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K^2
#2
Dec11-12, 03:50 AM
Sci Advisor
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.