Can “particle” waves break as ocean waves do?

I have heard about electrons surfing on wake fields which got me thinking. Are there analogs to reefs for these waves and can these waves break as ocean waves do when they hit a reef?

• The closest analogy would probably be called a "shock wave". – CuriousOne Jan 5 '15 at 3:29
• The breaking of waves on the shore is a nonlinear effect (in essence the top of the wave is further from the bottom of the tank, and therefore travels faster until it overtakes the rest of the wave). There are instances where high amplitude EM waves experience nonlinear phenomena when interacting with matter; that's the closest to "breaking" that I can think of. – Floris Jan 5 '15 at 4:10
• @Floris - No sir, he's asking about matter waves I think, not EM waves :) – User Anonymous Jan 5 '15 at 5:08
• @Floris has the right answer. Ocean waves break because of nonlinearities. The schödinger eqn is linear, and so cannot lead to 'breaking'. – QuantumDot Jan 5 '15 at 6:42
• I'm not exactly sure what you're talking about in terms of wake fields, so this might be off base. @Floris is describing a heuristic explanation of breaking in shallow water (governed by, eg the KdV equation). Deep water waves governed by the NLSE in 2+1D have blow up solutions due to a geometric instability (cf the lens/pseudo conformal transformation). This same equation governs the behavior of langmuir waves in plasmas (and hence suffers from the same blow up solutions), which might be related to what you are referring to. – Nick P Jan 5 '15 at 6:55

The wave nature of particles is only measured in probability distributions, as in the double slit single electron accumulation,

The first slide on top shows a few electrons going through the slit and hitting the screen. They hit it as individual particles.

The probability distribution accumulated shows an interference pattern , similar to the interference that a sea wave would show passing through two breaks of a barrier.

The distribution for the water waves is an energy distribution, more energy in the hills than in the troughs. The probability distribution of the electrons also shows in the end, as far as energy goes , more accumulated energy in the hills than in the troughs. The difference lies in the definition of what the "wave" represents. Macroscopically a wave is a collective phenomenon of energy distribution in an ensemble . Microscopically the individual electron is not a wave in an energy distribution. It is the probability of which way it will scatter through the slits that is described mathematically as a wave, and manifests in the interference pattern.

So there are similarities but also drastic differences, because if one took small water wave samples and threw them at the barrier, they would just go through the holes straight.

• Could time dilation and space contraction be regarded as a reef? As metioned in this paper time dilation is used to self accelerate electrons. dx.doi.org/10.1038/nphys3196 – Jitter Jan 21 '15 at 14:03

The short answer is yes. I have made comments about phenomena that manage to do this here, here, here, and here. A more detailed answer can be found here.

In electromagnetic waves, the concept of breaking is slightly different than in neutral fluids like water. However, the mathematical definition is still the same. Namely, that the amplitude of the wave reaches a point where the gradient scale length on the steepened edge approaches zero, called a gradient catastrophe. For instance, in magnetosonic waves the physical quantity responsible for the steepening is the current density. If the driving source (e.g., coronal mass ejection) and loss terms balance or the loss can exceed the source, then there will not be wave breaking. In general, if there is insufficient energy dissipation to limit any wave steepening, then the wave will break. You should note that were it not for energy dissipation, the sound waves produced by human speaking would steepen into shock waves.

In your example of wake field acceleration (WFA), I do not think that type of wave steepens. WFA is typically driven by a laser-like source of energy, which is effectively a quasi-static electric field. These types of fields do not really steepen, in the sense that I mentioned above, I don't think. The waves driven by the Weibel instability, whistler mode waves, and ion-acoustic waves can steepen because they can have a compressive longitudinal component. This term can lead to a finite $(\mathbf{V} \cdot \nabla) \mathbf{V}$ term, which is the nonlinear steepening term.

• In all your examples the wave doesn't actually break. Unsurprisingly so, because if it did, the wavefunction would become a multifunction (one part the lower hypersurface of wave, another two the hypersurfaces of the part which is breaking onto lower one). The waves you mention do steepen, indeed, but still never break (otherwise, what would be the next couple of frames in the animation in your first link?). So, I'm not really convinced that "The short answer is yes". – Ruslan Jan 5 '15 at 15:03
• Whistlers can break, see Krasnoselskikh et al., [2002], for example. Magnetosonic waves break, as evidenced by any work on shocklets or SLAMS, and shown in my Mathematica link (the wave breaks, which caused all sorts of numerical issues). Finally, ion-acoustic waves can form shocks and presumably break, though I will agree that I know of no work showing evidence for this. – honeste_vivere Jan 5 '15 at 15:06
• Wave breaking in magnetosonic shock waves is evidenced by non-stationarity and shock reformation, both produced in simulations and observed in nature. – honeste_vivere Jan 5 '15 at 15:07
• Well I guess this "breaking" is not the process I imagined. In particular, it doesn't mean "overturning", right? – Ruslan Jan 5 '15 at 15:09
• The waves can reach and pass gradient catastrophes, so they conform to the traditional definition of breaking. It will not, as you are implying, "look" like an overturning water wave in the simple sense (e.g., this might imply $\nabla \cdot \mathbf{B} \neq 0$). One of two things typically happen in EM wave steepening: the steepened edge begins to radiate like an antenna or they reflect/accelerate particles in a bursty fashion causing the front to effectively collapse. – honeste_vivere Jan 5 '15 at 15:15

You need something for the wave to break onto. These exist for things like water, and even light, but the wavelengths for particles is simply too short.

Alternately, one could consider nuclear collisions which produce different outputs as a kind of breaking of waves.

• I'm grasping at time dilation and space contraction for the waves to break. What are my chances? dx.doi.org/10.1038/nphys3196 – Jitter Jan 21 '15 at 14:09