”Quantum”-phenomena of bouncing droplets
Jelmer Doornenbal 10595759 July 19, 2016
Report Bachelor Project Physics and Astronomy, size 15 EC, conducted between April 2, 2016 and July 19, 2016.
FNWI, University of Amsterdam
Scientific summary
Experiments done by Y. Couder, S. Proti´ere and A. Boudaoud have shown that bouncing oil droplets exhibit numerous quantum-like phenomena. These oil droplets were made to float on a vibrating liquid bath of the same fluid. This is one of the few known example of a non-quantum system having non-quantum-like behaviour. Examples of these phenomena are single-slit and double-slit diffraction, tunneling, quantised energy levels, Anderson localisation and the creation/annihilation of droplet/bubble pairs. All these phenomena have been explained by R. Brady and R. Anderson in their article ’Why bouncing droplets are a pretty good model of quantum mechanics’. For this bachelor project, I have repeated the analysis done by them and I have concluded that all of these phenomena, apart from the creation/annihilation of droplet/bubble pairs, can be explained from a theoretical point of view.
Popular scientific summary
Tijdens dit bachelorproject is er onderzoek gedaan naar dansende oliedruppels. Deze oliedrup-pels kunnen op een vibrerend waterbad blijven springen. Door de frequentie van het vibrerend waterbad te verhogen kunnen oliedruppels naar links en rechts gaan bewegen. Er is door een onderzoeksgroep uit Parijs veel onderzoek gedaan naar dit systeem. Zij hebben onder andere gevonden dat dit systeem eigenschappen vertoont van zogenaamde quantum-particles, zoals elektronen. Quantum-particles zijn deeltjes die men zowel als een golf en als een deeltje kan beschrijven.
Aangezien dit een van de weinige niet-quantum systemen is dat quantum eigenschappen ver-toont, is er veel extra onderzoek naar gedaan. Uiteindelijk is er gevonden dat deze oliedruppels onder andere diffractie door een enkele spleet en door een dubbele spleet vertonen. Verder is er ontdekt dat de oliedruppels door een potentiaalbarri`ere heen kunnen tunnelen, dat ze gekwantiseerde energieniveaus hebben, dat er geen diffusie van golven plaatsvindt (Anderson localisatie) en dat er druppel paren gevormd en vernietigd kunnen worden.
Al deze bevindingen zijn door R. Brady en R. Anderson verklaard in hun artikel: ’Why bouncing droplets are a pretty good model of quantum mechanics’. Voor dit project heb ik analyse van R. Brady en R. Anderson herhaald. Ik ben daarbij begonnen met het oplossen van de golfvergelijking, een differentiaalvergelijking, die het verloop van een golf in ruimte en tijd beschrijft. Vervolgens heb ik de oplossing van deze vergelijking gebruikt om aan te tonen
Eerst heb ik laten zien dat de golven, die ontstaan door het neerkomen van de oliedruppel op het wateroppervlak, Lorentz invariant zijn. Dit betekent dat de resultaten van de experi-menten onveranderd blijven onder snelheidsverhoging of snelheids-verlaging in een bepaalde richting. Daarna zijn de krachten die de oliedruppels ondervinden onderzocht en is er vast-gesteld dat deze kracht als r12 afhangt van de afstand r. Dit is dezelfde afhankelijk als bij de
Couloumbkracht die onder andere elektronen ondervinden.
Deze eigenschappen zijn gebruikt om uiteindelijk ook te laten zien dat de oliedruppels voldoen aan de Klein-Gordon vergelijking en aan de Schr¨odinger vergelijking. Deze vergelijkingen beschrijven de verandering van een quantummechanisch systeem. Ten slotte heb ik ook laten zien dat de spin van de oliedruppels op precies dezelfde manier te beschrijven is als de spin van elektronen. Spin is een intrinsieke quantum mechanische eigenschap van atoomkernen, quarks en elementaire deeltjes. Deze eigenschappen zijn uiteindelijk gebruikt om de eerder genoemde eigenschappen te verklaren. Het is me helaas niet gelukt om te verklaren waarom het voor dit systeem zelf mogelijk is om druppel paren te vormen of te vernietigen.
Contents
1 Introduction 5
2 Description of my work and results 7
2.1 Solving the wave equation . . . 7
2.2 Lorentz covariance . . . 13
2.3 Forces between droplets . . . 17
2.4 Wavelength . . . 20
2.5 Diffraction . . . 21
2.6 The Klein-Gordon equation . . . 22
2.7 The Schr¨odinger equation . . . 24
2.8 Probability density . . . 26
2.9 Solutions to the wave equation depending on θ . . . 28
2.10 Orthogonal solutions . . . 31
2.11 Spin-1/2 symmetry . . . 31
2.12 Bloch’s sphere . . . 34
2.13 Pauli matrices . . . 35
2.14 Pauli exclusion prinicple . . . 41
3 Discussion 42
1
Introduction
Walker was in 1978 the first person to show that a droplet of soapy water could bounce for several minutes on a vibrating bath of water. [1] (Walker 1978) Almost 30 years later, in 2005, a french research group of Universit´e Paris Diderot consisting of Y. Couder, S. Protire and A. Boudaoud, showed with experiments that droplets can move horizontally by increasing the acceleration. Such a droplet is called a walker. A walker is a new type of localized state with particle-wave duality. The surface waves created by the bouncing droplet interact with the droplet itself. This causes the droplet to move so that it becomes a walker. [2] (Couder et al. 2005) This was publiced in an article in Nature.
After the publication of this article more research groups started to investigate this system. They found some more interesting quantum-like phenomena. They showed that the bounc-ing droplets exhibit sbounc-ingle-slit and double-slit diffraction, tunnelbounc-ing, quantised energy levels, Anderson localisation and the creation/annihilation of droplet/bubble pairs. [3] (Brady & An-derson 2014). This is the first known example of a non-quantum system having quantum-like behaviour. The experimental setup is shown in a picture below.
Figure 1: The experimental setup [3] (Brady & Anderson 2014)
All the experimental results have later been summarized in a more theoretical framework by Robert Brady and Ross Anderson. They have explained these quantum-like phenomena by starting from the wave equation and showing that the surface waves are Lorentz covariant,
The goal of this bachelorproject is to present the analysis of Robert Brady and Ross Anderson using the article they have publiced. This article is called ’Why bouncing droplets area a pretty good model of quantum mechanics’. In this article they argue that the bouncing droplet system is a good way to explain quantum mechanics without using complicated idea of an quantum mechanical wave-function.
2
Description of my work and results
2.1 Solving the wave equation
I will start repeating the analysis done by Anderson and Brady by solving the wave equation. The wave equation in cartesian coordinates is:
1 c2 ∂2h ∂t2 − ∂2h ∂x2 − ∂2h ∂y2 = 0 (1)
In this equation h is a function which describes the wave height of the bath near a droplet. I first transform the wave equation to polar coordinates using the following relations:
r =px2+ y2 (2)
θ = arctan(y
x) (3)
In order to transform to polar coordinates, the derivatives have to be calculated. In the first place, I calculate the second derivative with respect to x. By the chain rule, I obtain
hx = hrrx+ hθθx (4)
Differentiating once more with respect to x gives
hxx= (hrrx)x+ (hθθx)x (5)
= (hr)xrx+ hrrxx+ (hθ)xθx+ hθθxx (6)
= (hrrrx+ hrθθx)rx+ hrrxx+ (hθrrx+ hθθθx)θx+ hθθxx (7)
By differentiating r and θ twice, I find
rx= ∂ ∂x( p x2+ y2) = x p x2+ y2 = x r (8) rxx= ∂ ∂x( x p x2+ y2 = 1 p x2+ y2 − x2 p x2+ y23 = 1 r − x2 r3 = r2− x2 r3 = y2 r3 (9)
Substituting all these expressions into (7), while using that hrθ and hθr are equal, gives: hxx = x2 r2hrr− 2 xy r3hrθ+ y2 r4hθθ+ y2 r3hr+ 2 xy r4hθ (12)
Similarly, the second derivative of h with respect to y is
hyy = y2 r2hrr+ 2 xy r3hrθ+ x2 r4hθθ+ x2 r3hr− 2 xy r4hθ (13)
Adding hxx and hyy gives:
∇2h = hxx+ hyy = ∂2h ∂r2 + 1 r ∂h ∂r + 1 r2 ∂2h ∂θ2 (14)
This equation is also known as the Laplacian in polar coordinates.
Now substituting (14) into the wave equation (1) gives us the two-dimensional wave equation in polar coordinates. 1 c2 ∂2h ∂t2 − ( ∂2h ∂r2 + 1 r ∂h ∂r + 1 r2 ∂2h ∂θ2) = 0 (15)
The wave height obeys the following boundary conditions:
h(R, t) = 0 (16)
h(r, 0) = −h0 (17)
ht(r, 0) = 0 (18)
Where h0 is the maximum height and R is position of the bath’s boundary. In order to solve
this Partial Differential Equation (PDE), I have used the method of separation of variables by defining the following:
h(r, t) = W (r)G(t) (19)
Substituting h = W G and its derivatives into (15) and dividing by W G, we get ¨ G c2G = 1 W(W 0 +1 rW 00 ) (20)
Where the dots denote derivatives with respect to time and primes denote derivates with respect to r. The expressions on both sides must be equal to a negative constant, −κ2.
¨ G c2G = 1 W(W 0+1 rW 00) = −κ2 (21)
This gives the following two linear Ordinary Differential Equations (ODEs): ¨ G + c2κ2G = 0 (22) W00+1 rW 0 + κ2W = 0 (23)
First solving equation (22). The general solutions to this ODE are:
G(t) = A cos(cκt) + B sin(cκt) (24) At t = 0, the boundary conditions are G(0) = −h0 and ˙G(0) = 0. This yields
G(0) = A = −h0 (25)
˙
G(t) = cκ(B cos(cκt) − A sin(cκt)) (26) ˙
G(0) = cκB = 0 (27)
Hence the solution of this ODE is
G(t) = −h0cos(ω0t) (28)
with ω0 = cκ.
Now we will be solving equation (23) by subsituting s = κr. We obtain by using the chain rule W0 = dW dr = dW ds ds dr = dW ds κ (29) W00 = d 2W dr2 = d2W ds2 d2s dr2 = d2W ds2 κ 2 (30)
Now subsituting this back into (23) and dividing by κ2 gives d2W ds2 + 1 s dW ds + W = 0 (31)
This is the Bessel Differential Equation with n = 0, which can be solved by using Frobenius Theory.
This equation can be generalized to a Bessel Differential Equation of order n. This equation is as follows d2W ds2 + 1 s dW ds + (1 − n2 s2)W = 0 (32)
Frobenius theory assumes the following
W (s) =
∞
X
m=0
amsm+b (33)
The first and second derivative of this equation are
W0= ∞ X m=0 am(m + b)sm+b−1 (34) W00= ∞ X m=0 am(m + b)(m + b − 1)sm+b−2 (35)
Subsituting (33) and its derivatives back into (31) and multiplying by m2 gives
∞ X m=0 amsm+b−n2 ∞ X m=0 amsm+b−2+ ∞ X m=0 am(m + b)sm+b−2+ ∞ X m=0 am(m + b)(m + b − 1)sm+b−2 = 0 (36) In the first series we set m = v − 2 and in the second, third and fourth series we set m = v. Then we get ∞ X v=2 av−2sv+b−2− n2 ∞ X v=0 avsv+b−2+ ∞ X v=0 av(v + b − 2) + ∞ X v=0 av(v + b)(v + b − 1)sv+b−2= 0 (37)
In order to obtain a recursion relation, I collect the same powers of s. This gives
0 = ba0+ b(b − 1)a0− n2a0 (v = 0) (38)
0 = (b + 1)a1+ b(b + 1)a1− n2a1 (v = 1) (39)
From (38) the following equation can be found
(b + n)(b − n) = 0 (41)
The roots are b = n and b = −n. Subsituting b = n into the (39) gives
(n + 1 + n(n + 1) − n2)a1 = 0 (42)
(n + 1 + n2+ n − n2)a1 = 0 (43)
(2n + 1)a1 = 0 (44)
a1 = 0 (45)
By plugging in b = n into (40) a recursion formula can be found
av−2+ (v + n)av+ (v + n)(v + n − 1)av− n2av = 0 (46)
av−2+ (v + n)av+ (v2− v + 2nv + n2− n)av− n2av = 0 (47)
av−2+ v(v + 2n)av = 0 (48)
Since a1 = 0, it follows from (48) that all other odd-numbered coefficients are also zero. So
we only have to deal with even-numbered coefficents av with v = 2m.
Subsituting v = 2m into (48) gives
2m(2m + 2n)a2m+ a2m−2= 0 (49)
a2m= −
1
22m(m + v)a2m−2 (50)
Filling in some numbers for m in order to try to find the recursion formula
a2 = − a0 22(v + 1) (51) a4 = − a2 222(v + 2) = a0 242!(v + 2)(v + 1) (52) a6 = − a4 223(v + 3) = a2 243!(v + 3)(v + 2) = − a0 263!(v + 3)(v + 2)(v + 1) (53)
In general:
a2m=
(−1)ma0
22mm!(v + 1)(v + 2)....(v + m) (54)
We can choose a0 arbitrarily. In order to simplify we choose a0 to be
a0=
1
2nn! (55)
Hence the solution of the ODE (32) is
Wn(s) = sn ∞ X m=0 (−1)ms2m 22m+nm!(n + m)! = Jn(s) (56)
This equation is a Bessel function of the first kind of order m. Subsituting s = kr and w0 =
ck back into this equation gives
Jn(ω0r/c) = ( ω0r c ) n ∞ X m=0 (−1)m(ω0r/c)2m 22m+nm!(n + m)! (57)
Subsituting n = 0 in order to find the solution to equation (31) gives
W0(s) = J0(ω0r/c) = ∞ X m=0 (−1)m(ω0 c ) 2mr2m 22m(m!)2 (58)
Since (32) is a second order differential equation, it has two solutions. The second solution can be found using order reduction. This solution is a Bessel function of the second kind. This function can be defined as:
Yν(s) =
Jν(s)cos(νπ) − J−ν(s)
(sin(νπ) (59)
However this equation diverges to −∞ at the origin, so this equation can’t be used to describe the wave height.
Combining the two solutions of the linear ODEs gives us the final solution to the wave equation (15)
2.2 Lorentz covariance
Starting with the recently found solution for the wave equation I will try to derive a number of quantum-like phenonema shown by the oil droplets. However first I will show that the surface waves guiding the oil droplets are Lorentz covariant.
Applying the Lorentz transformation below to equation (60) changes h(x, y, t) to h(x0, y0, t0). This equation still obeys (1). The Lorentz transformation used is
x0 = γ(x − vt) (61) y0 = y (62) t0 = γ(t − vx c2) (63) γ = q 1 1 −vc22 (64)
If you multiply h(x0, y0, t0) with a constant α, it still obeys the wave equation (1). If you choose α = γ, it gives
x00= γ2(x − vt) (65)
y00= γy (66)
t00= γ2(t −vx
c2) (67)
Applying this Lorentz transformation to (60) yields
h = −h0cos(ω0t00)J0(ω0r00/c) (68) Subsituting t00 gives h = −h0cos(ω0(γ2t − γ2 vx c2))J0(ω0r 00/c) (69)
Rearranging this equation using (64)
h = −h0cos(ω0(t − γ2v c2 (x − vt))J0( ω0 c r 00) (70)
The first and second derivative of (70) are ∂h ∂x = −h0( γ2ω0v c2 sin(ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x)J0( ω0 c r 00 ) (71) − cos(ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x) γ4(x − vt) r00 ω0 c J1( ω0 c r 00 ) ∂2h ∂x2 = h0( γ4ω20v2 c4 cos(ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x)J0( ω0 c r 00) (72) + 2γ 2ω 0v c sin(ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x) γ4(x − vt) r00 ω0 c J1( ω0 c r 00) − 1 2cos(ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x) γ8(x − vt)2 r002 ω20 c2(J2( ω0 c r 00) − J 0( ω0 c r 00)) − cos(ω0t + γ 2ω 0v2 c2 t − γ2ω0v c2 x) γ6y2 r003 ω0 c J1( ω0 c r 00)
Solving ∂h∂x = 0, where the wave will be at its highest point, gives
γ2ω0v c2 sin(ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x)J0( ω0 c r 00) (73) = cos(ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x) γ4(x − vt) r00 ω0 c J1( ω0 c r 00)
Approximating cos(ω0t00) = 1 and sin(ω0t00) = ω0t00, assuming it takes little time before the
wave hits its highest point, gives γ2ω0v c2 (ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x)J0( ω0 c r 00) = γ4(x − vt) r00 ω0 c J1( ω0 c r 00) (74)
Dividing both sides by J0(ωc0r00) gives
γ2ω0v c2 (ω0t + γ2ω0v2 c2 t − γ2ω0v c2 x) = γ4(x − vt) r00 ω0 c J1(ωc0r00) J0(ωc0r00) (75) Rearanging this equation gives
γ2ω 0v c2 (ω0t + γ2ω 0v2 c2 t − γ2ω 0v c2 x) = γ 4(x − vt)ω02 c2 J1(ωc0r00) ω0 cr00J0( ω0 cr00) (76) Subsituting x − vt by ∆x gives γ2ω0v c2 (ω0t − γ2ω0v c2 ∆x) = γ 4∆xω02 c2 J1(ωc0r00) ω0 c r00J0( ω0 c r00) (77)
The right hand side of this equation can be simplified. First of all I subsitutue ω0 c r 00 by A. γ2ω0v c2 (ω0t − γ2ω0v c2 ∆x) = γ 4∆xω02 c2 J1(A AJ0(A) (78) Bessel functions are given by the following equation
Jα(A) = ∞ X m=0 (−1)m 22m+αm!(m + α)!A 2m+α (79)
I pick the highest point of the wave to be at (x00, y00) = (0, 0). Recalling that r =px002+ y002 means that A 1, gives
J0(A) ' J0(0) = 1 (80)
J1(A) '
A
2 (81)
Using these two formulas (78) can be simplified to γ2ω0v c2 (ω0t − γ2ω0v c2 ∆x) = 1 2γ 4∆xω02 c2 (82)
Rearranging this equation gives
γ2ω 2 0 c2vt − γ 4ω20 c2 v2 c2∆x = 1 2γ 4∆xω02 c2 (83)
Simplifying and rearranging this us the final equation at t = T , where T is the landing time gives vT = γ2(v 2 c2 + 1 2)∆x (84)
This equation can be simplified for the last time by realising that ∆x = vτ
γ2(v 2 c2 + 1 2) = T τ (85)
Experiments have shown that this linear relationship is remarkable accurate up to a Lorentz factor γ = 2.6. Hence, we can say that this system is Lorentz covariant up to this Lorentz factor. The experimental data is shown in the figure below. [3] (Brady & Anderson 2014)
Figure 2: The experimental data, which shows the linear relationship between γ2(vc22 +12) and
2.3 Forces between droplets
Experiments have shown that when a walker approaches the edge of the container, it does not actually touch the edge but is deflected away. [3] (Brady & Anderson 2014) There is a relation between the velocity normal to the boundary V⊥ and the distance to the boundary.
This relation is as follows
V⊥2= V02−B
r (86)
B in this equation is the slope of the graph.
Figure 3: The velocity normal to the boundary squared as a function of inverse distance to the boundary.[3] (Brady & Anderson 2014)
From this relation one can be deduce that there is an inverse square force between the droplets and the boundary
V⊥2 ∝ 1 r (87) V⊥∝ 1 √ r (88) a = dV⊥ dt = d dt 1 √ r (89) = −1 2 1 r√rV⊥ (90) = −1 2 1 r√r 1 √ r (91) a ∝ −1 r2 so F ∝ − 1 r2 (92)
This attractive force is created by bouncing droplets moving in antiphase. Each droplet drives radial flows in the liquid, decreasing velocity and increasing the Bernoulli pressure. This gives rise to a repulsion between the droplets. [3]
This repulsive force can be written in the following form
F = αβc
r2 (93)
The force on a droplet by the radial flows in the liquid is given by
F = ρ0U Q2= −ρ0
Q1Q2
4πr2 (94)
where U is the speed of the radial flow, Q2 is the flow of the liquid and ρ0 is the bubble
density. The flow is given by the area multiplied by the speed of the radial flow
Q = 4πr02vs= 4πr02
drb
dt (95)
Suppose the radius of a bubble is given by
rb = r0(1 + A cos(ωt)) (96)
Then the flow speed is given by
Subsituting this all into (94) yields F = −ρ0 (4πr03Aω cos(ωt))2 4πr2 (98) = −ρ04πr06A2ω2cos2(ωt) 1 r2 (99) = −A2ω2 1 2r2(4πρ0r 3 0)r30 (100) (101) In the above equation, cos2(ωt) has been replaced by its average value 12.
Noticing that the bubble density ρ0 is given by ρ0 = 4md 3πro3
, where mdis the displaced mass of
the bubble and that the inertial mass is approximately given by m = 12mdgives the equation
for the repulsive force
F = −3A2(r0ω c ) 3mc2 ω 1 r2 = −α¯b 1 r2 (102) α = 3A2(r0ω c ) 3 (103) ¯b = mc2 ω (104)
2.4 Wavelength
Defining the following functions
h = ψX (105)
ψ = cos(−ω0t) (106)
X = −h0J0(κr) (107)
Where X is the wave field and ψ is a wave. If we apply a Lorentz transformation to the wave height function h, we get the following two functions
ψ = cos(−ω0t0) (108)
X = −h0J0(κr0) (109)
Applying a Lorentz transformation to the wave height function h causes the solution to move. Hence ψ becomes a plane wave, which can be written as
ψ = cos(kx − ωt) = cos(−ω0t0) (110)
We can derive the values of k and ω by taking the derivative of S = −ω0t0 with respect to x
and t. k = ∂S ∂x = ∂S ∂t0 ∂t0 ∂x = γω0 c2 vx (111) ω = −∂S ∂t = − ∂S ∂t0 ∂t0 ∂t = γω0 (112) The wavelength of ψ is λ = 2π k = 2πc2 ωvx = b p (113)
In this equation (113) b = 2πmcω2 and p = mvx. This equation is equivalent to the Broglie
wavelength, except that Plank’s constant h has been replaced with b.
From these relations, we also find a formula for the momentum
p = ¯bk = mc
2
ω ω
2.5 Diffraction
Since by defining the equations (106) and (107), one can imagine this system to be a pilot wave like system. Such a system might show diffraction when interfering with a single or double slit. Experiments show that this is actually the case. The following histogram is found by Y. Couder ane E. Fort and show that there are indeed diffraction patterns.
Figure 4: A histogram of the deflection angle showing a clear diffraction pattern. [4] (Couder & Fort 2006)
2.6 The Klein-Gordon equation
ψ obeys the following relation
∂2ψ ∂t2 = −ω
2
0ψ (115)
However, this equation is not Lorentz covariant. To attempt to make this equation Lorentz covariant, we define a quantity with the dimensions of energy
E = ¯bω (116)
Starting with the basic relavistic formula which links energy to momentum and mass
E2− p2c2 = m20c4 (117) The next step is to define the following energy operator and impuls operator
E = −b i ∂ ∂t (118) p = b i∇ (119)
These operators are similiar to the energy and impuls operator in quantum mechanics, however Plank’s constant has been replaced be b. Subsituting these operators into (117) gives
(−b i ∂ ∂t) 2− c2(b i∇) 2= m2 0c4 (120)
After rearranging this formula, we get ∂2ψ
∂t2 − c
2∇2ψ = −m20c4
b2 ψ (121)
We can use equation (117) again to simplify this equation
1 b2m 2 0c4= 1 b2(E 2− p2c) (122) = E 2 b2 − p2c2 b2 (123) = ω2− c2k2 (124)
In the last step I have used equation (114). This last equation can again be simplified by subsituting the found values of ω and k
ω2− c2k2= γ2ω02−γ 2ω2 0 c2 v 2 x (125) = ω20γ2(1 − vx2/c2) (126) = ω20 (127)
Hence, the final Lorentz covariant form of equation (115) is ∂2ψ
∂t2 − c
2∇2ψ = −ω2
0ψ (128)
This equation is the same as the Klein-Gordon equation of quantum mechanics for a rela-tivistic particle except that the speed of light has been replaced with the characteristic speed c.
2.7 The Schr¨odinger equation
When (110) is given a small boost in velocity v in the x-direction, equation (110) transforms to ψ = cos(−ω0t0) = cos(vxωc20 − ω0t), where γ is approximated to be 1. By defining θ =
vxω0
c2 ,
I can rewrite this to
ψ = R cos(θ − ω0t) (129)
The function (110) can be extended to the complex plane by defining
ψs= Reiθ (130)
so that ψ = Re(e−iω0tψ
s). By subsituting this into the Klein-Gordon equation, we can find
a solution where both the real and the imaginary parts obey it.
∂2ψ ∂t2 − c 2∇2ψ = −ω2 0ψ (131) ∂ ∂t(−iω0e −iω0tψ s+ e−iω0t ∂ψs ∂t ) − c 2e−iω0t∇2ψ s= −ω20ψ (132)
−iω0e−iω0t(−iω
0ψs+ ∂ψs ∂t ) + e −iω0t(−iω 0 ∂ψs ∂t + ∂2ψs ∂t2 ) − c 2e−iω0t∇2ψ s= −ω20ψ (133) e−iω0t(−ω2 0ψs− 2iω0 ∂ψs ∂t + ∂2ψs ∂t2 ) − c 2e−iω0t∇2ψ s= −ω20ψ (134)
For slow velocities ∂2ψs
∂t2 is approximately 0. This gives us
e−iω0t(−ω2 0ψs− 2iω0 ∂ψs ∂t ) − c 2e−iω0t∇2ψ s= −ω02e−iω0tψs (135) −ω20ψs− 2iωo ∂ψs ∂t − c 2∇2ψ s= −ω02ψs (136) −2iωo ∂ψs ∂t = c 2∇2ψ s (137)
Rearranging the last equation gives us the final solution
i∂ψs ∂t = −
c2 2ω0
∇2ψs (138)
Multiplying both sides of equation (138) with ¯b gives
i¯b∂ψs ∂t = − ¯ bc2 2ω0 ∇2ψs (139)
Using the found formula for ¯b ((104) gives i¯b∂ψs ∂t = − ¯b2 2m0 ∇2ψ s (140)
This is the Schr¨odinger equation for a free particle with ~ replaced by ¯b.
I am not able to find the Schr¨odinger equation for particle experiencing a potential V . For example, if I make the following substitution: ¯bω0 = m0c2− V after multiplying both sides
by ¯b, we get i¯b∂ψs ∂t = − c2 2ω2 0 (m0c2− V )∇2ψs (141) i¯b∂ψs ∂t = (− c4 2ω2 0 m0+ c2 2ω2 0 V )∇2ψs (142) i¯b∂ψs ∂t = (− ¯b2 m0 ∇2+ c 2 2ω02V ∇ 2)ψ s (143)
This equation is definetely different than the Schr¨odinger equation with a potential term: i¯b∂ψs ∂t = (− ¯ b2 2m0∇ 2+ V )ψ s, since V c 2 2ω2 0 ∇2ψ s 6= V ψs.
2.8 Probability density
Starting with equation (138) and subsituting ψs = Reiθ gives the following equation for the
left hand side
i∂ψs ∂t = i
∂R ∂te
iθ (144)
Subsituting ψs= Reiθ in the right hand side gives
∇2ψs= ∇2(Reiθ) = ∇(∇(Reiθ) (145)
= ∇(∇R)eiθ+ i(∇θ)Reiθ) (146)
= ∇2Reiθ+ i∇2θReiθ+ i∇θ∇Reiθ+ i(∇θ)2Reiθ+ i∇R∇θeiθ (147) = ∇2Reiθ+ 2i∇θ∇Reiθ+ i∇2θReiθ− (∇θ)2Reiθ (148) Taking the imaginary part gives the following two equations
Im(i∂R ∂te
iθ) = ∂R
∂te
iθ (149)
Im(∇2(Reiθ) = 2∇θ∇Reiθ+ ∇2θReiθ (150) Subsituting these two equation back into equation (138)
∂R ∂te
iθ = − c2
2ω0
(2∇θ∇Reiθ+ ∇2θReiθ) (151) Rearranging this equation by dividing both sides by eiθ
∂R ∂t = −
c2 2ω0
(2∇θ∇R + ∇2θR) (152)
This equation can be further simplified by multiplying with 2R. This gives us the following left hand side
2R∂R ∂t = R ∂R ∂t + R ∂R ∂t = ∂R2 ∂t (153)
After multiplying with 2R the right hand side can be simplified as follows
−2R c 2 2ω0 (2∇θ∇R + ∇2θR) = −c 2 ω0 (R2∇2θ + 2R∇θ∇R) (154) = −∇(R2c 2 ω0 ∇θ) (155)
This gives us the final equation
∂R2
∂t + ∇(R
2v) = 0 (157)
This equation can be intrepeted as a continuity equation. In fluid dynamics the continuity equation is ∂ρ∂t + ∇(ρv) = 0. Comparing these two results tells us that in the found equation that ρ has been replaced by R2. Since R2 =| ψs |2, one can say that the configuration is
distributed according to | ψs |2. This is in agreement with the Copenhagen interpretation of
2.9 Solutions to the wave equation depending on θ
Up until this point only circular and spherical symmetric waves have been considered. Ex-perimental results suggest that we also have to take a look at solutions to the wave equation which depend on angle. [3] (Brady & Anderson 2014) The pictures in figure 5 demonstrates this suggestion.
Figure 5: Waves created by two bouncing droplets. These pictures suggest that there is a dependence on angle. [5] (Proti´ere et al. 2006)
Hence the solution of the wave equation is of the form h(r, t, θ). I will now try to solve the wave equation to try to find a solution of this form.
Starting with the wave equation in polar co¨ordinates (15): 1 c2 ∂2h ∂t2 − ( ∂2h ∂r2 + 1 r ∂h ∂r + 1 r2 ∂2h ∂θ2) = 0 (158)
The solution of this equation obeys the following conditions:
h(R, t, θ) = 0 (159)
h(r, 0, θ) = −h0 (160)
ht(r, 0, θ) = 0 (161)
h(r, t, 0) = h(r, t, n2π) (162) First we separate the function h(r, t, θ) into three parts
Subsituting this separation into (249) gives 1 c2R(r)Θ(θ) ∂2T (t) ∂t2 = T (t)Θ(θ) ∂2R(r) ∂r2 + 1 r2T (t)Θ(θ) ∂R(r) ∂r + 1 r2R(r)T (t) ∂2Θ(θ) ∂θ2 (164) Multiplying by R(r)T (t)Θ(θ) gives 1 c2 1 T (t) ∂2T (t) ∂t2 = 1 R(r) ∂2R(r) ∂r2 + 1 r 1 R(r) ∂R(r) ∂r + 1 r2 1 Θ(θ) ∂2Θ(θ) ∂θ2 (165)
The expressions on both sides must be equal to a negative constant −κ2. Hence 1 c2 1 T (t) ∂2T (t) ∂t2 = −κ 2 (166) 1 R(r) ∂2R(r) ∂r2 + 1 r 1 R(r) ∂R(r) ∂r + 1 r2 1 Θ(θ) ∂2Θ(θ) ∂θ2 = −κ 2 (167)
Rearranging (166) gives us the following ODE ∂2T (t)
∂t2 + c
2κ2T (t) = 0 (168)
We know the solution of equation (168), because it is the same equation as equation (22) with the same initial values. The solution is
T (t) = −h0cos(ω0t) (169)
with ω0 = cκ.
Now we will be solving equation (167). Multiplying by r12 and rearranging gives
r2 R(r) ∂2R(r) ∂r2 + r R(r) ∂R(r) ∂r + r 2κ2= − 1 Θ(θ) ∂2Θ(θ) ∂θ2 (170)
Both sides of this equation are equal to a positive constant n2. This gives us the following two ODEs. − 1 Θ(θ) ∂2Θ(θ) ∂θ2 = n 2 (171) r2 R(r) ∂2R(r) ∂r2 + r R(r) ∂R(r) ∂r + r 2κ2 = n2 (172)
The solution to this differential equation is
Θ(θ) = e±inθ (174)
Since we expect that Θ(θ) = Θ(θ + 2π), we can conclude that
ei2πn= 1 (175)
Hence n is indeed an integer. The final form of (174) is
Θ(θ) = einθ (176)
Now we still have to solve the third differential equatioin (171). Multiplying by rR2 and
rearraging gives ∂2R(r) ∂r2 + 1 r ∂R(r) ∂r + (κ 2−n2 r2)R(r) = 0 (177)
Now we will be solving this equation by subsituting s = κr. We obtain by using the chain rule R0 = dR dr = dR ds ds dr = dR dsκ (178) R00= d 2R dr2 = d2R ds2 d2s dr2 = d2R ds2κ 2 (179)
Now subsituting this back into (177) and dividing by κ2 gives d2R ds2 + 1 s dR ds + (1 − n2 s2)R = 0 (180)
This is the Bessel Differential Equation of order n. This equation has already been solved in the section 2.1. The solution is displayed in equation (57). Hence R(r) is
R(r) = Jn(ω0r/c) = ( ω0r c ) n ∞ X m=0 (−1)m(ω0 c )2mr2m 22m+nm!(n + m)! (181)
Combining the results of the three ODEs gives us the following solution
2.10 Orthogonal solutions
These rotating waves can be treated independently, because they are orthogonal. One can show that they are orthogonal by computing the following integral
Z 2π 0 h∗nhmdθ = h20cos2(ω0t)Jn(ω0r/c)Jm(ω0r/c) Z 2π 0 e−inθeimθdθ (183) If m 6= n, the integral is equal to zero. If m = n, the integral is equal to 2π. Hence hm and
hn are orthogonal.
Experiments have shown that we only have to look at n = 1 and n = −1. Since these two waves are orthogonal, one can write the function h, which describes the height of the wave, as a superposition of h1 and h−1.
h = cos(α)h1+ sin(α)h−1 (184)
2.11 Spin-1/2 symmetry
In order to show spin-1/2 symmetry, I will compute the expected angular momentum of a wave describes by the function h. The expected angular momentum can be calculated using
< h|L|h >= Z 2π
0
hLhdθ (185)
Angular momentum operator L can be written as
L = ¯b i
∂
∂θ (186)
Working with this operator on h gives
Lh = ¯b i ∂ ∂θh = ¯b i ∂ ∂θ(cos(α)h1+ sin(α)h−1) (187) Subsituting (182) in this equation gives
Lh = −h0cos(ω0t) ¯ b i ∂ ∂θ(cos(α)e iθJ 1(ω0r/c) + sin(α)e−iθJ−1(ω0r/c)) (188)
Now I write out the derivative with respect to θ.
Using the relation J1= −J−1 and gives
Lh = −h0cos(ω0t)J1(ω0r/c)¯b(cos(α)eiθ+ sin(α)e−iθ) (190)
We can also use this relation to rewrite equation (184)
h = −h0cos(ω0t)J1(ω0r/c)(cos(α)eiθ− sin(α)e−iθ) (191)
Subsituting equations (190) and (191) into (185) gives
< h|L|h > = h20cos2(ω0t)J1(ω0r/c)2¯b
Z 2π
0
(cos(α)eiθ+ sin(α)e−iθ) (192) (cos(α)e−iθ− sin(α)eiθ)dθ
Rearraning this equation by multiplying the terms in the parentheses gives
< h|L|h > = h20cos2(ω0t)J1(ω0r/c)2¯b
Z 2π 0
cos2(α) − sin2(α) (193) − cos(α) sin(α)(e2iθ− e−2iθ)dθ
This equation can be simplified by using that sin(2θ) =e2iθ−e2i−2iθ.
< h|L|h > = h20cos2(ω0t)J1(ω0r/c)2¯b
Z 2π
0
cos2(α) − sin2(α) (194) − cos(α) sin(α)2i sin(2θ)dθ
The first two terms do not depend on θ, so they just add an extra factor of 2π. The third term a sine function integrated twice over its period. Hence this integral is zero. This gives us the expactation value of L.
< h|L|h >= 2πh20cos2(ω0t)J1(ω0r/c)2¯b(cos2(α) − sin2(α)) (195)
By subsituting h1 into equation (185), you can find its angular momentum
Comparing this with the angular momentum of h gives us the relation between them
L = L0(cos2(α) − sin2(α)) = L0(cos(2α) (197)
where L0 is the angular momentum of h1.
The height of the wave field depends on paramater α. I compute the angular momentum for certain values of α in the table below.
α L/L0 h 0 1 h1 π 4 0 1 √ 2(h−1+ h1) π 2 -1 h−1 3π 4 0 1 √ 2(h−1− h1) π 1 −h1
Table 1: The angular momenteum and waveheight of the wave field for certain values of α. If you look at this table, you can see that the sign reverses once the system goes throught an entire cycle. Fermions also show this kind of behaviour. They reverse in sign under a rotation of 360 degrees.
2.12 Bloch’s sphere
A Bloch sphere can be constructed from equation (182) by extending it to the complex plane. This gives us the following equation
ζm = Ae−i(ω0t−mθ)Jm(ω0r/c) (198)
In this equation A is the amplitude. This equation can be split up into a part that describes the wave field χm and a part that describes the wave itself ψ.
ζ = ψχm (199)
ψ = e−iω0t (200)
χm = AeimθJm(ω0r/c) (201)
The factor ψ obeys the Schr¨odingers equation as seen in section 2.7. Hence it correctly describes the waves. Now we take a look at the wave field χm.
When we extend χm to the complex plane in the same way as (184), we get the following
expression
χ = eiS[cos(1
2β)χ1+ e
iφsin(1
2β)χ−1] (202)
where S is an arbritary phase and φ is the relative phase between χ1 and χ−1. This is the
formula which describes a Bloch’s sphere with χ1 being a spin-up particle and χ−1 being the
spin-down particle. β and φ describe the sphere as represented by the figure below.
2.13 Pauli matrices
The mapping of a droplet’s wave height onto the Bloch sphere can also be written as a dot product of two vectors a and χ.
χ = a · χ = (a1, a2) · (χ1, χ−1) (203)
The values for a1 and a2 can be obtained from equation (202). They are as follows
a1 = eiScos( 1 2β) (204) a2 = eiSeiθsin( 1 2β) (205)
We can compute the angular momentum of the first and the second component of this equa-tion. Using the following formula
< χ|L|χ >=¯b i Z 2π 0 χ∗ ∂ ∂θχdθ (206)
The angular momentum of the first component is
< a1χ1|L|a1χ1 >= |a1|2A2 ¯b iJ 2 1(ω0r/c) Z 2π 0 e−iθ ∂ ∂θe iθdθ (207)
Rearranging and taking the derivative with respect to θ gives
< a1χ1|L|a1χ1>= |a1|2A2¯bJ12(ω0r/c)
Z 2π
0
e−iθeiθdθ ∝ |a1|2 (208)
The angular momentum on the second component is
< a2χ1|L|a2χ2 >= |a2|2A2 ¯b iJ 2 −1(ω0r/c) Z 2π 0 eiθ ∂ ∂θe −iθdθ (209)
Rearranging and taking the derivative with respect to θ gives
< a2χ1|L|a2χ2>= −|a2|2A2¯bJ−12 (ω0r/c)
Z 2π
0
eiθe−iθdθ ∝ −|a2|2 (210)
Hence the angular momentum of the first component is equal to |a1|2, while the angular
mo-mentum of the second component is equal to −|a2|2. The normalized total angular momentum
This equation can be rearranged to
σi=
a ·bσia∗
a∗· a (212)
where ˆσi is a 2 by 2 matrix. These matrices are called the Pauli matrices. In order to make
sure that we can use the Pauli matrices to describe the angular momentum of these two components, I will compute ˆσi
First I will compute the matrix ˆσx. The angular momentum of spin-up in the x-direction on
the Bloch sphere can be found using β = 12π and φ = 0. If we subsitute these values into (202), we find (a1, a2) = √12(1, 1). The angular momentum of spin-down in the x-direction
on the Bloch sphere can be found using β = 12π and φ = π. If we subsitute these values into (202), we find (a1, a2) = √12(1, −1).
Plugging these found values for (a1, a2) into (212) and equating it to the eigenvalues of the
matrices, we find the following to expressions.
σx = a · ˆσxa∗ a∗· a = 1 √ 2(1, 1) · ˆσx 1 √ 2(1, 1) ∗ 1 √ 2(1, 1) ∗·√1 2(1, 1) = √1 2(1, 1) · ˆσx 1 √ 2(1, 1) ∗= 1 (213) σx = a · ˆσxa∗ a∗· a = 1 √ 2(1, −1) · ˆσx 1 √ 2(1, −1) ∗ 1 √ 2(1, −1) ∗·√1 2(1, −1) = √1 2(1, −1) · ˆσx 1 √ 2(1, −1) ∗= −1 (214) Assume ˆσx is ˆ σx= a b c d
Then using the above expressions we can find the coefficients of this 2 by 2 matrix. At first we look at the above spin-up expression.
1 √ 2(1, 1) · a b c d 1 √ 2 1 1 = 1 2(a + c, b + d) · 1 1 = 1 2(a + b + c + d) = 1 (215)
To find another constraint on ˆσx, we now look at the spin-down expression. 1 √ 2(1, −1) · a b c d 1 √ 2 1 −1 = 1 2(a − c, b − d) · 1 −1 = 1 2(a − b − c + d) = −1 (216)
This gives us the following equation: a − b − c + d = −2. These two equations have infinitely many solutions. However I choose the ones in which a = d and b = c, because a and d have a plus in front, while b and c have a minus in front of them. This simplifies the two equations to:
2a + 2b =2 (217)
2a − 2b = −2 (218)
The solution to this equation is: a = d = 0 and b = c = 1. With these results the matrix ˆσx
is ˆ σx = 0 1 1 0 (219)
Second I will compute the matrix ˆσy. The angular momentum of spin-up in the y-direction
on the Bloch sphere can be found using β = 12π and φ = 12π. If we subsitute these values into (202), we find (a1, a2) = √12(1, i). The angular momentum of spin-down in the y-direction on
the Bloch sphere can be found using β = 12π and φ = 32π. If we subsitute these values into (202), we find (a1, a2) = √12(1, −i).
Following the previous computation, we plug these found values for (a1, a2) into (212) and
equate them to the eigenvalues of the matrices. This gives us the following expressions.
σy = a · ˆσya∗ a∗· a = 1 √ 2(1, i) · ˆσy 1 √ 2(1, i) ∗ 1 √ 2(1, i) ∗·√1 2(1, i) = √1 2(1, i) · ˆσy 1 √ 2(1, i) ∗ = 1 (220) σy = a · ˆσya∗ a∗· a = 1 √ 2(1, −i) · ˆσy 1 √ 2(1, −i) ∗ 1 √ 2(1, −i) ∗·√1 2(1, −i) = √1 2(1, −i) · ˆσy 1 √ 2(1, −i) ∗= −1 (221) Assume ˆσy is ˆ σy = a b c d
Then using the above expressions we can find the coefficients of this 2 by 2 matrix. At first we look at the above spin-up expression.
1 √ 2(1, i) · a b c d 1 √ 2 1 −i = 1 2(a + ic, b + id) · 1 −i = 1 2(a − ib + ic + d) = 1 (222)
From here it follows that a − ib + ic + d = 2. To find another constraint on ˆσy, we now look
at the spin-down expression.
1 √ 2(1, −i) · a b c d 1 √ 2 1 i = 1 2(a − ic, b − id) · 1 i = 1 2(a + ib − ic + d) = −1 (223)
This gives us the following equation: a + ib − ic + d = −2. These two equations have infinitely many solutions. However I choose the ones in which a = d and b = −c, because a and d have a plus-sign in front, while b and c have an i in front of them. This simplifies the two equations to:
2a − 2ib =2 (224)
2a + 2ib = −2 (225)
The solution to this equation is: a = d = 0, b = −i and c = i. With these results the matrix ˆ σy is ˆ σy = 0 −i i 0 (226)
Finally I wil compute the matrix ˆσz. The angular momentum of spin-up in the z-direction
on the Bloch sphere can be found using β = 0 and φ = 0. If we subsitute these values into (202), we find (a1, a2) = (1, 0). The angular momentum of spin-down in the z-direction on
the Bloch sphere can be found using β = π and φ = 0. If we subsitute these values into (202), we find (a1, a2) = (0, 1).
Following the previous computation, we plug these found values for (a1, a2) into (212) and
equate them to the eigenvalues of the matrices. This gives us the following expressions.
σz= a · ˆσza∗ a∗· a = (1, 0) · ˆσz(1, 0)∗ (1, 0) · (1, 0) = (1, 0) · ˆσz(1, 0) ∗ = 1 (227) σz= a · ˆσza∗ a∗· a = (0, 1) · ˆσz(0, 1)∗ (0, 1) · (0, 1) = (0, 1) · ˆσz(0, 1) ∗ = −1 (228) Assume ˆσz is ˆ σz = a b c d
Then using the above expressions we can find the coefficients of this 2 by 2 matrix. At first we look at the above spin-up expression.
(1, 0) · a b c d 1 0 = (a, b) · 1 0 = a = 1 (229)
Hence a = 1. To find another constraint on ˆσz, we now look at the spin-down expression.
(0, 1) · a b c d 0 1 = (c, d) · 0 1 = d = −1 (230)
Hence d = −1. b and c can be chosen arbitrarily. I choose them to be 0. With these results the matrix ˆσz is ˆ σz = 1 0 0 −1 (231)
2.14 Pauli exclusion prinicple
The figure below is a schematic figure of two bouncing droplet pairs close to each other. Droplet pair A is a solution of (202) with the parameters (β, φ) = (12π, 0). If you subsitute (β, φ) = (12π, 0) into (202), you find (a1, a2) = √12(1, 1). This tells us the y-value of both
droplets are the same. However, the x-values are opposite, since J1= −J−1. The droplets in
A are in phase with one another. This means they will repel eachother.
Droplet pair B is solution of (202) with the parameters (β, φ) = (12π,12π). If you subsitute (β, φ) = (12π,12π) into (202), you find (a1, a2) = √12(i, −i). This tells us the y-value of both
droplets are the opposite. The x-value of both droplets are also opposite for the same reason as droplet pair A. In the figure below, x = 0 has been chosen. This means the droplets in B have opposite phases, which means they will attract each other.
Figure 7: Two droplet pairs A and B close to each other. The droplets in red are spin-up and the droplets in green are spin-down. [3] (Brady & Anderson 2014)
After B has rotated 12π counterclockwise, the solution only depends on x. The wave height is the real part of the superposition of the wave height near the droplet pairs.
ζ = ζa(x +
d
2, t) + ζb(x − d
2, t) (232)
In this equation d is the distance between the pairs A and B. Using the following relation: ζb(x − d, t) = −ζa(x − d, t), we find: ζ = ζa(x + d 2, t) − ζa(x − d 2, t) (233)
3
Discussion
The goal of this project was to present the analysis of R. Brady and R. Anderson. Their anal-sysis tempted to prove that bouncing droplets exhibit quantum-like behaviour. Experiments have shown that they exhibit single-slit and double-slit diffraction, tunneling, quantised en-ergy levels, Anderson localisation and the creation/annihilation of droplet/bubble pairs. They explained it by showing that the surface waves are Lorentz covariant, that pairs of droplets experience an inverse-square force of attraction depending on their relative phase, that the droplets obey an analogue of the Schr¨odinger equation and that the bouncing droplets exhibit spin-half symmetry and align antisymmetrically.
In this section I will compare my results with their results. I will also be discussing how my results explain the quantum-like behaviour. My analysis on the system, where a droplet was made to bounce on a liquid bath, has started in a similar way as the analysis done by R. Brady and R. Anderson. It also started with solving the wave equation in two dimensions. After transforming this equation to the wave equation in polar coordinates, I have found this result:
h(r, t) = −h0cos(ω0t)J0(ω0r/c) (234)
This result is the same as in the article published by R. Brady and R. Anderson [3], if you imply the boundary condition h(r, t = 0) = −h0. Since this computation was pretty
straightforward and our results are similar, I have no reason to doubt my solution to the wave equation.
After solving the wave equation, I have shown using experimental data, that the surface waves guiding the droplets are Lorentz covariant. I have tried to prove this by first subjecting the waveheight h(r, t) to a Lorentz transformation and multiplying it by γ. This resulting waveheight function has been differentiating once with respect to x. Solving for ∂h∂x = 0 and simplifies gives the following expression:
vT = γ2(v
2
c2 +
1
2)∆x (235)
This equation is different than the equation found by R. Brady and R. Anderson. They found the following equation:
vT = γ2ω0(
v2 c2 +
1
Dimensional analysis shows that this equation can’t possibly be right, because ω0 has the
dimension 1s. However, the equation I have found has the right dimensions. This means my equation could possibly be right. After simplyfying and nothing that ω0 is just a constant.
We all get the same final equation, which is shown to be Lorentz covariant by experiment.
γ2(v 2 c2 + 1 2) = T τ (237)
Experiments had shown that the velocity normal to the boundary is given by following fomula: V⊥2 = V02 −B
r. Using this formula, I have shown that the force is a repulsive inverse square
force. This force is given by the following equation:
F = −α¯b1 r2 (238) α = 3A2(r0ω c ) 3 (239) ¯b = mc2 ω (240)
This equation is very much alike the Coulomb force and is similar to the formula found by R. Brady and R. Anderson. [3]
By defining h = ψχ with χ being the wave field χ and ψ being the wave function, I have found the wavelength of the wavefunction ψ. The wavelength of ψ is
λ = 2π k = 2πc2 ωvx = b p (241)
In this equation (241) b = 2πmcω2 and p = mvx. This equation is equivalent to the Broglie
wavelength, except that Plank’s constant h has been replaced with b. I have found the same formula for the wavelength as R. Brady and R. Anderson did.
Since by defining h = ψχ, one can imagine this system to be a pilot wave-like system. In this system the droplet ψ is governed by the wave field χ. Such a system might show diffraction when interfering with a single or double slit. Experiments show that this is actually the case. The histogram in figure 4, section 2.5, is found by Y. Couder ane E. Fort and show that there are indeed diffraction patterns.
The wave function ψ obeys the following ODE: ∂2ψ
∂t2 = −ω 2
0ψ (242)
To make this equation Lorentz covariant, I have used the following basic relavistic relation: E2−p2c2= m2
0c4. After simplifying and rearranging I end up with the Klein-Gordon equation:
∂2ψ ∂t2 − c
2∇2ψ = −ω2
0ψ (243)
This equation is a bit different than the equation found by R. Brady and R. Anderson, because they found ∂∂t2ψ2 − c2∇2ψ = ω20ψ. Comparing both equations with the literature shows that
my equation is the Klein-Gordon equation. It is likely that R. Brady and R. Anderson did find the right formula, because the missing minus-sign could be a typo.
Starting from the Klein-Gordon equation, I have also found that a droplet obeys the Schr¨odinger equation for a free particle. This equation is the following
i¯b∂ψs ∂t = −
¯b2
2m0
∇2ψs (244)
This formula differs from the formula found by R. Brady and R. Anderson, since their formula also has a potential term in it. The found they have found is
i¯b∂ψs ∂t = (− ¯ b2 2m0 ∇2+ V )ψs (245)
I am unable to transform the Schr¨odinger equation for a free particle to the Schr¨odinger equation with a potential term.
Starting with equation (138) and subsituting ψs= Reiθ gives the following equation
i∂R ∂te
iθ = − c2
2ω0
∇2(Reiθ) (246)
Simplifying and taking derivatives gives us the following equation ∂R2
∂t + ∇(R
This equation can be interpreted as a continuity equation with R2 being the fluid density ρ. Since R2 = |ψ|2, you can say that this system is distributed according to |ψ|2. This is a postulate in the Copenhagen interpretation of quantum mechanics. This kind of probability distribution also indicates the possibility of tunneling. This tunneling have been seen in experiments done by A. Eddi, E. Fort and Y. Couder. The result of this experiment is shown in the figure below. [6] (Eddi et al. 2009)
Figure 8: The probability of a droplet tunneling a region of reduced depth depending on the barrier width. The width of the barrier is indicated on the x-axis, while the probability is indicated of the y-axis. [6] (Eddi et al. 2009)
Since the probability distrubition depends on |ψ|2, the interaction with the edge of the con-tainer falls off as r−2 and the potential is of the form of a 2d-square well, the probability distrubition remains localized:
X
n∈Z2
|ψ(t, n)|2|n| ≤ C (248)
This equation is true for any time t. In this equation |n| stands for the position of the droplet in the xy-plane. This phenomenon is called Anderson localization. It means that there is no diffusion of waves. [7] (Anderson 1958)
Up until this point only circular and spherical symmetric waves have been considered. Ex-perimental results suggest that we also have to take a look at solutions to the wave equation which depend on angle. Hence the solution is of the form h(r, t, θ). In order to find solutions depending on angle, I had to solve the wave equation in polar coordinates again:
1 c2 ∂2h ∂t2 − ( ∂2h ∂r2 + 1 r ∂h ∂r + 1 r2 ∂2h ∂θ2) = 0 (249)
The wave height function h can be seperated into three different functions:
h(r, t, θ) = R(r)T (t)Θ(θ) (250) Seperation of variables and simplificiation gives us the following three ODEs with solutions:
d2R ds2+ 1 s dR ds + (1 − n2 s2)R = 0 (251) R(r) = Jn(ω0r/c) = ( ω0r c ) n ∞ X m=0 (−1)m(ω0 c ) 2mr2m 22m+nm!(n + m)! (252)
Notice that this position dependant part is similar to the position dependant part of the spherical symmetric solution. The only difference is that the Bessel function is now of order n instead of order 0.
∂2T (t) ∂t2 + c
2κ2T (t) = 0 (253)
T (t) = − h0cos(ω0t) (254)
Notice that the time depence is the same for the spherical symmetric solution.
∂2Θ(θ) ∂t2 + n
2Θ(θ) = 0 (255)
Θ(θ) =einθ (256)
Combining the three solutions and simplifying them as discussed in section 2.9, gives us the wave height function depending on angle.
This is a slightly different solution than R. Brady and R. Anderson have found. First of all I have a minus-sign instead of plus-sign, because they have chosen for h(0, 0, 0) = h0 instead
of h(0, 0, 0) = −h0. The boundary conditions I chose to find angle dependant solutions are
the same as the boundary I have used to find the spherical symmetric solutions. Moreover, I have found a difference angle dependence. By taking the real part of the solution I have found, you can find their solution.
These rotating waves can be treated indepently, because they are orthogonal. Since the following equation is true
Z 2π
0
h∗mhndθ = 0 if m 6= n (258)
The wave height function is orthogonal and quantized by the parameter n. Hence the energy levels of this system will also be quantized by this parameter n.
Experiments have shown that we only have to look at n = 1 and n = −1. Since these two waves are orthogonal, one can write the function h, which describes the height of the wave, as a superposition of h1 and h−1.
h = cos(α)h1+ sin(α)h−1 (259)
This new wave height function can be used to compute the angular momentum of a droplet. The angular momentum can be computed using the following formula:
< h|L|h >= ¯b i Z 2π 0 h ∂ ∂θhdθ (260)
The results of this computation have been summarized in Table 1 in section 2.11. This table contains the same values found by R. Brady and R. Anderson. If you look at this table, you can see that the sign reverses once the system goes throught an entire cycle. Fermions also show this kind of behaviour. They reverse in sign under a rotation of 360 degrees.
A Bloch’s sphere can be constructed (257) as shown in section 2.12 by first extending it to the complex plane. This gives the following formula.
This Bloch’s sphere can described by writing it in a similar way as (259). The formula is as follows χ = eiS[cos(1 2β)χ1+ e iφsin(1 2β)χ−1] (262)
where S is an arbritary phase and φ is the relative phase between χ1 and χ−1. This is the
formula which describes a Bloch’s sphere with χ1 being a spin-up particle and χ−1 being the
spin-down particle. β and φ describe the sphere as represented by the figure 5.
The mapping of a droplet’s wave height onto the Bloch sphere can also be written as a dot product of two vectors a and χ.
χ = a · χ = (a1, a2) · (χ1, χ−1) (263)
The values for a1 and a2 can be obtained from equation (262). The equation for χ as well as
this equation has been also been found by R. Brady and R. Anderson. The angular momentum of χ1 is proportional to |a1|2, while the angular momentum of χ−1 is proportional to −|a2|2.
These relations can be used to find out what the normalized total angular momentum of chi is. This is given by the following equation
σi=
|a1|2− |a 2|2
|a1|2+ |a2|2
(264) which can be rearranged to
σi=
a · ˆσia∗
a∗· a (265)
In this equation σi stands for one of the three Pauli matrices. This equation is slightly
differ-ent from the equation found by R. Brady and R. Anderson. I have chosen to write it in this form, because it enables matrix multiplication.
This equation can be used to find the Pauli matrices by substituting the found eigenvectors of both spin-up and spin-down in a certain direction and equating them to their corresponding eigenvalues. The eigenvectors can be found by subsituting certain values for β and φ into (262). I have used this equation in order to succesfully find the Pauli matrices. Since the bouncing droplets have angular momenta described by the Pauli matrices, you can say that they are much alike spin-1/2 particles.
The bouncing droplets even obey the Pauli exclusion principle, because the wave height function ζ is antisymmetric under droplet exchange. This can be seen in the following formula:
ζ = ζa(x +
d
2, t) − ζa(x − d
2, t) (266)
In this formula d is the distance between droplet pairs A and B. If you exchange particles A and B, the sign of the wave field reverses. The fact that the angular momentum of the droplet pairs are described by Pauli matrices and that they even obey the Paulli exclusion principle means that these droplets are really similar to spin-1/2 fermions. Brady and Anderson get the same resulting equation and they draw the same conclusions.
Now that I have succesfully repeated the analsysis done by R. Brady and R. Anderson, I have yet to proof why my results explain the creation/annihilation of droplet/bubble pairs. I can only imagine that this has something to do with the similarities between the droplets and spin-1/2 fermions, because we know from quantum electrodynamics that an electron and a positron can be created by a photon and that they can annihilate to two photons. However this similarilty in no way proves why this should be possible for this system. R. Brady and R. Anderson don’t explain in their article. They just make the statement that they have succesfully explained it, which I find hard to believe.
4
Conclusions
I have repeated the analysis done by R. Brady and R. Anderson on a system where droplets where made to float on a liquid bath. This system has shown some quantum-like behaviour in experiments. In particular it exhibits single-slit and double-slit diffraction, tunneling, quan-tised energy levels, Anderson localisation and the creation/annihilation of droplet/bubble pairs. R. Brady and R. Anderson have explained in their paper: ’Why bouncing droplets are a pretty good model of quantum mechanics’ these particular phenomena.
Starting from the wave equation, I have shown that the wave height near a droplet is described by
h(r, t) = −h0cos(ω0t)J0(ω0r/c) (267)
By applying a Lorentz transformation to this function and multiplying by a factor γ, I have shown that the surface waves guiding the droplet are Lorentz covariant. According to exper-iments, the system is Lorentz covariant up to a factor γ = 2.6.
From experimental data, the equation V⊥2 = V02− B
r can be deduced. This means droplets
experience a inverse square force which pushes them away from the boundary. This inverse square force is given my the following formula
F = −3A2(r0ω c ) 3mc2 ω 1 r2 (268)
By defining h = ψχ, where ψ = cos(−ω0t) is the wave function of the droplet and χ =
−h0J0(kr) is the wave field, you can imagine this system to a pilot wave-like system with
wavelength λ = 2πk = 2πcωv2
x =
b
p. This system shows diffraction patterns when subjected to a
single- or double-sllit as shown in figure 4.
I have shown that the wave function ψ obeys the Klein-Gordon equation for relavistic particles. ∂2ψ
∂t2 − c
2∇2ψ = −ω2
0ψ (269)
ψ also obeys the Schr¨odinger equation for a free particle.
i¯b∂ψs ∂t = −
¯b2
2m0
∇2ψs (270)
ψ also obeys the contuinity equation. This equation is as follows ∂R2
∂t + ∇(R
2v) = 0 (271)
In fluid dynamics the continuity equation is ∂ρ∂t + ∇(ρv) = 0. Comparing these two results tells us that, in the found equation, ρ has been replaced by R2. Since R2 =| ψs |2, one can
say that the configuration is distrubuted according to | ψs|2. This is in agreement with the
Copenhagen interpretation of quantum mechanics as well as the Broglie-Bohm theory. This kind of probability distribution also indicates the possibility of tunneling. Experiments have proven this possibility, which is shown in figure 7.
Experimental results suggest that we also have to take a look at solutions to the wave equation which depend on angle. Hence the solution is of the form h(r, t, θ). The solution is
hn(r, t, θ) = −h0cos(ω0t)einθJn(ω0r/c) (272)
These rotating waves can be treated independently, because they are orthogonal. Experiments suggest that only h1 and h−1 are relevant. Hence hn(r, t, θ) can be written as
h = cos(α)h1+ sin(α)h−1 (273)
Using this formula the angular momentum can be calculated. The angular momentum for different values of α has been summarized in Table 1. You can see in this table that the angular momentum reverses under rotation of 360 degrees. This behaviour is also shown by fermions.
You can construct a Bloch’s sphere using (272) and extending it to the complex plane. The Bloch’s sphere is described by the following formula.
χ = eiS[cos(1
2β)χ1+ e
iφsin(1
2β)χ−1] (274)
where S is an abritary phase and φ is the relative phase between χ1 and χ−1. This is the
formula which describes a Bloch’s sphere with χ1 being a spin-up particle and χ−1 being the
The mapping of a droplet’s wave height onto the Bloch sphere can also be written as a dot product of two vectors a and χ.
χ = a · χ = (a1, a2) · (χ1, χ−1) (275)
I have deduced that the normalized angular momentum is given by the following formula
σi=
|a1|2− |a2|2
|a1|2+ |a2|2
(276) Rewriting this equation gives
σi=
a · ˆσia∗
a∗· a (277)
In this equation σi stands for one of the three Pauli matrices. I have deduced the correct
Pauli matrices using this formula. This means that the angular momentum of the droplets are described by the Pauli matrices.
The bouncing droplets even obey the Pauli exclusion principle, because the wave height function ζ is antisymmetric under droplet exchange. This can be seen in the following formula:
ζ = ζa(x +
d
2, t) − ζa(x − d
2, t) (278)
In this formula d is the distance between droplet pairs A and B. If you exchange particles A and B, the sign of the wave field reverses. The fact that the angular momentum of the droplet pairs are described by Pauli matrices and that they even obey the Paulli exclusion principle means that these droplets are really similar to spin-1/2 fermions.
Concluding, I have succesfully repeated the analysis of R. Brady and R. Anderson in order to show that this non-quantum system has quantum-like behaviour. Alll formulas found by me are the same as their formulas. I have shown that the surface waves are Lorentz covari-ant, that pairs of droplets experience an inverse-square force of attraction depending on their relative phase, that the droplets obey an analogue of the Schr¨odinger equation and that the bouncing droplets exhibit spin-half symmetry and align antisymmetrically.
Using these characteristics, I have explained why the droplets exhibit single-slit and double-slit diffraction, tunneling, quantised energy levels and Anderson localisation. However I have failed to explain why the creation/annihilation of droplet/bubble pairs is possible. I think further investigation is needed to properly explain this phenemenon.
References
[1] Jearl Walker. Drops of liquid can be made to float on liquid - what enables them to do so? Scientific American, 238, 1978.
[2] Yves Couder, Suzie Proti´ere, Emmanuel Fort, and Arezki Boudaoud. Dynamical phenom-ena: Walking and orbiting droplets. Nature, 437, 2005.
[3] Robert Brady and Ross Anderson. Why bouncing droplets are a pretty good model of quantum mechanics. arXiv, 1401.
[4] Yves Couder and Emmanuel Fort. Single-particle diffraction and interference at a macro-scopic scale. Physical review letters, 97, 2006.
[5] Suzie Proti`ere, Arezki Boudaoud, and Yves Couder. Particle–wave association on a fluid interface. Journal of Fluid Mechanics, 554:85–108, 2006.
[6] A Eddi, Emmanuel Fort, F Moisy, and Yves Couder. Unpredictable tunneling of a classical wave-particle association. Physical review letters, 102(24):240401, 2009.
[7] Philip W Anderson. Absence of diffusion in certain random lattices. Physical review, 109(5):1492, 1958.
