# Wave formula question

1. Aug 17, 2005

### asdf1

Why is the wave formula different in 3-D?

2. Aug 17, 2005

### Kruger

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

3. Aug 17, 2005

### asdf1

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

4. Aug 17, 2005

### 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.

5. Aug 17, 2005

### 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.

6. Aug 18, 2005

### asdf1

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

7. Aug 18, 2005

### Staff: Mentor

According to the three-dimensional Pythagorean formula:

$$r^2 = x^2 + y^2 + z^2[/itex] Something that might be causing confusion here is that the formula for a three-dimensional wave depends on the "shape" of the wave. For a plane wave (whose maxima form a series of planes marching through space), [tex]\psi(x, y, z, t) = A \cos (\vec k \cdot \vec r - \omega t) = A \cos (k_x x + k_y y + k_z z - \omega t)$$

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.

Last edited: Aug 18, 2005
8. Aug 18, 2005

### asdf1

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

9. Aug 18, 2005