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Part A Fluid Dynamics & Waves Draft date: 1 March 2010 4–
4 Vortex motion
4.1 Helmholtz’ Principle
So far in this course, we have seen how to calculate the velocity potential caused by
placing a vortex at a given location z = c in a flow, by using the method of images or
conformal mapping, for example. In practice, however, vortices in a fluid do not stay
still: they are convected by the flow, so that c will actually be a function of t.
We recall that the vorticity ω satisfies the equation
Dω
Dt
= (ω · ∇)u. (4.1)
In two-dimensional flow, the right-hand side of (4.1) is identically zero, so the vorticity
is conserved following the flow. This suggests that a vortex, which we can think of as a
line source of vorticity, should convect with the prevailing velocity of the flow in which
it sits.
Example 4.1 A vortex in a uniform flow
Suppose a vortex of strength Γ sits at the point z = c in an infinite expanse of fluid flowing uniformly in the x-direction with speed U. We superimpose the complex potentials due to the uniform flow and the vortex to obtain
w(z) = U z − iΓ 2 π log(z − c), (4.2)
and the resulting velocity components are given by
u − iv = dw dz
= U −
iΓ 2 π(z − c)
The first term on the right-hand side of equation (4.3) gives the velocity due to the uniform flow; the second term is the flow due to the vortex. When we now try to calculate the background flow experienced by the vortex, we encounter a problem, since (4.3) implies that the velocity is unbounded as z → c. The resolution of this difficulty is to ignore the velocity due to the vortex itself, i.e. the final term in (4.3). If we do so, we conclude that the vortex moves with velocity components given by
u − iv
vortex =^ U.^ (4.4)
Hence the vortex propagates with the uniform flow, in agreement with our physical intuition.
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Example 4.1 is a very simple illustration of Helmholtz’ Principle:
A vortex moves with the velocity field due to everything except itself. (4.5)
To make this principle more explicit, suppose we know the complex potential w(z) for
a flow containing a vortex at the point z = c. To find the velocity experienced by
the vortex, we first subtract off the velocity due to the vortex itself, then evaluate the
remaining flow (which should now be bounded) at the vortex location z = c, that is
u − iv
vortex = lim z→c
dw
dz
iΓ
2 π(z − c)
Helmholtz’ Principle implies that the vortex should move with velocity components
dc/dt = u + iv|vortex, and hence c(t) satisfies the differential equation
dc
dt
= lim z→c
dw
dz
iΓ
2 π(z − c)
4.2 Examples
Example 4.2 A vortex next to a wall
Suppose fluid occupies the half-plane Im z > 0 bounded by a rigid impermeable wall along the real-z-axis, and a vortex of strength Γ is at the point z = c. We find the velocity potential by using the method of images, placing a vortex of equal and opposite strength at the image point z = c:
w(z) = − iΓ 2 π log(z − c) + iΓ 2 π log (z − c). (4.8)
The velocity field is given by
u − iv = − iΓ 2 π(z − c)
and substitution into (4.7) tells us that the vortex location c(t) satisfies
dc dt = iΓ 2 π (c − c)
We can easily see that the right-hand side of (4.10) is purely real, implying that the vortex will move in the real-z-direction, parallel to the wall. To flesh this out further, let us write the components of the vortex location as c = x + iy, so that (4.10) becomes dx dt −^ i
dy dt =^
4 πy.^ (4.11)
Hence we see that the distance y of the vortex from the wall remains constant, while it propagates in the x-direction with speed Γ/ 4 πy, as shown schematically in Figure 4.1(i).
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G
0.0 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.
x
y
Figure 4.2: The path of a vortex confined to the quadrant x > 0, y > 0.
By taking the ratio of these equations, we find that the vortex follows a path in the (x, y)-plane on which dy dx
y^3 x^3
By integrating this equation, we find that
1 x^2
+^1
y^2 = constant. (4.17)
Such a path is plotted in Figure 4.2.
Example 4.4 Vortex in a channel
Suppose fluid fills the channel −a < Im z < a, in which a vortex of strength Γ > 0 sits at the point z = c. The channel is mapped to the half-space Re ζ > 0 by the conformal mapping ζ = eπz/^2 a and the complex potential
w(z) = − iΓ 2 π log
eπz/^2 a^ − eπc/^2 a
iΓ 2 π log
eπz/^2 a^ + eπc/^2 a
is then easily found by using the method of images. The velocity components are thus given by
u − iv = dw dz = iΓ 4 a
eπ(c−z)/^2 a^ − 1
eπ(c−z)/^2 a^ + 1
Part A Fluid Dynamics & Waves Draft date: 1 March 2010 4–
Re z
Im z
a z = c
a
Figure 4.3: Schematic of a vortex in the channel −a < Im z < a.
Now we wish to find the behaviour of the right-hand side of (4.19) as z → c. To this end, we expand the first term in braces to obtain
1 eπ(c−z)/^2 a^ − 1
π(c − z) 2 a
(^2) (c − z) 2 8 a^2
+ O
(z − c)^3
))−^1
2 a π(c − z)
π(c − z) 4 a
+ O
(z − c)^2
))−^1
2 a π(c − z)
- O (z − c) as z → c. (4.20)
Substituting (4.20) into (4.19), we find that
dc dt = lim z→c
dw dz +^
iΓ 2 π(z − c)
iΓ 4 a
2 +^
eπ(c−c)/^2 a^ + 1
and simplification leads to dc dt
8 a tan
π (c − c) 4ia
If we write the components of c as c = x + iy, then we find that x and y satisfy the differential equations
dx dt
8 a tan
( (^) πy 2 a
dy dt
Hence the vortex moves horizontally along the channel, in the negative x-direction if y > 0 or the positive x-direction if y < 0.
Example 4.4 illustrates how in general it may be necessary to expand dw/dz in a
Laurent expansion about z = c to evaluate the vortex velocity. A physical interpretation
of the result (4.23) is that the vortex moves under the influence of whichever wall is closer.
If y < 0 (for example), the bottom wall dominates and causes the vortex to propagate
to the right, as in Example 4.2. As the vortex approaches the lower wall y ↘ −a and
