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

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1. Jan 10, 2016

### 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 $z > vt$?

2. Jan 15, 2016

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

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