# 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