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Practically all sources I found through google search say that electromagnetic plane wave in vacuum is transverse. For example: https://en.wikipedia.org/wiki/Transverse_wave

I think the definition of transverse and longitudinal is hard to apply to EM waves because there is no medium that moves perpendicular or parallel to the direction of the advancement of the wave. But if we really want to apply the definition I think EM plane wave is longitudinal in vacuum.

I understand that E vectors and B vectors are perpendicular to each other and the direction of advancement (k). But E and B are just representations of quantities in one point of space. No medium moves in the direction perpendicular to k. On the other hand E and B is oscillating along the direction of k, which reminds me of longitudinal waves.

Please correct my logic. Thank you for your time.

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  • $\begingroup$ If i understand corectly, you have an issue with the direction part of the vector, not the magnitude. If so, for me the solution is to always think of having a test charge that when placed in a field gives the direction of the field. So mathematically you have a magnitude (how fast it is accelerated) and a direction. $\endgroup$ Commented Feb 16, 2023 at 19:49

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Transverse simply means that the polarization vector of the EM wave is orthogonal to the wave vector. That is the definition of the term. It applies to free EM waves because $\vec{E}$, $\vec{B}$, $\vec{k}$ are enforced to be orthogonal by imposing the constraints of Maxwell's equations on the decoupled EM wave equations.

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The statements I understand that $\bf E$ vectors and $\bf B$ vectors are perpendicular to each other and the direction of advancement ($\bf k$). and . . . $\bf E$ and $\bf B$ is oscillating along the direction of $\bf k$, . . . . . are contradictory statements with the second one being incorrect.

The direction of energy propagation or Poynting vector (for electromagnetic waves) can be used as the reference direction to differentiate between transverse and longitudinal waves.

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  • $\begingroup$ On this picture en.wikipedia.org/wiki/Transverse_wave#/media/…, we take a point in time, then we can say that E and B is oscillating along the Y axis, can't we? $\endgroup$
    – David
    Commented Feb 16, 2023 at 11:42
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    $\begingroup$ I think I get it. E and B are perpendicular to Poynting vector and we call this transverse wave, end of story. I shouldn't try to make this analogous to mechanical waves, where some medium actually moves. $\endgroup$
    – David
    Commented Feb 16, 2023 at 11:56
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As mentioned in another answer, it is incorrect that ${\bf E}$ and ${\bf B}$ are "oscillating along the direction of ${\bf k}$".

Blackbody radiation in a conducting cube is a nice way to see that the electric field must always point perpendicular to the wave vector. If you solve the electric field wave equation inside of a conducting cube of length $L$, you find that the electric field has components: $$ E_x({\bf r}, t) = E_{0x} \, \sin(\omega_{\bf n} \,t) \, \cos \left( \frac{n_x \pi x}{L} \right) \, \sin \left( \frac{n_x \pi y}{L} \right) \, \sin \left( \frac{n_z \pi z}{L} \right) $$ and similar expressions for $E_y$ and $E_z$, where ${\bf n}$ is a vector of non-negative integers. We get a countable set of solutions, labeled by ${\bf n}$, because of the finite box. The particular dependence on the $\cos$ and $\sin$ functions ensure that, at the conducting boundary, the component of the electric field parallel to that boundary vanishes (so that there is no moving charge in the conductor). An arbitrary electric field solution is found as a superposition of these ${\bf E}^{({\bf n})}$ solutions.

But there is one more constraint from Maxwell's equations that has not yet been applied. It turns out that the direction of the electric field vector, ${\bf E}_0 = \left( E_{0x}, E_{0y}, E_{0z} \right)$ is not allowed to be completely arbitrary. Applying Gauss's Law (in free space) to the above solution yields: $$ \nabla \cdot {\bf E} = 0 \quad \rightarrow \quad E_{0x} n_x + E_{0y} n_y + E_{0z} n_z = 0 \quad \rightarrow \quad {\bf E}_0 \cdot {\bf n} = 0 $$ Thus we see that for a particular solution labeled by ${\bf n}$, the electric field vector is always perpendicular to ${\bf n}$. It turns out that ${\bf n}$ is also the direction of propagation: it is the direction you get from ${\bf E} \times {\bf B}$.

Applying a similar analysis to a cubic mass-and-springs model of a solid yields almost the same solutions (because we are again seeking solutions to a wave equation). But in that case there exists no physical law that displacements be divergence-free, so longitudinal oscillations are allowed.

So, both EM waves and mechanical waves are determined by solving a wave equation. But EM waves must also satisfy the other parts of Maxwell's equations, namely Gauss's Law and $\nabla \cdot {\bf B} = 0$. These prohibit longitudinal oscillations of the the electric and magnetic fields.

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