V Hydrodynamics

V
Hydrodynamics

61. In the applications I am about to make in this I have practically nothing new to shew except the utility of Quaternion methods in the general theory of Hydrodynamics in all its parts.

I therefore take a treatise (Greenhill’s article in the Encyc. Brit. on this subject) and work out the general theory on the lines of the treatise. This is more necessary than at ﬁrst sight it would seem, for I believe mathematicians who have studied Quaternions are under the impression that the method does not lend itself conveniently to the establishment and treatment of such equations as the Lagrangian and those of Cauchy’s integrals. With our meaning of $\nabla$ , however, the Quaternion treatment of these equations is as much simpler than the Cartesian as in the case of the Eulerian equations.

Down to equation (13) below the subject has already been handled by Prof. Hicks³⁵ in his Quaternion treatment of Strains and Fluid motion (Quart. Journ. Math. xiv. [1877] p. 271). I do not hesitate to go over the ground again as my methods are diﬀerent from his.

15. Notation

62. For the vector velocity at any point we shall use $σ$ , for the density $D$ , for the force per unit mass $F$ , for its potential, when it has one, $v$ , and for the pressure $p$ . For time-ﬂux which follows a particle we shall use $d ∕ d t$ or the Newtonian dot, and for that which refers to a ﬁxed point of space $\partial ∕ \partial t$ ³⁶.

Thus

\begin{array}{l} d ∕ d t = \partial ∕ \partial t - S σ \nabla . & (1) \end{array}

This equation is given on p. 446 of Greenhill’s article already mentioned. In future we shall refer to this article simply as “Greenhill’s article.”

Euler’s equations

Euler’s Equations

63. To ﬁnd the equation of continuity, with Greenhill, we merely express symbolically that the rate of increase of the mass of the ﬂuid in any space equals the rate at which it is ﬂowing through the boundary. Thus $M$ being the mass in any space,

\begin{array}{l} \partial M ∕ \partial t = \iint D S σ d Σ . & (2) \end{array}

This is Greenhill’s equation (1) p. 445. By it and equation (9) Section 4 above we have

\begin{array}{l} \partial M ∕ \partial t = ∭ S \nabla (D σ) d 𝔰, \end{array}

whence reducing the volume to the element $d 𝔰$ ,

\begin{array}{l} \partial D ∕ \partial t = S \nabla (D σ) . & (3) \end{array}

This is Greenhill’s equation (2). Thus by equation (1) of last section,

\begin{array}{l} Ḋ ∕ D = S \nabla σ . & (4) \end{array}

64. To obtain Euler’s equations of motion we express that the vector sum of the impressed forces for any volume equals the vector sum of the bodily forces plus the vector sum of the pressures on the surface. Thus

\begin{array}{l} ∭ D \dot{σ} d 𝔰 & = ∭ D F d 𝔰 - \iint p d Σ \\ = ∭ (D F - \nabla p) d 𝔰, \end{array}

by equation (9) Section 4 above. Applying this to the element $d 𝔰$ and dividing by $D d 𝔰$ we have

\begin{array}{l} \dot{σ} = F - \nabla p ∕ D, & (5) \end{array}

or by equation (1) Section 15

\begin{array}{l} \partial σ ∕ \partial t - S σ \nabla · σ = F - \nabla p ∕ D . & (6) \end{array}

This is Greenhill’s equations (4) (5) (6).

If $F (ρ, t) = 0$ ( $F$ a scalar) be the equation of a surface always containing the same particles $Ḟ = 0$ , or by equation (1)

\begin{array}{l} \partial F ∕ \partial t - S σ \nabla F = 0 . & (7) \end{array}

This is Greenhill’s equation (7).

65. Let us now put

\begin{array}{l} \int d p ∕ D = P . & (8) \end{array}

This of course assumes that $D$ is a function of $p$ only, which is not always the case, for instance in a gas where diﬀusion of heat is taking place. If $F$ have a potential $v$ , $F = - \nabla v$ . Thus equation (5) becomes

\begin{array}{l} \dot{σ} = - \nabla (v + P), & (9) \end{array}

and equation (6)

\begin{array}{l} \partial σ ∕ \partial t - S σ \nabla · σ = - \nabla (v + P) . & (10) \end{array}

Now

\begin{array}{l} S σ \nabla_{1} · σ_{1} = V σ V \nabla_{1} σ_{1} + \nabla_{1} S σ σ_{1} = V σ V \nabla σ + \nabla (σ^{2}) ∕ 2 . \end{array}

Thus equation (10) becomes

\begin{array}{l} \partial σ ∕ \partial t + 2 V 𝜖 σ + \nabla R = 0, & (11) \end{array}

where the scalar $R$ is put for $P + v - σ^{2} ∕ 2$ and the vector $𝜖$ for $V \nabla σ ∕ 2$ . This is Greenhill’s equations (8) (9) (10).

If $𝜖 = 0$ , i.e. $V \nabla σ = 0$ , we have $σ = \nabla ϕ$ , whence our equation becomes

\begin{array}{l} \nabla (\partial ϕ ∕ \partial t + R) & = 0, \\ so that & \partial ϕ ∕ \partial t + R & = H, & (12) \end{array}

where $H$ is a function of $t$ only. In next section we shall obtain in the case of an inﬁnite ﬂuid a generalisation of this which I believe has not hitherto been obtained. Here we have made the assumption that if $V \nabla σ = 0$ at one epoch it will be so always. This we shall prove later.

66. Greenhill next considers the case of steady motion. In this case $\partial σ ∕ \partial t = 0$ , so that equation (11) becomes

\begin{array}{l} 2 V 𝜖 σ + \nabla R = 0, & (13) \end{array}

and therefore the surface $R = const.$ contains both vortices and stream-lines and the relation $d R ∕ d n = 2 q ω sin 𝜃$ given on p. 446 of Greenhill’s article is the natural interpretation of our equation.

So far we have been going over much the same ground as Hicks, but now we enter upon applications of Quaternions that I think have not been made before.

67. Greenhill next considers rotating axes and ﬁnds the form of the equations of motion when referred to these. Let $σ$ be the velocity referred to them; so that if the axes have at any time made the rotation $q () q^{- 1}$ the real velocity will be $q σ q^{- 1}$ . Thus³⁷, as always with rotating axes, if $α$ be any vector function of a particle the rate of increase of $α$ in space is

\begin{array}{l} \dot{α} + V ω α, \end{array}

where $ω$ is the angular velocity referred to the same system of rotating axes. Thus we see that the acceleration of a particle $= \dot{σ} + V ω σ$ . The velocity $σ = \dot{ρ}' + V ω ρ'$ , where $ρ'$ is the vector coordinate referred to our present axes (i.e. in the notation of the footnote $ρ' = q ρ q^{- 1}$ ). Thus our equation of motion is now

\begin{array}{l} \dot{σ} + V ω σ = F - ρ' \nabla p ∕ D, \end{array}

but now

\begin{array}{l} d ∕ d t = \partial ∕ \partial t - S \dot{ρ}' ρ' \nabla · = \partial ∕ \partial t - S (σ - ω ρ') ρ' \nabla, \end{array}

whence changing $ρ'$ into $ρ$ we have

\begin{array}{l} \partial σ ∕ \partial t - S (σ - ω ρ) \nabla · σ + V ω σ = F - \nabla p ∕ D, & (14) \end{array}

which is the equation Greenhill gives on p. 446.

The Lagrangian equations

The Lagrangian Equations³⁸

68. We now consider the history of a single particle and for this we require diﬀerent notation.

We consider the vector coordinate ( $ρ$ ) of a particle as a function of some other vector $α$ (say the initial value of $ρ$ ) and of $t$ .

We ﬁrst require the connection between $α \nabla$ and $ρ \nabla$ . We shall drop the aﬃx $α$ and retain $ρ$ , so that now not $Q (\nabla_{1}, ρ_{1})$ but $Q (\nabla_{1}, α_{1})$

\begin{array}{l} = Q (ζ, ζ) . \end{array}

\begin{array}{l} Now & S d α \nabla & = S d ρ ρ \nabla, \\ but & d ρ & = - ρ_{1} S d α \nabla_{1}, \\ ∴ S d α \nabla & = - S d α \nabla_{1} S ρ_{1} ρ \nabla, \end{array}

or since

d α

is perfectly arbitrary

\begin{array}{l} \nabla = - \nabla_{1} S ρ_{1} ρ \nabla . & (15) \end{array}

Thus operating upon equation (9) Section 15 by $\nabla_{1} S ρ_{1} ()$ and remembering that in that equation $\nabla$ must be changed to $ρ \nabla$ we get

\begin{array}{l} \nabla_{1} S ρ_{1} \dot{σ} = \nabla (v + P), & (16) \end{array}

and this is our new equation of motion (Greenhill’s (1) (2) (3) p. 448).

In equation (1) Section 7 above let us put for $ρ + η$ , $ρ$ ; and for $ρ$ , $α$ .

\begin{array}{l} Thus & χ ω & = - S ω \nabla · ρ . & (17) \end{array}

Hence by equation (6 $d$ ) Section 8 above,

\begin{array}{l} strained vol. of el./original vol. = S \nabla_{1} \nabla_{2} \nabla_{3} S ρ_{1} ρ_{2} ρ_{3} ∕ 6, \end{array}

whence we see that

\begin{array}{l} D S \nabla_{1} \nabla_{2} \nabla_{3} S ρ_{1} ρ_{2} ρ_{3} = 6 D_{0}, & (18) \end{array}

where $D_{0}$ is a constant which when $α$ is taken as the initial value of $ρ$ is the original density. This is the equation of continuity.

16. Cauchy’s integrals of these equations

69. To obtain these integrals we require to express $ρ \nabla$ in terms of $\nabla$ . Now equation (15) of last section expresses $\nabla$ as a linear function of $ρ \nabla$ . The converse we have already seen how to get. In fact from equation (6 $h$ ) Section 2 above

\begin{array}{l} 2 J ρ \nabla = - V ρ_{1} ρ_{2} S \nabla_{1} \nabla_{2} \nabla, & (19) \end{array}

where, with Greenhill, for brevity $J$ is put for $S \nabla_{1} \nabla_{2} \nabla_{3} S ρ_{1} ρ_{2} ρ_{3} ∕ 6$ . The only function we wish to apply this to is $V ρ \nabla σ$ . We have

\begin{array}{l} 2 J V ρ \nabla σ & = V σ_{3} V ρ_{1} ρ_{2} S \nabla_{1} \nabla_{2} \nabla_{3} \\ = (ρ_{2} S ρ_{1} σ_{3} - ρ_{1} S ρ_{2} σ_{3}) S \nabla_{1} \nabla_{2} \nabla_{3}, \end{array}

or interchanging the suﬃxes 1 and 2 in the last term

\begin{array}{l} J V ρ \nabla σ = ρ_{2} S ρ_{1} σ_{3} S \nabla_{1} \nabla_{2} \nabla_{3}, & (20) \end{array}

which gives the spin at any instant in terms of our present independent variables $α$ and $t$ .

70. In order to obtain Cauchy’s integral of equation (16) operate on it by $V \nabla ()$ . Thus

\begin{array}{l} 0 = V (\nabla_{1} + \nabla_{2}) \nabla_{1} S ρ_{1} {\dot{σ}}_{2} = S ρ_{1} {\dot{σ}}_{2} · V \nabla_{2} \nabla_{1} . \end{array}

Now

\begin{array}{l} d (S ρ_{1} σ_{2} · V \nabla_{2} \nabla_{1}) ∕ d t = S σ_{1} σ_{2} · V \nabla_{2} \nabla_{1} + S ρ_{1} {\dot{σ}}_{2} · V \nabla_{2} \nabla_{1} . \end{array}

The ﬁrst term of this last expression is zero since the sign is changed by interchanging the suﬃxes. From the last expression then

\begin{array}{l} S ρ_{1} σ_{2} · V \nabla_{2} \nabla_{1} & = constant \\ = initial value of S ζ σ_{2} V \nabla_{2} ζ = - V \nabla σ_{0}, \end{array}

where $σ_{0}$ is the initial value of $σ$ . Changing the suﬃxes and substituting in equation (20) we have

\begin{array}{l} J V ρ \nabla σ = - ρ_{1} S \nabla σ_{0} \nabla_{1}, \end{array}

or giving $J$ its value $D_{0} ∕ D$ from equation (18) and putting $V ρ \nabla σ = 𝜖$ , $V \nabla σ_{0} = 𝜖_{0}$ , we have

\begin{array}{l} 𝜖 ∕ D = - S (𝜖_{0} ∕ D_{0}) \nabla · ρ . & (21) \end{array}

This is Greenhill’s equations (4) (5) (6) p. 448.

The physical interpretation of this equation is quite easy. Consider a small vector $d α$ drawn in the ﬂuid initially. At the time $t$ this will have become $d ρ = - S d α \nabla · ρ$ . Thus we see that if a small vector $c 𝜖_{0} ∕ D_{0}$ be drawn in the ﬂuid initially it will at the time $t$ be $c 𝜖 ∕ D$ , from which we infer that an element of a vortex ﬁlament will always remain an element of a vortex ﬁlament; or, a vortex ﬁlament or tube always remains a vortex ﬁlament or tube. Again we see that $T 𝜖 ∕ D$ at any time varies directly with the elongation in the direction of the vortex ﬁlament so that $T 𝜖$ varies as that elongation $\times$ the density, i.e. inversely as the cross-section of a small vortex tube at the point. This is not the easiest way of arriving at these results, but it is well to show in passing how easy of interpretation are our results.

We see from equation (21) that if $𝜖_{0} = 0$ , $𝜖 = 0$ . In other words, if the motion have a velocity potential at one instant it will have one always.

17. Flow, circulation, vortex-motion

71. We are about to consider vortex-motion from another point of view, viz. that of circulation.

In Section 7 we saw that a strain due to a small displacement $η$ could be decomposed into a pure strain followed by a rotation, the vector rotation being $V \nabla η ∕ 2$ . If now for $η$ we put the small vector $σ δ t$ we see the propriety of calling $V \nabla σ ∕ 2$ the spin. This therefore is taken as a deﬁnition of spin. Greenhill does not take this (usual) course but uses the property we shall immediately prove concerning circulation and spin to lead to his deﬁnition.

It is not necessary here to deﬁne ﬂow and circulation. Putting $σ = \nabla ϕ$ we see that for irrotational motion the ﬂow $= - \int S σ d ρ = \int d ϕ$ from one point to another is the increment in the velocity potential. Thus for mutually reconcilable paths it is always the same.

Taking the circulation round a closed curve in the general case, the curve not inclosing any singular region of the ﬂuid, we may transform the line integral into a surface integral by equation (8) Section 4 above. Thus

\begin{array}{l} - \int S σ d ρ = - \iint S d Σ \nabla σ, & (22) \end{array}

so that the circulation round the curve equals twice the surface integral of the spin. Hence Greenhill’s deﬁnition of the spin.

72. From equation (9) Section 15 we see that

\begin{array}{l} - \frac{d}{d t} (\int_{A}^{B} S σ d ρ) & = - \int_{A}^{B} S σ d σ - \int_{A}^{B} S \dot{σ} d ρ \\ = - \frac{1}{2} \int_{A}^{B} d (σ^{2}) - \int_{A}^{B} d (v + P), \\ or & - \frac{d}{d t} (\int_{A}^{B} S σ d ρ) & = - {[v + P + σ^{2} ∕ 2]}_{A}^{B} . & (23) \end{array}

Therefore for a closed curve

\begin{array}{l} d (\int S σ d ρ) ∕ d t = 0, \end{array}

so that the circulation round the curve, and therefore the corresponding surface integral of the spin remains always constant. Taking the curve round a small vortex tube we once more arrive at the propositions enunciated in Section 16 about vortices.

Thus at any point of a vortex tube the strength which is deﬁned as the product of the spin into the cross-section is constant throughout all time. Also it is the same for all points of the tube, for by equation (9) Section 4 we have

\begin{array}{l} \iint S d Σ \nabla σ = ∭ S \nabla^{2} σ d 𝔰 = 0 \end{array}

for any portion of the tube. But the only part contributing to the surface integral is the ends of the part of the tube considered, so that the strength at these two ends is the same, and is therefore constant for the whole tube.

73. These propositions are proved in the paper by Hicks already referred to. There is yet a third method of proof which, like Hicks’s, is derived directly from the equation (9) of motion.

We have from equation (1) Section 15

\begin{array}{l} \frac{d}{d t} \nabla - \nabla \frac{d}{d t} = - S σ \nabla · \nabla + \nabla S σ \nabla · = \nabla_{1} S σ_{1} \nabla ·; & (24)39 therefore d V \nabla σ ∕ d t - V \nabla \dot{σ} = V (\nabla_{1} S σ_{1} \nabla_{2} · σ_{2}) . Now by equation (9) Section 15, V \nabla \dot{σ} = 0,
so that \begin{matrix} d V \nabla σ ∕ d t = V \nabla_{1} σ_{2} S \nabla_{2} σ_{1}, & (25) \\ ∴ d (ρ + x V \nabla σ) ∕ d t = σ + ẋ V \nabla σ + x V \nabla_{1} σ_{2} S \nabla_{2} σ_{1} . \end{matrix} Now σ_{1} S \nabla_{2} σ_{2} \nabla_{1} = V σ_{2} \nabla_{1} S \nabla_{2} σ_{1} + V \nabla_{1} \nabla_{2} S σ_{1} σ_{2} + V \nabla_{2} σ_{2} S \nabla_{1} σ_{1},
but V \nabla_{1} \nabla_{2} S σ_{1} σ_{2} = 0,
for an interchange of suﬃxes leads to a change of sign. Therefore \begin{array}{l} V \nabla_{1} σ_{2} S \nabla_{2} σ_{1} = V \nabla σ S \nabla σ - S \nabla σ \nabla · σ . \end{array} Put now x = c ∕ D where c is
some small constant. Thus \begin{array}{l} ẋ ∕ c = - Ḋ ∕ D^{2} & = - S \nabla σ ∕ D, \\ and we get & d (ρ + c V \nabla σ · ∕ D) d t & = σ - c S \nabla σ \nabla · σ · ∕ D . & (26) \end{array} This shews that ρ' \equiv ρ + c V \nabla σ ∕ D is the variable vector of a particle, for the equation asserts that the time-ﬂux
of ρ' = the
velocity at ρ' .
Thus again we get the laws of vortices given in Section 16 . 18. Irrotational Motion 74. We use this heading merely to connect what follows with what Greenhill has
under the same on p. 449. It is not very appropriate here. For irrotational motion we have seen that we may put \begin{array}{l} σ = \nabla ϕ . & (27) \end{array} If the ﬂuid be incompressible we have further \begin{array}{l} 0 = S \nabla σ = \nabla^{2} ϕ . & (28) \end{array} Let T be the kinetic energy of this liquid. Thus \begin{array}{l} - 2 T = D ∭ {(\nabla ϕ)}^{2} d 𝔰 = D \iint ϕ S d Σ \nabla ϕ - D ∭ ϕ \nabla^{2} ϕ d 𝔰 \end{array} by equation (9) Section 4 above. Thus by equation (28) \begin{array}{l} T = - \frac{1}{2} D \iint ϕ S d Σ \nabla ϕ . & (29) \end{array} Hence we see that if we have a vector σ which satisﬁes the equations V \nabla σ = 0 and S \nabla σ = 0 throughout any singly-connected space, or more generally satisﬁes the second
of these equations throughout any space, and also has zero circulation
round any closed curve in that space 40; and further satisﬁes the equation S d Σ σ = 0 at
the boundary; then the function \begin{array}{l} ∭ σ^{2} d 𝔰 = 0, \end{array} or σ = 0 at
all points of the space, for each element of the integral being essentially negative
must be zero. 75. The following important theorem now follows: If there are given; at every point of any region
the convergence S \nabla σ and the spin V \nabla σ,
at every point of the boundary the normal velocity (and therefore S d Σ σ),
and the circulation round every cavity which increases the
cyclomatic number of the region; then the motion (as given
by σ at every point) if possible, is unique. For
let σ be one possible
velocity, and if possible σ + τ another. Thus τ satisﬁes
all the conditions that σ does at the end of last section, and therefore τ = 0 at every
point or σ is unique. 19. Motion of a solid through a liquid 76. We assume that there is no circulation of the liquid for any cycle; in
other words, that if all the solids be brought to rest, so will be also the
liquid. We shall take axes of reference ﬁxed in the (one) moving solid. In the
foot-note to Section 15 is explained how the rotation of these axes is
taken account of. The eﬀect of the translation of the origin will be easily
found. Let σ, ω be
the linear and angular velocities respectively of the moving solid. Let us
put \begin{array}{l} ϕ = - S σ ψ - S ω χ, & (30) \end{array} where ϕ is the required velocity potential of the liquid, and ψ and χ are two vector functions of the position of a point independent of σ and ω . Let us
see whether ψ and χ can be found so as to satisfy these conditions. σ and ω are of course assumed as quite arbitrary. The conditions are ﬁrst the equation (28) of continuity \begin{array}{l} \nabla^{2} ϕ = 0, & (28) \end{array} which gives \begin{array}{l} \nabla^{2} ψ = 0, \nabla^{2} χ = 0, & (31) \end{array} and second, the equality of the normal velocity - S U d Σ \nabla ϕ of the
liquid at any point of the boundary with that of the boundary at the same point.
Thus for a ﬁxed boundary \begin{array}{l} S σ (S d Σ \nabla · ψ) + S ω (S d Σ \nabla · χ) = 0, \end{array} whence on account of the arbitrariness of σ and ω \begin{array}{l} S d Σ \nabla · ψ = S d Σ \nabla · χ = 0 . & (32) \end{array} For a moving boundary again we have \begin{array}{l} S d Σ \nabla ϕ & = S d Σ (σ + V ω ρ), \\ or & S σ (d Σ + S d Σ \nabla · ψ) & + S ω (V ρ d Σ + S d Σ \nabla · χ) = 0, \\ which gives \\ d Σ & + S d Σ \nabla · ψ = 0, & (33) \\ V ρ d Σ & + S d Σ \nabla · χ = 0 . & (34) \end{array} Now it is well known that ψ and χ can be determined to satisfy all these conditions. In fact x being a coordinate
of either ψ or χ these conditions
amount to:— \nabla^{2} x = 0 throughout
the space, and S d Σ \nabla x = given value at the boundary. 77. We do not propose to ﬁnd ψ and χ in any particular case. We leave p. 454 of Greenhill’s article and go on
to p. 455, i.e. we proceed to ﬁnd the equations of motion of the solid.
For this purpose we require the kinetic energy of the system. Calling
this T we shall have \begin{array}{l} T = - S σ Σ σ ∕ 2 - S ω Φ σ - S ω Ω ω ∕ 2, & (35) \end{array} where Σ, Ω and Φ are linear
vector functions and the ﬁrst two are self-conjugate. This is the most general 41 quadratic
function of ω and σ . It involves 21 independent
constants, six in Σ,
six in Ω and
nine in Φ .
When ψ and χ are
known Σ, Φ and Ω are all known. We will obtain the expressions for them, and for simplicity will
assume that the origin is at the centre of gravity of the moving solid.
Let M be the mass of
this last and μ ω (where,
as is well-known, μ is
a self-conjugate linear vector function) the moment of momentum. Thus by
equation (29) \begin{array}{l} 2 T = - D \iint ϕ S d Σ \nabla ϕ - M σ^{2} - S ω μ ω . \end{array} Putting S d Σ \nabla ϕ = S d Σ (σ + V ω ρ) at the moving boundary and zero at the ﬁxed, \begin{array}{l} 2 T = D \iint (S σ ψ + S ω χ) (S σ d Σ + S ω ρ d Σ) - M σ^{2} - S ω μ ω \end{array} where the surface integral must be taken only over the moving boundary. Thus noting
that Σ and Ω are self-conjugate \begin{array}{l} Σ λ & = M λ - \frac{1}{2} D \iint (ψ S λ d Σ + d Σ S λ ψ), & (36) \\ Ω λ & = μ λ - \frac{1}{2} D \iint (χ S λ ρ d Σ + V ρ d Σ S λ χ), & (37) \\ - S λ Φ λ' & = \frac{1}{2} D \iint (S λ χ S λ' d Σ + S λ ρ d Σ S λ' ψ) . & (38) \end{array} These surface integrals can be simpliﬁed, for in each of these equations the ﬁrst surface
integral equals the second. In equation (36) in order to prove this we have merely to
put for d Σ, - S d Σ \nabla · ψ (equation (33)) when we shall ﬁnd by equation (9) Section 4 above (since we may now suppose
the integrations to extend over the whole boundary), that \begin{array}{l} \iint ψ S λ d Σ = - ∭ ψ_{1} S \nabla_{1} \nabla_{2} S λ ψ_{2} d 𝔰 = \iint d Σ S λ ψ . \end{array} Similarly for equation (37) . Again in equation (38), \begin{array}{l} \iint S λ χ S λ' d Σ = - ∭ S \nabla_{1} \nabla_{2} S λ χ_{1} S λ' ψ_{2} d 𝔰 = \iint S λ ρ d Σ S λ' ψ . \end{array} Thus ﬁnally for Σ, Ω and Φ \begin{array}{l} Σ λ & = M λ - D \iint ψ S λ d Σ = M λ - D \iint d Σ S λ ψ, & (39) \\ Ω λ & = μ λ - D \iint χ S λ ρ d Σ = μ λ - D \iint V ρ d Σ S λ χ, & (40) \\ - S λ Φ λ' & = D \iint S λ χ S λ' d Σ = D \iint S λ ρ d Σ S λ' ψ . & (41) \end{array} This last may be put in the following four forms \begin{array}{l} Φ' λ & = - D \iint d Σ S λ χ = - D \iint ψ S λ ρ d Σ, & (42) \\ Φ λ & = - D \iint χ S λ d Σ = - D \iint V ρ d Σ S λ ψ . & (43) \end{array} Thus assuming that ψ and χ are determined
we have found T as a
quadratic function of σ and ω . 78. If P, G be the linear and angular impulses of the system respectively our equations of
motion are \begin{array}{l} Ṗ + V ω P & = F, \\ Ġ + V ω G & = M, \end{array} by the footnote to Section 15 above. Here F and M are the
                                                                          

                                                                          
external force and couple applied to the body. Now at the instant under consideration P = σ \nabla T, G = ω \nabla T . But if the origin had
been at the point - ρ, P would still
be σ \nabla T whereas G would be ω \nabla T + V ρ P . Thus diﬀerentiation
with regard to t does not
aﬀect the form of P but
does that of G . In fact using
the last stated values of P and G along with the last two equations and eventually putting \dot{ρ} = σ, ρ = 0 we
get \begin{array}{l} d σ \nabla T ∕ d t + V ω σ \nabla T = F, & (44) \\ d ω \nabla T ∕ d t + V ω ω \nabla T + V σ σ \nabla T = M . & (45) \end{array} Now from equation (35) we see that \begin{array}{l} σ \nabla T & = Σ σ + Φ' ω, & (46) \\ ω \nabla T & = Φ σ + Ω ω, & (47) \end{array} whence from equations (44) and (45) \begin{array}{l} Σ \dot{σ} + Φ' \dot{ω} + V ω Σ σ + V ω Φ' ω = F, & (48) \\ Φ \dot{σ} + Ω \dot{ω} + V σ Σ σ + (V σ Φ' ω + V ω Φ σ) + V ω Ω ω = M . & (49) \end{array} Thus we see that with Quaternion notation even the general equations of motion
are not too complicated to write down conveniently. We now leave Greenhill’s article and proceed to certain theorems not
contained therein. The Cartesian treatment of these subjects will be found in
Lamb’s treatise on The Motion of Fluids, Chapters vi. and ix., which are headed Vortex Motion and Viscosity respectively. 20. The velocity in terms of the convergences and spins 79. We have seen in Section 18 that if the spin and convergence are given for each
point of a bounded ﬂuid and the normal velocity at each point of the boundary,
there is if any, but one possible motion. We shall see directly that this unique
value always exists. At present we observe that we cannot ﬁnd a motion giving an assigned spin
and convergence for each point and any assigned velocity for each point of the
boundary. If however with such data a motion be possible we can ﬁnd the
velocity at any point explicitly. By equation (19) Section 5 above we
have \begin{array}{l} 4 π σ = - \nabla ∭ \nabla u σ d 𝔰, \end{array} but by equation (9) Section 4 \begin{matrix} ∭ \nabla u σ d 𝔰 = \iint u d Σ σ - ∭ u \nabla σ d 𝔰 \\ ∴ 4 π σ = \iint [V] \nabla u d Σ σ - ∭ [V] \nabla u \nabla σ d 𝔰, & (50) \end{matrix} where the square brackets indicate that we may or may not retain the V at
our convenience. This equation may be put \begin{array}{l} 4 π σ = \iint [V] (\nabla u d Σ σ - u d Σ \nabla σ) + ∭ u \nabla^{2} σ d 𝔰 . & (51) \end{array} Both of these equations solve the question now proposed. If the ﬂuid be considered inﬁnite the surface integral of equation (50) vanishes
if σ converges to zero at inﬁnity and also that of equation (51) if in addition the spin
and convergence converge to a quantity inﬁnitely small compared with the
reciprocal of the distance of the surface. 80. We may now consider the case when \nabla σ (spin and convergence) is given at all points and S d Σ σ (normal velocity) at the boundary. It must be observed that the given value of the
spin must be distributed in a solenoidal manner for \begin{array}{l} S \nabla V \nabla σ = 0 . \end{array} Whenever the quaternion \begin{array}{l} 4 π q = - ∭ \nabla u \nabla σ d 𝔰 = \nabla ∭ u \nabla σ d 𝔰, & (52) \end{array} is a vector, we see that all the conditions except the boundary ones are satisﬁed by putting σ = q . Let us then see
when q reduces
to a vector. By equation (9) Section 4 above \begin{array}{l} 4 π S q = - \iint u S d Σ \nabla σ + ∭ u S \nabla^{2} σ d 𝔰 . \end{array} Thus q is a vector if this surface integral vanish. The surface integral vanishes if all the
vortices form closed curves within the space. If they do not we must extend the
space and make them form closed curves outside the original space. Extending the
volume integral accordingly, we may put \begin{array}{l} 4 π σ' = - ∭ \nabla u \nabla σ d 𝔰, & (53) \end{array} and now \nabla σ' = \nabla σ .
Suppose then that \begin{array}{l} σ = σ' + σ'' . & (54) \end{array} We thus get \nabla σ'' = 0 and may therefore put \begin{array}{l} σ'' = \nabla ϕ, & (56) \end{array} where ϕ satisﬁes the equations \begin{array}{l} \nabla^{2} ϕ = 0, & (57) \\ and & S d Σ \nabla ϕ = S & d Σ (σ - σ') = known quantity. & (58) \end{array} Now it is well known that ϕ can be determined so as to satisfy these two equations. Therefore the problem
under discussion always admits of solution. The above is equivalent to Lamb’s §§ 128–131. His § 130 is the natural
interpretation of the equation \begin{array}{l} 4 π σ = - ∭ V \nabla u \nabla σ d 𝔰 . \end{array} His § 132 is seen at once from equation (50) above. For in the case
he considers we, in accordance with Section 4 above, take each
side of the surface of discontinuity as a part of the boundary. Now {[S d Σ σ]}_{a + b} = 0 so that for this part of the boundary we can leave out the part S d Σ σ and
                                                                          

                                                                          
we get \begin{array}{l} 4 π σ = \iint V \nabla u V d Σ σ - ∭ V \nabla u \nabla σ d 𝔰, & (59) \end{array} so that if we regard {[V d Σ σ]}_{a + b} as - 2 \times an
element of a vortex, we get the same law for these vortex sheets as for the vortices
in the rest of the ﬂuid. 81. The velocity potential due to a single vortex ﬁlament of
strength d 𝜃 is at once obtained by putting in equation (50) for V \nabla σ d 𝔰, 2 d 𝜃 d ρ . Thus
calling σ' the part of the velocity due to the ﬁlament \begin{array}{l} 2 π σ' = - d 𝜃 \int V \nabla u d ρ = - d 𝜃 \iint u_{1} V \nabla_{1} V d Σ \nabla_{1} \end{array} by equation (8) Section 4 above. Now \begin{array}{l} V \nabla_{1} V d Σ \nabla_{1} = - d Σ \nabla_{1}^{2} + \nabla_{1} S d Σ \nabla_{1} . \end{array} Hence, since \nabla^{2} u = 0 for all points not on the ﬁlament \begin{array}{l} 2 π σ' = \nabla (d 𝜃 \iint S d Σ \nabla u) . & (60) \end{array} Thus the velocity potential = {(2 π)}^{- l} \times the strength \times the solid angle subtended by the ﬁlament at the point considered. Kinetic Energy 82. Let us put \begin{array}{l} 4 π q & = ∭ u \nabla σ d 𝔰, & (61) \\ so that & σ & = \nabla q . & (62) \end{array} We may now ﬁnd expressions for T,
the kinetic energy, in one or two interesting forms. We have \begin{array}{l} 2 T & = - ∭ D σ^{2} d 𝔰, \\ whence & 2 T = - ∭ D S σ \nabla q d 𝔰 & = - \iint D S σ d Σ q + ∭ D_{1} S σ_{1} \nabla_{1} q d 𝔰 . \end{array} We assume that the ﬂuid extends to inﬁnity and that the vortices
and convergences are all at a ﬁnite distance, so that at inﬁnity σ is of
order 1 ∕ R^{2} and q of
order 1 ∕ R .
Thus the surface integral vanishes and we have \begin{array}{l} 2 T = ∭ D_{1} S σ_{1} \nabla_{1} q d 𝔰, & (63) \end{array} or, putting in the value of q from equation (61), \begin{array}{l} T = {(8 π)}^{- 1} ∭ ∭ u D_{1} S σ_{1} \nabla_{1} \nabla_{2} σ_{2} d 𝔰_{1} d 𝔰_{2} . & (64) \end{array} These are Lamb’s equations (28) and (29), p. 160, generalised. Similarly his equation 30 generalised is \begin{array}{l} 2 T & = ∭ D_{1} S ρ σ_{1} \nabla_{1} σ_{1} d 𝔰, & (65) \\ for & ∭ D_{1} S ρ σ_{1} \nabla_{1} σ_{1} d 𝔰 & = \iint D S ρ σ d Σ σ - ∭ D S ζ σ ζ σ d 𝔰, \end{array} by equation (9) Section 4 above. But \begin{array}{l} ζ σ ζ = 2 ζ S σ ζ - ζ^{2} σ = σ, \end{array} and the surface integral vanishes as before. 21. Viscosity 83. To consider viscosity the assumption is made that the shearing stress which causes it is μ \times the rate of shear of
the moving ﬂuid. μ is
assumed independent of the velocity and experiment seems to shew that it is also
independent of the density. This last we assume though the variation with density
can be easily treated. Consider a general strain χ ω .
Since \begin{matrix} m S λ μ ν = S χ λ χ μ χ ν = S λ χ' V χ μ χ ν \end{matrix} 42, V χ μ Putting μ and ν for any two vectors
                                                                          

                                                                          
perpendicular to ω we see
that the normal U ω to any
interface becomes U {χ'}^{- 1} ω by the
strain. Hence the interface ω experiences a shear (strain) which equals the resolved part of χ U ω perpendicular to the
vector {χ'}^{- 1} ω equals resolved
part of (χ - {χ'}^{- 1}) U ω perpendicular
to {χ'}^{- 1} ω . Now when χ is the strain function due
to a small displacement σ d t,
by Section 7 above \begin{array}{l} χ ω & = ω - S ω \nabla · σ d t, & (66) \\ whence & {χ'}^{- 1} ω & = ω + \nabla_{1} S ω σ_{1} d t, & (67) \end{array} and we see that the shear is the resolved part of \begin{array}{l} - (S U ω \nabla · σ + \nabla_{1} S U ω σ_{1}) d t \end{array} perpendicular to ω,
i.e. parallel to the interface ω .
From this we see by our assumption concerning viscosity that any element of the ﬂuid is subject
to a stress ϕ given by \begin{array}{l} ϕ ω = - R ω - μ (S ω \nabla · σ + \nabla_{1} S ω σ_{1}) \end{array} R being a scalar. Now we
deﬁne the pressure p by putting \begin{matrix} 3 p = S ζ ϕ ζ = 3 R + 2 μ S \nabla σ \\ ∴ ϕ ω = - p ω + \frac{2}{3} μ ω S \nabla σ - μ (S ω \nabla · σ + \nabla_{1} S ω σ_{1}) . & (68) \end{matrix} The equation of motion is \begin{array}{l} D \dot{σ} = D F + ϕ Δ, \end{array} or \begin{array}{l} D (\partial σ ∕ \partial t - S σ \nabla · σ) = D \dot{σ} = D F - \nabla p - μ \nabla S \nabla σ ∕ 3 - μ \nabla^{2} σ . & (69) \end{array} If μ be not as stated
independent of D this equation must be made to contain certain space ﬂuxes
of μ which are quite easy to insert. 84. In § 179 of Lamb’s treatise he considers the dissipation function due to the
viscosity of a ﬂuid in a way that seems to me misleading if not wrong. It appears
as if he should add to the ﬁrst expression of that section \begin{array}{l} u d p_{x x} ∕ d x + v d p_{x y} ∕ d x + w d p_{x z} ∕ d x . \end{array} He refers to Stokes, Camb. Trans. vol. ix. p. 58. Let us give the quaternion
treatment of Stokes’s method. If T is the kinetic energy of any portion of the ﬂuid \begin{array}{l} 2 T & = - ∭ D σ^{2} d 𝔰 . \\ Thus & ∵ & d (D d 𝔰) ∕ d t = 0, \\ Ṫ & = - ∭ D S σ \dot{σ} d 𝔰, \\ but & D \dot{σ} & = D F + ϕ_{1} \nabla_{1}, \\ so that & Ṫ & = - ∭ D S σ F d 𝔰 - ∭ S σ ϕ_{1} \nabla_{1} d 𝔰 \\ or & Ṫ = - ∭ D S σ F d 𝔰 & - \iint S σ ϕ d Σ + ∭ S σ_{1} ϕ \nabla_{1} d 𝔰 . & (70) \end{array} If now ϕ be expressed
in terms of p and μ we have Ṫ depending
on F, p and μ . The
part of Ṫ depending on F and p represents energy stored up as potential energy of position and
potential energy of strain respectively, but the part depending
on μ represents a loss of energy to the system we are considering the energy being
converted into heat. Thus putting ϕ = p + ϖ we have by equation (68) \begin{array}{l} \begin{aligned} ϖ ω & = \frac{2}{3} μ ω S \nabla σ - μ (S ω \nabla · σ + \nabla_{1} S ω σ_{1}) \\ = \frac{2}{3} μ ω S ζ ψ ζ + 2 μ ψ ω, \end{aligned}\} & (71) \end{array} where ψ is
given by equation (75) below. Thus we see that the rate of loss of energy is \begin{array}{l} \iint S σ ϖ d Σ - ∭ S σ_{1} ϖ \nabla_{1} d 𝔰 . & (72) \end{array} The surface integral is the work done by viscosity against the moving ﬂuid at the
boundary and the volume integral is considered due to the work done against the
straining of the ﬂuid. Thus we put \begin{array}{l} F = - S σ_{1} ϖ \nabla_{1}, & (73) \end{array} and call F the “dissipation function.” By equation (6) Section 2 above \begin{array}{l} F = - S ψ ζ ϖ ζ, & (74) \end{array} where ψ is
the rate of pure strain of the ﬂuid, i.e. \begin{array}{l} 2 ψ ω = - S ω \nabla · σ - \nabla_{1} S ω σ_{1} . & (75) \end{array} Again by equation (18) Section 10 above because F is quadratic
in ψ \begin{array}{l} F = - S ψ ζ ψ D F ζ ∕ 2, & (76) \end{array} so that from equation (74) we have, by Section 2 above, \begin{array}{l} ϖ = ψ D F ∕ 2 . & (77) \end{array} Substituting for ϖ from equation (71) in equation (74) \begin{array}{l} F = - \frac{2}{3} μ {(S ζ ψ ζ)}^{2} - 2 μ ψ ζ ψ ζ, & (78) \end{array} which gives F in terms of ψ . up next prev ptail top \end{array}

VHydrodynamics

15. Notation

Euler’s equations

The Lagrangian equations

16. Cauchy’s integrals of these equations

17. Flow, circulation, vortex-motion

18. Irrotational Motion

19. Motion of a solid through a liquid

20. The velocity in terms of the convergences and spins

Kinetic Energy

21. Viscosity

V
Hydrodynamics