Graphics Reference
In-Depth Information
Thus for a rigid body rotation, the gradient has no symmetric part—there's
no deformation after all—and the skew-symmetric part lets us read out the
components of angular velocity directly.
Take a look at the skew-symmetric part of the velocity gradient (in
general, not just for rigid body motions):
⎛
⎝
⎞
⎠
∂u
∂y
−
∂v
∂x
∂u
∂z
−
∂w
∂x
0
∂u
∂x
−
T
=
1
2
1
2
∂u
∂x
∂v
∂x
−
∂u
∂y
∂v
∂z
−
∂w
∂y
.
0
∂w
∂x
−
∂u
∂z
∂w
∂y
−
∂v
∂z
0
Reading off the local measure of angular velocity this represents, just as
we saw in the rigid case, we get
∂w
∂y
−
.
Ω(
x
)=
1
2
∂v
∂z
,
∂u
∂z
−
∂w
∂x
,
∂v
∂x
−
∂u
∂y
This is exactly half the curl of the velocity field.
In fact we define
vorticity
ω
to be the curl of the velocity field, which
will then be twice the local angular velocity. Again, in three dimensions
this is a vector:
ω
=
u
=
∂w
∇×
.
∂v
∂z
,
∂u
∂z
−
∂w
∂x
,
∂v
∂x
−
∂u
∂y
∂y
−
In two dimensions it reduces to a scalar:
∂v
∂x
−
∂u
∂y
.
ω
=
∇×u
=
This turns out to be one of the most useful “derived” quantities for a fluid
flow.
Take the curl of the momentum equation (1.1), assuming constant vis-
cosity:
1
ρ
∇
p
=
∂u
∂t
+
∇×
∇×
(
u
·∇
u
)+
∇×
∇×
g
+
∇×
ν
∇·∇
u.
Switching the order of some derivatives, and assuming that density
ρ
is
constant so it can be brought outside the curl for the pressure term, gives
∂
∇×
u
u
)+
1
+
∇×
(
u
·∇
ρ
∇×∇
p
=
∇×
g
+
ν
∇·∇
(
∇×
u
)
.
∂t
