## Hydrodynamics in Flatland

Two-dimensional motion of an incompressible fluid

Edward Abbott's Flatland was a novel set in a two-dimensional world. There is also a hydrodynamics in Flatland, and it actually has practical applications, since it is the same as three-dimensional hydrodynamics where the motion is the same in parallel planes, and can be described in terms of two coordinates, x and y, and the time t. Essentially, a slice of unit thickness is being described. In Flatland, a fluid would have a density ρ in g/cm2. Let up be the direction +y, if up means anything in Flatland. We suppose that the fluid is incompressible, and are interested in steady motions.

Now take any fixed point O, and consider a second point P(x,y) that can take any position. Draw any contour between O and P, and consider the flux across it. This is the line integral of the normal component of the velocity along the contour. If two such contours are drawn, the flux must be the same across each, because the fluid is incompressible. Therefore, the flux is a function of the position x,y of P only, so we can represent it by a function ψ(x,y) called the stream function. The lines ψ = constant are the streamlines, a very handy result.

Now let us take two neighboring streamlines, and consider the fluid that flows between them in the stream tube thus established. Since the fluid is incompressible, the rate of flow at any point must be the same, so q ds = constant, where q is the velocity and ds the distance between streamlines. The closer the streamlines are, the greater the velocity. Consider a small volume of fluid dV that enters a portion of the stream tube. Work pdV is done by pressure, its kinetic energy is ρq2dV/2, and its potential energy is ρdVgy, where we imagine a gravitational acceleration in the -y direction. Any other potential function would do as well. Since the total energy is conserved, p + ρq2/2 + ρgy must be a constant along the stream tube. This is, of course, Bernoulli's Equation, which, we see, also holds in Flatland.

The normal component of the velocity v = dψ/ds, so we can write u = -dψ/dy and v = dψ/dx, where u and v are the x- and y-components of the velocity, and we have defined the stream function so that a positive velocity is to the left as we trace the curve. The derivatives are, of course, partials, that are hard to represent in HTML. u and v then satisfy du/dx + dv/dy = 0, which is the equation of continuity expressing the incompressibility of the fluid.

Another sort of line integral is the integral of the component of velocity along the curve, or u(dx/ds)+v(dy/ds). If this is taken around a closed curve, the result, the integral of udx + vdy, is called the circulation around the curve. Any such closed curve can be imagined to be made up of many small closed curves with common boundaries, so the circulation around the finite curve is the integral of the circulation around a small area dS. If we calculate that circulation for a small rectangle of sides dx and dy, we find (dv/dx - du/dy)dS. The quantity ζ = (dv/dx - du/dy) is called the vorticity. In terms of the stream function, ζ = d2ψ/dx2 + d2ψ/dy2. If a small disc is rotating at angular velocity ω, the circulation is 2πr(rω) = ζπr2, so ζ = ω/2. If ζ = 0 everywhere, the motion is said to be irrotational.

Now that we know about vorticity, we can analyze stream tube motion further. When a stream tube is curved, with radius of curvature R, there is a centripetal acceleration q2/R that must be counteracted by the pressure and external forces. In fact, d(p + ρgy)/dn = ρq2/R, where n is the direction directly outwards from the centre of curvature. The circulation around a small area of sides ds and Rdθ is ζ = q/R + dq/dn. Our force equation becomes d(p + ρq2/2 + ρgy)/dn = ρqζ. This reveals that the difference in the constant in Bernoulli's Equation between two streamlines is proportional to the vorticity. Since the difference is constant in steady motion, we see that the vorticity must be constant along a streamline as well. Vorticity is conserved during the motion, an important result.

If a cylinder of fluid is rotating rigidly with angular velocity ω, the velocity at any point is ωr, the radius of curvature R = r, and ζ = 2ω, where the origin is taken at the centre of rotation. The equation, integrated from 0 to r, gives p = p0 + ρω2r2/2, where p0 is the pressure at the centre. Suppose the rotating cylinder of fluid is of radius a, and outside it ζ = 0. This means that q/r = -dq/dr, or qr = constant. The constant is ωa2. In this outside region, p + ρq2/2 is constant, equal to p0 + ρω2a2/2. So now we know the velocity and pressure at all points of this Flatland vortex. In our world, a vortex usually ends at a free surface, and causes a dimple there, with gravity cooperating.

In irrotational motion, d2ψ/dx2 + d2ψ/dy2 = 0, or ψ satisfies Laplace's Equation, and is, therefore, a harmonic function in two dimensions. In polar coordinates, these include rn or r-n times cos nθ or sin nθ. We can find new solutions by superimposing old ones. Moreover, the constant in Bernoulli's Equation is the same throughout the irrotational region.

As a simple example, ψ = -Uy = -Ur sin θ describes a uniform flow with velocity U in the +x direction. A solution ψ = (-Ur + C/r) sin θ can satisfy the condition that the normal velocity, dψ/rdθ, vanishes on the surface of a cylinder of radius a: -U -C/a2 = 0, or C = -Ua2, so that ψ = -U(r - a2/r) sin θ. From this result, so easily obtained, we can find the pressure and velocity everywhere, and the streamlines as well. It is easy to see, from the symmetry, that there is no resultant pressure force on the cylinder. If we add Ur to ψ, we have a cylinder moving with velocity U in a fluid at rest. If we calculate the kinetic energy of the fluid in this case, we get M'U2/2, where M' = ρπa2 is the mass of the fluid displaced by the cylinder. The effect of the fluid is to increase the apparent mass of the cylinder by this amount. These results are similar to those derived for a sphere in three dimensions, but slightly different.

Let's go back to the two routes from point O to point P. We found that the line integral of the velocity perpendicular to the path was the same on both, if only the fluid was incompressible, and from this got the stream function ψ, on which our analysis has been based so far. If the flow is irrotational, the line integral of the velocity parallel to the path is also the same on both routes, since it must be zero on a closed circuit, say from O to P and back to O. The negative of the value of the line integral of udx + vdy from O to P is called the velocity potential φ, and u = -dφ/dx, v = -dφ/dy. We use the negative simply to make the results like those for the electrostatic potential. The equation of continuity, du/dx + dv/dy = 0, then gives d2φ/dx2 + d2φ/dy2 = 0, so the velocity potential, like the stream function, satisfies Laplace's Equation, and the solutions are harmonic functions.

The velocity is normal to the equipotential lines φ(x,y) = constant. Since it is also parallel to the lines ψ(x,y) = constant, these two families of lines must form an orthogonal net, and the one determines the other. This property of orthogonality is shared by the real and imaginary parts of an analytic complex function Φ(x,y) = φ(x,y) + iψ(x,y). This provides a powerful tool for finding two-dimensional flow patterns. An analytic function is one whose derivative is the same in any direction, such as any function formed by combining elementary functions.

For example, consider the function w = C/z = C(cos θ - i sin θ)/r. If C = -Ua2, we find ψ = Ua2sin θ/r, the stream function for a cylinder moving with velocity U, and ψ = -Ua2cos θ, which must be the velocity potential for the same problem, which can easily be checked by seeing that the normal velocity at the surface of the cylinder is zero. The assignments of ψ and φ could be interchanged, but the problem would be a different one, usually not as interesting. If w = C log z, however, both are interesting. One is a line source, and the other a line vortex.

We superimposed the solutions for a uniform flow and for a double line source at the orgin to find the flow around a cylinder. If we add a line vortex, ψ = &kappa log r/2π, to the solution, we have the flow around a rotating cylinder. There is now a net force on the cylinder, but it is at right angles to the direction of flow, similar to the lift of an airfoil.

There is a hydrodynamics in Flatland, and the methods available for its description are powerful.

### References

H. Lamb, in C. V. Drysdale, et. al., The Mechanical Properties of Fluids (London: Blackie and Son, 1925), Chapter II, pp. 54-68, gives a lucid account of two-dimensional hydrodynamics.