## Appendix A The Doppler Shift  
For completeness, the shift in the frequency of a propagating wave due to relative motion between source and receiver is reviewed. Consider a stationary point at \(x=0\) with a wave incident on it at its propagation velocity, \(c\), along the direction of the \(x\)-axis. Define the phase of the wave to be 0 there. With the help of Figure 11, the phase of any other point \(x(t)\) along the \(x\)-axis at the same instant in time is given by  
\[\psi(x)=2\pi\bigg{(}\frac{x(t)}{\lambda}\bigg{)} \tag{11}\]  
Next, let the point \(x(t)\) move along the \(x\)-axis at a constant velocity \(v\). Because the phase is linear with position, this will cause a corresponding linear change in its phase as it moves, given by  
\[\frac{d}{dt}\psi(x(t))=\frac{d\psi(x(t))}{dx}\frac{dx(t)}{dt}=\bigg{(}\frac{2 \pi}{\lambda}\bigg{)}\frac{dx(t)}{dt} \tag{12}\]  
The rate of change of phase with respect to time is defined as radian frequency, \(2\pi F_{d}\), while the rate of change of distance with time is simply velocity, \(v\), therefore  
\[\frac{d}{dt}\psi(x(t))=2\pi F_{d}=\bigg{(}\frac{2\pi}{\lambda}\bigg{)}v \tag{13}\]or simply  
\[F_{d}=\left(\frac{v}{\lambda}\right)=f\left(\frac{v}{c}\right) \tag{10}\]  
where \(F_{d}\) is recognized as the (axial) Doppler frequency and \(f\) is the frequency of the incident wave. Although the incident wave is moving at the same time that point \(x(t)\) is moving, all this is calculated at a single point in time, as if time was frozen, so the position and velocity of the point \(x\) and the incident wave are assumed known at the instant \(t\), Heisenberg notwithstanding. One can simply visualize point \(x(t)\) moving along the frozen wave, at each position reading its corresponding phase; the frozen wavelength constitutes a phase gradient from \(0\) to \(2\pi\) radians. When a wave with OAM is incident on a point, the received phase can experience a frequency shift even as it moves in a direction transverse to the direction of propagation. Specifically, the observation point here will be moving in the direction of \(\theta\) along the circumference of the peak of the conical OAM beam in Figure 7 of the main text, where a phase gradient also exists at a given point in time. It is shown below as the \"unwrapped\" footprint of Figure 4 of the main text. Again, the electrical phase at the observation point \(R\theta=d\) is \(d\) times the phase gradient \(k/R\), equal to the proportion that \(d\) constitutes of the circumference \(2\pi R\), times \(k\) cycles of \(2\pi\) that the \(k\)-th order OAM beam imposes on that circumference of its footprint. It will vary in time if its position is changing in time, because it is passing through the phase gradient, so  
\[\frac{d}{dt}\psi(R\theta(t))=\frac{d\psi(R\theta(t))}{d\theta}\frac{d\theta(t )}{dt}=\left(\frac{2\pi k}{2\pi R}\right)R\frac{d\theta(t)}{dt}\]  
Simplifying in terms of the physical variables, the rate of change of electrical phase with time again constituting an angular frequency,  
\[\frac{d}{dt}\psi(R\theta(t))=2\pi\left(\frac{kR}{2\pi R}\right)\frac{d\theta(t )}{dt}=k\frac{d\theta(t)}{dt}=2\pi F_{T}\]  
Recognizing that \(v=Rd\theta/dt\), the above reduces to  
\[2\pi F_{T}=\frac{kv}{R} \tag{11}\]  
where \(F_{T}\) is the transverse Doppler frequency in Hz. Note that there is now no dependence on RF carrier frequency, \(f\), of the OAM beam. Note that this is not the transverse Doppler shift that is sometimes derived in physics context using relativity arguments. One may wonder why in Figure 10, the wave fronts are not propagating in the same direction as the beam axis but are angled relative to it. (The angle is shown exaggerated.) The reason is that the beam has OAM, which causes its Poynting vector to follow a helical \"corkscrew\" path, orthogonal to the wave fronts which form helical paths around the beam axis. For an OAM beam of order \(k\), one can visualize the phase fronts (wave fronts) as \(k\) parallel threads wrapped around the beam axis like the threads of a machine screw.