Meaning of the Vector Wave Equation So I thought I would try my luck here on physics stack exchange about an intuitive meaning of the Vector Wave Equation. I know there are a lot of resources out there that explain this equation, but none of them try to give you an idea of what this means intuitively, and in my philosophy of solving complicated mathematics without actually knowing what's going on is just about a waste of time whether you can do it or not. 
I know how to derive the wave equation using Maxwell's equation, and I know what the wave function means.. I am just having a really hard time putting the equation into.. pseudo-math.. Can anyone try to explain it in a manner that is easy to understand?
$$\nabla^2\mathbf E=u_o\epsilon_o\frac{\partial^2\mathbf E}{\partial t^2}$$
Here is a gander at what both sides of the equations are telling me, and if someone can comment whether I am correct or not would be greatly appreciated.
The left hand side states that the change in the change of the vector electric field in position equals the change in change of the electric field in respect to time with a few constants multiplied. 
Even then all I did was just explain the relationship in English words, I still struggle to visualise this concept in my head. It almost sounds like the acceleration of the electric field in the position realms equals the acceleration of the electric field in the time realm.
Thanks.
 A: Here is the way I usually discuss the wave equation from a heuristic, semi-mathematical viewpoint. We'll focus on the simpler version of a wave on a string rather than electromagnetic waves, because the physics is more understandable. We are then considering the equation
$$
\frac{\partial^2u}{\partial x^2} = \frac{\partial^2u}{\partial t^2}.
$$
Imagine a string on a guitar that has been stretched (somehow) into the shape $u(x)$ shown in the figure below, and we have taken a snapshot of the string at this very moment before we release the string to see what it does.

Here's what it does:

Concentrate on the regions where I have drawn red arrows. The red arrow represents the acceleration
$$
a(x,t) = \frac{d^2u}{dt^2}
$$
of that point on the string as a function of time. This is exactly the right-hand side of the equation. Notice that the acceleration always points back towards the axis and decreases in magnitude as the point gets closer to the axis. We'll come back to this.
Now, what causes an acceleration? A force. When you stretch a string by pulling one point $x_0$ of the string away from its equilibrium position, there is a tension in the string, and the pieces of the string neighboring the point in question exert a force on the piece at $x_0$. The fact that we stretch the string away from the equilibrium position implies that it is the curvature of the string that determines this force, and it sort of feels right that a larger concavity means that the tension force is larger (the string is more stretched).
More quantitatively, it can be shown (for small amplitudes) that the force exerted by neighboring chunks of the string depends on the slope of the string at nearby-points, and so the net force on the chunk in question (at $x_0$) depends on the difference of nearby slopes, which exactly means that the force depends on the local concavity of $u$, which is of course exactly the left-hand side of the wave equation.
A: There are a few interpretations. 
Method 1: Discretizing
I find it helpful to discretize the equation on a square lattice. 
The "vector wave equation" is actually just three uncoupled wave equations, so we might as well just talk about the wave equation $\nabla^2 \phi=\frac{1}{c^2} \ddot{\phi}$. For the bit of intuition, the number of dimensions is a bit superfluous, so just consider one dimension, giving: $\partial_x^2 \phi=\frac{1}{c^2} \partial_t^2\phi$. If we discretize this in the x direction, we get: $$\partial_t^2\phi(x_i,t)= c^2 \frac{\phi(x_{i-1},t)+\phi(x_{i+1},t)-2\phi(x_{i},t)}{\Delta x^2}$$
Through some algebra you can see this is like two Hookean springs, connecting $x_i$ to $x_{i+1}$ and $x_{i-1}$:
$$\partial_t^2\phi(x_i,t)=\frac{c^2}{\Delta x^2}(\phi(x_{i-1},t)-\phi(x_i,t))+\frac{c^2}{\Delta x^2}(\phi(x_{i+1},t)-\phi(x_i,t))$$
Therefore, the wave equation is the continuum limit of a lattice of springs all connected to adjacent springs. 
Method 2: Lorentz Invariance
If you're trying to come up with physics in the form of partial differential equations, you will probably want to make sure the form of your equation doesn't change when you rotate your coordinates. In regular everyday 3-space, this tells you that your equation can only have terms like $\phi$, $\phi^2$, $\cdots$ as well as terms like $\nabla^2 \phi$, $(\nabla \phi)^2$, etc. If you just worry about invariance under rotations, this already tells you to look for equations like the heat equation $\partial_t \phi=\nabla^2 \phi$. Ignoring time, the simplest equation you can make, besides $\phi=0$, is $\nabla^2 \phi=0$, which already gives you tons of neat mathematics and neat physics.
But if you know the universe is relativistic, you actually want equations that are invariant under the Lorentz transformation. Now the heat equation doesn't cut it any more. You still can have terms like $\phi$, $\phi^2$, $\cdots$ in your equations, but now $\nabla^2\phi$ doesn't cut it any more. Just as $\nabla^2$ is invariant under spatial rotations, $\Box \phi=\frac{1}{c^2}\partial_t^2 \phi-\nabla^2 \phi$ is invariant under Lorentz transformations. This $\Box$ operator is called the d'Alembertian. The d'Alembertian is to Lorentz geometry as the Laplacian is to Euclidean geometry.
Whereas the simplest interesting equation invariant under rotations is $\nabla^2 \phi=0$, the simplest interesting equation invariant under Lorentz transformations is $\Box \phi=0$, the wave equation.
Now that it's clear it's invariant under Lorentz transformations, this tells you lots of things, most importantly: Causality is obeyed. So things will only ever spread out in circles/spheres, and points can only effect each other if they're in each other's light cones.
This may seem awfully mathematical, but I really just am trying to hijack your intuition about special relativity to apply it to the wave equation.
A: I have been pondering your question overnight and it strikes me that you might be able to interpret the left hand side as a measure of "curvature" in a sense, and the right hand side as (sort of) an "acceleration."
This analogy just came to mind from second time derivatives of position being accelerations, and second space derivatives being related to curvature.
Of course this is a very fuzzy idea so far but I'll think about it some more.  It was your mention of the acceleration idea that got me going in this direction. Your question really made me think.  Thanks.
