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This depends on the wavelength. Essentially a wave hitting an object that has features that are smaller than roughly the wavelength will see those features as a
by cycomanic 5y ago
This depends on the wavelength. Essentially a wave hitting an object that has features that are smaller than roughly the wavelength will see those features as an average. So we therefore don't need to simulate the infinite resolution fractal. This is somewhat more complicated in the video because the waves are pulses, i.e. they contain many frequencies.
The video actually illustrates this effect when simulating the "narrow" ring waves and the "wide" ones. Those two contain different frequency components and thus see more or less of the fractals.
- ermir 5y agoMy initial thought was that was two particles that are infinitesimally close at the start of the simulation will have radically different reflections and paths when they hit the fractal, since the fractal has no local similarity, therefore the choice of the number of particles to use in the simulation will greatly affect the shape of the reflected wave. However by the same logic that the fractal has no local similarity we can also conclude that because the fractal has no derivative, and therefore no tangent, and therefore no way to calculate the reflection angle for any point at all.
- maqflp 5y agoWould that have then something in common with butterfly effect?
- cycomanic 5y agoGenerally it is better to think of waves as waves, not as being composed of particles as they behave quite differently, for example particles can't interfere (the dark bright patterns we see). I assume the simulation to create the video simulates the wave equations and the are not composed of "particles".
- maqflp 5y agoprecisely, this is solution to wave equation (I found this thread as suggested by youtube - you people here gave me quite a lot of traffic recently ;)