Do lightwaves exist? and what exactly is light?

[the4thamigo_uk]

1) We draw the classic light wavetrain as a sine(x) function, but is it possible (even classically) for a 'single' wavetrain of light to exist. By 'single' I mean a varying electric/magnetic field occurring only along a one dimensional line in space. Or is there a fundamental reason why light must always exist as a wavefront over an extended region of space?

2) If it could exist, what would the electric field be on points just to either side of the line? zero? I normally think of electric fields being 'caused' by a charged particle. In this case the electric field reduces as you move away from the particle, in other words the particle is the source of a field that extends some distance away. For a light wave though, it would seem that the fields exist only 'along' the one dimensional line itself? Is this true?

3) In fact, for a single wavetrain (or photon of light), what is 'causing' the electric magnetic fields to vary exactly? Is it simply an unexplained physical phenomena on which we base other physics or is the existence of the varying electric/magnetic fields explainable in more fundamental terms. I imagine a magnetic field being produced by an accelerating charge. When we think of a 'single' light wave, do we imagine a more fundamental charged particle wiggling up and down somehow to produce the magnetic field?Does this have any relationship to quantum spin or not?

 

[Dale]

1) We draw the classic light wavetrain as a sine(x) function, but is it possible (even classically) for a 'single' wavetrain of light to exist. By 'single' I mean a varying electric/magnetic field occurring only along a one dimensional line in space. Or is there a fundamental reason why light must always exist as a wavefront over an extended region of space?

If you read the caption for those drawings usually they are referring to an infinite plane wave. The drawing is shown that way because of the limitations of drawing a 4D tensor field on a 2D piece of paper. Any arbitrary EM wave can be decomposed into a superposition of plane waves. So, you can think of a plane wave as being a "single wavetrain" although there is no need to do so and no particular significance to it.

I do not believe that it is possible for a field to remain along a single one-dimensional line and still satisfy Maxwell's equations, but I could be wrong.

Is it simply an unexplained physical phenomena on which we base other physics or is the existence of the varying electric/magnetic fields explainable in more fundamental terms.

My understanding is that the existence of the EM fields is explainable in terms of an underlying symmetry. Per Noether's theorem each symmetry corresponds to some conserved quantity so a symmetry and a conservation law are two sides of the same coin. In the case of EM, the symmetry is called "gauge invariance" meaning roughly that you can change the potential without changing the field, and the conservation law is the conservation of charge.

 

[GRDixon]

1) We draw the classic light wavetrain as a sine(x) function, but is it possible (even classically) for a 'single' wavetrain of light to exist. By 'single' I mean a varying electric/magnetic field occurring only along a one dimensional line in space. Or is there a fundamental reason why light must always exist as a wavefront over an extended region of space?

The electric and magnetic field vectors cannot vary discontinuously in charge-free space. Thus your "single ray" wave is in violation of Maxwell's equations. From a slightly different perspective, the electric field of a charge with arbitrary motion can be computed, and there is no source charge motion that can create your "surgical" one-dimensional wave.

 

[htg]

Any arbitrary EM wave can be decomposed into a superposition of plane waves.

Really? Can you show me how to decompose a narrow (naturally, somewhat divergent) beam into plane waves? Just give me the idea.

 

[Dale]

Sure. Just write the expression for the scalar and vector potentials and then do a Fourier transform in all four dimensions. Each point of the potentials in the Fourier domain represents a plane wave.

 

[htg]

No. If you do the Fourier transform of the H and E fields as functions of the spatial coordinates (x,y,z) and time (t), then you get a decomposition into functions of the form exp(iwt+i(k1 x + k2 y + k3 z), which is very different from decomposing a beam into plane waves, presumably of the same frequency as the frequency of the EM wave considered.

 

[Dale]

No. If you do the Fourier transform of the H and E fields as functions of the spatial coordinates (x,y,z) and time (t), then you get a decomposition into functions of the form exp(iwt+i(k1 x + k2 y + k3 z)

What is the expression for a plane wave? Hint:
http://en.wikipedia.org/wiki/Plane_wave

 

[htg]

The problem is that this way you get a decomposition into plane waves of all possible wavelengths, not just the wavelength of the wave considered.

 

[Dale]

Yes, this is correct. A localized wave, in time or space, will have an infinite range of frequencies or wavelengths respectively.

 

[htg]

You should not call it a decomposition into plane waves. It suggests that it is a decomposition into principally physically realizable plane EM waves, which is not the case.

 

[Dale]

Since when is being "physically realizable" a requirement for a set of basis functions or even implied?

The fact is that each point in the Fourier domain represents a plane wave and any EM field can be Fourier transformed. Therefore, any arbitrary EM wave can be decomposed into a superposition of plane waves.

 

[Born2bwire]

Since when is being "physically realizable" a requirement for a set of basis functions or even implied?

The fact is that each point in the Fourier domain represents a plane wave and any EM field can be Fourier transformed. Therefore, any arbitrary EM wave can be decomposed into a superposition of plane waves.

One should also note the fact that any plane wave is physically unrealizable as it requires an infinite source.

 

[htg]

As I said, decomposition of an EM wave into plane waves suggests that it is decomposition into EM waves, which is not the case.
You can calculate the Fourier transform of almost anything (using distribution theory), but this is not a decomposition into WAVES.

 

[AJ Bentley]

As I said, decomposition of an EM wave into plane waves suggests that it is decomposition into EM waves, which is not the case.

I don't understand what you mean.
We have a theoretical model of a physical process that we call an EM wave.
That theoretical model allows us to de-construct it (the model) into fundamental parts and also to reassemble it.
The fundamental parts are sometimes called plane waves.

Are you saying that because the physical corollary is never observed in nature, we cannot call the fundamental parts of the model by the same name as the full construct?

 

[htg]

Waves are actual physical phenomena - there are E and H fields, and EM waves propagate with the velocity c in vacuum. There is a clear relation between frequency and wavelength.
This relation is violated by elements of the Fourier decomposition, so they are not EM waves.

 

[maxverywell]

This relation is violated by elements of the Fourier decomposition, so they are not EM waves.

No it's not violated.

 

[htg]

Yes it is, the Fourier transform of a wave of given frequency, which is localized in space (e.g. a beam) contains elements corresponding to all wavelengths, not just the wavelength of the physical EM wave.

 

[AJ Bentley]

This relation is violated by elements of the Fourier decomposition, so they are not EM waves.

That's just semantics. The distinction is too subtle to be worth losing sleep over. If you shave the hair off a cat, it's still a cat. Why should we go to the trouble of inventing new name for 'whatever-it-is'?
I'm perfectly happy with calling them waves. I've never met anyone else who has a problem with it.

 

[Born2bwire]

Yes it is, the Fourier transform of a wave of given frequency, which is localized in space (e.g. a beam) contains elements corresponding to all wavelengths, not just the wavelength of the physical EM wave.

That's not true. Look at the Weyl Identity which decomposes a point source into a superposition of plane waves. The Fourier transform of the spherical potential here is an integral over all directions of plane waves, not frequencies, as there are inherent restrictions on the wavenumbers over the coordinates.

Even for the situations where we do have decompositions over a range of frequencies, each dk_a element of the integrand represents and independent plane wave. The integral is taking a superposition of these plane waves and the physics are perfectly valid. This is a vital part of electrodynamic theory because we rely on the change of basis to solve a myriad of problems.

 

[htg]

Look at the Fourier transform of a propagating beam of EM wave of given wavelength.
The Fourier transform contains all wavelengths, so the Fourier components are not EM waves.

 

[AJ Bentley]

Fourier components are not EM waves.

What do you want to call them?

The components have the same mathematical form as the original wave. They are solutions of the same wave equation.

Looks like a duck, walks like a duck, smells like a duck...

 

[Born2bwire]

Look at the Fourier transform of a propagating beam of EM wave of given wavelength.
The Fourier transform contains all wavelengths, so the Fourier components are not EM waves.

That cannot be strictly true, unless you are talking about something like a Gaussian beam that requires a bandwidth. The Weyl Identity allows you to decompose a point source into plane waves. You can thus decompose any source into a superposition of these point sources and thus any monochromatic wave can be reproduced as a superposition of monochromatic plane waves.

For the Weyl Identity,

$$\frac{e^{ik_0r}}{r} = \frac{i}{2\pi} \int\int^{\infty}_{-\infty} dk_x dk_y \frac{e^{ik_xx + ik_yy + ik_z|z|}}{k_z}$$

Both sides are valid electromagnetic waves. The left hand side is a spherical wave while the right hand side is a superposition of plane waves with direction k_x\hat{x}+k_y\hat{y}+k_z\hat{z} and total wavenumber of k_0. The ability to decompose the electromagnetic waves is an underlying principle when dealing with many problems. For example, we can solve for the scattering and propagation of waves in a planarly layered medium by analyzing the point source response. We expand the point source response, the aforementioned spherical wave, using the Sommerfeld Identity which decomposes the spherical wave into a superposition of conical wavefronts (cylindrical waves multiplied by plane waves). This allows us to use the boundary conditions for electromagnetic waves and apply them on the appropriate part of the decomposition (in this case, we allow the plane waves to propagate normally to the inhomogeneity and the cylindrical waves parallel. That allows us to apply the reflections to the plane waves and leave the cylindrical waves alone.). We use the same physics that we apply to any electromagnetic waves. This allows us to find the Green's function for planarly, cylindrical and spherically layered medium. It also allows us to also trivialize the solving of scattering from say a PEC sphere (Mie series which relies on decomposing a plane wave into a superposition of spherical waves via spherical harmonics and addition theorem).

I think you are misunderstanding two points. The first is that the Fourier transform is not going from a time domain to frequency domain. We are already in the frequency domain by considering time-harmonic problems. The transform is trivial since we are already working with discrete frequency problems (ignoring signals that are spread over a bandwidth). Instead, we use the Fourier transform to decompose it over the k-space and apply the restriction that magnitude of k must be related back to the monochromatic frequency of our problem. When we decompose using the Weyl Identity, Sommerfeld Identity, Addition Theorem, etc., we are not going over frequency space but we are making a change of basis. How this basis is changed can be done using a Fourier transform in k-space (Weyl or Sommerfeld) or it can be done using an explicit expansion of the Green's function (Addition Theorem/Mie Series).

The second point is that these are still valid electromagnetic waves. They still solve the vector wave equations and satisfy Maxwell's Equations. I'm not sure how else I can convince you of this fact but perhaps if you understand the previous point you would be more acceptable of this fact.

 

[AJ Bentley]

I think the duck argument has more going for it.

 

[Born2bwire]

I think the duck argument has more going for it.

Look, I didn't want to have to say anything. BUT, we were hungry, the duck was there, we voted, we ate the duck. We slow roasted in a brick oven and served it Peking style. It was delicious. So until we can get ourselves a replacement duck I think we're stuck with math. If it's any consolation, we have a lamb. I don't know how much longer we'll have him though; we have a large surplus of apple mint jelly.

 

[htg]

Consider a point source that starts to radiate at the time t=0. Clearly you cannot decompose this wave into plane waves, because plane waves have the full amplitude at t=0 as far from the point source as we wish.

 

[Born2bwire]

Consider a point source that starts to radiate at the time t=0. Clearly you cannot decompose this wave into plane waves, because plane waves have the full amplitude at t=0 as far from the point source as we wish.

Sure we can. First we decompose the time-dependent signal into it's frequency Fourier components. This will give us a superposition of point sources over varying frequencies. Then we expand each of these point sources into a superposition of plane waves. The Fourier transform of the time signal just gives us the scaling of the set of point source plane wave decompositions over varying frequencies.

This is no different then decomposing any time dependent signal into its frequency components.

 

[htg]

The components of such a decomposition are NOT WAVES.
Just like Fourier components of a 1-dimensional signal are not actual signals - if they were, causality would be violated.

 

[maxverywell]

The components of such a decomposition are NOT WAVES.

I think that they are monochromatic standing EM WAVES at time t=0. Then you multiplie every component by cos(w_n*t) to take time evolution of the wave.
(We just take Fourier series of the EM wave at t=0, in 1-D)

 

[htg]

At time t=0 there isn't anything to transform or decompose. I was thinking of the spherical wave at some time t1. Fourier decomposition leads to objects that violate causality and frequency-wavelength relation.

 

[maxverywell]

Fourier decomposition leads to objects that violate causality and frequency-wavelength relation.

Why? I don't think so...

 

[Dale]

Consider a point source that starts to radiate at the time t=0. Clearly you cannot decompose this wave into plane waves, because plane waves have the full amplitude at t=0 as far from the point source as we wish.

This does not follow. The superposition of the waves can sum to 0 even though the individual basis functions are generally non-zero.

 

[htg]

This does not follow. The superposition of the waves can sum to 0 even though the individual basis functions are generally non-zero.

We are talking about an L^2 (R^n) space here, and either about an orthonormal basis or about Fourier transform.
When you take a linear combination of basis elements, the L^2 norm of this combination is equal to SQRT(sum of squares of moduli of coefficients), so if at least one coefficient is non-zero, you get a non-zero linear combination.
Probably what you meant is that a non-trivial linear combination of basis functions can be zero on some proper subset of R^n.
And you still confuse waves and basis functions - IT IS NOT THE SAME.

 

[htg]

Why? I don't think so...

Consider the Fourier transform of restriction of sin(w*t) to an interval [a,b].
Ther Fourier transform decomposes this time-limited signal into a continuum of complex exponential functions defined on the whole real line. So if we want to say that they are signals, we must be able to detect them before our time-restricted signal began to be non-zero. Obviously it would violate causality.

 

[Dale]

Probably what you meant is that a non-trivial linear combination of basis functions can be zero on some proper subset of R^n.

Yes, I understood that to be your objection. That the basis functions were non-zero in a region far away from the origin at t=0 despite the fact that the function was 0 in that region.

 

[htg]

Fourier transform (as distinguished from Fourier series) does not decompose into basis functions. L^2 (R^n) is a countably dimensional space and has a countable basis.