# Why is the electric field phase shifted in this circular plane wave?

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.

• 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. Nov 20, 2018 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: (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.

• 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 Nov 20, 2018 at 18:24