Planck constant is Lorentz invariant?

[keji8341]

Attached is my code, completely open for inspection.I did not do that. I told you exactly the conditions I used:
"Here is a contour plot of the lines of constant phase for t=1, z=0, and v=-.6."

There is no problem. The plot is correct and accurately reflects the familiar behavior of Doppler-shifted spherical wavefronts. This familiar behavior emerges naturally from the formalism of four-vectors and how they transform.

You are calling me a liar? Here is my code, you can check for yourself that it is as I say.

Yes it is.

\left(<br /> \begin{array}{cccc}<br /> \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &amp; -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} &amp; 0 &amp; 0 \\<br /> -\frac{v}{c \sqrt{1-\frac{v^2}{c^2}}} &amp; \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 1<br /> \end{array}<br /> \right)

Compare to http://en.wikipedia.org/wiki/Lorentz_transformation#Matrix_form

It is completely standard.

I don't know what would lead you to believe this. The Lorentz transformations will not decouple two dependent quantities. In the frame where the point source is at rest k' depends on r', so I don't know why you would think that the Lorentz transform would decopule them in the moving frame so that k would be independent of r. Your understanding of the Lorentz transform seems to be incorrect.

I believe your calculations, but I realized that your Lorentz transformation is not a "standard" Lorentz transformation. Let me explain.

The wave 4-vector (k',w'/c) in the source-rest frame and (x',ct') are completely independent originally, belonging to two different spaces. You first redefine the wave 4-vector by setting k'=|k'|(x'/|x'|), this is a kind of "transformation". Then you set (|k'|(x'/|x'|),w'/c) to follow Lorentz transformation. So from the original (k',w'/c) to (k,w/c), something more than Lorentz transformation is imposed.

What difference between "standard transformation" and "not standard transformation"?

1. Standard: (k,w/c) and (x,ct) are completely independent. Since (x,ct) must follows Lorentz transformation, the invariance of phase phi=wt-k.x and the covariance of (k,w/c) are equivalent; that is, a sufficient and necessary condition for the invariance of phase is the covariance of (k,w/c).

2. Not standard: If (k',w'/c) and (x',ct') are not independent, like (|k'|(x'/|x'|), w'/c) which you used, the invariance of phase and the covariance of (k',w'/c) are not equivalent; in other words, the covariance of (k',w'/c) is a sufficient condition for the invariance of phase, but not a necessary one!

 

[Dale]

1. My main question is: your Doppler formula has a sigularity at the overlap-point (you realize that), which is not physical at all.

Classical point particles themselves are also not physical at all, and many other valid formulas suffer the same problem. So I see no issue here.

2. Your Doppler formula is different from Einstein's formula.

It reduces to his in the appropriate situation.

3. From your Doppler formula, the observed frequency in the lab frame changes with locations, as you indicated in your post #73:

"Off of the x-axis the Doppler shift depends on both position and time..."

That means a photon's frequency changes during propagation

No, it doesn't. Different photons go to different locations. No photon changes frequency.

You are really good at making incorrect conclusions from correct premises. Here is another example:

The wave 4-vector (k',w'/c) in the source-rest frame and (x',ct') are completely independent originally, belonging to two different spaces.

Many physics equations express a dependence between vectors in two different spaces. The fact that they are in different spaces has nothing to do with whether or not two vectors are independent. In this case the dependency between x and k is given by the equation \phi=g_{\mu\nu}x^{\mu}k^{\nu}

 

[Dale]

I realized that your Lorentz transformation is not a "standard" Lorentz transformation. Let me explain.

Then kindly point out in my code exactly where my code deviates from a standard Lorentz transform.

I assert that it is standard and as evidence I have posted the formula I used, a link to the standard formula for comparison, and my code. You are either saying I am a liar and deliberately misrepresenting my code or you are saying I am stupid and don't realize I am misrepresenting it.

So either point out the exact line in my code where I deviate from the standard Lorentz transform or stop impugning my integrity.

 

[PeterDonis]

No, it doesn't. Different photons go to different locations. No photon changes frequency.

I had been trying to figure out how to describe that particular error keji8341 was making, but this sums it up perfectly.

 

[keji8341]

Then kindly point out in my code exactly where my code deviates from a standard Lorentz transform.

I assert that it is standard and as evidence I have posted the formula I used, a link to the standard formula for comparison, and my code. You are either saying I am a liar and deliberately misrepresenting my code or you are saying I am stupid and don't realize I am misrepresenting it.

So either point out the exact line in my code where I deviate from the standard Lorentz transform or stop impugning my integrity.

Dear DaleSpam, you misunderstood my comments. You are a very kind and serious scientist; I am very happy that I had a chance to discuss the question I raised. I completely believe your calculations based on your own math model. Although we have different academic viewpoints, I find that you have a very deep understanding in relativity, you have quick reflections to new physical problems, and you are a very strong discussion rival who I have never met in this field. All my comments are just pointing to academic issues, nothing else. If any of my words make you unhappy, that is never my original intention, and I apologize to you.

 

[Dale]

I completely believe your calculations based on your own math model. ... If any of my words make you unhappy, that is never my original intention, and I apologize to you.

Again, it is not my model, it is the standard Lorentz transform, but I appreciate and accept the apology, provided you don't return to accusing me of falsifying the Lorentz transform.

 

[Phrak]

The transformation matrix to obtain (k_x',w') from (k_x,w) should directly derive from the Lorentz matrix transforming (x,t) to (x',t'), where k_x = 1/x and w = 1/t. What are the matrix elements?

This matrix seems to be something that appears to be a "reciprical" matrix 1/L, from 1/x' = 1/[L(x)], in very schematic notation.

This problem of showing that h is lorentz invariant, given E=hv, amounts to showing that 1/L is also a lorentz transform.

 

[Dale]

The transformation matrix to obtain (k_x',w') from (k_x,w) should directly derive from the Lorentz matrix transforming (x,t) to (x',t'), where k_x = 1/x and w = 1/t.

No, k is a four-vector, so the transformation matrix is the Lorentz matrix. That is essentially the definition of a vector, that it transforms according to the Lorentz matrix.

 

[Phrak]

No, k is a four-vector, so the transformation matrix is the Lorentz matrix. That is essentially the definition of a vector, that it transforms according to the Lorentz matrix.

"No, k is a four vector", or "yes, k is a four vector?" There's nothing in error in what I said that I can see.

Have you demonstrated that it transforms as a four vector or given a reference? This is a very long thread.

 

[Dale]

"No, k is a four vector", or "yes, k is a four vector?" There's nothing in error in what I said that I can see.

The error was that you needed to derive a different transformation matrix. You simply use the standard Lorentz transformation matrix.

Have you demonstrated that it transforms as a four vector or given a reference? This is a very long thread.

Yes, I demonstrated it for the case of a spherical source.

 

[Phrak]

The error was that you needed to derive a different transformation matrix. You simply use the standard Lorentz transformation matrix.

Yes, I demonstrated it for the case of a spherical source.

I'll look into these two points. In which post did you demonstrate a spherical source?

This is interesting in its own right, that 1/x^mu should transform as x^mu to necessitate h as a scalar.

 

[Dale]

Post 74, with the code posted in 97. If you have Mathematica then I would start with 97 since it is more complete.

 

[Phrak]

Post 74, with the code posted in 97. If you have Mathematica then I would start with 97 since it is more complete.

I no longer have mathematica. It would help if variables were identified.

 

[Phrak]

by the way, in your definition of k'

k&#039;=\left(1,\frac{x&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}},\frac{y&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}},\frac{z&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}}\right)

you will want to remove the square root operators. Instead,

k_x = x/r²

for spatial dimensions.

 

[Dale]

No, that is not correct. That would mean that the wavelength increases as the wave gets further from the source. Why would you think that?

 

[PeterDonis]

Phrak, I think you may be assuming a different convention for the units of k. In the convention DaleSpam is using, the phase \phi has the same units as the "position vector" r; that means k is dimensionless (see the formulas in post #73). You may be thinking of a different convention where k is a "wavenumber vector" and has dimensions of inverse length, so that the phase is dimensionless. The latter is the convention I learned when I took wave mechanics in school; the reason for adopting it was that if you describe a wave as a complex exponential (or sines and cosines), the argument of the exponential (or the sines and cosines), which is the dot product k . r, has to be dimensionless.

Even in that convention, though, I don't think k = 1/r would be correct, since, as DaleSpam points out, that would mean the wavelength increases with distance from the source. I'm not sure how you would modify the formulas in post #73 for the unit convention where k has dimensions of inverse length.

 

[Phrak]

No, that is not correct. That would mean that the wavelength increases as the wave gets further from the source. Why would you think that?

A wave front at r, centered at the origin, has wave numbers k_xⁱ = xⁱ/r.

 

[DrGreg]

Please bear in mind that DaleSpam stated in post #73 that

...I will use units of time such that in the primed frame w=1 and units of distance such that c=1.

This means you can't use dimensional analysis on any equations derived from this assumption. To avoid confusion you would have to reinsert symbols for \omega and c into the formulae.

 

[Dale]

A wave front at r, centered at the origin, has wave numbers k_xⁱ = xⁱ/r.

Yes, as you suggest here x/r is correct, not x/r² as you suggested above. My formula for k is correct.

 

[Dale]

Phrak, I think you may be assuming a different convention for the units of k. In the convention DaleSpam is using, the phase \phi has the same units as the "position vector" r; that means k is dimensionless (see the formulas in post #73). You may be thinking of a different convention where k is a "wavenumber vector" and has dimensions of inverse length, so that the phase is dimensionless. The latter is the convention I learned when I took wave mechanics in school; the reason for adopting it was that if you describe a wave as a complex exponential (or sines and cosines), the argument of the exponential (or the sines and cosines), which is the dot product k . r, has to be dimensionless.

I am using the standard convention. Phase is dimensionless, x has units of distance, and k therefore has units of inverse distance. As DrGreg points out I am using units of time such that w=1 and units of distance such that c=1. So you need to plug the appropriate factors back in wherever the units don't make sense. I apologize for the confusion that has caused. It is a pretty common thing to do, but it is somewhat sloppy and definitely confusing if you aren't looking out for it.

 

[PeterDonis]

I am using the standard convention. Phase is dimensionless, x has units of distance, and k therefore has units of inverse distance. As DrGreg points out I am using units of time such that w=1 and units of distance such that c=1. So you need to plug the appropriate factors back in wherever the units don't make sense. I apologize for the confusion that has caused. It is a pretty common thing to do, but it is somewhat sloppy and definitely confusing if you aren't looking out for it.

Ah, forgot about that! Thanks for the clarification.

 

[PeterDonis]

Just to expand a bit and make sure I've got this right, if we rewrite DaleSpam's formulas from post #73 in conventional units (i.e., putting back in the factors of \omega and c so that everything is in conventional units of length), we get (in the primed frame):

r&#039; = (ct&#039;, x&#039;, y&#039;, z&#039;)

k&#039; = \left( \frac{\omega&#039;}{c}, \frac{\omega&#039; x&#039;}{c \sqrt{x&#039;^{2} + y&#039;^{2} + z&#039;^{2}}}, \frac{\omega&#039; y&#039;}{c \sqrt{x&#039;^{2} + y&#039;^{2} + z&#039;^{2}}}, \frac{\omega&#039; z&#039;}{c \sqrt{x&#039;^{2} + y&#039;^{2} + z&#039;^{2}}} \right)

\phi = \eta_{\mu \nu} k&#039;^{\mu} r&#039;^{\nu} = \omega&#039; t&#039; - \frac{\omega&#039;}{c} \sqrt{x&#039;^{2} + y&#039;^{2} + z&#039;^{2}}

 

[rickrherbert]

I found it really fascinating to browse I might love to find out you create far more on this topic.

 

[sciencewatch]

OK keji8341, here we go. Without loss of generality I will use two reference frames in the standard configuration with the point source at rest at the origin in the primed frame, and I will use units of time such that in the primed frame w=1 and units of distance such that c=1. Then, in the primed frame we have:
r&#039;=\left( t&#039;,x&#039;,y&#039;,z&#039; \right)
k&#039;=\left(1,\frac{x&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}},\frac{y&#039;}{\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}},\frac{z&#039;}{\sqrt{x&#039;^2+<br /> y&#039;^2+z&#039;^2}}\right)
\eta_{\mu\nu}k&#039;^{\mu}k&#039;^{\nu}=0
\phi=\eta_{\mu\nu}k&#039;^{\mu}r&#039;^{\nu}=t&#039;-\sqrt{x&#039;^2+y&#039;^2+z&#039;^2}

Boosting to the unprimed frame we get
r^{\mu}=\Lambda^{\mu}_{\nu&#039;}r&#039;^{\nu&#039;}=\left(t,x,y,z\right)
k^{\mu}=\Lambda^{\mu}_{\nu&#039;}k&#039;^{\nu&#039;}=\left(<br /> \begin{array}{c}<br /> \frac{v (t v+x)}{\left(v^2-1\right) \sqrt{-\frac{(t<br /> v+x)^2}{v^2-1}+y^2+z^2}}+\frac{1}{\sqrt{1-v^2}} \\<br /> \frac{t v+x}{\left(1-v^2\right) \sqrt{-\frac{(t<br /> v+x)^2}{v^2-1}+y^2+z^2}}-\frac{v}{\sqrt{1-v^2}} \\<br /> \frac{y}{\sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}} \\<br /> \frac{z}{\sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}}<br /> \end{array}<br /> \right)
\eta_{\mu\nu}k^{\mu}k^{\nu}=0
\phi=\eta_{\mu\nu}k^{\mu}r^{\nu}=\frac{-\sqrt{1-v^2} \sqrt{-\frac{(t v+x)^2}{v^2-1}+y^2+z^2}+t+v x}{\sqrt{1-v^2}}

k behaves as you would expect for the wave four-vector. E.g. for y=0 and z=0 we get
k^t=\frac{1-v \; \text{sgn}(t v+x)}{\sqrt{1-v^2}}
which is the standard expression for the relativistic Doppler effect including the sign change as the point source passes a given location on the x axis. Off of the x-axis the Doppler shift depends on both position and time, as you would expect from everyday experience. Also, as I stated above, the spacelike part of k is not generally parallel to the spacelike part of r.

phi also behaves as you would expect for the phase of a moving point source. Surfaces of constant phase form light cones centered on the location of the point source at t=\phi. The formula is valid as close to the point source as you like.

PS k is written as a colum vector just so that it would fit on the screen width easily

I found a paper, saying the wave vector and frequency for a moving point light source CANNOT form a Lorentz covariant 4-vector. I am not sure if it is correct.

 

[Dale]

Hi sciencewatch, welcome to PF!

Does the paper come from a mainstream scientific source? If so, can you link to it?

 

[sciencewatch]

Hi sciencewatch, welcome to PF!

Does the paper come from a mainstream scientific source? If so, can you link to it?

seems in arxiv.org. I will check later.

 

[sciencewatch]

Hi sciencewatch, welcome to PF!

Does the paper come from a mainstream scientific source? If so, can you link to it?

http://arxiv.org/ftp/arxiv/papers/1007/1007.0980.pdf

seems not published; everybody can post their own stuffs there.

 

[Dale]

Hi sciencewatch, I looked at the reference. You are correct, it is not published in a mainstream science journal. The arguments presented in the paper are exactly the same as those debunked here in this thread. In fact, the arguments and language are so similar that this paper must be the source of keji8341's misunderstanding.

 

[Phrak]

Yes, as you suggest here x/r is correct, not x/r² as you suggested above. My formula for k is correct.

Sorry, that was a typo. I meant to write x/r². But without your defintion of r, it's difficult to know what your equations mean. But that's ok.

 

[Phrak]

Is Plank’s constant Lorentz invariant?

To solve, I use planar waves.

k \cdot \lambda = 1
First, for k to be a vector, it must be invaiant over the group of continuous rotational transformations SO[3].
A plane including the origin has phase \phi. A second plane at distance r_0 has phase \phi + 2\pi.
The equation of a plan at a normal distance \lambda = r_0 is given by
\hat{n} \cdot (r-r_0)=0
Also
r_0 \cdot (r-r_0)=0
r_0 is the distance to the plane from the origin and r is the vector (x,y,z).
(The r term could be upgraded to its 2-form directed area to produce a well behaved tensor equation good in any unaccelerated coordinate system, i.e.: *(*r (r-r_0))=0.)
(x_{0}^2+y_{0}^2+z_{0}^2) = x x_0+ y y_0+z z_0
where r_{0}^2 = x_{0}^2 + y_{0}^2 + z_{0}^2.

The plane intersects the x, y and z axis at:
x = \lambda^2 / x_0
y = \lambda^2 / y_0
z = \lambda^2 / z_0
(x,y,z) = (\lambda_x, \lambda_y, \lambda_z) are the directed wavelengths, and together, do *not* transform as a vector. (\lambda_x, \lambda_y, \lambda_z) is not a vector.

The normal wave number k, however, is proportional to the reciprical of the normal wavelength.
x^i is a vector, \lambda is a constant; k^i = x_0^{i} / \lambda^2
k is a vector.

The same strategy can be followed for a Lorentz boost in the x direction.
r_0 \cdot (r-r_0)=0
r_0 = (ct_0, x_0)
r = (ct, x)
This obtains
k^t = ct_0 / \lambda^2
k^x = x_0 / \lambda^2
k^t is in units of inverse length.
\nu /c = ct_0 / \lambda^2
(\nu /c, k^x, 0, 0)is a 4-vector.
One boost and SO(3) uniquely define the Lorentz group; k is a Lorentz invariant vector. If (E,c/p) is a Lorentz invariant vector then Plank’s constant, h is a scalar constant under the Lorentz group.

The full Poincare group would be a different matter.

 

[sciencewatch]

Is Plank’s constant Lorentz invariant?

...(\nu /c, k^x, 0, 0)is a 4-vector.
... k is a Lorentz invariant vector. If (E,c/p) is a Lorentz invariant vector then Plank’s constant, h is a scalar constant under the Lorentz group.

The full Poincare group would be a different matter.

Do you mean:

(\nu /c, k^x, 0, 0) is a 4-vector + h(\nu /c, k^x, 0, 0) is a 4-vector ---->h =a scalar constant under the Lorentz group ?

 

[keji8341]

...3. From your Doppler formula, the observed frequency in the lab frame changes with locations, as you indicated in your post #73:

"Off of the x-axis the Doppler shift depends on both position and time..."

That means a photon's frequency changes during propagation, clearly challenging the energy conservation of Einstein's light-quantum hypothesis if the Planck constant is a "universal constant".

Therefore, actually it is you yourself who is chanlleging the invariance of Planck constant.

Actually the point-source effect only takes place at the microscale. The breaking of the invariance of Planck constant in the microscale might be used for explaining why the Planck's blackbody radiation law breaks down in a nanoscale (http://web.mit.edu/newsoffice/2009/heat-0729.html ).

 

[Dale]

The breaking of the invariance of Planck constant in the microscale might be used for explaining why the Planck's blackbody radiation law breaks down in a nanoscale (http://web.mit.edu/newsoffice/2009/heat-0729.html ).

That is a speculation piled on top of a speculation. Do you have any mainstream scientific references to support the suggestions that:
1) the Planck constant is not frame invariant at small scales
2) that has anything to do with the increased thermal transfer at small distances

If not, then your unsupported speculations have no place on this forum and are a violation of the rules you agreed to when you signed up.

 

[Phrak]

Do you mean:

(\nu /c, k^x, 0, 0) is a 4-vector + h(\nu /c, k^x, 0, 0) is a 4-vector ---->h =a scalar constant under the Lorentz group ?

No. From the original post, it is implicitely assume (E/c, p) is a 4-vector. It is also assumed, by implication, that

E = h \nu
p_x = h k_x
p_y = h k_y
p_z = h k_z

in some inertial frame of reference.

I think I have demonstrated, roughly, withing the confines of special relativity, if (E/c,p) is a 4-vector of some wave phenomena, there is an associated 4-vector wave number in invariant proportions in its elements. It's a rough demonstration because it's all rubbish, anyway. It just happens to work because it assumes coordinate axes are orthonormal where it is meaningful to use the dot product between vectors. I'm fairly confident it still works in curvilinear coordinates in Minkowski spacetime, but it would be a lot more work to show it, and no one would understand the archane notation anyway.

 

[keji8341]

...k behaves as you would expect for the wave four-vector. E.g. for y=0 and z=0 we get
k^t=\frac{1-v \; \text{sgn}(t v+x)}{\sqrt{1-v^2}}
which is the standard expression for the relativistic Doppler effect including the sign change as the point source passes a given location on the x axis. Off of the x-axis the Doppler shift depends on both position and time, as you would expect from everyday experience. ...

ZT: "Most scientists insist that Einstein’s plane wave Doppler formula be applicable to any cases, no matter whether the observer is close to a moving source or not. A strong argument is that the spherical wave produced by a moving point source can be decomposed into plane waves. "

“The plane wave decomposition is mathematically universal.”

Do you think so?

 

[Dale]

No, I have never seen any survey of scientists which would indicate that most insist on that point. I personally don't hold an opinion on what most scientists insist. Do you have a reference supporting that claim?

 

[keji8341]

No, I have never seen any survey of scientists which would indicate that most insist on that point. I personally don't hold an opinion on what most scientists insist. Do you have a reference supporting that claim?

By a lot of private communications. They are surprised when I mentioned a paper taking about moving point-source Doppler effect.

 

[Dale]

Private communications with several scientists does not constitute a mainstream scientific reference concerning the opinion of a majority of scientists. Would you care to rephrase your question without the attempt to give it an artificial level of credibility by pretending that it is a position known to be held by a majority of scientists?

 

[keji8341]

Private communications with several scientists does not constitute a mainstream scientific reference concerning the opinion of a majority of scientists. Would you care to rephrase your question without the attempt to give it an artificial level of credibility by pretending that it is a position known to be held by a majority of scientists?

To tell the truth, you are the first one I met, who realized there is something different.

Some one said to me, “The plane wave decomposition is mathematically universal.” The point-source spherical wave can be decomposed into plane waves. So should be the same.

I don't know much about the math principle: “The plane wave decomposition is mathematically universal.” That is a famous principle?

 

[Dale]

I don't know if it is "universal", but this much is correct.

The point-source spherical wave can be decomposed into plane waves.

A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves. I don't know what the result of an infinite sum of Doppler shifts would be.

 

[keji8341]

I don't know if it is "universal", but this much is correct. A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves. I don't know what the result of an infinite sum of Doppler shifts would be.

Any references? Please.

 

[Dale]

I would start here, particularly the introductory paragraph.
http://en.wikipedia.org/wiki/Fourier_optics

 

[keji8341]

I would start here, particularly the introductory paragraph.
http://en.wikipedia.org/wiki/Fourier_optics

I did not find the theorem as you mentioned: "A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves. "

Be specific, Please.

 

[Dale]

Did you not even read the first paragraph? "In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself."

If you want something more in depth you can get a textbook:
http://www.wiley.com/WileyCDA/WileyTitle/productCd-0780334116.html

Or Google "Fourier optics" or "plane wave spectrum".

 

[keji8341]

Did you not even read the first paragraph? "In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself." ...

Thanks. Yes, I read these words:

“In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself.”

Did you get your conclusion from the above words? It seems to me that, the above ambiguous words cannot result in your conclusion:

"A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves."

Probably that is your own work, related to your intellectual property. If so, I won’t ask.

I don’t have much math. But to my knowledge, whether a function f(x) can be represented as a Fourier integral depends on the property which the function has. Here is a theorem I got from
http://mathworld.wolfram.com/FourierTransform.html

A function f(x) has a Fourier transform if
1. The integral of absolute f(x) exists.
2. There are a finite number of discontinuities.
3. The function has bounded variation. A sufficient weaker condition is fulfillment of the Lipschitz condition.

I think the field produced by a point source has a singularity. I am not sure if the first and the third math conditions can be satisfied.

 

[Dale]

“In Fourier optics, by contrast, the wave is regarded as a superposition of plane waves which are not related to any identifiable sources; instead they are the natural modes of the propagation medium itself.”

Did you get your conclusion from the above words? It seems to me that, the above ambiguous words cannot result in your conclusion:

"A spherically symmetric wave can indeed be expanded as an infinite sum of plane waves."

Why not? They are saying the same thing.

 

[PhilDSP]

A function f(x) has a Fourier transform if
1. The integral of absolute f(x) exists.
2. There are a finite number of discontinuities.
3. The function has bounded variation. A sufficient weaker condition is fulfillment of the Lipschitz condition.

I think the field produced by a point source has a singularity. I am not sure if the first and the third math conditions can be satisfied.

Probably even more basic considerations are what domain you are interested into evaluate or define the Fourier Transform. For the field at a single point you only need a one dimensional Fourier Transform and at that point the transform (and wave equation) will be the same whether the wave is a plane wave or whether it is spherical.

It's only when you wish to evaluate a spatially extended region that you will see a difference between values for a plane wave or spherical wave originating from a particular point. In that case you'll need to define and evaluate a Fourier Transform in at least 2 dimensions. You will see a phase variation across the extended region which will be visible in the FT.

 

[keji8341]

Probably even more basic considerations are what domain you are interested into evaluate or define the Fourier Transform. For the field at a single point you only need a one dimensional Fourier Transform and at that point the transform (and wave equation) will be the same whether the wave is a plane wave or whether it is spherical.

It's only when you wish to evaluate a spatially extended region that you will see a difference between values for a plane wave or spherical wave originating from a particular point. In that case you'll need to define and evaluate a Fourier Transform in at least 2 dimensions. You will see a phase variation across the extended region which will be visible in the FT.

Thanks. I think you are talking about numerical computations about FT. Actually my question is a pure theoretical problem. The point source has a sigularity at r=0; the function to be transformed is ~expi[wt-(w/c)*r]/r with r=sqrt(x**2+y**2+z**2), the frequency w is given, monochromatic wave. I am not sure if it can be represented by uniform plane waves, although some scientists often firmly claim “The plane wave decomposition is mathematically universal.”

The point-source solution actually is closely related so-called "invariant Green's function" in classical electrodynamic.

 

[Dale]

You may be interested in this page:
http://lmb.informatik.uni-freiburg.de/papers/download/wa_report01_08.pdf

 

[PhilDSP]

Thanks. I think you are talking about numerical computations about FT. Actually my question is a pure theoretical problem. The point source has a sigularity at r=0; the function to be transformed is ~expi[wt-(w/c)*r]/r with r=sqrt(x**2+y**2+z**2), the frequency w is given, monochromatic wave. I am not sure if it can be represented by uniform plane waves, although some scientists often firmly claim “The plane wave decomposition is mathematically universal.”

The point-source solution actually is closely related so-called "invariant Green's function" in classical electrodynamic.

I'm not sure what you really mean by "mathematically universal". Sure, very often EM problems or SR problems are expressed and evaluated in terms of plane waves for simplicity. It seems that a spherical wave decomposition of a plane wave is much more commonly described than the converse. But even that is rather complicated and potentially fraught with technical problems such as in this description:

http://farside.ph.utexas.edu/teaching/jk1/lectures/node102.html

In short, it seems far more complicated to force a spherical wave solution to a problem described in terms of a plane wave than to re-describe the problem in terms of spherical waves.

By the term "evaluate" I mean evaluate analytically (not numerically). Usually to evaluate a FT numerically you need to use a Discrete Fourier Transform which at best only approximates a FT (Continuous Fourier Transform)

If you could rewrite your function so that it doesn't already have terms for frequency then it might be simple to determine the FT analytically. Can you somehow replace the frequency term with a partial derivative (involving dr/dt with a constant c) ?