

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer

Suppose $\vec{F}(x,y,z)=\vec{f}(x,y)\exp(ikz)$ and $\vec{F}$ satisfies the equations $\nabla \cdot \Re{\vec{F}}=0$ where $\Re{\vec{F}}$ is the real part of $\vec{F}$. It also satisfies $\nabla \times \Re\vec{F}=\Re\vec{G}$ for $\Re\vec{G}=\vec{g}(x,y)\exp(ikz)$

If I want to express these equations in terms of $\vec{f},\vec{g},k$, Is it possible to simplify do better than $\Re\left[\left({\partial \vec{f}\over \partial x}+{\partial \vec{f}\over \partial y}+ik\vec{f}\right)\exp(ikz)\right]=0$ for the first one? As for the second one I really have no idea how to express it in the desired form except trivially substituting the entire expression of $\vec{F}$ into the equation.

Could anyone help?

Added: $F,f$ are both 3-D vectors the $x,y,z$ are just their arguments.


share|cite|improve this question
The LHS suggests that F is a 3-dimensional vector field, however, on the RHS you have a 2-dimensional vector f in both real and imaginary parts, multiplied by a scalar. If you split real and imaginary parts and wrote the Laplacian clearly in components, perhaps it would become clearer – Valentin May 27 '12 at 23:26
@Valentin: Thank you, actually $f$ is a 3-D vector but it only depends on $x,y$ – hatch May 28 '12 at 7:03
up vote 0 down vote accepted

There are so many chances to get confused due to possibly ambiguous notation in this example (e.g. $i$ and $\vec{\boldsymbol{i}}$, $k$ and $\vec{\boldsymbol{k}}$) that I would suggest carefully writing out the components of each expression: $$\vec{F}=\vec{f}\cos kz+i\vec{f}\sin kz$$ $$\vec{f}=P\left(x,y\right)\vec{\boldsymbol{i}}+Q\left(x,y\right)\boldsymbol{\vec{j}+}R\left(x,y\right)\boldsymbol{\vec{k}}$$ $$\Re\vec{F}=\vec{f}\cos kz=P\left(x,y\right)\cos kz\vec{\boldsymbol{i}}+Q\left(x,y\right)\cos kz\boldsymbol{\vec{j}+}R\left(x,y\right)\cos kz\vec{\boldsymbol{k}}$$ $$\nabla\Re\vec{F}=\frac{\partial P}{\partial x}\cos kz+\frac{\partial Q}{\partial y}\cos kz-kR\left(x,y\right)\sin kz$$

Similar approach holds for the second one if you write $\nabla \times \vec{F}$ in components. If that is not what you are looking for, you have to specify the "desired form" more rigidly. Hope that helps.

share|cite|improve this answer
Thank you,Valentin – hatch May 28 '12 at 22:02

Your Answer


