The plane wave decomposition is mathematically universal?

[keji8341]

The plane wave decomposition is mathematically universal?

1. My questions is: Can "expi[wt-(w/c)*r]/r with r=sqrt(x**2+y**2+z**2), the frequency w is given, monochromatic wave" be represented as a sum of uniform plane waves? Note: r=0 is a singularity.

This is actually the potential function produced by an ideal radiation electric dipole. [J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), Chapter 9, p. 410, Eq. (9.16).]

(i) A spherical-wave decomposition of a plane wave is presented in textbook; for example, J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), p. 471, Eq. (10.44) in Chapter 10.

(ii) The converse: A plane-wave decomposition of spherical waves is given by MacPhie and Ke-Li Wu, “A Plane Wave Expansion of Spherical Wave Functions for Modal Analysis of Guided Wave Structures and Scatterers”, IEEE Trans. Antennas and Propagation 51, 2801 (2003). Note: The spherical waves are analytical at r=0.

2. Where can I find the conclusions:
(a) The plane wave decomposition is mathematically universal.
(b) Any spherical wave may be decomposed in a plane wave basis.
Are they math theorems?

Thanks a lot.

 

[BruceW]

I doubt that any mathematical function can be expressed as a linear sum of plane waves - there are a lot of weird functions out there.

If we're talking about electromagnetic waves only, then according to wikipedia: "In linear uniform media, a wave solution can be expressed as a superposition of plane waves. This approach is known as the Angular spectrum method. The form of the planewave solution is actually a general consequence of translational symmetry."

So I am guessing that if the media is not uniform, then you can't always express the electromagnetic field as a sum of plane waves.

 

[kith]

The decomposition of a function in plane waves is just the Fourier transform. So every function for which the Fourier transform exists, is decomposable in plane waves.

 

[chrisbaird]

Yes, I believe any physical wave (in other words, finite and continuous) can be represented by a sum of plane waves (including spherical waves). The question really comes down to a mathematical one about Fourier series representations of functions: for what type of functions does its Fourier series representation converge? I believe the answer is: functions that are continuous and finite, which includes all physical electromagnetic waves.

 

[BruceW]

I think it depends on how we are defining plane waves. In EM, plane waves usually mean something like: $e^{if(r,t)}$ where f is a real number.
But the Fourier transforms give a sum of plane waves where f can be a complex number.

 

[keji8341]

I think it depends on how we are defining plane waves. In EM, plane waves usually mean something like: $e^{if(r,t)}$ where f is a real number.
But the Fourier transforms give a sum of plane waves where f can be a complex number.

Can "expi[wt-(w/c)*r]/r with r=sqrt(x**2+y**2+z**2), the frequency w is given, monochromatic wave" be represented as a sum of uniform plane waves?

Physically speaking, the answer is “no”.
Mathematically speaking, the answer is “yes”.
Why? Here are my explanations.

There are two kinds of 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.

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.

Therefore, "expi[wt-(w/c)*r]/r with r=sqrt(x**2+y**2+z**2), the frequency w is given, monochromatic wave" CANNOT represented as a sum of REAL uniform plane waves, because its Fourier integration must be carried out in a complex plane by designation a contour for poles, and its Fourier transform has no physical meaning, just a kind of math correspondence.

 

[vanhees71]

The plane wave can be expanded in terms of spherical waves and vice versa. The expansion is convergent in the sense of generalized functions (distributions).

 

[keji8341]

I think it depends on how we are defining plane waves. ...But the Fourier transforms give a sum of plane waves where f can be a complex number.

1. In the plane-wave factor expi(wt-k.x), if w and k is complex, then the plane wave decays or grows exponentally with time and position. In free space, such a plane wave is not consistent with energy conservation law in the sense of classical electrodynamics.

2. In the sense of quantum mechanics, when w is complex, the Planck constant should be proportional to the conjugate complex of w, if E=hbar*w is real..., ha, a lot of new results...

Of course, for many theories, it is not required for every intermediate math operation to have physical meaning, especially in quantum mechanics. Assigning a physical explanation is just for being easy to remenber sometimes.

 

[Dale]

You are confusing the Laplace transform and the Fourier transform. Also, it would be helpful if you would stop double posting. Decide if you want to carry on this conversation here or in the relativity forum.

 

[keji8341]

You are confusing the Laplace transform and the Fourier transform. Also, it would be helpful if you would stop double posting. Decide if you want to carry on this conversation here or in the relativity forum.

No, I am not confusing, I am talking about Fourier transform, never talking about Laplace transform before. Probably you misunderstood something.

Because this is an electromagnetic problem, maybe this topic is more suitable here according to the forum classifications. OK, I'll post here.

Here is an example to carry out Fourier integartion (inverse Fourier transform) in the complex plane by designating a contour for poles: http://www.bnl.gov/atf/exp/Dielectric/2-Theory_of_Wakes.pdf
on p. 1270, Fig. 1, corresponding to the integral in Eq. (2.22).

To help reading, let me give some explanations.
The above paper [ "Theory of wakefields in a dielectric-lined waveguide" Phys. Rev. E 62, 1266–1283 (2000)]
uses Fourier-transform approach to solve wave equation excited by a moving electron in a dielectric-loaded cylindrical waveguide. The radiation field is Cerenkov radition field, usually called wake-field because the radiation field is always after the electron.

 

[BruceW]

For a Fourier transform, we have:
$$\hat{f}(\xi) = \int_{-\infty}^\infty \ f(x) e^{-2 \pi ix \xi} \ dx$$
Where x and $\xi$ are real numbers. And f(x) is generally a complex function of x.
You could then say that the plane waves are $e^{-2 \pi ix \xi}$ and the f(x) is just a complex coefficient to go with the plane waves. In this case, the Fourier transform is trivially a sum over plane waves, since the plane waves are multiplied by some other (generally non-planar) function.

 

[keji8341]

For a Fourier transform, we have:
$$\hat{f}(\xi) = \int_{-\infty}^\infty \ f(x) e^{-2 \pi ix \xi} \ dx$$
Where x and $\xi$ are real numbers. And f(x) is generally a complex function of x.
You could then say that the plane waves are $e^{-2 \pi ix \xi}$ and the f(x) is just a complex coefficient to go with the plane waves. In this case, the Fourier transform is trivially a sum over plane waves, since the plane waves are multiplied by some other (generally non-planar) function.

I am trying to understand your question.
For example, f(x)=exp(-|x|), which is not a wave.
But it can be expressed as f(x)=exp(-|x|) = Inte{F(k)*exp(-ik*x)*dk} by FT method,
where F(k)*exp(-ik*x)*dk is a "component plane wave", the amplitude is F(k)*dk, and the phase factor is exp(-ik*x)=cos(kx)-i*sin(kx). What do mean for "trivially"?

 

[Dale]

Here is an example to carry out Fourier integartion (inverse Fourier transform) in the complex plane by designating a contour for poles: http://www.bnl.gov/atf/exp/Dielectric/2-Theory_of_Wakes.pdf
on p. 1270, Fig. 1, corresponding to the integral in Eq. (2.22).

That is a Laplace transform. And he is working in cylindrical coordinates, in a dielectric medium, an with non-uniform material properties. Due to the interaction with matter in this problem decaying and growing modes are indeed physical, hence the relevance of the Laplace transform.

I don't really see the relevance to your problem where we have uniform free-space propagation of spherical waves.

 

[BruceW]

Keji - I said trivially because F(k) is any general complex function of k, so to call it the amplitude of a plane wave trivially means that the Fourier transform is an integral of plane waves.

I usually think of the amplitude of a plane wave as some constant. But in this case it is used to mean a general function.

I don't have any objection, I would simply prefer to say that the Fourier transform gives a function of x as an integral of plane waves multiplied by a general function of k.

 

[keji8341]

That is a Laplace transform. And he is working in cylindrical coordinates, in a dielectric medium, an with non-uniform material properties. Due to the interaction with matter in this problem decaying and growing modes are indeed physical, hence the relevance of the Laplace transform.

I don't really see the relevance to your problem where we have uniform free-space propagation of spherical waves.

I just want to answer your question: Fourier integrals have to be carried out in the complex plane for some cases. That is Fourier integral in that paper, the integration limits are from -infiniy to +infinity, while for Laplace transform, from zero to +infinity.

Another example: Do you remember that the Lorentz invariant Green function in the relativistic electrodynamics is obtained by Fourier-transform approach? The Fourier integration is carried out in the complex plane by designating a contour for poles. Please check with the well-known textbook by J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), Chapter 12, p. 612, Eq. (12.129).

 

[Claude Bile]

Plane waves cannot be used to describe some functions if the wave velocity is finite.

If we try to decompose certain functions (such as infinitesimal points), we get some imaginary wave-vectors which represent evanescent waves, not plane waves.

Plane waves can only be used to completely describe functions that do not violate the diffraction limit.

Claude.

 

[Dale]

I just want to answer your question: Fourier integrals have to be carried out in the complex plane for some cases. That is Fourier integral in that paper, the integration limits are from -infiniy to +infinity, while for Laplace transform, from zero to +infinity.

Another example: Do you remember that the Lorentz invariant Green function in the relativistic electrodynamics is obtained by Fourier-transform approach? The Fourier integration is carried out in the complex plane by designating a contour for poles. Please check with the well-known textbook by J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), Chapter 12, p. 612, Eq. (12.129).

The Fourier transform transforms a complex-valued function of a real number (time or space) into another complex-valued function of a real number (frequency or wavenumber). That is the definition of the Fourier transform.

If you are dealing with a complex-valued function of a complex number then it is simply not a Fourier transform. It is a Laplace transform. The Laplace transform may certainly be useful also, but that does not make it a Fourier transform.

http://en.wikipedia.org/wiki/Fourier_transform
http://www.math.ucla.edu/~tao/preprints/fourier.pdf

In addition, I thought that your interest was in the decomposition of a spherical wave into an infinite sum of plane waves. If so, then you are not interested in other types of transforms, but only the standard Fourier decomposition with a real domain. I don't know why you are bringing in unrelated topics that you are not even interested in.

 

[keji8341]

Plane waves cannot be used to describe some functions if the wave velocity is finite.

If we try to decompose certain functions (such as infinitesimal points), we get some imaginary wave-vectors which represent evanescent waves, not plane waves.

Plane waves can only be used to completely describe functions that do not violate the diffraction limit.

Claude.

Interesting! Could you please show some references? I mean textbooks or journal papers. Thanks in advance.

 

[keji8341]

The plane wave can be expanded in terms of spherical waves and vice versa. The expansion is convergent in the sense of generalized functions (distributions).

Questions for you:

1. "The plane wave can be expanded in terms of spherical waves"
Do you mean "A spherical-wave decomposition of a plane wave presented in textbook by J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), p. 471, Eq. (10.44) in Chapter 10" ? That is spherical harmonics expansion, r=0 is analytic.

If not, please give your references.

2. "and vice versa" means Fourier integral or what? Please give your references.

Thanks a lot.

 

[vanhees71]

Sure, you find all this in any textbook on quantum theory or in the more advanced ones on classical em.

 

[keji8341]

Sure, you find all this in any textbook on quantum theory or in the more advanced ones on classical em.

Be specific for the two cases you mentioned above, please. Which book, pages., Eqs...
and the plane-wave decomposition of spherical wave is Fourier integral or Laplace integral? Thanks a lot.

 

[keji8341]

...
In addition, I thought that your interest was in the decomposition of a spherical wave into an infinite sum of plane waves. If so, then you are not interested in other types of transforms, but only the standard Fourier decomposition with a real domain. I don't know why you are bringing in unrelated topics that you are not even interested in.

Because I need the critical comments on my views from outstanding scientists like you. Sometimes your criticism might make me not so comfortable, but I can really learn something new from here.

 

[vanhees71]

Plane-wave decomposition in spherical waves: [corrected after hint to typos in #24 thanks!]

$$\exp(\mathrm{i} \vec{k} \cdot \vec{r}) = \sum_{l=0}^{\infty} \mathrm{i}^l (2l+1) \mathrm{j}_l(k r) \mathrm{P}_l(\cos \vartheta).$$

Here, $\mathrm{j}_l$ are the spherical Bessel functions, $\mathrm{P}_l$ are the Legendre polynomials, and $\vartheta$ is the angle between $\vec{k}$ and $\vec{r}$.

From this, one obtains the spherical Bessel functions by generalized Fourier transformation with respect to $u=\cos \vartheta$

$$\int_{-1}^1 \mathrm{d} u \exp(\mathrm{i} \rho u) \mathrm{P}_l(u)=2 \mathrm{i}^l \mathrm{j}_l(\rho).$$

I've collected this from many textbooks when I wrote a manuscript for mathematical methods for theoretical physics:

http://fias.uni-frankfurt.de/~hees/publ/maphy.pdf

This, however, is in German.

 

[keji8341]

Plane-wave decomposition in spherical waves:

$$\exp(\mathrm{i} \vec{k} \cdot \vec{r}) = 4 \pi \sum_{l=0}^{\infty} \mathrm{i}^l (2l+1) \mathrm{j}_l(k r) \mathrm{P}_l(\cos \vartheta).$$

Here, $\mathrm{j}_l$ are the spherical Bessel functions, $\mathrm{P}_l$ are the Legendre polynomials, and $\vartheta$ is the angle between $\vec{k}$ and $\vec{r}$.

From this, one obtains the spherical Bessel functions by generalized Fourier transformation with respect to $u=\cos \vartheta$

$$\int_{-1}^1 \mathrm{d} u \exp(\mathrm{i} \rho u)=2 \mathrm{i}^l \mathrm{j}_l(\rho).$$

I've collected this from many textbooks when I wrote a manuscript for mathematical methods for theoretical physics:

http://fias.uni-frankfurt.de/~hees/publ/maphy.pdf

This, however, is in German.

Thanks a lot.
Except for 4*pi, that is the same as in the book by J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), p. 471, Eq. (10.44) in Chapter 10.

I think you had two typo, no 4*pi, according to Eq. (3.4.43) in your book, and pl(u) lost in the expression for spherical Bessel functions.
----------------
One more question for you.

The plane-wave decomposition of the poit-source spherical wave is a Fourier integral (namely Green's function) usually presented in quantum mechanics for approximately solving Schrodinger wave equation.

Here is my understanding of this decompostion. This Fourier integral converges in the sense of taking a limit for a designated contour in the k-complex plane. Thus the component plane wave has a complex wave vector k, and this is not a physical plane wave, because a plane wave with a complex wave number in free space is not consistent with energy conservation law in the sense of classical electrodynamics, and it is also not consistent with photon momentum hypothesis in the sense of quantum mechanics. Therefore, the plane-wave decomposition of a point-source-generated spherical wave is not a physical-plane-wave decomposition, just a kind of mathematical correspondence (treatment).

What do you think of my argument? I hope you can refute my argument.

 

[Dale]

a plane wave with a complex wave number in free space is not consistent with energy conservation law in the sense of classical electrodynamics

It is consistent if there is a matching wave which is providing/taking the missing/extra energy. Similarly, some plane waves will require currents in free space, which is ok as long as there is another wave with the opposite current. The individual waves need not be free-space solutions to Maxwell's equations as long as the sum is.

Again, this is another reason why the answer to this question is not relevant to the Doppler shift.

I still disagree that a transform involving complex wave-numbers fits the definition of a Fourier transform.

 

[vanhees71]

In the Fourier transform the wave numbers are real. Convergence is always meant in the sense of some function space. In physics, most clearly in quantum theory, the most simple and still mathematically correct structure is the convergence in the sense of the Hilber space of square (Lebesgue-)integrable functions.

The sometimes sloppy physicists' approach to the theory of unbound operators in Hilbert space can be justified by the use of a "rigged Hilbert space", where one defines a dense subspace where the usual differential operators used in physics (like gradient, Laplacian, angular-momentum operator etc.) and their powers of any order are well defined and don't lead out of this subspace. Then one considers the operators on this "test-function space" and it's dual and/or anti-dual. These linear and antilinear functionals then lead to the definition of distributions, and all "generalized eigenstates" of self-adjoint operators live as distributions in these spaces. A very nicely written overview of this with a simple but illuminating example (square-well potential and 1D-scattering states of this potential) for the rigged-Hilbert-space formalism can be found in

Rafael de la Madrid, Eur. J. Phys. 26 (2005) 287–312
http://stacks.iop.org/EJP/26/287
http://de.arxiv.org/abs/quant-ph/0502053

 

[keji8341]

It is consistent if there is a matching wave which is providing/taking the missing/extra energy. Similarly, some plane waves will require currents in free space, which is ok as long as there is another wave with the opposite current. The individual waves need not be free-space solutions to Maxwell's equations as long as the sum is.

Your argument sounds having something reasonable, but on second thoughts, the validation of your argument is questionable.

(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.

(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.

 

[keji8341]

Again, this is another reason why the answer to this question is not relevant to the Doppler shift.

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.

 

[keji8341]

I still disagree that a transform involving complex wave-numbers fits the definition of a Fourier transform.

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.

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.

This kind of Fourier transform is based on generalized-function theory. It is widely used in physics literature. For example, Lorentz invariant Green function in the relativistic electrodynamics is obtained by this kind of Fourier-transform; see J. D. Jackson, Classical Electrodynamics, 3rd edition, (John Wiley & Sons, NJ, 1999), Chapter 12, p. 612, Eq. (12.129).

In fact, it doesn't matter what you call it. My original question is: The plane-wave decomposition is mathematically universal? The physical plane-wave decompostion is applicable to the spherical wave generated by a point source, so that the Einstein's Doppler formula can be used for each of the component plane waves?

 

[vanhees71]

A uniform plane wave with a complex wave number in free space cannot exist independently, and it is not physical.

A uniform plane wave with a complex wave number indeed seems to contradict Maxwell's equations. They are equivalent to the wave equation for each component of the em. four-potential in Lorenz gauge, and thus plane-wave solutions with complex wave numbers are not possible. Also such a solution would contradict the boundary condition that the em. field should not diverge at infinity.

Of course, plane waves in the literal sense are not physical either. You can have wave packets of the character of a long wave train, but not infinitely extended plane waves, because they contain an infinite amount of energy.

Another thing are em. waves when matter is present. There you often have damping: energy of the em. wave is deposited in the medium. Examples are dispersion in continuous media, evanescent modes in wave guides, etc.

 

[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".