# Intro Electromag Question - Wave Equation

1. Apr 17, 2012

### Kyle91

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

b) I could figure it out if kz was changed to kx...

Double Derivative of E(r, t) with respect to x is = 0

Double Derivative of E(r, t) with respect to t is = -ω2*E0*cos(kz - wt + ∅0)

Multiply the second term by k22 doesn't help.

2. Apr 18, 2012

### collinsmark

For this problem, you probably could do that without any loss of generality if you wanted to (however, see below for a better way). The z in $$\vec E(\vec r, t) = (1,0,0) E_0 \cos (kz -\omega t + \phi_0)$$ just means that the wave is traveling along the z direction. If it happened to be traveling along the x direction it wouldn't change this particular problem in any important way (although you might want to change the "(1,0,0)" to ensure that the electric field is always perpendicular to the direction of propagation -- but again, see below for a better approach).

The 'x' in the

$$\frac{\partial^2 \psi}{\partial x^2} - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} = 0$$

wave equation simply refers to the spacial dimension (instead of time dimension). The wave equation doesn't imply that all waves *must* travel in the North-South direction only and that East-West and Up-Down directions are prohibited. It just means "spacial". That's because the version of the wave equation that you gave is the "one dimensional" wave equation.

For a three denominational version, use this version of the wave equation:

$$\nabla^2 \psi - \frac{1}{c^2} \frac{\partial^2 \psi}{\partial t^2} = 0$$

where $\nabla^2$ is the Laplacian operator. In Cartesian coordinates, the Laplacian is

$$\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$

---------------
Edit: It doesn't matter much for this particular problem, but I couldn't tell if the (1,0,0) notation was Cartesian (x,y,z) or cylindrical (r,θ,z).

In cylindrical coordinates, the Laplacian of a function, f is

$$\nabla^2 f = \frac{1}{r} \frac{\partial}{\partial r} \left(r \frac{\partial f}{\partial r} \right) + \frac{1}{r^2}\frac{\partial^2 f}{\partial \theta^2} + \frac{\partial^2 f}{\partial z^2}$$

Last edited: Apr 18, 2012
3. Apr 19, 2012

### Kyle91

Hey thank you so much for this, it certainly makes a lot more sense with the three-d wave equation!