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Can someone explain why this two equation are equivalent?
enter image description here

$\nabla_T$ 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 :

enter image description here

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 $H_T$ : enter image description here

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

after using this vector formulas: enter image description here

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!

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  • $\begingroup$ 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... $\endgroup$ – Ryan Unger Feb 4 '15 at 23:17
  • $\begingroup$ $0celo7 where you were until now? I already answered him. The many formulas he presents is repelling people from getting into the issue, but the calculi are very simple. $\endgroup$ – Sofia Feb 4 '15 at 23:20
  • $\begingroup$ @NumLock don't write products in the form $\hat {\vec z} \times \hat {\vec z} \times ∂^2A_T/∂z^2 $ because $\hat {\vec z} \times \hat {\vec z} = 0$. Put the parentheses to indicate what with what you multiply first. $\endgroup$ – Sofia Feb 4 '15 at 23:34
  • $\begingroup$ @NumLock - An improvement $\vec C \times (\vec B \times \vec A)= (\vec C \cdot \vec B) \vec A − \vec B (\vec C \cdot \vec A)$. $\endgroup$ – Sofia Feb 5 '15 at 0:53
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You know of course how to calculate a vectorial product. Then, let's calculate the two vectorial products with which you have a problem, according to the formula that you indicated for $\nabla _T$:

$ \ (I) \ -j \omega \mu \nabla _T \times \hat z H_z = \begin {bmatrix} {\hat {\vec x}} & {\hat {\vec y}} & {\hat {\vec z}} \\ {\frac {∂}{∂x}} & {\frac {∂}{∂y}} & {\frac {∂}{∂z}} \\ 0 & 0 & {H_z} \end {bmatrix} = -j \omega \mu \left( \hat {\vec x} \frac {∂H_z}{∂y} - \hat {\vec y} \frac {∂H_z}{∂x} \right). $

Now, in the same way we will calculate the 2nd equality

$ \ (II) \ j \omega \mu \hat {\vec z} \times \nabla _T H_z = \begin {bmatrix} {\hat {\vec x}} & {\hat {\vec y}} & {\hat {\vec z}} \\ 0 & 0 & 1 \\ {\frac {∂H_z}{∂x}} & {\frac {∂H_z}{∂y}} & 0 \end {bmatrix} = j \omega \mu \left( - \hat {\vec x} \frac {∂H_z}{∂y} + \hat {\vec y} \frac {∂H_z}{∂x} \right). $

Well, please compare the two results.

I will calculate one more vector product, and you will be able to do analogue calculi.

$ \ (III) \ \nabla _T \times H_T = \begin {bmatrix} {\hat {\vec x}} & {\hat {\vec y}} & {\hat {\vec z}} \\ {\frac {∂}{∂x}} & {\frac {∂}{∂y}} & {\frac {∂}{∂z}} \\ {H_x} & {H_y} & 0 \end {bmatrix} = \hat {\vec z} \left( \frac {∂H_y}{∂x} - \frac {∂H_x}{∂y} \right). $

From this result you get easily your equality (3) by taking into account that in the RHS of the 2nd Maxwell equation the only term along $\hat {\vec z}$ is $j\epsilon E_z$. Recall, for easiness of calculus, that any vector product in which one of the factors is along $\hat {\vec z}$, has the result perpendicular to $\hat {\vec z}$.

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  • $\begingroup$ 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... $\endgroup$ – Ryan Unger Feb 4 '15 at 23:18

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