gave earlier, the Fermi Contact interaction, the
strength of the interaction is proportional to |Psi(r=0)|^2 and is
usually interpreted as involving the probability that the electron is
to be found within the nucleus. (It's |Psi(0)|^2, rather than an
integral over nuclear volume, because the interaction term in the
Hamiltonian is usually approximated as a delta function in r.)
>ps _do_ stop throwing the names of Physical Chemistry books around!
I disagree. When I cite references in usenet posts, I try to cite
something that people will be able to find easily; if possible,
something that has a good chance of being on someone's bookshelf.
Most of the issues discussed in this thread were elementary and the
treatment in P.Chem. books is fine.
------
Robert
G.K. Chan
> Ah, but it is only the *Hamiltonian* that is spherically symmetric; the
> L>0 probability densities are not. So the atom need not be spherically
> symmetric; indeed, people study collisions between "aligned atoms."
Admittedly. In bonding which is clearly along axes, it would be strange
to think of overlap of spherically averaged rdf's! But I think the previous
posts were referring to L=0 probability densities as they were having all
the discussions about finite probabilities at the nucleus! I think people
probably have less difficult with L>0 probability densities as they look
qualitatively similar in both three dimensional and one dimensional forms.
Garnet
is there any future in theoretical chemistry . . .
Lloyd R. Parker
: L. Parker writes in article: <49la0r$l...@larry.cc.emory.edu>
: >From de Broglie's eq, lambda = h/(m*v), you
: >can get lambda = h/p (where p = momentum), and therefore p = h/lambda.
: >So the momentum of a photon should be Planck's constant divided by its
: >wavelength. The units do work out to kg-m.
: And what is a kg? I always thought it was a unit of measurement defined
: exclusively for mass. But I could be wrong, as you are implying that a
: particle with no mass has the units kg-m. By following this logic, must we
: then assume that the momentum will always be zero?
A joule is a kg-m^2/s^2. So an object with 10 J of added energy has an
added 10 kg-m^2/s^2, even though you haven't added any mass to it (well,
certainly not 10 kg of mass).
Tonek Jansen
> I once asked a similar question on a P.Chem. exam: "What would the
> periodic table look like if electrons had spin zero?"
I guess that all elements would have very similar properties. Assuming that
an independent-particle model is still a good approximation, the electronic
wavefunction would not be a Slater-determinant, but a Slater-permanent. This
you get by changing all minus-signs into plus-signs. This wavefunction would
be even under permutation of electrons, because S=0 particles are bosons. The
ground state has all electrons in the 1s-orbital.
Tonek
tgt...@chem.tue.nl