# Determine the direction of electric and magnetic field in for plane EM wave

A problem states that

Measurement of the electric field (E) and the magnetic field (B) in a plane-polarized electromagnetic wave in vacuum led to the following: $$\begin{array}{ll} \frac{\partial E}{\partial x}=\frac{\partial E}{\partial y}=0 & \frac{\partial E}{\partial z}=-\frac{\partial B}{\partial t} \\ \frac{\partial B}{\partial x}=\frac{\partial B}{\partial y}=0 & \frac{\partial B}{\partial z}=+\frac{\partial E}{\partial t} \end{array}$$ ,then what can one conclude about the directions of the electric and magnetic fields?

The two equations on the left side imply that the electric and magnetic fields are constant in the xy-plane. So our wave is a plane wave in the XY plane and therefore the direction of propagation is perpendicular to the xy plane. But this does not specify the precise direction of the electric and magnetic fields.Neither am I able to use the other time derivative equations to conclude anything. Could anyone please help me or hint me .

Also this is not a homework problem as I am self studying the subject and I will be grateful for any help. Thank you.

• Even if you can't say precisely what direction the E field is pointing in, can you say anything about its components? For example, is there a direction in which the component of the E field has to be zero? – Andrew Nov 14 '20 at 13:03
• Thank you Andrew for your time. The $z$ component of E is zero but still that doesn't specify the direction of E field. – Kashmiri Nov 14 '20 at 14:04
• That's right. We could ask if it is surprising that there is no direction specified. What happens to the equations if you apply a rotation about the z axis? – Andrew Nov 14 '20 at 14:12
• I think they remain the same though I'm not 100 percent sure. – Kashmiri Nov 14 '20 at 15:03
• Exactly. From this fact you can conclude it's impossible for you to pick out a unique direction for the E field. Let's say you found the E field pointed in the x direction. But now I can decide to rotate my head by 90 degrees, so that the E field now points in the y direction. The equations are the same in my new frame of reference (as you said, after a rotation the equations are the same). So E pointing in the y direction must also be a valid solution. Therefore, these equations do not have a unique solution for the direction of E. – Andrew Nov 14 '20 at 15:09

You can determine that the direction of the $$E$$ and $$B$$ fields are perpendicular to the direction of motion, and you can conclude that $$E$$ and $$B$$ are perpendicular. But you can't determine the specific polarization (ie in which direction $$E$$ is pointing) based on the information given.

This follows since the equations are invariant under rotations about the $$z$$ axis. Let's say you found the $$E$$ field pointed in the $$x$$ direction. But now I can decide to rotate my head by 90 degrees, so that the $$E$$ field now points in the $$y$$ direction. The equations are the same in my new frame of reference (as you said, after a rotation the equations are the same). So $$E$$ pointing in the $$y$$ direction must also be a valid solution. Therefore, these equations do not have a unique solution for the direction of $$E$$.

Let's work out a solution to the equations.

First, let's start with the vector form of Maxwell's equations in vacuum (setting $$c=1$$).

$$\begin{eqnarray} \frac{\partial \vec{E}}{\partial t} &=& \nabla \times \vec{B} \\ \frac{\partial \vec{B}}{\partial t} &=& -\nabla \times \vec{E} \\ \nabla \cdot \vec{E} &=& 0 \\ \nabla \cdot \vec{B} &=& 0 \end{eqnarray}$$ Now we assume a plane wave moving in the $$z$$ direction, so that $$\frac{\partial \vec{E}}{\partial x}=\frac{\partial \vec{E}}{\partial y}=\frac{\partial \vec{B}}{\partial x}=\frac{\partial{\vec B}}{\partial y}=0$$. The last two of Maxwell's equations then imply $$\frac{\partial E_z}{\partial z}=\frac{\partial B_z}{\partial z}=0$$ (the fields have to be perpendicular to the propagation direction).

Now let's assume $$\vec{E}$$ points in the $$x$$ direction and $$\vec{B}$$ points in the $$y$$ direction. (For linearly polarized waves, we don't lose any generality here since we can always rotate our coordinate system). Then the non-trivial components of the first two equations become $$\begin{eqnarray} \frac{\partial E_x}{\partial t} &=& (\nabla \times \vec{B})_x = \frac{\partial B_z}{\partial y} - \frac{\partial B_y}{\partial z} = -\frac{\partial B_y}{\partial z} \\ \frac{\partial B_y}{\partial t} &=& -(\nabla \times \vec{E})_y = -\left[\frac{\partial E_x}{\partial z} - \frac{\partial E_z}{\partial x} \right]= -\frac{\partial E_x}{\partial z} \\ \end{eqnarray}$$ Then a solution like $$E_x = A \cos\left(k(t-z)\right)\hat{e}_x, B_y=A\cos\left(k(t-z)\right)\hat{e}_y$$ will solve the equations.

Another interesting example of a solution is circular polarization $$\begin{eqnarray} \vec{E} &=& \cos(k(t-z)) \hat{e}_x + \sin(k(t-z))\hat{e}_y \\ \vec{B} &=& -\sin(k(t-z)) \hat{e}_x + \cos(k(t-z))\hat{e}_y \end{eqnarray}$$ It's a good exercise to see how this solves Maxwell's equations.

• If you have some time to spare then could you give me an example that satisfies the above equations. I tried to fit in $E=cos(wt-kz)$ and other variations of it but they dont satisfy the time dependent equations, they differ by a negative sign. – Kashmiri Nov 15 '20 at 4:57
• I added some comments. I'm not quite sure how your book gets the equations they wrote down. I do recommend working through the circular polarization example, it's a good exercise. – Andrew Nov 15 '20 at 7:23

The direction of the fields (perpendicular to the direction of propagation) is determined by the source of the wave. If you consider a positive charge oscillating back and forth, the motion of the charge introduces a transverse component into the (preexisting) electric field (in the direction of motion). The motion also produces a magnetic field (wrapped around the direction of motion, as predicted by a right-hand-rule). Both of these disturbances are at a maximum when the velocity is a maximum, and they move away from the charge with a velocity predicted by Maxwell's equations. It would appear that the cross product BxE shows the direction of propagation.