# Transverse and Longitudinal Energy Flows in a Sharp Focus of Vortex and Cylindrical Vector Beams
## 3 Spin and Energy Flow in the Focal Plane for an Optical Vortex with Cylindrical Polarization
\begin{split}\overline{P}_{z}=\int\limits_{0}^{\infty}\int \limits_{0}^{2\pi}P_{z}rdrd\varphi=\frac{1}{2}\int\limits_{0}^{\infty}\int \limits_{0}^{2\pi}rdrd\varphi\left(\dot{J}_{0,m+n}^{2}+\dot{J}_{0,m-n}^{2}- \dot{J}_{2,m+n-2}^{2}-\dot{J}_{2,m-n+2}^{2}\right)=\\ \frac{1}{2}(W_{0}+W_{0}-W_{2}-W_{2})=W_{0}-W_{2}=W-2W_{2}-2W_{1}. \end{split}  \tag{26}\]  
\[ \begin{split}\overline{S}_{z}&=\int\limits_{0}^{ \infty}\int\limits_{0}^{2\pi}S_{z}rdrd\varphi=\frac{1}{2}\int\limits_{0}^{ \infty}\int\limits_{0}^{2\pi}rdrd\varphi\left(\dot{J}_{0,m-n}^{2}-\dot{J}_{0,m+ n}^{2}+\dot{J}_{2,m+n-2}^{2}-\dot{J}_{2,m-n+2}^{2}\right)\\ +&(-1)^{n}\int\limits_{0}^{\infty}\int\limits_{0}^{2 \pi}rdrd\varphi\cos(2(n-1)\varphi)(J_{0,m-n}\dot{I}_{2,m+n-2}-\dot{J}_{0,m+n} \dot{I}_{2,m-n+2})=\frac{1}{2}(W_{0}-W_{0}+W_{2}-W_{2})=0.\end{split}  \tag{27}\]  
It is clear from the comparison that expression (25) coincides with (7), and expression (26) coincides with (8). But (27) does not coincide with (9) since field (21) has no spin in the initial plane. It should not be in any other plane, including the plane of focus. The fact that the total beam energies (1) and (21) coincide follows from the fact that the initial amplitudes of both beams are the same, \(B(\theta)\), and the vortex component \(\exp(im\varphi)\) and the polarization distribution (\(\cos\varphi\), \(\sin\varphi\)) do not affect the beam energy value. The total longitudinal energy flux of (26) and (8) is the same for both beams (1) and (21) because both beams have the same vortex component \(\exp(im\varphi)\). And, although the azimuthal energy flows (2) and (22) for beams (1) and (21) are different, the part of the energy going to the formation of the transverse azimuthal flow is the same and is equal to expression (15). Note that, when \(m=0\) (there is no optical vortex), the azimuthal projection of the energy flow is zero: \(P_{\varphi}=\Omega(r)=0\). Therefore, the following question arises: where does the part of the energy flow in (26) go which previously went to the formation of azimuthal transverse rotation of energy (18) at the focus of the beam (1)? In order to deal with this issue, let us write expressions for the projections of the electric field strength in the focal plane of the initial beam (21):  
\[ \begin{split} F_{x}&=\frac{\dot{i}^{m+n-1}}{2}e^{ i(m+n)\varphi}\left(J_{0,m+n}+e^{-2i\varphi}J_{2,m+n-2}\right)\\ +&\frac{\dot{i}^{m-n-1}}{2}e^{i(m-n)\varphi}\left(J_{ 0,m-n}+e^{2i\varphi}J_{2,m-n+2}\right),\\ F_{y}&=\frac{\dot{i}^{m+n}}{2}e^{i(m+n)\varphi} \left(-J_{0,m+n}+e^{-2i\varphi}J_{2,m+n-2}\right)\\ +&\frac{\dot{i}^{m-n}}{2}e^{i(m-n)\varphi}\left(J_{0,m -n}-e^{2i\varphi}J_{2,m-n+2}\right),\\ F_{z}&=\dot{i}^{m+n}e^{i(m+n-1)\varphi}J_{1,m+n-1} -\dot{i}^{m-n}e^{i(m-n+1)\varphi}J_{1,m-n+1}.\end{split}  \tag{28}\]  
The initial beam (21), even in the absence of an optical vortex (\(m=0\)) in the focal plane, is seen from (28) to form four new optical vortices, \(\exp(i(n-2)\varphi)\), \(\exp(i(2-n)\varphi)\), \(\exp(i(n-1)\varphi)\) and \(\exp(i(1-n)\varphi)\) with topological charges \((n-2)\), \(-(n-2)\), \((n-1)\) and \(-(n-1)\). A part of the light energy is spent on their formation. There is a transverse azimuthal energy flow (15) if \(m>0\) in the focal plane since \(P_{\varphi}=\Omega(r)\
eq 0\). In this case, the initial optical vortex defines the rotation direction of the energy flow (for \(m>0\), the energy rotation in the focal plane is counterclockwise). And the situation for field (21) is similar to that formed at the focus for field (1). But when \(m=0\), the azimuthal energy flow disappears: \(P_{\varphi}=\Omega(r)=0\). Which direction is the transverse flow facing at the focus if it is present before the focus (the beam converges) and after the focus (the beam diverges)? In the focal plane, when \(m\) equals 0 and 1, the longitudinal component of the energy flow is shown by numerical simulation to not change, on average (see Section 3).