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Photon is electromagnetic field, right?

by Barry_G
Tags: electromagnetic, field, photon
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Barry_G
#1
Dec11-12, 02:45 AM
P: 68
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.

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.
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K^2
#2
Dec11-12, 03:50 AM
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P: 2,470
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 (EČ+BČ)/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.
DaleSpam
#3
Dec11-12, 06:11 AM
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Quote Quote by Barry_G View Post
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/Maxwell...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.

andrien
#4
Dec11-12, 07:27 AM
P: 1,020
Photon is electromagnetic field, right?

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/Quantiz...magnetic_field
Barry_G
#5
Dec11-12, 09:41 AM
P: 68
Quote Quote by andrien View Post
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/Quantiz...magnetic_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?
Barry_G
#6
Dec11-12, 09:51 AM
P: 68
Quote Quote by DaleSpam View Post
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/Maxwell...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?
Barry_G
#7
Dec11-12, 10:49 AM
P: 68
Quote Quote by K^2 View Post
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.
K^2
#8
Dec11-12, 01:21 PM
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Quote Quote by Barry_G View Post
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.

Quote Quote by Barry_G
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.
DaleSpam
#9
Dec11-12, 03:14 PM
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Quote Quote by Barry_G View Post
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.
Drakkith
#10
Dec11-12, 04:18 PM
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Quote Quote by Barry_G View Post
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!
andrien
#11
Dec11-12, 11:41 PM
P: 1,020
Quote Quote by Barry_G View Post
Do you know if photon spin can be measured?
yes,the spin of photon can be determined on pure theoretical basis.For this you may see this thread
http://www.physicsforums.com/showthread.php?t=654329
Barry_G
#12
Dec12-12, 02:58 AM
P: 68
Quote Quote by K^2 View Post
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.
K^2
#13
Dec12-12, 03:20 AM
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Quote Quote by Barry_G View Post
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.
Barry_G
#14
Dec12-12, 03:46 AM
P: 68
Quote Quote by K^2 View Post
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?
Barry_G
#15
Dec12-12, 04:03 AM
P: 68
Quote Quote by DaleSpam View Post
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?
Barry_G
#16
Dec12-12, 04:23 AM
P: 68
Quote Quote by Drakkith View Post
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.

Barry_G
#17
Dec12-12, 04:28 AM
P: 68
Quote Quote by andrien View Post
yes,the spin of photon can be determined on pure theoretical basis.For this you may see this thread
http://www.physicsforums.com/showthread.php?t=654329
So you mean "no", photon spin can not be experimentally measured?
K^2
#18
Dec12-12, 04:36 AM
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Quote Quote by Barry_G View Post
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.


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