[~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