**2.1 Vector potential**

The solution to wave equations in vector potential form is given by [64].

$$
\nabla^2 \overrightarrow{A} - \frac{\mathbf{1}}{c^2} \frac{\partial^2 \overrightarrow{A}}{\partial t^2} = \mathbf{0} \tag{3}
$$

Expanding it in the cylindrical coordinate ð Þ *ρ*, *ϕ*, *z* yields

$$\begin{aligned} & \left[ \frac{\partial^2 A\_\rho}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial A\_\rho}{\partial \rho} + \frac{1}{\rho^2} \left( \frac{\partial^2 A\_\rho}{\partial \phi^2} - A\_\rho \right) + \frac{\partial^2 A\_\rho}{\partial z^2} - \frac{2}{\rho^2} \frac{\partial A\_\phi}{\partial \phi} - \frac{1}{c^2} \frac{\partial^2 A\_\rho}{\partial t^2} \right] \hat{\phi} \\ &+ \left[ \frac{\partial^2 A\_\phi}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial A\_\phi}{\partial \rho} + \frac{1}{\rho^2} \left( \frac{\partial^2 A\_\phi}{\partial \phi^2} - A\_\phi \right) + \frac{\partial^2 A\_\varphi}{\partial z^2} + \frac{2}{\rho^2} \frac{\partial A\_\rho}{\partial \phi} - \frac{1}{c^2} \frac{\partial^2 A\_\phi}{\partial t^2} \right] \hat{\phi} \\ &+ \left[ \frac{\partial^2 A\_z}{\partial \rho^2} + \frac{1}{\rho} \frac{\partial A\_x}{\partial \rho} + \frac{1}{\rho^2} \frac{\partial^2 A\_z}{\partial \phi^2} + \frac{\partial^2 A\_x}{\partial z^2} - \frac{1}{c^2} \frac{\partial^2 A\_z}{\partial t^2} \right] \hat{z} \\ &= 0 \end{aligned} \tag{4}$$

where vector potential *A* ! is defined in the cylindrical coordinate system ð Þ *ρ*, *ϕ*, *z* and *Aρ*, *Aϕ*, *Az* � � are the scalar function. For the scenario when *<sup>A</sup><sup>ρ</sup>* <sup>¼</sup> *<sup>A</sup><sup>ϕ</sup>* <sup>¼</sup> 0 and *Az* ¼ *ψ*, Eq. (4) reduces to

$$\frac{d^2\psi}{d\rho^2} + \frac{1}{\rho}\frac{d\psi}{d\rho} + \left(k\_\psi^2 - k\_x^2 - \frac{l^2}{\rho^2}\right)\psi = 0\tag{5}$$

Hence, a set of Bessel beams is obtained by vector potential as

$$A\_{\rho} = A\_{\varphi} = 0; A\_{\mathfrak{z}}\left(\stackrel{\rightarrow}{r}, t\right) = J\_l\left(k\_{\rho}\rho\right)e^{j\left(k\_{\mathfrak{z}}x - at \pm l\phi\right)}\tag{6}$$

where *kz* ¼ *k<sup>ψ</sup>* cosð Þ *α* , *k<sup>ρ</sup>* ¼ *k<sup>ψ</sup>* sin ð Þ *α* , which can be deduced from **Figure 2**.

#### **Figure 2.**

*(a) Plane-wave normal vectors making nondiffracting Bessel beam along the cone. A group of plane waves makes an angle α with the direction of propagation kz to form an OAM wave. (b) 3-D phase front plot of OAM mode number (i) l* ¼ *0 (plane wave), (ii) l* ¼ 1*, (iii) l* ¼ 2*, (iv) l* ¼ 3*.*

#### **2.2 Electric and magnetic field**

The electric and magnetic fields can be obtained from vector potential *A* ! in the frequency domain form as

$$
\overrightarrow{E} = -\nabla\Phi - \frac{1}{c}\frac{\partial\mathcal{A}}{\partial t}; \overrightarrow{H} = \frac{1}{\mu}\nabla \times \overrightarrow{\mathcal{A}}\tag{7}
$$

After manipulating Eqs (6), (7), and using Lorentz condition, we obtain the electric and magnetic fields in a cylindrical coordinate system for TM and TE modes.

*2.2.1 TM modes*

$$\begin{pmatrix} E\_{\rho} \\ E\_{\phi} \\ E\_{x} \end{pmatrix} = \begin{pmatrix} jk\_{\rho}k\_{x}\cos aI\_{l}'(k\_{\rho}\rho) \\\\ \pm \frac{l}{\rho}k\_{x}\cos aI\_{l}(k\_{\rho}\rho) \\\\ k\_{\rho}^{2}\sin aI\_{l}(k\_{\rho}\rho) \end{pmatrix} e^{j(k\_{x}x - at \pm l\phi)} \tag{8}$$

$$\begin{Bmatrix} H\_{\rho} \\ H\_{\phi} \\ H\_{z} \end{Bmatrix} = \begin{pmatrix} \pm \frac{l}{\rho} \text{coeff}\_{l}(\mathbb{k}\_{\rho}\rho) \\\\ jk\_{\rho}\omega\epsilon f\_{l}^{\prime}(\mathbb{k}\_{\rho}\rho) \\\mathbf{0} \end{pmatrix} e^{j(\mathbb{k}\_{\mathrm{s}}x - \alpha t \pm l\phi)}\tag{9}$$

#### *2.2.2 TE modes*

Using the principle of duality ð Þ *ETE* ¼ *HTM*; *HTE* ¼ �*ETM*

$$
\begin{pmatrix} E\_{\rho} \\ E\_{\phi} \\ E\_{z} \end{pmatrix} = \begin{pmatrix} \pm \frac{l}{\rho} \alpha \varepsilon f\_{l}(k\_{\rho}\rho) \\ \rho \\ jk\_{\rho}\alpha \varepsilon f\_{l}'(k\_{\rho}\rho) \\ \mathbf{0} \end{pmatrix} e^{j(k\_{z}x - \alpha t \pm l\phi)} \tag{10}
$$

$$\begin{aligned} \begin{bmatrix} H\_{\rho} \\ H\_{\phi} \\ H\_{z} \end{bmatrix} = \begin{pmatrix} jk\_{\rho}k\_{\rho}\cos aI\_{l}'(k\_{\rho}\rho) \\ \pm \frac{l}{\rho}k\_{x}\cos aI\_{l}(k\_{\rho}\rho) \\ k\_{\rho}^{2}\sin aI\_{l}(k\_{\rho}\rho) \end{pmatrix} e^{j(k\_{x}x - at \pm l\phi)} \end{aligned} \tag{11}$$

Clearly, all the electric and magnetic field components in TE and TM modes show azimuthal phase dependence with the OAM index number *l* and addition of extra elements to the exponential term. This is one of the possible solutions when *Az* ¼ *ψ* and *A<sup>ρ</sup>* ¼ *A<sup>ϕ</sup>* ¼ 0, and other possible solution can be considered as *A<sup>ρ</sup>* ¼ *ψ*, *A<sup>ϕ</sup>* ¼ �*jA<sup>ρ</sup>* and *Az* ¼ 0. In this work, our focus is on the first case as it deals with linearly polarized OAM waves and the latter case corresponds to circularly polarized OAM waves, which is not the scope of this chapter as the antenna designed in further sections is also focused on a linearly polarized OWPA.

#### **2.3 Key performance parameters for plane wave and OAM wave**

In OAM, the wave travels helically around its longitudinal propagation direction, and this leads to define the pitch of OAM waves similar to the screw thread pitch

$$\text{Pitch of the OAM wave} \ \left( T\_p \right) = \frac{2\pi l}{k\_\Psi \cos \left( a \right)} = \frac{l\lambda\_\Psi}{\cos \left( a \right)} \tag{12}$$

where *λψ* <sup>¼</sup> <sup>2</sup>*<sup>π</sup> <sup>k</sup><sup>ψ</sup>* <sup>¼</sup> <sup>2</sup>*π<sup>c</sup> <sup>ω</sup>* ¼ *λplane*

As *l* = 0 the pitch of vortex waves vanishes, which can be seen from **Figure 2(b)**, indicating the formation of a plane wave phase front.

The wave vector is the position gradient of the phase of the Bessel beam

$$\overrightarrow{k} = \nabla(k\_x z \pm l\phi) = k\_x \hat{z} \pm \frac{l}{\rho} \hat{\phi} \tag{13}$$

where

$$k\_{OAM}^2 = k\_x^2 + \frac{l^2}{\rho^2} \to k = \sqrt{k\_x^2 + \frac{l^2}{\rho^2}}\tag{14}$$

When Bessel beams with zero orbital angular momentum (*l* ¼ 0; plane wave), the angular velocity is given by

$$
\omega a = k\_{\varphi} c = \frac{k\_{x}c}{\cos a} \tag{15}
$$

So, the group velocity of OAM waves is obtained by the phase gradient of plane wave angular velocity, using Eq. (15)

$$
\vec{V}\_{\mathcal{g}} = \nabla\_k \boldsymbol{\sigma} = \frac{c}{\cos a} \hat{\boldsymbol{z}} \tag{16}
$$

The intrinsic impedance of OAM waves in both TM and TE cases is obtained using Eq. (8) to (11)

$$\eta\_{\text{OAM\\_TM}} = \frac{E\_\rho}{H\_\phi} = \frac{-E\_\phi}{H\_\rho} = \frac{k\_x}{a\epsilon\epsilon} \cos a = \eta\_{plane} \cos a \tag{17}$$

$$
\eta\_{OAM\\_TE} = \frac{\eta\_{plane}}{\cos a} \tag{18}
$$

For plane-wave case *α* ¼ 0, which gives the classical value of the impedance of 377 Ω. This suggests that the generic form of the wave is actually an OAM wave, whereas the plane wave is a special case of OAM wave when *l* ¼ 0 and *α* ¼ 0.

#### **2.4 Propagation constant of plane wave and OAM wave in free space**

The wavenumber in free space for OAM as in Eq. (14) and *kz* ¼ *k<sup>ψ</sup>* cos *α*; This leads to

$$k\_{OAM} = \sqrt{k\_{\nu}^2 \cos^2(a) + \frac{l^2}{\rho^2}} = \sqrt{a^2 \mu \epsilon \cos^2(a) + \frac{l^2}{\rho^2}}\tag{19}$$

In free space, the attenuation constant and phase constant of OAM waves are obtained as

$$a\_{OAM} = 0; \quad \beta\_{OAM} = \sqrt{\alpha^2 \mu\_o \varepsilon\_o \cos^2(a) + \frac{l^2}{\rho^2}}\tag{20}$$

Special case: *α* ¼ 0; *l* ¼ 0 (plane wave)

$$a\_{plane} = 0; \; \beta\_{plane} = a \sqrt{\mu\_o \varepsilon\_o} = \frac{a}{\varepsilon} \tag{21}$$

This gives us the attenuation and phase constant of the plane waves as a special case of the OAM waves.

We have derived all the physical quantities associated with wave propagation such as electric field, magnetic field, intrinsic impedance, wave vector, propagation constant, velocity, wavelength, and pitch of the OAM wave and compared it with the plane wave. By virtue of the derivation presented above, it brings us to the theoretical


#### **Table 1.**

*Comparison of plane wave and OAM wave propagation.*

*Orbital Angular Momentum Wave and Propagation DOI: http://dx.doi.org/10.5772/intechopen.104477*

comparison of plane wave propagation (Gaussian beam propagation; zero-order Bessel beam) and OAM wave propagation (Bessel beam propagation; higher-order Bessel beam) in **Table 1**. After careful investigation and comparison, it is understandable that the plane wave is the special case of the OAM wave when *α* ¼ 0, *l* ¼ 0. **Figure 2(a)** shows that the conical beam pattern formed by the Bessel beam propagating along the z-axis and group of plane waves at an angle combines to create an OAM wave. Moreover, once we have all the derivations of physical quantities for the OAM wave, respective physical quantities of the plane wave can be deduced by substituting *α* ¼ 0, *l* ¼ 0 as presented in **Table 1**. The phase front plot of the OAM wave for different index numbers is shown in **Figure 2(b)**. This plot clearly indicates that the OAM wave of zero-order has no pitch, but as the order of OAM increases, the number of pitches increases same as the order of OAM per wavelength. The number of helixes formed for the OAM wave depends on the OAM order.
