1
$\begingroup$

I understand that in vacuum, monochromatic plane waves are transverse. I suppose that a non-monochromatic plane wave would also have this property, as it can be decomposed using the Fourier transform and each of its components would be transverse.

However, what happens if the waves are not plane? Are they still transverse? The only demonstrations I've found are used for plane waves, so I don't know if this is a necessary condition or is just done for simplicity but can be generalized.

$\endgroup$
  • 1
    $\begingroup$ Read this, in particular "Maxwell's equations do lead to the appearance of longitudinal waves under some circumstances, for example, in plasma waves or guided waves". $\endgroup$ – Carucel Jan 8 '17 at 13:26
  • $\begingroup$ @Carucel Is there a demonstration for that (i.e. all waves being transverse in vacuum)? For example, in Griffiths' Introduction to Electrodynamics it is asumed that the wave is plane. $\endgroup$ – Tendero Jan 8 '17 at 13:29
  • $\begingroup$ I don't exactly get your point, sorry. It is assumed there that "we are in free space". Also, the first comment there by Mark K Cowan, says "EM waves can have longitudinal components if we avoid the plane wave approximation. See MJ Cliffe, S Jamison for more" $\endgroup$ – Carucel Jan 8 '17 at 13:32
  • $\begingroup$ @Carucel My question is if, in free space, all EM waves are transverse or if they have to be plane in order to be transverse. $\endgroup$ – Tendero Jan 8 '17 at 13:33
  • 1
    $\begingroup$ Aha, I misunderstood your question. I do not know the proof, but "electromagnetic waves in free-space are always transverse", see this. $\endgroup$ – Carucel Jan 8 '17 at 13:37
5
$\begingroup$

It depends on what you mean by "EM waves are transverse". If you mean "the $\mathbf E$ and $\mathbf B$ fields are always orthogonal to the direction of propagation", where you're willing to take a case-by-case approach to the direction of propagation (or you want to use e.g. the time-averaged Poynting vector or some similar measure for the direction of propagation) then the assertion is false.

For most working physicists, however, that's a bit of a wonky understanding of what 'transverse' means, particularly for fields with a complex spatial dependence where you do not have a well-defined direction of propagation. In those cases, as AccidentalFourierTransform points out, the correct understanding of the phrase is purely in terms of the Maxwell scalar equations: $$ \nabla\cdot\mathbf E = 0 = \nabla\cdot\mathbf B. $$ This gives you your transversality constraint: it's phrased in a wonky differential-equation language, but at each point it essentially gives you a direction in which each field cannot point, based on its local coordinate dependence.

On the other hand, this can give some weird results if you want to stick to some more intuitive understanding of the "propagation direction" of the field. The clearest example of this is in a waveguide, which has a well-defined "direction" in the axis of the waveguide, but for which the electric or magnetic fields can have nonzero components in that direction. These components are typically small (and they vanish in the limit where the wavelength is much smaller than the waveguide width) but they are nonzero, and they can be important for applications. (For an example of these components in action (and/or shameless plug), see this paper.)

In essence, the way this works is that a confined waveguide mode, such as $$ \mathbf E(\mathbf r,t) = E_0\hat{\mathbf e}_y \sin(x/L)\cos(kz-\omega t), $$ is only confined because it is a standing wave in the $x$ direction, and this is best seen as the superposition of two travelling waves: $$ \mathbf E(\mathbf r,t) = \mathrm{Re}\mathopen{}\left( E_0\hat{\mathbf e}_y \frac{1}{2i}\left(e^{i(x/L+kz-\omega t)}-e^{i(-x/L+kz-\omega t)}\right) \right)\mathclose{}. $$ Here both waves are transverse, and (assuming $kL\ll 1$) they both have $\mathbf k$ vectors close to but not quite on the waveguide axis $\hat{\mathbf e}_z$. Thus, the electric and magnetic fields for each component are orthogonal to the propagation direction of each plane wave, but since there is more than one such propagation direction, the global field gets a bit confused about what that term actually means.

The specific example above is a transverse electric (TE) mode, with the electric field orthogonal to the waveguide axis, so it may sound a bit confusing, but with a bit of thought you can see that the corresponding magnetic field needs to point in the $y,z$ plane and therefore it's a bit more complicated. Finding this magnetic field is easy, since we can just plop in the corresponding field for each plane-wave component, but now you need to account for the fact that the two magnetic components are not parallel: \begin{align} \mathbf B(\mathbf r,t)& = \mathrm{Re}\mathopen{}\left( B_0 \frac{1}{2i}\left( \frac{-\hat{\mathbf e}_x-kL\hat{\mathbf e}_z}{\sqrt{1+k^2L^2}} e^{i(x/L+kz-\omega t)} -\frac{-\hat{\mathbf e}_x+kL\hat{\mathbf e}_z}{\sqrt{1+k^2L^2}} e^{i(-x/L+kz-\omega t)}\right) \right)\mathclose{} \\& = -\frac{B_0\hat{\mathbf e}_x}{\sqrt{1+k^2L^2}}\sin(x/L)\cos(kz-\omega t) -B_0\frac{kL\hat{\mathbf e}_z}{\sqrt{1+k^2L^2}}\cos(x/L)\sin(kz-\omega t) . \end{align} In particular, note that now you have a nonzero $\hat{\mathbf e}_z$ component (though again this is small when $kL\ll 1$). Does this make the magnetic field stop being "transverse"? That's up to what you want to make of the term, really.

While this is the simplest example, it should be clear that the situation is pretty generic any time that vector optics come into play. Thus, you have equivalent effects in a tight Gaussian focus (example), general waveguides (example), spherical waves (cf. Jackson 3rd ed. §9.7), and so on. These admit a similar explanation to the above in terms of a continuous superposition of plane waves (i.e. they can be decomposed as Fourier transforms of a bunch of transverse plane waves with $\mathbf k\cdot\mathbf E=0 = \mathbf k \cdot\mathbf B$), but anytime a point gets significant contributions from multiple different directions of $\mathbf k$ vectors then you will have some trouble establishing a unique 'direction of propagation'.

$\endgroup$
6
$\begingroup$

As discussed in the comments, the plane wave approximation is mostly irrelevant. The only important ingredient is the fact that the homogeneous vacuum Maxwell equations read $$ \nabla\cdot\boldsymbol E=\nabla\cdot\boldsymbol B= 0 $$ and therefore, after a Fourier transform $\boldsymbol x\to\boldsymbol k$ we get $$ \boldsymbol k\cdot\boldsymbol E=\boldsymbol k\cdot\boldsymbol B= 0 $$ which means that the fields are transverse.

In practice, the most simple treatment of electromagnetic waves is through the vector potential. The equations are simplest in the Lorenz gauge $$ \partial_\mu A^\mu=0 $$ which, again, in Fourier space becomes $k\cdot A=0$, i.e., the vector potential is transverse. Other very useful gauge is the radiation gauge $$ \nabla\cdot \boldsymbol A=0 $$ which, again, implies that $\boldsymbol A$ is transverse: $$ \boldsymbol k\cdot\boldsymbol A=0 $$

$\endgroup$
  • $\begingroup$ What if the sources are inside the integration domain? I mean, for EM waves to exist there have to be sources somewhere, and our decision of letting them be inside or outside the enclosure being analyzed shouldn't change the physical reality of the problem. So how can one prove that, if the sources are taken into account, the waves are still plane? And if they are not, how can that make sense if it would be the same problem as before but with a different mathematical approach? $\endgroup$ – Tendero Jan 8 '17 at 17:45
  • $\begingroup$ If you have sources, then EM waves are no longer transverse - you're no longer in vacuum, and div E is no longer zero. (Thus e.g. plasmons can be longitudinal.) However, your question explicitly asks for results in vacuum, and it's a bit late to turn the tables. That said, the presence of sources has nothing to do with whether the waves are plane waves or not. Most waves are not plane, but you can have plans waves with sources (a flat, infinite plane of current) and longitudinal plans waves (in the presence of a free charge bull density). $\endgroup$ – Emilio Pisanty Jan 9 '17 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.