And there I go drifting off topic again
ap_cycles said:
Hi,
My question is as per the title. Why 90 deg and no other angle else? I did a thorough Net Search and could find no satisfactory answer. Can it be explained using Maxwell's Equation? (Sorry, i don't really understand Maxwell's equations)
I am a teacher who is apprehensive about teaching this topic due to its abstractness (to me anyway).
Happy New Year to all!
There is a derivation of the mathematical description of electromagnetic waves on Wikipedia:
http://en.wikipedia.org/wiki/Electromagnetic_waves
Part of the derivation shows why the electric and magnetic fields have to be perpendicular to the direction of propagation. Basically, by putting together two of Maxwell's equations, you obtain the wave equation
\biggl[\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\biggr]f(x, y, z, t) = 0
where
f is any component of either an electric or magnetic field. By solving this equation, you find that the solutions have to be proportional to
f(k_x x + k_y y + k_z z + \omega t)
where k_x^2 + k_y^2 + k_z^2 = \omega^2/c^2. (Any function that depends on x, y, z, and t in that particular combination, and only that particular combination, satisfies the wave equation.) The vector with components k_x,k_y,k_z, the wavevector, points in the direction of propagation.
Now, the other two of Maxwell's equations say that the fields are divergence-free; that is,
\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} = 0
and similarly for the magnetic field. Remember that each of these components has to take a form like
E_x = E_{(0)x}f(k_x x + k_y y + k_z z + \omega t)
where E_{(0)x} is some amplitude. Anyway, if you put this type of function into the divergence-free condition from a couple lines above, you find that
E_x k_x + E_y k_y + E_z k_z = 0
which is just saying that
\vec{E}\cdot\vec{k} = 0
This tells you that the electric field is perpendicular to the direction of propagation.
By the way, I commend you on taking the time to learn the material well before you teach it. It's disappointing how often people seem to not do that.