# Finding Transversal Components from Longitudinal component for Electric and Magnetic Field in a cylindrical coordinates system

Can someone explain why this two equation are equivalent?

$\nabla_T$ denotes the transverse two-dimensional nabla operator: $\nabla_T=\hat{x}\frac{\partial}{\partial x}+\hat{y}\frac{\partial}{\partial y}$
$\hat{z}$ is unit vector of $z$ axis in the cylindrical coordinates system.
$H_z$ come from Helmholtz scalar equation for longitudinal component of magnetic field in cylindrical coordinates system : $\nabla^2H_z+k^2H_z=0$

$E_T = \hat{x}E_x+\hat{y}E_y$
$H_T = \hat{x}H_x+\hat{y}H_y$
$\epsilon\mu\omega^2 = k^2$

Electric and magnetic fields may be decomposed into a two-dimensional transverse component (a vector function) and one longitudinal component (a scalar function).
$E = E_T + \hat{z}E_z$
$H = H_T + \hat{z}H_z$

The Laplacian operator in any cylindrical coordinate system is:
$\nabla^2A=\nabla^2A_T + \hat{z}\nabla^2A_z, \nabla = \nabla_T + \hat{z}\frac{\partial}{\partial z}$

From Maxwell's Equation :

Until here I understand . But I don't understand how to make calculus in the above equation. They say that :
$\nabla_T\times E_T = -j\mu\omega H_z\hat{z} (1)$
$\nabla_T\times \hat{z}E_z + \hat{z} \frac{\partial E_T}{\partial z} = -j\mu\omega H_T (2)$
$\nabla_T\times H_T = j\mu\epsilon E_z \hat{z} (3)$
$\nabla_T\times \hat{z}H_z + \hat{z} \frac{\partial H_T}{\partial z} = j\mu\epsilon E_T (4)$

Aplying $\hat{z} \times \frac{\partial}{\partial z}$ to (2) and multiplying (4) by $-j\mu\omega$ then adding up them and canceling $H_T$ :

is the the same with this ($E_T$ is the two-dimensional transverse vector for electric field in the cylindrical coordinate system):

after using this vector formulas:

Can someone explain this using simple vectors calculus (identities)? I am interested in the equality of the first two expressions but also I need some explications for the two formulas they used .
Why $\nabla_T \times \hat{z}A_z= -\hat{z}\times \nabla_TA_z$ ?

I am stuck on this vector identities with partial derivative and scalars. I can't memorize,I want to understand because it would be easier to take my exam.

Thanks!

• Is this not a simple case of the cross product identity $\vec a\times\vec b=-\vec b\times\vec a$? Since $\nabla_T\hat z=0$, I think this is trivial... Commented Feb 4, 2015 at 23:17