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Most diagrams of EM waves show arrows pointing away from (perpendicular to) the axial line of propagation, but I am not sure if this just shows the amplitude of the B and E fields or their lines of force as well...

wiki em wave

I saw one diagram that showed the electric and magnetic field lines pointing parallel to the direction of propagation, instead of orthogonally, but reversing direction every half-wave...

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    $\begingroup$ There is no line of propagation (although there is a direction of propagation). This is a plane wave and the wavefronts are planes. $\endgroup$
    – G. Smith
    Dec 26, 2020 at 19:56

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I believe that in this diagram we are presented with a plane electromagnetic wave whose $\vec{E}$ field is given by $$\vec{E} = E_0 \cos(kz-\omega t) \hat{e}_x$$

Now let's try to understand this equation. First you will notice that the $\vec E$ is directed only in the $\hat e_x$ direction. So all the arrows representing $\vec E$ must be parallel to the $x$-axis. So it is actual vectors of electric field that are pointed as shown in the diagram and their length represents their intensity.

Next we can do is to fix $t$ to analyze the field in space at some specific moment.For $t=0$ the equation becomes: $$\vec E = E_0 \cos(kz) \hat e_x$$ This means that if you move along the $z$-axis, the intensity of the field will vary. Cosine has maxima at $2n\pi, n \in \mathbb Z$ and minima at $(2n+1)\pi, n \in \mathbb Z$. This means that $\vert \vec E \vert$ has maxima at $z= \frac{2\pi}{k}\cdot n$ and minima ($\vec E$ is directed along negative $\hat e_x$ direction) at $z = \frac{(2n+1)\pi}{k}$. This is exactly what you can see in the diagram. Even though $\vec E \ || \ \hat e_x$ at every point in space, the intensity of $\vec E$ is distributed like cosine along $z$-axis. Another thing to be aware of is that because nothing depends on $y$, you can translate this cosine along $y$-axis to get something like this: blue = x; red = y; green = z

This shows only the intensity of field at all points on the gray plane. Exact same result you would get for any plane parallel to it.

And finally, at some fixed point $z=0$, field changes with time as follows: $$\vec E(t) = E_0 \cos(-\omega t) \hat e_x$$ So again, just like a cosine.

Maybe it would be helpful to think of each point in space as a seat on a stadium, and every fan is the electric field vector. Now the fans together decide to perform a stadium wave.

  • every fan only stands up or sits down remaining in the vertical position, just like the $\vec E$ field is always directed along $x$-axis here
  • all the fans having seats one above the other always stand and sit at the same time, just like $\vec E$ is totally independent of $y$ direction
  • because of coordination of all fans, the wave travels along the stadium, just like the field wave does along $z$-axis.

I saw one diagram that showed the electric and magnetic field lines pointing parallel to the direction of propagation, instead of orthogonally, but reversing direction every half-wave...

As far as I know, Maxwell's equations tell us that EM waves can only be transverse, never longitudinal in vacuum at least. Suppose we had $\vec E$ field pointing in $z$ direction while also traveling in the same direction:

$$\vec E = E_0 \cos(kz-\omega t) \hat e_z \\ \implies \nabla \cdot \vec E = -E_0 k \sin(kz-\omega t) \not \equiv 0$$

But in vacuum where there are no charges floating around, Gauss' law gives us: $$\nabla \cdot \vec E = \frac{\rho}{\epsilon_0} = 0$$

Hence this wave contradicts the first Maxwell equation in vacuum (though not necessarily elsewhere).

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