Wave Formula Differs in 3-D | Why?

[asdf1]

Why is the wave formula different in 3-D?

 

[Kruger]

Eh? Because it dependants from four values x,y,z,t and the one dimensional only on two value x,y.

 

[asdf1]

opps~ i think i didn't make myself clear...
what i meant was why in 3-D does kx become k*r?

 

[Galileo]

Because that's the natural extension. It reduces to one kx in 1D and is rotationally invariant: the predictions does not change by rotating your coordinate axes.

 

[Kruger]

you can rewrite k*x as k*x=kx*x+ky*y+kz*z where kx, ky and kz are the corresponding wave vectors of x, y and z.

 

[asdf1]

@@a
isn't r the radius? what does that have to do with the wave vectors, x, y, z?

 

[jtbell]

According to the three-dimensional Pythagorean formula:

[tex]r^2 = x^2 + y^2 + z^2[/itex]

Something that might be causing confusion here is that the formula for a three-dimensional wave depends on the "shape" of the wave. For a plane wave (whose maxima form a series of planes marching through space),

$$\psi(x, y, z, t) = A \cos (\vec k \cdot \vec r - \omega t) = A \cos (k_x x + k_y y + k_z z - \omega t)$$

where the $\vec k$ vector and its components are constant.

For a spherical wave (whose maxima form a series of concentric spheres spreading out from a central point, let's say the origin),

$$\psi(x, y, z, t) = A \cos (kr - \omega t) = A \cos (k \sqrt{x^2 + y^2 + z^2} - \omega t)$$

At each point in a spherical wave the $\vec k$ vector points radially outward from the origin, so the direction is different everywhere but the magnitude $k = \sqrt {k_x^2 + k_y^2 + k_z^2}$ is constant.

 

[asdf1]

so k and r is different depending on the wave's shape?
how many different kinds of different shapes are there?

 

[selfAdjoint]

so k and r is different depending on the wave's shape?
how many different kinds of different shapes are there?

Infinitely many. As many as there are combinations of simple oscillators of different frequency and amplitude.

 

[asdf1]

thanks! :)