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Electromagnetic waves are generally depicted like this:

enter image description here

Where the electric fields and magnetic fields exist in the planes perpendicular to the direction of propagation. I also realize that as the electric field changes while the wave is propagating, a magnetic field is induced and vice versa (by faraday's and maxwell's laws of induction). But, those laws predict that the fields will be circular. So, won't the electric and magnetic fields look different? Won't they be circles along arrows that are drawn in the figure? I haven't seen anything written about this anywhere.

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    $\begingroup$ We need clarification. Why do you say that Maxwell's equations predict that the fields will be circular? $\endgroup$ – garyp Jul 9 '14 at 13:24
  • $\begingroup$ @garyp: Because take for instance Faraday's Law: $$\oint \vec{E}\cdot d\vec{s} = -\frac{d\Phi}{dt}$$ When we integrate around the closed loop, by symmetry, won't the fields be circular? $\endgroup$ – Gerard Jul 9 '14 at 13:29
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    $\begingroup$ More on E and B fields in EM wave: physics.stackexchange.com/q/20331/2451 and links therein. $\endgroup$ – Qmechanic Jul 9 '14 at 13:53
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    $\begingroup$ @Gerard You've got it the other way around. If the fields have circular symmetry, then those integrals are easy to do. But if they don't, then the laws are still valid. The problem is that the line integrals will become much harder. $\endgroup$ – Javier Jul 9 '14 at 13:59
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    $\begingroup$ @Gerard: By studying vector calculus and applying it to the differential form of Maxwell's equations, as in Kyle's answer. $\endgroup$ – Javier Jul 9 '14 at 14:12
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You write the integral formulation of Faraday's law, but there is also the equivalent differential formalism: $$ \nabla\times\mathbf E=-\frac{\partial\mathbf B}{\partial t}\tag{1} $$ which can be proven in a straight-forward manner (and ought to be done in your standard E&M textbooks).

Using standard planar waves equations, $$ E_y=E_0\sin\left(kx-\omega t\right) \\ B_z=B_0\sin\left(kx-\omega t\right), \\ $$ then Equation (1) gives us that $$ kE_0\cos(kx-\omega t)=\omega B_0\cos(kx-\omega t) $$ Which enforces the well-known relation that $E_0=cB_0$. So clearly plane waves do satisfy Faraday's law.

A similar situation exists with Ampere's law, usually written as $$ \oint\mathbf B\cdot d\mathbf l=\frac{1}{c^2}\frac{d\phi_E}{dt} $$ which leads to $$ \nabla\times\mathbf B=\frac{1}{c^2}\frac{\partial\mathbf E}{\partial t} $$ (I'm ignoring the current density here).

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  • $\begingroup$ I'm not familiar with curl and divergence operations. I would be grateful if you could answer without using them. Thanks. $\endgroup$ – Gerard Jul 9 '14 at 13:56
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    $\begingroup$ I am not sure how you can be familiar with line integrals but not with curl. This website provides a good tutorial on the connection of the line integral with the curl. $\endgroup$ – Kyle Kanos Jul 9 '14 at 13:59
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    $\begingroup$ The derivation of the wave equation starting at the integral form of Maxwell's equations (doesn't use div or curl) which leads to the figure shown can be found in most intro textbooks. It's worth a close study. It's also longer than the derivation from the differential form, which probably explains why I can't find a version of the derivation on the web. Perhaps someone can find the derivation and post the link here. $\endgroup$ – garyp Jul 9 '14 at 14:37
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    $\begingroup$ Here's a derivation, starting on page 13-7. The figure you show pops up as Figure 13.4.5. You can see why the differential form is more commonly used. (I found a few other derivations, but they were either typographically garbled, or lacked figures.) $\endgroup$ – garyp Jul 9 '14 at 14:55
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You shouldn't think of electromagnetic wave as the simple figure depicted as an arrow and two planes. That's too simplified illustration and surely misleading in your case.

Electromagnetic waves fill certain 3D space. It's actually hard to accurately visualize it with a picture. In mathematical and a bit more rigorous language, at every point (x, y, z) in the space where electromagnetic waves exist, there's electric and magnetic field with value E(x, y, z) and H(x, y, z). Also they vary with time, so in the end E(x, y, z, t) and H(x, y, z, t) represent the wave. These two functions satisfy Maxwell's equations.

As for the familiar picture of electromagnetic wave, it's trying to describe a plane wave in the whole space (let's say propagating in x direction), which means fields are varying in sinusoidal and cosinunoidal way with x, but invariant with y and z. In this sense, it allows to use the oversimplified arrow-plane illustration.

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  • $\begingroup$ Is there any approximate illustration available that might provide a better understanding of the wave structure? $\endgroup$ – Gerard Jul 9 '14 at 14:35
  • $\begingroup$ well... strictly speaking you will need 2 arrows at every point to illustrate it. So You'd better just use your imagination :) $\endgroup$ – Pu Zhang Jul 9 '14 at 15:10

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