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.


  • $\begingroup$ 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, 2015 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, 2015 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, 2015 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, 2015 at 0:53

1 Answer 1


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 $\hat {\vec z}$.

  • $\begingroup$ 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, 2015 at 23:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.