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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.

Thanks.

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    @Valentin: Thank you, actually $f$ is a 3-D vector but it only depends on $x,y$2012-05-28

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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.