The plane wave decomposition is mathematically universal?

[Dale]

(1) A uniform plane wave in free space does NOT produce conducting current. Because of symmetry, EM fields E and B must be perpendicular to the wave vector k, -----> div E = 0 -----> electric charge density = 0 -----> div (J= conducting current) = 0 -----> J = 0.

It does if it has components not moving at c. A general Fourier decomposition is not limited to null wave four-vectors.

(2) A physical wave usually means that it satisfies Maxwell equations and boundary conditions, and consequently, it can exist independently. For example, in the spherical-wave decomposition of a plane wave [J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), p. 471, Eq. (10.44) in Chapter 10], each of the component spherical waves is physical and can exist independently. If an “EM wave” cannot exist independently, then this “EM” wave is not physical. A uniform plane wave with a complex wave number in free space cannot exist independently, and it is not physical.

None of these waves are claimed to exist independently. They are all claimed to exist only as an infinite sum. It is not uncommon that a solution to some specified differential equation and boundary conditions can be expanded in a basis where the basis functions are not solutions to the specified differential equation and boundary conditions. Again, this is exactly why I think this whole line of reasoning is completely irrelevant to the discussion of Doppler shifts.

 

[Dale]

It IS relevant to the Doppler shift. Because some scientists insist that the Einstein’s plane-wave Doppler formula is applicable to any cases, and of course, it is applicable to the spherical wave produced by a moving point source. So the first thing we have to check is whether the component plane wave in the plane-wave decomposition of the point-source generated spherical wave is physical or not.

Again, if you have a problem with someone's claims, it is only reasonable to take it up with the person making the claim.

 

[Dale]

As I mentioned in my opinion, there are two kinds of (inverse) Fourier transforms:

1. “Sum of real plane waves”. 1D-example: f(x)=Inte{F(k)*exp(-ikx)*dk}, where F(k)*exp(-ikx)*dk is real plane wave component, with k is real. The integration is carried out from –infinity<k<+infinity, and the integral is convergent.

This kind of Fourier transform is based on classical analysis. What you referred to is this one.

That is the DEFINITION of the Fourier transform.

2. “Math correspondence”. 1D-example: f(x)=Inte{F(k)*exp(-ikx)*dk}, where F(k)*exp(-ikx)*dk is NOT a real plane wave, because k is set to be complex to make the integral converge. The integration is carried out in a complex plane by designating a contour for poles. Such Fourier transform is usually used to solve differential equations.

That is the DEFINITION of the Laplace transform. And indeed, the purpose of the Laplace transform is to solve differential equations

In fact, it doesn't matter what you call it.

Then call it the Lapalce transform. Proper use of terminology is important for communication.

Suppose that you told me that horses were unacceptable animals for riding because they are too small and as evidence you showed me a dog on a scale weighing 30 kg. I might reasonably reply, "That's a dog, dogs are not horses, and I agree that dogs are not suitable for riding, but horses are". Would you then reply, "It doesn't matter what you call it".