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William Kingdon Clifford.

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we tilt the plane a little round q in any direction, so that
it cuts the ellipsoid E, the surface of the system which
touches the tilted plane will be an ellipsoid wholly inside
of E. Hence k will be less for the tilted plane, because
the axes of the touching ellipsoid are diminished. There-
fore P is the plane through q which has the greatest
swing-radius. Now the plane through any point which
has the greatest swing-radius, is normal to the longest
axis of the swing-ellipsoid at the point. Therefore the
longest axis of the swing-ellipsoid (and consequently the
shortest axis of the ellipsoid of gyration) at the point q,
is normal to the ellipsoid E, confocal to the null-quadric,
which passes through q.

Let R be the tangent plane at q to the two-sheeted
hyperboloid, H, of the system, which passes through q.
This surface, like the ellipsoid, lies entirely on one side of
the tangent plane in the neighbourhood of the point of
contact. Hence if we tilt the plane a little it will cut the
surface in a small oval curve; and therefore the surface
of the system which touches the new position of the plane
will be a two-sheeted hyperboloid lying further away from
the centre than H, and therefore having a larger trans-
verse axis. Consequently k is increased by the tilting;
whence it follows that R is the plane of least swing-
radius that can be drawn through q. This plane is normal
to the shortest axis of the swing-ellipsoid, or longest axis
of the ellipsoid of gyration.

We may put these results together by saying that at
any point the axis of least moment is normal to the ellip-
soid, and the axis of greatest moment to the two-sheeted
hyperboloid, which can be drawn through the point con-
focal to the null-quadric.

It follows that the ellipsoids and the two-sheeted hyper-
boloids of the system cut each other everywhere at right
angles. We shall now prove that the third principal axis
is normal to the one-sheeted hyperboloid through q, so
that the three systems of surfaces cut each other every-
where at right angles. [Cetera desunt.]



DYNAMIC.



CORE OF A SOLID.

The poles of all those planes which do not cut the
boundary of a solid fill up a certain space which is called
its core. If the solid has indentations we must suppose
a membrane to be tightly wrapped round it; then the
boundary of the core will be traced out by the pole of a
plane which moves about so as always to touch this
membrane.

The main use of knowing the core of the simpler class
of solids is that it is more easily remembered than a
formula, and supplies a ready means of finding the swing-
radius about any axis.

We have seen that the core of a finite line, which is
of one dimension, is the middle third ; and that the core
of a triangle or ellipse, which may be regarded as typical
figures of two dimensions, is the middle fourth. We shall
shew that the core of a tetrahedron or ellipsoid, which are
figures in three dimensions, is the middle fifth; that is, it
is a figure similar to the original, having the same mass-
centre, and of one-fifth of the linear dimensions.

A parallelepiped is to be taken as a figure of one
dimension in regard to each of the lines joining the centres
of opposite faces. Its core is accordingly
an octahedron having the middle thirds
of those lines for its three diagonals.
This figure is the reciprocal of the
parallelepiped, in the sense that the
vertices, edges and faces of one of
them are polar to, the faces, edges, and
vertices of the other. Thus the paral-
lelepiped has six four-sided faces and
eight three-legged vertices, while the
octahedron has six four-legged vertices and eight three-
sided faces.

For a triangular prism, or elliptic cylinder, we must as
before take the middle third of the line joining the mass-
centres of the opposite plane ends, and then join the ends




CORE OF A SOLID. 45

of this middle third to the middle fourth of a section of
the solid made by a plane parallel to the two ends and
midway between them. We shall thus form a core which
is in one case a pentacron, like that considered on p. 8,
and in the other case is made of two cones with a common
base and vertices on opposite sides of it. The pentacron,
having two three-legged and three four-legged vertices,
with six three-sided faces, is the reciprocal of the triangular
prism, which has two three-sided and three four-sided
faces, with six three-legged vertices.

In general, we may obtain the core of any prism or
cylinder, bounded partly by two parallel plane faces or
ends, and partly by a surface made up of parallel straight
lines, by taking the ends of the middle third of the line
which joins the mass-centres of the plane ends, and joining
them to the core of the middle section, which is parallel
to the plane ends and midway between them. The figure
so formed consists always of two pyramids or cones, having
a common base, and vertices on either side of it.

To determine the core of a tetrahedron, it is sufficient
to find the pole of a plane through one vertex a parallel
to the opposite face, bed. If the tetrahedron when loaded
from this plane be divided into slices of equal thickness
by planes parallel to it, the areas of the slices will vary as
the squares of their distances from a, and their densities
directly as those distances ; therefore their masses will be
proportional to the cubes of the distances from a. If g be
the mass-centre of the tetrahedron, a of the face bed, the
mass-centres of all these slices will be in aa, and conse-
quently the tetrahedron may be replaced by a rod aa.
whose density varies as the cube of the distance from a.
Thus the mass-centre of the loaded tetrahedron (which is
the pole of the plane through a parallel to bed) is | of
the way from a towards a, and its distance from a is
therefore | . ag = || ag. Thus the squared swing-radius,
measured parallel to ag, of a plane through g parallel to
bed is j-^ag 2 . Therefore the pole of bed, whose distance
from g measured along ag is $ag, must be at a point
whose distance is ^ga. Now the core of the tetrahedron
is clearly a tetrahedron whose vertices are the poles of



46 DYNAMIC.

the faces ; and we see that these poles are on the lines
ga, gb, gc, gd, at distances from g equal to one-fifth of
those lines respectively. Thus the core is the middle
fifth.

For a sphere we must proceed in the same way as for
a circle. The squared swing-radius of the centre in regard
to the sphere is the sum of the squared swing-radii of
three mutually perpendicular planes through it, and is
therefore three times that of either of them. If we
suppose the sphere made of concentric shells of equal
thickness, the masses of these shells are as the squares of
their distances from the centre ; the problem is therefore
the same as that of finding the squared swing-radius of a
rod, whose density varies as the squared distance from one
end, in regard to that end. If a is the length of the rod,
the second moment is ^a 5 , and the mass is ^a 3 ; thus the
squared swing-radius is fa 2 . This is therefore the squared
swing-radius of a sphere of radius a in regard to its
centre. In regard to a diametral plane it is one-third of
this, or ^a 2 ; from which it follows at once that the core
is a concentric sphere of the radius \a.

If any body be transformed by a homogeneous strain,
the core of the strained body is the strained position of the
core of the unstrained body. For let x be the distance of
a particle Sm of the body from a given plane, measured
in a given direction, and k the swing-radius in regard to
that plane, measured in the same direction. Then

mk 2 =fx 2 8m.

If the body receive a homogeneous strain, so that the
masses of corresponding volumes are unaltered, the lines
x and k, which are all parallel, will remain parallel and
be altered in the same ratio; so that if every x becomes
\x, k will become \k. Therefore m. (\kf =/(X#) 2 Sm, or
\k is the swing-radius measured parallel to the \x in the
strained body. Hence it follows that the pole of the given
plane, the swing-ellipsoid at every point, the null-quadric,
and the core of the body, all undergo the same homo-
geneous strain as the body does. It should be noticed
that this is not true of the ellipsoid of gyration.



CORE OF A SOLID. 47

Hence it follows, by homogeneous strain of a sphere,
that the core of an ellipsoid is a similar, similarly situated,
and concentric ellipsoid, of one-fifth the linear dimensions;
or, as we may say, it is the middle fifth of the body.

We may exemplify the use of the core by finding the
swing- radius of a sphere about certain axes. For an axis
through the centre the squared swing-radius is twice that
of a diametral plane, and therefore fa 2 , since the core is
the middle fifth. Hence about an axis touching the. sphere
the squared swing-radius is |a 2 .

Similarly in the case of a right circular cylinder,
height 2h, radius a, the squared swing-radius of its middle
section is |7i 2 , and of a plane through its axis a 2 . Hence
about the axis it is ^-a 2 , about a line through the centre
perpendicular to it a 2 + ^A 2 , and about a diameter of
either plane face Ja 2 + |& 2 .



CHAPTER III. MOMENTUM.



MOMENTUM OF TRANSLATION-VELOCITY.

THE product of the mass of a particle by its velocity is
called the momentum of the particle. Like the velocity, it
is a directed quantity ; but whereas the velocity is a vector,
and may be specified by a line of proper length and direc-
tion drawn anywhere in space, the momentum is what we
have called a rotor, and is situated in the straight line
through the particle along which it is moving. In fact,
the velocity is a magnitude associated with a direction,
and the mass is a magnitude associated with a definite
point ; thus the momentum, which is the product of these
two, is the product of their magnitudes associated with a
line drawn through the point in the given direction.

To compound together the momenta of two or more
particles, therefore, we must treat them as if they were
rotations about axes through the particles [p. 124]. If, for
example, particles of masses I, m situated at a, b, have the
same velocity, their momenta are proportional to the
masses I, m, and may be represented by lengths I, m
measured on lines through a, b in the direction of the
common velocity. If the lengths represented spins about
these lines as axes, the resultant spin would have a mag-
nitude equal to their sum, and would be about a parallel
line through the point c, which divides ab in the ratio




MOMENT OF MOMENTUM. 49

of m : I. Now c is the mass-centre
of the particles at a. b, and its velo-
city is the same as that of a and b.
Hence the resultant of the momenta
of two particles which have the same
velocity is the momentum of their re-
sultant mass. c 6

If we add to these a third particle, with the same
velocity, we may compound its momentum with that of the
resultant mass of the first two, and so find that the result-
ant of all the momenta is the momentum of the resultant
mass of the three ; and so on. Hence if a body have a
translation-velocity, the resultant of the momenta of its par-
ticles is equal to the momentum of its resultant mass ; that
is to say, its magnitude is the mass of the whole body
multiplied by the magnitude of the given velocity, and acts
in a straight line through the mass-centre in the direc-
tion of the given velocity.



MOMENT OF MOMENTUM.

The moment of the momentum of a particle in
regard to any point is defined in the same way as the
moment of any similar magnitude [p. 92]. Namely, if p
be the position of the particle, pt a
line representing its momentum, the
moment of the momentum about o is
twice the area of the triangle opt.
This is to be regarded as a vector
perpendicular to the plane opt, but not
localized. If m is the mass of a particle whose position-
vector is p, and velocity p, the moment of its momentum
in regard to the origin is mVpp.

Since momenta are to be compounded like spins, it
follows that the moment of the resultant of any number of
momenta in regard to any point is equal to the sum of the
moments of the components in regard to the same point.

We know that two spins about parallel axes, of the
same magnitude but of opposite senses, compound into a

c. 4




50 DYNAMIC.

translation-velocity perpendicular to the plane containing
them. This translation- velocity is equal to the sum of
their moments in regard to any point in space, and its
magnitude is the product of the magnitude of either spin
into the distance between their axes. Two such spins are
therefore equivalent to any other two equal and opposite
spins about parallel axes in any plane parallel to that of
the first pair, provided that this product is the same for
the two pairs.

In the same way, if the momenta of two particles are
on two parallel lines, equal in magnitude, but opposite in
direction, the sum of their moments is the same in regard
to every point of space, and the two momenta are equiva-
lent to any other such pair the sum of whose moments in
regard to all points is the same as that of the first pair.
They must therefore be regarded as together constituting
a vector quantity, of the nature of a moment of momentum ;
namely, it is precisely this constant sum of their moments.

If we start with two opposite momenta along parallel
lines, which are only nearly equal in magnitude, the effect
of makingithem more nearly equal is" to send their result-
ant farthest off and to diminish its magnitude. Hence a
moment of momentum may be regarded as the limit of a
very small momentum along a line which is very far off;
just as a translation-velocity may be regarded as an in-
finitely small spin about an infinitely distant axis.

Every momentum may be resolved into an equal and
parallel momentum through the origin, together with its
moment in regard to the origin. Thus if the particle m at
the end of the vector p have the velocity p, its momentum
is equivalent to mp through the origin together with the
moment of momentum m Vpp.

It is sometimes convenient to indicate a rotor through
the origin by prefixing the letter ft to the symbol of a
vector having the same magnitude and direction. Thus if
a is any vector, ft a means the rotor through the origin
which is one way of representing that vector. In this
notation, the momentum of the particle whose mass-vector
is mp will be denoted by ft mp + mVpp. And similarly the
velocity of a rigid body which has a spin 6 about an axis



S



ROTOR PART OF MOMENTUM. 51

passing through the end of the vector p will be denoted by
16+ VpO. In general, since any system of rotors and
vectors is equivalent to a rotor through the origin and a
vector, the expression for the resultant motor of such a
system is Ha -f- ft.

The moment of a vector in regard to any point is
simply the vector itself. Hence the moment of the system
Ha + fi, in regard to the point whose position-vector is 7,
is Fa7 + .

ROTOR PART OF MOMENTUM.

When we resolve the momentum of every particle of a
body into a parallel momentum through the origin, together
with a moment of momentum, and then add the results for
the whole body ; the rotor part of this is independent of
the particular origin we take. For it is simply the sum
of all the momenta regarded as vectors.

We shall prove that this rotor part is equal (in magni-
tude and direction) to the momentum of the resultant mass ;
that is to say, it is the same as if the whole mass were
collected at the mass-centre, and moving with its velocity.
To prove this, it is only necessary to remember that the
mass-vector of the resultant mass is the sum of the mass-
vectors of all the particles, or it is fpdm. The momentum
of the resultant mass is the rate of change of this, or
fpdm; but this is the vector-sum of the momenta of all the
particles.

Now every motion of a body can be resolved into two ;
a translation equal to the velocity of the mass-centre, and
the motion relative to the mass-centre. The momentum
of the whole body is compounded of the momenta due to
these motions. The momentum of the translation is a
pure rotor, passing through the mass-centre ; the proposi-
tion just proved shews us that the momentum of the motion
relative to the mass-centre is a pure vector. For the rotor
part of the whole momentum has been shewn to be equal
to the momentum of the translation.

In calculating the momentum of any motion, therefore,
it is sufficient to find the moment of momentum about the

42



52 DYNAMIC.

mass-centre ; for since the momentum of motion relative
to the mass-centre is a pure vector, its moment is the
same in regard to every point of space.

MOMENTUM OF SPINS ABOUT FIXED POINT.

We now consider a rigid body rotating about a fixed
point, and propose to calculate its moment of momentum
in regard to the point.

Every spin about an axis through the point may be
resolved into three component spins about the principal
axes ; and its moment of momentum will be the vector-
sum of the moments of momenta due to these. The first
thing to be done, therefore, is to find the moment of
momentum of a spin about a principal axis.

In the first place we may remark that the component
along the axis of the moment of momentum of a spin about
any straight line, is equal to the angular velocity multi-
plied by the second moment of the body about that line.
For let a be the origin, ab the spin, p a
rotating particle. The moment of its mo-
mentum about a is i . ap multiplied by the
magnitude of the momentum. Hence the
component of this along ab is i . mp multi-
plied by the same magnitude, which is
ab . mp into the mass of the particle. Thus
the component is ab . mp 2 into the mass of
the particle, or the angular velocity multi-
plied by the second moment. Adding this result for all
the particles, we have the proposition as stated for the
whole body.

The component in any direction perpendicular to the
axis is the angular velocity multiplied by the rnixed mo-
ment of two planes, one of which is the plane through a
perpendicular to the axis, and the other is the plane
through the axis perpendicular to the given direction.
For the whole component perpendicular to the axis is
i . am x magnitude of momentum ; that is, it is parallel to
pm, and of magnitude ab . am .pm. Now am is the
distance of p from the normal plane through a, and the



MOMENTUM OF SPINS ABOUT FIXED POINT. 53

component of pm in any direction perpendicular to ab is
the distance of p from a plane through ab perpendicular
to that direction ; whence the proposition follows.

Now if the axis is a principal axis, the mixed moment
of a plane through it and a plane perpendicular to it is
zero. Hence the moment about any point of momentum
due to a spin about a principal axis through the point is
directed along the axis, and is equal in magnitude to the
angular velocity multiplied by the second moment.

We may put these propositions together as follows.
Let oX, o Y, oZ be three rectangular lines through o, and
let us consider a unit spin about oZ, which is denoted by
k. The velocity of the end of p, = xi + yj + zTc, due to this
spin, is Vk (xi + yj + zk), = xj yi. Let Sm be the mass of
the particle there situated, then the moment of its mo-
mentum is

V (xi + yj + zk) (xj yi) 8m ={ xz.i yz.j+(x* + y z ) k} 8m.
Hence the moment of momentum of the whole body is

Jxzdm . i fyzdm . j + /(a; 2 + y 2 ) dm . k,
which gives the components as above stated.

Now let the axes be principal axes, and let the swing-
radii about them be a, b, c. Then if the body have a spin
0,=pi+qj + rk, its moment of momentum about the
origin will be m (a 2 pi + b 2 qj + c 2 rk) ; since it is the result-
ant of the moments of momentum due to the three spins
pi, qj, rk. The components of the spin must therefore be
respectively multiplied by ma 2 , mb 2 , me 2 , to give the com-
ponents of the moment of momentum.

Thus the moment of momentum is a pure linear function
of the spin. The ellipsoid representing this function must
have its semi-axes inversely proportional to the square
roots of the second moments ma 2 , mb 2 , me 2 , that is, in-
versely proportional to a, b, c. Their actual values, in ac-
cordance with the formula [p. 176], are mbc, mca, mab
each divided by TT. Thus the relation between x, y, z for
any point on this ellipsoid is

7T 2 (aV + b*y 2 + cV) = (mabcf.



54 DYNAMIC.

The ellipsoid thus determined is called the momental
ellipsoid,

There is no great convenience in thus fixing the size of
the momental ellipsoid, on account of the two scales of
representation involved ; angular velocity is represented by
a line, and mass by the inverse of a line (since mbc is one
semi-axis). If we take any ellipsoid similar to this one,
we may so choose the scale on which angular velocities are
represented that the momentum due to a spin represented
by any radius vector of the ellipsoid shall be equal to the
mass of the body multiplied by the area of the conjugate
section. We shall therefore give the name " momental
ellipsoid" to any ellipsoid having axes along oX, oY, oZ
inversely proportional to a, b, c ; so that we may determine
its size in particular cases by other considerations of con-
venience.

The momental ellipsoid is the reciprocal of the ellipsoid
of gyration. For the ellipsoid of gyration has semi-axes
a, 6, c, and therefore if p be a perpendicular from the
origin on a tangent plane to it, and make angles a/3y with
the axes, we must have p 2 = a* cos 2 a + 6 2 cos' 2 /3 + c 2 cos 2 7.
Now let us measure on the line p a distance r from the
origin, such that pr = C, and let s be the point at the end
of this distance, so that os = r. Then the locus of s will be
the reciprocal of the ellipsoid of gyration [p. 101]. The
coordinates of s are

x = r cos ot, y = r cos /3, z = r cos 7,
or px = G cos ai,py=G cos /3, pz = C cos 7.

Therefore

p 2 (aV + by + cV) = <7 2 (a 2 cos 2 a + 6 2 cos 2 /9 + c 2 cos 2 7 ) = C*p*
and consequently aV + 6 2 y 2 + cV = C 2

so that the locus of s is an ellipsoid whose axes are in-
versely proportional to a, b, c, that is, an ellipsoid of
momentum.



MOMENTUM OF TWIST. 55



MOMENTUM OF TWIST.

Let us now take the mass-centre of a rigid body for
origin. The instantaneous state of motion of the body is
made up of a translation-velocity equal to the velocity of
the mass-centre, and a spin about some axis through it.
Let these, regarded as vectors, be denoted by a and /3 re-
spectively ; then to indicate that the latter is really a rotor
passing through the origin, we may denote the whole
motion of the body by ftyS -,'- a. This is the general symbol
for a twist-velocity.

It is important to observe that the vector part of this
velocity gives rise to the rotor part of the momentum, and
vice versa. The translation is not localized, but its mo-
mentum has to pass through the mass-centre. The rota-
tion is about a definite axis ; but its momentum is a mere
moment of momentum, which has only magnitude and
direction, no definite position.

Thus the momentum due to the translation a is ftraa,
where m is the mass of the body. The momentum due
to the spin ft/3 is 0/3, where is the pure function which
converts any spin into its moment of momentum. Hence
the whole momentum due to the twist ft/3 +a is

ftma + 0/3.

Given any screw, the momentum due to a twist about
it may be associated with another screw. Namely the
momentum is equivalent to a rotor part along a certain
axis, together with a vector part parallel to the axis ; the
proof of this is precisely like that for twists [p. 126].

We now propose to determine in what cases these two
screws are identical. When this is so, the momentum
ftraa + 0/3 is a numerical multiple of the velocity ft/3 +a ;

say ftraa + 0/3 = x (ft/3 + a),

whence ma. = #/3, 0/3 = #a,

and therefore ra/3 = x 2 ft.



56 DYNAMIC.

Hence the spin /3 must be about one of the principal axes
at the mass-centre. Suppose it to be oX, then </3 = ma 2 /3,
and x = ma. Consequently a = + a/3, and the pitch is
+ a. There are therefore six screws having this property ;
they are called the principal screws of the body. These
axes are the principal axes at the mass-centre, and their
pitches are the swing-radii about them.



57



APPENDIX I. (A.)



ACCELERATION DEPENDING ON STRAIN.

Imagine a spiral spring, like that of a spring-balance,
with circular disks fastened to its ends and a small hole
through one of them. Let it be partially compressed, and
held in that position by a string attaching one of the two


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