Maxwell's equations and time-harmonic solutions

[fog37]

Hello,
For a linear medium, the two curl Maxwell's equations
$$\nabla \times \bf{E} = - \frac {\partial \bf{B}} {\partial t}$$
$$\nabla \times \bf{H} = \frac {\partial \bf{D}} {\partial t}$$
change to
$$\nabla \times \bf{E} = i \omega \bf{B}$$
$$\nabla \times \bf{H} = - i \omega\bf{D}$$

whose all solutions are space-dependent vector functions of the form $\tilde{\bf{E}} (x,y,z) e^{-i \omega t}$. The last two equations actually have the complex-functions $\tilde{\bf{E}} (x,y,z)$ as solutions.

• Is it correct to assume that if the field solution $\tilde{\bf{E}} (x,y,z)$ is real-valued (zero imaginary part), the overall solution field $\tilde{\bf{E}} (x,y,z) e^{-i \omega t}$ would then represent a standing wave while if $\tilde{\bf{E}} (x,y,z)$ has both nonzero $Re$ and $Im$ parts the field solution would be a traveling wave?
• The spatial complex-valued function $\tilde{\bf{E}} (x,y,z)$ is truly the three scalar functions: $$ \tilde{E}_{x} (x,y,z)$$ $$\tilde{E}_{y} (x,y,z)$$ $$ \tilde{E}_{z} (x,y,z)$$ Is it possible for any of these scalar functions to be complex-valued while the others are real-valued? What kind of field would that be? Hybrid, i.e. traveling and stationary at the same time?

 

[vanhees71]

Usually the idea with the ansatz with the exponential function is that the physical fields are the real parts of these complex fields. The $\tilde{\boldsymbol{E}}$ and $\tilde{\boldsymbol{B}}$ fields are then general complex solutions of your equations.

You have a standing wave, wenn the field is a product of a space-dependent factor with a time dependent function. Your solutions are indeed of this kind if these fields are purely real or imaginary. Usually you get standing waves in a cavity, e.g., the interior of a hollow sphere (or more easy to calculate a cuboid). If you have a wave guide (e.g., a cylinder or a coaxial cable) you have both traveling waves along the axis of the wave guide and standing waves in any plane perpendicular to it.

 

[fog37]

Thank you vanhees71. always appreciate your comments.

A few points:

• I am familiar with the fact that, at the very end of all calculations, we must alway take the real part of the solution field to obtain the final answer since physical fields are always real-valued because measuring devices produce numerical answers with real values.
• In the case of a purely real-valued function $f(x,t)$ is separable, it always represents a standing wave because the spatial and temporal coordinates pertain to the two different functions. For example, in 1D, the real field $E(x,t) = g(x) p(t)$ is separable and does not represent a traveling field.
• However, if the functions $g(x)$ and $p(t)$ are both complex valued, the field could also be a traveling wave and not necessarily a standing wave. I would state that the field $E(x,t) = g(x) p(t)$ is a standing wave when one of the function $g(x)$ and $p(t)$ or both of them are real-valued.

Thanks!

 

[vanhees71]

exactly!

 

[fog37]

Great, thank you!

Here some (possibly correct) examples of 1D field functions representing standing and traveling waves:

Real-valued standing wave: $E(x,y) = (2t^2) (3x)$
Complex valued standing wave: $E(x,t) = (x^3-4) e^{i\omega t}$
Complex valued standing wave: $E(x,t) = e^i(x^3-4) cos(\omega t)$
Complex-valued traveling wave: $E(x,t) = (2x+ i 4x^3) (5t + i t^2)$
Real valued traveling wave: $E(x,t)=3(x-t)^4$