# Spherical wave

1. May 1, 2007

### Repetit

Hey!

Im quite confused about spherical waves. I mean, I understand that a spherical wave can be described by

$$\Psi = \frac{1}{r} e^{i r},$$

because the intensity of such a wave decreases as $$1/r^2$$. The intensity of such a wave is given by $$I = 1/r^2$$ which makes sense to me. But a spherical wave can also be described by

$$\Psi = \frac{1}{r} \cos r,$$

which gives a much different behaviour of the intensity because the intensity of such a wave is $$1/r^2 cos^2(r)$$. If these two expressions both describe a spherical wave, how come they don't have the same intensity?

2. May 1, 2007

### christianjb

At a guess, I'd say that they're the same thing if you take time averages.

<cos^2(kx-wt)>=1/2

3. May 1, 2007

### Dick

Those are spherical functions as in that they are angle independent. As they have no time dependence, in what sense are they waves?

4. May 2, 2007

### Repetit

Okay, so if they both had time dependence $$-i \omega t$$ so that

$$\Psi = \frac{1}{r} e^{i ( k r - \omega t)}$$

and

$$\Psi = \frac{1}{r} \cos( k r - \omega t)},$$

but they still don't have the same intensity, since the intensity of the second one is an oscillating function of r and t whereas the first one takes off as 1/r^2 and is therefore not oscillating.

5. May 2, 2007

### christianjb

Again, <cos^2(kr-wt)>=1/2 at any particular value of r and averaging over time.

6. May 2, 2007

### Dick

???? Not oscillating? The second equation is the real part of the first.

7. May 2, 2007

### Repetit

But isn't the intensity given by $$\Psi \Psi^*$$? This gives an intensity equal to 1/r^2 for the wave described by a complex exponential function but an intensity equal to $$1/r^2 cos^2(k r - w t)$$ for the other one.

8. May 2, 2007

### Dick

Ah, that's what you're trying to say. Yes, the cos one has a time-dependent 'intensity' and the other doesn't. But as christianjb pointed out, <cos^2>=1/2. So in an average sense one is 1/2 of the other. Not surprising since it's also the 'real half'.