| Topic: |
Science > Physics |
| User: |
"" |
| Date: |
03 Mar 2005 12:55:55 AM |
| Object: |
Why can't waves cause the photoelectric effect |
Now we all know that light has been proven to consist of particles due
to the photoelectric effect, but do we really? Based on descriptions
found on websites and books such as "The Evolution of Physics" by
Albert Einstein/Leopold Infeld, not much thought has been put into what
waves ought to do when shined on a metal surface. It is surmised
(without any further reasoning) that the intensity of the light wave
would correspond to its amplitude or energy level, so we would expect
that higher intensity would increase the speed of the electrons from
the surface. We don't see this experimentally, so we must conclude that
light is a particle.
However, based upon what we know about how light is generated, this is
not a foregone conclusion. Light is generated by an electron
transistioning between energy levels within the atom. As such, the
duration of the wave generated is limited - probably to only a single
wave and its amplitude is also limited since the range of motion from
energy level to energy level is limited. This kind of wave would be
like what you would get if you took a taught rope and jerked it once
very quickly. The single wave would travel down the rope. A beam of
light would be a series of these small waves travelling through space.
It would not be a continuous wave of variable amplitude. The intensity
would not be due to the amplitude of the individual waves, but rather
the density of waves hitting an object. A bright object is simply
emitting more of these "single" waves per unit time than a dim object.
Now think of how such a wave would interact with the surface of metal.
Since the amplitude of the wave doesn't change, electrons are ejected
from the surface with the same energy no matter how many hit the
surface. The speed of the ejected electron would depend on the amount
of kinetic energy imparted by the wave which would correspond to the
frequency of the wave as is demonstrated by the fact that you can shake
out nuts with greater speed out of a tree if you shake it faster. This
is exactly the behavior seen in the photoelectric effect without
resorting to thinking of light as particles.
This seems terribly obvious based upon what we know about light
generation, so why is this argument not brought up?
fhuphoto
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| User: "Franz Heymann" |
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| Title: Re: Why can't waves cause the photoelectric effect |
03 Mar 2005 10:15:37 AM |
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<franklinhu@yahoo.com> wrote in message
news:1109832955.451534.297400@g14g2000cwa.googlegroups.com...
[snip]
transistioning
The English language does not have a verb "to transition".
The rest of your lengthy note is not worth commenting on.
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
.
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
03 Mar 2005 04:46:00 AM |
|
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wrote:
Now we all know that light has been proven to consist of particles due
to the photoelectric effect, but do we really?
Not "proven to be". With some assumptions (e.g. "light is only
absorbed by atoms in discrete quantities, but is continuous in free
flight), one could still explain this with waves. But the point is
that such assumptions would be totally farfetched and unjustified,
themselves in need of an explanation, and further are contradicted by
other evidence like e.g. the Compton effect. In other words, the photo
effect (and other effects) provided *strong* *evidence* for light
consisting of particles - but not proof.
Based on descriptions
found on websites and books such as "The Evolution of Physics" by
Albert Einstein/Leopold Infeld, not much thought has been put into what
waves ought to do when shined on a metal surface. It is surmised
(without any further reasoning) that the intensity of the light wave
would correspond to its amplitude or energy level,
Please define what exactly you mean with "intensity" and "energy
level" here.
so we would expect
that higher intensity would increase the speed of the electrons from
the surface. We don't see this experimentally, so we must conclude that
light is a particle.
That's only part of the reasoning. You conveniently ignore the other
part: that the speed of the electrons indeed increases when the
frequency is increased.
However, based upon what we know about how light is generated, this is
not a foregone conclusion. Light is generated by an electron
transistioning between energy levels within the atom.
That's one of several possibilities only.
As such, the
duration of the wave generated is limited - probably to only a single
wave
"single wave" makes little sense. Do you mean "single period"? If yes,
you are utterly wrong. Light frequencies are around 10^15 Hz, light
emission takes about 10^(-10) s. Hence there are 10^5 periods in every
wave train emitted.
and its amplitude is also limited since the range of motion from
energy level to energy level is limited.
The amplitude of the emitted wave has precisely nil to do with
the "range of motion from energy level to energy level".
This kind of wave would be
like what you would get if you took a taught rope and jerked it once
very quickly.
Wrong, false analogy and false numbers.
The single wave would travel down the rope.
Again, "single wave" makes no sense here.
A beam of
light would be a series of these small waves travelling through space.
No, it isn't. See the numbers above.
And even if it were: why couldn't several of these small wave trains
be at the same place and add up?
It would not be a continuous wave of variable amplitude. The intensity
would not be due to the amplitude of the individual waves, but rather
the density of waves hitting an object.
I can't comment on that as long as you don't tell me how exactly
you define "intensity".
A bright object is simply
emitting more of these "single" waves per unit time than a dim object.
Care to tell me what the big difference (conceptually) between your
concept of "single waves" and the concept of photons is? You *do*
know that photons are often described as *wave packets*, don't you?
[snip more of that]
This seems terribly obvious based upon what we know about light
generation, so why is this argument not brought up?
Essentially, it is. Photon = wave packet is an idea which has
been around for 70-80 years now. It works for many applications
- but ultimately, it fails if you look closer at the idea.
Bye,
Bjoern
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| User: "" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 12:53:19 AM |
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Bjoern Feuerbacher wrote:
franklinhu@yahoo.com wrote:
Now we all know that light has been proven to consist of particles
due
to the photoelectric effect, but do we really?
Not "proven to be". With some assumptions (e.g. "light is only
absorbed by atoms in discrete quantities, but is continuous in free
flight), one could still explain this with waves. But the point is
that such assumptions would be totally farfetched and unjustified,
themselves in need of an explanation, and further are contradicted by
other evidence like e.g. the Compton effect. In other words, the
photo
effect (and other effects) provided *strong* *evidence* for light
consisting of particles - but not proof.
Based on descriptions
found on websites and books such as "The Evolution of Physics" by
Albert Einstein/Leopold Infeld, not much thought has been put into
what
waves ought to do when shined on a metal surface. It is surmised
(without any further reasoning) that the intensity of the light
wave
would correspond to its amplitude or energy level,
Please define what exactly you mean with "intensity" and "energy
level" here.
I would say that this is indeed a major problem with describing this
experiment. The explanations I've seen leave this ambiguous and seem to
correspond to the idea that if it is brighter to our eyes, it has more
energy and more intensity. However, a more precise description would be
that the intensity could correspond to both a combination of the wave
amplitude and the number of wave cycles hitting an object per unit
time.
so we would expect
that higher intensity would increase the speed of the electrons
from
the surface. We don't see this experimentally, so we must conclude
that
light is a particle.
That's only part of the reasoning. You conveniently ignore the other
part: that the speed of the electrons indeed increases when the
frequency is increased.
No, I did explain in a commonsense way, that the faster you shake
something (increase frequency) If there are parts which are loosely
bound (like nuts in a tree), these parts will be ejected faster with an
increase in frequency.
However, based upon what we know about how light is generated,
this is
not a foregone conclusion. Light is generated by an electron
transistioning between energy levels within the atom.
That's one of several possibilities only.
As such, the
duration of the wave generated is limited - probably to only a
single
wave
"single wave" makes little sense. Do you mean "single period"? If
yes,
you are utterly wrong. Light frequencies are around 10^15 Hz, light
emission takes about 10^(-10) s. Hence there are 10^5 periods in
every
wave train emitted.
That is very interesting that it takes that long to eject light from a
single electron transition. How on earth did they determine such a tiny
time period? Do we have any theories on how it produces this 10^5 wave
packet from an electron moving from one energy level to another? Has
this length ever been experimentally measured?
In any case, my argument did not rely on it being a single period, but
rather that it have a definite and limited duration. This makes each of
the "wave packets" nearly identical to each other as they are
generated. It is this identical nature which explains why when it hits
the metal surface, it ejects electrons with a consistent speed.
and its amplitude is also limited since the range of motion from
energy level to energy level is limited.
The amplitude of the emitted wave has precisely nil to do with
the "range of motion from energy level to energy level".
A more precise description would be that the change of electron energy
from level to level is fairly fixed, which determines the frequency and
amplitude of the wave generated to be a fixed value.
This kind of wave would be
like what you would get if you took a taught rope and jerked it
once
very quickly.
Wrong, false analogy and false numbers.
OK, how about you jerk the rope 10^5 times and then let it go. The wave
train still travels down the rope as before and an identifiable entity.
The single wave would travel down the rope.
Again, "single wave" makes no sense here.
A beam of
light would be a series of these small waves travelling through
space.
No, it isn't. See the numbers above.
And even if it were: why couldn't several of these small wave trains
be at the same place and add up?
They could, but more likely they are all generated out of phase since
light generation would tend to be a rather random process. I would
think that whether a wave packet travels alone or along with other
waves, it's actual impact is similar to if had travelled alone.
It would not be a continuous wave of variable amplitude. The
intensity
would not be due to the amplitude of the individual waves, but
rather
the density of waves hitting an object.
I can't comment on that as long as you don't tell me how exactly
you define "intensity".
I think the idea of intensity is the key issue here. The usual theory
says that intensity corresponds to amplitude of the wave. The wave
itself is thought to be continuous like a radio carrier wave. My
interpretation is that intensity is measured as number of wave cycles
hitting a surface per unit time. The amplitude of all of the waves is
nearly identical.
A bright object is simply
emitting more of these "single" waves per unit time than a dim
object.
Care to tell me what the big difference (conceptually) between your
concept of "single waves" and the concept of photons is? You *do*
know that photons are often described as *wave packets*, don't you?
As you describe it, there is no difference. What I am describing is
exactly a wave packet - which is a wave phenomenon, not a particle
phenomenon. A photon described as a wave packet makes perfect sense in
the photolectric effect. A photon does not need to be a corpuscular
particle with a particular volume, diameter, etc. and occupies a
particular quanta of space.
[snip more of that]
This seems terribly obvious based upon what we know about light
generation, so why is this argument not brought up?
Essentially, it is. Photon = wave packet is an idea which has
been around for 70-80 years now. It works for many applications
- but ultimately, it fails if you look closer at the idea.
Then you would agree that the photon = wave packet is a reasonable
explanation for the photoelectric effect and that the photoelectric
effect by itself is not a great piece of evidence for the particle
nature of light since a wave explanation cannot be ruled out?
Why did Einstein get a Nobel prize for the photoelectric effect when it
had a fairly simple counterargument?
What are these other closer ideas which make the wave picture of
photons fail?
Bye,
Bjoern
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| User: "Old Man" |
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| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 04:44:11 AM |
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<franklinhu@yahoo.com> wrote in message
news:1109919199.567903.193540@o13g2000cwo.googlegroups.com...
Bjoern Feuerbacher wrote:
franklinhu@yahoo.com wrote:
I would say that this is indeed a major problem with describing this
experiment. The explanations I've seen leave this ambiguous and seem to
correspond to the idea that if it is brighter to our eyes, it has more
energy and more intensity. However, a more precise description would be
that the intensity could correspond to both a combination of the wave
amplitude and the number of wave cycles hitting an object per unit
time.
Ambiguous. Wrong. Where's the physics ? Physics is
unambiguous on this matter.
The energy flux ( J / s - m^2) of an electromagnetic wave is
proportional to E x B, where E = E_0*sin(w*t) is the electric
field strength, and B = B_0*sin(w*t) is the magnetic field
strength, whereof |E_0| = |B_0|.
Thus, the time averaged energy flux is proportional to (E_0)^2
and doesn't depend upon the frequency, w. The photon flux,
proportional to (E_0)^2 / hbar*w, is inversely proportional to
the frequency.
[Old Man]
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| User: "FrediFizzx" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 06:46:05 PM |
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"Old Man" <nomail@nomail.net> wrote in message
news:18KdnewMvNpiprXfRVn-tQ@prairiewave.com...
|
| <> wrote in message
| news:1109919199.567903.193540@o13g2000cwo.googlegroups.com...
| >
| > Bjoern Feuerbacher wrote:
| >> wrote:
| >
| > I would say that this is indeed a major problem with describing this
| > experiment. The explanations I've seen leave this ambiguous and seem
to
| > correspond to the idea that if it is brighter to our eyes, it has
more
| > energy and more intensity. However, a more precise description would
be
| > that the intensity could correspond to both a combination of the
wave
| > amplitude and the number of wave cycles hitting an object per unit
| > time.
|
| Ambiguous. Wrong. Where's the physics ? Physics is
| unambiguous on this matter.
|
| The energy flux ( J / s - m^2) of an electromagnetic wave is
| proportional to E x B, where E = E_0*sin(w*t) is the electric
| field strength, and B = B_0*sin(w*t) is the magnetic field
| strength, whereof |E_0| = |B_0|.
|
| Thus, the time averaged energy flux is proportional to (E_0)^2
| and doesn't depend upon the frequency, w. The photon flux,
| proportional to (E_0)^2 / hbar*w, is inversely proportional to
| the frequency.
What the heck is E_0? Your analysis is flawed.
E_0 = +,- sqrt(hbar*w^4*n/(4pi^2*c^3))
Where n is the number of identical photons. In natural units,
E_0 = +,- (w^2/2pi)(sqrt n)
So we can see it is both dependent on frequency and the number of
photons.
FrediFizzx
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| User: "Old Man" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
05 Mar 2005 08:24:49 AM |
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|
"FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:38sdhdF5pgu4sU1@individual.net...
"Old Man" <nomail@nomail.net> wrote in message
news:18KdnewMvNpiprXfRVn-tQ@prairiewave.com...
|
| < > wrote in message
| news:1109919199.567903.193540@o13g2000cwo.googlegroups.com...
| >
| > Bjoern Feuerbacher wrote:
| >> wrote:
| >
| > I would say that this is indeed a major problem with describing this
| > experiment. The explanations I've seen leave this ambiguous and seem
to
| > correspond to the idea that if it is brighter to our eyes, it has
more
| > energy and more intensity. However, a more precise description would
be
| > that the intensity could correspond to both a combination of the
wave
| > amplitude and the number of wave cycles hitting an object per unit
| > time.
|
| Ambiguous. Wrong. Where's the physics ? Physics is
| unambiguous on this matter.
|
| The energy flux ( J / s - m^2) of an electromagnetic wave is
| proportional to E x B, where E = E_0*sin(w*t) is the electric
| field strength, and B = B_0*sin(w*t) is the magnetic field
| strength, whereof |E_0| = |B_0|.
|
| Thus, the time averaged energy flux is proportional to (E_0)^2
| and doesn't depend upon the frequency, w. The photon flux,
| proportional to (E_0)^2 / hbar*w, is inversely proportional to
| the frequency.
What the heck is E_0? Your analysis is flawed.
FizzX suffers from delusions of competence.
E_0 = +,- sqrt(hbar*w^4*n/(4pi^2*c^3))
Where n is the number of identical photons.
No. "n" is the photon number density. (E_0)^2 is
proportional to the energy density .
In natural units,
E_0 = +,- (w^2/2pi)(sqrt n)
So we can see it is both dependent on frequency and the number of
photons.
As speed is dependent upon distance and time.
FrediFizzx
[Old Man]
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| User: "" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
11 Mar 2005 05:32:55 PM |
|
|
Old Man wrote:
"FrediFizzx" <fredifizzx@hotmail.com> wrote in message
news:38sdhdF5pgu4sU1@individual.net...
"Old Man" <nomail@nomail.net> wrote in message
news:18KdnewMvNpiprXfRVn-tQ@prairiewave.com...
|
| < > wrote in message
| news:1109919199.567903.193540@o13g2000cwo.googlegroups.com...
| >
| > Bjoern Feuerbacher wrote:
| >> wrote:
| >
| > I would say that this is indeed a major problem with describing
this
| > experiment. The explanations I've seen leave this ambiguous and
seem
to
| > correspond to the idea that if it is brighter to our eyes, it
has
more
| > energy and more intensity. However, a more precise description
would
be
| > that the intensity could correspond to both a combination of
the
wave
| > amplitude and the number of wave cycles hitting an object per
unit
| > time.
|
| Ambiguous. Wrong. Where's the physics ? Physics is
| unambiguous on this matter.
|
| The energy flux ( J / s - m^2) of an electromagnetic wave is
| proportional to E x B, where E = E_0*sin(w*t) is the electric
| field strength, and B = B_0*sin(w*t) is the magnetic field
| strength, whereof |E_0| = |B_0|.
|
| Thus, the time averaged energy flux is proportional to (E_0)^2
| and doesn't depend upon the frequency, w. The photon flux,
| proportional to (E_0)^2 / hbar*w, is inversely proportional to
| the frequency.
What the heck is E_0? Your analysis is flawed.
FizzX suffers from delusions of competence.
Check Milonni's book "The Quantum Vacuum" and tell me what E_0 is.
E_0 = +,- sqrt(hbar*w^4*n/(4pi^2*c^3))
Where n is the number of identical photons.
No. "n" is the photon number density. (E_0)^2 is
proportional to the energy density .
The is no such thing as (E_0)^2 physically. Try again.
In natural units,
E_0 = +,- (w^2/2pi)(sqrt n)
So we can see it is both dependent on frequency and the number of
photons.
As speed is dependent upon distance and time.
?? So what?
FrediFizzx
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| User: "PD" |
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| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 08:29:17 AM |
|
|
wrote:
I think the idea of intensity is the key issue here. The usual theory
says that intensity corresponds to amplitude of the wave. The wave
itself is thought to be continuous like a radio carrier wave. My
interpretation is that intensity is measured as number of wave cycles
hitting a surface per unit time. The amplitude of all of the waves is
nearly identical.
I see what your problem is. You are mixing up two notions: those of
waves and those of particles. This in itself is not a bad thing, but it
does indicate that you're not using the same language that the
physicists are.
In a wave picture, the "medium", if you like, is the region of space
between the source and -- in this case -- the metal surface. A wave is
taken to be a function at every point in space. The key phrase is "a
function", not "one of many functions". The value of the function at
that place is the strength of *the* electromagnetic field at that point
in space. The intensity of the light is (roughly) the averaged square
of that strength. The point is, at a given point in space (say, just
above the surface), there are not multiple waves. There are not
multiple electric fields. There is only a *single* function at each
point in space.
In the wave picture, anyway.
In the wave picture, then, the energy delivered is indeed dependent on
the amplitude of that wave (the maximum field strength), and indeed on
the frequency of the wave. (Note that the frequency of the wave does
NOT mean how many independent waves are hitting the surface; it means
how fast the oscillation of the single wave is.) In this picture, then,
the predicted behavior of the photoelectric effect would be that there
should be no minimum threshold of frequency of light for electrons to
be emitted (a stronger intensity of a lower frequency should eject them
as well); that for low-intensity light there should be a delay in time
before electrons are ejected (until enough energy is accumulated in the
metal); and that the energy of the ejected electrons should increase
with the intensity of the light (the more energy delivered, the faster
the electrons). What was astounding experimentally is that NONE of
these predictions held true. This is what led to the notion that light
must not be interpretable as a wave in this experiment.
You seem to be casting a picture that says that there are multiple,
locally confined waves, each containing a finite amount of energy,
impinging on the surface. That is not what the classical wave idea is.
Indeed, what you're proposing is essentially what photons are!
PD
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| User: "Franz Heymann" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 04:20:39 PM |
|
|
"PD" <pdraper@yahoo.com> wrote in message
news:1109946557.262723.183700@l41g2000cwc.googlegroups.com...
franklinhu@yahoo.com wrote:
I think the idea of intensity is the key issue here. The usual
theory
says that intensity corresponds to amplitude of the wave. The wave
itself is thought to be continuous like a radio carrier wave. My
interpretation is that intensity is measured as number of wave
cycles
hitting a surface per unit time. The amplitude of all of the waves
is
nearly identical.
I see what your problem is. You are mixing up two notions: those of
waves and those of particles. This in itself is not a bad thing, but
it
does indicate that you're not using the same language that the
physicists are.
In a wave picture, the "medium", if you like, is the region of space
between the source and -- in this case -- the metal surface. A wave
is
taken to be a function at every point in space. The key phrase is "a
function", not "one of many functions". The value of the function at
that place is the strength of *the* electromagnetic field at that
point
in space. The intensity of the light is (roughly) the averaged
square
of that strength. The point is, at a given point in space (say, just
above the surface), there are not multiple waves. There are not
multiple electric fields. There is only a *single* function at each
point in space.
In the wave picture, anyway.
In the wave picture, then, the energy delivered is indeed dependent
on
the amplitude of that wave (the maximum field strength), and indeed
on
the frequency of the wave.
In the wave picture, the power flow per unit area is E x H
There is no frequency involved in that expression.
{:-((
[snip]
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
.
|
|
|
| User: "PD" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 04:24:12 PM |
|
|
Franz Heymann wrote:
"PD" <pdraper@yahoo.com> wrote in message
news:1109946557.262723.183700@l41g2000cwc.googlegroups.com...
franklinhu@yahoo.com wrote:
I think the idea of intensity is the key issue here. The usual
theory
says that intensity corresponds to amplitude of the wave. The
wave
itself is thought to be continuous like a radio carrier wave. My
interpretation is that intensity is measured as number of wave
cycles
hitting a surface per unit time. The amplitude of all of the
waves
is
nearly identical.
I see what your problem is. You are mixing up two notions: those of
waves and those of particles. This in itself is not a bad thing,
but
it
does indicate that you're not using the same language that the
physicists are.
In a wave picture, the "medium", if you like, is the region of
space
between the source and -- in this case -- the metal surface. A wave
is
taken to be a function at every point in space. The key phrase is
"a
function", not "one of many functions". The value of the function
at
that place is the strength of *the* electromagnetic field at that
point
in space. The intensity of the light is (roughly) the averaged
square
of that strength. The point is, at a given point in space (say,
just
above the surface), there are not multiple waves. There are not
multiple electric fields. There is only a *single* function at each
point in space.
In the wave picture, anyway.
In the wave picture, then, the energy delivered is indeed dependent
on
the amplitude of that wave (the maximum field strength), and indeed
on
the frequency of the wave.
In the wave picture, the power flow per unit area is E x H
There is no frequency involved in that expression.
{:-((
Indeed, a typo. I meant to say "not on the frequency of the wave".
Sorry.
[snip]
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
.
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| User: "G=EMC^2 Glazier" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 04:33:41 PM |
|
|
If the wave is very short it can Bert
.
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 04:28:54 AM |
|
|
wrote:
Bjoern Feuerbacher wrote:
wrote:
[snip]
Please define what exactly you mean with "intensity" and "energy
level" here.
I would say that this is indeed a major problem with describing this
experiment. The explanations I've seen leave this ambiguous and seem to
correspond to the idea that if it is brighter to our eyes, it has more
energy and more intensity.
Well, what explanations *have* you read? Must I remind you again that
you shouldn't rely on popular science sources for learning something
about physics, but try to get a real education?
However, a more precise description would be
that the intensity could correspond to both a combination of the wave
amplitude and the number of wave cycles hitting an object per unit
time.
I don't think that any physics text would actually define it that
way.
A much more sensible assumption would be, IMO, that they used the
power consumption of the light source for determining the light
intensity, i.e. "intensity" refers to the energy content of the
light (and according to Maxwell's equations, therefore to the
square of its amplitude).
so we would expect
that higher intensity would increase the speed of the electrons
from
the surface. We don't see this experimentally, so we must conclude
that light is a particle.
That's only part of the reasoning. You conveniently ignore the other
part: that the speed of the electrons indeed increases when the
frequency is increased.
No, I did explain in a commonsense way, that the faster you shake
something (increase frequency) If there are parts which are loosely
bound (like nuts in a tree), these parts will be ejected faster with
an increase in frequency.
I must have missed that part of your explanation.
Anyway, this conclusion does not follow in general. For starters,
electrons bound in a metal have little to do with nuts in a tree.
[snip]
As such, the
duration of the wave generated is limited - probably to only a
single wave
"single wave" makes little sense. Do you mean "single period"? If
yes,
you are utterly wrong. Light frequencies are around 10^15 Hz, light
emission takes about 10^(-10) s. Hence there are 10^5 periods in
every wave train emitted.
That is very interesting that it takes that long to eject light from a
single electron transition. How on earth did they determine such a tiny
time period?
There are various possibilities, I don't remember all the details.
For the 100th time: open a book on atomic and molecular physics,
where all these things are explained.
IIRC, one method is to measure the "intrinsic line width" Gamma, which
is connected to the time of light emission by
t = h / Gamma,
where h is Planck's constant.
Do we have any theories on how it produces this 10^5 wave
packet from an electron moving from one energy level to another?
Yes. QM.
Has this length ever been experimentally measured?
What one can measure is e.g. the so-called "coherence length", AFAIK.
That gives a good estimate for the length of the light trains.
For the 101th time: open a book on atomic and molecular physics,
where all these things are explained.
In any case, my argument did not rely on it being a single period, but
rather that it have a definite and limited duration. This makes each of
the "wave packets" nearly identical to each other as they are
generated. It is this identical nature which explains why when it hits
the metal surface, it ejects electrons with a consistent speed.
As already mentioned, the idea of "photons = wave packets" is
anything but new.
and its amplitude is also limited since the range of motion from
energy level to energy level is limited.
The amplitude of the emitted wave has precisely nil to do with
the "range of motion from energy level to energy level".
A more precise description would be that the change of electron energy
from level to level is fairly fixed, which determines the frequency and
amplitude of the wave generated to be a fixed value.
Yes, that would be better. And even better would be if you talked
about "wave packet" here instead of "wave".
[snip]
It would not be a continuous wave of variable amplitude. The
intensity
would not be due to the amplitude of the individual waves, but
rather the density of waves hitting an object.
I can't comment on that as long as you don't tell me how exactly
you define "intensity".
I think the idea of intensity is the key issue here. The usual theory
says that intensity corresponds to amplitude of the wave.
That theory would be called "Maxwell's equation". You did not bother
to study them in the meantime, did you?
The wave
itself is thought to be continuous like a radio carrier wave.
No, that's not necessary for associating intensity with the amplitude.
[snip]
A bright object is simply
emitting more of these "single" waves per unit time than a dim
object.
Care to tell me what the big difference (conceptually) between your
concept of "single waves" and the concept of photons is? You *do*
know that photons are often described as *wave packets*, don't you?
As you describe it, there is no difference. What I am describing is
exactly a wave packet - which is a wave phenomenon, not a particle
phenomenon.
Err, ever heard of "wave-particle duality"? That is exactly the point
- that photons have partly wave properties and partly particle properties!
You are free to try explaining the Compton effect also with your idea
of wave packets.
A photon described as a wave packet makes perfect sense in
the photolectric effect.
As a qualitative picture, yes. But as usual, as soon as you
try to get quantitative, this won't work out.
A photon does not need to be a corpuscular
particle with a particular volume, diameter, etc. and occupies a
particular quanta of space.
Err, no particle physicist ever pictured a photon as having a volume,
diameter etc.
And what exactly is "a quanta of space" supposed to mean? First,
"quanta" is plural; the singular would be "quantum". Secondly, we
don't know if space is quantized.
This seems terribly obvious based upon what we know about light
generation, so why is this argument not brought up?
Essentially, it is. Photon = wave packet is an idea which has
been around for 70-80 years now. It works for many applications
- but ultimately, it fails if you look closer at the idea.
Then you would agree that the photon = wave packet is a reasonable
explanation for the photoelectric effect
A *qualitative* explanation, yes. Feel free to try to *quantify* this
and see if it still works.
and that the photoelectric
effect by itself is not a great piece of evidence for the particle
nature of light since a wave explanation cannot be ruled out?
Again, there is not such a big difference between "wave packet"
and "particle".
Why did Einstein get a Nobel prize for the photoelectric effect when it
had a fairly simple counterargument?
Because what you presented is not a counterargument to the particle
interpretation. You simply assumed that physicists mean "something
with a definite volume, diameter etc." when they say "particle". But
they don't.
The point of Einstein's interpretation was *not* that "light consists
of things which have a definite volume, diameter etc.". It was
"light consists of discrete things with definite energy, the energy
is not distributed continuously over the wave". And nothing you said
contradicts that.
What are these other closer ideas which make the wave picture of
photons fail?
1) Doing quantitative analysis instead of just handwavings for the
photo effect.
2) Compton effect.
3) Essentially every single effect predicted by QED.
Bye,
Bjoern
.
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| User: "Franz Heymann" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
04 Mar 2005 04:20:38 PM |
|
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"Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in
message news:d09d96$h72$1@news.urz.uni-heidelberg.de...
franklinhu@yahoo.com wrote:
Bjoern Feuerbacher wrote:
franklinhu@yahoo.com wrote:
[snip]
Please define what exactly you mean with "intensity" and "energy
level" here.
I would say that this is indeed a major problem with describing
this
experiment. The explanations I've seen leave this ambiguous and
seem to
correspond to the idea that if it is brighter to our eyes, it has
more
energy and more intensity.
Well, what explanations *have* you read? Must I remind you again
that
you shouldn't rely on popular science sources for learning something
about physics, but try to get a real education?
However, a more precise description would be
that the intensity could correspond to both a combination of the
wave
amplitude and the number of wave cycles hitting an object per unit
time.
I don't think that any physics text would actually define it that
way.
A much more sensible assumption would be, IMO, that they used the
power consumption of the light source for determining the light
intensity, i.e. "intensity" refers to the energy content of the
light (and according to Maxwell's equations, therefore to the
square of its amplitude).
so we would expect
that higher intensity would increase the speed of the electrons
from
the surface. We don't see this experimentally, so we must
conclude
that light is a particle.
That's only part of the reasoning. You conveniently ignore the
other
part: that the speed of the electrons indeed increases when the
frequency is increased.
No, I did explain in a commonsense way, that the faster you shake
something (increase frequency) If there are parts which are
loosely
bound (like nuts in a tree), these parts will be ejected faster
with
an increase in frequency.
I must have missed that part of your explanation.
Anyway, this conclusion does not follow in general. For starters,
electrons bound in a metal have little to do with nuts in a tree.
[snip]
As such, the
duration of the wave generated is limited - probably to only a
single wave
"single wave" makes little sense. Do you mean "single period"? If
yes,
you are utterly wrong. Light frequencies are around 10^15 Hz,
light
emission takes about 10^(-10) s. Hence there are 10^5 periods in
every wave train emitted.
That is very interesting that it takes that long to eject light
from a
single electron transition. How on earth did they determine such a
tiny
time period?
There are various possibilities, I don't remember all the details.
For the 100th time: open a book on atomic and molecular physics,
where all these things are explained.
IIRC, one method is to measure the "intrinsic line width" Gamma,
which
is connected to the time of light emission by
t = h / Gamma,
where h is Planck's constant.
Yes. This was originally done by Unsold (umlaut on the o) in a series
of papers in Z. f. Phys. during the years 1930 - 1932. The
measurements wee not very precise, because of the background width due
to doppler broadening.
Do we have any theories on how it produces this 10^5 wave
packet from an electron moving from one energy level to another?
Yes. QM.
Has this length ever been experimentally measured?
What one can measure is e.g. the so-called "coherence length",
AFAIK.
That gives a good estimate for the length of the light trains.
I don't think the coherence length of incoherent light has ever been
measured, but I may be mistaken. I think the time and spatial lengths
of photon waves have only ever been measured by measuring natural line
widths in spontaneous emission. Tannoudji-Cohen and his colleagues
developed powerful techniques in the early eighties for avoiding the
doppler broadening background, and their measured line widths are in
excellent agreement with the values predicted from the computed atomic
wave functions..
[snip]
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
.
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| User: "Bjoern Feuerbacher" |
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| Title: Re: Why can't waves cause the photoelectric effect |
08 Mar 2005 10:33:41 AM |
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Franz Heymann wrote:
"Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in
message news:d09d96$h72$1@news.urz.uni-heidelberg.de...
[snip]
IIRC, one method is to measure the "intrinsic line width" Gamma,
which is connected to the time of light emission by
t = h / Gamma,
where h is Planck's constant.
Yes. This was originally done by Unsold (umlaut on the o) in a series
of papers in Z. f. Phys. during the years 1930 - 1932. The
measurements wee not very precise, because of the background width due
to doppler broadening.
Interesting.
BTW, is this the same Unsoeld who wrote this book on astronomy?
<http://www.amazon.com/exec/obidos/tg/detail/-/3540678778/qid=1110299500/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-6206700-7016708?v=glance&s=books&n=507846>
If yes, then he was born in a small village just about 10 kilometers
from my home town. ;-)
[snip]
What one can measure is e.g. the so-called "coherence length",
AFAIK.
That gives a good estimate for the length of the light trains.
I don't think the coherence length of incoherent light has ever been
measured, but I may be mistaken.
I vaguely remember that I measured this even in my lab work during
my study, but I also may be mistaken. ;-)
[snip]
Bye,
Bjoern
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| User: "Franz Heymann" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
08 Mar 2005 04:30:03 PM |
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"Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in
message news:d0kk55$g95$1@news.urz.uni-heidelberg.de...
Franz Heymann wrote:
"Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in
message news:d09d96$h72$1@news.urz.uni-heidelberg.de...
[snip]
IIRC, one method is to measure the "intrinsic line width" Gamma,
which is connected to the time of light emission by
t = h / Gamma,
where h is Planck's constant.
Yes. This was originally done by Unsold (umlaut on the o) in a
series
of papers in Z. f. Phys. during the years 1930 - 1932. The
measurements wee not very precise, because of the background width
due
to doppler broadening.
Interesting.
BTW, is this the same Unsoeld who wrote this book on astronomy?
<http://www.amazon.com/exec/obidos/tg/detail/-/3540678778/qid=11102995
00/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-6206700-7016708?v=glance&s=books
&n=507846>
If yes, then he was born in a small village just about 10 kilometers
from my home town. ;-)
I am talking about A. Unsoeld.
Does it gel?
[snip]
What one can measure is e.g. the so-called "coherence length",
AFAIK.
That gives a good estimate for the length of the light trains.
I don't think the coherence length of incoherent light has ever
been
measured, but I may be mistaken.
I vaguely remember that I measured this even in my lab work during
my study, but I also may be mistaken. ;-)
Are you sure it was not laser light?
Actually, come to think of it, one could do it by doing any simple
interference observation, and introducing extra path lengths in one of
the paths until the interference fringes become washed out. So I have
probably been barking up the wrong tree.
Perhaps some other reader can confirm the position?
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
.
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| User: "Bjoern Feuerbacher" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
09 Mar 2005 04:01:19 AM |
|
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Franz Heymann wrote:
"Bjoern Feuerbacher" <feuerbac@thphys.uni-heidelberg.de> wrote in
message news:d0kk55$g95$1@news.urz.uni-heidelberg.de...
Franz Heymann wrote:
[snip]
Yes. This was originally done by Unsold (umlaut on the o) in a
series
of papers in Z. f. Phys. during the years 1930 - 1932. The
measurements wee not very precise, because of the background width
due to doppler broadening.
Interesting.
BTW, is this the same Unsoeld who wrote this book on astronomy?
<http://www.amazon.com/exec/obidos/tg/detail/-/3540678778/qid=11102995
00/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/104-6206700-7016708?v=glance&s=books
&n=507846>
If yes, then he was born in a small village just about 10 kilometers
from my home town. ;-)
I am talking about A. Unsoeld.
Does it gel?
Albrecht Unsoeld, indeed.
[snip]
What one can measure is e.g. the so-called "coherence length",
AFAIK.
That gives a good estimate for the length of the light trains.
I don't think the coherence length of incoherent light has ever
been measured, but I may be mistaken.
I vaguely remember that I measured this even in my lab work during
my study, but I also may be mistaken. ;-)
Are you sure it was not laser light?
IIRC, I did some interference experiments with laser light, and
afterwards some measurements with incoherent light in order to
measure the coherence length.
Actually, come to think of it, one could do it by doing any simple
interference observation, and introducing extra path lengths in one of
the paths until the interference fringes become washed out.
Yes, I remember something like that. That was about 9 years ago,
so I'm not sure. I'll try if I can find that stuff again somewhere...
[snip]
Bye,
Bjoern
.
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| User: "bz" |
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| Title: Re: Why can't waves cause the photoelectric effect |
09 Mar 2005 08:22:24 AM |
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Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in news:d0mhhg
$7ki$1@news.urz.uni-heidelberg.de:
IRC, I did some interference experiments with laser light, and
afterwards some measurements with incoherent light in order to
measure the coherence length.
Does this imply that even light from incoherent sources, such as a light
bulb, when properly filtered, displays some coherence, when studied at a
particular wave length? I am assuming that you were NOT measuring the
coherence lenght of white light from an incoherent source.
I guess white lasers must have 3 coherence lengths.
[quote http://www.edinst.com/whitelaser.htm ]
Wavelength (nm) red at 671nm, blue at 473nm, green at 532nm
Transverse mode TEM00
Operating mode CW
Output power (mW) 100, 150, 200
Operating temperature (?) 15-35
Beam divergence, full angle (mrad) <1.5
Beam deflection (mrad) 1
Expected lifetime (hours) 10000
[unquote]
I want one!
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
.
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| User: "Franz Heymann" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
10 Mar 2005 02:19:48 AM |
|
|
"bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message
news:Xns9614552F7FAB3WQAHBGMXSZHVspammote@130.39.198.139...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in
news:d0mhhg
$7ki$1@news.urz.uni-heidelberg.de:
IRC, I did some interference experiments with laser light, and
afterwards some measurements with incoherent light in order to
measure the coherence length.
Does this imply that even light from incoherent sources, such as a
light
bulb, when properly filtered, displays some coherence, when studied
at a
particular wave length? I am assuming that you were NOT measuring
the
coherence lenght of white light from an incoherent source.
I guess white lasers must have 3 coherence lengths.
[quote http://www.edinst.com/whitelaser.htm ]
Wavelength (nm) red at 671nm, blue at 473nm, green at 532nm
Transverse mode TEM00
Operating mode CW
Output power (mW) 100, 150, 200
Operating temperature (?) 15-35
Beam divergence, full angle (mrad) <1.5
Beam deflection (mrad) 1
Expected lifetime (hours) 10000
[unquote]
I want one!
It can only be three lasers operating at three different wavelengths.
I suppose it can all be built into one container.
Two different wavlengths cannot be coherent.
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
.
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| User: "bz" |
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| Title: Re: Why can't waves cause the photoelectric effect |
10 Mar 2005 06:43:15 AM |
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in news:d0ovv3$1rs
$10@nwrdmz01.dmz.ncs.ea.ibs-infra.bt.com:
['white lasers']
It can only be three lasers operating at three different wavelengths.
of course.
I suppose it can all be built into one container.
It is being done.
Two different wavlengths cannot be coherent.
I beg to differ, no insult intended.
Each can be independently coherent.
It should be easy to test; take a prism or grating and separate the three
different wave lengths. Test the coherence length of each.
The separation of the 3 beams didn't suddenly create coherence, where none
existed before. It just made it easier to measure the coherence.
Another test:
Three different radio transmitters on different frequencies can feed the
same antenna system. The three different signals radiate independently.
Once caution: any nonlinear element in the antenna or feed lines will allow
the signals to 'mix' and will produce spurious 'sum and difference'
signals.
--
bz
please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.
bz+sp@ch100-5.chem.lsu.edu remove ch100-5 to avoid spam trap
.
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| User: "Franz Heymann" |
|
| Title: Re: Why can't waves cause the photoelectric effect |
11 Mar 2005 01:43:48 AM |
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"bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message
news:Xns96154461A49F8WQAHBGMXSZHVspammote@130.39.198.139...
"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in
news:d0ovv3$1rs
$10@nwrdmz01.dmz.ncs.ea.ibs-infra.bt.com:
['white lasers']
It can only be three lasers operating at three different
wavelengths.
of course.
I suppose it can all be built into one container.
It is being done.
Two different wavlengths cannot be coherent.
I beg to differ, no insult intended.
Each can be independently coherent.
Of course.
It should be easy to test; take a prism or grating and separate the
three
different wave lengths. Test the coherence length of each.
I am not querying the coherence of each laser separately, so you are
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blacketttilting at windmills.
{:-((
[snip]
.
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| User: "Franz Heymann" |
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| Title: Re: Why can't waves cause the photoelectric effect |
11 Mar 2005 05:46:49 AM |
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"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in message
news:d0ri7j$k3o$1@nwrdmz02.dmz.ncs.ea.ibs-infra.bt.com...
"bz" <bz+sp@ch100-5.chem.lsu.edu> wrote in message
news:Xns96154461A49F8WQAHBGMXSZHVspammote@130.39.198.139...
"Franz Heymann" <notfranz.heymann@btopenworld.com> wrote in
news:d0ovv3$1rs
$10@nwrdmz01.dmz.ncs.ea.ibs-infra.bt.com:
['white lasers']
It can only be three lasers operating at three different
wavelengths.
of course.
I suppose it can all be built into one container.
It is being done.
Two different wavlengths cannot be coherent.
I beg to differ, no insult intended.
Each can be independently coherent.
Of course.
It should be easy to test; take a prism or grating and separate
the
three
different wave lengths. Test the coherence length of each.
I am not querying the coherence of each laser separately, so you are
The words which disappeared magically are "tilting at windmills"
--
Franz
"A first-rate laboratory is one in which mediocre scientists can
produce outstanding work"
P.M.S. Blackett
.
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| User: "Dr. Photon" |
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| Title: Re: Why can't waves cause the photoelectric effect |
11 Mar 2005 12:03:09 PM |
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bz <bz+sp@ch100-5.chem.lsu.edu> wrote in message news:<Xns9614552F7FAB3WQAHBGMXSZHVspammote@130.39.198.139>...
Bjoern Feuerbacher <feuerbac@thphys.uni-heidelberg.de> wrote in news:d0mhhg
$7ki$1@news.urz.uni-heidelberg.de:
IRC, I did some interference experiments with laser light, and
afterwards some measurements with incoherent light in order to
measure the coherence length.
Does this imply that even light from incoherent sources, such as a light
bulb, when properly filtered, displays some coherence, when studied at a
particular wave length? I am assuming that you were NOT measuring the
coherence lenght of white light from an incoherent source.
You can measure the coherence length of pretty much any source you
like using a Michelson interferometer. For most purposes the term
coherent or incoherent source only refers to narrow linewidth or broad
linewidth source, with no hard definition of the difference.
Intriguingly, coherence length *does* increase when you filter the
spectrum, and is not related to whether the source is coherent or
incoherent. So if you take a white light bulb, pass the light through
a 0.01nm notch filter, then you end up with quite coherent light. You
can increase the coherence further by narrowing the filter. The
penalty you pay is that the intensity drops, so a white light bulb
filtered down to 0.00001nm will be as coherent as a medium quality
laser, but hardly any light will pass!
I guess white lasers must have 3 coherence lengths.
[quote http://www.edinst.com/whitelaser.htm ]
Wavelength (nm) red at 671nm, blue at 473nm, green at 532nm
Transverse mode TEM00
Operating mode CW
Output power (mW) 100, 150, 200
Operating temperature (?) 15-35
Beam divergence, full angle (mrad) <1.5
Beam deflection (mrad) 1
Expected lifetime (hours) 10000
[unquote]
I want one!
There are other white light lasers which are continuous cover the
whole visible spectrum. They are pulsed lasers which attain the broad
spectrum due to emitting *very* short pulses (a few fs). I am not
certain, but that laser used in the "electron uncertainty in time"
measurement probably counts as a white light laser. Unfortunately it
would also be *much* more expensive to use and maintain than the
three-colour laser you refer to.
BR
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| User: "RP" |
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| Title: Re: Why can't waves cause the photoelectric effect |
03 Mar 2005 11:42:47 PM |
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wrote:
Now we all know that light has been proven to consist of particles due
to the photoelectric effect, but do we really? Based on descriptions
found on websites and books such as "The Evolution of Physics" by
Albert Einstein/Leopold Infeld, not much thought has been put into what
waves ought to do when shined on a metal surface. It is surmised
(without any further reasoning) that the intensity of the light wave
would correspond to its amplitude or energy level, so we would expect
that higher intensity would increase the speed of the electrons from
the surface. We don't see this experimentally, so we must conclude that
light is a particle.
However, based upon what we know about how light is generated, this is
not a foregone conclusion. Light is generated by an electron
transistioning between energy levels within the atom. As such, the
duration of the wave generated is limited - probably to only a single
wave and its amplitude is also limited since the range of motion from
energy level to energy level is limited. This kind of wave would be
like what you would get if you took a taught rope and jerked it once
very quickly. The single wave would travel down the rope. A beam of
light would be a series of these small waves travelling through space.
It would not be a continuous wave of variable amplitude. The intensity
would not be due to the amplitude of the individual waves, but rather
the density of waves hitting an object. A bright object is simply
emitting more of these "single" waves per unit time than a dim object.
Now think of how such a wave would interact with the surface of metal.
Since the amplitude of the wave doesn't change, electrons are ejected
from the surface with the same energy no matter how many hit the
surface. The speed of the ejected electron would depend on the amount
of kinetic energy imparted by the wave which would correspond to the
frequency of the wave as is demonstrated by the fact that you can shake
out nuts with greater speed out of a tree if you shake it faster. This
is exactly the behavior seen in the photoelectric effect without
resorting to thinking of light as particles.
This seems terribly obvious based upon what we know about light
generation, so why is this argument not brought up?
fhuphoto
Your argument, taken in context, is reasonable. My version involved
window panes. A series of sonic waves impinging on a wall of identical
window panes would result in some breaking and others not, even though
all identical. Now add to this resonance effect a preexisting
oscillation of the panes (analogous to thermal activity) and in this
case only a single incident wave will cause some panes to shatter and
some not, all still identical but differing in phase superposition
with the incident wave.
IOW, Carlip's argument that the energy must *build up* is not even
wrong. The electrons, like the agitated window panes, already have
energy, and some are already ejecting from the surface even without
incident radiation. Thus one can expect that an addition of energy
from an outside source would increase the probability of ejection. It
isn't that simple, but it isn't so very complicated either. Thermions
are evidence enough of the general idea.
Photoelectrons, OTOH, are linked to radiant energy above a given
frequency. In order to explain this relationship lets go back to the
wall of window panes. Note first that they have a natural resonant
frequency, and thus we can expect that impinging waves of this
frequency will lead to a drastic change in the probability of pane
breakage over that incurred from lesser frequencies, the latter of
which will tend to flex all of the windows simultaneously, thus
increasing pressure inside the building, which in turn counters the
inward flexing of the windows, i.e. these lower frequency waves are
dampened in the material. The wall of windows will act essentially as
a low pass filter. Higher frequencies will be absorbed because a
single pane cannot oscillate at these frequencies in its entirety;
it's dimensions are greater than the wavelength, and thus local
oscillations cause counter forces on the pane as a whole. Internal
pressures fluctuations in the structure are dampened as well.
The electronic version is that internal electromagnetic "pressures"
superpose over the radiation pressures, canceling local effects on a
given electron, on average. Higher frequency incident radiation isn't
dampened locally, because the interference nodes are local, i.e.
affecting single electrons differently than their neighbors.
Richard Perry
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| User: "RP" |
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| Title: Re: Why can't waves cause the photoelectric effect |
03 Mar 2005 11:48:57 PM |
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RP wrote:
franklinhu@yahoo.com wrote:
Now we all know that light has been proven to consist of particles due
to the photoelectric effect, but do we really? Based on descriptions
found on websites and books such as "The Evolution of Physics" by
Albert Einstein/Leopold Infeld, not much thought has been put into what
waves ought to do when shined on a metal surface. It is surmised
(without any further reasoning) that the intensity of the light wave
would correspond to its amplitude or energy level, so we would expect
that higher intensity would increase the speed of the electrons from
the surface. We don't see this experimentally, so we must conclude that
light is a particle.
However, based upon what we know about how light is generated, this is
not a foregone conclusion. Light is generated by an electron
transistioning between energy levels within the atom. As such, the
duration of the wave generated is limited - probably to only a single
wave and its amplitude is also limited since the range of motion from
energy level to energy level is limited. This kind of wave would be
like what you would get if you took a taught rope and jerked it once
very quickly. The single wave would travel down the rope. A beam of
light would be a series of these small waves travelling through space.
It would not be a continuous wave of variable amplitude. The intensity
would not be due to the amplitude of the individual waves, but rather
the density of waves hitting an object. A bright object is simply
emitting more of these "single" waves per unit time than a dim object.
Now think of how such a wave would interact with the surface of metal.
Since the amplitude of the wave doesn't change, electrons are ejected
from the surface with the same energy no matter how many hit the
surface. The speed of the ejected electron would depend on the amount
of kinetic energy imparted by the wave which would correspond to the
frequency of the wave as is demonstrated by the fact that you can shake
out nuts with greater speed out of a tree if you shake it faster. This
is exactly the behavior seen in the photoelectric effect without
resorting to thinking of light as particles.
This seems terribly obvious based upon what we know about light
generation, so why is this argument not brought up?
fhuphoto
Your argument, taken in context, is reasonable. My version involved
window panes. A series of sonic waves impinging on a wall of identical
window panes would result in some breaking and others not, even though
all identical. Now add to this resonance effect a preexisting
oscillation of the panes (analogous to thermal activity) and in this
case only a single incident wave will cause some panes to shatter and
some not, all still identical but differing in phase superposition with
the incident wave.
IOW, Carlip's argument that the energy must *build up* is not even
wrong. The electrons, like the agitated window panes, already have
energy, and some are already ejecting from the surface even without
incident radiation. Thus one can expect that an addition of energy from
an outside source would increase the probability of ejection. It isn't
that simple, but it isn't so very complicated either. Thermions are
evidence enough of the general idea.
Photoelectrons, OTOH, are linked to radiant energy above a given
frequency. In order to explain this relationship lets go back to the
wall of window panes. Note first that they have a natural resonant
frequency, and thus we can expect that impinging waves of this frequency
will lead to a drastic change in the probability of pane breakage over
that incurred from lesser frequencies, the latter of which will tend to
flex all of the windows simultaneously, thus increasing pressure inside
the building, which in turn counters the inward flexing of the windows,
i.e. these lower frequency waves are dampened in the material. The wall
of windows will act essentially as a low pass filter. Higher frequencies
will be absorbed because a single pane cannot oscillate at these
frequencies in its entirety; it's dimensions are greater than the
wavelength, and thus local oscillations cause counter forces on the pane
as a whole.
Pardon me, I accidentally omitted a sentence here:
"Stresses in the glass are amplified".
Internal pressure fluctuations in the structure are
dampened as well.
And I should have added here something like, "which is analogous to
the restriction of the effect to surface electrons"
The electronic version is that internal electromagnetic "pressures"
superpose over the radiation pressures, canceling local effects on a
given electron, on average. Higher frequency incident radiation isn't
dampened locally, because the interference nodes are local, i.e.
affecting single electrons differently than their neighbors.
Richard Perry
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| User: "" |
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| Title: Re: Why can't waves cause the photoelectric effect |
03 Mar 2005 01:52:28 PM |
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wrote:
Now we all know that light has been proven to consist of particles due
to the photoelectric effect, but do we really? Based on descriptions
found on websites and books such as "The Evolution of Physics" by
Albert Einstein/Leopold Infeld, not much thought has been put into what
waves ought to do when shined on a metal surface. It is surmised
(without any further reasoning) that the intensity of the light wave
would correspond to its amplitude or energy level, so we would expect
that higher intensity would increase the speed of the electrons from
the surface. We don't see this experimentally, so we must conclude that
light is a particle.
There's more to it than that, though the conclusive experiments are
relatively recent. The basic reason to conclude from the photoelectric
effect that light is quantized comes from the appearance of "prompt
electrons," electrons that are kicked out of the surface before a wave
could have built up enough energy to eject them. Unless you're ready
to give up energy conservation, the observation of such prompt electrons
is pretty conclusive evidence against a wave picture.
The best experiment I know of is described in Davis and Mandel, in the
book _Coherence and Quantum Optics_ (ed. Mandel and Wolf, 1973). This
is a very careful observation of prompt electrons, emitted before there
would be time, in a wave model, for enough energy to build up even over
the entire cathode to overcome the potential barrier. The experiment
shows that the energy is *not* delivered continuously to a photodetector.
They do more, as well -- they look at the statistics of prompt electron
emission, and show that it matches the quantum mechanical predictions
for the distribution of photons.
Steve Carlip
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| User: "RP" |
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| Title: Re: Why can't waves cause the photoelectric effect |
03 Mar 2005 10:51:20 PM |
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wrote:
franklinhu@yahoo.com wrote:
Now we all know that light has been proven to consist of particles due
to the photoelectric effect, but do we really? Based on descriptions
found on websites and books such as "The Evolution of Physics" by
Albert Einstein/Leopold Infeld, not much thought has been put into what
waves ought to do when shined on a metal surface. It is surmised
(without any further reasoning) that the intensity of the light wave
would correspond to its amplitude or energy level, so we would expect
that higher intensity would increase the speed of the electrons from
the surface. We don't see this experimentally, so we must conclude that
light is a particle.
There's more to it than that, though the conclusive experiments are
relatively recent. The basic reason to conclude from the photoelectric
effect that light is quantized comes from the appearance of "prompt
electrons," electrons that are kicked out of the surface before a wave
could have built up enough energy to eject them. Unless you're ready
to give up energy conservation, the observation of such prompt electrons
is pretty conclusive evidence against a wave picture.
Conclusive? Persuasive maybe, but not conclusive. The electron has a
probability of absorbing energy and of emitting energy. It must be
receptive to the incident frequency (in phase). The electron has an
instantaneous momentum at the time of coincidence with some portion of
the sinusoidal field impinging on it, a momentum that will either
enhance or diminish its probability of ejection.
I was impressed with the recent article on the time version of the
double-slit experiment, not because it supported QM per se, but
because it further supported the fact that the electron has a
probability of interacting with incident fields that is independent of
the probability of the field to impinge on it.
Richard Perry
The best experiment I know of is described in Davis and Mandel, in the
book _Coherence and Quantum Optics_ (ed. Mandel and Wolf, 1973). This
is a very careful observation of prompt electrons, emitted before there
would be time, in a wave model, for enough energy to build up even over
the entire cathode to overcome the potential barrier. The experiment
shows that the energy is *not* delivered continuously to a photodetector.
They do more, as well -- they look at the statistics of prompt electron
emission, and show that it matches the quantum mechanical predictions
for the distribution of photons.
Steve Carlip
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