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3 0 0 0 Z − mm mm δ z 3 − 1 ⎤ ⎥ ⎥ ⎥ ⎥ ⎦ • Can be applied to any point on the end effector x y z ' ' ' p p p 1 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ = 0 T 3 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ − 0 0 300 1 ⎫ ⎪ ⎪ ⎬ ⎪ ⎪ ⎭ Homework #5 • Short answers on TRIZ and probability • Error budgeting – Two tasks are to be done with the robot – Analyze the tasks ...
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7.3.6 Homogeneous and Inhomogeneous Broadening Laser media are also distinguished by the line broadening mechanisms in- volved. Very often it is the case that the linewidth observed in the absorption or emission spectrum is not only due to dephasing process that are acting on 310 CHAPTER 7. LASERS Laser Medium Wave- l...
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THz) T2 0.210 1.2 0.390 0.300 3 4 0.06 100 80 65 80 0.0015 0.0035 0.000060 5 25 Refr. index n 1.82 1.47 (ne) 1.82 (ne) 2.19 (ne) 1.5 1.46 1.76 1.76 1.4 1.4 1.4 1 ∼ 1 ∼ 1 ∼ 1.33 3 - 4 Typ H H H H H/I H/I H H H H H I I H H H/I Table 7.1: Wavelength range, cross-section for stimulated emission, upper- state lifetime, line...
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ously broadened laser medium (Nd:phosphate glass), [14] broadened medium the loss or gain saturates homogenously, i.e. the whole line is reduced. In an inhomogenously broadened medium a spectral hole burning occurs, i.e. only that sub-group of atoms that are sufficiently in resonance with the driving field saturate and th...
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− f0 Then the lineshape is a Gaussian g(f ) = mc 2πkT f0 exp mc2 2kT f 2 f0 − f0 ¶ . # µ "− r The full width at half maximum of the line is ∆f = 8 ln(2) r kT mc2 f0. (7.7) (7.8) (7.9) 7.4 Laser Dynamics (Single Mode) In this section we want study the single mode laser dynamics. The laser typically starts to lase in a f...
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. LASERS Figure 7.20: Laser gain and cavity loss spectra, longitudinal mode location, and laser output for multimode laser operation. where νg is the group velocity in the cavity in the frequency range considered, and L is the cavity length of the linear or ring cavity and 2∗ = 1 for the ring cavity and 2∗ = 2 for the ...
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1 2∗ hfLnLvg, (7.15) where hfL is the photon energy. 2∗ = 2 for a linear laser resonator (then only half of the photons are going in one direction), and 2∗ = 1 for a ring 316 CHAPTER 7. LASERS laser. In this first treatment we consider the case of space-independent rate equations, i.e. we assume that the laser is oscil...
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the inversion in the absence of a field, 1/τ L, is not only due to spontaneous emission, but is also a result of non radiative decay processes. See for example the four level system shown in Fig. 7.6. So the two rate equations are d dt d dt N2 = nL = − − N2 τ L − nL τ p + 2∗ 2∗σvgN2nL + Rp σvg V N2 nL + µ 1 V . ¶ (7.19)...
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¿ ¿ neglecting Pvac P or Psat = Esat/τ L, than g = g0 and we obtain from Eq.(7.22), dP P = 2 (g0 dt TR l) − P (t) = P (0)e2(g0 − l) t TR . (7.27) (7.28) The laser power builts up from vaccum fluctuations, see Figure 7.23 until it reaches the saturation power, when saturation of the gain sets in within the built-up time ...
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u p t u O g= =lgth P=0 0 gth g = =l Small signal gain g0 Figure 7.24: Output power and gain of a laser as a function of pump power. Substitution into Eqs.(7.21-7.22) and linearization leads to = +2 = − Ps TR gs Esat ∆g ∆P 1 τ stim − ∆g (7.34) (7.35) d∆P dt d∆g dt 1 + Ps P sat 1 τ stim = 1 τ L where is the inverse stimu...
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1 2τ stim à − j ± s 4 (r − r 1) τ stim τ p − , 1 ! = r 2τ L ± j − s (r 1) τ Lτ p − − 2 r 2τ L ¶ µ (7.40) (7.41) There are several conclusions to draw: • • (i): The stationary state (0, g0) for g0 < l and (Ps, gs) for g0 > l are si always stable, i.e. Re { < 0. } (ii): For lasers pumped above threshold, r > 1, and long ...
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Psat = IsatAef f = 1.8 W, Ps = 91.5W = 24μs, τ p = 1μs, ωR = 1 τ stimτ p = 2 · s 105s− 1. Figure 7.25 shows the typically observed fluctuations of the output of a solid- Figure 7.25: Relaxation oscillations in the time and frequency domain. state laser in the time and frequency domain. Note, that this laser has a long u...
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�στ L (7.46) (7.47) The power losses of lasers are due to the internal losses 2lint and the trans- mission T through the output coupling mirror. The internal losses can be a significant fraction of the total losses. The output power of the laser is Pout = T Psat · µ 2g0 2lint + T − 1 ¶ The pump power of a laser is minim...
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longer be neglected and we should use the full rate equations (7.21) and (7.22) d dt d dt g = P = gP Esat g g0 − τ L − 2g TR P + − − 1 τ p (P + Pvac) , (7.54) (7.55) where Pvac is the power of a single photon in the mode. The steady state conditions are gs = g0 (1 + Ps/Psat) , 0 = (2gs − 2l) P + 2gsPvac. (7.56) (7.57) ...
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2.0 Pump parameter r 1 -1 10 -2 10 -3 p v =10 10 -10 0.5 1.0 1.5 2.0 Pump parameter r p , r e w o p y t i v a c a r t n I p , r e w o p y t i v a c a r t n I Figure 7.26: Intracavity power as a function of pump parameter r on a linear scale (a) and a logarithmic scale (b) for various values of the normalized vacuum pow...
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7.91 / 20.490 / 6.874 / HST.506 7.36 / 20.390 / 6.802 C. Burge Lecture #10 March 11, 2014 Markov & Hidden Markov Models of Genomic & Protein Features 1 Modeling & Discovery of Sequence Motifs • Motif Discovery with Gibbs Sampling Algorithm • Information Content of a Motif • Parameter Es...
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0.8 1.0 0.1 0.0 0.0 1.0 0.4 0.1 0.1 0.1 0.8 0.0 0.2 0.5 S = S1 S2 S3 S4 S5 S6 S7 S8 S9 Ex: TAGGTCAGT P(S|+) = P-3(S1)P-2(S2)P-1(S3) ••• P5(S8)P6(S9) ‘Inhomogeneous’, assumes independence between positions What if this is not true? 4 Inhomogeneous 1st-Order Markov Model ...
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This content is excluded from our Creative Commons license. or more information, see http://ocw.mit.edu/help/faq-fair-use/. 6 Estimating Parameters for a Markov Model -3 -2 -1 1 2 3 4 5 6 P-2(A |C) = CA N (−3,−2) N (−3) C -3 -2 -1 1 2 3 4 5 6 What about longer-range depen...
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10 ML = maximum likelihood (of generating the observed data) Bayes est. = Bayesian posterior relative to Dirichlet prior Good treatment of this in appendix of: Biological Sequence Analysis by Durbin, Eddy, Krogh, Mitchison See also: Probability and Statistics Primer (under Materials > Resources) 9 ...
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A/a Phenotype - LDL cholesterol (observed) 150 Suppose that we can’t observe genotype directly, only some phenotype related to the A locus, and this phenotype depends probabilistically on the genotype. Then we have a Hidden Markov Model. a/a 250 Probability A/a a/a a/a 200 Bart 100 150 200 250 300 L...
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999 Pgi = 0.00001 Island … A C T C G A G T A “Emission Probabilities” bj(k) C CpG Island: 0.3 0.2 Genome: G 0.3 0.2 A 0.2 0.3 T 0.2 0.3 18 CpG Island HMM III Want to infer … A C T C G A G T A Observe But HMM is written in the other direction (observable depends on hidden) 19 Reversing the Condition...
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= h1,...,hn) P(O = o1, ...,on) P(O = o1 ,...,on) a bit tricky to calculate, but is independent of h1,…, hn so can treat as a constant and simply maximize = P( H = h1,..., hn, O = o1, ...,on) P(O = 1o , ..., no ) 22 ...
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” aij Pii = 0.999 Pgi = 0.00001 Island … A C T C G A G T A “Emission Probabilities” bj(k) C CpG Island: 0.3 0.2 Genome: G 0.3 0.2 A 0.2 0.3 T 0.2 0.3 δt(i) probability of optimal parse of the subsequence 1..t ending in state i ψt(i) the state at t-1 that resulted in the optimal parse of 1..t endin...
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past exams/Psets 1st, textbook 2nd Midterm exams from previous years are posted on course web site Note: there is some variation in topics from year to year 30 Midterm 1 Exam will cover course topics from Topics 1, 2 and 3 through Hidden Markov Models (but will NOT cover RNA Secondary Structure) R F...
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18.354 Nonlinear Dynamics II: Continuum Systems Spring 2015 Lecture Notes Instructor: J¨orn Dunkel Authors of Lecture Notes: Michael Brenner, Tom Peacock, Roman Stocker, Pedro Reis, J¨orn Dunkel 1 Contents 1 Math basics 1.1 Derivatives and differential equations . . . . . . . . . . . . . . . . . . . . . . . ....
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Taylor’s blast . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 The drag on a sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 Kepler’s problem and Hamiltonian dynamics 4.1 Kepler’s laws of planeta...
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. . 5.2 Derivation using probabilities . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Suggestions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Solving the diffusion equation 6.1 Fourier method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Green’s funct...
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dimension 7.4.2 Lotka-Volterra model 6 6 7 7 8 9 9 9 10 10 11 12 12 13 14 15 15 15 16 18 18 19 20 20 22 23 24 25 25 28 30 31 31 32 33 36 37 37 2 8 Variational Calculus 8.1 What is the shortest path between two points? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...
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. . . . . 10.3 Minimal surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Helfrich’s model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 Elasticity 11.1 Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ....
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13.1 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.2 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.3 Some simple solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13.4 The Reynolds number . . . . . . . . . . . . . ...
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. . . . . . . . . . . . . . . . . . . . . . . . . . . 14.6.3 Hele-Shaw flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 38 39 41 41 43 43 43 44 46 46 47 49 50 51 51 52 53 54 55 56 57 57 58 59 64 64 67 68 69 70 71 71 72 73 75 76 76 78 79 3 15 The coffee cup 1...
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. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.4 Flow around a cylindrical wing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18.5 Forces on the circular wing 19 Stream functions and conformal maps 19.1 The Cauchy-Riemann equations . ....
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. . . . . . . . . . . . . . . . . . . . . 98 99 21.2 Steady, inviscid flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Taylor columns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 21.4 More on rotating flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ...
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. . . . . . . . . . . . . . . . . . . . . 108 23.4 Flow created by a 1D ‘boat’ . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 4 24 Solitons 110 24.1 History . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 24.2 Korteweg-de Vries (KdV) equation . . . . . . . . . . . . . ...
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one variable. Depending on context, we will denote partial derivatives by ∂f ∂x i = ∂x f = ∂if = fx = f,i i i (2) In standard 3D Cartesian coordinates (x1, x2, x3) defined with respect to some global orthonormal frame Σ, spanned by the basis vectors (e1, e2, e3), the gradient-operator ∇ is defined by ∇ = ∂xe1 + ∂ye2 + ∂z...
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2x(t) (7) (8) which has fundamental sin- and cos-solutions, that can be used to construct more general solutions by superposition. An important (homogeneous) linear PDE, is Laplace’s equation ∇2f (t, x) = 0. (9) Functions f satisfying this equation are called harmonic. Later on, we will often try to approximate nonline...
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uchy-Riemann equations ∂xu = ∂yv , ∂yu = −∂xv (15a) 7 By differentiating again, we find ∂2 xu = ∂x∂yv = −∂2 y v = ∂y∂xu = −∂2 ∂2 y u = 0 xv = 0; this means that analytic functions f = (u, v) are harmonic ∇2f = 0. (15b) (15c) (15d) (15e) An analytic function that we will frequently encounter is the exponential function E...
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∞ yielding the useful Fourier representation (cid:90) ∞ dt eiωt δ(t) = 1 √ 2π , δ(t) = 1 (cid:90) ∞ 2π −∞ dω e−iωt (20a) (20b) (21) (22) These definitions and properties extend directly to higher dimensions. A main advantage of Fourier transformations is that they translate differential equations into simpler algebraic e...
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practice, however, we will rarely find useful exact solutions to the Navier-Stokes equations, and so dimensional analysis will often give us insight before diving into the mathematics or numerical simulations. Before formalising our approach, let us consider a few examples where simple dimensional arguments intuitively ...
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φ is the same for the two little triangles as the big triangle). Moreover, since these are all right triangles, the function f is the same for each. Therefore, since the area of the big triangle is just the sum of the areas of the little ones, we have or L2f = a2f + b2f, L2 = a2 + b2. (26) 2.3 The gravitational oscilla...
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tension γ 10 provides the relevant restoring force and we can neglect gravity. γ has dimensions of en- ergy/area, so that [γ] = M T −2. The only quantity we can now make with the dimensions of T −1 using our physical variables is (cid:114) ω = c γ ρR3 , (29) which is not independent of the radius. For water γ = 0.07Nm...
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when the wave- length gets small. In this case we replace g with γ in our list of physically relevant variables. Given that [γ] = M T −2, the dispersion relation must be of the form (cid:112) ω = c γk3/ρ, (33) which is very different to that for gravity waves. If we look for a crossover, we find that the frequencies of g...
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theorem does not predict the functional form of F or G, and this must be determined experimentally. The n − d dimensionless groups Πi are independent. A dimensionless group Πi is not independent if it can be formed from a product or quotient of other dimensionless groups in the problem. (36b) 3.1 The pendulum To develo...
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fied information, so he resorted to dimensional analysis. In a nuclear explosion there is an essentially instantaneous release of energy E in a small region of space. This produces a spherical shock wave, with the pressure inside the shock wave several thousands of times greater than the initial air pressure, that can b...
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. 3.3 The drag on a sphere Now what happens if you have two dimensionless groups in a problem? Let’s consider the problem of the drag on a sphere. We reason that the drag on a sphere D will depend on the relative velocity, U , the sphere radius, R, the fluid density ρ and the fluid viscosity µ. Thus or D = D(U, R, ρ, µ) ...
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∝ µU R. (52) This is Stokes drag, caused by the viscosity of the fluid. The power of taking this approach can now be seen. Without dimensional analysis, to determine the functional dependence of the drag on the relevant physical variables would have required four sets of experiments to determine the functional depe...
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1687 that the origins of this motion were understood. Newton proposed that “Every object in the Universe attracts every other object with a force directed along a line of centres for the two objects that is proportional to the product of their masses and inversely proportional to the square of the separation of the...
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plane polar coordinates dr dt d2r dt2 m d2r dt2 = f (r)rˆ. = r˙rˆ + rθθˆ, ˙ = (r¨ − ˙2 ¨ rθ )ˆr + (rθ + 2r˙θ)θˆ. ˙ In component form, equation (56) therefore becomes ˙ m(r¨ − rθ2) = f (r), ˙ m(rθ + 2r˙θ) = 0. ¨ Putting (57a) into (53) gives Thus (cid:12) (cid:12) L = (cid:12) (cid:12) r ∧ m r d dt (cid:12) (cid:12) (ci...
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63) This is the differential equation governing the motion of a particle under a central force. Conversely, if one is given the polar equation of the orbit r = r(θ), the force function can be derived by differentiating and putting the result into the differential equation. According to Newtons law of gravitation f (r) = −...
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17 4.1.3 Third law To prove Keplers third law go back to equation (55). Integrating this area law over time gives 0 where τ is the period of the orbit. The area A of an ellipse can also be written as A = πab where a and b are the semi-major and semi-minor axes, yielding A(τ ) = dA = , (70) (cid:90) τ lτ 2 lτ 2 = πab T...
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(xn, xk). (76b) 18 Given H, Newton’s equations can be compactly rewritten as x˙ n = ∇pnH , p˙ n = −∇xnH. (77) That this so-called Hamiltonian dynamics is indeed equivalent to Newton’s laws of motion can be seen by direct insertion, which yields x˙ n = n , p m n p˙ n = mnx¨ n = −∇x nU. (78) An important observation is ...
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ing the sun. If we consider a more realistic problem in which several planets orbit the sun, all in- teracting with each other via gravity, the problem becomes analytically intractable. Indeed, for just two planets orbiting the sun one encounters the celebrated ‘three-body problem’, for which there is no general analyt...
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out to be of fundamental importance, but was not recognised universally until the 1920’s. (iii) The mathematics of how to solve ‘macroscopic equations’, which are nonlinear partial differential equations, is non-trivial. We will need to introduce many new ideas. In tackling these problems we will spend a lot of time...
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0 at position x = 0 and execute a random walk according to the following rules: (i) Each particle steps to the right or the left once every τ seconds, moving a distance dxi = ±δ. 20 (ii) The probability of going to the right at each step is 1/2, and the probability of going to the left is 1/2, independently of ...
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:88) (cid:42)(cid:34) 1 N xi(n) (cid:35) (cid:34) 1 N N (cid:88) k=1 (cid:35)(cid:43) xk(n) = = = (cid:104)xi(n)xk(n)(cid:105) i=1 N (cid:88) N 1 (cid:88) N 2 i=1 k=1 N (cid:88) (cid:104)xi(n)2(cid:105), 1 N 2 i=1 (83) where we have used that, by virtue of assumption (iii), (cid:104)xi(n)xk(n)(cid:105) = (cid:104)xi(n)...
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85) (86) If we write n in terms of time such that t = nτ , where τ is the time in between each step, then (cid:112) (cid:104)xi(n)2(cid:105) = (cid:19) 1 2 (cid:18) t τ δ = (cid:19) 1 2 (cid:18) δ2t τ = (cid:18) δ2 τ (cid:19) 1 2 √ t. (87) The root-mean-square displacement of each particle is proportional to the square...
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where D = δ2/2τ is the diffusion coefficient and n(x, t) = N (x, t) Aδ is the particle density (i.e., the number of particles per unit volume at position x at time t). If δ is assumed to be very small, then in the limit δ → 0, the flux becomes ∂n Jx = −D , ∂x (92) where we have ignored higher-order derivatives in making th...
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��usion equation using a different approach, involving probabilities. Assuming non-interacting particles, the number of particles in the interval [x − δ/2, x + δ/2] at any given time is N (x, t) = N0P (x, t) = N0p(x, t)δ, (95) where N0 is the total number of particles in the sample, P (x, t) = p(x, t)δ the probability o...
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we note that the number density of particles is given by n = N0p, and our derivation is complete. Note that we have ignored higher-order terms in the Taylor expansion. Additional terms would give us δ2 2τ Are we allowed to ignore these extra terms? To see if we are, compare the ratio of the neglected terms on the right...
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, the ideas we have discussed are applied to a number of biological phenomena, including the motion of bacteria (which would make a good course project). 24 6 Solving the diffusion equation We have shown, through two different arguments, that the density of random walkers on a one dimensional lattice obeys the diffusion ...
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ives (cid:90) ∞ ∂n −∞ ∂t e−ikxdx = (cid:90) ∞ −∞ D ∂2n ∂x2 e−ikxdx ∂nˆ(k, t) ∂t = (ik)2Dnˆ(k, t) = −k2Dnˆ(k, t). Given the initial condition n0(x) → nˆ0(k) the solution of (107) is nˆ(k, t) = nˆ0(k)e−Dk t. 2 (106) (107) (108) 3An intuitive way of thinking is to note that a plane wave can be written as eikx = cos(kx) + ...
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(cid:90) ∞ nˆ0(k) = √ 1 2πσ2 −∞ e−ikx− x2 2σ2 dx = √ 1 2πσ2 −∞ − (cid:16) x2 2σ e (cid:17) 2 +ikx dx. (111) Completing the square for the exponent x2 2σ2 + ikx = 1 2σ2 (cid:0)x2 + 2σ2ikx(cid:1) = 1 (cid:104)(cid:0) 2σ2 (cid:1) x + ikσ2 2 (cid:105) + k2σ4 . enables (111) to be rewritten as nˆ0(k) = 2 2 2 e− k σ √ 2πσ2 (...
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2σ2 dz + −R+ikσ2 R+ikσ2 (cid:90) −R R e− z2 22σ dz + (cid:90) −R+ikσ2 −R − z e 2 22σ dz (116) 26 In the limit R → ∞, the second as well as the last integral vanish due to the exponential damping with large R, leading to (cid:90) R+ikσ2 lim R→∞ −R+ikσ2 2 − z e 2σ2 dz = lim R→∞ (cid:90) R −R e− z2 22σ dz . Inserting thi...
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σ2 2 (cid:18) − ikx = Dt + (cid:19) (cid:40)(cid:20) σ2 2 k − ix 2(Dt + σ2/2) (cid:21)2 + x2 2 4 (Dt + σ2/2) (cid:41) , (122) so that the integral becomes n(x, t) = 2 x 4(Dt+σ2/2) (cid:90) ∞ − e 2π −∞ (cid:104) −(Dt+σ2/2) e k− ix (cid:105)2 2 2(Dt+σ /2) dk. (123) This is essentially the same integral that we had before...
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method relies on another trick for representing the solution, that is somewhat more intuitive. Now, instead of representing n in a basis of plane wave states, we will express it as a basis of states which are localised in position. This is done by using the so-called Dirac delta function, denoted δ(x − x0). You should ...
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2π −∞ eikx dk. 28 spike to find the final density distribution. We define the Green’s function G(x − x(cid:48), t) so that G(x − x(cid:48), 0) = δ(x − x(cid:48)), and n(x, t) = (cid:90) ∞ −∞ G(x − x(cid:48), t) n0(x(cid:48)) dx(cid:48). Plugging this into the diffusion equation we see that n0 x(cid:48) ∂G(x − x(cid:48), t...
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diffusion equation, taking a single time derivative of the function gives a number of about the same size as when you take two spatial derivatives. This means that the characteristic length scale over which n varies is of order t. Now, since the initial distribution δ is perfectly localised, we expect that at time t, G(...
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the time factors and integrating this equation once gives − F y = DF (cid:48). 1 2 This equation can be immediately integrated to give F (y) = F 0e−y 2 / 4D, and thus G(x − x(cid:48) F0 , t) = √ t e− (x−x(cid:48))2 4Dt , where the constant F0 = 1/ √ 4πD is determined by requiring that (cid:82) dx G = 1. (134) (135) (13...
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a constant drift velocity u in one dimension, described by the conservation law with current ∂n ∂t = − ∂ ∂x Jx Jx = un − D n. ∂ ∂x 30 (140) (141) Interpreting x > 0 as the distance from the bottom of a vessel and assuming reflective boundaries at x = 0, we may think of u arising from the effects of gravity u = −gm∗ ...
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governed by the nonlinear ODE d dt v(t) = −(α + βv2)v =: f (v). (148) We assume that the parameter β is strictly positive, but allow α to be either positive or negative. The fixed points of Eq. (148) are, by definition, velocity values v∗ that satisfy 0. For α < 0, the condition f (v ) = 0. For α > 0, there exists ...
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(v∗)t. (cid:48) (149) (150) (151) (152) If f (cid:48)(v∗) > 0, then the perturbation will grow and the fixed point is said to be linearly unstable. whereas for f (cid:48)(v∗) < 0 the perturbation will decay implying that the fixed point is stable. For our specific example, we find f (cid:48)(v0) = (cid:0)α , f (cid:48)(v±)...
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155) (156) To evaluate its stability, we can consider wave-like perturbations n(x, t) = n0 + δn(x, t) , δn = (cid:15) eσt−ikx. Inserting this perturbation ansatz into (154) gives the dispersion relation σ(k) = (cid:0)Dk2 (cid:21) 0, (157) (158) signaling that n0 is a stable solution, because all modes with jkj > 0 bec...
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more ‘nonlocal’ than the diffusion equation (154). The field ψ could, for example, quantify local energy fluctuations, local alignment, phase differences, or vorticity. In general, it is very challenging to derive the exact functional de- pendence between macroscopic transport coefficients (a, b, c, γ1, γ2) and microscopic i...
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as long as γ0 > 0. If, however, physical or biological mechanisms can cause γ0 to become negative, then higher-order damping terms, such as the γ2-term in (159), cannot be neglected any longer as they are essential for ensuring stability at large wave-numbers, as we shall see next. Linear stability analysis The fixed po...
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affect the motion of individual microorganisms or cells can be implemented into macroscopic field equations that describe large collections of such cells. To demonstrate this, we interpret ψ as a vorticity- like 2D pseudo-scalar field that quantifies local angular momentum in a dense microbial suspension, assumed to be con...
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a swimming cell can exhibit a preferred handedness. For example, the bacteria Escherichia coli and Caulobacter have been observed to swim in circles when confined near to a solid surface. More precisely, due to an intrinsic chirality in their swimming apparatus, these organisms move on circular orbits in clockwise (anti...
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viewed from the bulk fluid. (cid:54)→ − 7.4 Reaction-diffusion (RD) systems RD systems provide another generic way of modeling structure formation in chemical and biological systems. The idea that RD processes could be responsible for morphogenesis goes back to a 1952 paper by Alan Turing (see class slides), and it seems...
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�(cid:15) eσt−ikx = (cid:18) (cid:19) (cid:15)ˆ ηˆ eσt−ikx, σˆ(cid:15) = (cid:0) (cid:18) k2Du 0 (cid:19) 0 k2Dv ˆ(cid:15) + (cid:18) (cid:19) u F ∗ F ∗ v u G∗ ˆ(cid:15) (cid:17) M ˆ(cid:15), G∗ v (169a) (169b) (170) where F ∗ u = ∂uF (u∗, v∗) , F ∗ v = ∂ vF (u∗, v∗) , ∗ Gu = ∂uG(u∗, v∗) , G∗ v = ∂vG(u∗, v∗). Solving t...
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x) measures the concentration of prey and v(t, x) that of the predators. The model has two fixed points (u0, v0) = (0, 0) , (u , v ) = (C/E, A/B), ∗ ∗ with Jacobians and Fu(u0, v0) Fv(u0, v0) Gu(u0, v0) Gv(u0, v0) = A 0 0 −C ∗ Fu(u v ) F , v ) ∗ Gu(u , v ) Gv(u , v ) ∗ v(u , ∗ ∗ ∗ ∗ ∗ = − E C B A C A − −C...
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this case the state of the system is described not by discrete variables xi but by functions fi(x). A good example of the difference is to consider a mass-spring system. If there are n masses connected by springs then we would want to know xi(t), the position of the ith mass. The limit of infinitely many small masses ...
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h) . (cid:18) d dx (177) Taylor-expanding the integrand for small δh and keeping only terms of linear order in δh(cid:48) = dδh/dx, we find Integrating by parts gives x1 L[h + δh] − L[h] (cid:39) dx √ (cid:90) x2 h(cid:48)δh(cid:48) 1 + h(cid:48)2 . L[h + δh] − L[h] (cid:39) √ (cid:20) h(cid:48) 1 + h(cid:48)2 (cid:21)x...
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related to the delta function by ∆x0[f ] = (cid:90) ∞ −∞ dx δ(x − x0)f (x) = f (x0), (181a) or integrals J[f ] = (cid:90) ∞ −∞ dx f (x) c(x). A functional I is called linear, if it satisfies I[af + bg] = aI[f ] + bI[g] 39 (181b) (182) for arbitrary numbers a, b and functions f, g. Obviously, both examples in Eq. (181) ...
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{F [f ]G[f ]} = δF δf G[f ] + F δG δf δ δf (F [g(f )]) = δ δf g(F [f ]) = δ(F [g(f )]) δg dg(F [f ]) dF dg(x) df (x) δ(F [f ]) δf (187b) (187c) (187d) As a nice little exercise, you can use the above properties to prove that the exponential functional satisfies the functional differential equation (cid:82) F [f ] = e dx ...
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− y) dx − d dx ∂f (cid:21) ∂Y (cid:48) δ(x − y)dx. Equating this to zero, yields the Euler-Lagrange equations 0 = ∂f ∂Y − d dx ∂f ∂Y (cid:48) (191) (192) In fact, the relation might even indicate a maximum. It should be noted that the condition δI/δY = 0 alone is not a sufficient condition for It is often possible, a min...
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from A to B is (cid:90) B A dt = (cid:90) B A ds 2g[h0 − h(x)] . (cid:112) We know that ds = 1 + h(cid:48)2, so that the time taken is √ The integrand T [h] = (cid:115) dx (cid:90) B A 1 + h(cid:48)2 2g(h0 − h) . (cid:115) f = 1 + h(cid:48)2 2g(h0 − h) (195) (196) (197) determines the time of descent. We can insert f d...
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x0 + a(θ − sin θ), h = h0 − a(1 − cos θ). (203) These are the equations of a cycloid generated by the motion of a fixed point on the circumference of a circle of radius a, which rolls on the negative side of the given line h = h0. By adjustments of the arbitrary constants, a and x0 it is always possible to construct one...
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2 − λ (cid:90) B dx y the second integral being the volume of the bubble. We combine this into (cid:90) E[y] = (cid:104) (cid:112) dx γ 1 + y(cid:48)2 − λy (cid:105) Inserting the integrand into the Euler-Lagrange equations gives (cid:33)(cid:48) (cid:32) γ (cid:112) y(cid:48) 1 + y(cid:48)2 = λ. (206) (207) (208) The ...
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enoid 8 r = a cosh(x/a), (213) which corresponds to the special case R/a = cosh(1). For other ratios, we need to solve (212) numerically. 9.3 Rayleigh-Plateau Instability In the previous section, we considered the shape of a soap film, stretched between two hoops. We know, however, that the film breaks after a certain ex...
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surfaces in more detail. 44 For the perfect cylinder y(x) = r0 is a solution with λ = γ/r0. Note that λ has dimensions of force/area, and is in fact the pressure. We now consider a perturbation such that r(x) = r0 + (cid:15) cos kx. (216) This perturbation is in some sense arbitrary, because any perturbation can be de...
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determine the energy per wavelength. We know that (cid:90) 2π 0 dx sin2 kx = (cid:90) 2π 0 dx cos2 kx = π k , so that the energy change, ∆E, is ∆E = (cid:15)2γπ2 (cid:18) r0k − (cid:19) . 1 r0k (220) (221) √ So now we see that if r0k > 1 the system is stable, because ∆E is positive. However, if r0k < 1 then ∆E is negat...
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there is a pinch off?). What would happen if the ends of the two hoops were closed, so that the volume constraint did then apply? Our stability analysis would now seem more relevant. The only way to check is to now go do some experiments (a good course project!). 10 Some basic differential geometry In the next secti...
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local torsion τ (t) by κ(t) = t˙ · n , || r˙ || τ (t) = n˙ b · . || r˙ || 46 (226) (227) Plane curves satisfy, by definition, b = const. or, equivalently, τ = 0. Given ||r˙ ||, κ(t), τ (t) and the initial values {t(0), n(0), b(0)}, the Frenet frames along the curve can be obtained by solving the Frenet-Serr...
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imizing this expression with respect to the polymer shape h yields the Euler-Lagrange equation hxx = 0. 10.2 Two-dimensional surfaces We now consider an orientable surface in R3. Possible local parameterizations are F (s1, s2) ∈ R3 (231) (232) where (s1, s2) ∈ U ⊆ R2. Alternatively, if one chooses Cartesian coordinates...
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fxfy 2 1 + fy and its determinant |g| = 1 + f 2 x + f 2 y , (236) (237a) (237b) where fx = ∂xf and fy = ∂yf . For later use, we still note that the inverse of the metric tensor is given by g−1 = (g−1 ij ) = (cid:18) 1 x + f 2 −y 1 + f 2 1 + f 2 y −fxfy 1 + f 2 fyf x x (cid:19) . (237c) Assuming the surface is regular a...
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x =  0  , F yy =  0  (242a) fxx fxy fyy yielding the curvature tensor (Rij) = (cid:18) N · F xx N · F xy N · F yx N · F yy (cid:19) = (cid:113) 1 1 + f 2 x + f 2 y (cid:19) (cid:18) fxx fxy fyy fyx (242b) Denoting the eigenvalues of the matrix g−1 · R by κ1 and κ2, we obtain for the mean curvature H = 1 2 (κ1 + κ2)...
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245) implies that, for any closed surface, the integral over K is always a constant. That is, for closed membranes, the first integral in Eq. (245) represents just a trivial (constant) energetic contribution. 10.3 Minimal surfaces Minimal surfaces are surfaces that minimize the area within a given contour ∂M , A(M |∂M )...
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following geometric curvature energy per unit area for a closed membrane (cid:15) = (2H − c0)2 + kG K, kc 2 (252) where constants kc, kG are bending rigidities and c0 is the spontaneous curvature of the membrane. The full free energy for a closed membrane can then be written as (cid:90) Ec = dA (cid:15) + σ (cid:90) dA...
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-trivial set of boundary conditions. 11 Elasticity What shape does a piece of paper take when we push it in at the ends? To answer this question let’s acquaint ourselves with another continuum approximation, used to describe the deformation of elastic solids (we might actually have studied this before our work on ...
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body is dx(cid:48) 2 + dx2 1 + dx2 i = dxi + dui. This distance between the points was originally dl = 1 + dx(cid:48)2 3 . Using the summation convention, which tells us to sum over repeated indices (i.e., aibi = a1b1 + a2b2 + a3b3), and substituting in that dui = (∂ui/∂xk)dxk, we get 3 and is now dl(cid:48) = dx(cid:4...
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a volume element V can be obtained from stresses by transforming a surface integral into a volume integral σik = 0. (264) (cid:90) ∂V (cid:90) σikdAk = (∂kσik)dV = V (cid:90) V fidV Hence, the vector fi must be given by the divergence of the stress tensor σik. fi = ∂σik ∂xk . (265) (266) We recognize that σikdAk is the...
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