[psi] = -½Uy²a^4[mu]/r^4, [psi]´ = ½Uy²(1 - a^4[mu]/r^4), (22)
[phi]´ = Ux + [phi], [phi] = -1/3 U(a^4/r³)P2, P2 = 3/2 [mu]² - ½, (23)
representing a stream past the surface r^4 = a^4[mu].
35. A circular vortex, such as a smoke ring, will set up motion
symmetrical about an axis, and provide an illustration; a half vortex
ring can be generated in water by drawing a semicircular blade a short
distance forward, the tip of a spoon for instance. The vortex advances
with a certain velocity; and if an equal circular vortex is generated
coaxially with the first, the mutual influence can be observed. The
first vortex dilates and moves slower, while the second contracts and
shoots through the first; after which the motion is reversed
periodically, as if in a game of leap-frog. Projected perpendicularly
against a plane boundary, the motion is determined by an equal opposite
vortex ring, the optical image; the vortex ring spreads out and moves
more slowly as it approaches the wall; at the same time the molecular
rotation, inversely as the cross-section of the vortex, is seen to
increase. The analytical treatment of such vortex rings is the same as
for the electro-magnetic effect of a current circulating in each ring.
36. _Irrotational Motion in General._--Liquid originally at rest in a
singly-connected space cannot be set in motion by a field of force due
to a single-valued potential function; any motion set up in the liquid
must be due to a movement of the boundary, and the motion will be
irrotational; for any small spherical element of the liquid may be
considered a smooth solid sphere for a moment, and the normal pressure
of the surrounding liquid cannot impart to it any rotation.
The kinetic energy of the liquid inside a surface S due to the
velocity function [phi] is given by
_ _ _ _ _
/ / / | /d[phi]\² /d[phi]\² /d[phi]\² |
T = ½[rho] | | | | ( ------ ) + ( ------ ) + ( ------ ) | dx dy dz,
_/_/_/ |_ \ dx / \ dy / \ dz / _|
_ _
/ / d[phi]
= ½[rho] | | [phi] ------ dS (1)
_/_/ d[nu]
by Green's transformation, d[nu] denoting an elementary step along the
normal to the exterior of the surface; so that d[phi]/d[nu] = 0 over
the surface makes T = 0, and then
/d[phi]\² /d[phi]\² /d[phi]\² d[phi] d[phi] d[phi]
( ------ ) + ( ------ ) + ( ------ ) = 0, ------ = 0, ------ = 0, ------ = 0 (2)
\ dx / \ dy / \ dz / dx dy dz
If the actual motion at any instant is supposed to be generated
instantaneously from rest by the application of pressure impulse over
the surface, or suddenly reduced to rest again, then, since no natural
forces can act impulsively throughout the liquid, the pressure impulse
[~[omega]] satisfies the equations
1 d[~omega] 1 d[~omega] 1 d[~omega]
----- --------- = -u, ----- --------- = -v, ----- --------- = -[~omega], (3)
[rho] dx [rho] dy [rho] dz
[~omega] = [rho][phi] + a constant, (4)
and the constant may be ignored; and Green's transformation of the
energy T amounts to the theorem that the work done by an impulse is
the product of the impulse and average velocity, or half the velocity
from rest.
In a multiply connected space, like a ring, with a multiply valued
velocity function [phi], the liquid can circulate in the circuits
independently of any motion of the surface; thus, for example,
[phi] = m[theta] = m tan^(-1) y/x (5)
will give motion to the liquid, circulating in any ring-shaped figure
of revolution round Oz.
To find the kinetic energy of such motion in a multiply connected
space, the channels must be supposed barred, and the space made
acyclic by a membrane, moving with the velocity of the liquid; and
then if k denotes the cyclic constant of [phi] in any circuit, or the
value by which [phi] has increased in completing the circuit, the
values of [phi] on the two sides of the membrane are taken as
differing by k, so that the integral over the membrane
_ _ _ _
/ / d[phi] / / d[phi]
| | [phi] ------ dS = k | | ------ dS, (6)
_/_/ d[nu] _/_/ d[nu]
and this term is to be added to the terms in (1) to obtain the
additional part in the kinetic energy; the continuity shows that the
integral is independent of the shape of the barrier membrane, and its
position. Thus, in (5), the cyclic constant k = 2[pi]m.
In plane motion the kinetic energy per unit length parallel to Oz
_ _ _ _ _ _ _ _
/ / | /d[phi]\² /d[phi]\² | / / | /d[psi]\² /d[psi]\² |
T = ½[rho] | | | ( ------ ) + ( ------ ) | dx dy = ½[rho] | | | ( ------ ) + ( ------ ) | dx dy
_/_/ |_ \ dx / \ dy / _| _/_/ |_ \ dx / \ dy / _|
_ _
/ d[phi] / d[phi]
= ½[rho] | [phi] ------ ds = ½[rho] | [psi] ------ ds. (7)
_/ d[nu] _/ d[nu]
For example, in the equilateral triangle of (8) § 28, referred to
coordinate axes made by the base and height,
[psi]´ = -2R[alpha][beta][gamma]/h = -½Ry[(h - y)² - 3x²]/h (8)
[psi] = [psi]´ - ½R [(1/3h - y)² + x²]
= -½R [½h³ + 1/3 h²y + h) (x² - y²) - 3x²y + y³] /h (9)
and over the base y = 0,
dx/d[nu] = -dx/dy = + ½R(1/3 h² - 3x²)/h, [psi] = -½R(1/9 h² + x²). (10)
Integrating over the base, to obtain one-third of the kinetic energy
T,
_
/ h/[root]3
1/3 T = ½[rho] | ¼R²(3x^4 - 1/27 h^4) dx/h
_/ -h/[root]3
= [rho]R²h^4/135[root]3 (11)
so that the effective k² of the liquid filling the triangle is given
by
k² = T/½[rho]R²A = 2h²/45
= 2/5 (radius of the inscribed circle)², (12)
or two-fifths of the k² for the solid triangle.