Съдържание

Условие Part A. Steady filament (0.75 points) (0,75 т.)Part B. Vortex motion (1.4 points) (1,4 т.)Part C. Momentum and energy (1.75 points) (1,75 т.)Part D. Trapped charges (2.85 points) (2,85 т.)Part E. Influence of the boundaries (3.25 points) (3,25 т.)Отговори Решение

APhO 2017, Теория — Задача 1. Vortices in superfluid

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Theory APhO YAKUTSK 2017 RUSSIA Q1 English (Official)

Vortices in Superfluid

MODD-Problems

May 5, 2017

Условие

Introduction

Superfluidity is a property of flowing without friction. Everyday experience tells us that motion of an ordinary fluid (say, water at room temperature) is always accompanied by viscous dissipation of energy, so that the flow gradually becomes slower unless it is maintained by external forces. In contrast, superfluid exhibits no loss of kinetic energy: once excited the motion of superfluid can continue indefinitely. Superfluidity was originally discovered experimentally in liquid helium.

We study properties of superfluid helium at zero temperature. It will be treated as an incompressible fluid with density $\rho$ . Flow continuity (the fact that the mass flowing into and the mass flowing out of a given infinitesimal volume are equal) implies that the flux of helium velocity $\vec{v}$ through a closed surface is always zero. Superfluid velocity in this aspect is analogous to the magnetic field intensity. By analogy with the magnetic field lines, “streamlines” are tangential to the fluid velocity at each point and their density is proportional to its magnitude.

True superflow has an important property of being irrotational: circulation of superfluid velocity $\vec{v}$ along any closed path within helium is zero

\int\limits_L \vec{v}\cdot d\vec{l} = 0 \qquad (1)

This statement must be amended if superfluidity is absent along a thin “vortex filament”. The thickness of the filament itself is of approximately atomic dimensions $a$ , but the vortex induces long range velocity field in surrounding superfluid: velocity circulation around such filament is the circulation quantum $^1$

\left|\int\limits_L \vec{v}\cdot d\vec{l}\right| = 2\pi\kappa, \qquad (2)

and zero if the path can be contracted to a single point without crossing the vortex, see Fig. 1. This supports the analogy between superflow and the magnetic field created by wires carrying current: superposition of two valid velocity distributions is a valid velocity distribution and the velocity at any point is equal (up to a dimensional factor) to the magnetic field produced by the unit currents running through a system of wires representing vortex filaments.

$^1$ Circulation quantization is a macroscopic quantum effect and corresponds to the angular momentum quantization in Bohr model. The circulation quantum can be expressed as $\kappa = \hbar/m_{\mathrm{He}}$ , where $m_{\mathrm{He}}$ is the mass of helium atom.

A light-blue rectangle of superfluid crossed from top to bottom by a curved red vortex filament of thickness a (marked by a dimension at the top). Six closed paths with arrows are drawn: L1 and L3 loop near the filament, L2 is a small loop at the upper right, L4 is a large loop, L5 is a loop around part of the filament, L6 loops across the filament near the bottom. — Fig. 1: Vortex filament (red) in superfluid (light blue). Velocity circulations along paths $L_1$ , $L_2$ , $L_5$ , and $L_6$ are all zero, but those for $L_3$ and $L_4$ are equal to $\pm 2\pi\kappa$ . Note that circulations along $L_3$ and $L_4$ have opposite signs.

Part A. Steady filament (0.75 points) (0,75 т.)

Consider a cylindrical beaker (radius $R_0 \gg a$ ) of superfluid helium and a straight vertical vortex filament in its center Fig. 2.

A cylindrical beaker filled with light-blue helium; a red vertical vortex line along its axis with a circular arrow around it; a radius arrow labelled R0 at the bottom; a downward arrow labelled g on the right; coordinate axes x, y, z at the lower left. — Fig. 2: Straight vortex along the axis of a beaker.

A.1 Plot the streamlines. Find out the velocity $v$ at a point $\vec{r}$ . [0,25 т.]

A.2 Work out the free surface shape (height as a function of coordinate $z(\vec{r})$ ) around the vortex. Free fall acceleration is $g$ . Surface tension can be neglected. [0,5 т.]

Part B. Vortex motion (1.4 points) (1,4 т.)

Free vortices move about in space with the flow $^2$ . In other words each element of the filament moves with the velocity $\vec{v}$ of the fluid at the position of that element.

$^2$ This is a consequence of momentum conservation, see next section.

As an example, consider a pair of counter-rotating straight vortices placed initially at distance $r_0$ from each other, see Fig. 3. Each vortex produces velocity $v_0 = \kappa/r_0$ at the axis of another. As a result, the vortex pair moves rectilinearly with constant speed $v_0 = \kappa/r_0$ so that the distance between them remains unchanged.

Two red dots (vortex cross-sections) a distance r0 apart, marked by a double arrow between dashed lines. Each has an upward velocity arrow labelled v0; the left one has a counter-clockwise curved arrow, the right one a clockwise curved arrow. — Fig. 3: Parallel vortex filaments with opposite circulations.

Two red dots a distance r0 apart, marked by a double arrow between dashed lines; each has a clockwise curved arrow beside it. — Fig. 4: Parallel vortex filaments with equal circulations.

Top view of a circular beaker of diameter 2R0 filled with a regular array of small vortex symbols; arrows mark vortex A near the wall, vortex B midway and vortex C at the center. A green square region is magnified at the right, showing the triangular lattice with spacing u marked along three directions by dash-dotted lines. — Fig. 5: Triangular lattice of vortices in a beaker. The view from above.

B.1 Consider two identical straight vortices initially placed at distance $r_0$ from each other as shown in Fig. 4. Find initial velocities of the vortices and draw their trajectories. [0,25 т.]

A beaker of helium (see Part A) is filled with triangular lattice ( $u \ll R_0$ ) of identical vertical vortices, see Fig. 5.

B.2 Draw the trajectories of vortices A, B, and C (located in the center). [0,15 т.]

B.3 Find velocity $v(\vec{r})$ of a vortex positioned at $\vec{r}$ . [0,4 т.]

B.4 Find the distance AB $(t)$ between the vortices A and B at time $t$ . Treat AB $(0)$ as given. [0,35 т.]

B.5 Work out the "smoothed out" (omitting the lattice structure) free helium surface shape $z(\vec{r})$ . [0,25 т.]

Part C. Momentum and energy (1.75 points) (1,75 т.)

The long range velocity field is the major contribution to the energy of a system of vortices, it is insensitive to exact structure of the filament. The filament itself can not be properly described by the macroscopic theory and apparent singularities (infinities) are insignificant. Real physical quantities, such as the energy, of the region inside a thin tube of radius $a$ around the filament should be neglected. Outside of this tube the density of superflow kinetic energy $\rho v^2/2$ (where $\rho = \mathrm{const}$ ) is analogous to the energy density of the magnetic field $B^2/(2\mu_0)$ — they are both quadratic in respective variables. This analogy together with the correspondence between magnetic field and superfluid velocity generated by vortices (currents) facilitates calculation of the flow energy for a given system. For instance, given the inductance of a circular wire loop $L \approx \mu_0 R \log(R/a)$ , where $R$ is the loop radius and $a$ is wire radius, we get the superfluid vortex loop energy $^3$

U \approx 2R\rho\pi^2\kappa^2 \log(R/a) \qquad (3)

Total fluid momentum is also determined by the long range velocity distribution. It is obtained by integration of the momentum density $\rho\vec{v}$ . Again, consider a flow generated by a circular vortex loop placed in $xy$ plane. It is obvious from the symmetry considerations, that total momentum has only $z$ component:

P = \int \rho v_z dV = \rho \int\int \underbrace{\left(\int v_z dz\right)}_{q(x,y)} dxdy \qquad (4)

The innermost integration is in fact an integration along appropriate paths parallel to $z$ -axis, see Fig. 6. From the circulation identity (2) it follows that

q(x,y) = \int\limits_{L(x,y)} \vec{v}\cdot d\vec{l} \qquad (5)

is piecewise constant. Particularly, it is zero for paths passing outside the ring and $2\pi\kappa$ for paths inside it. Total momentum is therefore

P = \rho\cdot\pi R^2\cdot 2\pi\kappa = 2\pi^2\rho R^2\kappa \qquad (6)

$^3$ This expression is also valid only if $\log R/a \gg 1$ .

Left: a red circular vortex loop in perspective with black streamlines looping around it and passing upward through its interior. Right: cross-section in the x–z plane with the two red cross-section points of the loop, black closed and open streamlines around them, and seven vertical green lines with upward arrows marking the integration paths; axes z and x. — Fig. 6: Velocity field of a circular vortex loop and integration paths (green) for $q(x,y)$ calculation.

A red elongated rounded rectangle of length d (dimension line above) and height b (dimension at the right), with small curved arrows at the top and bottom sides showing the circulation. — Fig. 7: A nearly rectangular vortex loop, $b \ll d$ .

Axes x, y, z in perspective; a red straight vortex filament along z with a circulation arrow, and a faded copy shifted by a distance b in the x direction (arrow labelled b). — Fig. 8: Momentum changes whenever the vortex shifts with respect to the fluid.

C.1 Consider a nearly rectangular vortex loop $b \times d$ , $b \ll d$ , Fig. 7. Indicate the direction of its momentum $\vec{P}$ . Find out the momentum magnitude. [0,3 т.]

C.2 Calculate its energy $U$ . [0,7 т.]

C.3 Suppose we shift a long straight vortex filament by a distance $b$ in $x$ direction, see Fig. 8. How much does the fluid momentum change? Indicate the momentum change direction. The filament length (constrained by the vessel walls) is $d$ . [0,75 т.]

Part D. Trapped charges (2.85 points) (2,85 т.)

Electrons, if injected in helium, get trapped in the vortex filaments. Here and below polarizability of helium can be neglected ( $\epsilon = 1$ ).

Axes x and y; a red dot (vortex cross-section) with a curved circulation arrow beneath it, and to the right a horizontal arrow labelled E pointing in the +x direction. — Fig. 9: Straight vortex in a uniform electric field.

Left: a red vortex ring seen obliquely with its center C and a downward arrow labelled E. Right: axes x and z with the two red cross-section points of the ring, each with a circulation arrow, the center C between them, and a downward arrow labelled E. — Figure 10: (left) Vortex ring in a uniform electric field. (right) Cross section of the ring.

D.1 Consider a straight vortex charged with uniform linear density $\lambda < 0$ in a uniform electric field $\vec{E}$ . Draw the vortex trajectory. Find its velocity as a function of time. [0,5 т.]

A circular vortex loop of radius $R_0$ initially charged with uniform linear density $\lambda < 0$ is placed in a uniform electric field $\vec{E}$ perpendicular to its plane, opposite to its momentum $\vec{P}_0$ .

D.2 Draw the trajectory of the loop center $C$ . Find the radius of the loop as a function of time. [0,6 т.]

D.3 Find its velocity $v(t)$ as a function of time. [1,5 т.]

D.4 The field is switched off at a time $t^*$ when the velocity reaches the value $v^* = v(t^*)$ . Find the loop velocity $v(t)$ at a later time $t > t^*$ . [0,25 т.]

Part E. Influence of the boundaries (3.25 points) (3,25 т.)

Solid walls alter the velocity field created by a vortex filament, because the fluid cannot flow through them. Mathematically this means that the wall-normal velocity component vanishes at the wall surface.

A red dot (vortex cross-section) with a clockwise curved arrow above it, at height h0 above a flat grey wall, the distance marked by a vertical double arrow labelled h0. — Fig. 11: Straight vortex filament near a flat wall.

A right-angle corner formed by a grey vertical wall on the left and a grey horizontal wall at the bottom; a red dot (vortex) with a curved circulation arrow, at distance h0 from the vertical wall and h0 from the bottom wall, both marked by double arrows labelled h0. — Fig. 12: Straight vortex filament in a corner.

E.1 Draw the trajectory of a straight vortex, initially placed at a distance $h_0$ from a flat wall. Find its velocity as a function of time. [0,5 т.]

Consider a straight vortex placed in a corner at a distance $h_0$ from both walls.

E.2 What is the initial velocity $v_0$ of the vortex? [0,75 т.]

E.3 Draw the trajectory of the vortex. [0,5 т.]

E.4 What is the velocity of the vortex $v_\infty$ after very long time? [1,5 т.]

Отговори

Part A. Steady filament (0.75 points), A.1 $v = \kappa/r$

Streamlines are circles around the vortex.
Part A. Steady filament (0.75 points), A.2 $z(r) = [z_0] - \frac{\kappa^2}{2gr^2}$
Part B. Vortex motion (1.4 points), B.1 $v_0 = \kappa/r_0$

The filaments rotate around the point halfway between them.
Part B. Vortex motion (1.4 points), B.3 $v = \frac{2\pi\kappa r}{u^2\sqrt{3}}$
Part B. Vortex motion (1.4 points), B.4 $\mathrm{AB}(t) = \mathrm{AB}(0)$
Part B. Vortex motion (1.4 points), B.5 $z(r) = [z_0] + \frac{2\pi^2\kappa^2 r^2}{3gu^4}$
Part C. Momentum and energy (1.75 points), C.1 $P = 2\pi\kappa\rho b d$

The momentum is perpendicular to the plane of the loop.
Part C. Momentum and energy (1.75 points), C.2 $U = 2\pi\kappa^2\rho d\log\frac{b}{a}$
Part C. Momentum and energy (1.75 points), C.3 $\Delta P = 2\pi\kappa\rho b d$

The momentum change is parallel to the y axis.
Part D. Trapped charges (2.85 points), D.1 $v = \frac{E\lambda}{2\pi\kappa\rho}$

Constant velocity perpendicular to $\vec{E}$ ; the trajectory is a straight line parallel to the y axis.
Part D. Trapped charges (2.85 points), D.2 $R = \sqrt{R_0^2 + \frac{ER_0|\lambda|t}{\pi\rho\kappa}}$
Part D. Trapped charges (2.85 points), D.3 $v \approx \frac{\kappa\log(R_0/a)}{2\sqrt{R_0^2 + ER_0|\lambda|t/(\pi\rho\kappa)}}$
Part D. Trapped charges (2.85 points), D.4 $v(t) = v^*$
Part E. Influence of the boundaries (3.25 points), E.1 $v = \frac{\kappa}{2h_0}$

The vortex moves along the wall with constant velocity.
Part E. Influence of the boundaries (3.25 points), E.2 $v_0 = \frac{\kappa}{2\sqrt{2}h_0}$
Part E. Influence of the boundaries (3.25 points), E.4 $v_\infty = \frac{\kappa}{h_0\sqrt{2}}$

Решение

A. Steady filament (0.75) (0,75 т.)

Consider a cylindrical beaker (radius $R_0 \gg a$ ) of superfluid helium and a straight vertical vortex filament in its center Fig. 2.

A1 (0.25)

Plot the streamlines. Find out the velocity $v$ at a point $\vec{r}$ .

The streamlines are circular. From the circulation identity (1) it is obvious that $v = \kappa/r$ .

Streamlines are plotted correctly (one at least) .......... 0.1
$v = \frac{k}{r}$ .......... 0.15

A2 (0.5)

Work out the free surface shape (height as a function of coordinate $z(\vec{r})$ ) around the vortex. Free fall acceleration is $g$ . Surface tension can be neglected.

Consider a thin circular layer of the radius $r$ . Equilibrium condition for its surface is given by the requirement

g\frac{dz}{dr} = \frac{v^2}{r} = \frac{\kappa^2}{r^3}. \qquad (1)

This equation is satisfied by the surface profile

z(r) = [z_0] - \frac{\kappa^2}{2gr^2}. \qquad (2)

$\tan\alpha = \frac{k^2}{gr^3}$ or equivalent .......... 0.25
$z = [z_0] - \frac{k^2}{2gr^2}$ .......... 0.25

Top view of the beaker: a light-blue disc with a red dot at the center, a thick curved arrow around it, and three concentric circular streamlines with arrowheads.

B. Vortex motion (1.4) (1,4 т.)

B1 (0.25)

Consider two identical straight vortices initially placed at distance $r_0$ from each other as shown in Fig. 4. Find initial velocities of the vortices and draw their trajectories.

Being advected by each other’s flow field, filaments will rotate around a point halfway between them. The velocity is given by $v_0 = \kappa/r_0$ .

Trajectories are plotted correctly .......... 0.15
Correct expression for velocity .......... 0.1

B2 (0.15)

Draw the trajectories of vortices A, B, and C (located in the center).

Trajectories are plotted correctly .......... 0.15

B3 (0.4)

Find velocity $v(\vec{r})$ of a vortex positioned at $\vec{r}$ .

Consider a circular path of radius $r \gg u$ around the beaker center. The circulation along this path is given by the number of vortices within it (vortex density per unit area is $(u^2\sqrt{3}/2)^{-1}$ ):

2\pi r v = 2\pi\kappa\frac{\pi r^2}{u^2\sqrt{3}/2}. \qquad (3)

The velocity field

v = \frac{2\pi\kappa r}{u^2\sqrt{3}}. \qquad (4)

Expression for vortex density .......... 0.2
Correct expression for $v(r)$ .......... 0.2

B4 (0.35)

Find the distance AB $(t)$ between the vortices A and B at time $t$ . Treat AB $(0)$ as given.

This velocity pattern corresponds to the rotation of the lattice as a whole around the beaker center with angular velocity

\omega = \frac{2\pi\kappa}{u^2\sqrt{3}}. \qquad (5)

AB $(t) =$ AB $(0)$

Correct answer .......... 0.35

B5 (0.25)

Work out the “smoothed out” (omitting the lattice structure) free helium surface shape $z(\vec{r})$ .

The surface shape is

z(r) = [z_0] + \frac{\omega^2 r^2}{2g} = [z_0] + \frac{2\pi^2\kappa^2 r^2}{3gu^4}. \qquad (6)

Correct answer .......... 0.25

Two red dots a distance r0 apart on a circle; the circle has an arrow showing both vortices moving around the midpoint; curved circulation arrows beside each vortex.

Top view of the beaker filled with the vortex lattice; vortices A, B and C are marked by blue arrows; A and B move along concentric circles about the center (arrowheads on the circles), C sits at the center.

C. Momentum and Energy (1.75) (1,75 т.)

C1 (0.3)

Consider a nearly rectangular vortex loop $b \times d$ , $b \ll d$ , Fig. 7. Indicate the direction of its momentum $\vec{P}$ . Find out the momentum magnitude.

Momentum of a flat loop (see Introduction) is perpendicular to its plane and proportional to its area. For a rectangular loop the magnitude is $P = 2\pi\kappa\rho b d$ .

Correct direction of momentum .......... 0.15
Correct expression for momentum magnitude .......... 0.15

C2 (0.7)

Calculate its energy $U$ .

To produce equal magnetic and kinetic energy densities $B^2/(2\mu_0) = \rho v^2/2$ , the magnetic field has to be $B = v\sqrt{\mu_0\rho} = \kappa\sqrt{\mu_0\rho}/r$ . This field is generated by a current $I = 2\pi\kappa\sqrt{\rho/\mu_0}$ . Energy of the wire loop can be found from the inductance $U = LI^2/2$ . Inductance of a nearly rectangular wire loop:

L = \frac{\Phi}{I} = 2dI^{-1}\int_a^b \frac{\mu_0 I}{2\pi r}dr = \frac{\mu_0 d}{\pi}\log\frac{b}{a}. \qquad (7)

This gives for the energy

U = 2\pi\kappa^2\rho d\log\frac{b}{a} \qquad (8)

Integration limits are $\sim a$ and $b$ .......... 0.2
Analogy with a magnitude field is used ( $U = \frac{LI^2}{2}$ , $L = \frac{\Phi}{I}$ ) or energy is calculated as $W = \int F\,dr$ , where $F = \frac{dP}{dt}$ .......... 0.2
Correct expression for energy .......... 0.3

C3 (0.75)

Suppose we shift a long straight vortex filament by a distance $b$ in $x$ direction, see Fig. 8. How much does the fluid momentum change? Indicate the momentum change direction. The filament length (constrained by the vessel walls) is $d$ .

The momentum change is equal to the momentum of a long rectangular loop $P = 2\pi\kappa\rho b d$ .

The result of C1 used .......... 0.3
Momentum change is parallel to Y axis .......... 0.1
Correct direction of momentum change .......... 0.15
Correct expression for momentum change magnitude .......... 0.2

Interestingly, this provides an alternative approach to find the energy of such a loop. Namely, if we slowly move one straight vortex in the velocity field of another, then we apply a force

F = 2\pi\kappa\rho d v = 2\pi\kappa\rho d\frac{\kappa}{r} = \frac{2\pi\kappa^2\rho d}{r}. \qquad (9)

The work

W = \int_a^b \frac{2\pi\kappa^2\rho d}{r}dr = 2\pi\kappa^2\rho d\log\frac{b}{a} \qquad (10)

has to be performed to move it from distance $a$ to $b$ .

The red nearly rectangular loop of length d and height b with circulation arrows; below it the symbol ⊙ P showing the momentum pointing out of the page.

Axes x and y; a black vortex symbol with a red dot and a faded grey copy shifted to the right by b (arrow labelled b); to the right an upward arrow labelled ΔP.

D. Trapped charges (2.85) (2,85 т.)

D1 (0.5)

Consider a straight vortex charged with uniform linear density $\lambda < 0$ in a uniform electric field $\vec{E}$ . Draw the vortex trajectory. Find its velocity as a function of time.

Electric force $F = E\lambda d$ moves the vortex with velocity

v = \frac{F}{2\pi\kappa\rho d} = \frac{E\lambda}{2\pi\kappa\rho} \qquad (11)

perpendicular to $\vec{E}$ .

Trajectory is straight line parallel to Y axis .......... 0.1
Correct direction of velocity .......... 0.2
Correct expression for velocity magnitude .......... 0.2

D234

D2 (0.6)

Draw the trajectory of the loop center $C$ . Find the radius of the loop as a function of time.

Electric force upon the loop $F = -2\pi ER_0|\lambda|$ is constant and fluid momentum linearly depends on time

P = P_0 + 2\pi ER_0|\lambda|t = 2\pi^2\rho R^2\kappa. \qquad (12)

The loop is growing and its radius is increasing with time $t$

R = \sqrt{R_0^2 + \frac{ER_0|\lambda|t}{\pi\rho\kappa}}. \qquad (13)

Trajectory is straight line along $y$ .......... 0.1
Correct velocity direction .......... 0.15
$P(t)$ .......... 0.15
$2\pi^2\rho R^2\kappa$ .......... 0.15
Correct expression for $R(t)$ .......... 0.05

D3 (1.5)

Find its velocity $v(t)$ as a function of time.

The loop velocity $v$ can be easily found from a relationship between the energy change rate and the momentum change rate

\frac{dU}{dt} = Fv = \frac{dP}{dt}v. \qquad (14)

This gives for the velocity

v = \frac{dU}{dP} \approx \frac{\kappa}{2R}\log\frac{R}{a} = \frac{\kappa\log\left(\sqrt{R_0^2 + ER_0|\lambda|t/(\pi\rho\kappa)}/a\right)}{2\sqrt{R_0^2 + ER_0|\lambda|t/(\pi\rho\kappa)}} \approx \frac{\kappa\log(R_0/a)}{2\sqrt{R_0^2 + ER_0|\lambda|t/(\pi\rho\kappa)}}. \qquad (15)

This means that the vortex is moving in the direction of the force but its velocity is decreasing.

Expression for $v \propto \frac{1}{R}\log(R)$ .......... 1.0
Correct expression for $v(t)$ .......... 0.5

D4 (0.25)

The field is switched off at a time $t^*$ when the velocity reaches the value $v^* = v(t^*)$ . Find the loop velocity $v(t)$ at a later time $t > t^*$ .

When $E = 0 \Rightarrow P = \mathrm{const} \Rightarrow R = \mathrm{const} \Rightarrow v = \mathrm{const} \Rightarrow v(t) = v^*$ .

Correct expression for $v(t)$ .......... 0.25

Axes x and y; the charged vortex (red dot with circulation arrow) with a long vertical arrow pointing up (+y) from it; a horizontal arrow labelled E pointing right.

Axes x and z; the two red cross-section points of the ring with circulation arrows and the center C between them; a vertical arrow pointing up (+z) from C; a downward arrow labelled E on the right.

E. Influence of the boundaries (3.25) (3,25 т.)

Draw the trajectory of a straight vortex, initially placed at a distance $h_0$ from a flat wall. Find its velocity as a function of time.

E1 (0.5)

Well known technique of image charges (currents) in electrostatics (magnetostatics) can be directly used to solve this problem. Namely, the wall can be “substituted” with a reflected fictitious vortex on the other side of the wall. The velocity distribution of two vortices together in the upper semi-space is identical to the one produce by a single vortex above the wall. Indeed, the symmetry of the problem ensures that there is no flow through the plane of symmetry. Thus, a straight vortex line situated a distance $h_0$ above a flat wall with its image behave as a pair of vortices of opposite circulation a distance $2h_0$ apart. This means that the vortex moves along the wall with velocity

v = \frac{\kappa}{2h_0}. \qquad (16)

Illustration of the image method for the straight vortex filament near a flat wall

Trajectory is plotted correctly .......... 0.15
Correct direction of velocity .......... 0.1
Correct expression for velocity magnitude .......... 0.25

E234

Consider a straight vortex placed in a corner at a distance $h_0$ from both walls.

E2 (0.75)

What is the initial velocity $v_0$ of the vortex?

The velocity of the filament is given by superposition of the velocities $\vec{v}_1$ , $\vec{v}_2$ and $\vec{v}_3$ induced by the image vortices 1, 2 and 3, respectively (see Fig. in E3 solution). One readily obtains

v_1 = \frac{\kappa}{2h_0}, \quad v_2 = \frac{\kappa}{2\sqrt{2}h_0}, \quad v_3 = \frac{\kappa}{2h_0}.

The modulus of the filament velocity at the initial moment is

v_0 = |\vec{v}_1 + \vec{v}_2 + \vec{v}_3| = \sqrt{2}v_1 - v_2 = \boxed{\frac{\kappa}{2\sqrt{2}h_0}}

Ideas of using superposition principal and technique of image charges .......... 0.25
Correct expression for initial velocity magnitude .......... 0.5

E3 (0.5)

Draw the trajectory of the vortex.

Image vortices in the corner.

The trajectory has correct form .......... 0.3
Correct direction of initial velocity .......... 0.2

E4 (1.5)

What is the velocity of the vortex $v_\infty$ after very long time?

Energy for the system of vortices is proportional to

U_{\mathrm{tot}} \propto \log\frac{\sqrt{x^2+y^2}}{a} - \log\frac{x}{a} - \log\frac{y}{a}. \qquad (17)

The energy conservation implies that

C = \frac{x^2+y^2}{x^2y^2} = \frac{2}{h_0^2} \qquad (18)

is constant along the trajectory. After very long time $y \to h_0/\sqrt{2}$ and the vortex velocity is

v_\infty = \frac{\kappa}{h_0\sqrt{2}}. \qquad (19)

$E = \mathrm{const}$ .......... 0.5
Correct expression for velocity after very long time .......... 1.5

A grey flat wall; a red vortex at height h0 above it with a clockwise circulation arrow and a velocity arrow v pointing left along the wall; its mirror image vortex at h0 below the wall with the opposite circulation arrow. — Illustration of the image method for the straight vortex filament near a flat wall

The corner formed by a grey vertical wall and a grey horizontal wall; the real vortex at distance h0 from both walls and three image vortices numbered 1 (mirrored across the vertical wall), 2 (diagonally opposite) and 3 (mirrored across the horizontal wall), all distances labelled h0, with circulation arrows; a green trajectory starts at the real vortex, bends toward the bottom wall and runs parallel to it to the right, with an arrow. — Image vortices in the corner.

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