On May 19, 11:24 pm, Pentcho Valev wrote:
Eric Gisse wrote:
On May 19, 11:05 am, "Y.Porat" wrote:

[...]

So what is the mass of the photon, Porat?

Such that the speed of the photon varies with the gravitational
potential in accordance with Einstein's 1911

[snip remainder, unread]

Idiot. Stop obsessing about the irrelevant. Relativity does not depend
on Einstein.

sci.physics.cond-matter, sci.philosophy.tech, sci.astro removed. You
can't even figure out how to post to the right newsgroups, ****up.

On May 19, 6:50 pm, The_Man wrote:
On May 19, 4:42 pm, RP wrote:





On May 19, 12:30 pm, The_Man wrote:

On May 19, 1:11 pm, RP wrote:

On May 19, 11:43 am, The_Man wrote:

On May 19, 12:25 pm, RP wrote:

On May 19, 10:00 am, The_Man wrote:

On May 19, 9:31 am, RP wrote:

On May 18, 1:58 am, Pentcho Valev wrote:

http://groups.google.com/group/sci.p...hread/d28e668a...

On Oct 25, 1995, John Baez wrote in sci.physics:

"Nonetheless, it's a fact that a photon has a nonzero momentum."

A curious person asked:

"Are you therefore asserting that it has nonzero mass? If not, why
not?"

John Baez replied:

"You can see that I did not assert anything about the photon's mass. I
know what the photon's mass is, but I never talk about it around here
because the endless discussion of the photon's mass is boring, boring,
boring."

While I don't agree that the topic is boring, I do agree that it isn't
easy to understand, so I'll try to explain it. Photons don't exist.
Nothing that doesn't exist has mass.

You state as a fact what seems more like an opinion.

If special relativity is opinion, then I retract my statement

Photons have (spin) angular mometum. If photons "don't exist", the law
of conservation of angular momentum falls apart. Along with it, the
understanding of which electronic transitons occur, and how often they
occur (what we call "selection rules")

Photons don't have anything, because they don't exist. Momentum is
globally conserved, not locally conserved. The solution to the
radiation reaction problem is that there isn't a reaction, just an
action. Electrons are foward acting particles. The past light cone
contains events that influence the electron, the foward light cone
those events that it influences. There is no experimental evidence of
electron recoil under the influence of its own emitted radiation.
Before replying please note a distinction between a single electron
and an array of electrons, the latter of which produces feedback
effects.

If you can explain why s- d transitions don't occur, or why p-p
transitions don't occur, or why d-d transitions for high spin Fe 3+
occur with such a small probability, I'm all ears.

The atom is an electronic structure in equilibrium. Some states are
unstable and others impossible.

A sodium atoms is electronically excited by heating it. Sodium emits
light at two very specific wavelengths. These correspond to
transitions from 2P 3/2 - 1S0, or 2P 1/2 - 1S0. In the proces, the
atom "loses" 1 unit of orbital angular momentum. Where did it go? This
is a simple question...

It goes to all other electrons via interaction with the source
electrons, directly, at c.

So the angular momentum of the other electrons change? Yet, if that
happened, you would violate the law of conservation of energy. There
is an even easier refutation - suppose we talking about the spectrum
of hydrogen, which has NO other electrons.

Now the ball is in your court, to explain the selection rules for
hydrogen: Delta l = +/1 1, delta s =0; delta m sub l = 0, +/-1.

Because of the quantum states of recieving
atoms, they absorb energy by the same amount, and that you assume that
this absorbed energy all derived from the same single point source is
for you to prove, not me.

Proven LONG ago. Lasers do single photo absorption experiments all the
time. This is the so-called "resonance condition" - It is the basis of
all spectroscopy.

That theory goes against all the known laws
of em. Statistically your version works "because" of global momentum
conservation.

The 1 unit of orbital angular momentum is carried away by the photon,
which has a spin angular momentum of 1. We never see electronic
transitions involving changes of anything OTHER than 1 unit of angular
momentum.

If you can tell us how you can derive all the selection rules (for
both 1 electron and multi-electron atoms) WITHOUT a photon, all of
chemistry and physics would be glad to hear it. Not the least, that
you would also have to explain the photoelectric effect, and how TV's
work.

See above. It's more or less a holographic process.

No. There is a one-to-one correspondence between the number of photns
absorbed and the number of electrons emitted during the photoelectric
effect.

Incorrect. You can't count the photons, only the photoelectrons. At
any given moment a surface is being impinged upon by radiation from
very many sources.

Actually, photon counting is relatively (no pun intended) simple. It
is done all the time - it is necesary to determine the quantum
efficiency of such processes as fluorescence.

As for the "surface being impinged upon by radiation from very many
sources", this suggests a very incompetent experimentalist. The
surface would be sputtered with Ar+ to clean crap off the surface, the
surface would be kept in high vacuum, and all radiation sources other
than the source would be shielded against.

If I'm allowed to speak, for arguments sake, of
this radiation in terms of photons, then the number of photons
striking the target is far greater than the photoelectrons.

I didn't say "striking the target", I said "absorbed". The number of
photons NOT absorbed strikes a detector, and are counted. The number
of electrons ejected is also easily counted. So long as the photons
had the necessry energy, yadda yadda yadda, the # photons launched = #
photons NOT absorbed + # electrons emitted.

You simply
disgard the fact that the surface is pumped, because it suits your
argument. Again, it is still for you to prove the one-to-one
correspondence that you assume.

It is more straightforward to analyze in terms of absorption by atoms,
rather surfaces. You are confusing the fact of the probability of
absorption of a particular photon, which is usually less than 1, with
the probability that upon absorption, the photon will cause a change
in orbital angular momentum of 1 unit, which is exactly1. Of course,
the photon has to have the right energy to cause an electronic
transition, blah,blah, blah, which you don't understand anyway.

The emission spectrum of hydrogen was explained by Bohr - it is quite
simple. The spectra of heavier elements is more complicated (electron
electron repulsion complicates the analysis). Each line corresponds to
the energy difference between two electronic levels in hydrogen, ALL
of which differ in orbital angular momentum by sqrt(2) h bar.

If that isn't good enough for you, too bad. I don't have time to
repeat the results of 100 years of experiments.

You're the contrary one. Just show which lines of ANY atom result from
a change in other than 1 unit of angular momentum. If you understood L-
S or j-j coupling, I would put it in the requisite clarifications, but
I am sure that it is over your head. Just as the normal or anomalous
Zeeman effect would further prove my point.

Or show that the absorption of one photon by a surface ejects more
or,less than one electron.

Given that you've already decided that I know nothing about quantum
processes, and especially that you've made Gisse your bed-partner in
ad hominem, you aren't deserving of any more of my time. You're just
another loser hanging out here becuase its the only place where you
feel any self-worth, that is, amongs the kooks and trolls. But from
my perspective you are nothing more than one of the latter.

On May 20, 8:18 am, RP wrote:
On May 19, 6:50 pm, The_Man wrote:





On May 19, 4:42 pm, RP wrote:

On May 19, 12:30 pm, The_Man wrote:

On May 19, 1:11 pm, RP wrote:

On May 19, 11:43 am, The_Man wrote:

On May 19, 12:25 pm, RP wrote:

On May 19, 10:00 am, The_Man wrote:

On May 19, 9:31 am, RP wrote:

On May 18, 1:58 am, Pentcho Valev wrote:

http://groups.google.com/group/sci.p...hread/d28e668a...

On Oct 25, 1995, John Baez wrote in sci.physics:

"Nonetheless, it's a fact that a photon has a nonzero momentum."

A curious person asked:

"Are you therefore asserting that it has nonzero mass? If not, why
not?"

John Baez replied:

"You can see that I did not assert anything about the photon's mass. I
know what the photon's mass is, but I never talk about it around here
because the endless discussion of the photon's mass is boring, boring,
boring."

While I don't agree that the topic is boring, I do agree that it isn't
easy to understand, so I'll try to explain it. Photons don't exist.
Nothing that doesn't exist has mass.

You state as a fact what seems more like an opinion.

If special relativity is opinion, then I retract my statement

Photons have (spin) angular mometum. If photons "don't exist", the law
of conservation of angular momentum falls apart. Along with it, the
understanding of which electronic transitons occur, and how often they
occur (what we call "selection rules")

Photons don't have anything, because they don't exist. Momentum is
globally conserved, not locally conserved. The solution to the
radiation reaction problem is that there isn't a reaction, just an
action. Electrons are foward acting particles. The past light cone
contains events that influence the electron, the foward light cone
those events that it influences. There is no experimental evidence of
electron recoil under the influence of its own emitted radiation.
Before replying please note a distinction between a single electron
and an array of electrons, the latter of which produces feedback
effects.

If you can explain why s- d transitions don't occur, or why p-p
transitions don't occur, or why d-d transitions for high spin Fe 3+
occur with such a small probability, I'm all ears.

The atom is an electronic structure in equilibrium. Some states are
unstable and others impossible.

A sodium atoms is electronically excited by heating it. Sodium emits
light at two very specific wavelengths. These correspond to
transitions from 2P 3/2 - 1S0, or 2P 1/2 - 1S0. In the proces, the
atom "loses" 1 unit of orbital angular momentum. Where did it go? This
is a simple question...

It goes to all other electrons via interaction with the source
electrons, directly, at c.

So the angular momentum of the other electrons change? Yet, if that
happened, you would violate the law of conservation of energy. There
is an even easier refutation - suppose we talking about the spectrum
of hydrogen, which has NO other electrons.

Now the ball is in your court, to explain the selection rules for
hydrogen: Delta l = +/1 1, delta s =0; delta m sub l = 0, +/-1.

Because of the quantum states of recieving
atoms, they absorb energy by the same amount, and that you assume that
this absorbed energy all derived from the same single point source is
for you to prove, not me.

Proven LONG ago. Lasers do single photo absorption experiments all the
time. This is the so-called "resonance condition" - It is the basis of
all spectroscopy.

That theory goes against all the known laws
of em. Statistically your version works "because" of global momentum
conservation.

The 1 unit of orbital angular momentum is carried away by the photon,
which has a spin angular momentum of 1. We never see electronic
transitions involving changes of anything OTHER than 1 unit of angular
momentum.

If you can tell us how you can derive all the selection rules (for
both 1 electron and multi-electron atoms) WITHOUT a photon, all of
chemistry and physics would be glad to hear it. Not the least, that
you would also have to explain the photoelectric effect, and how TV's
work.

See above. It's more or less a holographic process.

No. There is a one-to-one correspondence between the number of photns
absorbed and the number of electrons emitted during the photoelectric
effect.

Incorrect. You can't count the photons, only the photoelectrons. At
any given moment a surface is being impinged upon by radiation from
very many sources.

Actually, photon counting is relatively (no pun intended) simple. It
is done all the time - it is necesary to determine the quantum
efficiency of such processes as fluorescence.

As for the "surface being impinged upon by radiation from very many
sources", this suggests a very incompetent experimentalist. The
surface would be sputtered with Ar+ to clean crap off the surface, the
surface would be kept in high vacuum, and all radiation sources other
than the source would be shielded against.

If I'm allowed to speak, for arguments sake, of
this radiation in terms of photons, then the number of photons
striking the target is far greater than the photoelectrons.

I didn't say "striking the target", I said "absorbed". The number of
photons NOT absorbed strikes a detector, and are counted. The number
of electrons ejected is also easily counted. So long as the photons
had the necessry energy, yadda yadda yadda, the # photons launched = #
photons NOT absorbed + # electrons emitted.

You simply
disgard the fact that the surface is pumped, because it suits your
argument. Again, it is still for you to prove the one-to-one
correspondence that you assume.

It is more straightforward to analyze in terms of absorption by atoms,
rather surfaces. You are confusing the fact of the probability of
absorption of a particular photon, which is usually less than 1, with
the probability that upon absorption, the photon will cause a change
in orbital angular momentum of 1 unit, which is exactly1. Of course,
the photon has to have the right energy to cause an electronic
transition, blah,blah, blah, which you don't understand anyway.

The emission spectrum of hydrogen was explained by Bohr - it is quite
simple. The spectra of heavier elements is more complicated (electron
electron repulsion complicates the analysis). Each line corresponds to
the energy difference between two electronic levels in hydrogen, ALL
of which differ in orbital angular momentum by sqrt(2) h bar.

If that isn't good enough for you, too bad. I don't have time to
repeat the results of 100 years of experiments.

You're the contrary one. Just show which lines of ANY atom result from
a change in other than 1 unit of angular momentum. If you understood L-
S or j-j coupling, I would put it in the requisite clarifications, but
I am sure that it is over your head. Just as the normal or anomalous
Zeeman effect would further prove my point.

Or show that the absorption of one photon by a surface ejects more
or,less than one electron.

Given that you've already decided that I know nothing about quantum
processes, and especially that you've made Gisse your bed-partner in
ad hominem, you aren't deserving of any more of my time.

Translation: You can't answer any of the questions I posed.

You're just
another loser hanging out here becuase its the only place where you
feel any self-worth, that is, amongs the kooks and trolls.

Translation: You feel inferior because you don't even know what spin-
orbit coupling is, or Russell-Saunders or j-j coupling, nor selection
rules, yet you want to impress yourself that you know something.

But from
my perspective you are nothing more than one of the latter.

Translation: Boo-hoo, WAAAAAAAAAA!

- Hide quoted text -

- Show quoted text -

On May 20, 7:31 am, The_Man wrote:
On May 20, 8:18 am, RP wrote:





On May 19, 6:50 pm, The_Man wrote:

On May 19, 4:42 pm, RP wrote:

On May 19, 12:30 pm, The_Man wrote:

On May 19, 1:11 pm, RP wrote:

On May 19, 11:43 am, The_Man wrote:

On May 19, 12:25 pm, RP wrote:

On May 19, 10:00 am, The_Man wrote:

On May 19, 9:31 am, RP wrote:

On May 18, 1:58 am, Pentcho Valev wrote:

http://groups.google.com/group/sci.p...hread/d28e668a...

On Oct 25, 1995, John Baez wrote in sci.physics:

"Nonetheless, it's a fact that a photon has a nonzero momentum."

A curious person asked:

"Are you therefore asserting that it has nonzero mass? If not, why
not?"

John Baez replied:

"You can see that I did not assert anything about the photon's mass. I
know what the photon's mass is, but I never talk about it around here
because the endless discussion of the photon's mass is boring, boring,
boring."

While I don't agree that the topic is boring, I do agree that it isn't
easy to understand, so I'll try to explain it. Photons don't exist.
Nothing that doesn't exist has mass.

You state as a fact what seems more like an opinion.

If special relativity is opinion, then I retract my statement

Photons have (spin) angular mometum. If photons "don't exist", the law
of conservation of angular momentum falls apart. Along with it, the
understanding of which electronic transitons occur, and how often they
occur (what we call "selection rules")

Photons don't have anything, because they don't exist. Momentum is
globally conserved, not locally conserved. The solution to the
radiation reaction problem is that there isn't a reaction, just an
action. Electrons are foward acting particles. The past light cone
contains events that influence the electron, the foward light cone
those events that it influences. There is no experimental evidence of
electron recoil under the influence of its own emitted radiation.
Before replying please note a distinction between a single electron
and an array of electrons, the latter of which produces feedback
effects.

If you can explain why s- d transitions don't occur, or why p-p
transitions don't occur, or why d-d transitions for high spin Fe 3+
occur with such a small probability, I'm all ears.

The atom is an electronic structure in equilibrium. Some states are
unstable and others impossible.

A sodium atoms is electronically excited by heating it. Sodium emits
light at two very specific wavelengths. These correspond to
transitions from 2P 3/2 - 1S0, or 2P 1/2 - 1S0. In the proces, the
atom "loses" 1 unit of orbital angular momentum. Where did it go? This
is a simple question...

It goes to all other electrons via interaction with the source
electrons, directly, at c.

So the angular momentum of the other electrons change? Yet, if that
happened, you would violate the law of conservation of energy. There
is an even easier refutation - suppose we talking about the spectrum
of hydrogen, which has NO other electrons.

Now the ball is in your court, to explain the selection rules for
hydrogen: Delta l = +/1 1, delta s =0; delta m sub l = 0, +/-1.

Because of the quantum states of recieving
atoms, they absorb energy by the same amount, and that you assume that
this absorbed energy all derived from the same single point source is
for you to prove, not me.

Proven LONG ago. Lasers do single photo absorption experiments all the
time. This is the so-called "resonance condition" - It is the basis of
all spectroscopy.

That theory goes against all the known laws
of em. Statistically your version works "because" of global momentum
conservation.

The 1 unit of orbital angular momentum is carried away by the photon,
which has a spin angular momentum of 1. We never see electronic
transitions involving changes of anything OTHER than 1 unit of angular
momentum.

If you can tell us how you can derive all the selection rules (for
both 1 electron and multi-electron atoms) WITHOUT a photon, all of
chemistry and physics would be glad to hear it. Not the least, that
you would also have to explain the photoelectric effect, and how TV's
work.

See above. It's more or less a holographic process.

No. There is a one-to-one correspondence between the number of photns
absorbed and the number of electrons emitted during the photoelectric
effect.

Incorrect. You can't count the photons, only the photoelectrons. At
any given moment a surface is being impinged upon by radiation from
very many sources.

Actually, photon counting is relatively (no pun intended) simple. It
is done all the time - it is necesary to determine the quantum
efficiency of such processes as fluorescence.

As for the "surface being impinged upon by radiation from very many
sources", this suggests a very incompetent experimentalist. The
surface would be sputtered with Ar+ to clean crap off the surface, the
surface would be kept in high vacuum, and all radiation sources other
than the source would be shielded against.

If I'm allowed to speak, for arguments sake, of
this radiation in terms of photons, then the number of photons
striking the target is far greater than the photoelectrons.

I didn't say "striking the target", I said "absorbed". The number of
photons NOT absorbed strikes a detector, and are counted. The number
of electrons ejected is also easily counted. So long as the photons
had the necessry energy, yadda yadda yadda, the # photons launched = #
photons NOT absorbed + # electrons emitted.

You simply
disgard the fact that the surface is pumped, because it suits your
argument. Again, it is still for you to prove the one-to-one
correspondence that you assume.

It is more straightforward to analyze in terms of absorption by atoms,
rather surfaces. You are confusing the fact of the probability of
absorption of a particular photon, which is usually less than 1, with
the probability that upon absorption, the photon will cause a change
in orbital angular momentum of 1 unit, which is exactly1. Of course,
the photon has to have the right energy to cause an electronic
transition, blah,blah, blah, which you don't understand anyway.

The emission spectrum of hydrogen was explained by Bohr - it is quite
simple. The spectra of heavier elements is more complicated (electron
electron repulsion complicates the analysis). Each line corresponds to
the energy difference between two electronic levels in hydrogen, ALL
of which differ in orbital angular momentum by sqrt(2) h bar.

If that isn't good enough for you, too bad. I don't have time to
repeat the results of 100 years of experiments.

You're the contrary one. Just show which lines of ANY atom result from
a change in other than 1 unit of angular momentum. If you understood L-
S or j-j coupling, I would put it in the requisite clarifications, but
I am sure that it is over your head. Just as the normal or anomalous
Zeeman effect would further prove my point.

Or show that the absorption of one photon by a surface ejects more
or,less than one electron.

Given that you've already decided that I know nothing about quantum
processes, and especially that you've made Gisse your bed-partner in
ad hominem, you aren't deserving of any more of my time.

Translation: You can't answer any of the questions I posed.

You're just
another loser hanging out here becuase its the only place where you
feel any self-worth, that is, amongs the kooks and trolls.

Translation: You feel inferior because you don't even know what spin-
orbit coupling is, or Russell-Saunders or j-j coupling, nor selection
rules, yet you want to impress yourself that you know something.

But from
my perspective you are nothing more than one of the latter.

Translation: Boo-hoo, WAAAAAAAAAA!

I rest my case. You're developmentally challenged, maybe even a bit
schizophrenic.

The point you missed, is that spin-orbit coupling is imaterial to
answering the question "what is light"? Further proof that you aren't
capable of maintaining a mature conversation. This is the favorite
tactic of the kooks and trolls that you mock, changing the subject
when you're losing an argument.

On May 20, 8:43 am, RP wrote:
On May 20, 7:31 am, The_Man wrote:





On May 20, 8:18 am, RP wrote:

On May 19, 6:50 pm, The_Man wrote:

On May 19, 4:42 pm, RP wrote:

On May 19, 12:30 pm, The_Man wrote:

On May 19, 1:11 pm, RP wrote:

On May 19, 11:43 am, The_Man wrote:

On May 19, 12:25 pm, RP wrote:

On May 19, 10:00 am, The_Man wrote:

On May 19, 9:31 am, RP wrote:

On May 18, 1:58 am, Pentcho Valev wrote:

http://groups.google.com/group/sci.p...hread/d28e668a...

On Oct 25, 1995, John Baez wrote in sci.physics:

"Nonetheless, it's a fact that a photon has a nonzero momentum."

A curious person asked:

"Are you therefore asserting that it has nonzero mass? If not, why
not?"

John Baez replied:

"You can see that I did not assert anything about the photon's mass. I
know what the photon's mass is, but I never talk about it around here
because the endless discussion of the photon's mass is boring, boring,
boring."

While I don't agree that the topic is boring, I do agree that it isn't
easy to understand, so I'll try to explain it. Photons don't exist.
Nothing that doesn't exist has mass.

You state as a fact what seems more like an opinion.

If special relativity is opinion, then I retract my statement

Photons have (spin) angular mometum. If photons "don't exist", the law
of conservation of angular momentum falls apart. Along with it, the
understanding of which electronic transitons occur, and how often they
occur (what we call "selection rules")

Photons don't have anything, because they don't exist. Momentum is
globally conserved, not locally conserved. The solution to the
radiation reaction problem is that there isn't a reaction, just an
action. Electrons are foward acting particles. The past light cone
contains events that influence the electron, the foward light cone
those events that it influences. There is no experimental evidence of
electron recoil under the influence of its own emitted radiation.
Before replying please note a distinction between a single electron
and an array of electrons, the latter of which produces feedback
effects.

If you can explain why s- d transitions don't occur, or why p-p
transitions don't occur, or why d-d transitions for high spin Fe 3+
occur with such a small probability, I'm all ears.

The atom is an electronic structure in equilibrium. Some states are
unstable and others impossible.

A sodium atoms is electronically excited by heating it. Sodium emits
light at two very specific wavelengths. These correspond to
transitions from 2P 3/2 - 1S0, or 2P 1/2 - 1S0. In the proces, the
atom "loses" 1 unit of orbital angular momentum. Where did it go? This
is a simple question...

It goes to all other electrons via interaction with the source
electrons, directly, at c.

So the angular momentum of the other electrons change? Yet, if that
happened, you would violate the law of conservation of energy. There
is an even easier refutation - suppose we talking about the spectrum
of hydrogen, which has NO other electrons.

Now the ball is in your court, to explain the selection rules for
hydrogen: Delta l = +/1 1, delta s =0; delta m sub l = 0, +/-1.

Because of the quantum states of recieving
atoms, they absorb energy by the same amount, and that you assume that
this absorbed energy all derived from the same single point source is
for you to prove, not me.

Proven LONG ago. Lasers do single photo absorption experiments all the
time. This is the so-called "resonance condition" - It is the basis of
all spectroscopy.

That theory goes against all the known laws
of em. Statistically your version works "because" of global momentum
conservation.

The 1 unit of orbital angular momentum is carried away by the photon,
which has a spin angular momentum of 1. We never see electronic
transitions involving changes of anything OTHER than 1 unit of angular
momentum.

If you can tell us how you can derive all the selection rules (for
both 1 electron and multi-electron atoms) WITHOUT a photon, all of
chemistry and physics would be glad to hear it. Not the least, that
you would also have to explain the photoelectric effect, and how TV's
work.

See above. It's more or less a holographic process.

No. There is a one-to-one correspondence between the number of photns
absorbed and the number of electrons emitted during the photoelectric
effect.

Incorrect. You can't count the photons, only the photoelectrons. At
any given moment a surface is being impinged upon by radiation from
very many sources.

Actually, photon counting is relatively (no pun intended) simple. It
is done all the time - it is necesary to determine the quantum
efficiency of such processes as fluorescence.

As for the "surface being impinged upon by radiation from very many
sources", this suggests a very incompetent experimentalist. The
surface would be sputtered with Ar+ to clean crap off the surface, the
surface would be kept in high vacuum, and all radiation sources other
than the source would be shielded against.

If I'm allowed to speak, for arguments sake, of
this radiation in terms of photons, then the number of photons
striking the target is far greater than the photoelectrons.

I didn't say "striking the target", I said "absorbed". The number of
photons NOT absorbed strikes a detector, and are counted. The number
of electrons ejected is also easily counted. So long as the photons
had the necessry energy, yadda yadda yadda, the # photons launched = #
photons NOT absorbed + # electrons emitted.

You simply
disgard the fact that the surface is pumped, because it suits your
argument. Again, it is still for you to prove the one-to-one
correspondence that you assume.

It is more straightforward to analyze in terms of absorption by atoms,
rather surfaces. You are confusing the fact of the probability of
absorption of a particular photon, which is usually less than 1, with
the probability that upon absorption, the photon will cause a change
in orbital angular momentum of 1 unit, which is exactly1. Of course,
the photon has to have the right energy to cause an electronic
transition, blah,blah, blah, which you don't understand anyway.

The emission spectrum of hydrogen was explained by Bohr - it is quite
simple. The spectra of heavier elements is more complicated (electron
electron repulsion complicates the analysis). Each line corresponds to
the energy difference between two electronic levels in hydrogen, ALL
of which differ in orbital angular momentum by sqrt(2) h bar.

If that isn't good enough for you, too bad. I don't have time to
repeat the results of 100 years of experiments.

You're the contrary one. Just show which lines of ANY atom result from
a change in other than 1 unit of angular momentum. If you understood L-
S or j-j coupling, I would put it in the requisite clarifications, but
I am sure that it is over your head. Just as the normal or anomalous
Zeeman effect would further prove my point.

Or show that the absorption of one photon by a surface ejects more
or,less than one electron.

Given that you've already decided that I know nothing about quantum
processes, and especially that you've made Gisse your bed-partner in
ad hominem, you aren't deserving of any more of my time.

Translation: You can't answer any of the questions I posed.

You're just
another loser hanging out here becuase its the only place where you
feel any self-worth, that is, amongs the kooks and trolls.

Translation: You feel inferior because you don't even know what spin-
orbit coupling is, or Russell-Saunders or j-j coupling, nor selection
rules, yet you want to impress yourself that you know something.

But from
my perspective you are nothing more than one of the latter.

Translation: Boo-hoo, WAAAAAAAAAA!

I rest my case. You're developmentally challenged, maybe even a bit
schizophrenic.

The point you missed, is that spin-orbit coupling is imaterial to
answering the question "what is light"?

Spin-orbit coupling is completely material to answering the question
"what is light". That you don't see that is either a function of
unwillingness on your part, or lack of understanding.

Further proof that you aren't
capable of maintaining a mature conversation.

"Mature" ="flatters you that you actually know what you are talking
about." It is clear that you haven't even passed freshman chemistry,
physics, or biology, yet you believe that you have a deeper
understanding of very subtle phenomena than the sum of all the great
theoreticians and experimentalists.

You believe this in spite of that fact you've never done an
experiment, and are incapable of understandng the results of
experiments.

You are incapable of understanding the theory, or even how to
manipulate equations, yet you fancy yourself more intelligent than
those who do.

Moreever, when someone who does know what they are talking about (i.e,
me) takes his time to teach you for free (when I get paid a very good
living to do this for pay) you get snippy. I can afford to be snippy -
I know my stuff. You don't.

This is the favorite
tactic of the kooks and trolls that you mock, changing the subject
when you're losing an argument.

You lost the argument long ago. If you don't understand the selection
rules for electronic transitions, nor even know what they are, you are
incapable of understanding the topic, period.

I couldn't win an argument with a dog, since a dog is simply too
stupid to comprehend. I am in the same situation with you.

- Hide quoted text -

- Show quoted text -

On May 20, 8:43 am, RP wrote:
On May 20, 7:31 am, The_Man wrote:





On May 20, 8:18 am, RP wrote:

On May 19, 6:50 pm, The_Man wrote:

On May 19, 4:42 pm, RP wrote:

On May 19, 12:30 pm, The_Man wrote:

On May 19, 1:11 pm, RP wrote:

On May 19, 11:43 am, The_Man wrote:

On May 19, 12:25 pm, RP wrote:

On May 19, 10:00 am, The_Man wrote:

On May 19, 9:31 am, RP wrote:

On May 18, 1:58 am, Pentcho Valev wrote:

http://groups.google.com/group/sci.p...hread/d28e668a...

On Oct 25, 1995, John Baez wrote in sci.physics:

"Nonetheless, it's a fact that a photon has a nonzero momentum."

A curious person asked:

"Are you therefore asserting that it has nonzero mass? If not, why
not?"

John Baez replied:

"You can see that I did not assert anything about the photon's mass. I
know what the photon's mass is, but I never talk about it around here
because the endless discussion of the photon's mass is boring, boring,
boring."

While I don't agree that the topic is boring, I do agree that it isn't
easy to understand, so I'll try to explain it. Photons don't exist.
Nothing that doesn't exist has mass.

You state as a fact what seems more like an opinion.

If special relativity is opinion, then I retract my statement

Photons have (spin) angular mometum. If photons "don't exist", the law
of conservation of angular momentum falls apart. Along with it, the
understanding of which electronic transitons occur, and how often they
occur (what we call "selection rules")

Photons don't have anything, because they don't exist. Momentum is
globally conserved, not locally conserved. The solution to the
radiation reaction problem is that there isn't a reaction, just an
action. Electrons are foward acting particles. The past light cone
contains events that influence the electron, the foward light cone
those events that it influences. There is no experimental evidence of
electron recoil under the influence of its own emitted radiation.
Before replying please note a distinction between a single electron
and an array of electrons, the latter of which produces feedback
effects.

If you can explain why s- d transitions don't occur, or why p-p
transitions don't occur, or why d-d transitions for high spin Fe 3+
occur with such a small probability, I'm all ears.

The atom is an electronic structure in equilibrium. Some states are
unstable and others impossible.

A sodium atoms is electronically excited by heating it. Sodium emits
light at two very specific wavelengths. These correspond to
transitions from 2P 3/2 - 1S0, or 2P 1/2 - 1S0. In the proces, the
atom "loses" 1 unit of orbital angular momentum. Where did it go? This
is a simple question...

It goes to all other electrons via interaction with the source
electrons, directly, at c.

So the angular momentum of the other electrons change? Yet, if that
happened, you would violate the law of conservation of energy. There
is an even easier refutation - suppose we talking about the spectrum
of hydrogen, which has NO other electrons.

Now the ball is in your court, to explain the selection rules for
hydrogen: Delta l = +/1 1, delta s =0; delta m sub l = 0, +/-1.

Because of the quantum states of recieving
atoms, they absorb energy by the same amount, and that you assume that
this absorbed energy all derived from the same single point source is
for you to prove, not me.

Proven LONG ago. Lasers do single photo absorption experiments all the
time. This is the so-called "resonance condition" - It is the basis of
all spectroscopy.

That theory goes against all the known laws
of em. Statistically your version works "because" of global momentum
conservation.

The 1 unit of orbital angular momentum is carried away by the photon,
which has a spin angular momentum of 1. We never see electronic
transitions involving changes of anything OTHER than 1 unit of angular
momentum.

If you can tell us how you can derive all the selection rules (for
both 1 electron and multi-electron atoms) WITHOUT a photon, all of
chemistry and physics would be glad to hear it. Not the least, that
you would also have to explain the photoelectric effect, and how TV's
work.

See above. It's more or less a holographic process.

No. There is a one-to-one correspondence between the number of photns
absorbed and the number of electrons emitted during the photoelectric
effect.

Incorrect. You can't count the photons, only the photoelectrons. At
any given moment a surface is being impinged upon by radiation from
very many sources.

Actually, photon counting is relatively (no pun intended) simple. It
is done all the time - it is necesary to determine the quantum
efficiency of such processes as fluorescence.

As for the "surface being impinged upon by radiation from very many
sources", this suggests a very incompetent experimentalist. The
surface would be sputtered with Ar+ to clean crap off the surface, the
surface would be kept in high vacuum, and all radiation sources other
than the source would be shielded against.

If I'm allowed to speak, for arguments sake, of
this radiation in terms of photons, then the number of photons
striking the target is far greater than the photoelectrons.

I didn't say "striking the target", I said "absorbed". The number of
photons NOT absorbed strikes a detector, and are counted. The number
of electrons ejected is also easily counted. So long as the photons
had the necessry energy, yadda yadda yadda, the # photons launched = #
photons NOT absorbed + # electrons emitted.

You simply
disgard the fact that the surface is pumped, because it suits your
argument. Again, it is still for you to prove the one-to-one
correspondence that you assume.

It is more straightforward to analyze in terms of absorption by atoms,
rather surfaces. You are confusing the fact of the probability of
absorption of a particular photon, which is usually less than 1, with
the probability that upon absorption, the photon will cause a change
in orbital angular momentum of 1 unit, which is exactly1. Of course,
the photon has to have the right energy to cause an electronic
transition, blah,blah, blah, which you don't understand anyway.

The emission spectrum of hydrogen was explained by Bohr - it is quite
simple. The spectra of heavier elements is more complicated (electron
electron repulsion complicates the analysis). Each line corresponds to
the energy difference between two electronic levels in hydrogen, ALL
of which differ in orbital angular momentum by sqrt(2) h bar.

If that isn't good enough for you, too bad. I don't have time to
repeat the results of 100 years of experiments.

You're the contrary one. Just show which lines of ANY atom result from
a change in other than 1 unit of angular momentum. If you understood L-
S or j-j coupling, I would put it in the requisite clarifications, but
I am sure that it is over your head. Just as the normal or anomalous
Zeeman effect would further prove my point.

Or show that the absorption of one photon by a surface ejects more
or,less than one electron.

Given that you've already decided that I know nothing about quantum
processes, and especially that you've made Gisse your bed-partner in
ad hominem, you aren't deserving of any more of my time.

Translation: You can't answer any of the questions I posed.

You're just
another loser hanging out here becuase its the only place where you
feel any self-worth, that is, amongs the kooks and trolls.

Translation: You feel inferior because you don't even know what spin-
orbit coupling is, or Russell-Saunders or j-j coupling, nor selection
rules, yet you want to impress yourself that you know something.

But from
my perspective you are nothing more than one of the latter.

Translation: Boo-hoo, WAAAAAAAAAA!

I rest my case. You're developmentally challenged, maybe even a bit
schizophrenic.

Yet I have a Ph.D., done a post-doc, and been granted a professorship.
And you? "Would you like fries with that?"


The point you missed, is that spin-orbit coupling is imaterial to
answering the question "what is light"? Further proof that you aren't
capable of maintaining a mature conversation. This is the favorite
tactic of the kooks and trolls that you mock, changing the subject
when you're losing an argument.- Hide quoted text -

- Show quoted text -

You lost the argument long ago. If you don't understand the selection
rules for electronic transitions, nor even know what they are, you are
incapable of understanding the topic, period.

I couldn't win an argument with a dog, since a dog is simply too
stupid to comprehend. I am in the same situation with you.- Hide quoted text -

I hope you've managed to impress somebody, because it wasn't me.
You're an idiot.

On May 20, 9:57 am, RP wrote:
You lost the argument long ago. If you don't understand the selection
rules for electronic transitions, nor even know what they are, you are
incapable of understanding the topic, period.

I couldn't win an argument with a dog, since a dog is simply too
stupid to comprehend. I am in the same situation with you.- Hide quoted text -

I hope you've managed to impress somebody, because it wasn't me.

Woof! Woof! Aarf! Aarf!

You're an idiot.

You're the one who doesn't understand angular momenta. Go play fetch,
or roll over and play dead.

On May 20, 9:09 am, The_Man wrote:
On May 20, 9:57 am, RP wrote:

You lost the argument long ago. If you don't understand the selection
rules for electronic transitions, nor even know what they are, you are
incapable of understanding the topic, period.

I couldn't win an argument with a dog, since a dog is simply too
stupid to comprehend. I am in the same situation with you.- Hide quoted text -

I hope you've managed to impress somebody, because it wasn't me.

Woof! Woof! Aarf! Aarf!

You're an idiot.

You're the one who doesn't understand angular momenta. Go play fetch,
or roll over and play dead.

http://focus.aps.org/story/v13/st21

Clearly light propagates in wave form.

And you're still an idiot.

Now define the g factor for us, as if that actually has any bearing on
the above article.

On May 20, 10:59 am, RP wrote:
On May 20, 9:09 am, The_Man wrote:





On May 20, 9:57 am, RP wrote:

You lost the argument long ago. If you don't understand the selection
rules for electronic transitions, nor even know what they are, you are
incapable of understanding the topic, period.

I couldn't win an argument with a dog, since a dog is simply too
stupid to comprehend. I am in the same situation with you.- Hide quoted text -

I hope you've managed to impress somebody, because it wasn't me.

Woof! Woof! Aarf! Aarf!

You're an idiot.

You're the one who doesn't understand angular momenta. Go play fetch,
or roll over and play dead.

http://focus.aps.org/story/v13/st21

Clearly light propagates in wave form.

Clearly, in this experiment
Clearly, in the photoecletric effect experiment, it doesn't.
Wow! Sometimes a photon acts like a WAVE, and sometimes it acts like a
PARTICLE! Maybe I should PUBLISH this discovery!!!!!!!!!


And you're still an idiot.

I'm an idiot with a Ph.D., and you are an idiot with? High school
diploma? GED?


Now define the g factor for us, as if that actually has any bearing on
the above article.

There is more than one g factor? Which do you want?

- Hide quoted text -

- Show quoted text -