Reverse engineering formulas for directing a plane wave

[yefj]

Homework Statement

plane wave

Relevant Equations

plane wave

Hello,There is a model which tunes a plane wave using certain expressions which resembles the spherical to cartesian coordinates.
There are two types of definitions:
propagation normal and electric field vector.
Why they put in propagation normal exactly the spherical to cartesian to represent propagation normal?
What is the link between propagation normal and E-field formulas?
Thanks.

 

[kuruman]

Why they put in propagation normal exactly the spherical to cartesian to represent propagation normal?

It looks like you have a spherical wave. The direction of propagation is radially out and a wavefront is a sphere of radius $R$. This means that the direction of propagation at any point on the sphere is radially out, i.e. normal to the surface. A unit vector normal to the surface of a sphere is $$\mathbf{\hat n}=\sin\!\theta\cos\!\phi~\mathbf{\hat x} +\sin\!\theta\sin\!\phi~\mathbf{\hat y}+\cos\!\theta~\mathbf{\hat z}$$ which is the same as the radial unit vector $\mathbf{\hat r}$ along which the wave propagates.

 

[yefj]

how does E-field formulas were derived from the normal?
Thanks.

 

[kuruman]

how does E-field formulas were derived from the normal?
Thanks.
View attachment 355731

The E-field vector is in the plane perpendicular to the normal. In that plane there is an infinity of directions that the E-field vector can have. Here you are given one of many. You can verify that this is a valid E-field by showing that $~\mathbf E\cdot \mathbf{\hat n}=0.$

For a given normal you can get all the possible electric fields by varying angle $\alpha$ which rotates the electric field in the plane perpendicular to the normal. You can easily see how this works for specific choices of the angles.

For example, if I choose $\theta=\dfrac{\pi}{2}$ and $\phi=0$, I get
(a)$~~\mathbf{\hat n}= \mathbf{\hat x}$,
(b)$~~\mathbf{\hat E}= \sin\!\alpha~\mathbf{\hat y}-\cos\!\alpha~\mathbf{\hat z}$
The above says that (a) the normal is along the $x$-axis and (b) a unit vector in the direction of the electric field is in the $yz$-plane and changes direction as $\alpha$ changes.

 

[yefj]

Hello Kuruman, I understand that E-field can have endless options.
What is the logic for choosing the expression for e-field?
Thanks.

 

[kuruman]

Hello Kuruman, I understand that E-field can have endless options.
What is the logic for choosing the expression for e-field?
Thanks.

If you understand that the E-field has infinitely many options and that it must be perpendicular to the direction of propagation, there is no logic for choosing a particular direction as opposed to another. Pick a direction and move on with your life. That's what the person that made the simulation that you posted most probably did understanding that it is necessary to make a choice before drawing the surface, the normal and the E-field.