# Photon is electromagnetic field, right?

by Barry_G
Tags: electromagnetic, field, photon
Mentor
P: 12,010
 Quote by Barry_G 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_%28physics%29
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.
P: 68
 Quote by K^2 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?
P: 1,020
 Quote by Barry_G 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
P: 2,470
 Quote by Barry_G 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.
P: 68
 Quote by Drakkith 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_%28physics%29 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.
P: 68
 Quote by andrien 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.
P: 2,470
 Quote by Barry_G 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.
Mentor
P: 3,967
 Quote by Barry_G 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.
P: 68
 Quote by K^2 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.
P: 68
 Quote by K^2 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?
P: 2,470
 Quote by Barry_G 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 lets walk through it. Let $E_{\gamma}$ and $B_{\gamma}$ be electric and magnetic fields of some beam of light respectively. They, therefore, satisfy charge-free Maxwell's Equations.

$$\nabla \cdot E_{\gamma} = 0$$
$$\nabla \times E_{\gamma} = -\frac{\partial B_{\gamma}}{\partial t}$$
$$\nabla \cdot B_{\gamma} = 0$$
$$\nabla \times B_{\gamma} = \mu_0 \epsilon_0 \frac{\partial E_{\gamma}}{\partial t}$$

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 $E$ and $B$. These fields must satisfy Maxwell's Equations with charges and currents.

$$\nabla \cdot E = \frac{\rho}{\epsilon_0}$$
$$\nabla \times E = -\frac{\partial B}{\partial t}$$
$$\nabla \cdot B = 0$$
$$\nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}$$

Here, $\rho$ and $J$ 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 $E + E_{\gamma}$ and new magnetic field is simply $B + B_{\gamma}$. We show that by demonstrating that these sums satisfy the Maxwell's equations with the same charge and current densities.

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

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.
P: 68
 Quote by Nugatory "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?
 Sci Advisor P: 2,470 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.
PF Gold
P: 3,225
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.
 Quote by Barry_G 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.

 Quote by Barry_G 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.
P: 261
 Quote by Barry_G 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:

$$\nabla \cdot E=0$$
$$\nabla \times E = -\frac{\partial B}{\partial t}$$
$$\nabla \cdot B=0$$
$$\nabla \times B = \frac{1}{c^2} \frac{\partial E}{\partial t}$$

Now, using the following "curl of the curl" identity: $\nabla \times \nabla \times A=\nabla(\nabla\cdot A)-\nabla^2A$
$$\nabla \times \nabla \times E=\nabla(0)-\nabla^2E=-\nabla \times \frac{\partial B}{\partial t}=-\frac{\partial }{\partial t}[\nabla \times B]=-\frac{\partial }{\partial t}[\frac{1}{c^2} \frac{\partial E}{\partial t}]$$
Simplifying you get:
$$\nabla^2E=\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}$$
You can apply the same identity to the curl of the magnetic field to get the following:
$$\nabla^2B=\frac{1}{c^2}\frac{\partial^2 B}{\partial t^2}$$
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.
P: 68
 Quote by K^2 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.
P: 261
 Quote by Barry_G 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 $E_{\gamma}$ and $B_{\gamma}$. Maxwell's Equations aren't just for the narrow range of applications you list above. They completely describe how electric and magnetic fields behave!

 Quote by Barry_G 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.
Mentor
P: 17,543
 Quote by Barry_G Does electric field not imply electric charge?
No, electric field does not imply electric charge. See Maxwell's vacuum equations.

 Quote by Barry_G 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?
Maxwell's equations. The key property of Maxwell's equations that lead to this is the linearity. The linearity of Maxwell's equations shows that EM follows the principle of superposition which in turn implies that an EM field won't be altered by passing through an external static electric or magnetic field.

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

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