Photon is electromagnetic field, right?

In summary: Electromagnetic_waveThe article says the theorem is a purely mathematical one.So is the rest of physics.The article says the invariants reveal that the electric and magnetic fields are perpendicular...Yes, they are perpendicular.
  • #1
Barry_G
68
0
This thread is to move this discussion away from another thread in order to talk about it in more detail, so here is a brief recapitulation of how that went to make an opening for the discussion...

You do realize that a photon IS electromagnetic field, right?

A photon is quanta of electromagnetic radiation, and despite the name, despite there are, I mean could very well be, magnetic and electric fields constituting a photon, it is still electrically and magnetically neutral, which means its electric and magnetic field is measured to be zero.



http://en.wikipedia.org/wiki/Electromagnetic_field

Electrons are not electromagnetic fields, so no, this doesn't apply to them.

The article says electromagnetic field is a physical field produced by moving electrically charged objects. I'd say "electrically charged object" describes electron, positron, proton or some ion, but I don't see how it describes photon at all since photons are not electrically charged.


The EM field associated with a photon (more precisely, associated with a classical electromagnetic wave, which is the best classical approximation to a photon) is a particular type of EM field called a "null electromagnetic field", as described, for example, here:

http://en.wikipedia.org/wiki/Electromagnetic_field

The article says the theorem is a purely mathematical one.


How many electric fields and how many magnetic fields a single photon has, exactly?

Um, one of each?

In regards to "null electromagnetic field" mentioned above the article says: "the invariants reveal that the electric and magnetic fields are perpendicular...", which seems to imply there is at least two electric and two magnetic fields, otherwise they should have said "electric field is perpendicular to magnetic field", if that's what is supposed to be the meaning. But then how could they be perpendicular if there is only one of each, what property of each field would define its vector orientation to judge such geometrical relation?


What is the the strength of those fields?

It depends on the energy of the photon.

340px-Light-wave.svg.png

http://en.wikipedia.org/wiki/Photon

According to this picture there are indeed only one electric and one magnetic field and what's perpendicular about them seems to be the plane of their oscillation. However, it doesn't seem to me energy of the photon defines the strength of those fields, but rather the other way around. That is the amplitude/wavelength of their oscillation is what defines energy of the photon, where the strength of the fields remains constant.

Furthermore, how could there be a single magnetic field on its own, wouldn't that be a monopole? And also, how could there be an electric field in motion without creating yet another magnetic field directly around itself, which can't be the same one that is oscillating perpendicularly to it?


So, if net electric charge of a photon is zero, does that mean it contains both positive and negative electric fields, or what?

It means the electric and magnetic fields satisfy the source-free Maxwell's Equations:

http://en.wikipedia.org/wiki/Maxwell's_equations

I don't see any reference to "source-free" equation or anything similar that would relate to zero net charge of a photon.
 
Physics news on Phys.org
  • #2
The article says electromagnetic field is a physical field produced by moving electrically charged objects.
And a photon has to be emitted by a moving charge. So the electromagnetic field of a photon can be traced back to such a charge. No problem there. Though, it is a somewhat confusing definition.
The article says the theorem is a purely mathematical one.
So is the rest of physics.
"the invariants reveal that the electric and magnetic fields are perpendicular...", which seems to imply there is at least two electric and two magnetic fields
No, there are two fields. One electric and one magnetic makes two fields.
However, it doesn't seem to me energy of the photon defines the strength of those fields, but rather the other way around. That is the amplitude/wavelength of their oscillation is what defines energy of the photon, where the strength of the fields remains constant.
The amplitude remains constant, but not the fields themselves. Fields oscillate. They change sign. That definitely is not constant. Keep in mind that this is a linearly polarized field. The energy here is non-uniform. For uniform energy distribution, you have to go to circular polarization. Animation on that page shows how the E field behaves in a circularly-polarized EM wave, which is the proper basis for photon in QFT.

At any rate, energy is related to the square of the amplitude, because energy density is given by (E²+B²)/2.
I don't see any reference to "source-free" equation or anything similar that would relate to zero net charge of a photon.
Source-free Maxwell's Equations are these.
[tex]\nabla \cdot E = 0[/tex]
[tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]
[tex]\nabla \cdot B = 0[/tex]
[tex]\nabla \times B = \mu_0 \epsilon_0 \frac{\partial E}{\partial t}[/tex]

These are given in electromagnetic wave article you've already linked. As well as derivation of following from the above.

[tex]\nabla^2E = \frac{1}{c^2}\frac{\partial^2B}{\partial t^2}[/tex]

Similar equation is derived for the B field. The solution is the electromagnetic wave given in the article.
 
  • #3
Barry_G said:
I don't see any reference to "source-free" equation or anything similar that would relate to zero net charge of a photon.
It was on that page that you were already given. Here is the direct link so you cannot miss it.
http://en.wikipedia.org/wiki/Maxwel...s.2C_Electromagnetic_waves_and_speed_of_light

Btw, none of your questions are actually about photons. They are about classical EM. Specifically, you want to know how you can have EM fields in the absence of a charge. Maxwells equations in vacuum are the answer to that question.
 
Last edited:
  • #5
andrien said:
photon arises when one quantizes the electromagnetic field using creation and annihilation operator.All properties described to electromagnetic field can be ascribed to photon,if one quantizes it.you can see here for this quantization
http://en.wikipedia.org/wiki/Quantization_of_the_electromagnetic_field

After my browser crashed and destroyed my response to K^2 I'll start from the bottom now. Uh! That's a nice link, thank you. Do you know if photon spin can be measured?
 
  • #6
DaleSpam said:
It was on that page that you were already given. Here is the direct link so you cannot miss it.
http://en.wikipedia.org/wiki/Maxwel...s.2C_Electromagnetic_waves_and_speed_of_light

Btw, none of your questions are actually about photons. They are about classical EM. Specifically, you want to know how you can have EM fields in the absence of a charge. Maxwells equations in vacuum are the answer to that question.

I did not ask about photons propagating through region without charge, but how do you explain they are themselves electrically and magnetically neutral, considering they are (made of) electric and magnetic fields. Photon is electromagnetic field, right?
 
  • #7
K^2 said:
And a photon has to be emitted by a moving charge. So the electromagnetic field of a photon can be traced back to such a charge. No problem there. Though, it is a somewhat confusing definition.

I don't see how can it be traced nor what is that supposed to explain. An electron in an atom jumps an orbit and a photon is emitted, but neither electric charge, magnetic charge, nor dipole magnetic moment of that electron changes, so what does that tells us about photon electromagnetic fields?

In any case the question was about this sentence: "electromagnetic field is a physical field produced by moving electrically charged objects". So ok, I'll take it from here that photons are indeed electromagnetic fields, but the question is then how come electron is not electromagnetic fields, even more so since it fits that description better?


So is the rest of physics.

C'mon, "purely mathematical" means there is no experimental measurements relating to that specific theory and set of equations. Most of the physics can be experimentally confirmed.


No, there are two fields. One electric and one magnetic makes two fields.

I thought electron and positron are carriers of the smallest amount of charge, both electric and magnetic, so if these two fields are not electrons or positrons, do they even have a carrier? I guess they just "are", but do we see such electric or magnetic fields, devoid of any carrier particles, anywhere else, and is there any explanation how can that be?


The amplitude remains constant, but not the fields themselves. Fields oscillate. They change sign.

So, for example when electric field goes up it's positive and when it comes down it's negative? Do they change magnitude as they transfer from one sign to another?


That definitely is not constant. Keep in mind that this is a linearly polarized field. The energy here is non-uniform. For uniform energy distribution, you have to go to circular polarization.

That's a nice link, thank you. So when they are not circularly polarized why is it we can not deflect them with external electric or magnetic fields since one side would be negatively and the other side positively charged?


Source-free Maxwell's Equations are these.
[tex]\nabla \cdot E = 0[/tex]
[tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]
[tex]\nabla \cdot B = 0[/tex]
[tex]\nabla \times B = \mu_0 \epsilon_0 \frac{\partial E}{\partial t}[/tex]

I don't see how any of those equations can explain why photons have zero electric and magnetic charge.
 
  • #8
Barry_G said:
I did not ask about photons propagating through region without charge, but how do you explain they are themselves electrically and magnetically neutral, considering they are (made of) electric and magnetic fields. Photon is electromagnetic field, right?
In QED, photon can be described as electron + positron. They are virtual, of course, but if you feel like there absolutely has to be a charge, maybe it will help you sleep at night.

Barry_G said:
I don't see how can it be traced nor what is that supposed to explain. An electron in an atom jumps an orbit and a photon is emitted, but neither electric charge, magnetic charge, nor dipole magnetic moment of that electron changes, so what does that tells us about photon electromagnetic fields?
That's because you've only heard the explanation of transition that's given to chemists so that they don't have to think about physics too hard.

In reality, electron does not simply jump from one state to another. An atom that's emitting EM radiation as electron transitions from one energy level to another exists in superposition of two energy levels. While pure states have zero dipole moment, the superposition of any two states has a dipole expectation which rotates. So you actually do have an oscillating charge in an atom while it emits radiation.

Unfortunately, there aren't many good references on that. Pretty much the only thing I can suggest is taking the hydrogen atom solutions, composing a superposition state, say (1s+2p)/√2, and computing the dipole expectation. There is nothing really difficult about it, but it's about a page of notes.

C'mon, "purely mathematical" means there is no experimental measurements relating to that specific theory and set of equations.
Electric field in electromagnetic wave of low enough frequency can be measured directly. That's basically what radio is all about. So yes, we have very good experimental background on Maxwell's Equations, including wave solutions. Not to mention that they are used in just about every aspect of optics and electrical engineering.

Everything in physics that is part of standard theory has overwhelming experimental support. If you think something doesn't, you are simply ignorant of that branch of physics. Feel free to ask, though. Keep in mind that a lot of things are confirmed indirectly.

I thought electron and positron are carriers of the smallest amount of charge, both electric and magnetic
Neither. First of all, there is no such thing as magnetic charge. There are only dipoles, and these can be arbitrarily small. Smallest electric charge is held by down, strange, and bottom quarks at ±1/3 that of the electron. These are followed by the up, charmed, and top quarks at ±2/3. Charged leptons come in in 3rd with ±1, with electron being one of them.

What's interesting is that charge can only change by a unit of 1e, which is why electron charge is also known as elementary charge. It's like spin. You can have a fraction, but you have to change it by a unit.
I guess they just "are", but do we see such electric or magnetic fields, devoid of any carrier particles, anywhere else, and is there any explanation how can that be?
Photons ARE the carriers of electromagnetic field. I'll let you think on that for a while. Of course, again, that is QED. In classic field theory, carrier particle is not required.
So, for example when electric field goes up it's positive and when it comes down it's negative? Do they change magnitude as they transfer from one sign to another?
In linearly polarized, yes. Magnitude and direction oscillate as a sine wave. In circularly polarized, only direction changes, with magnitude remaining constant.

I don't see how any of those equations can explain why photons have zero electric and magnetic charge.
Because ρ and j are taken to be zero in that set of equations. That means there are no charges or currents present. No sources for electrostatic or magnetostatic fields. If you only consider the static solution to these equations, you will get trivial solution E=B=0 as the only possibility.

Maybe you would understand EM propagation better via retarded potentials. You can think of EM wave as being caused by electric field of the distant charge that oscillates, and because "information" about position of the charge is delayed by speed of light, you get an oscillating EM field. This is really a carriage before horse explanation, because speed of light is hand-waved into it, but again, if it makes it easier for you to understand, maybe it's worth for you to take a look. Retarded Potential.
 
  • #9
Barry_G said:
how do you explain they are themselves electrically and magnetically neutral, considering they are (made of) electric and magnetic fields.
Using Maxwell's equations for vacuum, as I already stated and linked you to. The point of those equations is that you DO NOT NEED a charge in order to have an EM field. I.e. it is perfectly consistent with Maxwell's equations to have something which is uncharged and yet is made of electric and magnetic fields.
 
  • #10
Barry_G said:
I did not ask about photons propagating through region without charge, but how do you explain they are themselves electrically and magnetically neutral, considering they are (made of) electric and magnetic fields. Photon is electromagnetic field, right?

Charge is a property of matter that has a curious effect of causing a charged particle to make other charged particles either scramble away from or rush towards it. We describe all this rushing and scrambling mathematically using a "field". A photon is how waves in this field interact with other objects. The waves themselves, and as such the photons, are not charged. They cannot be, for the EM wave is a disturbance in the field itself, not a charged particle.

Imagine for a moment that ships in the ocean could cause other ships to be pushed away by pushing all nearby water outwards or pulled in by pulling all nearby water towards themselves. Would you say the water itself has this property? No! Clearly it's just the ships themselves that do this. The water far away from any ships does nothing. How about a wave in this water, would it have this curious pushing or pulling property? Of course not!
 
  • #12
K^2 said:
In QED, photon can be described as electron + positron.

And there I thought I was the first one to figure that out, uh! So you see what I meant when I said two electric and six magnetic fields. I'd like to read more about it, but all I find when searching for it is annihilation, pair production and Dirac Sea, so if you could point some links that talk specifically about that description of a photon as electron + positron, that would be marvelous.


In reality, electron does not simply jump from one state to another. An atom that's emitting EM radiation as electron transitions from one energy level to another exists in superposition of two energy levels. While pure states have zero dipole moment, the superposition of any two states has a dipole expectation which rotates. So you actually do have an oscillating charge in an atom while it emits radiation.

Ok, but that's not quite what I was originally asking about.

http://en.wikipedia.org/wiki/Electromagnetic_field : An electromagnetic field is a physical field produced by moving electrically charged objects. Does that mean electron, for example, is considered to be "electromagnetic field", or should they have better said: "electromagnetic radiation is a physical field emitted by moving electrically charged objects"?


Electric field in electromagnetic wave of low enough frequency can be measured directly. That's basically what radio is all about.

Is that before or after collision? I mean can it be measured, or sensed in whatever way, before it actually collides with, or gets absorbed by, something?


Everything in physics that is part of standard theory has overwhelming experimental support. If you think something doesn't, you are simply ignorant of that branch of physics. Feel free to ask, though. Keep in mind that a lot of things are confirmed indirectly.

Experiments, like words, can be interpreted in more than one way, but I'm not asking about measurements because I doubt it, which I do, it's mostly that I'd just like to know about it.


First of all, there is no such thing as magnetic charge.

I thought "charge" is susceptibility to some force, so I use "charge" as a sort of synonym for "field". For example, if a particle has measurable electric field, I say it is electrically charged. Or if combination of particles, like some ion atom or molecule, has nonuniform distribution or uneven number of positive and negative fields, then I say it's electrically charged, or that it has net electric charge greater than zero. I apply the same logic to magnetic fields, so that magnetic dipole moment has both negative and positive magnetic charge. You think something is wrong with such use of the terminology?


There are only dipoles, and these can be arbitrarily small. Smallest electric charge is held by down, strange, and bottom quarks at ±1/3 that of the electron. These are followed by the up, charmed, and top quarks at ±2/3. Charged leptons come in in 3rd with ±1, with electron being one of them.

What's interesting is that charge can only change by a unit of 1e, which is why electron charge is also known as elementary charge. It's like spin. You can have a fraction, but you have to change it by a unit.

Ok, but I would not call that "arbitrarily" small.


Photons ARE the carriers of electromagnetic field. I'll let you think on that for a while. Of course, again, that is QED. In classic field theory, carrier particle is not required.

That's strange thing to say. Carrier particle is something that is in the center of a field, where the field is strongest. But all those diagrams show photon is rather a composition of those fields, made of them, not really a source of them. So if QED, which I confirm, can describe a photon as electron-positron interaction, well that's how it looks like, as two separate oscillating charges interacting with each other, and superposition would then explain why would those two appear to macroscopically have zero net charge.


Because ρ and j are taken to be zero in that set of equations. That means there are no charges or currents present. No sources for electrostatic or magnetostatic fields. If you only consider the static solution to these equations, you will get trivial solution E=B=0 as the only possibility.

How can you say there is electric field but there is no electric charge? Electric field implies electric charge, doesn't it? And if there is electric field, then it should be influenced by an external electric field, unless... why not explain it by the change of sign, so that they on AVERAGE, or macroscopically, appear to have zero charge, but internally you could still have those fields?


Maybe you would understand EM propagation better via retarded potentials. You can think of EM wave as being caused by electric field of the distant charge that oscillates, and because "information" about position of the charge is delayed by speed of light, you get an oscillating EM field. This is really a carriage before horse explanation, because speed of light is hand-waved into it, but again, if it makes it easier for you to understand, maybe it's worth for you to take a look. Retarded Potential.

Can I stick with QED and electron-positron description? I like that one.
 
Last edited:
  • #13
Barry_G said:
Can I stick with QED and electron-positron description? I like that one.
That's the one description you have least hope of understanding any time soon. Seriously, forget all of this stuff, and go learn electrodynamics. Just open up a classical ED textbook and start reading. Then you'll have grasp of at least one description we can actually discuss, and we can build on that.

How can I explain to you that electron emitting a photon and photon being electron-positron pair are exactly the same thing, when I can't rely on your understanding of basic, non-relativistic field theory? This is advanced stuff that's built on more advanced stuff that's built on several different branches of fundamental stuff. You can't just jump to the end and hope to understand any of it.
 
  • #14
K^2 said:
That's the one description you have least hope of understanding any time soon. Seriously, forget all of this stuff, and go learn electrodynamics. Just open up a classical ED textbook and start reading. Then you'll have grasp of at least one description we can actually discuss, and we can build on that.

I find your condescending assumptions are amusing, and if you haven't realized by now I came up to that conclusion myself, without any prior knowledge about it, which I'd say quite well justifies my curiosity about QED interpretation of the same thing. Anyway, give me some links or go mind your own business, your display of vanity and personal comments about me are out of the topic and very unnecessary.


How can I explain to you that electron emitting a photon and photon being electron-positron pair are exactly the same thing, when I can't rely on your understanding of basic, non-relativistic field theory? This is advanced stuff that's built on more advanced stuff that's built on several different branches of fundamental stuff. You can't just jump to the end and hope to understand any of it.

I ask specific questions to which I'd like to get specific answers. Don't you worry about me, just answer, if you can, or get off my back, will ya?
 
  • #15
DaleSpam said:
Using Maxwell's equations for vacuum, as I already stated and linked you to. The point of those equations is that you DO NOT NEED a charge in order to have an EM field. I.e. it is perfectly consistent with Maxwell's equations to have something which is uncharged and yet is made of electric and magnetic fields.

Does electric field not imply electric charge? All I am asking is if photon is electromagnetic field why can not be influenced by an external electric or magnetic field. What equation explains that?
 
  • #16
Drakkith said:
Charge is a property of matter that has a curious effect of causing a charged particle to make other charged particles either scramble away from or rush towards it. We describe all this rushing and scrambling mathematically using a "field".

Yes, they are synonyms. Electron has electric field, therefore it is electrically charged.

A photon is how waves in this field interact with other objects. The waves themselves, and as such the photons, are not charged. They cannot be, for the EM wave is a disturbance in the field itself, not a charged particle.

I disagree. As all those diagrams show photon is not waves in some field, but rather waves of those fields. It's not photon that waves, it is magnetic and electric fields that move, and their combination is what photon is.

440px-Circular.Polarization.Circularly.Polarized.Light_With.Components_Right.Handed.svg.png
 
  • #17
  • #18
Barry_G said:
I find your condescending assumptions are amusing, and if you haven't realized by now I came up to that conclusion myself, without any prior knowledge about it, which I'd say quite well justifies my curiosity about QED interpretation of the same thing.
Yet you still don't realize that the charge density at every point in space in and around the photon is precisely zero. It never becomes non-zero. You made an assumption that EM field requires a charge density, and that's why you cling onto QED interpretation. But QED interpretation does not give you a charge. It gives you virtual particle production, but no charge.

Does electric field not imply electric charge?
No. How many times do several people have to tell you that? Maxwell's equations with zero charge still allow for EM wave propagation.

All I am asking is if photon is electromagnetic field why can not be influenced by an external electric or magnetic field.
Because photon is neutral. Neutral particles are not influenced by electric field. Photon also has zero magnetic moment, so it is not influenced by magnetic field either.

I disagree. As all those diagrams show photon is not waves in some field, but rather waves of those fields.
It's the same thing. Photon is the field. Field is the photon. Waves in the field or waves of the field are exactly the same thing in this case.

So you mean "no", photon spin can not be experimentally measured?
Photon spin is measured experimentally by transferring the angular momentum from photon to a charged particle and measuring resonance in magnetic field. Is that good enough for you? Or do you need a probe that you would like to stick into a photon? Because if it's the later case, I have to disappoint you. All particle physics measurements are indirect.
 
  • #19
Barry_G said:
Yes, they are synonyms. Electron has electric field, therefore it is electrically charged.

No, they are not synonyms. Charge is a property of a particle, a field is not. The field describes how particles will react to each other based on their charge, distance, and relative motion.

http://en.wikipedia.org/wiki/Field_(physics)
http://en.wikipedia.org/wiki/Electric_charge

I disagree. As all those diagrams show photon is not waves in some field, but rather waves of those fields. It's not photon that waves, it is magnetic and electric fields that move, and their combination is what photon is.

Please, that's like arguing that a water wave is a wave OF water instead of a wave IN water. It means the same thing.
 
  • #20
K^2 said:
Yet you still don't realize that the charge density at every point in space in and around the photon is precisely zero. It never becomes non-zero. You made an assumption that EM field requires a charge density, and that's why you cling onto QED interpretation. But QED interpretation does not give you a charge. It gives you virtual particle production, but no charge.

So the charge of the electric field in a photon is zero? Then what is the magnitude of that electric field, say at peaks of its amplitude?


No. How many times do several people have to tell you that?

Insistence does not prove or solve anything. Can you point some reference that distinguishes electric field from electric charge?


Because photon is neutral. Neutral particles are not influenced by electric field. Photon also has zero magnetic moment, so it is not influenced by magnetic field either.

Is magnetic or electric field of a photon neutral?



It's the same thing. Photon is the field. Field is the photon. Waves in the field or waves of the field are exactly the same thing in this case.

It's not the same thing. Like gravitational waves are not the same thing as spatial motion of gravity field as whole.



Photon spin is measured experimentally by transferring the angular momentum from photon to a charged particle and measuring resonance in magnetic field.

So photon spin is has something do with orientation of its magnetic field?
 
  • #21
Barry_G said:
So you mean "no", photon spin can not be experimentally measured?
no,it can be measured experimetally.Here is some old experiment on it
http://prola.aps.org/abstract/PR/v50/i2/p115_1
 
  • #22
Barry_G said:
Can you point some reference that distinguishes electric field from electric charge?
Maxwell's. Equations. There are no other equations for electromagnetic field.
So the charge of the electric field in a photon is zero?
There is no such thing as charge of the electric field.
 
  • #23
Drakkith said:
No, they are not synonyms. Charge is a property of a particle, a field is not. The field describes how particles will react to each other based on their charge, distance, and relative motion.

http://en.wikipedia.org/wiki/Field_(physics)
http://en.wikipedia.org/wiki/Electric_charge

Ok, they are not synonyms. Field is a concept that describes geometrical distribution of charge, and charge just refers to the magnitude. For example, a field can have shape, described by gradients, while charge is a scalar number, a measure of the magnitude at some point in the field.

The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably.


Please, that's like arguing that a water wave is a wave OF water instead of a wave IN water. It means the same thing.

Water wave is effect caused by the motion OF water molecules. You have to distinguish cause and effect, what is moving and what is the product of that motion.
 
  • #24
andrien said:
no,it can be measured experimetally.Here is some old experiment on it
http://prola.aps.org/abstract/PR/v50/i2/p115_1

I'm afraid that's about polarization, not spin, unless those two are related, but I don't see how since polarization is about planes of oscillation of em fields and seems it can be arbitrary, while spin is supposed to be intrinsic and just 1 or -1.
 
  • #25
Barry_G said:
I'm afraid that's about polarization, not spin, unless those two are related, but I don't see how since polarization is about planes of oscillation of em fields and seems it can be arbitrary, while spin is supposed to be intrinsic and just 1 or -1.
If you actually bothered to read the article on circular polarizaion that has been previously linked, you would know the answer to this question.
 
  • #26
Barry_G said:
Insistence does not prove or solve anything. Can you point some reference that distinguishes electric field from electric charge?

"Electricity and Magnetism" by Edward Purcell would be a good start. Google will find you a copy somewhere for cheap.
 
  • #27
K^2 said:
Maxwell's. Equations. There are no other equations for electromagnetic field.

But there are charges, electric and magnetic fields outside of photon, so why are we having this silly semantic argument, why can not we find nice specific definition that would point out differences so we can distinguish what is field and what is charge?

Maxwell's equations do not explain how can photon have electric and magnetic field and yet we can not bend a beam of light by external electric or magnetic fields.

Actually polarization should answer this, if we can change polarization then that would be the way to influence these fields, right? Perhaps we can not bend a beam of light, but if we can change the plane of em fields oscillation, well that's pretty good too.


There is no such thing as charge of the electric field.

Depends on definition. We need a technical dictionary, or something that would make it clear how they relate and how they differ. Until then I think my definition is better than yours.
 
Last edited:
  • #28
K^2 said:
If you actually bothered to read the article on circular polarizaion that has been previously linked, you would know the answer to this question.

It says almost nothing about it, just rises many questions. What am I to conclude, left and right spin corresponds to 1 and -1? Ok, so why then Wikipedia says spin of a photon is 1, why does it not say 1 or -1? And what would be the spin of non-spinning photon? Also, how does that compare to electron spin since it seems photon can be spinning at arbitrary speed?
 
Last edited:
  • #29
Barry_G said:
Maxwell's equations do not explain how can photon have electric and magnetic field and yet we can not bend a beam of light by external electric or magnetic fields.
Yes they do. They do precisely that. This is absolutely obvious, but let's walk through it. Let [itex]E_{\gamma}[/itex] and [itex]B_{\gamma}[/itex] be electric and magnetic fields of some beam of light respectively. They, therefore, satisfy charge-free Maxwell's Equations.

[tex]\nabla \cdot E_{\gamma} = 0[/tex]
[tex]\nabla \times E_{\gamma} = -\frac{\partial B_{\gamma}}{\partial t}[/tex]
[tex]\nabla \cdot B_{\gamma} = 0[/tex]
[tex]\nabla \times B_{\gamma} = \mu_0 \epsilon_0 \frac{\partial E_{\gamma}}{\partial t}[/tex]

However, instead of sending that beam of light through the vacuum, we send it through some system of charges and currents that have already created some fields [itex]E[/itex] and [itex]B[/itex]. These fields must satisfy Maxwell's Equations with charges and currents.

[tex]\nabla \cdot E = \frac{\rho}{\epsilon_0}[/tex]
[tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]
[tex]\nabla \cdot B = 0[/tex]
[tex]\nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}[/tex]

Here, [itex]\rho[/itex] and [itex]J[/itex] are charge and current densities, respectively.

We wish to demonstrate that this does not alter the trajectory of the light beam. In other words, that the new electric field is simply [itex]E + E_{\gamma}[/itex] and new magnetic field is simply [itex]B + B_{\gamma}[/itex]. We show that by demonstrating that these sums satisfy the Maxwell's equations with the same charge and current densities.

[tex]\nabla \cdot (E + E_{\gamma}) = \nabla \cdot E + \nabla \cdot E_{\gamma} = \frac{\rho}{\epsilon_0} + 0 = \frac{\rho}{\epsilon_0}[/tex]
[tex]\nabla \times (E + E_{\gamma}) = \nabla \times E + \nabla \times E_{\gamma} = -\frac{\partial B}{\partial t} -\frac{\partial B_{\gamma}}{\partial t} = -\frac{\partial(B + B_{\gamma})}{\partial t}[/tex]
[tex]\nabla \cdot (B + B_{\gamma}) = \nabla \cdot B + \nabla \cdot B_{\gamma} = 0 + 0 = 0[/tex]
[tex]\nabla \times (B + B_{\gamma}) = \nabla \times B + \nabla \times B_{\gamma} = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t} + \mu_0 \epsilon_0 \frac{\partial E_{\gamma}}{\partial t} = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial (E + E_{\gamma})}{\partial t}[/tex]

The equations are satisfied. Therefore, the electromagnetic field of a beam of light simply adds on top of electromagnetic fields already present without any kind of distortion. In other words, photons are not affected by electric or magnetic fields.

Are we finally done with your "charged photon" nonsense?

It says almost nothing about it, just rises many questions. What am I to conclude, left and right spin corresponds to 1 and -1? Ok, so why then Wikipedia says spin of a photon is 1, why does it not say 1 or -1? And what would be the spin of non-spinning photon? Also, how does that compare to electron spin since it seems photon can be spinning at arbitrary speed?
That's only because you have no idea what spin is. I strongly suggest reading an article on that. Pay attention to quantization axes and connection to angular momentum.
 
  • #30
Nugatory said:
"Electricity and Magnetism" by Edward Purcell would be a good start. Google will find you a copy somewhere for cheap.

Do you answer all the question like that? It's not very helpful. I can not justify spending all that time or any money to just get definition of two terms which I already believe to know what they mean. Do you have that book, why not just write it down for everyone? Does it not say field is a description of geometrical distribution of charge, where charge is just a magnitude at some point of that field?
 
  • #31
You aren't just confused on a couple of definitions. You have no clue what you are talkinga bout at all. You need to read a textbook. That was probably the most helpful suggestion you got.
 
  • #32
Barry, you seem to miss the point that K^2 tried so hard to show you.
There can be an electric field without any charge. An EM wave is just this.
For example the light coming from stars contain no charge at all, yet it is made of an electromagnetic field. All the light is like this.
You can take a region in space with an electric field and find no charge.
Barry_G said:
Ok, they are not synonyms. Field is a concept that describes geometrical distribution of charge, and charge just refers to the magnitude. For example, a field can have shape, described by gradients, while charge is a scalar number, a measure of the magnitude at some point in the field.
I'm not really understanding this.

Barry_G said:
The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably.
No, absolutely not as explained above.
 
  • #33
Barry_G said:
Ok, they are not synonyms. Field is a concept that describes geometrical distribution of charge, and charge just refers to the magnitude. For example, a field can have shape, described by gradients, while charge is a scalar number, a measure of the magnitude at some point in the field.

The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably.

After reading through this and the thread you hijacked in the Relativity section, I must be the fifth or sixth person to tell you what I'm about to tell you:

There does not need to be any charges for there to be an electric field. Maxwell's Equations tell us that a changing magnetic field will also produce an electric field. Similarly, you don't need any currents for there to be a magnetic field; a changing electric field will produce a magnetic field. Essentially what happens in an EM wave is that a changing electric field produces a changing magnetic field which produces a changing electric field which... This allows the electric and magnetic fields to propagate without any charges or currents. I'll even go through how this is derived, step by step:

As you've been shown numerous times, Maxwell's Equations in a vacuum are:

[tex]\nabla \cdot E=0[/tex]
[tex]\nabla \times E = -\frac{\partial B}{\partial t}[/tex]
[tex]\nabla \cdot B=0[/tex]
[tex]\nabla \times B = \frac{1}{c^2} \frac{\partial E}{\partial t}[/tex]Now, using the following "curl of the curl" identity: [itex]\nabla \times \nabla \times A=\nabla(\nabla\cdot A)-\nabla^2A[/itex]
[tex]\nabla \times \nabla \times E=\nabla(0)-\nabla^2E=-\nabla \times \frac{\partial B}{\partial t}=-\frac{\partial }{\partial t}[\nabla \times B]=-\frac{\partial }{\partial t}[\frac{1}{c^2} \frac{\partial E}{\partial t}][/tex]
Simplifying you get:
[tex]\nabla^2E=\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}[/tex]
You can apply the same identity to the curl of the magnetic field to get the following:
[tex]\nabla^2B=\frac{1}{c^2}\frac{\partial^2 B}{\partial t^2}[/tex]
The solution to both of these differential equations is a sinusoidal wave which moves at a velocity c.
You accuse K^2 of being condescending, yet he's absolutely correct. You have no idea how basic electrodynamics works. What you need to do is stop pretending you know what you're talking about and go pick up a textbook. You're arguing about things you don't understand.
 
Last edited:
  • #34
K^2 said:
The equations are satisfied. Therefore, the electromagnetic field of a beam of light simply adds on top of electromagnetic fields already present without any kind of distortion. In other words, photons are not affected by electric or magnetic fields.

Are we finally done with your "charged photon" nonsense?

That's ridiculous. There are no any photons, nor external fields in those equations except for Faraday's law. Gauss's law is about electric flux through closed surface, it's analogous to Coulomb's law. Gauss's law for magnetism describes geometry of magnetic field. Faraday's law of induction is about generating voltage by external magnetic field in closed circuit loops. Ampère's circuital law is about magnetic field generated around close circuit loop.


That's only because you have no idea what spin is. I strongly suggest reading an article on that. Pay attention to quantization axes and connection to angular momentum.

Pay attention to what I said and if you have any idea what spin is you might understand.


You aren't just confused on a couple of definitions. You have no clue what you are talkinga bout at all. You need to read a textbook. That was probably the most helpful suggestion you got.

You aren't just confused about actual application of Maxwell's equations, but you also have no clue what you are talking about at all. You need to wake up.
 
  • #35
Barry_G said:
That's ridiculous. There are no any photons, nor external fields in those equations except for Faraday's law. Gauss's law is about electric flux through closed surface, it's analogous to Coulomb's law. Gauss's law for magnetism describes geometry of magnetic field. Faraday's law of induction is about generating voltage by external magnetic field in closed circuit loops. Ampère's circuital law is about magnetic field generated around close circuit loop.

Your responses are quite painful to read. The "photon" in question consists of an electric and a magnetic field, which he denoted as [itex]E_{\gamma}[/itex] and [itex]B_{\gamma}[/itex]. Maxwell's Equations aren't just for the narrow range of applications you list above. They completely describe how electric and magnetic fields behave!
Barry_G said:
You aren't just confused about actual application of Maxwell's equations, but you also have no clue what you are talking about at all. You need to wake up.

I really don't see how you could possibly be under the impression that you're the one here with the most understanding of E&M. You understand practically nothing. In fact, I'm starting to suspect this entire thread is just a giant troll. I don't see how you could realistically be this delusional.
 
<h2>1. What is a photon?</h2><p>A photon is the smallest unit of light and is considered both a particle and a wave. It carries energy and has no mass.</p><h2>2. How is a photon related to the electromagnetic field?</h2><p>A photon is an electromagnetic wave that carries energy and momentum through space. It is the fundamental unit of the electromagnetic field.</p><h2>3. Is a photon always an electromagnetic field?</h2><p>Yes, a photon is always an electromagnetic field. It is the fundamental unit of the electromagnetic field and cannot exist without it.</p><h2>4. How does a photon behave?</h2><p>A photon behaves both as a particle and a wave. It can travel through space in a straight line like a particle, but also exhibits wave-like properties such as interference and diffraction.</p><h2>5. Can a photon be detected?</h2><p>Yes, a photon can be detected using specialized equipment such as a photomultiplier tube or a photodiode. These devices can measure the energy and intensity of a photon when it interacts with matter.</p>

1. What is a photon?

A photon is the smallest unit of light and is considered both a particle and a wave. It carries energy and has no mass.

2. How is a photon related to the electromagnetic field?

A photon is an electromagnetic wave that carries energy and momentum through space. It is the fundamental unit of the electromagnetic field.

3. Is a photon always an electromagnetic field?

Yes, a photon is always an electromagnetic field. It is the fundamental unit of the electromagnetic field and cannot exist without it.

4. How does a photon behave?

A photon behaves both as a particle and a wave. It can travel through space in a straight line like a particle, but also exhibits wave-like properties such as interference and diffraction.

5. Can a photon be detected?

Yes, a photon can be detected using specialized equipment such as a photomultiplier tube or a photodiode. These devices can measure the energy and intensity of a photon when it interacts with matter.

Similar threads

  • Electromagnetism
Replies
4
Views
811
  • Electromagnetism
Replies
11
Views
598
Replies
8
Views
946
Replies
10
Views
2K
Replies
14
Views
1K
Replies
4
Views
1K
  • Electromagnetism
Replies
25
Views
20K
  • Electromagnetism
Replies
1
Views
265
  • Electromagnetism
Replies
8
Views
2K
Replies
17
Views
967
Back
Top