Weyls representation of a propagating (z-v*t) spherical wave

[Rumo]

Hello!

The following wave solves the 3D wave equation:
$$ \frac{\sin\left(k\sqrt{x^2+y^2+\frac{(z-vt)^2}{1-\frac{v^2}{c^2}}}\right)}{\sqrt{x^2+y^2+\frac{(z-vt)^2}{1-\frac{v^2}{c^2}}}}\cos\left(w\frac{t-\frac{vz}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}\right) $$

This is a propagating standing spherical wave. I want to represent this as a sum of plane waves. Concerning this topic I found Weyls representation of spherical waves:
$$ \frac{\exp(ikr)}{r} = \frac{ik}{2\pi}\int\int\frac{1}{m}\exp(ik(px+qy+m|z|))\text{d}p\text{d}q $$
with $m = (1-p^2-q^2)^{\frac{1}{2}}$ or $i(p^2+q^2-1)^{\frac{1}{2}}$.

Can Weyls representation of a spherical wave be modified to represent the propagating standing spherical wave?

My ideas:
The propagating standing spherical wave is still a sum of outgoing and incoming propagating waves, so we can write:
$$ \frac{\exp\left(ik\sqrt{x^2+y^2+\frac{(z-vt)^2}{1-\frac{v^2}{c^2}}}\pm w\frac{t-\frac{vz}{c^2}}{\sqrt{1-\frac{v^2}{c^2}}}\right)}{\sqrt{x^2+y^2+\frac{(z-vt)^2}{1-\frac{v^2}{c^2}}}} $$
Deriving Weyls representation, they look at the plane $z \to 0$. Can we look at the moving plane $z \to vt$ to first find a representation of:
$$ \frac{\exp(ik\sqrt{x^2+y^2}\pm wt\sqrt{1-\frac{v^2}{c^2}})}{\sqrt{x^2+y^2}} $$?

Given the ansatz (I found the calculation in "Optical Coherence and Quantum Optics" by Leonard Mandel, Emil Wolf):
$$ \frac{\exp(ik\sqrt{x^2+y^2}\pm wt\sqrt{1-\frac{v^2}{c^2}})}{\sqrt{x^2+y^2}} = \int\int a(p,q)\exp(ik(px+qy))\text{d}p\text{d}q $$
we find with the inverse Fourier transformation:
$$ a(p,q) = \left(\frac{k}{2\pi}\right)^2 \int\int \frac{\exp(ik\sqrt{x^2+y^2}\pm wt\sqrt{1-\frac{v^2}{c^2}})}{\sqrt{x^2+y^2}}\exp(-ik(px+qy))\text{d}x\text{d}y $$
Following Weyl again, we can use polar coordinates:
$$ x = R\cos{\varphi}, y = R\sin{\varphi}, p = \rho\cos{\chi}, q = \rho\sin{\chi} $$
and find:
$$ a(p,q) = \left(\frac{k}{2\pi}\right)^2 \int\int \exp\left(ikR\pm wt\sqrt{1-\frac{v^2}{c^2}}\right)\exp(-ikR\rho\cos(\varphi-\chi))\text{d}R\text{d}\varphi = \left(\frac{k}{2\pi}\right)^2\exp\left(\pm wt\sqrt{1-\frac{v^2}{c^2}}\right) \int\int \exp(ikR)\exp(-ikR\rho\cos(\varphi-\chi))\text{d}R\text{d}\varphi $$
This is now the same as in Weyls representation apart of one factor. So we find
$$ a(p,q) = \frac{ik}{2\pi}\frac{1}{m}\exp\left(\pm wt\sqrt{1-\frac{v^2}{c^2}}\right) $$
Is this correct? But at this point I do not know how to proceed. Could you help me out? How can I expand this to $z > vt$?

 

[jvicens]

Hello,

Thank you for sharing your thoughts and calculations on representing the propagating standing spherical wave as a sum of plane waves. Your approach seems to be correct and you have successfully derived the expression for the plane wave coefficients. To extend this to $z>vt$, we can use the fact that the spherical wave can be represented as a sum of outgoing and incoming waves. The outgoing wave can be represented as a sum of plane waves with positive frequency, while the incoming wave can be represented as a sum of plane waves with negative frequency. Therefore, we can modify our expression for the plane wave coefficients to include the positive and negative frequency components:

$$a(p,q) = \frac{ik}{2\pi}\left(\frac{1}{m}\exp\left(\pm wt\sqrt{1-\frac{v^2}{c^2}}\right) + \frac{1}{m}\exp\left(\mp wt\sqrt{1-\frac{v^2}{c^2}}\right)\right)$$

We can then use this modified expression for the plane wave coefficients in the integral form of Weyl's representation to represent the propagating standing spherical wave as a sum of plane waves. I hope this helps. Let me know if you have any further questions or if you need clarification on any part of the process.