The arrows point in the positive

x

direction, with largest speed

v_{0}

y = 0

and rapidly decreasing length away from that line. Since only

v_{x}

is nonzero,

\nabla \times v = - \frac{\partial v _{x}}{\partial y} e_{z} = 2 v_{0} y e^{- y^{2}} e_{z} .

The vorticity points towards positive

z

above the centre line and negative

z

below it. A small fluid element therefore has local angular velocity

(\nabla \times v) /2

: anticlockwise above and clockwise below when viewed from positive

z

. At

y = 0

, its speed is maximal but its local vorticity is zero. Translation speed and local rotation are different properties.

b

Solution

Using

r = (x, y, z)

and constant

a

\partial_{i} (ϵ_{ijk} r_{j} a_{k}) = ϵ_{ijk} δ_{ij} a_{k} = 0.

For the curl, the curl of a cross product identity gives

\nabla \times (r \times a) = r (\nabla \cdot a) - a (\nabla \cdot r) + (a \cdot \nabla) r - (r \cdot \nabla) a = 0 - 3 a + a - 0.

Thus

\nabla \cdot (r \times a) = 0, \nabla \times (r \times a) = - 2 a .

For a quick sign check, taking

a = e_{z}

gives

r \times a = (y, - x, 0)

, whose curl is

(0, 0, - 2)

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