Cylindrical coordinates
edit
A point plotted with cylindrical coordinates.
Consider a cylindrical coordinate system ( ρ , φ , z ), with the z –axis the line around which the incompressible flow is axisymmetrical, φ the azimuthal angle and ρ the distance to the z –axis. Then the flow velocity components uρ and uz can be expressed in terms of the Stokes stream function
Ψ
{\displaystyle \Psi }
by:[1]
u
ρ
=
−
1
ρ
∂
Ψ
∂
z
,
u
z
=
+
1
ρ
∂
Ψ
∂
ρ
.
{\displaystyle {\begin{aligned}u_{\rho }&=-{\frac {1}{\rho }}\,{\frac {\partial \Psi }{\partial z}},\\u_{z}&=+{\frac {1}{\rho }}\,{\frac {\partial \Psi }{\partial \rho }}.\end{aligned}}}
The azimuthal velocity component uφ does not depend on the stream function. Due to the axisymmetry, all three velocity components ( uρ , uφ , uz ) only depend on ρ and z and not on the azimuth φ .
The volume flux, through the surface bounded by a constant value ψ of the Stokes stream function, is equal to 2π ψ .
Spherical coordinates
edit
A point plotted using the spherical coordinate system
In spherical coordinates ( r , θ , φ ), r is the radial distance from the origin , θ is the zenith angle and φ is the azimuthal angle . In axisymmetric flow, with θ = 0 the rotational symmetry axis, the quantities describing the flow are again independent of the azimuth φ . The flow velocity components ur and uθ are related to the Stokes stream function
Ψ
{\displaystyle \Psi }
through:[2]
u
r
=
+
1
r
2
sin
θ
∂
Ψ
∂
θ
,
u
θ
=
−
1
r
sin
θ
∂
Ψ
∂
r
.
{\displaystyle {\begin{aligned}u_{r}&=+{\frac {1}{r^{2}\,\sin \theta }}\,{\frac {\partial \Psi }{\partial \theta }},\\u_{\theta }&=-{\frac {1}{r\,\sin \theta }}\,{\frac {\partial \Psi }{\partial r}}.\end{aligned}}}
Again, the azimuthal velocity component uφ is not a function of the Stokes stream function ψ . The volume flux through a stream tube, bounded by a surface of constant ψ , equals 2π ψ , as before.
The vorticity is defined as:
ω
=
∇
×
u
=
∇
×
∇
×
ψ
{\displaystyle {\boldsymbol {\omega }}=\nabla \times {\boldsymbol {u}}=\nabla \times \nabla \times {\boldsymbol {\psi }}}
, where
ψ
=
−
Ψ
r
sin
θ
ϕ
^
,
{\displaystyle {\boldsymbol {\psi }}=-{\frac {\Psi }{r\sin \theta }}{\boldsymbol {\hat {\phi }}},}
with
ϕ
^
{\displaystyle {\boldsymbol {\hat {\phi }}}}
the unit vector in the
ϕ
{\displaystyle \phi \,}
–direction.
Derivation of vorticity
ω
{\displaystyle {\boldsymbol {\omega }}}
using a Stokes stream function
Consider the vorticity as defined by
ω
=
∇
×
u
.
{\displaystyle {\boldsymbol {\omega }}=\nabla \times {\boldsymbol {u}}.}
From the definition of the curl in spherical coordinates :
ω
r
=
1
r
sin
θ
(
∂
∂
θ
(
u
ϕ
sin
θ
)
−
∂
u
θ
∂
ϕ
)
r
^
,
ω
θ
=
1
r
(
1
sin
θ
∂
u
r
∂
ϕ
−
∂
∂
r
(
r
u
ϕ
)
)
θ
^
,
ω
ϕ
=
1
r
(
∂
∂
r
(
r
u
θ
)
−
∂
u
r
∂
θ
)
ϕ
^
.
{\displaystyle {\begin{aligned}\omega _{r}&={1 \over r\sin \theta }\left({\partial \over \partial \theta }\left(u_{\phi }\sin \theta \right)-{\partial u_{\theta } \over \partial \phi }\right){\boldsymbol {\hat {r}}},\\\omega _{\theta }&={1 \over r}\left({1 \over \sin \theta }{\partial u_{r} \over \partial \phi }-{\partial \over \partial r}\left(ru_{\phi }\right)\right){\boldsymbol {\hat {\theta }}},\\\omega _{\phi }&={1 \over r}\left({\partial \over \partial r}\left(ru_{\theta }\right)-{\partial u_{r} \over \partial \theta }\right){\boldsymbol {\hat {\phi }}}.\end{aligned}}}
First notice that the
r
{\displaystyle r}
and
θ
{\displaystyle \theta }
components are equal to 0. Secondly substitute
u
r
{\displaystyle u_{r}}
and
u
θ
{\displaystyle u_{\theta }}
into
ω
ϕ
.
{\displaystyle \omega _{\phi }.}
The result is:
ω
r
=
0
,
ω
θ
=
0
,
ω
ϕ
=
1
r
(
∂
∂
r
(
r
(
−
1
r
sin
θ
∂
Ψ
∂
r
)
)
−
∂
∂
θ
(
1
r
2
sin
θ
∂
Ψ
∂
θ
)
)
.
{\displaystyle {\begin{aligned}\omega _{r}&=0,\\\omega _{\theta }&=0,\\\omega _{\phi }&={1 \over r}\left({\partial \over \partial r}\left(r\left(-{\frac {1}{r\sin \theta }}{\frac {\partial \Psi }{\partial r}}\right)\right)-{\partial \over \partial \theta }\left({\frac {1}{r^{2}\sin \theta }}{\frac {\partial \Psi }{\partial \theta }}\right)\right).\end{aligned}}}
Next the following algebra is performed:
ω
ϕ
=
1
r
(
−
1
sin
θ
(
∂
∂
r
(
∂
Ψ
∂
r
)
)
−
1
r
2
∂
∂
θ
(
1
sin
θ
∂
Ψ
∂
θ
)
)
=
1
r
(
−
1
sin
θ
(
∂
2
Ψ
∂
r
2
)
−
sin
θ
r
2
sin
θ
∂
∂
θ
(
1
sin
θ
∂
Ψ
∂
θ
)
)
=
−
1
r
sin
θ
(
∂
2
Ψ
∂
r
2
+
sin
θ
r
2
∂
∂
θ
(
1
sin
θ
∂
Ψ
∂
θ
)
)
.
{\displaystyle {\begin{aligned}\omega _{\phi }&={1 \over r}\left(-{\frac {1}{\sin \theta }}\left({\partial \over \partial r}\left({\frac {\partial \Psi }{\partial r}}\right)\right)-{\frac {1}{r^{2}}}{\partial \over \partial \theta }\left({\frac {1}{\sin \theta }}{\frac {\partial \Psi }{\partial \theta }}\right)\right)\\&={1 \over r}\left(-{\frac {1}{\sin \theta }}\left({\frac {\partial ^{2}\Psi }{\partial r^{2}}}\right)-{\frac {\sin \theta }{r^{2}\sin \theta }}{\partial \over \partial \theta }\left({\frac {1}{\sin \theta }}{\frac {\partial \Psi }{\partial \theta }}\right)\right)\\&=-{\frac {1}{r\sin \theta }}\left({\frac {\partial ^{2}\Psi }{\partial r^{2}}}+{\frac {\sin \theta }{r^{2}}}{\partial \over \partial \theta }\left({\frac {1}{\sin \theta }}{\frac {\partial \Psi }{\partial \theta }}\right)\right).\end{aligned}}}
As a result, from the calculation the vorticity vector is found to be equal to:
ω
=
(
0
0
−
1
r
sin
θ
(
∂
2
Ψ
∂
r
2
+
sin
θ
r
2
∂
∂
θ
(
1
sin
θ
∂
Ψ
∂
θ
)
)
)
.
{\displaystyle {\boldsymbol {\omega }}={\begin{pmatrix}0\\[1ex]0\\[1ex]\displaystyle -{\frac {1}{r\sin \theta }}\left({\frac {\partial ^{2}\Psi }{\partial r^{2}}}+{\frac {\sin \theta }{r^{2}}}{\partial \over \partial \theta }\left({\frac {1}{\sin \theta }}{\frac {\partial \Psi }{\partial \theta }}\right)\right)\end{pmatrix}}.}
Comparison with cylindrical
edit
The cylindrical and spherical coordinate systems are related through
z
=
r
cos
θ
{\displaystyle z=r\,\cos \theta \,}
and
ρ
=
r
sin
θ
.
{\displaystyle \rho =r\,\sin \theta .\,}
Alternative definition with opposite sign
edit
As explained in the general stream function article, definitions using an opposite sign convention – for the relationship between the Stokes stream function and flow velocity – are also in use.[3]
In cylindrical coordinates, the divergence of the velocity field u becomes:[4]
∇
⋅
u
=
1
ρ
∂
∂
ρ
(
ρ
u
ρ
)
+
∂
u
z
∂
z
=
1
ρ
∂
∂
ρ
(
−
∂
Ψ
∂
z
)
+
∂
∂
z
(
1
ρ
∂
Ψ
∂
ρ
)
=
0
,
{\displaystyle {\begin{aligned}\nabla \cdot {\boldsymbol {u}}&={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}{\Bigl (}\rho \,u_{\rho }{\Bigr )}+{\frac {\partial u_{z}}{\partial z}}\\&={\frac {1}{\rho }}{\frac {\partial }{\partial \rho }}\left(-{\frac {\partial \Psi }{\partial z}}\right)+{\frac {\partial }{\partial z}}\left({\frac {1}{\rho }}{\frac {\partial \Psi }{\partial \rho }}\right)=0,\end{aligned}}}
as expected for an incompressible flow.
And in spherical coordinates:[5]
∇
⋅
u
=
1
r
sin
θ
∂
∂
θ
(
u
θ
sin
θ
)
+
1
r
2
∂
∂
r
(
r
2
u
r
)
=
1
r
sin
θ
∂
∂
θ
(
−
1
r
∂
Ψ
∂
r
)
+
1
r
2
∂
∂
r
(
1
sin
θ
∂
Ψ
∂
θ
)
=
0.
{\displaystyle {\begin{aligned}\nabla \cdot {\boldsymbol {u}}&={\frac {1}{r\,\sin \theta }}{\frac {\partial }{\partial \theta }}(u_{\theta }\,\sin \theta )+{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}{\Bigl (}r^{2}\,u_{r}{\Bigr )}\\&={\frac {1}{r\,\sin \theta }}{\frac {\partial }{\partial \theta }}\left(-{\frac {1}{r}}{\frac {\partial \Psi }{\partial r}}\right)+{\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left({\frac {1}{\sin \theta }}{\frac {\partial \Psi }{\partial \theta }}\right)=0.\end{aligned}}}
Streamlines as curves of constant stream function
edit
From calculus it is known that the gradient vector
∇
Ψ
{\displaystyle \nabla \Psi }
is normal to the curve
Ψ
=
C
{\displaystyle \Psi =C}
(see e.g. Level set#Level sets versus the gradient ). If it is shown that everywhere
u
⋅
∇
Ψ
=
0
,
{\displaystyle {\boldsymbol {u}}\cdot \nabla \Psi =0,}
using the formula for
u
{\displaystyle {\boldsymbol {u}}}
in terms of
Ψ
,
{\displaystyle \Psi ,}
then this proves that level curves of
Ψ
{\displaystyle \Psi }
are streamlines.
Cylindrical coordinates
In cylindrical coordinates,
∇
Ψ
=
∂
Ψ
∂
ρ
e
ρ
+
∂
Ψ
∂
z
e
z
{\displaystyle \nabla \Psi ={\partial \Psi \over \partial \rho }{\boldsymbol {e}}_{\rho }+{\partial \Psi \over \partial z}{\boldsymbol {e}}_{z}}
.
and
u
=
u
ρ
e
ρ
+
u
z
e
z
=
−
1
ρ
∂
Ψ
∂
z
e
ρ
+
1
ρ
∂
Ψ
∂
ρ
e
z
.
{\displaystyle {\boldsymbol {u}}=u_{\rho }{\boldsymbol {e}}_{\rho }+u_{z}{\boldsymbol {e}}_{z}=-{1 \over \rho }{\partial \Psi \over \partial z}{\boldsymbol {e}}_{\rho }+{1 \over \rho }{\partial \Psi \over \partial \rho }{\boldsymbol {e}}_{z}.}
So that
∇
Ψ
⋅
u
=
∂
Ψ
∂
ρ
(
−
1
ρ
∂
Ψ
∂
z
)
+
∂
Ψ
∂
z
1
ρ
∂
Ψ
∂
ρ
=
0.
{\displaystyle \nabla \Psi \cdot {\boldsymbol {u}}={\partial \Psi \over \partial \rho }(-{1 \over \rho }{\partial \Psi \over \partial z})+{\partial \Psi \over \partial z}{1 \over \rho }{\partial \Psi \over \partial \rho }=0.}
Spherical coordinates
And in spherical coordinates
∇
Ψ
=
∂
Ψ
∂
r
e
r
+
1
r
∂
Ψ
∂
θ
e
θ
{\displaystyle \nabla \Psi ={\partial \Psi \over \partial r}{\boldsymbol {e}}_{r}+{1 \over r}{\partial \Psi \over \partial \theta }{\boldsymbol {e}}_{\theta }}
and
u
=
u
r
e
r
+
u
θ
e
θ
=
1
r
2
sin
θ
∂
Ψ
∂
θ
e
r
−
1
r
sin
θ
∂
Ψ
∂
r
e
θ
.
{\displaystyle {\boldsymbol {u}}=u_{r}{\boldsymbol {e}}_{r}+u_{\theta }{\boldsymbol {e}}_{\theta }={1 \over r^{2}\sin \theta }{\partial \Psi \over \partial \theta }{\boldsymbol {e}}_{r}-{1 \over r\sin \theta }{\partial \Psi \over \partial r}{\boldsymbol {e}}_{\theta }.}
So that
∇
Ψ
⋅
u
=
∂
Ψ
∂
r
⋅
1
r
2
sin
θ
∂
Ψ
∂
θ
+
1
r
∂
Ψ
∂
θ
⋅
(
−
1
r
sin
θ
∂
Ψ
∂
r
)
=
0.
{\displaystyle \nabla \Psi \cdot {\boldsymbol {u}}={\partial \Psi \over \partial r}\cdot {1 \over r^{2}\sin \theta }{\partial \Psi \over \partial \theta }+{1 \over r}{\partial \Psi \over \partial \theta }\cdot {\Big (}-{1 \over r\sin \theta }{\partial \Psi \over \partial r}{\Big )}=0.}
^ Batchelor (1967), p. 78.
^ Batchelor (1967), p. 79.
^ E.g. Brenner, Howard (1961). "The slow motion of a sphere through a viscous fluid towards a plane surface". Chemical Engineering Science . 16 (3–4): 242–251. doi :10.1016/0009-2509(61)80035-3 .
^ Batchelor (1967), p. 602.
^ Batchelor (1967), p. 601.