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Course summary 18.354 w simplest Kepler’s to derive, and is a statement tha problem er a central force, such as gravity, is con a particle with mass m and velocity u is dr L=r⇥m , dt ition of the particle. The rate of change dL = r ⇥ f = r ⇥ f (r)r̂ = 0, dt where we have ignored higher order derivatives in making the approximati Now consider a single box with boundaries at x /2 and x + /2. In a Jx (x /2, t)A⌧ particles will enter from the left and Jx (x + /2, t)A⌧ pa through the right boundary. The number of particles in the box increases Random walks & diffusion N (x, t + ⌧ ) N (x, t) = [Jx (x /2, t) Jx (x + /2, t)] A By dividing both sides by A ⌧ the number of particles per unit volume in is seen to increase at the rate1 n(x, t + ⌧ ) ⌧ In the limit ⌧ ! 0 and n(x, t) = /2, t)] @Jx @2n = D 2, @x @x For strictly one-dimensional systems, the boundary area is just a point and, hence, 7 Mark Haw David Walker Jx (x ! 0, this becomes @n = @t 1 [Jx (x + /2, t) e V is Z Z d/dt inside the integral sign we must take account pndS = rpdV. (7) of this. The Reyno ⇢u · ndS, ⇢dV.so, and it can be shown that for a deforming, (1)incompressible flu does V Z S Z S(t) are being deformed by the motion of the fluid, so if we want to take the d Du eaving this volume through the bounding surface S is where u(x, t) is the velocity of the fluid and=n is the norm ⇢udV ⇢ outward dV ntegral sign we must take account of this. The Reynolds transport theorem dt V (t) V (t) Dt Z Z Z can be shown that for a deforming, incompressibleZfluid element @⇢ ⇢u · ndS, Here dV = ⇢u · ndS(2) = r · (⇢u)d Z Z S V @t S V d Du Hydrodynamics D @ (8) ⇢udV = ⇢ dV = have + (u · r) Dt (t) V (t) the velocity ofdttheV fluid and is thehold outward normal. Hence we Thisn must for any arbitrary fluid element dV , thus Dt @t Z Z Z is called the convective derivative, and @⇢we shall discuss it’s significance @⇢ + r · (⇢u) dV = ⇢u · assuming ndS = thatr (3) = 0. F ·is(⇢u)dV. solely given by gravity, @t D [email protected] V @t V = + (u · r) Z Z(9) Dt @tThis is called the continuity equation. Du ⇢ dV = ( rp + ⇢g)dV for any arbitrary fluid element dV , thus V (t) Dt (t) For fluids like water, the density does not Vchange very much and vective derivative, and we shall discuss it’s significance in a moment. Hence, @⇢ to neglect the density variations. If we make this approximation F is solely given by gravity, + r Since · (⇢u)this = 0.must hold for any arbitrary fluid element (4) we arrive at to the incompressibility condition @t reduces Z Z Du Du (10) rp ⇢ dV = ( rp + ⇢g)dV = + g. e continuity Vequation. Dt Dt r · u⇢ = 0. (t) V (t) e water, the density does not change very much and we will often be tempted This, combined with equation (4), constitutes Eu hold forvariations. any arbitrary fluid element we arrive at the the ensity If we make this approximation thecontinuity continuity equation Like all approximations, this one is sometimes very good andthesom can be tidied up a little if we realise that the gravitational force, being ncompressibility condition will have to figure out where it fails. Du rp = written + g.as the gradient of a scalar potential r (11). It is therefore usual t Dt p⇢+ ! p. This implies that gravity simply modifies the pressure di r6.1.2 ·u = 0.Momentum (5) equations and does nothing to change the velocity. However, we cannot do this with the the continuity equation (4),have constitutes the Euler equations. Things if we a free surface (as we shall see later with water waves). mations, this one is sometimes very good and sometimes not so good. We So far we have more unknowns than equations (three velocity co p a little if we realise that the gravitational force, being conservative, can be Assuming the density is constant means we now have four equatio Dimensional analysis Elasticity Calculus of variations What is the di↵erential equation satisfied by Y (x)? To answer this question, we compute What is di↵erential equation satisfied by Y (x)? To answer this question, we compute thethe functional derivative he functional derivative I[Y ] 1 = lim {I[f (x) + ✏ (x y)] I[f (x)]} ✏!0 I[Y ] Y 1 ✏ Z = lim x2{I[f I[f (x)]} @f (x) + ✏ (x @f y)] 0 Y =✏!0 ✏ Z x2x1 @Y (x y) + @Y 0 (x y) dx @f 0 Z x2 @f = (x y) dx @f (x d y) @f+ 0 = x1 (x y)dx. @Y @Y 0 Z x2x1 @Y dx @Y @f d @f = the Euler-Lagrange equations (x y)dx. Equating this to zero, yields 0 @Y dx @Y x1 @f d @f Equating this to zero, yields the Euler-Lagrange equations @Y dx @Y 0 0 = (17) (17) (18) It should be noted that the condition I/ @f Y = d0 alone @f is not a sufficient condition for 0 might = even indicate 0 a maximum. It is often possible, (18) a minimum. In fact, the relation @Y dx @Y however, to convince oneself that no maximum exists for the integral (e.g., the distance t should noted that thebecondition I/asYwe=like), 0 alone is not a sufficient condition for alongbea smooth path can made as long and that our solution is a minimum. To be rigorous, however, one should also consider the possibility that the minimum is merely minimum. In fact, the relation might even indicate a maximum. It is often possible, a local minimum, oneself or perhaps the no relation I/ Y = 0 indicates a point of inflexion. owever, to convince that maximum exists for the integral (e.g., the distance It is easy to check thatmade Eq. (18) yieldsasthe equations long a smooth path can be as long weNewton like), and that our solution is a minimum. Surface tension Singular perturbations tary di↵erential equation (Acheson, pp. 2 ↵erential equation 2 d u du ✏ 2+ = 1. dx dx might argue that we can neglect this term, the solut u = x + C. onsider the full problem the solution is e,the using the Cauchy-Riemann we see that Z-plane uniform flow pastequations a more wing-like shape? (Note that we have d some technical details here, such2as the requirement that dF/dz 6= 0 at any 2 @ @ @ @ would cause+a blow-up of the + velocity).= 0. = (6) 2 2 @x @y @[email protected] @[email protected] Conformal mappings le conformal maps e used for . We can therefore consider any analytic function (e.g., he real map is and imaginary parts and both of them satisfy Laplace’s equaZ = F (z) = z + b, (12) onents u and v are directly related to dw/dz, which is conveniently onds to a translation. Then there is dw @ @ = +Zi = F= u= ze iv.i↵ , (7) (13) (z) dz @x @x consider at anangle angle↵.↵ to The corresponding onds to auniform rotationflow through In the this x-axis. case, the complex potential for i↵ . Using the above relation, wast = au0cylinder ze i↵ . In this case dw/dz = u e 0 making angle ↵ with the stream is u0 cos ↵ and v = u0 sin✓↵. ◆ 2 R past i mine the complex potential for since we know that i↵ flow i↵ a cylinder W (Z) = u0 Ze + e ln Z. (14) Z 2⇡ ✓ ◆ R2 = u r + cos ✓, (8) i↵ 0 s expression could also include the term ln e = i↵ which I have neglected. r constant however and doesn’t change the velocity. al the complex potential re part is theofnon-trivial Joukowski transformation, ✓ ◆ 2 2 completely controls the dynamics. In the process of deriving this result we rn about a rather remarkable phenomenon in rotating fluid dynamics. Rotating flows The Taylor-Proudman theorem r a fluid rotating with angular velocity ⌦. The equation of motion in the fram e rotating with the fluid is @u 1 + u · ru + ⌦ ⇥ (⌦ ⇥ r) = rp⌦ + ⌫r2 u @t ⇢ r · u = 0. 2⌦ ⇥ u, re two additional terms: the first ⌦ ⇥ (⌦ ⇥ r) is the centrifugal acceleration, w e discussed before. This can be thought of as an augmentation to the pre tion, using the identity Taylor columns, 1 etc ⌦ ⇥ (⌦ ⇥ r) = 2 r(⌦ ⇥ r)2 . rth, we will simply absorb this into the pressure by writing p = p⌦ ⇢ r(⌦ ⇥ r)2 . Solitons KdV equation Active matter FIG. 12: (a) Hedgehog defect. Magnets have no line defects (you can’t lasso a basketball), but do have point defects. ⃗ (x) = M0 x̂. You can’t Here is shown the hedgehog defect, M surround a point defect in three dimensions with a loop, but you can enclose it in a sphere. The order parameter space, remember, is also a sphere. The order parameter field takes the enclosing sphere and maps it onto the order parameter space, wrapping it exactly once. The point defects in magnets are categorized by this wrapping number: the second Homotopy group of the sphere is Z, the integers. (b) Defect line in a nematic liquid crystal. You can’t lasso the sphere, but you can lasso a hemisphere! Here is the defect corresponding to the path shown in figure 5. As you pass clockwise around the defect line, the order parameter ro- Two parallel defects can coalesce each one individually is stable: ea the sphere, and the whole loop ca Π1 (RP 2 ) = than the 1-D curve we used in th You might get the impression t is really just seven strength 1 de You’d be quite right: occasional but usually big ones decompose doesn’t mean, though, that addi gives a bigger one. In nematic li defects are as good as none! M line defects: a loop in real space thing it can’t smooth out. Forma group of the sphere is zero: you c For a nematic liquid crystal, tho ter space was a hemisphere (figu on the hemisphere in figure 5 th by twisting and stretching. It d but you have to remember that t on the equater really represent t tation. The corresponding defect which rotates 180◦ as the defect shows one typical configuration ( fect). Now, if you put two of thes cancel. (I can’t draw the pictures, lenging exercise in geometric visua defects add modulo 2, like clock ar school: Topological defects