Source of Magnetic component of light

[Danyon]

Does the magnetic component of light emitted by an antenna arise from the effects of length contractions or from the fundamental magnetic moment of the electron?

 

[Dale]

I don't think that the magnetic moment of the electron is in any way relevant to radio antennas.

 

[phyzguy]

Neither. The source of the magnetic field in EM radiation is the changing electric field. This is how EM radiation works - the changing electric field gives rise to a changing magnetic field which gives rise to a changing electric field which gives rise to ...

 

[Drakkith]

Neither. The source of the magnetic field in EM radiation is the changing electric field. This is how EM radiation works - the changing electric field gives rise to a changing magnetic field which gives rise to a changing electric field which gives rise to ...

Wouldn't that explanation require that the magnetic and electric field vectors be out of phase with one another?

 

[PeterDonis]

Wouldn't that explanation require that the magnetic and electric field vectors be out of phase with one another?

Yes. They are.

 

[Dale]

Yes. They are.

Not in a plane wave in vacuum.

 

[phyzguy]

I think Dale Spam is correct. In Plane-polarized light the E and B fields are in phase. However, this doesn't change the fact that it is the changing electric field that is the source of the magnetic field and the changing magnetic field that is the source of the electric field. We can see this by looking at Maxwell's equations in vacuum.
\rm \nabla x E = -\frac{\partial B}{\partial t}
\rm \nabla x B = \frac{1}{c^2} \frac{\partial E}{\partial t}
Drakkith's intuition (and mine!) says they should be out of phase, but when you work through the math, you find they are not. For example, the following vectors are solutions:
\rm E = E_x cos(kz - \omega t)
\rm B = B_y cos(kz - \omega t)

 

[andresB]

Jefimenko equations might be relevant for that

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

Jefimenko says, "...neither Maxwell's equations nor their solutions indicate an existence of causal links between electric and magnetic fields. Therefore, we must conclude that an electromagnetic field is a dual entity always having an electric and a magnetic component simultaneously created by their common sources: time-variable electric charges and currents."

 

[PeterDonis]

Not in a plane wave in vacuum.

the following vectors are solutions:

$$
E = E_x \cos (kz - \omega t)
$$
$$
B = B_y \cos (kz - \omega t)
$$​

Apparently we are using "in phase" to mean different things. To me, "in phase" means each of the components are in phase. The above solution has $E_x$ in phase with $B_y$, not $B_x$. For this solution, which is a linearly polarized wave, $B_x = E_y = E_z = B_z = 0$, so I suppose one could use "in phase" to refer to the only nonzero components. But for a wave with any other polarization, this won't work. For example, try a circularly polarized wave: the $E$ and $B$ vectors are both rotating, but at any instant of time they point at directions 90 degrees apart, i.e., they are 90 degrees out of phase; if you compare individual components of $E$ and $B$ (such as $E_x$ and $B_x$), you see that one will be a cosine and one will be a sine, both with the same argument.

And note that the "90 degrees apart" condition is general; it just says $E \cdot B = 0$, which is always true for an EM wave in vacuum (note that the linearly polarized solution given above satisfies this condition). So the interpretation of "out of phase" I am using can be applied to any EM wave; whereas the interpretation according to which the E and B fields of a linearly polarized wave are "in phase" only works for that particular kind of polarization.

 

[pervect]

I've usually heard the magnetic field of light being described as coming from the displacement current.

http://en.wikipedia.org/w/index.php?title=Displacement_current&oldid=651189963

In electromagnetism, displacement current is a quantity appearing in Maxwell's equations that is defined in terms of the rate of change of electric displacement field. Displacement current has the units of electric current density, and it has an associated magnetic field just as actual currents do. However it is not an electric current of moving charges, but a time-varying electric field.

 

[Danyon]

When the electrons accelerate up to the top of the transmission antenna the electric field emits one wavelength, all the while the magnetic field produced by the current rotates clockwise around the wire, Which is orthogonal to the electric field change. When the electrons accelerate downward the magnetic field rotates anti-clockwise and is orthogonal to the next wavelength of electric field change, the magnetic fields produced by the wire travel outward with the electric fields forming electromagnetic waves

 

[PeterDonis]

I've usually heard the magnetic field of light being described as coming from the displacement current.

This is true in the sense that it is the displacement current term in Maxwell's equations that makes electromagnetic waves a possible solution with zero source.

 

[phyzguy]

I've usually heard the magnetic field of light being described as coming from the displacement current.

Yes. The displacement current is the \frac{1}{c^2}\frac{\partial E}{\partial t} term on the right side of the \rm \nabla x B equation. So when I say "The source of the magnetic field is the changing electric field," and when you say, "The source of the magnetic field is the displacement current," we are saying the same thing.

 

[phyzguy]

Apparently we are using "in phase" to mean different things.

Yes, apparently we did mean different things. I interpreted "out of phase" as "out of phase in time", meaning that the peak magnitude of the magnetic field occurs when the magnitude of the electric field magnitude is crossing zero, and the peak magnitude of the electric field occurs when the magnetic field magnitude is crossing zero. In fact, for a plane wave in a vacuum, both the electric field magnitude and the magnetic field magnitude peak at the same time, so they are "in phase" in that sense. So I don't think we disagree at all.

 

[pervect]

This is true in the sense that it is the displacement current term in Maxwell's equations that makes electromagnetic waves a possible solution with zero source.

Yes, that's the sense I meant it.

It doesn't really answer the question "why do antenna's in particular radiate", it just answer the question of why radiation exists. A quick answer to the later question is that in special relativity, accelerated charges radiate, and the electrons in the wire of antenna are accelerating, hence antenna's radiate. This simple answer may not be complete, but I don't want to get too advanced, I just want to point out / emphasize jthat the process has nothing to do with the magnetic moment of the electron.

The actual field of the antenna can be derived from maxwell's equations, and consists of a radiative part (the far field), and a non-radiative part (the near field).

T

 

[PeterDonis]

the process has nothing to do with the magnetic moment of the electron.

Yes, agreed.

 

[Dale]

Apparently we are using "in phase" to mean different things.

Yes, it appears so. I refer to the phase shift in time. What you are calling "out of phase" I would call "perpendicular", but your point is good and I can see your usage is reasonable although it is different from what I am used to.

 

[Dale]

When the electrons accelerate up to the top of the transmission antenna the electric field emits one wavelength, all the while the magnetic field produced by the current rotates clockwise around the wire, Which is orthogonal to the electric field change. When the electrons accelerate downward the magnetic field rotates anti-clockwise and is orthogonal to the next wavelength of electric field change, the magnetic fields produced by the wire travel outward with the electric fields forming electromagnetic waves

Yes, although at this level I prefer to not even speak of electrons, but just current and charge.