Since the solution of the wave function in vaccum gives two progressive plane waves $f(x-ct)+g(x+ct)$ depending on $x$ the direction of propagation, in the other side we have the $\operatorname{div}(Ex)=0$ so the $E$ field is static so the wave is not propagating!!!. I can't understand the reasoning here and How this explain that the wave is transverse
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1$\begingroup$ $\nabla \cdot \vec{E} = 0$ Does not mean the field is static, which is evident mathematically. $\endgroup$– jensen paullCommented Dec 13, 2022 at 9:27
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$\begingroup$ Electric field is not zero, but it is purely rotational: $\nabla\times \mathbf{E}\neq0$, but $\nabla\cdot\mathbf{E}=0$. Perhaps this answer provides some necessary math background: physics.stackexchange.com/a/597691/247642 $\endgroup$– Roger V.Commented Dec 13, 2022 at 9:27
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$\begingroup$ @RogerVadim I understood your answer. Just to remove the doubt what makes these waves transverse waves? $\endgroup$– Student-qeùtfCommented Dec 13, 2022 at 13:50
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$\begingroup$ If $\nabla\cdot\mathbf{E}=0$ the waves can be only transverse. This is always the case in free space, but in media sometimes can be longitudinal EM waves, due to polarization charges. $\endgroup$– Roger V.Commented Dec 13, 2022 at 14:16
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2 Answers
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A plane wave solution propagating in the $x$-direction has $E_y$ (or $E_z$) equal to $f(x - ct) + g(x+ct)$, and $E_x = 0$. If this is the case, then we have $$ \nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} $$ and we can see that the first term vanishes because $E_x = 0$ and the second two terms vanish because the field does not depend on $y$ or $z$.
If it was the case that $E_x$ depended on $x$, then we would have $\nabla \cdot \vec{E} \neq 0$. So this means that $E_x$ cannot depend on $x$. This means that the electric field in a plane wave must be perpendicular to the direction of propagation — i.e., the wave must be transverse.
answered Dec 13, 2022 at 17:48
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The solution of the wave function in vacuum gives two progressive plane waves $f(x-ct)+g(x+ct)$, which both represent solutions to the wave equation. This can also be written in terms of the electric field, $E(x,t)$, which is related to the wave function by $E=-(1/c)∂ψ/∂t$. Since $E$ is static, the wave is not propagating, but it is still transverse, as the two progressive plane waves are perpendicular to each other. This means that the electric field is oscillating in the perpendicular direction, and thus is transverse (i.e. it has an oscillating component in the x-direction, but no oscillating component in the y-direction).
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$\begingroup$ What does make f(x+ct) and g(x-ct) perpendicular to each other ? $\endgroup$ Commented Dec 13, 2022 at 11:48