Established Member
That's again not what I asked, Mike. Here:
WHAT PARTICLES CARRY AWAY THE ENERGY OF THIS 'HEAT WAVE?'
Sorry about the emphasis, but you haven't answered a single question of mine, EVER. So, if you miss that one, I'm done...
[edit to add]
Well, in all truth, you have answered my questions, but not addressed any of the content of said questions to my satisfaction...
Order of Kilopi
Where is this proven? Why can't "orthodox" photons take part in the Compton effect? (You might have noticed that all of the theoretical work on the Compton effect to date has managed to do it all without your VNCPs.)Originally Posted by cyrek1
No. The square root of the expectation value of the velocity is the same as the velocity that you quote, which is not the same thing at all. This is a statistical quantity and represents the mean value of a large set of measurements carried out a set of identically prepared hydrogen atoms. Your GS velocity, on the other hand, corresponds to a single magnitude for the velocity. If you carried out the measurements, you would get the same result every time assuming that your measurement systems are up to the job. Note that QM doesn't prevent you from measuring the velocity as accurately as you like, it only sets limits in the simultaneous measurement of velocity and position.fortis
After all that work with Schroedingers equations(?), you derive the same answer for the GS velocity?
By a similar procedure, you can also work out the expectation value for the square of the orbital angular momentum. To your surprise you would(if you did the calculations) discover that the expectation value for the square of the angular momentum in the ground state is zero. Not only that, but the ground state wavefunction is an eigenfunction of the L^2 operator, which means that the "actual" value for the square of the angular momentum is zero. Tucked away in this, of course, is the fact that the orbital magnetic moment due to the electron in the ground state is also zero.
This, though correct, is completely at odds with the picture of the hydrogen atom that you are using.
Just because you can get an answer that looks right, it doesn't mean that you really understand what is going on.
As you can see from the wavefunction, the probability distribution for finding an electron extends to infinity. This means that there is no such thing as a diameter for the "cloud". In fact this means that there is a tiny, but finite, probability of finding the electron out beyond Pluto. Again, this seems ludicrous (and the probability is tiny), but is a natural consequence of a theory that has a far greater predictive capability than the Bohr model. Sticking with more sensible radii, you can see that the Bohr model is incapable of explaining the detection of an electron beyond the Bohr radius. It's a nice model as far as it goes, and it feels intuitive, but it is wrong.OK, so you create a cloud around the proton called an orbital or a wave function?.
Now what is the diameter of that cloud?
Will it be the same diameter as I have given for the GS radius times two?
By the way, you still haven't commented on the points that I make with respect to your model of the photon.
Established Member
Cyrek reply
Travis wrote
That's again not what I asked, Mike. Here:
WHAT PARTICLES CARRY AWAY THE ENERGY OF THIS 'HEAT WAVE?'
Sorry about the emphasis, but you haven't answered a single question
of mine, EVER. So, if you miss that one, I'm done...
Well, in all truth, you have answered my questions, but not addressed
any of the content of said questions to my satisfaction...
Fortis wrote
Cyrek quote
Travis
I said that the electron field composition that is composed of
'negative charged virtual particles', constitute the photon condensed
congregate. When this photon hits a surface that absorbs it, these
virtual particles 'rattle' the electrons in that surface to generate
the heat waves.
Remember, these are 'negative charged particles'. This is proven
because these photons can bounce electrons into outer orbits or in
space (Compton effect).
cyrek added reply
Travis....the particles that radiate the heat waves is probably the
negative virtual particles (remember, this is purely hypothetical amd
my opinion).
Maybe fortis might want to provide an answer also.
fortis wrote
Where is this proven? Why can't "orthodox" photons take part in the
Compton effect? (You might have noticed that all of the theoretical
work on the Compton effect to date has managed to do it all without
your VNCPs.)
reply....what are orthodox photoms?
To me, there are real photons that we see and the Feynman virtual
photons that he uses. I do not know what a VP is. Fortis....can you
explain what a virtual photon is?
cyrek quote
After all that work with Schroedingers equations(?), you derive the
same answer for the GS velocity?
fortis wrote
No. The square root of the expectation value of the velocity is the
same as the velocity that you quote, which is not the same thing at
all. This is a statistical quantity and represents the mean value
of a large set of measurements carried out a set of identically
prepared hydrogen atoms. Your GS velocity, on the other hand,
corresponds to a single magnitude for the velocity. If you carried
out the measurements, you would get the same result every time
assuming that your measurement systems are up to the job. Note that
QM doesn't prevent you from measuring the velocity as accurately as
you like, it only sets limits in the simultaneous measurement of
velocity and position.
By a similar procedure, you can also work out the expectation value
for the square of the orbital angular momentum. To your surprise you
would(if you did the calculations) discover that the expectation value
for the square of the angular momentum in the ground state is zero.
Not only that, but the ground state wavefunction is an eigenfunction
of the L^2 operator, which means that the "actual" value for the
square of the angular momentum is zero. Tucked away in this, of course,
is the fact that the orbital magnetic moment due to the electron in
the ground state is also zero.
This, though correct, is completely at odds with the picture of the
hydrogen atom that you are using.
Just because you can get an answer that looks right, it doesn't mean
that you really understand what is going on.
cyrek Quote:
OK, so you create a cloud around the proton called an orbital or a
wave function?.
Now what is the diameter of that cloud?
Will it be the same diameter as I have given for the GS radius times
two?
cyrek added reply
Fortis....will you explain what 'orbital angular momentum' is in your
own words?
fortis quote
As you can see from the wavefunction, the probability distribution
for finding an electron extends to infinity. This means that there
is no such thing as a diameter for the "cloud". In fact this means
that there is a tiny, but finite, probability of finding the electron
out beyond Pluto. Again, this seems ludicrous (and the probability is
tiny), but is a natural consequence of a theory that has a far greater
predictive capability than the Bohr model. Sticking with more sensible
radii, you can see that the Bohr model is incapable of explaining the
detection of an electron beyond the Bohr radius. It's a nice model as
far as it goes, and it feels intuitive, but it is wrong.
reply
The HA binary is explained and it has formulas for its solutions for
'standing waves' only.
QM comes up with these clouds that have thousands of solutions in
complex HA environments. No wonder it is correct with all these
possible solutions.
fortis
By the way, you still haven't commented on the points that I make with
respect to your model of the photon.
cyrek reply
The electric field is real. It is composed of VNCP's that normally
distribute themselves uniformally through mutual repulsion around the
electron (my hypothesis). When the electron transits (returns) to its
original position, its magnetic pulse compacts these VP's to a
condensed congregate that moves through the field at the VoL. The
magnatic field can do this because the motion of the electron is
moving these VP's.
It has momentum and charge (condensed V P's). This is not an intrinsic
charge like an electron has. But this 'condensed congregate' does have
charge.
The photon is magnetic during the electrons transition and it is
directional.
The electrons electric field is omni-directional, so its magnitude daes
not vary like the magnetic pulse where its magnitude varies fron zero
to maximum to zero in one direction only relative to the observer.
This magnitude reaches its highest point that is perpendicular to the
electrons motion and velocity. Also, this point would be in line with
the plane of the electrons motion.
I hope you understand what I say here.
Established Member
Okay, Mike, I'm starting to get it. You don't know quite yet how all of this works mathematically, but you've got a grasp on the 'image' of it in your mind. I am not knocking you here, nor did I ever mean to intentionally (well, maby once, for which I appologize). I so want to understand your viewpoint, at least so I can accurately and to your satisfaction will or compell you to see that such an alteration is only to achieve an end goal of yours: SSU.
Correct me if I am wrong.
Travis
Order of Kilopi
Well, I've already answered the first bit, so I will now define orbital angular momentum for you. For a particle with linear momentum given by the vector p, the orbital angular momentum around a point is defined by
L = rxp
where r is the position vector for the particle, and the "x" is the symbol for the vector cross product.
The key thing for you to notice here is that the cross product of two parallel vectors is zero, and the only component of the instantaneous linear momentum that comes into the angular momentum is that which is perpendicular to the position vector, r. Perhaps this may help you to understand what you thought was the apparently paradoxical behaviour of a particle without an orbital angular momentum.
By the way I use the word orbital to differentiate the quantity from the angular momentum due to the intrinsic spin of the particles.
I'll respond to your other comments shortly.
Order of Kilopi
I'm finding it tricky following the order of your nested quotes (it may help if you use the "quote" functionality of the message board, but that is a stylistic issue
Anyway, orthodox photons are the ones that are described by current, orthodox, physics, rather than the ones consisting of VNCPs etc. They are the quanta of the electromagnetic field (both the real and virtual ones.) The difference between real and virtual photons really comes down to the origin of their energy. "Real" photons are typically produced when charged particles radiate energy, or through the annihilation of particle anti-particle pairs. Virtual photons could be thought of as real photons living on borrowed time. They are quantum fluctuations, given birth by the time-energy uncertainty principle. The greater the time interval, the smaller the uncertainty in the energy. Other than that, they not really any different to "real" photons. That's part of the beauty of QED. In the same way that "real" photons can impart momentum to a charged particle, virtual photons transfer momentum between charged particles, hence how they mediate the EM interaction.
Needless to say, QED has been an incredibly successful theory, and when unified with the theory of the weak interaction (the one that deals with things like neutrinos) in the electroweak theory, it is capable of predicting the value of the anomalous magnetc moment of the electron to an accuracy better than
This seems to be pretty good evidence that is a good model of what is going on. (If you want to propose an alternative, you need to at least do as well as this.)
QM only has a single ground state for the hydrogen atom. What do you mean by "thousands of solutions"? (If it's the total number of states that are available, then that is also true of the Bohr model...)reply
The HA binary is explained and it has formulas for its solutions for
'standing waves' only.
QM comes up with these clouds that have thousands of solutions in
complex HA environments. No wonder it is correct with all these
possible solutions.
So this is analogous to the way that virtual photons mediate the EM force in QED? i.e. light is photons, and the EM field consists of photons as well?cyrek reply
The electric field is real. It is composed of VNCP's that normally
distribute themselves uniformally through mutual repulsion around the
electron (my hypothesis).
Are the VPs already moving? If not then they won't interact with a magnetic field. (They would interact with an electric field, however.) Why do you think that it is down to the magnetic field? Also wouldn't they find themselves attracted to the proton, but repelled by the electron?When the electron transits (returns) to its
original position, its magnetic pulse compacts these VP's to a
condensed congregate that moves through the field at the VoL. The
magnatic field can do this because the motion of the electron is
moving these VP's.
Another thing to consider is that, as far as I can see, in this model a charge (such as an electron) moving at a uniform velocity, will cause the VNCPs to accelerate (they are repelled when the electron moves through them.) You now have accelerating charges. Wouldn't this lead to the radiation of photons, effectively by non-accelerating charges? Is there any evidence that this has been observed?
Really not sure what you mean here. If the charge is not intrinsic, then where does it come from? What does this mean? Is charge conservation broken?It has momentum and charge (condensed V P's). This is not an intrinsic
charge like an electron has. But this 'condensed congregate' does have
charge.
EM waves possess both a magnetic and electric field component.The photon is magnetic during the electrons transition and it is
directional.
If you have a time varying magnetic field, then you also have an electric field (which may also be time varying.)The electrons electric field is omni-directional, so its magnitude daes
not vary like the magnetic pulse where its magnitude varies fron zero
to maximum to zero in one direction only relative to the observer.
This magnitude reaches its highest point that is perpendicular to the
electrons motion and velocity. Also, this point would be in line with
the plane of the electrons motion.
To be honest, I'm not sure I do, but thanks for trying. Hopefully in the not too distant future we will both be able to understand what the other means.I hope you understand what I say here.
By the way, you didn't say if the VNCPs possessed a rest mass, or if they are massless particles.
Established Member
Ooh, ooh, I know. This might be meaning, sorry if I'm out of bounds here Mike; eather. VCNP that exisit, and a collection of them is his "condensed congregate," which, taken as a collection, is the electron?It has momentum and charge (condensed V P's). This is not an intrinsic
charge like an electron has. But this 'condensed congregate' does have
charge.
I don't even want to comment on the right/wrong aspect, just wanting to understand your idea Mike...
Trav
Established Member
cyrek reply
Travis
Yes, that is axactly what my objective is. To promote the SSU.
There has to be another reason for the cosmological redshift since the EoS redshift is not supported by 'empirical evidence'.
Your last post on this thread commenting about the nature of the photon pulse has to be structured from the use of the 'electric field surrounding the electron.
The current orthodox photon pulse has momentum and energy. I still do not know what the composition of this photon is that represents the energy.
My version has charge (not intrinsic) and momentum.
fortis wrote
cyrek Quote:
OK, so you create a cloud around the proton called an orbital or a
wave function?.
Now what is the diameter of that cloud?
Will it be the same diameter as I have given for the GS radius times
two?
cyrek added reply
Fortis....will you explain what 'orbital angular momentum' is in your
own words?
fortis
Well, I've already answered the first bit, so I will now define orbital
angular momentum for you. For a particle with linear momentum given by
the vector p, the orbital angular momentum around a point is defined by
L = rxp
where r is the position vector for the particle, and the "x" is the
symbol for the vector cross product.
The key thing for you to notice here is that the cross product of two
parallel vectors is zero, and the only component of the instantaneous
linear momentum that comes into the angular momentum is that which is
perpendicular to the position vector, r. Perhaps this may help you to
understand what you thought was the apparently paradoxical behaviour
of a particle without an orbital angular momentum.
By the way I use the word orbital to differentiate the quantity from
the angular momentum due to the intrinsic spin of the particles.
I'll respond to your other comments shortly.
reply
From what I understand here is that the cross product vector is zero.
Then the linear momentum is still intact and represented by the
electron velocity.
In other words, what I think you say is that the orbital radius is
stabalized in position as a standing wave.
With these cirumstances, I can agree that the position vector is zero.
Fortis - I will get back to your last post tomorrow.
Order of Kilopi
As has been pointed out, the energy of a particle with zero rest mass is given by E=p.c, where p is the momentum. Have a read up on special relativity. (Can't think of a good intro to SR, but you might want to find one as it will give you some of the background that you will need to understand what is happening here.cyrek reply
What is the formula for this photon energy? From what I understand,
there has to be mass for the object to have energy.
They possess spin. In the way that Fermions, such as electrons, are described as spin 1/2 particles, photons are spin-1 bosons. If you want a semi-classical picture of what this spin is, think about circularly polarised light. (Note of caution, this is a semi-classical picture, and if you take it too far you can break it.Also, how does the angular momentum come into this photon of yours?
Photons move in a straight line as far as I know.) In this rather curious polarisation state, the electric vector rotates around the direction of motion of the photon. (If you're happier working with linear polarisation you can think of this as the linear combination of two perpendicular linear states that are 90 degrees out of phase.
Anyway, this gives you something that looks like it is spinning around the direction of motion. (Though this is only a semi-classical picture to help you think about how it may work.)
How do we know that photons possess angular momentum? Simple. One way is to look at the nature of the transitions in the hydrogen atom. Electronic transitions that lead to a change in angular momentum, mean that the angular momentum must go somewhere. (Even in the Bohr model, there is a change in angular momentum after a transition. Unless angular momentum is not conserved, that angular momentum must go somewhere.
So is a photon a cluster of these virtual particles moving through space, or is it a travelling disturbance in the density of these particles? (Something akin to the aether.)Normandy Quote
I think the real question is, how would you explain the experimental
results of the Compton effect if the photon did have a charge?
cyrek reply
My concept of the photon is that the 'condensed virtual particle
charge' is what causes the bumping. The photon does not have a charge
of its own but uses the fields charge that is condensed as the bumper.
An electron doesn't possess any measurable size (at least as far as anyone has been able to measure.) Each of these virtual particles may possess only tiny charges (which would make them intersting as so far we've not seen charges smaller than 1/3 of the charge on an electron,) but what is the total charge of this "condensed congregate"? Is it the smae sort of magnitude as that of an electron, or is it a lot smaller, or a lot larger?These virtual particles are extremely tiny negative charges. The extent
of this condensed congregate is huge, however, in comparison to the
size of an electron.
Where do you get the wavelength of a photon from? Do you know what the wavelength of a gamma ray can be, or and x-ray, or UV photon? How about a microwave photon? The visible spectrum is in the range ~450-700 nm, but wavelength of other photons can be orders of magnitude different.Compare the wavelength of a photon (10^-7) to the size of an electron
(10^-14) and you can see the enormous difference of the two objects(?).
Add these VNCP's together that constitute the photon and you could have
a combined charge equal to the electron.
Also where did you get the size of an electron from? Do you have a reference for this, because a finite size for the electron is news to me.
Order of Kilopi
There isn't an "orbital radius". Particularly in the case of the 1s ground state, the notion of an orbit doesn't make physical sense. After all, a particle in a classical orbit would possess angular momentum, whereas we know that the value of L^2 for the electron is zero. A different way to think of the ground state might be in terms of the uncertainty principle. As the Coulomb field of the nucleus tries to confine the electron into smaller and smaller volumes (i.e. in effect reducing the uncertainty in position), the uncertainty in the linear momentum must increase, and this increase prevents the electron from falling any further. This is a hand waving explanation, but at least it provides us with another picture to try to understand what is going on, and which avoids having to talk about orbits. (The best picture, of course, is to just do the QM calculation.
I'm not sure if this is a typo, or if you misunderstood what I was saying, but I don't believe I claimed that the position vector was zero. It is the orbital angular momentum (or more precisely L^2) that is zero in the 1s ground state.With these cirumstances, I can agree that the position vector is zero.
Established Member
cyrek reply (excerpts)
fortis wrote
As has been pointed out, the energy of a particle with zero rest mass
is given by E=p.c, where p is the momentum. Have a read up on special
relativity. (Can't think of a good intro to SR, but you might want to
find one as it will give you some of the background that you will need
to understand what is happening here. )
cyrek reply - Since when is SR relavant to QM?
cyrek Quote:
Also, how does the angular momentum come into this photon of yours?
Photons move in a straight line as far as I know.
fW
They possess spin. In the way that Fermions, such as electrons, are
described as spin 1/2 particles, photons are spin-1 bosons. If you
want a semi-classical picture of what this spin is, think about
circularly polarised light. (Note of caution, this is a semi-classical
picture, and if you take it too far you can break it. ) In this rather
curious polarisation state, the electric vector rotates around the
direction of motion of the photon. (If you're happier working with
linear polarisation you can think of this as the linear combination
of two perpendicular linear states that are 90 degrees out of phase. )
CR - Bosons have mass. How can they be linked to photons? Spiraling
electrons are moving through a strong magnetic ield. Your replies are
using data that is irrelavant to the discussion.
Anyway, this gives you something that looks like it is spinning around
the direction of motion. (Though this is only a semi-classical picture
to help you think about how it may work.)
fW
Do you know that photons possess angular momentum? Simple. One way is
to look at the nature of the transitions in the hydrogen atom. Electronic
transitions that lead to a change in angular momentum, mean that the
angular momentum must go somewhere. (Even in the Bohr model, there is
a change in angular momentum after a transition. Unless angular
momentum is not conserved, that angular momentum must go somewhere. )
CR - Changes in electron momentum is a two way thing when it absorbs
and than radiates. There is no loss or gain in AM.
Normandy Quote
I think the real question is, how would you explain the experimental
results of the Compton effect if the photon did have a charge?
CR
My concept of the photon is that the 'condensed virtual particle
charge' is what causes the bumping. The photon does not have a charge
of its own but uses the fields charge that is condensed as the bumper.
fW
So is a photon a cluster of these virtual particles moving through
space, or is it a travelling disturbance in the density of these
particles? (Something akin to the aether.)
cyrek Quote:
These virtual particles are extremely tiny negative charges. The extent
of this condensed congregate is huge, however, in comparison to the
size of an electron.
fortis quote
An electron doesn't possess any measurable size (at least as far as
anyone has been able to measure.) Each of these virtual particles may
possess only tiny charges (which would make them intersting as so far
we've not seen charges smaller than 1/3 of the charge on an electron,)
but what is the total charge of this "condensed congregate"? Is it the
smae sort of magnitude as that of an electron, or is it a lot smaller,
or a lot larger?
cyrek Quote:
Compare the wavelength of a photon (10^-7) to the size of an electron
(10^-14) and you can see the enormous difference of the two objects(?).
Add these VNCP's together that constitute the photon and you could have
a combined charge equal to the electron.
fortis quote
where did you get the size of an electron from? Do you have a
reference for this, because a finite size for the electron is news to
me.
CR - The reference I read about the Electrons size is lost in
antiquity. Do not remember but there was a size mentioned.
cyrek1 quote:
From what I understand here is that the cross product vector is zero.
Then the linear momentum is still intact and represented by the
electron velocity.
In other words, what I think you say is that the orbital radius is
stabalized in position as a standing wave.
fortis quote
There isn't an "orbital radius". Particularly in the case of the 1s
ground state, the notion of an orbit doesn't make physical sense.
After all, a particle in a classical orbit would possess angular
momentum, whereas we know that the value of L^2 for the electron is
zero. A different way to think of the ground state might be in terms
of the uncertainty principle. As the Coulomb field of the nucleus
tries to confine the electron into smaller and smaller volumes (i.e.
in effect reducing the uncertainty in position), the uncertainty in
the linear momentum must increase, and this increase prevents the
electron from falling any further. This is a hand waving explanation,
but at least it provides us with another picture to try to understand
what is going on, and which avoids having to talk about orbits.
(The best picture, of course, is to just do the QM calculation. )
CR - With that answer above, I agree. It adds up to zero.
cyrek Quote:
With these cirumstances, I can agree that the position vector is zero.
fortis quote
I'm not sure if this is a typo, or if you misunderstood what I was
saying, but I don't believe I claimed that the position vector was
zero. It is the orbital angular momentum (or more precisely L
CR - Your wording of the 'position vector' would have to be zero if
it is interpreted as the radial(?) vector?
Established Member
Yeah Mike, you're gonna have to learn how to use the quote function on here. Your threaded-threads are unbearable.
SR and QM have a bunch to do with each other. Little tiny particles going relavistic velocities? Are you jokin? :wink:
I love IT:
Schodinger's equation was found to be an approximation of microscopic physics. Quantum electrodynamics was found by merging special relativity with quantum electromagnetic force and its interaction with matter.
It's quantum because all of the probabilistic and uncertainty issues are incorporated from the outset; it's a field theory because it merges the quantum principles into the previous classical notion of a force field - in this case, Maxwell's electromagnetic field. And finally, it's relavtivistic because special relativity is also incorporated from the outset.
It's full name is Relativistic Quantum Field Theory, it's QM for short. Look things up some time bud. 8) It's so much more rewarding than making stuff up without checking into it first...
Order of Kilopi
Historically, the first attempt at QM by Schrodinger QM was a relativistic formulation that we now call the Klein-Gordon equation. The problem for Erwin was that whilst it looked good (effectively a QM equivalent of the equation that Normandy keeps explaining to you), it was only good for particles without spin, i.e. spin-0 bosons. So Schrodinger had to make do with a non-relativistic formulation. Not too long after, however, Dirac managed to construct a relativistic equation that described particles with spin, by effectively taking the square-root of the Klein-Gordon equation. So a relativistic equation that could describe the QM of the hydrogen atom appeared only 2 years after the non-relativistic one. Hope you enjoyed the little history lesson.
Where did you find this bit of information? Please read some decent intros to QM, EM, and particle physics, with perhaps a bit of QED, and SR thrown in for good measure. Why? Because your statement is wrong. The only thing that defines a boson is that it possesses integer spin.CR - Bosons have mass. How can they be linked to photons?
You had asked about photons and angular momentum. I had answered your question. How does your model of the photon handle all of the different types of polarisation that we can create?Spiraling electrons are moving through a strong magnetic ield. Your replies are using data that is irrelavant to the discussion.
Are you saying that the time between absorption and emission is zero? If not, then the excited state possesses a different angular momentum to the ground state and hence angular momentum is not conserved unless the photon carries it away.CR - Changes in electron momentum is a two way thing when it absorbs
and than radiates. There is no loss or gain in AM.
You haven't answered this question and it would seem to be a pretty fundamental part of your theory. You see, what I'm trying to get my head around is how your photons manage to propagate at the speed of light. Do these virtual particles possess mass, for instance?fW
So is a photon a cluster of these virtual particles moving through
space, or is it a travelling disturbance in the density of these
particles? (Something akin to the aether.)
Well you might be happy to discover that the electron is, to all intents and purposes, a point particle, with no size. There is no experimental evidence to the contrary.CR - The reference I read about the Electrons size is lost in
antiquity. Do not remember but there was a size mentioned.
So are you happy with the idea that the magnitude of the orbital angular momentum of the ground state is zero? Because that is what this says, and it also contradicts the prediction from the Bohr model.CR - With that answer above, I agree. It adds up to zero.
the radial position vector is the vector that points from the centre of the atom to the location of interest. I'm still not sure exactly what you're trying to get at here.CR - Your wording of the 'position vector' would have to be zero if
it is interpreted as the radial(?) vector?
Established Member
Neither is Mike. I applaude your effort Cyrek1, but it's like you found a bucket of 'scientifical' words and a sack of equations from 1899 and are combing them at random...
Order of Kilopi
Mike, I was having a read through the current copy of Nature and thought you might find the following bits of information to be useful.
Current upper limit on the mass of the photon is 10^-50 kg.
(Compare this with the mass of an electron, which is ~9.1x10^-31 kg, i.e. the mass of the photon is less than 10^-20 of the mass of the electron.)
Current upper limit on the charge of the photon is 10^-17 time the charge of an electron. Again this is tiny. I can perhaps give you a feel for how tiny a fraction this is by saying that it is roughly twice the ratio of the wavelength of red light to the distance between the earth and the sun.
(I can provide references for this if you like.
Doesn't sound good news for your model.
Established Member
Cyrek reply
Triavis wrote
Schodinger's equation was found to be an approximation of microscopic
physics. Quantum electrodynamics was found by merging special
relativity with quantum electromagnetic force and its interaction with
matter.
It's quantum because all of the probabilistic and uncertainty issues
are incorporated from the outset; it's a field theory because it
merges the quantum principles into the previous classical notion of a
force field - in this case, Maxwell's electromagnetic field. And
finally, it's relavtivistic because special relativity is also
incorporated from the outset.
It's full name is Relativistic Quantum Field Theory, it's QM for short.
Look things up some time bud. It's so much more rewarding than making
stuff up without checking into it first...
Fortis wrote
cyrek1 wrote:
Since when is SR relavant to QM?.... Well, Travis answered above.
fortis
Historically, the first attempt at QM by Schrodinger QM was a
relativistic formulation that we now call the Klein-Gordon equation.
The problem for Erwin was that whilst it looked good (effectively a
QM equivalent of the equation that Normandy keeps explaining to you),
it was only good for particles without spin, i.e. spin-0 bosons.
So Schrodinger had to make do with a non-relativistic formulation.
Not too long after, however, Dirac managed to construct a relativistic
equation that described particles with spin, by effectively taking the
square-root of the Klein-Gordon equation. So a relativistic equation
that could describe the QM of the hydrogen atom appeared only 2 years
after the non-relativistic one. Hope you enjoyed the little history
lesson.
Quote:
CR - Bosons have mass. How can they be linked to photons?
fortis Quote
Where did you find this bit of information? Please read some decent
intros to QM, EM, and particle physics, with perhaps a bit of QED, and
SR thrown in for good measure. Why? Because your statement is wrong.
The only thing that defines a boson is that it possesses integer spin.
CR - In one encyclopedia, there were two types of bosons listed. The guage bosons (massless(?) and the 'weak bosons' (W, Z) that have the mass.
cyrek Quote:
Spiraling electrons are moving through a strong magnetic field. Your
replies are using data that is irrelavant to the discussion.
fortis quote
You had asked about photons and angular momentum. I had answered your question. How does your model of the photon handle all of the different types of polarisation that we can create?
Quote:
CR - Changes in electron momentum is a two way thing when it absorbs
and than radiates. There is no loss or gain in AM.
fortis quote
Are you saying that the time between absorption and emission is zero?
If not, then the excited state possesses a different angular momentum
to the ground state and hence angular momentum is not conserved unless
the photon carries it away.
CR - The excited state (absorption) is temporary. The electron returns to its original lower energy state (emission). I do not unnderstand your questioning conservation of momentum here?
FW
So is a photon a cluster of these virtual particles moving through
space, or is it a travelling disturbance in the density of these
particles? (Something akin to the aether.)
CR - It is a disturbance that uses the electric field for transferring
the congregate momentum as though it was an aether. The front of the
photon VP's transfer their momentum to the VP's in front of them and
this transfer moves at the VoL. Remember, I said these are negative
charges propelling themselves.
fortis
You haven't answered this question and it would seem to be a pretty
fundamental part of your theory. You see, what I'm trying to get my
head around is how your photons manage to propagate at the speed of
light. Do these virtual particles possess mass, for instance?
CR - These are virtual particles. If they do have mass, it would be,
no doubt, non-measurable. Something akin to the neutrino, but with
charge. Maxwells equations might have some data on this. It appears
he used them to explain the EM field patterns.
Quote:
CR - The reference I read about the Electrons size is lost in
antiquity. Do not remember but there was a size mentioned.
fortis
Well you might be happy to discover that the electron is, to all
intents and purposes, a point particle, with no size. There is no
experimental evidence to the contrary.
CR - I hate to embarass you fortis, but after you posted this,
I discovered a reference for the electron size. I posted it on
another thread but here is a repeat:
Radius = 2.81794x10^-15 meters. Source: McGraw Hill Encyclopedia
of Physics - 2nd edition.
fortis
So are you happy with the idea that the magnitude of the orbital
angular momentum of the ground state is zero? Because that is what
this says, and it also contradicts the prediction from the Bohr model.
Quote:
CR - Your wording of the 'position vector' would have to be zero if
it is interpreted as the radial(?) vector?
fortis
the radial position vector is the vector that points from the centre
of the atom to the location of interest. I'm still not sure exactly
what you're trying to get at here.
CR - Because in the GS, the electron would not be radiating any photons
(but a continuous wave only) and its radial vector would be unchanging,
or zero.
Order of Kilopi
I hate to embarass you as well, but the upper bound on the size of the electron is rather less than this. One reference on the
UK Science Museum website (hardly a peer reviewed journal, but it was written in collaboration with the Institute of Physics, so one would hope that it would be reasonably accurate), says that all attempts to measure the size of the electron have failed but that we know that it is less than 10^-18 m.
A paper out of Berkley claims that experimentally the "size" of the electron is less that 10^-19 m.
These upper limits are rather less than the value that you quote.
Order of Kilopi
And, by any chance, are these massless gauge bosons, bosons? How does your encyclopedia define "boson"? Does it say that bosons must possess mass?
I still think that you might want to answer my question about polarisation.fortis quote
You had asked about photons and angular momentum. I had answered your question. How does your model of the photon handle all of the different types of polarisation that we can create?
Ask yourself what the total angular momentum is before the absorption, what the total angular momentum of the excited state is, and finally what the total angular momentum of the final state is after emission. At each of these three steps do you get the same answer. If you do, then angular momentum is conserved. If you don't then angular momentum is not conserved.fortis quote
Are you saying that the time between absorption and emission is zero?
If not, then the excited state possesses a different angular momentum
to the ground state and hence angular momentum is not conserved unless
the photon carries it away.
CR - The excited state (absorption) is temporary. The electron returns to its original lower energy state (emission). I do not unnderstand your questioning conservation of momentum here?
So it is like a sound wave propagating through the negative charged particles? It should be possible to work out the velocity of a disturbance propagating through this medium. Have you done that and proved to yourself that the disturbance travels at the velocity of light? If so, would you like to show it to us?CR - It is a disturbance that uses the electric field for transferring
the congregate momentum as though it was an aether. The front of the
photon VP's transfer their momentum to the VP's in front of them and
this transfer moves at the VoL.
I read his paper a few years ago. (In a volume of the "Philosophical Transactions of the Royal Society....") IIRC while he tried to build a mechanical aether it was rather more complex than the one you're describing, and is also unecessary. I'm curious to know why you say that if they possessed mass (and on the other thread you refer to the electron as being lighter than the photon?) that the mass would not be measurable. Why?CR - These are virtual particles. If they do have mass, it would be,
no doubt, non-measurable. Something akin to the neutrino, but with
charge. Maxwells equations might have some data on this. It appears
he used them to explain the EM field patterns.
Your reply means that in your model the electron is either static ("radial vector is unchanging" or it is sat in the nucleus ("radial vector is zero".) How do you square this with an electron orbiting the nucleus (which is also part of your model)?the radial position vector is the vector that points from the centre
of the atom to the location of interest. I'm still not sure exactly
what you're trying to get at here.
CR - Because in the GS, the electron would not be radiating any photons
(but a continuous wave only) and its radial vector would be unchanging,
or zero.
By the way, do you notice how the use of the "quote" facility makes it easier to follow what is going on in my post?
Established Member
cyrek reply
fortis
These long posts are taking too much of my time. So I will not answer any more.
I plan to post a new article on the SSU in a few days and that will be it.
I will answer on that new post only.
Thanks for your comments.
Order of Kilopi
You're welcome.
Surely you at least have a comment about how you reconcile your claim that the photon has a charge roughly equal to that of an electron, when experimentally it has been found that if it has a charge, then it is less than 10^-17 that of the electron?
You claim that your model is based on the "real world" and yet when it is brought up against that hard light of reality it seems to fall apart.
If you choose not to answer any more questions, please at least consider finding out a bit more about our current understanding of reality. (Books on QED, EM, QM, etc.) It may help you to construct a model that is more consistent with the "real world".
Established Member
Most people who embark on a 'scientific' 'fact-finding' mission miss the point that they have an agenda. I don't know about you, Fortis, but I can't wait for this SSU thread!
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