# Photon Polarization from Electro-dynamic Potential ?

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Roughly speaking, an Electro-dynamic potential, $\vec{A}$, can generate both $\vec{E}$ & $\vec{B}$, from $\vec{E} \approx \frac{\partial \vec{A}}{\partial t}$ & $\vec{B} = \nabla \times \vec{A}$. Thus, an EDP, oscillating up & down along a single axis, would generate both the electric & magnetic fields, of a light wave. Is, then, the direction of $\vec{A}$ the polarization axis of light ?

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What is the curl of a vector with a constant direction?

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Roughly speaking, an Electro-dynamic potential, $\vec{A}$, can generate both $\vec{E}$ & $\vec{B}$, from $\vec{E} \approx \frac{\partial \vec{A}}{\partial t}$ & $\vec{B} = \nabla \times \vec{A}$. Thus, an EDP, oscillating up & down along a single axis, would generate both the electric & magnetic fields, of a light wave. Is, then, the direction of $\vec{A}$ the polarization axis of light ?

Yes, that is right. A(x,t) is not a constant vector in space and time so both E and B may be different from zero. In particular, if A(x,t) = a⋅exp(ikx-iωt), you have a plane EMW.

Edited by Bob_for_short
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• 2 weeks later...

(Thanks for the replies)

Can you create a "scalar photon", w/o using an electro-dynamic potential (A), but only a time-varying electro-static potential (V) ?

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Can you create a "scalar photon", w/o using an electro-dynamic potential (A), but only a time-varying electro-static potential (V) ?

Such a field exists indeed next to a vibrating dipole (along its axis) but this variable field does not propagate too far - it decays rapidly with the distance and is called a "near field". Anyway it's an axial (radial) electric field E = -grad(V).

Edited by Bob_for_short
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Thanks for the response

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Hi Widdekind, Bob. Can either of you explain the difference between A and V?

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Hi Widdekind, Bob. Can either of you explain the difference between A and V?

V is written upside down and does not have a "-" in the middle.

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The reason I'm asking is related to the Aharonov-Bohm effect and electromagnetic four-potential. And it's related to the OP, particularly since Widdekind has been talking about this elsewhere. Take a look at http://www.chem.yale.edu/~chem125/125/xray/diffract.html and this depiction of an electromagnetic wave: Look at one complete wavelength, starting from the left. There's an electromagnetic field variation here, but no charged particle present. So there has to be some sort of current, and it isn't conduction current. So surely it's displacement current? Orthogonal to the propagation direction, and giving the polarization axis? Assuming a plane-polarized light wave, does the sinusoidal waveform denote the strength of this current? It increases to a positive peak, but continues "flowing upwards" weaker and weaker as the waveform returns to zero. If so the displacement current presumably then turns downwards, reaching a peak strength at the negative sinusoidal peak before weakening on the return to zero. So halfway along the waveform, where the electromagnetic field is zero, the displacement is at a peak, and the photon is essentially a pulse, as per http://arxiv.org/abs/0803.2596. In the Aharonov-Bohm effect there's no electric or magnetic field outside the solenoid, but we know there's something there. Is this the same sort of thing, and is the evanescent wave a standing displacement? Clearing up those As and Vs would help here.

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Look at one complete wavelength, starting from the left. There's an electromagnetic field variation here, but no charged particle present. So there has to be some sort of current, and it isn't conduction current. So surely it's displacement current?

Why? How do you know there's no charged particle?

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...Look at one complete wavelength, starting from the left. There's an electromagnetic field variation here, but no charged particle present. ...

This is a graph of the electric force F(t) = q*E(t). Without a charge, there is no force to act on.

If you speak of the field E source, it is far away, an antenna, for example.

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In the Aharonov-Bohm effect there's no electric or magnetic field outside the solenoid, but we know there's something there. Is this the same sort of thing, and is the evanescent wave a standing displacement? Clearing up those As and Vs would help here.

I understand, that the AB effect relies directly upon the Potentials, which "underlie" the so-called E&B fields. It is the Potentials (V, A) which appear in the Hamiltonian, of the SWE. Quantum 'particles' respond directly to the Potentials. (The "fields" derived therefrom are only useful in the Classical limit, of Ehrenfest Theorem expectation values, d<x>/dt = -<p>/m, etc.)

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Why? How do you know there's no charged particle?
Because I'm talking about an electromagnetic wave here. See this alternative picture. Forget about the magnetic field variation because it's an electromagnetic field variation, and consider one wavelength starting from the origin rather than the peak: This is a graph of the electric force F(t) = q*E(t). Without a charge, there is no force to act on. If you speak of the field E source, it is far away, an antenna, for example.

I understand, that the AB effect relies directly upon the Potentials, which "underlie" the so-called E&B fields. It is the Potentials (V, A) which appear in the Hamiltonian, of the SWE. Quantum 'particles' respond directly to the Potentials. (The "fields" derived therefrom are only useful in the Classical limit, of Ehrenfest Theorem expectation values, d<x>/dt = -<p>/m, etc.)
I'm scratching my head about them. I don't actually understand what the potentials are. The wiki page on four-potential defines it as:

$A^{\alpha} = \left(\frac{\phi}{c}, \mathbf A \right) \qquad \left(A^{\alpha} = ( \phi, \mathbf A)\right)$

..in which "φ is the electrical potential, and A is the magnetic potential, a vector potential." I'm confused about the A mentioned in the AB effect. Looking at the photon in terms of displacement current says the displacement is at a maximum midway along the sinusoidal wavelength where the electromagnetic field variation is zero. That's like the AB effect. This midway point is where the potential is at a maximum, which means the sinusoidal waveform is telling you the slope of the potential. Ever read Ehrenberg and Siday's The Refractive Index in Electron Optics and the Principles of Dynamics? They wrote it in 1949, it's all classical, and figure 3 shows what's now known as the Aharonov-Bohm effect. I found this paper interesting too: http://arxiv.org/abs/quant-ph/0604169. Not that I understand much of it.

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Because I'm talking about an electromagnetic wave here. See this alternative picture. Forget about the magnetic field variation because it's an electromagnetic field variation, and consider one wavelength starting from the origin rather than the peak:

The question still stands.

Where in Maxwell's equations is it required that there be a charge co-located with the field? IOW, a field, even a varying one, present in a region doesn't mean the charge is there. It could be somewhere else. Bob already mentioned the example of which I was thinking — an antenna.

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It's an electromagnetic wave Swanson. There is no charged particle present. But the field varies, so there is a current. It isn't conduction current. It's a displacement current.

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• 1 month later...

Such a field exists indeed next to a vibrating dipole (along its axis) but this variable field does not propagate too far - it decays rapidly with the distance and is called a "near field". Anyway it's an axial (radial) electric field E = -grad(V).

The field of an electric dipole exists throughout all space. When you vibrate a dipole, a very strange effect occurs. You can prove this to yourself by considering the zero-field plane of the dipole, which must change shape with vibration.

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