# Spin Hall Effect before and after the Focus of a High-Order Cylindrical Vector Beam
## 3 Transverse Energy Flow before and beyond the Focus  
We further derive the transverse components of the energy flow vector. The Poynting vector reads as [19]:  
\[\mathbf{P}=\frac{c}{4\pi}\text{Re}(\mathbf{E}^{*}\times\mathbf{H}). \tag{9}\]  
In Equation (9), vectors \(\mathbf{E}\) and \(\mathbf{H}\) stand, respectively, for the electric and magnetic fields, the asterisk * means the complex conjugation, \(\times\) is the cross product, and \(c\) is the free-  
In Equation (11), the following designations are introduced for the real and imaginary parts of the integrals (3):  
\[ \begin{array}{ll}R_{0}=I_{0,n}(z=0),&I_{0}=\overline{I}_{0,n},&R_{1}=I_{1,n-1}( z=0),\\ I_{1}=\overline{I}_{1,n-1},&R_{2}=I_{2,n-2}(z=0),&I_{2}=\overline{I}_{2,n-2}, \\ \overline{I}_{\
u,\mu}=\left(\frac{4\pi f}{\lambda}\right)\int\limits_{0}^{ \theta_{0}}\sin^{\
u+1}\left(\frac{\theta}{2}\right)\cos^{3-\
u}\left(\frac{ \theta}{2}\right)\cos^{3/2}(\theta)A(\theta)J_{\mu}(\xi)d\theta.\end{array}  \tag{12}\]  
For demonstrative purposes, we now move to the polar coordinates. Then, instead of Equation (11), we derive the radial and azimuthal components of the transverse Poynting vector near the focus:  
\[ \begin{array}{ll}P_{r}=2 kzQ(r),&\\ P_{\varphi}=0,&\\ Q(r)=R_{1}\big{(}\overline{I}_{0}-\overline{I}_{2}\big{)}+\overline{I}_{1}(R _{2}-R_{0}).&\end{array}  \tag{13}\]  
The radial and azimuthal components of the transverse Poynting vector describe, respectively, the convergence/divergence of the light field. Expression (13) illustrates that, regardless of the beam order \(n\), the energy flow near the focus of the beam (1) is diverging or converging from the optical axis along the radial paths. At \(z=0\) (in the focus), the energy flow is parallel to the optical axis. Before the focus (\(z<0\)), the transverse energy flow is converging along the radii to the optical axis, whereas after the focus (\(z>0\)), it is diverging. Since the sign of the function \(Q(r)\) in Equation (13) can change at some distances from the optical axis, then at certain circles centered on the propagation axis, the transverse energy flow is diverging before the focus and converging after the focus. Now we show for the field with polarization (1) that its longitudinal component of the angular momentum vector is equal to zero near the focus, as in the initial plane. Indeed, the angular momentum vector of light is given by [23]:  
\[\mathbf{J}=\frac{1}{2c}\mathrm{Re}(\mathbf{r}\times(\mathbf{E}^{*}\times \mathbf{H}))=\frac{2\pi}{c^{2}}(\mathbf{r}\times\mathbf{P}), \tag{14}\]  
whereas its longitudinal component (without the insignificant constant factor) is  
\[J_{z}=rP_{\varphi}=0. \tag{15}\]  
Expression (15) follows from Equation (13), since \(P_{\varphi}=0\). The angular momentum (14) can be represented [24] as a sum of the spin angular momentum \(\mathbf{S}\) and the orbital angular momentum \(\mathbf{L}\):  
\[\mathbf{J}=\mathbf{S}+\mathbf{L}=\frac{1}{8\pi\omega}\mathrm{Im}(\mathbf{E}^ {*}\times\mathbf{E})+\frac{1}{8\pi\omega}\sum_{p=x,y,z}\mathrm{Im}\Big{(}E_{ p}^{*}(\mathbf{r}\times\
abla)E_{p}\Big{)}. \tag{16}\]  
Without the constant factor \(1/(8\pi\omega)\), the longitudinal SAM component, as follows from Equation (7), is equal to \(S_{z}\simeq 2 kz\sin(2(n-1)\varphi)(I_{0}R_{2}-I_{2}R_{0})\). The longitudinal OAM component from Equation (16) is given by:  
\[L_{z}=\mathrm{Im}\bigg{[}E_{x}^{*}\frac{\partial}{\partial\varphi}E_{x}+E_{y} ^{*}\frac{\partial}{\partial\varphi}E_{y}+E_{z}^{*}\frac{\partial}{\partial \varphi}E_{z}\bigg{]}. \tag{17}\]  
Substituting the expressions (2) for the electric field components into Equation (17), we have that near the focus:  
\[L_{z}\simeq-2 kz\sin(2(n-1)\varphi)(I_{0}R_{2}-I_{2}R_{0}). \tag{18}\]  
Summation of Equations (7) and (18) prove the validity of expression (15). Thus, near the focus, longitudinal SAM and OAM components are equal by magnitude and directed oppositely (\(S_{z}=-L_{z}\)), and, therefore, the longitudinal component of the angular momentum is equal to zero.