# 3D spherical vs 2D radial waves

1. Apr 14, 2012

### jjustinn

The Green's functions for a 3d wave are like δ(r - ct)/r -- so if you have static source at the origin that is turned on at t=0, you get an expanding ball around it of radius ct, with strength 1/r. If you look just at the XY plane, you see an expanding disc of value 1/r.

Similarly, if you turned the source on at t = 0 and off at t=1, you would get a an expanding spherical shell of radius ct and thickness c, or an expanding annulus in the XY plane.

However, for 2d waves, there is an "afterglow" -- I don't recall the exact Green function, but rather than a δ-function, it's a lograthmic or exponential decay, and there's something similar for 1-D waves -- e.g. If someone turns a light on for one second, you will still see lit for longer than 1s, though it will get dimmer as time goes on.

This is often discussed around Huygens' principle -- IIRC the principle is the "no afterglow" rule for 3D, which holds in odd dimensions > 1.

However, because of symmetry, the spherical wave equation satisfies the one-dimensional wave equation: (rV),tt = (rV),rr -- where r is the distance from the origin (√xx+yy+zz), V(r, t) is the amplitude a distance r from the origin, and t is the time.

So, then, shouldn't our point source at the origin - which is obviously spherically-symmetrical -- exhibit the afterglow,and therefore *not* give the simple constant 1/r dependence? E.g. Since there is an afterglow, the value at r is not just affected by the source at t=r/c, but also all previous times, leading to (apparently) V -> infinity as t-> infinity?

Similarly, it would seem th XY plane (or any plane through the origin) would satisfy a 2D wave equation, by symmetry...but I'm not so sure there.

2. Apr 15, 2012

### vanhees71

The wave equation in any spatial dimension reads (setting the phase speed of the wave to unity)

$$(\partial_t^2-\Delta_D) \Phi(t,\vec{x})=0,$$

where $\Delta_D$ is the Laplace(-Beltrami) operator of Euclidean space in $D$ dimensions. For isotropic problems, we have $\Phi(t,\vec{x})=\Phi(t,r)$ with $r=|\vec{x}|$, and the equation reads

$$\left [\partial_t^2 -\frac{1}{r^{D-1}}\frac{\partial}{\partial r} \left (r^{D-1} \frac{\partial}{\partial r} \right ) \right ] \Phi(t,r)=0.$$

This shows that the equations are different for 2 or 3 dimensions. The equation also holds for the Green's function of the wave operator (except at the origin),

$$(\partial_t^2-\Delta_{D}) G(t,\vec{x})=\delta(t) \delta^{(D)}(\vec{x}).$$

This explains why the Green's functions are different for different space dimensions.

3. Apr 15, 2012

### haruspex

A light is in 3D, so why do you expect afterglow?
The afterglow from a doused incandescent lamp is because it takes a while to cool down.
You won't see it with a LED.

You've lost me. There's no afterglow in one or three dimensions, right? So why do you think there should be afterglow?
And the reason the spherical wave equation can be transformed into a 1-D equation does not follow from symmetry (or it would happen in all dimensions). Read http://bigbro.biophys.cornell.edu/~toombes/Science_Education/Laser_Diffraction/Resources/Huygens_Principle.htm [Broken].

Last edited by a moderator: May 5, 2017
4. Apr 15, 2012

### jjustinn

I didn't mean to imply that the was any observed evidence of the "afterglow" in 3D -- there's clearly not (to take the incandescent lamp, the source itself is what has a finite turn-off time).

For some reason, I recalled reading that there was an afterglow in 1D / that Huygens' principle held in 2n+1, n > 0 dimensions. If I'm recalling correctly, isn't the "afterglow" actually equal to the original impulse in 1D? E.g. a unit origin impulse at t=0 is felt at x for *all* t > x/c? If I'm retroactively hallucinating, forgive me.

I think that's what was confusing me -- why it doesn't happen in all dimensions. Ill have to check out the links you and vanhees posted, and see if those clear anything up.

Thanks,
Justin

Last edited: Apr 15, 2012