The $x$-component of a circular polarized plane wave is

$$ E_x(\vec r,t)=E_0\cos\left(\frac{w}{c}(0.6y-0.8z)-wt\right) $$

With only this given, we can devise the total electric field as

$$ \vec E(\vec r,t)=E_0 \left[\cos\left(\frac{w}{c}(0.6y-0.8z)-wt\right) \hat x \pm \sin\left(\frac{w}{c}(0.6y-0.8z)-wt\right)(0.8 \hat y + 0.6 \hat z) \right]$$

When looking for the total electric field, we first need to define the wave vector, which is $\vec k = \frac{w}{c}(0.6\hat y - 0.8\hat z)$. We know that $\vec k \cdot \vec E = 0$, which is already satisfied for the $x$-component of our electric field.

Since we want $\vec k \cdot \vec E = 0$ to be satisfied in the $y,z$-directions aswell, we need to add a term to our total electric field which becomes zero when multiplied by $\vec k$, this is represented by $(0.8 \hat y + 0.6 \hat z)$ in our answer, since $\vec k \cdot (0.8 \hat y + 0.6 \hat z) =0$. What I don't understand in this question is why the second term needs to be a sine-term, and not just be attached to the cosine? The answer would then look like

$$ \vec E(\vec r,t)=E_0 \left[\cos\left(\frac{w}{c}(0.6y-0.8z)-wt\right) (\hat x + 0.8 \hat y + 0.6 \hat z) \right]$$

But this is not a correct answer, because according to my lecture notes, $E_{y,z}$ needs to be phase shifted 90 degrees, which is done using sine instead of cosine. Any help understanding why this is would be greatly appreciated.

  • $\begingroup$ Hint: calculate the magnitude of the total electric field at a fixed point as a function of time. For a circularly polarized field this should be constant; for a linear polarization it will oscillate all the way down to zero. $\endgroup$ – Emilio Pisanty Nov 20 '18 at 19:40

Your solution gives a linearly polarized plane wave, not a circularly polarized one.

, i don't understand why we need to phase shift our electric field (the y and z components of E) through the sine-term

This is what it means to be circularly polarized.

Think about a simpler situation where the field only has x and z components. Instead of the x and z components both going to zero at the same time, one is 90 degrees out of phase from the other, so $\vec{E}$ traces a circle over time:

enter image description here

(image source)

The wave in your example is the same as this, only the basis vectors are rotated so that propagation is not exactly along one particular axis.

As a key point, the magnitude of $\vec{E}$ is constant in time for a circularly polarized wave, but it varies between 0 and $E_0$ for a linearly polarized wave.

  • $\begingroup$ ok, thanks. if you have any intuition in regards to why that is, i would love to understand more. i've done some research but i can't seem to get a hang of it, i don't understand why we need to phase shift our electric field (the y and z components of E) through the sine-term $\endgroup$ – armara Nov 20 '18 at 18:24

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