# The amplitude for the electric field of a particular electromagnetic plane is given in the direction $\mathbf{\hat{j}}$?

Example 2.6 of Optics, Fifth Edition, by Hecht, is as follows:

We will see in the next chapter, electric field ($$\mathbf{E}$$) of a particular electromagnetic plane wave can be given by the expression

$$\mathbf{E} = (100 V / m)\mathbf{\hat{j}} e^{i(kz + \omega t)}$$

(a) What is the amplitude of this wave in the electric field?

For reference/clarification, the following are illustrations from the same section of the textbook (although, not directly referenced in the aforementioned example):

The amplitude is said to be $$100 V/m$$.

If my understanding is correct, the equation of the complex form of a harmonic wavefunction is

$$\psi(\mathbf{r}) = Ae^{i \mathbf{k \cdot r}},$$

where $$\mathbf{r}$$ is the position vector of the plane wave and is perpendicular to a given vector $$\mathbf{k}$$, and $$A$$ is the amplitude of the wave. The resemblance between this form of the wavefunction and the electric field of a particular electromagnetic plane wave $$\mathbf{E} = (100 V / m)\mathbf{\hat{j}} e^{i(kz + \omega t)}$$ is obvious. However, one difference is that the amplitude for the electric field of a particular electromagnetic plane is given in the direction $$\mathbf{\hat{j}}$$.

I don't understand why the amplitude is given a direction, nor do I understand what this means physically. I would greatly appreciate it if people could please take the time to explain this.

## 2 Answers

$$\mathbf{\hat{j}}$$ is the direction of the electric field, which is a vector and has a direction. In an electromagnetic wave the direction is perpendicular to the direction of the wave propagation, which in this case is in the $$\mathbf{\hat{k}}$$,that is, z-axis, direction. The scalar equation for a wave that you wrote does not need a direction because it is a scalar wave, in which A is some scalar quantity, such as temperature.

Also, it is not uncommon to write the wavefunction of a vector quantity as a scalar one, because you might be interested only in the intensity, not in the direction

As you know the electric field is a vector quantity, so:

$$\vec{E} = (E_x, E_y, E_z)^T$$

Your notation is a bit confusing as you use z and j as directions, so let

$$\hat{i} = \hat{e}_x$$ $$\hat{j} = \hat{e}_y$$ $$\hat{k} = \hat{e}_z$$

Also let $$E_0 = 100 V/m$$. You then have

$$\vec{E} = E_0 e^{i(kz-\omega t)} \hat{e}_y$$

for the electric field. As you may know $$\vec{k}$$ tells you in which direction your plane wave is moving and the unit vector $$\hat{e}_y$$ in which direction your electric field is oscillating.

You may also write:

$$\vec{E} = \vec{E}_0 e^{i(kz-\omega t)} \;\text{where}\; \vec{E}_0 = E_0 \hat{e}_y$$

which means the same. It just tells you in which direction the electric field is pointing/oscillating.

This is important for example for polarized light where your electro-magnetic-field may oscillate in the (x-y)-plane while propagating in the z-direction. Note that you mostly just work with the electric field. Maybe the following illustration helps you.

On the far left side you can see the vector arrows of the field for different angles. For example in the case of the red wave the field only oscillates in the y direction while as for the pink its oscillating in both the x and the y direction.

I hope that helps. If you have any question or something is unclear feel free to ask.

• I have posted an image from the same section of the textbook. This might clarify the choice of coordinates. Dec 12, 2019 at 23:22
• Where is the j direction in your picture? And as far as I know its either ijk or xyz.
– Tera
Dec 12, 2019 at 23:34
• I think $y$ is the $j$-direction. I posted another image from the same section of the textbook, but presented earlier. Dec 12, 2019 at 23:38
• Seems like it. I don't know why they did it so unnecessarily complicated . But it's essentially what I wrote with the unit vectors.
– Tera
Dec 13, 2019 at 8:22