# A Electromagnetism skin depth

1. Jul 24, 2016

### mertcan

hi, nowadays I try to get involved electromagnetic waves equations, and I came across that $$\nabla$$ is treated like $$ik$$ and $$\frac {\partial } {\partial t}$$ is treated like $$-iw$$ And if you look at the part painted with orange in my attachment, afterwards you can see that $$\nabla$$ and $$\frac {\partial } {\partial t}$$ is equated to the $$ik$$ and $$-iw$$ respectively. It is so weird because those operators are equated to a number. How is it possible??? What is the logic of this situation ??? Is there a proof of that situation???

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2. Jul 24, 2016

### vanhees71

You should relabel your question as category B. I guess, it is also good advice to learn vector calculus first. A very good short summary can be found in many textbooks in electrodynamics, as in the textbooks by Abraham and Becker, Schwartz, or Griffiths.

In Cartesian coordinates the nabla symbol represents the differential operator
$$\vec{\nabla}=\vec{e}_x \partial_x + \vec{e}_y \partial_y + \vec{e}_z \partial_z=\sum_{j=1}^3 \vec{e}_j \partial_j.$$
For a plane-wave ansatz of an arbitrary field
$$\phi(t,\vec{x})=\phi_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x}),$$
you have, e.g.,
$$\partial_x \phi=\mathrm{i} k_x \phi_0 \exp(-\mathrm{i} \omega t+\mathrm{i} \vec{k} \cdot \vec{x})=\mathrm{i} k_x \phi(t,\vec{x}),$$
i.e.,
$$\vec{\nabla} \phi=\mathrm{i} \vec{k} \phi.$$
Thus for such fields you can indeed set $\vec{\nabla} \rightarrow \mathrm{i} \vec{k}$.

The time derivative in the same sense is obviously equivalent to the substitution $\partial_t=-\mathrm{i} \omega$.

This underlines the merit of the Fourier decomposition of fields: The vector operators in time-position representation become algebraic relations with $\omega$ and $\vec{k}$ (i.e., in frequency-wavevector space), i.e., the field equations become algebraic equations, and often you can solve problems by transforming the solutions of these algebraic equations back to time-position space.