Photon, not the electron

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Electric energy = electromagnitism. In other words, electric energy is composed of an electromagnetic field. And in the electric curcuit, electrical energy doesn't flow inside the conductor, but inbetween the empty are surrounding the conductor. So it is safe to say that photons are our charge carriers of electrical energy, not the electron.

Now, I may be restating a well known fact, or maybe I'm on to something, but wouldn't it (given the facts above) be logical to state that the energy we use to power our everyday electrical devices, are given rise by photons, and not presently thought electrons?

Would this not also make it easier to define an electromagnetic field. Since the electromagnetic field is made by the buildup of an electric charge, it is easier to see that the electric charge is created by the photon and not the electron, because the particle form of EM radiation is a photon.

"There is no inbetween." -This statement could be used in many different interpretations here.
Paden Roder
 

jcsd

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Originally posted by PRodQuanta
Electric energy = electromagnitism. In other words, electric energy is composed of an electromagnetic field. And in the electric curcuit, electrical energy doesn't flow inside the conductor, but inbetween the empty are surrounding the conductor. So it is safe to say that photons are our charge carriers of electrical energy, not the electron.

Now, I may be restating a well known fact, or maybe I'm on to something, but wouldn't it (given the facts above) be logical to state that the energy we use to power our everyday electrical devices, are given rise by photons, and not presently thought electrons?

Would this not also make it easier to define an electromagnetic field. Since the electromagnetic field is made by the buildup of an electric charge, it is easier to see that the electric charge is created by the photon and not the electron, because the particle form of EM radiation is a photon.

"There is no inbetween." -This statement could be used in many different interpretations here.
Paden Roder
photons are the force carriers for EM, but they are chargeless so they certainly are not the charge carriers in an eletric circuit. In a circuit the electrical enrgy is created by the movement of electrons and hence charge.
 
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But the electrons give rise to energy as photons. I'm saying that these photons actually power our electrical devices, not the electron.
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jcsd

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Photons produced by an electric circuit are just electrical interference.
 
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How so?
 

jcsd

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Cos any accelrating charge produces photons, so for example in an AC circuit photons are produced of very long wavelengths and mainfest themselves as electrical interference.
 

Tom Mattson

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Originally posted by PRodQuanta
Now, I may be restating a well known fact, or maybe I'm on to something, but wouldn't it (given the facts above) be logical to state that the energy we use to power our everyday electrical devices, are given rise by photons, and not presently thought electrons?
No, it wouldn't, because there is a mountain of experimental evidence that states uniquivocally that the photon has no charge and that the electron has a charge of 1.6×10-19C.

Would this not also make it easier to define an electromagnetic field. Since the electromagnetic field is made by the buildup of an electric charge, it is easier to see that the electric charge is created by the photon and not the electron, because the particle form of EM radiation is a photon.
No, it wouldn't make it easier to define the EM field. The EM field is currently very well defined both classically and quantum mechanically, and in such a way as to fit perfectly with the rest of physics.

This thread has nothing to do with the physics of particles and nuclei. It belongs in Theory Development.
 
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I know a photon has no charge, but it does carry energy. Right? A photon's energy is equivilent to it's momentum (I may be mistaken). Is not energy needed to establish an electical current in the first place?
Paden Roder
 

Robert Zaleski

A recent experiment has determined that if a photon does have mass, it would have to be less than 1.2x10^-51 grams and the photons charge would have to be less than 8.5x10^-17e.

www.aps.org/media/tips/tips32.html - 14k - Sep 19, 2003
 
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Very interesting. Thanks for the post.

And also to prove my point. Electrons constantly emit certain quanta of energy when excited. This emitting and reabsorbing would accound for our positive and negative state of atoms, without which , the concept of electricity wouldn't make sense.

And we also know that when electrons are excited, they emit photons. And the frequencies vary, depending on the level of excitation. So it is logical to assume that EMF are given rise by the HUGE amount of photons which excitation of the electrons in the wire gives rise to and that each such photon has its own EMF. So it is reasonable to assume that the EMF is associated w/ the photon, NOT the other way around.
Paden Roder
 

Tom Mattson

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Originally posted by PRodQuanta
I know a photon has no charge, but it does carry energy. Right?
Right.

A photon's energy is equivilent to it's momentum (I may be mistaken).
No, it is not. The photon's 4-momentum is given by:

pμ=(p,E/c).

The momentum is a 3-vector, and the energy is the timelike piece of a 4-vector. They behave differently under Lorentz transformations.

Is not energy needed to establish an electical current in the first place?
What is your point?

Just because light carries energy, and energy is needed to have an electric current, that does not even remotely imply that light is the charge carrier in a wire. Water waves carry energy, too, but you wouldn't suggest that they are responsible for carrying electrical current (would you?!).
 

Tom Mattson

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Originally posted by PRodQuanta
And also to prove my point. Electrons constantly emit certain quanta of energy when excited.
No, all experimental evidence indicates that electrons do not become excited, and no electron has ever been observed to emit quanta. Only structured entities can be found in excited states, and by deexcitation, emit quanta.

This emitting and reabsorbing would accound for our positive and negative state of atoms, without which , the concept of electricity wouldn't make sense.
No, electricity makes perfect sense apart from atoms. Just put a bunch of electrons in a glass tube and apply a potential difference. Similarly, the electrons in a wire are not localized to any particular nucleus, and their wavefunctions are not atomic wavefunctions.

And we also know that when electrons are excited, they emit photons. And the frequencies vary, depending on the level of excitation. So it is logical to assume that EMF are given rise by the HUGE amount of photons which excitation of the electrons in the wire gives rise to and that each such photon has its own EMF. So it is reasonable to assume that the EMF is associated w/ the photon, NOT the other way around.

Since the premises are wrong, so is the conclusion.
 
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1)But isn't it to that when electrons absorb photons, (because photons directly relate to energy) the electron takes on more energy and thus it's kinetic molecular motion is increased (excitation)? Or am I off here?

2) Are you saying that when electron/electron interaction occurs, It DOESN'T result in photons?

3) And the resulting photon DOESN'T depend on the level of excitation the electrons were in?

I may be wrong in all of those questions, but I always have been under the impression that when there is an electron/electron interaction, the electrons give off energy known as a virtual photons, which are very short lived, and then these photons are then very quickly absorbed once again by the electrons.
Paden Roder
 

Tom Mattson

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Originally posted by PRodQuanta
1)But isn't it to that when electrons absorb photons, (because photons directly relate to energy) the electron takes on more energy and thus it's kinetic molecular motion is increased (excitation)? Or am I off here?
My point was precisely that electrons do not "absorb photons". All experimental evidence indicates that photons scatter off of electrons. Structured bodies such as atoms and nuclei, however, do absorb and emit photons.

2) Are you saying that when electron/electron interaction occurs, It DOESN'T result in photons?
That's right.

3) And the resulting photon DOESN'T depend on the level of excitation the electrons were in?
Since my point was that electrons do not have excited states, this question doesn't even make it to the table.

I may be wrong in all of those questions, but I always have been under the impression that when there is an electron/electron interaction, the electrons give off energy known as a virtual photons, which are very short lived, and then these photons are then very quickly absorbed once again by the electrons.
It is true that the electron-electron scattering is mediated by virtual photons. As you further note, it is also true that the energy of the photon is short lived. In fact it is precisely so short lived as to not be observable, so there is no way it serve as the source for a current in a wire.
 
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Just to give addition on Tom's statement with virtual photons:Electron emitts photons butthat photons carry momentum,and not energy so the electro doesn't lose energy or absorb im meaning gets.So Einstein equation E^2-p^2c^2=m^2c^4 for REAL photo is (because rest mass of photon =0) E^2-p^2c^2=0,but as that photon (virtual) doesn't carry energy E=0 so we get that for VIRTUAL photon:
E^2-p^2c^2<0 or to generalize for or VIRTUAL particles:
E^2-p^2c^2 different of m^2c^4..
Best wishes
SVRadic
 

Tom Mattson

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Originally posted by Radic
Just to give addition on Tom's statement with virtual photons:Electron emitts photons butthat photons carry momentum,and not energy so the electro doesn't lose energy or absorb im meaning gets.
In general, virtual photons have both energy and momentum. In, say, an e-&mu;- scattering reaction, let pn and pn' be the incoming and outgoing electron momenta, respectively, and let qn and qn' be the incoming and outgoing muon momenta, respectively. The momentum of the virtual photon can be expressed in terms of the particle momenta as follows:

kn=qn'-qn=pn-pn'

In general, the timelike pieces of q and q', or p and p', are not the same. In the initial rest frame of the muon, for instance, the muon definitely gains energy, so the timelike piece of the virtual photon's momentum is nonzero.

On more general grounds, we do not expect the virtual photon energy to always be zero because energy is not a Lorentz invariant.

edit: fixed a typo
 
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Ok, let's look at this from a theoretical standpoint. I just read an article on how the photon actually has a rest mass. (Small, but a rest mass) This mass makes it impossible for a photon to travel as speed c . *For your info, the conclusion made that a photon has a small rest energy was derived from Maxwell's equations.

But if the photon were to have a mass, even if small, would it not imply it had a charge?

*Also note, I'm really trying to prove this theory, I'm looking at this as a nonprejudice veiw point. Pointing out the downfalls of this theory is very helpful, but also, maybe if you were to point me in the right direction (things that I say are logical) that would be muched appreciated.

Paden Roder
 

Tom Mattson

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Originally posted by PRodQuanta
Ok, let's look at this from a theoretical standpoint.
I have been! The theoretical standpoint I have been taking is the Standard Model (Electroweak+QCD).

I just read an article on how the photon actually has a rest mass. (Small, but a rest mass) This mass makes it impossible for a photon to travel as speed c .
Do you have a reference?

*For your info, the conclusion made that a photon has a small rest energy was derived from Maxwell's equations.
That is not possible. The massless photon is precisely what allows the gauge invariance of Maxwell's equations. In order to accomodate a massive photon, the EM Lagrangian:

LEM=j&mu;A&mu;+(1/4&mu;0)F&mu;&nu;F&mu;&nu;

would have to be replaced by the Proca Lagrangian:

LProca=LEM+m2A&mu;A&mu;,

where m is the photon mass. A massive vector field (aka "Proca field") has completely different equations. Since the 4-potential appears quadratically in the Lagrangian, the potential itself appears in the field equations, giving rise to the aforementioned spoiling of gauge freedom that the Maxwell field enjoys.

But if the photon were to have a mass, even if small, would it not imply it had a charge?
No, it would not.

*Also note, I'm really trying to prove this theory, I'm looking at this as a nonprejudice veiw point. Pointing out the downfalls of this theory is very helpful, but also, maybe if you were to point me in the right direction (things that I say are logical) that would be muched appreciated.
The "right direction" would be to start looking into the things I am telling you and to rid yourself of these misconceptions you have.
 
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Yep, thanks. That's why I'm asking the expert.

Also, for the reference of that book. Just today I was doing some research on it in Iowa State's library. When having a discussion with the professor, he also asked for the book, and I drew a blank. Um.... I know the book's main title was "PHOTON" then there was a subtitle: (something like) "The old and new conception" I dunno, if you're really interested, maybe try looking that up. If not, what can I say?

Thanks
Paden Roder
 

Tom Mattson

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Originally posted by PRodQuanta
I know the book's main title was "PHOTON" then there was a subtitle: (something like) "The old and new conception" I dunno, if you're really interested, maybe try looking that up.
I searched at Google and Amazon, but came up empty. :frown:
 

Robert Zaleski

Could this be the book your referring to:

Photon : Old Problems in Light of New Ideas (A Volume in Contemporary Fundamental Physics - Valerie V. Dvoeglazov - Editor)
Nova Science Publishers, Inc.; ISBN 1560728108
 

Eyesee

I am a little confused with the fact that photons
have no charge. Electromagnetic waves are represented
as self regenerating waves with an electric field and a
magnetic field component- how can an EM wave be charge-less
if it has an E-field component, which is assumed to
be a product/characteristic of electric charges?
 
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Good question. My ears are open..
Paden Roder
 

Tom Mattson

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Originally posted by Eyesee
I am a little confused with the fact that photons
have no charge. Electromagnetic waves are represented
as self regenerating waves with an electric field and a
magnetic field component- how can an EM wave be charge-less
if it has an E-field component, which is assumed to
be a product/characteristic of electric charges?
The electric field you envision would be the real electric field if only Gauss' law holds (setting c=&epsilon;0=&mu;0=1):

[nab].E=&rho;

The above (very roughly) says that an electric field E arises from a charge density &rho;. However, that is not the only law that describes the electric field. There is also Faraday's law:

[nab]&times;E=-&part;B/&part;t

The above says that a changing magnetic field can also act as a source for an electric field. So, where does the changing magnetic field come from? Let's look at Ampere's law:

[nab]&times;B=&part;E/&part;t

The above says that a changing electric field produces a magnetic field (note the asymmetry with the "-" signs).

So, there we have it. A charge has an electric field, as per Gauss' law. When it oscillates, its electric field changes in time and a magnetic field is produced, as per Ampere's law. And the magnetic field is not constant in time. In fact, its time dependence is precisely the same as that of the electric field. So, the magnetic field is changing, too, which in turn produces an electric field, as per Faraday's law. The changing electric and magnetic fields "feed" each other, and a propagating EM wave is born.
 

Eyesee

Originally posted by Tom
The electric field you envision would be the real electric field if only Gauss' law holds (setting c=&epsilon;0=&mu;0=1):

[nab].E=&rho;

The above (very roughly) says that an electric field E arises from a charge density &rho;. However, that is not the only law that describes the electric field. There is also Faraday's law:

[nab]&times;E=-&part;B/&part;t

The above says that a changing magnetic field can also act as a source for an electric field. So, where does the changing magnetic field come from? Let's look at Ampere's law:

[nab]&times;B=&part;E/&part;t

The above says that a changing electric field produces a magnetic field (note the asymmetry with the "-" signs).

So, there we have it. A charge has an electric field, as per Gauss' law. When it oscillates, its electric field changes in time and a magnetic field is produced, as per Ampere's law. And the magnetic field is not constant in time. In fact, its time dependence is precisely the same as that of the electric field. So, the magnetic field is changing, too, which in turn produces an electric field, as per Faraday's law. The changing electric and magnetic fields "feed" each other, and a propagating EM wave is born.
Thx for the explanation but I still don't understand why
if a photon has an e-field component, it wouldn't affect
an electron or a proton since the Couloumb force between
charges is supposed to be a result of the E-field.
What is the difference between the E-field used to describe
the force between charged particles and the E-field used to describe the photon?
 
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