As for the composition of rotations about nonintersecting axes, and generally
of an arbitrary succession of displacements of a solid system, given by
individual displacements $\Delta^{\prime},\Delta^{\prime\prime},etc.$ whose
analytic form is known, we shall have

$\displaystyle\Delta
x=\Delta^{\prime}x+\Delta^{\prime\prime}x+\Delta^{\prime\prime\prime}x+...=A+2\tan(\Theta/2)(Y\cos
L-Z\cos H),$ $\displaystyle\Delta
y=\Delta^{\prime}y+\Delta^{\prime\prime}y+\Delta^{\prime\prime\prime}y+...=B+2\tan(\Theta/2)(Z\cos
G-X\cos L),$ $\displaystyle\Delta
z=\Delta^{\prime}z+\Delta^{\prime\prime}z+\Delta^{\prime\prime\prime}z+...=C+2\tan(\Theta/2)(X\cos
H-Y\cos G).$ (124)

The constants $A,B,C\;etc.$ (pertaining to the resultant displacement) are to
be found in terms of analogous constants
$A^{\prime},B^{\prime},C^{\prime}etc.$; ;
$A^{\prime\prime},B^{\prime\prime},C^{\prime\prime}etc.$; … belonging to each
of the consecutive displacements to be combined, and of the other elements of
these displacements. The expressions beginning $A$,$B$,$C$ are closely related
to those seen in (15) and (15). But it is evident that the elements
$\Theta,G,H,L$ of the resultant rotation depend only on the _rotational_
elements of these displacements, just as we have seen earlier from geometric
considerations.

Having posed a problem of sweeping generality, the author proceeds to solve
only the simplest case, that of composing two rotations about nonintersecting
axes. Moreover, he will select a made-to-order orientation of the coordinate
system similar to the one he introduced to treat two intersecting axes.
Whereas in the previous section both axes passed through the point $(0,0,0)$,
and (119) gave their direction cosines as $(1,0,0)$ and $(\cos\nu,\sin\nu,0)$,
in the nonintersecting case (119) will still give the direction cosines but
the second axis will pass not through $(0,0,0)$ but through $(0,0,u)$ where
$u$ is geometrically the closest distance of approach between the two axes.

Let us take, for example, the composition of rotations about two fixed
nonintersecting axes, one identified with the $x$-axis and the other normal to
the $z$-axis and cutting that axis at a distance $u$ from the origin. We
shall, as before, denote by $\nu$ the angle between these two rotation axes
and by $\theta^{\prime},\theta^{\prime\prime}$ the amplitudes of the
respective rotations. To begin with, we shall have for the amplitude of the
resultant rotation and the direction of its axis, the same formulas as
obtained in the previous section. To fix the position of the central (i.e.
composite) axis, as well as to find the displacements of the coordinates, one
needs only to calculate the displacements $\alpha,\beta,\gamma$ from the
origin of coordinates. Now, those displacements arising from the first
rotation about the $x$ axis are null; hence it suffices to calculate those
arising from the rotation $\theta^{\prime\prime}$, for which one has in
general (cf. (15) near the end of Section 15)

$\displaystyle\Delta^{\prime\prime}x=\alpha+2\tan(\theta^{\prime\prime}/2)[(Y^{\prime\prime}-(1/2)\beta)\cos
l^{\prime\prime}-(Z^{\prime\prime}-(1/2)\gamma)\cos h^{\prime\prime}]$
$\displaystyle\Delta^{\prime\prime}y=\beta+2\tan(\theta^{\prime\prime}/2)[(Z^{\prime\prime}-(1/2)\gamma)\cos
g^{\prime\prime}-(X^{\prime\prime}-(1/2)\alpha)\cos l^{\prime\prime}]$
$\displaystyle\Delta^{\prime\prime}z=\gamma+2\tan(\theta^{\prime\prime}/2)[(X^{\prime\prime}-(1/2)\alpha)\cos
h^{\prime\prime}-(Y^{\prime\prime}-(1/2)\beta)\cos g^{\prime\prime};$ (125)

these displacements must vanish for all points on the axis of rotation
$\theta^{\prime\prime}$, for which one has

$\cos l^{\prime\prime}=0,\;\cos h^{\prime\prime}=\sin\nu,\;\cos
g^{\prime\prime}=\cos\nu,\;Y^{\prime\prime}=X^{\prime\prime}\tan\nu,\;Z^{\prime\prime}=u.$
(126)

From this there result the following values for $\alpha,\beta,\gamma$, which
one could easily derive from the construction itself,

$\alpha=u\sin\nu\sin\theta^{\prime\prime},\;\;\beta=-u\cos\nu\sin\theta^{\prime\prime},\;\;\gamma=2u\sin^{2}(\theta^{\prime\prime}/2),$
(127)

in agreement with the theorems of Section 13.

Thus the resultant axis and the two given divergent axes become more nearly
parallel to a single plane, the greater the distance between the two latter
axes in comparison with the absolute translation of the resultant
displacement. That is, $\cos L$ approaches zero as $u>>T$.

Since the two rotation axes do not intersect, Euler’s fixed point theorem
(Section 3) does not hold, and the resultant displacement is not a pure
rotation but contains an absolute translation $\vec{\,T}=T\hat{T}$.

In (20) the coordinates of the point $\xi,\eta,\zeta$ defined in (16) have
been replaced in (15) by $X,Y,Z$ as given in (126). Note that in the case
$u=0$ (two intersecting axes) $X,Y,Z$ would be strictly proportional to $\cos
g^{\prime\prime},\cos h^{\prime\prime},\cos l^{\prime\prime}$ and consequently
the terms in these quantities would vanish; this would leave only
$(\alpha,\beta,\gamma)$ on the right side of (20) so that the left side is
made to vanish by setting $(\alpha,\beta,\gamma)=0$. This is why
$(\alpha,\beta,\gamma)$ do not appear in Section 19.

For $u\neq 0$, only the terms in $X,Y$ cancel out, and to solve for
$\alpha,\beta,\gamma$ it suffices to make $\Delta^{\prime\prime}(x,y,z)$
vanish when $(X,Y,Z)=(0,0,u)$. This yields (using (126) as well)

$\displaystyle
0=\alpha+2\tan(\theta^{\prime\prime}/2)[-(u-(1/2)\gamma)\sin\nu]$
$\displaystyle 0=\beta+2\tan(\theta^{\prime\prime}/2)[(u-(1/2)\gamma)\cos\nu]$
$\displaystyle
0=\gamma+2\tan(\theta^{\prime\prime}/2)[-(1/2)\alpha\sin\nu-(-(1/2)\beta\cos\nu]$
(128)

The first two lines give

$\alpha=(2u-\gamma)\tan(\theta^{\prime\prime}/2)\sin\nu,\;\;\;\beta=-(2u-\gamma)\tan(\theta^{\prime\prime}/2)\cos\nu$
(129)

and the third gives

$\gamma=(2u-\gamma)\tan^{2}(\theta^{\prime\prime}/2)(\sin^{2}\nu+\cos^{2}\nu)=2u\tan^{2}(\theta/2)-\gamma\tan^{2}(\theta/2).$
(130)

Transposing the last term to the left, and multiplying the equation by
$\cos^{2}(\theta^{\prime\prime}/2$, we have

$\gamma=2u\tan^{2}(\theta^{\prime\prime}/2)\cos^{2}(\theta^{\prime\prime}/2)=2u\sin^{2}(\theta^{\prime\prime}/2),$
(131)

and substituting (131) into (129) gives

$\alpha=u\sin\nu\sin\theta^{\prime\prime},\;\;\;\beta=-u\cos\nu\sin\theta^{\prime\prime}$
(132)

as in $\eqref{alphbetgammnonint}$. (The text has a mistake: $\sin\nu$ instead
of $\cos\nu$ in $\beta$.)

The analogy between this calculation and the one discussed in the last
paragraphs of Section 15 may be obscured by the fact that then only a single
rotation was under consideration, whereas now there are two. But there is no
difference if one makes the appropriate correspondences. The role played by
the _central_ axis in Section 15 is here played by the _second_ of the two
axes, having direction $\hat{q}^{\prime\prime}$ and rotation angle
$\theta^{\prime\prime}$. Call this axis the _active_ axis. In both cases the
active axis does not pass through the origin of coordinates. In both cases a
_substitute_ axis is introduced, parallel to the active axis but passing
through the origin of coordinates. In both cases
$(\alpha,\beta,\gamma)=\vec{\delta}$ is the displacement of a point starting
at the origin. In the present case there is an additional axis, the first,
which plays no part in determining $\vec{\delta}$ because it lies on the
substitute axis and its action on the point of origin is null.

In the present case, the values of $\alpha,\beta,\gamma$ can be easily
obtained because the direction $\hat{q}^{\prime\prime}=(\cos\nu,\sin\nu,0)$ is
given. Starting at the origin $(0,0,0)$, the trajectory of rotation about the
active axis describes an arc $\theta^{\prime\prime}$ of the circle whose
center is at $(0,0,u)$ and whose plane is normal to $\hat{q}^{\prime\prime}$.
This gives immediately
$\gamma=u(1-\cos\theta^{\prime\prime})=2u\sin^{2}(\theta^{\prime\prime}/2)$ as
well as $\sqrt{\alpha^{2}+\beta^{2}}=u\sin\theta^{\prime\prime}$. From the
direction of $\hat{q}^{\prime\prime}$ we infer that
$\alpha:\beta::\sin\nu:(-\cos\nu)$ and so
$\alpha=u\sin\theta^{\prime\prime}\sin\nu^{\prime\prime},\beta=-u\sin\theta^{\prime\prime}\cos\nu$.
In this way (131) and (132) are derived from geometry alone.

From (127) one derives, for the value $T$ of the absolute translation
resulting from two rotations about two nonintersecting fixed axes:

$\displaystyle T$ $\displaystyle=$ $\displaystyle\alpha\cos G+\beta\cos
H+\gamma\cos L$ (133) $\displaystyle=$
$\displaystyle\frac{2u\sin\nu\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)}{\sin(\Theta/2)}$
$\displaystyle=$ $\displaystyle 2u\cos L.$

This array contains three equalities. To prove the first equality, one notes
that $T=\vec{\,T}\cdot\hat{T}$ where $\vec{\,T}$ is the displacement of the
composite axis along its length due to the successive actions of the two
partial rotations. We now treat the composite axis just as we treated the
second partial axis in deriving (131) and (132): we introduce a _substitute
composite axis_ parallel to the true composite axis but passing through the
origin. Referring rotations to this substitute axis, we deduce as before that
$(\alpha,\beta,\gamma)=\vec{\delta}$ is the displacement undergone by the
point originally at $(0,0,0)$, that is, $\vec{\delta}$ is the translation,
common to all points, that must be added to rotation about the substitute axis
in order to simulate the original composite displacement $\vec{\,T}$. But
rotation about the substitute axis, applied to an arbitrary point, produces a
trajectory perpendicular to its direction $\hat{T}$, so that
$\vec{\,T}\cdot\hat{T}=\vec{\delta}\cdot\hat{T}$. Hence
$T=\delta\cdot\hat{T}$, which is the first equality.

The second equality is obtained by takng $(\alpha,\beta,\gamma)$ from (127)
and $\hat{T}$ from (19). This gives

$\displaystyle\alpha\cos G\sin(\Theta/2)$ $\displaystyle=$ $\displaystyle
u\sin\nu\sin\theta^{\prime\prime}[\sin(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)+\cos(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)\cos\nu]$
$\displaystyle\beta\cos H\sin(\Theta/2)$ $\displaystyle=$
$\displaystyle-u\cos\nu\sin\theta^{\prime\prime}[\cos(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)\sin\nu]$
$\displaystyle\gamma\cos L\sin(\Theta/2)$ $\displaystyle=$ $\displaystyle
2u\sin^{2}(\theta^{\prime\prime}/2)[\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)\sin\nu]$
(134)

The second term in the first line cancels the second line so that

$\alpha\cos G\sin(\Theta/2)+\beta\cos
H\sin(\Theta/2)=u\sin\nu\sin\theta^{\prime\prime}[\sin(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)]=2u\sin\nu\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)\cos^{2}(\theta^{\prime\prime}/2)$
(135)

and adding the third line we obtain the second equality of (133). The third
equality is obtained by simply comparing the numerator of the second line of
(133) with the third line of (19).

The equations of the resultant axis are obtained by substituting for
$\alpha,\beta,\gamma$ their values from (127) in the general equations

$\displaystyle\frac{x-(1/2)\alpha-(1/2)\cot(\Theta/2)(\beta\cos L-\gamma\cos
H)}{\cos G}$ $\displaystyle=\frac{y-(1/2)\beta-(1/2)\cot(\Theta/2)(\gamma\cos
G-\alpha\cos L)}{\cos H}$
$\displaystyle=\frac{z-(1/2)\gamma-(1/2)\cot(\Theta/2)(\alpha\cos H-\beta\cos
G)}{\cos L}.$ (136)

These equations are the same as (52) with $g,h,l,\theta$ capitalized.

In the case of infinitely small rotations these formulas are considerably
simplified, and it is found that the (resultant) central axis is parallel to
the plane of the two contributing axes and intersects their shortest distance.
Effectively, in this case, with neglect of infinitesimals of second order, one
has

$\displaystyle\cos G$ $\displaystyle=$
$\displaystyle\frac{\theta^{\prime}+\theta^{\prime\prime}\cos\nu}{\Theta},\;\cos
H=\frac{\theta^{\prime\prime}\sin\nu}{\Theta},\;\cos
L=\frac{\theta^{\prime}\theta^{\prime\prime}\sin\nu}{2\Theta},$
$\displaystyle\Theta^{2}$ $\displaystyle=$ $\displaystyle\theta^{\prime
2}+\theta^{\prime\prime 2}+2\theta^{\prime}\theta^{\prime\prime}\cos\nu,$
$\displaystyle\alpha$ $\displaystyle=$ $\displaystyle
u\theta^{\prime\prime}\sin\nu,\;\;\beta=-u\sin\theta^{\prime\prime}\cos\nu,\;\;\gamma=0,\;\;T=\frac{u\theta^{\prime}\theta^{\prime\prime}\sin\nu}{\Theta},$
(137)

(I have corrected three errors in the French text of this array: (i) top line,
$\cos H$ numerator, $\sin\nu$ incorrectly given as $\sin^{2}\nu$; (ii) second
line, left side of equation, $\Theta^{2}$ incorrectly given as $\Theta$; (iii)
bottom line, $T$ numerator incorrectly given as $u\theta\theta\theta\sin\nu$)

and for the equations of the (resultant) central axis,

$y=\frac{x\theta^{\prime\prime}\sin\nu}{\theta^{\prime}+\theta^{\prime\prime}\cos\nu},\;\;\;z=\frac{u\theta^{\prime\prime}(\theta^{\prime\prime}+\theta^{\prime}\cos\nu)}{\Theta^{2}}.$
(138)

##  21 Composition of successive rotations about three perpendicular axes.

We shall end this subject by giving the following formulas for the composition
of three successive rotations $\theta,\theta^{\prime},\theta^{\prime\prime}$
about the three coordinate axes $x,y,z$. The elements $\Theta,G,H,L$ of the
composite rotation are expressed as follows:

$\cos(\Theta/2)=\cos(\theta/2)\cos(\theta^{\prime}/2)\cos(\theta^{\prime\prime}/2)-\sin(\theta/2)\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2),$
(139)

$\displaystyle\sin^{2}G$ $\displaystyle=$
$\displaystyle\frac{1-\cos\theta^{\prime}\cos\theta^{\prime\prime}}{2\sin^{2}(\Theta/2)},$
$\displaystyle\sin^{2}H$ $\displaystyle=$
$\displaystyle\frac{1-\cos\theta\cos\theta^{\prime\prime}+\sin(\theta/2)\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)}{2\sin^{2}(\Theta/2)},$
$\displaystyle\sin^{2}L$ $\displaystyle=$
$\displaystyle\frac{1-\cos\theta\cos\theta^{\prime}}{2\sin^{2}(\Theta/2)}.$
(140)

These formulas become symmetric with respect to each of the three successive
rotations only when, the rotations being infinitely small, the term
$\sin(\theta/2)\sin(\theta^{\prime}/2)\sin(\theta^{\prime\prime}/2)$ vanishes
in $\sin^{2}H$. This vanishing also makes the order of these rotations
indifferent; one then finds, in accordance with the law of composition of
infinitesimally small rotations,

$\Theta^{2}=\theta^{2}+\theta^{\prime 2}+\theta^{\prime\prime 2},\cos
G=\theta/\Theta,\cos H=\theta^{\prime}/\Theta,\cos
L=\theta^{\prime\prime}/\Theta.$ (141)

The problem inverse to the one we have just solved would have for its object
the _decomposition_ of a finite rotation about a given axis into three
rotations about the three coordinate axes. This amounts to solving the above
equations for $\theta,\theta^{\prime},\theta^{\prime\prime}$ in terms of
$\Theta,G,H,L$, which cannot be done for finite rotations but is utterly
simple when the rotations are infinitely small.

##  22 On the composition of successive infinitesimal displacements of a solid
system.

We shall now consider, directly and with a special extension, the laws of
composition and decomposition of successive _infinitesimal_ displacements,
drawn from the analytic expression for the infinitely small coordinate changes
of a solid system.

These changes are generally expressed in the following way,

$\delta x=\alpha+py-nz,\;\;\delta y=\beta+mz-px,\;\;\delta z=\gamma+nx-my,$
(142)

($\vec{\Delta}=\vec{\delta}+\vec{r}\times\vec{q}$, where $\vec{\Delta}$ and
$\vec{q}$ are infinitesimal)

as _linear_ functions of the infinitesimal elements of the displacement,
$\alpha,\beta,\gamma,m,n,p$ (that is, $\vec{\delta}$ and $\vec{q}$). It
results that the changes arising from several successive infinitesimal
displacements combine by adding together the changes due separately to each of
these successive displacements, referred to the original situation of the
system. The elements of the composite displacement are the sums of the
analogous elements of the partial displacements.

This is the (usual) way in which the complete differential of a function of
several variables is formed by adding the partial differentials relative to
each of those variables. And since one neglects infinitesimals of the second
order, it makes no difference whether the differentials are expressed in terms
of the starting values of the finite variables, or in terms of their values
successively augmented by the infinitesimal increments they undergo.

Hence if one denotes by
$\vec{\delta}^{\prime},\vec{q}^{\prime};\vec{\delta}^{\prime\prime},\vec{q}^{\prime\prime};\vec{\delta}^{\prime\prime\prime},\vec{q}^{\prime\prime\prime}$
etc. the elements of the successive displacements to be composed, for each of
which one has

$\vec{\Delta}^{\prime}=\vec{\delta}^{\prime}+\vec{q}^{\prime}\times\vec{r},\;\;\vec{\Delta}^{\prime\prime}=\vec{\delta}^{\prime\prime}+\vec{q}^{\prime\prime}\times\vec{r},etc.,$
(143)

the elements of the composite displacement, represented by $A.B,C,M,N,P$ (that
is, by $\vec{\Gamma},\vec{Q}$), will be respectively the sums of the given
partial elements; one will have

$A=\alpha^{\prime}+\alpha^{\prime\prime}+...=\Sigma\alpha,B=\Sigma\beta,C=\Sigma\gamma,M=\Sigma
m,N=\Sigma n,P=\Sigma p,$ (144)

and for the expressions of the composite (infinitesimal) variations of the
coordinates $x,y,z$,

$\vec{\Delta}=\Gamma+\vec{r}\times\vec{Q}=\Gamma+\Theta\vec{r}\times\hat{Q}$
(145)

where we have introduced the (infinitesimal) rotation $\Theta$ and the
direction $\hat{Q}$.

The author now points out that since the displacements are infinitesimal, one
might equally decompose the same total displacement (resulting from pure
rotations about arbitrarily many axes arbitrarily placed and directed) into
exactly _three_ partial rotations taken successively about the $x$,$y$,$z$
axes, with infinitesimal angles $m,n,p$, each accompanied by the appropriate
“screw” translation
$\vec{t}\cdot\hat{x},\vec{t}\cdot\hat{y},\vec{t}\cdot\hat{z}$.

##  23 Geometry and analytical mechanics.

These successive rotations $m,n,p$ are known in mechanics by the name of
elementary rotations, and considered as simultaneous in the passage from the
geometric to the mechanical laws of the displacement of bodies,
notwithstanding that geometry cannot take account of them except by supposing
them to be successive. For it is evident that the system, in turning about the
axis whose rotation is $\theta$ and whose angles with the $x,y,z$ axes are
$g,h,l$, does not achieve at the same time the three rotations $m,n,p$ about
those coordinate axes: this would require four axes of rotation instead of a
single one (Mécanique Analytique, vol. 1, p. 52).[4]

This reference is to the epoch-making two-volume treatise (1788-9) by Joseph-
Louis Lagrange, drawing on discoveries and insights from Euler, D’Alembert,
the Bernoulli brothers, and others, which for the first time showed that all
mechanical properties of a system could be derived from a single algebraic
formula and embodied in a single differential equation, without (according to
Lagrange himself) requiring “either geometrical or mechanical constructions or
reasoning”.

We encounter here a fundamental point in the philosophy of mathematics, that
which separates geometry from mechanics, and the importance of which it is the
object of this Memoir to establish in its totality.

The rotation $\theta$ results from the successive composition of the rotations
$\theta\cos g,\theta\cos h,\theta\cos l$, because the displacement due to this
rotation $\theta$ is for each coordinate axis the sum of the displacements
which would be due separately to each of the elementary rotations. In fact, if
the system were to turn only about the $x$-axis with a rotation $\theta\cos
g$, one would have for this displacement

$\delta^{\prime}x=0,\;\;\delta^{\prime}y=\theta z\cos
g,\;\;\delta^{\prime}z=-\theta y\cos g;$ (146)

if on the contrary the system were to turn only about the $y$-axis with a
rotation $\theta\cos h$, one would have for this displacement

$\delta^{\prime\prime}x=-\theta z\cos
h,\;\;\delta^{\prime\prime}y=0,\;\;\delta^{\prime\prime}z=\theta x\cos h;$
(147)

(the expressions for $\delta^{\prime\prime}x$ and $\delta^{\prime\prime}z$ in
the French text contain slight errors)

and finally the rotation $\theta\cos l$ about the $z$-axis. considered alone,
would give

$\delta^{\prime\prime\prime}x=\theta y\cos
l,\;\;\delta^{\prime\prime\prime}y=-\theta x\cos
l,\;\;\delta^{\prime\prime\prime}z=0.$ (148)

The sum of these composite displacements relative to each axis will therefore
give for the whole displacement, effected definitively by the rotation
$\theta$ about the axis $(g,h,l)$, the following expression:

$\delta\vec{r}=\theta\vec{r}\times\hat{t}.$ (149)

This is the sum, in modern notation, of
$\delta^{\prime}\vec{r}+\delta^{\prime\prime}\vec{r}+\delta^{\prime\prime\prime}\vec{r}$
as given by the three previous equations, where $\hat{t}=(\cos g,\cos h,\cos
l)$. It appears that the author is continuing to make all displacements
infinitesimal as in the previous section.

##  24 Successive infinitesimal displacements

Let us return to the composition of arbitrary displacements, or changes, given
successively for the same system. The formula for a single such displacement
can be written, in conformity with paragraph 15, as

$\displaystyle\delta x=\alpha+\theta(y\cos l-z\cos h)=t\cos g+\theta u\cos G$
$\displaystyle\delta y=\beta+\theta(z\cos g-x\cos l)=t\cos h+\theta u\cos H$
$\displaystyle\delta z=\gamma+\theta(x\cos h-y\cos g)=t\cos l+\theta u\cos L.$
(150)

These equations, in which $u$ denotes the distance from the point $(x,y,z)$ to
the central axis of the displacement, and $(G,H,L)$ are the angles formed with
the coordinate axes by the direction of the infinitely small arc $u\theta$,
describe a rotation about that central axis.

In modern form the above array becomes

$\delta\vec{r}=\vec{\delta}+\theta\vec{r}\times\hat{t}=t\hat{t}-\theta\vec{u}\times\hat{t}.$
(151)

We are dealing again with a substitute axis having the same direction
$\hat{t}$ as the central axis, but displaced so as to pass through the origin
$\vec{r}=0$. The rotation about the central axis is equivalent to the same
rotation about the substitute axis, augmented by a translation $\vec{\delta}$
that is common to all points $\vec{r}$. In the case $\vec{r}=0$ the substitute
rotation has no effect and $\vec{\delta}$ is the whole displacement. For
general $\vec{r}$ one has also the substitute rotation
$\theta\vec{r}\times\hat{t}$.

The expression after the second = sign is obtained by decomposing
$\vec{\delta}$ into a part parallel and a part perpendicular to the central
axis. The parallel part $\vec{\delta}\cdot\hat{t}\hat{t}=t\hat{t}$ is the
translation of the central axis along itself, which in general accompanies the
rotation about the central axis in accordance with the “screw” principle. The
perpendicular part describes the rotation of the origin of coordinates about
the central axis; since $\vec{u}$ is the perpendicular from the point
$\vec{r}$ to the central axis, $\vec{r}+\vec{u}$ is the perpendicular from the
origin to the central axis. Therefore the perpendicular part of $\vec{\delta}$
is $-\theta(\vec{r}+\vec{u})\times\hat{t}$. Combining the two parts, we obtain
$\vec{\delta}=t\hat{t}-\theta\vec{r}\times\hat{t}-\theta\vec{u}\times\hat{t}$
or

$\vec{\delta}+\theta\vec{r}\times\hat{t}=t\hat{t}-\theta\vec{u}\times\hat{t}$
(152)

in keeping with (151).

In general, the displacement relative to an arbitrary direction $s$, making
angles $a,b,c$ with the $x,y,z$ coordinates, is given by

$\delta s=t\cos(t,s)+\theta u\cos(tu,s)$ (153)

where $\cos(t,s)$ , $\cos(tu,s)$ are the cosines of the angles made by this
direction with the central axis and with the infinitely small arc of rotation
$\theta$. [[If we put this equation in modern form,

$\hat{s}\cdot\delta\vec{r}=t\hat{s}\cdot\hat{t}+\theta
u\hat{s}\cdot(-\hat{u}\times\hat{t}),$ (154)

we see that it results from taking the dot product of (151) with $\hat{s}$. We
can also recover the three equations (24) by replacing $s$ with $x$, $y$, or
$z$.

But if two lines are given in the space, one knows that the (shortest)
distance from a (particular) point of one line to the other line is reciprocal
to the sine of the angle formed by the first line with the plane containing
the second line and the point (of the first line) under consideration. Or
equivalently, the product of this sine and this distance is constantly equal
to the product of the shortest distance between the two lines and the sine of
their inclination (that is, of the angle between their directions). If,
therefore, we define $D$ as the distance between the central axis and the line
passing through the point of the system under consideration (that is, the
point $\vec{r}$) in the direction $s$, and $\nu$ as the angle between this
line and the central axis, we shall have

$u\cos(tu,s)=D\sin\nu,\;\;\;\delta s=t\cos\nu+D\theta\sin\nu.$ (155)

Here the author liberates himself from the given point $\vec{r}$ and refers in
his description to whatever point of the $s$-line through $\vec{r}$ is closest
to the central axis. This completes his analysis of a single displacement.

If we now consider the successive (infinitesimal) displacements of the system
about central axes whose elements are $t,\theta,g,h,l$;
$t^{\prime},\theta^{\prime},g^{\prime},h^{\prime},l^{\prime}$;
$t^{\prime\prime},\theta^{\prime\prime},g^{\prime\prime},h^{\prime\prime},l^{\prime\prime}$;
etc., and denote by $\delta S$ the resultant displacement of a point of the
system relative to that same fixed direction $s$ and by $T,\Theta,G,H,L$ the
elements of the resultant central axis, we shall have

$\delta S=\Sigma t\cos\nu+\Sigma D\theta\sin\nu=T\cos V+\Theta{\cal D}\sin V,$
(156)

${\cal D}$ being the distance from the resultant axis to the line $s$ drawn
through the point $(x,y,z)$.

This equation, owing to the indeterminate parameters $a,b,c$ implicit in it
and to the fact that it must hold for all points of the system, is equivalent
to the following six equations which give the position of the resultant
central axis and the resultant translation and rotation:

$\displaystyle\Sigma\alpha-T\cos G+\Theta Y\cos L-\Theta Z\cos H=0,$
$\displaystyle\Sigma\beta-T\cos H+\Theta Z\cos L-\Theta X\cos L=0,$
$\displaystyle\Sigma\gamma-T\cos L+\Theta X\cos L-\Theta Y\cos G=0,$
$\displaystyle\Theta\cos G=\Sigma\theta\cos g,\Theta\cos H=\Sigma\theta\cos
h,\Theta\cos L=\Sigma\theta\cos l,$ (157)

where $X,Y,Z$ are the coordinates of an arbitrary point on the resultant
central axis.

Without loss of rigor, one could consider only pure rotations in this
analysis, since the translations $t,t^{\prime},...$ can always be represented
by couples of rotations; and simply put

$\delta S=T\cos V+\Theta{\cal D}\sin V=\Sigma\theta D\sin\nu.$ (158)

Here I omit some commentary that seems to me both tedious and repetitious.

##  25 Conditions for equilibrium from many infinitely small successive
displacements.

We are now led to seek out what conditions need to be satisfied by the
elements of the successive displacements proposed for a solid system, in order
that the system, passing successively through various infinitesimally
neighboring situations, should return to its initial position; this would
amount to a condition of equilibrium, or neutralization, on the totality of
the successive displacements. Now, it is evident that all the relevant
conditions are contained in a single equation, which can be decomposed into
six others on account of the indeterminate quantities implicit in it, to wit:

$\delta S=0,$ (159)

since this equation expresses that each point of the system has returned to
its initial position.

The six equations hidden in $\delta S=0$ are

$\displaystyle\delta x_{0}$ $\displaystyle=$ $\displaystyle\Sigma\alpha=0,$
$\displaystyle\delta y_{0}$ $\displaystyle=$ $\displaystyle\Sigma\beta=0,$
$\displaystyle\delta z_{0}$ $\displaystyle=$ $\displaystyle\Sigma\gamma=0,$
$\displaystyle\Sigma\theta\cos g=0,\;\;\Sigma\theta\cos h$ $\displaystyle=$
$\displaystyle 0,\;\;\Sigma\theta\cos l=0,$ (160)

where $\delta x_{0},\delta y_{0},\delta z_{0}$ stand for the resultant
variations of the coordinates of the origin. The first three equations express
the immobility of the origin of the coordinates; and the the three others,
that no resultant rotation has occurred in the displaced system (i.e. no
rotation about a resultant fixed axis through the origin, such as Euler’s
theorem permits). This double condition excludes the possibility of any
resultant displacement whatever.

The double condition is an immediate consequence of the relations (158)
and(159), which cannot be satisfied for every point in the system unless

$T=0,\Theta=0.$ (161)

For $\Theta=0$ implies the three last equations of(25), while the first three
follow from (21) of Section 15.

These six equations of equilibrium are analytically contained in a single
equation which expresses the _general_ law of this equilibrium in the simplest
way (in terms of the elements of the successive displacements), namely

$\Sigma\theta D\sin\nu=0.$ (162)

But the equilibrium of these infinitesimal successive displacements, all else
remaining the same, will continue to hold no matter how rapidly they succeed
one another. Passing to the limit, one arrives at the identity of the laws of
equilibrium due to successive infinitesimal displacements with those due to
simultaneous infinitesimal displacements. This equivalence is needed in order
to justify the analogy between geometry and mechanics, to be presented in the
following section.

##  26 Analogy of these (geometric) laws of composition and equilibrium with
those of composition and equlibrium of forces applied to an immovable system.

The analogy between this general (geometric) law and that of the equilibrium
of forces applied to an immovable system is striking. Let the applied forces
follow the axes of rotation and suppose them proportional to those rotations;
then the moment of a force on the system is exactly proportional to that of
the corresponding rotation, and each translation is replaced by a couple (in
the sense of earlier sections) of applied forces following the axes of the
couple of rotations equivalent to the translation. The analogy, however,
extends to the laws of composition and may be stated thus:

A system of successive displacements being given to be composed into a
resultant displacement, and at the same time a system of forces proportional
to the successive rotations given for each displacement and applied following
the same axes as the rotations, the translations of the successive
displacements, supposing that they are not implicitly included in the
rotations by being represented by couples of rotations, being then represented
in the system of forces under consideration by couples of forces whose moments
would be equal to those of the translations, relatively to the three
coordinate axes, the system of displacements will amount to a resultant
displacement composed of a rotation and an absolute translation relative to
the central axis of rotation; just as the system of forces will resolve itself
by the successive composition of its elements into a single force and a single
couple of forces situated in a plane normal to the resultant force. This
resultant force will be applied at the central axis of the resultant
displacement, which will be at the same time the central axis of the static
system; it will be proportional to the resultant rotation, and the moment of
the couple normal to this force will be proportional to the absolute
translation of the system that operates in a way parallel to the central axis.
Should the axes of the rotations to be combined be all parallel and pass
through determined points, the resultant central axis is also parallel to them
and passes through a certain point that corresponds to the center of the
parallel forces, the same point no matter what be the direction of the axes of
rotation, and which is nothing other than the center of gravity of the points
on the composing axes of rotation that are determined when all the rotations
are equal.

This passage, as translated faithfully above, is somewhat in need of a
retranslation or decipherment. The basic key to the passage is that the phrase
“moment d’une force suivant un axe” means, at least in modern French, the
torque $\vec{r}\times\vec{F}$ about that axis, supposing that the force
$\vec{F}$ is applied to or through a point whose position vector relative to
some point on the axis is $\vec{r}$. Thus, in general, the force is not
applied along the axis, nor even toward some point on the axis; if it were,
the “moment” would be zero. Of course the words “torque” and “moment” were not
as cleanly defined in the author’s time as now, but this identification does
clarify the intention of the passage.

With this key in hand, we can appreciate what otherwise would be perplexing:
that a single force is repeatedly associated with a geometric rotation, while
a geometric translation corresponds to the moment of a pair or couple of
forces acting in a plane normal to the geometric translation. As to the
geometric meaning of the word “moment”, it may be helpful to refer to the
discussion following (21) in 15. Of course the modern form of a cross-product
using the right-hand rule was not available; instead the author measures the
“moment” of a rotation by an area, an idea introduced independently by H.
Grassmann almost at the same time.

In fact, the equations of the resultant central axis of the composition of the
fixed axes of rotation reduce in this case to

$\vec{W}\times\hat{T}+\frac{\Sigma\vec{\delta}}{\Sigma\vec{\theta}}=0,$ (163)

and we have

$\displaystyle T$ $\displaystyle=$ $\displaystyle 0,\;\;\Theta=\Sigma\theta$
$\displaystyle=\Sigma\alpha$ $\displaystyle=$ $\displaystyle\Sigma\theta(Z\cos
h-Y\cos h),$ $\displaystyle=\Sigma\beta$ $\displaystyle=$
$\displaystyle\Sigma\theta(X\cos h-Z\cos h),$ $\displaystyle=\Sigma\gamma$
$\displaystyle=$ $\displaystyle\Sigma\theta(Y\cos h-X\cos h)$ (164)

where $X,Y,Z$ denote the coordinates of the axis of rotation $\theta$. Hence
the resultant axis passes through the point whose coordinates are

$x=\frac{\Sigma\theta X}{\Sigma\theta},\;\;y=\frac{\Sigma\theta
Y}{\Sigma\theta},\;\;z=\frac{\Sigma\theta Z}{\Sigma\theta}.$ (165)

##  27 Determination of the changes in coordinates of a solid due to an
arbitrary displacement, deduced analytically from the conditions of
invariability of the system.

We consider the displacement of the coordinate axes, given that they are
rigidly attached to the system as it is displaced. This leads immediately to
the algebraic expression for the changes in the coordinates of any point, and
there remains only to reduce to a minimum the number of arbitrary constants
that enter the calculation, as we shall now see.

The author now proposes to derive the preceding results without using the
formulas derived from (49), but by a different method.

Let us denote by
$a,b,c;\;a^{\prime},b^{\prime},c^{\prime};\;a^{\prime\prime},b^{\prime\prime},c^{\prime\prime}$
the cosines of the angles formed by the displaced coordinate axes with their
original directions. Then the new coordinates $x+\Delta x,y+\Delta y,z+\Delta
z$, of a point of the system after displacement relative to the “old” axes,
will be expressed as a function of the same displaced coordinates relative to
the new axes. But this will be the same function, already well known, that
gives the new axes in terms of the old, that is

$\displaystyle x+\Delta x=\alpha+ax+by+cz,$ $\displaystyle y+\Delta
y=\beta+a^{\prime}x+b^{\prime}y+c^{\prime}z,$ $\displaystyle z+\Delta
z=\gamma+a^{\prime\prime}x+b^{\prime\prime}y+c^{\prime\prime}z.$ (166)

The idea is that since the axes are rigidly attached to the system, they
suffer the same displacement from old to new as does the point under
consideration. So if one compares the final position of the point to its old
position, both relative to the old axes, one finds the same transformation
formulas as in comparing the coordinates of the same new point, relative to
the old axes, with its coordinates relative to the new ones.

These formulas express $x+\Delta x,y+\Delta y,z+\Delta z$ as linear functions
of the initial coordinates $x,y,z$ for any displacement of the solid whatever.
They contain 12 arbitrary coefficients, but really only 6 of these are
independent. The first three, $\alpha,\beta,\gamma$, determine the movement of
the origin, and the three “diagonal” coefficients
$a,b^{\prime},c^{\prime\prime}$ serve to define the directions of the three
displaced axes relative to the old ones. In terms of the latter, one may
eliminate the six off-diagonal elements by the formulas of Monge.

But the reduction of twelve constants to six can be achieved even more simply,
without using the formulas of Monge (complicated by radicals) and by a route
that leads to the simplest possible expressions for $\Delta x,\Delta y,\Delta
z$ in terms of $x,y,z$. We introduce into (27) the coordinates
$\xi,\eta,\zeta$ of the midpoint of the line joining the initial to the final
position of the point in question. This gives us

$x=\xi-(1/2)\Delta x,\;\;y=\eta-(1/2)\Delta y,\;\;z=\zeta-(1/2)\Delta z.$
(167)

Then we unite these three equations into one by multiplying them respectively
by three indeterminate factors $\mu,\nu,\pi$ and adding together the results.
We now have the single equation

$\displaystyle(1/2)\Delta x(\bar{a}+\mu)$ $\displaystyle+$
$\displaystyle(1/2)\Delta y(\bar{b}+\nu)+(1/2)\Delta
z(\bar{c}+\pi)=(\alpha\mu+\beta\nu+\gamma\pi)$ (168) $\displaystyle+$
$\displaystyle\xi(\bar{a}-\mu)+\eta(\bar{b}-\nu)+\zeta(\bar{c}-\pi),$

where we define

$\bar{a}=a\mu+a^{\prime}\nu+a^{\prime\prime}\pi,\;\bar{b}=b\mu+b^{\prime}\nu+b^{\prime\prime}\pi,\;\bar{c}=c\mu+c^{\prime}\nu+c^{\prime\prime}\pi.$
(169)

To determine (for example) $\Delta x$, we must assign to $\mu,\nu,\pi$ those
values that cause the coefficients of $\Delta y$ and $\Delta z$ to vanish;
that is, we must set

$\bar{b}+\nu=\bar{c}+\pi=0,$ (170)

from which we find

$\mu=(1+b^{\prime})(1+c^{\prime\prime})-c^{\prime}b^{\prime\prime},\nu=b^{\prime\prime}c-b(1+c^{\prime\prime}),\pi=bc^{\prime}-c(1+b^{\prime}).$
(171)

(French text has incorrectly $a^{\prime\prime}$ instead of $b^{\prime\prime}c$
in $\nu$).

Now, the nine cosines
$\vec{a}=(a,a^{\prime},a^{\prime\prime}),\vec{b}=(b,b^{\prime},b^{\prime\prime}),\vec{c}=(c,c^{\prime},c^{\prime\prime})$
satisfy, as is well known, the relations

$\displaystyle\vec{a}=\vec{c}\times\vec{b},$
$\displaystyle\vec{b}=\vec{a}\times\vec{c},$
$\displaystyle\vec{c}=\vec{b}\times\vec{a}.$ (172)

These can be deduced from six others,

$\displaystyle|\vec{a}|^{2}=|\vec{b}|^{2}=|\vec{c}|^{2}=1,$
$\displaystyle\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=0,$
(173)

which say that the coordinate axes, both the old and the new, form orthonormal
systems. The ambiguity of signs that enter into the deduction is resolved by
another condition altogether necessary in considering the displacement of a
solid system, namely that the system of new axes arising from the displacement
must always remain in the condition of superposition with the old axes that is
possible for each respective axis and its correspondent.

I have adhered to my usual convention of reversing cross-products so as to
maintain right-handedness (see 15). I hope that by doing so I satisfy the
author’s condition above.

This being understood, it is clear that the above relations lead to

$\mu=1+a+b^{\prime}+c^{\prime\prime},\nu=a^{\prime}-b,\pi=a^{\prime\prime}-c,$
(174)

from which

$\bar{a}=1+a+b^{\prime}+c^{\prime\prime}=\mu$ (175)

and hence

$\Delta
x-\alpha=\frac{2[(\eta-(1/2)\beta)(b-a^{\prime})-(\zeta-(1/2)\gamma)(a^{\prime\prime}-c)]}{1+a+b^{\prime}+c^{\prime\prime}}.$
(176)

In eqs (170), (171),(174),(175), and (176) , the author has broken the
threefold symmetry of (168) by singling out $\Delta x$ as the component to be
determined. He now retreats from that choice by considering in turn each of
the other components $\Delta y,\Delta z$.

Similarly one can obtain

$\Delta
y-\beta=\frac{2[(\zeta-(1/2)\gamma)(c^{\prime}-b^{\prime\prime})-(\xi-(1/2)\alpha)(b-a^{\prime})]}{1+a+b^{\prime}+c^{\prime\prime}}.$
(177)

$\Delta
z-\gamma=\frac{2[(\xi-(1/2)\alpha)(a^{\prime\prime}-c)-(\eta-(1/2)\beta)(c^{\prime}-b^{\prime\prime})]}{1+a+b^{\prime}+c^{\prime\prime}}.$
(178)

These formulas are identical to those of 15], provided that we set

$m=\frac{2(c^{\prime}-b^{\prime\prime})}{1+a+b^{\prime}+c^{\prime\prime}},\;\;n=\frac{2(a^{\prime\prime}-c)}{1+a+b^{\prime}+c^{\prime\prime}},\;\;p=\frac{2(b-a^{\prime})}{1+a+b^{\prime}+c^{\prime\prime}}.$
(179)

##  28 Infinitesimal version of 27.

In the case of infinitesimal displacements, we neglect angular displacements
of second order and directly obtain $a=b^{\prime}=c^{\prime\prime}=1$, from
which

$\delta x=\alpha+by+cz,\;\;\delta y=\beta+c^{\prime}z+a^{\prime}x,\;\;\delta
z=\gamma+a^{\prime\prime}x+b^{\prime\prime}y.$ (180)

Also, the distance from any point to the origin, when the latter is drawn
along with the displacement of the system, is invariable, so that
(vectorially) $\vec{r}\cdot(\delta\vec{r}-\vec{\delta})=0$ for any arbitrary
point $\vec{r}$. This necessitates the following relations among the “off-
diagonal” first-order infinitesimal cosines:

$b+a^{\prime}=c+a^{\prime\prime}=c^{\prime}+b^{\prime\prime}=0.$ (181)

Hence by setting
$m=c^{\prime}=-b^{\prime\prime},n=a^{\prime\prime}=-c,p=b=-a^{\prime}$, we
have finally

$\delta x=\alpha+py-nz,\;\;\delta y=\beta+mz-px,\;\;\ \delta z=\gamma+nx-my.$
(182)

##  29 Algebraic deduction of coordinate changes.

And now, yet another method.

But it is interesting to arrive at these same formulas for finite or
infinitesimal coordinate changes by a purely algebraic route, independent of
any geometric consideration (other than the Pythagorean formula for the
distance between two points) starting from the invariability of the distances
between points of the solid.

So let

$\vec{r}_{0},\vec{r}_{1},\vec{r}_{2},\vec{r}$ (183)

be the positions of four points invariably linked together and belonging to
the solid system. The first three points and their displacements are to be
considered as known, and the displacement of the fourth point is to be
calculated as a function of its starting position and the known quantities.

The distances between these four points will remain constant under an
arbitrary displacement of the system; this condition, when expressed
algebraically, will give the following six equations:

$\displaystyle|\vec{r}_{1}+\Delta\vec{r}_{1}-\vec{r}_{0}-\Delta\vec{r}_{0}|^{2}=|\vec{r}_{1}-\vec{r}_{0}|^{2}$
$\displaystyle|\vec{r}_{2}+\Delta\vec{r}_{2}-\vec{r}_{0}-\Delta\vec{r}_{0}|^{2}=|\vec{r}_{2}-\vec{r}_{0}|^{2}$
$\displaystyle|\vec{r}_{2}+\Delta\vec{r}_{2}-\vec{r}_{1}-\Delta\vec{r}_{1}|^{2}=|\vec{r}_{2}-\vec{r}_{1}|^{2}$
$\displaystyle|\vec{r}+\Delta\vec{r}-\vec{r}_{0}-\Delta\vec{r}_{0}|^{2}=|\vec{r}-\vec{r}_{0}|^{2}$
$\displaystyle|\vec{r}+\Delta\vec{r}-\vec{r}_{1}-\Delta\vec{r}_{1}|^{2}=|\vec{r}-\vec{r}_{1}|^{2}$
$\displaystyle|\vec{r}+\Delta\vec{r}-\vec{r}_{2}-\Delta\vec{r}_{2}|^{2}=|\vec{r}-\vec{r}_{2})|^{2}.$
(184)

I am rendering equations when possible in vector form for brevity and ease of
reading. The Pythagorean formula for distance is implicit in the vectors; thus
$|\vec{r}-\vec{r}_{0}|^{2}=(x-x_{0})^{2}+(y-y_{0})^{2}+(z-z_{0})^{2}$, etc.

We may think of a tetrahedron with vertex at $\vec{r}$ and base having the
three points $\vec{r}_{0},\vec{r}_{1},\vec{r}_{2}.$ Then the first three
equations of (29) pertain to the edges of the base triangle, and the last
three to the edges that meet at the vertex.

The six equations are quadratic in the displacements
$\Delta\vec{r_{0}}$,$\Delta\vec{r}_{1}$,$\Delta\vec{r}_{2}$,$\Delta\vec{r}$
but can be made linear by introducing the midpoint position
$\vec{\omega}=\vec{r}+(1/2)\Delta\vec{r}$, and likewise for the three base
points. One then obtains

$\displaystyle
2(\vec{\omega}-\vec{\omega}_{0})\cdot(\Delta\vec{r}-\Delta\vec{r}_{0})=0,$
$\displaystyle
2(\vec{\omega}_{1}-\vec{\omega}_{0})\cdot(\Delta\vec{r}_{1}-\Delta\vec{r}_{0})=0,$
$\displaystyle
2(\vec{\omega}_{2}-\vec{\omega}_{0})\cdot(\Delta\vec{r}_{2}-\Delta\vec{r}_{0})=0,$
$\displaystyle(\vec{\omega}-\vec{\omega}_{0})\cdot(\Delta\vec{r}_{1}-\Delta\vec{r}_{0})+(\Delta\vec{r}-\Delta\vec{r}_{0})\cdot(\vec{\omega}_{1}-\vec{\omega}_{0})=0,$
$\displaystyle(\vec{\omega}-\vec{\omega}_{0})\cdot(\Delta\vec{r}_{2}-\Delta\vec{r}_{0})+(\Delta\vec{r}-\Delta\vec{r}_{0})\cdot(\vec{\omega}_{2}-\vec{\omega}_{0})=0,$
$\displaystyle(\vec{\omega}_{1}-\vec{\omega}_{0})\cdot(\Delta\vec{r}_{2}-\Delta\vec{r}_{0})+(\Delta\vec{r}_{1}-\Delta\vec{r}_{0})\cdot(\vec{\omega}_{2}-\vec{\omega}_{0})=0.$
(185)

These equations may be derived as follows. We note that each linear factor in
(29) contains either $\vec{\omega}_{0}$ or $\Delta\vec{r}_{0}$. Let
$(\vec{R},\vec{\Omega})$ and $(\vec{R}^{\prime},\vec{\Omega}^{\prime})$ be any
of the pairs $(\vec{r},\vec{\omega})$, $(\vec{r}_{1},\vec{\omega}_{1})$,
$(\vec{r}_{2},\vec{\omega}_{2})$. From the definition
$\vec{\Omega}=\vec{R}+(1/2)\Delta\vec{R}$ we have

$\vec{R}=\vec{\Omega}-(1/2)\Delta\vec{R},\;\;\vec{R}+\Delta\vec{R}=\vec{\Omega}+(1/2)\Delta\vec{R},$
(186)

and likewise with $\vec{R}^{\prime},\vec{\Omega}^{\prime}$ in place of
$\vec{R},\vec{\Omega}$. Therefore

$\displaystyle[(\vec{R}+\Delta\vec{R})-(\vec{r}_{0}+\Delta\vec{r}_{0})]\cdot[(\vec{R}^{\prime}+\Delta\vec{R}^{\prime})-(\vec{r}_{0}+\Delta\vec{r}_{0})]-(\vec{R}-\vec{r}_{0})\cdot(\vec{R}^{\prime}-\vec{r}_{0})$
(187) $\displaystyle=$
$\displaystyle[(\vec{\Omega}+(1/2)\Delta\vec{R})-(\vec{\omega}_{0}+(1/2)\Delta\vec{r}_{0})]\cdot[(\vec{\Omega}^{\prime}+(1/2)\Delta\vec{R}^{\prime})-(\vec{\omega}_{0}+(1/2)\Delta\vec{r}_{0})]$
$\displaystyle-$
$\displaystyle[(\vec{\Omega}-(1/2)\Delta\vec{R})-(\vec{\omega}_{0}-(1/2)\Delta\vec{r}_{0})]\cdot[(\vec{\Omega}^{\prime}-(1/2)\Delta\vec{R}^{\prime})-(\vec{\omega}_{0}-(1/2)\Delta\vec{r}_{0})]$
$\displaystyle=$
$\displaystyle(\vec{\Omega}-\vec{\omega}_{0})\cdot(\Delta\vec{R}^{\prime}-\Delta\vec{r}_{0})+(\Delta\vec{R}-\Delta\vec{r}_{0})\cdot(\vec{\Omega}^{\prime}-\vec{\omega}_{0})$

and since the top line of (187) vanishes by (29), the bottom line vanishes,
giving us the first three lines of (29) if $(\vec{R},\vec{\Omega})$ and
$(\vec{R}^{\prime},\vec{\Omega}^{\prime})$ are the same, and the last three if
they are different.

It will be observed that the author has selected the pair
$(\vec{r}_{0},\vec{\omega}_{0})$ to play the starring rôle in (29), inasmuch
as it is the only pair that is represented in every linear factor in every
equation. Other equations could be written down, but they can be deduced from
the six displayed. For example, the equation
$(\vec{\omega}_{0}-\vec{\omega})\cdot(\Delta\vec{r}_{1}-\Delta\vec{r})+(\Delta\vec{r}_{0}-\Delta\vec{r})\cdot(\vec{\omega}_{1}-\vec{\omega})=0$,
obtained by interchanging $\vec{r},\vec{\omega}$ with
$\vec{r}_{0},\vec{\omega}_{0}$ in the fourth line of (29), can be deduced by
subtracting that line from the top line.

We now multiply these six equations by the respective six factors
$\mu^{2},\nu^{2},\pi^{2},\mu\nu,\mu\pi,\nu\pi$ and add them together. This
results in the following single equation which (when $\mu,\nu,\pi$ vary
independently) contains the foregoing six:

$[\mu(\vec{\omega}-\vec{\omega}_{0})+\nu(\vec{\omega_{1}}-\vec{\omega}_{0})+\pi(\vec{\omega_{2}}-\vec{\omega}_{0})]\cdot[\mu(\Delta\vec{r}-\Delta\vec{r}_{0})+\nu(\Delta\vec{r}_{1}-\Delta\vec{r}_{0})+\pi(\Delta\vec{r}_{2}-\Delta\vec{r}_{0})]=0.$
(188)

When a similar operation was performed in 27, the coefficients $\mu,\nu,\pi$
were attached to the $x,y,z$-components and the resulting manipulations could
not be written vectorially. Here, the three coefficients are attached to three
different vectors.

By a suitable choice of $\mu,\nu,\pi$, this equation can be made to contain
only $\Delta\vec{r}$ [[and not $\Delta\vec{r_{1}}$ or $\Delta\vec{r_{2}}$]]
and the value of this displacement can be found in the simplest way.

Having thus far allowed us the luxury of our vector notation, the author is
now going to forbid it by singling out one component $x$ for study. What makes
this necessary is that he wishes to use the principle that the vanishing of a
product of two scalar quantities implies the vanishing of at least one of the
factors, and this is not true of the dot product of two vectors.

To wit, suppose we set

$\displaystyle\mu(\eta-\eta_{0})+\nu(\eta_{1}-\eta_{0})+\pi(\eta_{2}-\eta_{0})=0,$
$\displaystyle\mu(\zeta-\zeta_{0})+\nu(\zeta_{1}-\zeta_{0})+\pi(\zeta_{2}-\zeta_{0})=0:$
(189)

then (188) reduces to

$[\mu(\xi-\xi_{0})+\nu(\xi_{1}-\xi_{0})+\pi(\xi_{2}-\xi_{0})][\mu(\Delta
x-\Delta x_{0})+\nu(\Delta x_{1}-\Delta x_{0})+\pi(\Delta x_{2}-\Delta
x_{0})]=0.$ (190)

He has arranged $\mu,\nu,\pi$ to make the first factor of the $y$ and $z$
parts of (188) vanish, and so the corresponding products vanish; therefore the
product in the $x$ part must also vanish. The principle of factorization then
dictates that at least one of the $x$ factors vanish, but he claims (deferring
proof to the following section) that the first factor cannot vanish if the
second does not, and so infers that the second factor vanishes:

Now, the first factor of (190) cannot be zero unless the second is also, as we
shall demonstrate hereafter; we therefore have the simultaneous equations

$\displaystyle\mu(\Delta x-\Delta x_{0})+\nu(\Delta x_{1}-\Delta
x_{0})+\pi(\Delta x_{2}-\Delta x_{0})=0,$
$\displaystyle\mu(\eta-\eta_{0})+\nu(\eta_{1}-\eta_{0})+\pi(\eta_{2}-\eta_{0})=0,$
$\displaystyle\mu(\zeta-\zeta_{0})+\nu(\zeta_{1}-\zeta_{0})+\pi(\zeta_{2}-\zeta_{0})=0.$
(191)

These equations evidently imply the following ones:

$\displaystyle\Delta x_{1}-\Delta
x_{0}=p(\eta_{1}-\eta_{0})-n(\zeta_{1}-\zeta_{0}),$ $\displaystyle\Delta
x_{2}-\Delta x_{0}=p(\eta_{2}-\eta_{0})-n(\zeta_{2}-\zeta_{0}),$
$\displaystyle\Delta x-\Delta x_{0}=p(\eta-\eta_{0})-n(\zeta-\zeta_{0}),$
(192)

(French text has mistakenly $p(\xi_{1}-\xi_{0})$ in third equation) where $n$
and $p$ are two constants bound to the displacements of the three first points
($\vec{r}_{1},\vec{r}_{2},\vec{r}_{0}$, the base of the tetrahedron) by the
first two of these three equations.

The three simultaneous equations (29) are homogeneous in $\mu,\nu,\pi$;
therefore the determinant must vanish. Interchanging rows and columns, we
deduce that the three homogeneous equations

$\displaystyle(\Delta x-\Delta
x_{0})\bar{m}+(\eta-\eta_{0})\bar{n}+(\zeta-\zeta_{0})\bar{p}=0,$
$\displaystyle(\Delta x_{1}-\Delta
x_{0})\bar{m}+(\eta_{1}-\eta_{0})\bar{n}+(\zeta-\zeta_{0})\bar{p}=0,$
$\displaystyle(\Delta x_{2}-\Delta
x_{0})\bar{m}+(\eta_{2}-\eta_{0})\bar{n}+(\zeta_{2}-\zeta_{0})\bar{p}=0$ (193)

have a solution $\bar{m},\bar{n},\bar{p}$. Defining $n=\bar{p}/\bar{m}$,
$p=-\bar{n}/\bar{m}$, we obtain (29).

The same analysis (if $y$ or $z$ instead of $x$ had been singled out) would
give

$\displaystyle\Delta y-\Delta
y_{0}=m^{\prime}(\zeta-\zeta_{0})-p^{\prime}(\xi-\xi_{0}),$
$\displaystyle\Delta z-\Delta
z_{0}=n^{\prime\prime}(\xi-\xi_{0})-m^{\prime\prime}(\eta-\eta_{0}).$ (194)

[[French text has $n^{\prime},m^{\prime}$ in third line]] But we have also
(from the top line of (29))

$(\xi-\xi_{0})(\Delta x-\Delta x_{0})+(\eta-\eta_{0})(\Delta y-\Delta
y_{0})+(\zeta-\zeta_{0})(\Delta z-\Delta z_{0})=0;$ (195)

it follows that $m=m^{\prime}=m^{\prime\prime}$,
$n=n^{\prime}=n^{\prime\prime}$, $p=p^{\prime}=p^{\prime\prime}$, and so we
have at last

$\displaystyle\Delta x-\Delta x_{0}=A+p\eta-n\zeta,$ $\displaystyle\Delta
y-\Delta y_{0}=B+m\zeta-p\xi,$ $\displaystyle\Delta z-\Delta
z_{0}=C+n\xi-m\eta,$ (196)

where the six constants $A,B,C,m,n,p$ are functions of the displacements
$\Delta\vec{r}_{0},\Delta\vec{r}_{1},\Delta\vec{r}_{2}$.

Substituting (29) and the third line of (29) into (195), we find for example
that the product $(\xi-\xi_{0})(\eta-\eta_{0})$ appears with coefficient
$p-p^{\prime}$ so that $p^{\prime}=p$. This is one of six equations that
together justify dropping all the primes in (29).

##  30 Proof of prior claim.

But it remains to prove what we have claimed, that in (190) the first factor
$[\mu(\xi-\xi_{0})+\nu(\xi_{1}-\xi_{0})+\pi(\xi_{2}-\xi_{0})]$ cannot vanish
unless the second factor $[\mu(\Delta x-\Delta x_{0})+\nu(\Delta x_{1}-\Delta
x_{0})+\pi(\Delta x_{2}-\Delta x_{0})]$ does so as well.

The three simultaneous equations (they are easily expressed as one, in vector
notation)

$\mu(\vec{\omega}-\vec{\omega}_{0})+\nu(\vec{\omega}_{1}-\vec{\omega}_{0})+\pi(\vec{\omega}_{2}-\vec{\omega}_{0})=0$
(197)

express that the midpoints of the lines traversed by the four points we are
considering lie in the same plane. But this unusual condition can be fulfilled
only if either, on the one hand, the four points themselves also lie in one
plane, or on the other hand, the pyramid formed by these points after
displacement is not genuinely superposable on the one formed initially, but is
only symmetric to it. The second hypothesis is not admissible in our problem,
but the first must be examined.

Consider the phrase “midpoint plane” as referring to the plane containing the
four midpoints $\vec{\omega}$, etc. If we examine the four points of the
tetrahedron in relation to the midpoint plane, we see that the only way to
make the actual midpoints lie in this plane is to make the displacement take
each vertex of the tetrahedron to the opposite side of the plane. But the
resulting pyramid is not “superposable” on the original one (see 1, second
paragraph) unless all four points already lie in the midpoint plane, so that
the tetrahedron is identical to its mirror image. This exceptional case is
precisely the “first hypothesis” advanced by the author.

We shall now prove that if the four points of the tetrahedron, as well as
those to which they are displaced, all lie in one plane, then both factors of
(190) vanish.

Let $(x,y,z)$, $(x_{1},y_{1},z_{1})$, $(x_{2},y_{2},z_{2})$,
$(x_{0},y_{0},z_{0})$ denote the initial positions of the four points, and
$(x^{\prime},y^{\prime},z^{\prime})$, etc. their positions after displacement.
If the four points are initially in the same plane, there must be coefficients
$g,h,l$ such that

$\displaystyle g(x-x_{0})+h(x_{1}-x_{0})+l(x_{2}-x_{0})=0,$ $\displaystyle
g(y-y_{0})+h(y_{1}-y_{0})+l(y_{2}-y_{0})=0,$ $\displaystyle
g(z-z_{0})+h(z_{1}-z_{0})+l(z_{2}-z_{0})=0.$ (198)

Likewise, if the four points after displacement are in the same plane, there
must be $g^{\prime},h^{\prime},l^{\prime}$ such that

$\displaystyle
g^{\prime}(x^{\prime}-x^{\prime}_{0})+h^{\prime}(x^{\prime}_{1}-x^{\prime}_{0})+l^{\prime}(x^{\prime}_{2}-x^{\prime}_{0})=0,$
$\displaystyle
g^{\prime}(y^{\prime}-y^{\prime}_{0})+h^{\prime}(y^{\prime}_{1}-y^{\prime}_{0})+l^{\prime}(y^{\prime}_{2}-y^{\prime}_{0})=0,$
$\displaystyle
g^{\prime}(z^{\prime}-z^{\prime}_{0})+h^{\prime}(z^{\prime}_{1}-z^{\prime}_{0})+l^{\prime}(z^{\prime}_{2}-z^{\prime}_{0})=0.$
(199)

Now, from the invariability of distances we must have

$\displaystyle|\vec{r}-\vec{r}_{0}|^{2}=|\vec{\,r}^{\prime}-\vec{\,r}^{\prime}_{0}|^{2},$
$\displaystyle|\vec{r}_{1}-\vec{r}_{0}|^{2}=|,\vec{\,r}^{\prime}_{1}-\vec{\,r}^{\prime}_{0}|^{2},$
$\displaystyle|\vec{r}_{2}-\vec{r}_{0}|^{2}=|\vec{\,r}^{\prime}_{2}-\vec{\,r}^{\prime}_{0}|^{2}$
(200)

as well as

$\displaystyle(\vec{r}-\vec{r}_{0})\cdot(\vec{r}_{1}-\vec{r}_{0})=(\vec{\,r}^{\prime}-\vec{r}^{\prime}_{0})\cdot(\vec{\,r}^{\prime}_{1}-\vec{\,r}^{\prime}_{0}),$
$\displaystyle(\vec{r}-\vec{r}_{0})\cdot(\vec{r}_{2}-\vec{r}_{0})=(\vec{\,r}^{\prime}-\vec{r}^{\prime}_{0})\cdot(\vec{\,r}^{\prime}_{2}-\vec{\,r}^{\prime}_{0}),$
$\displaystyle(\vec{r}_{1}-\vec{r}_{0})\cdot(\vec{r}_{2}-\vec{r}_{0})=(\vec{\,r}_{1}^{\prime}-\vec{r}^{\prime}_{0})\cdot(\vec{\,r}^{\prime}_{2}-\vec{\,r}^{\prime}_{0}).$
(201)

But the coefficients $g^{\prime},h^{\prime},l^{\prime}$ satisfy the same
relations (30) to
$\vec{\,r}^{\prime}-\vec{\,r}^{\prime}_{0},\vec{\,r}^{\prime}_{1}-\vec{r}^{\prime}_{0},\vec{\,r}^{\prime}_{2}-\vec{r}^{\prime}_{0}$
as do the coefficients $g,h,l$ (see (30)) to
$\vec{\,r}-\vec{\,r}_{0},\vec{\,r}_{1}-\vec{r}_{0},\vec{\,r}_{2}-\vec{r}_{0}$.
Consequently $g^{\prime},h^{\prime},l^{\prime}$ are proportional to $g,h,l$
and we may as well set the former equal to the latter.

In vector notation, (30) and (30) become

$\displaystyle
g(\vec{r}-\vec{r}_{0})+h(\vec{r}-\vec{r}_{0})+l(\vec{r}-\vec{r}_{0})=0$
$\displaystyle
g(\vec{\,r}^{\prime}-\vec{\,r}^{\prime}_{0})+h(\vec{\,r}^{\prime}_{1}-\vec{\,r}^{\prime}_{0})+l(\vec{\,r}^{\prime}_{2}-\vec{\,r}^{\prime}_{0})=0.]]$
(202)

Putting $(\vec{r},\vec{\,r}^{\prime})$, etc. in terms of
$(\vec{\omega},\Delta\vec{r})$, etc. we have finally

$\displaystyle
g(\vec{\omega}-\vec{\omega}_{0})+h(\vec{\omega}_{1}-\vec{\omega}_{0})+l(\vec{\omega}_{2}-\vec{\omega}_{0})=0,$
$\displaystyle
g(\Delta\vec{r}-\Delta\vec{r}_{0})+h(\Delta\vec{r}_{1}-\Delta\vec{r}_{0})+l(\Delta\vec{r}_{2}-\Delta\vec{r}_{0})=0.$
(203)

Here the upper line, compared with (197), shows that $g,h,l$ are proportional
to $\mu,\nu,\pi$. Therefore the former can be replaced by the latter in the
lower line, yielding

$\mu(\Delta\vec{r}-\Delta\vec{r}_{0})+\nu(\Delta\vec{r}_{1}-\Delta\vec{r}_{0})+\pi(\Delta\vec{r}_{2}-\Delta\vec{r}_{0}),$
(204)

Q. E. D.

##  31 Coordinate changes from infinitesimal analysis.

And a third way!

The two analytic methods that we have just presented for determining the
formulas for the coordinate changes of solid system are based on purely
algebraic procedures. From infinitesimal analysis we can derive an even
simpler proof of these formulas, resembling that given by Lagrange in his
_Analytic Mechanics_ , and describing in the same analysis both the expression
for finite changes and that for infinitesimal ones. Here is that
demonstration.

Given a point $\vec{r}$ in the undisplaced system, consider a neighboring
point $\vec{r}+d\vec{r}$ also before displacement, $d\vec{r}$ being
infinitesimal. Let the changes due to displacment be denoted by a prefix
$\Delta$ or $\delta$ according as the displacement is finite or infinitesimal.
We wish to determine the general expression for $\Delta\vec{r}$ by means of
the equation $\Delta|d\vec{r}|^{2}=0$.

Now, letting $\vec{\omega}=\vec{r}+(1/2)\Delta\vec{r}$, and using the
commutation of the signs $d$ and $\Delta$, the foregoing equation becomes

$d\vec{\omega}\cdot d\Delta\vec{r}=0,$ (205)

which is to be satisfied in the most general way (that is, for arbitrary
choices of $d\vec{r}$).

The proof of (205) is made easier by defining
$\vec{r\,}^{\prime}=\vec{r}+\Delta\vec{r}$. Then

$\Delta|d\vec{r}|^{2}=d\vec{r\,}^{\prime}\cdot d\vec{r^{\prime}}-d\vec{r}\cdot
d\vec{r}=(d\vec{r\,}^{\prime}+d\vec{r})\cdot(d\vec{r\,}^{\prime}-d\vec{r})=2d\vec{\omega}\cdot\Delta\vec{r},$
(206)

which establishes (205).

To this end, let us consider $\Delta\vec{r}$ as a function of $\vec{\omega}$;
then (205) can be written, taking $d\vec{\omega}$ as constant (that is, as
unchanged by the displacement), as $\Delta(d\vec{\omega}\cdot d\vec{r})=0$,
from which follows

$\frac{d\vec{\omega}\cdot\Delta\vec{r}}{|d\vec{\omega}|}=constant$ (207)

(in the same sense of “constant”).

This equation (207) represents algebraically the property of a quadrilateral
two of whose sides are equal, that these two sides, and the two other sides as
well, project equally (that is, each opposed pair makes a pair of equal
projections) on the line joining the midpoints of the second pair of sides.

This geometrical theorem needs some interpretation as well as a proof. The two
sides that are equal (in length) are the initial and final vectors $d\vec{r}$
and $d(\vec{r}+\Delta\vec{r})$, since the two neighboring points are rigidly
connected during the displacement. The other two sides are the lines traversed
by the two neighboring points, which may be finite; these two lines need not
be of equal lengths (although the difference $d|\Delta\vec{r}|$ is
infinitesimal) since different points can move differently under a given
displacement. There is no requirement that the quadrilateral be all in one
plane - this makes visualization even more challenging. I have been unable to
find a geometric proof simpler than the algebraic one already given leading to
(207).

However, it may help to restate the theorem without infinitesimals, but with
vector notation. Let the four vertices of the quadrilateral be called
$\vec{r}_{1},(\vec{r}_{1})^{\prime},\vec{r}_{2},(\vec{r}_{2})^{\prime}$, and
let the opposite lengths $|\vec{r}_{1}-\vec{r}_{2}|$ and
$|(\vec{r}_{1})^{\prime}-(\vec{r}_{2})^{\prime}|$ be equal. Let the second
pair of opposite sides be called
$\vec{\Delta}_{1}=(\vec{r}_{1})^{\prime}-\vec{r}_{1}$ and
$\vec{\Delta}_{2}=(\vec{r}_{2})^{\prime}-\vec{r}_{2}$. Let
$\vec{\omega}_{1},\vec{\omega}_{2}$ be the midpoints of these two sides. Then
the four vertices can be renamed
$\vec{\omega}_{1}-(1/2)\vec{\Delta}_{1},\vec{\omega}_{1}+(1/2)\vec{\Delta}_{1},\vec{\omega}_{2}-(1/2)\vec{\Delta}_{2},\vec{\omega}_{2}+(1/2)\vec{\Delta}_{2}$.
The two equal lengths are now
$|(\vec{\omega}_{1}-\vec{\omega}_{2})-(1/2)(\vec{\Delta}_{1}-\vec{\Delta}_{2})|$
and
$|(\vec{\omega}_{1}-\vec{\omega}_{2})+(1/2)(\vec{\Delta}_{1}-\vec{\Delta}_{2})|$,
from which it follows trivially that $(\vec{\omega}_{1}-\vec{\omega}_{2})$ is
orthogonal to $(\vec{\Delta}_{1}-\vec{\Delta}_{2})|$ as well as to
$\vec{r}_{1}-\vec{r}_{2}$, as stated by the theorem.

Consider now the equation

$d\xi d\Delta x+d\eta d\Delta y+d\zeta d\Delta z=0.$ (208)

(This is eq (205) written out in components.) We shall expand the complete
differentials $d\Delta x$, etc. in terms of partials $\partial$. (That is,
$d\Delta x=d\xi\frac{\partial\Delta x}{\partial\xi}+d\eta\frac{\partial\Delta
x}{\partial\eta}+d\zeta\frac{\partial\Delta x}{\partial\zeta}$., etc.) (208)
thus becomes

$\displaystyle(d\xi)^{2}\frac{\partial\Delta
x}{\partial\xi}+(d\eta)^{2}\frac{\partial\Delta
y}{\partial\eta}+(d\zeta)^{2}\frac{\partial\Delta z}{\partial\zeta}+d\xi
d\eta(\frac{\partial\Delta x}{\partial\eta}+\frac{\partial\Delta
y}{\partial\xi})$ $\displaystyle+d\xi d\zeta(\frac{\partial\Delta
x}{\partial\zeta}+\frac{\partial\Delta z}{\partial\xi})+d\eta
d\zeta(\frac{\partial\Delta y}{\partial\zeta}+\frac{\partial\Delta
z}{\partial\eta})=0.$ (209)

Inasmuch as the differentials $d\xi,d\eta,d\zeta$ are independent, (31)
implies

$\displaystyle\frac{\partial\Delta x}{\partial\xi}=\frac{\partial\Delta
y}{\partial\eta}=\frac{\partial\Delta z}{\partial\zeta}=0,$
$\displaystyle\frac{\partial\Delta x}{\partial\eta}+\frac{\partial\Delta
y}{\partial\xi}=\frac{\partial\Delta x}{\partial\zeta}+\frac{\partial\Delta
z}{\partial\xi}=\frac{\partial\Delta y}{\partial\zeta}+\frac{\partial\Delta
z}{\partial\eta}=0.$ (210)

By setting $d\eta=d\zeta=0$, one establishes the vanishing of
$\frac{\partial\Delta x}{\partial\xi}$, similarly $\frac{\partial\Delta
y}{\partial\eta}$ and $\frac{\partial\Delta z}{\partial\zeta}$. Then by
constraining only $d\zeta$ to vanish, one isolates $\frac{\partial\Delta
x}{\partial\eta}+\frac{\partial\Delta y}{\partial\xi}$, and so forth.

This system of six equations can be easily integrated. The first shows that
$\Delta x$ is independent of $\xi$, but then its derivatives
$\frac{\partial\Delta x}{\partial\xi}$, $\frac{\partial\Delta
x}{\partial\eta}$, $\frac{\partial\Delta x}{\partial\zeta}$ are also
independent of $\xi$, and likewise starting with the second or third term, so
that each of the nine partial derivatives is independent of each of the
independent variables $\xi.\eta,\zeta$; (i.e., each is constant.) Moreover,
the matrix of derivatives $\frac{\partial(\Delta
x,y,z)}{\partial(\xi,\eta,\zeta)}$ is an antisymmetric matrix of constants,
and this leads finally to the expressions at which we have previously arrived,

$\Delta x=A+p\eta-n\zeta,\;\;\Delta y=B+m\zeta-p\xi,\;\;\Delta
z=C+n\xi-m\eta.$ (211)

Here the sense of “constant” is that the quantity so called vanishes under the
operator $d$, unlike its sense in (207). The point is that the equations (205)
and (208) remain true with no change in form, if the “neighboring point”
$\vec{r}+d\vec{r}$ is replaced by a different neighboring point. Therefore all
the consequences can be differentiated again and again if one wishes. In
particular, (31) can be differentiated so as to show that all 27 second
partial derivatives are zero, and consequently all nine first partial
derivatives $\frac{\partial(\Delta x,y,z)}{\partial(\xi,\eta,\zeta)}$ are
constant. Then appealing again to (31), one sees that this constant matrix is
antisymmetric.

We shall not return to the transformations undergone by these formulas in
reestablishing the variables $x,y,z$; it suffices to have shown how the method
of variations applies to the study of these formulas and gives in a single
algebraic form both finite and infinitesimal coordinate changes in displaced
points, these points being replaced in the case of finite displacements by the
midpoints of the interval traversed in a straight line by them.

##  32 Recapitulation

To wind up this work it remains to deduce quickly, from the expression for
these changes, the geometric laws for the displacement of solid bodies that we
developed synthetically in the first place, and took as point of departure for
our first analysis.

The formula

$\vec{\Delta}=\vec{\Gamma}+\vec{q}\times\vec{\omega}$ (212)

(this is the vectorial form of (211)) immediately gives the following
fundamental relation:

$\vec{q}\cdot\vec{\Delta}=\vec{q}\cdot\vec{\Gamma},$ (213)

from which one sees that the lines actually traversed by all the points of the
system, in passing from one situation to the other, all have equal projections
onto a particular direction $\hat{q}$. Recall that the components of
$\vec{\Gamma}$ and those of $\vec{q}$ are those six constants
$A,B,\Gamma,m,n,p$ which characterize the displacement as a whole. Therefore,
(213) tells us that although different points $\vec{r}$ traverse different
lines $\vec{\Delta}$, these lines all project equally on the fixed direction
$\hat{q}$. Denoting this projection by $t$, one will have for all the points
of the displaced system

$\hat{q}\cdot\vec{\Delta}=t$ (214)

and for two different points,

$\hat{q}\cdot(\vec{\Delta}_{1}-\vec{\Delta}_{2})=0.$ (215)

Here $\vec{\Delta}_{1}-\vec{\Delta}_{2}$ is the chord of the arc that would be
described by the first point about an axis of rotation drawn through the
second point and parallel to the direction $\hat{q}$; the two points can
thereafter be brought to their final positions by translating them both by
$\vec{\Delta}_{2}$.

If $\theta$ represents the angle of that rotation, and $u$ the distance from
that chord to that axis of rotation, one has evidently

$4u^{2}\tan^{2}(\theta/2)=|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}=\vec{\,q}^{2}[|\vec{\omega}_{1}-\vec{\omega}_{2}|^{2}-((\vec{\omega}_{1}-\vec{\omega}_{2}))\cdot\hat{q})^{2}],$
(216)

$u^{2}=|\vec{\omega}_{1}-\vec{\omega}_{2}|^{2}-[(\vec{\omega}_{1}-\vec{\omega}_{2}))\cdot\hat{q})]^{2}$
(217)

and therefore, regardless of what two points are being considered,

$4\tan^{2}(\theta/2)=\vec{\,q}^{2}.$ (218)

The author has worked his way back to what followed immediately from the
definition $\vec{q}=(m,n,p)$ given in Section 15. It should be noted that
$\hat{q}$ is the same unit vector as $\hat{t}$.

Thus, the displacement given to a solid, from one situation to another, can
always be resolved into two consecutive displacements, one of rotation and one
of translation, just as has been explained at the start of this treatise.

Furthermore, let $\nu$ represent the amplitude of the angular displacement of
a _line_ within the solid, and $\phi$ the angle formed by the axis of rotation
with this line, then if $\vec{r}_{1},\vec{r}_{2}$ are two points on the line
we have

$\cos\phi=\frac{(\vec{r}_{1}-\vec{r}_{2})\cdot\hat{q}}{|\vec{r}_{1}-\vec{r}_{2}|},$
(219)

and, in view of the invariablity of the distances between points of the solid,

$\Delta\cos\phi=\frac{(\vec{\Delta}_{1}-\vec{\Delta}_{2})\cdot\hat{q}}{|\vec{r}_{1}-\vec{r}_{2}|}=0.$
(220)

The angle $\phi$ is the same before and after the displacement. As for the
angle $\nu$, we have

$\cos\nu=\frac{(\vec{r}_{1}-\vec{r}_{2})\cdot(\vec{r}_{1}-\vec{r}_{2}+\vec{\Delta}_{1}-\vec{\Delta}_{2})}{|\vec{r}_{1}-\vec{r}_{2}|^{2}},$
(221)

whence finally the remarkable relation

$\sin(\nu/2)=\sin\phi\sin(\theta/2)$ (222)

expressing the theorem stated in 5. (Actually the relation is given in 6.)

It is not trivial to derive (222) from the preceding equations. An essential
preliminary step is to derive the identity

$(\vec{\omega}_{1}-\vec{\omega}_{2})\cdot(\vec{\Delta}_{1}-\vec{\Delta}_{2})=0.$
(223)

This is easily done by noting that for any point $\vec{r}$ we have
$\vec{r}=\vec{\omega}-(1/2)\vec{\Delta}$ and
$\vec{r}+\vec{\Delta}=\vec{\omega}+(1/2)\vec{\Delta}$, so that the equivalence
of the two lengths $|\vec{r}_{1}-\vec{r}_{2}|$ and
$|(\vec{r}_{1}+\vec{\Delta}_{1})-(\vec{r}_{2}+\vec{\Delta}_{2})|$ can be
written as

$|(\vec{\omega}_{1}-(1/2)\vec{\Delta}_{1})-(\vec{\omega}_{2}-(1/2)\vec{\Delta}_{2})|^{2}=|(\vec{\omega}_{1}+(1/2)\vec{\Delta}_{1})-(\vec{\omega}_{2}+(1/2)\vec{\Delta}_{2})|^{2},$
(224)

which is equivalent to (223).

From (223) there follow the two useful relations

$(\vec{\omega}_{1}-\vec{\omega}_{2})\cdot\vec{\Delta}_{1}=(\vec{\omega}_{1}-\vec{\omega}_{2})\cdot\vec{\Delta}_{2}$
(225)

and

$|\vec{r}_{1}-\vec{r}_{2}|^{2}=|\vec{\omega}_{1}-\vec{\omega}_{2}|^{2}+(1/4)|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}.$
(226)

Now consider (221). The numerator can be written as

$[(\vec{\omega}_{1}-(1/2)\vec{\Delta}_{1})-(\vec{\omega}_{2}-(1/2)\vec{\Delta}_{2})]\cdot[(\vec{\omega}_{1}+(1/2)\vec{\Delta}_{1})-(\vec{\omega}_{2}+(1/2)\vec{\Delta}_{2})]=|\vec{\omega}_{1}-\vec{\omega}_{2}|^{2}-(1/4)|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}$
(227)

in view of (223), and the denominator as

$|\vec{\omega}_{1}-\vec{\omega}_{2}|^{2}+(1/4)|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}$
(228)

by (226). Then

$\sin^{2}(\nu/2)=(1/2)(1-\cos\nu)=(1/4)\frac{|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}}{|\vec{\omega}_{1}-\vec{\omega}_{2}|^{2}+(1/4)|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}}=(1/4)\frac{|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}}{|\vec{r}_{1}-\vec{r}_{2}|^{2}}.$
(229)

.

In dealing with $\phi$, we recall that it is the angle between the line
containing $\vec{r}_{1}$ and $\vec{r}_{2}$ and the fixed direction $\hat{q}$,
also called the axis of rotation. If we think of $\hat{q}$ as vertical,
$\vec{r}_{1}-\vec{r}_{2}$ may be resolved into a vertical part
$(\vec{r}_{1}-\vec{r}_{2})\cdot\hat{q}$ and a horizontal part which we may
call $(\vec{r}_{1}-\vec{r}_{2})_{\perp}$. Then $\cos\phi$ is given by (219),
and for $\sin\phi$ we have

$\sin\phi=\frac{|\vec{r}_{1}-\vec{r}_{2}|_{\perp}}{|\vec{r}_{1}-\vec{r}_{2}|}.$
(230)

Combining (230) with (229), we find

$\frac{\sin^{2}(\nu/2)}{\sin^{2}\phi}=(1/4)\frac{|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}}{|\vec{r}_{1}-\vec{r}_{2}|_{\perp}^{2}}.$
(231)

To obtain $\sin^{2}(\theta/2)$, it is not sufficient to proceed directly from
(218), as this will involve a factor $\vec{\,q}^{2}=m^{2}+n^{2}+p^{2}$ which
does not enter into (231). Instead, consider the arc swept out by rotating
$\vec{r}_{1}$ an angle $\theta$ about the “vertical” axis passing through
$\vec{r}_{2}$. The radius of this arc is $|\vec{r}_{1}-\vec{r}_{2}|_{\perp}$,
and the chord subtended by $\theta$ has length
$|\vec{\Delta}_{1}-\vec{\Delta}_{2}|$. Therefore

$\sin^{2}(\theta/2)=(1/4)\frac{|\vec{\Delta}_{1}-\vec{\Delta}_{2}|^{2}}{|\vec{r}_{1}-\vec{r}_{2}|_{\perp}^{2}}.$
(232)

Combining (232) with (231), we obtain (222).

—

The lines parallel to the direction of the axis of rotation are therefore
transported parallel to themselves. Among all these lines there is one that
simply glides upon itself; for this line the change $\Delta\vec{r}$ is
evidently in the direction $\hat{q}$. The equation of this line is therefore

$\Delta\vec{r}=\hat{q}t=\vec{t};$ (233)

and since

$\vec{\omega}=\vec{r}+(1/2)\Delta\vec{r},$ (234)

we obtain the same equation already given in 16 for the _central_ axis of the
displacement:

$\Gamma+\vec{q}\times\vec{r}=t\hat{q}=\vec{t}.$ (235)

##  33 Conclusion - General law of Statics.

Geometry considers the displacements, finite or infinitesimal, of _solid_
bodies, brought about by the _successive_ action of causes or of forces
capable of producing them.

Mechanics considers consecutive displacements of solid bodies, and more
generally of _arbitrary_ systems of points, brought about by the
_simultaneous_ and prolonged action of causes or of forces capable of
producing them.

Statics is that most elementary part of Mechanics in which one considers only
the possibility of infinitesimal or _virtual_ displacements of these systems,
resulting from the simultaneous and discontinuous action of those same causes.

The idea seems to be that the distinction between Geometry and Mechanics
disappears when one considers only infinitesimal displacements in Geometry, or
instantaneous applications of force in Mechanics.

Geometry teaches that the displacement of a solid body reduces to a turning
about one or two fixed axes. (Two, if one wishes to eliminate translations.)

It follows that if the forces that act simultaneously on a solid system cannot
impress on it any rotation about any fixed axis whatever, these forces
equilibrate or neutralize one another, and the body remains at rest.

These forces, considered separately, can act only in two ways, either by
tending to turn this solid body about a fixed axis, or by tending to displace
a certain point of the system, or more expressively to change the coordinates
of that point. That is the most general way to consider and to examine, in
Mechanics, the action of forces.

At any rate the law of equilibrium is identical in these two modes of thought,
as we shall see. (Probably the two modes are that of Geometry and that of
Mechanics.)

If any forces or causes of displacement tend successively, or else
simultaneously in passing to the limit (that is to say in passing from
Geometry to Mechanics), to impress on a solid elementary or virtual rotations
$\theta,\theta^{\prime},\theta^{\prime\prime},...$, about given fixed axes,
the law of equilibrium of these forces is that the sum of the moments of these
rotations should vanish relatively to any axis whatever, This law is rendered
algebraically by the equation

$\Sigma\theta D\sin{\nu}=0,$ (236)

implying, on account of the indeterminacy of that arbitrary axis, six special
equations, which reduce to three when the solid system reduces to a point.

Let us now examine what happens when the forces acting simultaneously on the
solid are applied individually to various points of the system. As any
displacement refers virtually to a fixed axis of rotation, it will suffice to
consider for each point the change that can result, from the action of the
forces affecting that point, in the coordinate of that point orthogonal to
that fixed axis (and also orthogonal to the perpendicular dropped from the
point to the axis), whose resistance is opposed to any change in its (the
point’s) other rectangular coordinates. The fixed axis is understood to be
mechanically fixed: it cannot move either along its own length or
perpendicularly to it. The only way the point can move without the axis moving
is on the tangent to the circle it would describe if the system rotates about
the axis.

Now, it is evident that (wordy redundant passage omitted).

From another point of view, on account of the rigidity of the system, it is
evident that two forces equal (in magnitude) will, in their action on any
given point, be in equilibrium (1) if they are applied through the point in
opposite directions; (2) if they are applied in opposite sense through the
extremities of a fixed line segment; (3) if they tend to turn in opposite
sense a circumference whose center is fixed, in the plane of which they are
applied tangentially; (4) if, more generally, they tend to turn oppositely a
right cylinder whose axis is fixed, at whose surface they are applied
tangentially and orthogonal to its axis.

It results from these propositions that if one considers all the forces that
tend to displace the individual points of a solid system which contains a
fixed axis, and acting in given directions, and to impress on them
individually given virtual translations, there will be equilibrium among all
these forces if, supposing that they are all applied to points equidistant
from the fixed axis - which is always possible - the sum of the moments of the
virtual translations that measure the effect of these forces is zero relative
to that axis. For “moment”, read “torque: the total torque about the fixed
axis should vanish, in order to produce equilibrium. I don’t understand the
language about points equidistant from the fixed axis.

Passing from a fixed axis to an arbitrary axis, it will follow necessarily
that the general equation (236) is the algebraic expression of the equilibrium
of a set of forces capable of producing virtual or infinitesimal translations
proportional to the rotations
$\theta,\theta^{\prime},\theta^{\prime\prime},...$, those forces being applied
about the axes of these rotations positively or negatively according to the
signs of the rotations.

This explains the remarkable analogy between the laws of equilibrium (and
consequently of the composition) of infinitesimal rotations and the laws of
equilibrium and composition of forces, considered in Statics. This is the
analogy discussed in 26.

If one denotes these forces by their finite magnitudes $P,P^{\prime},...,$,the
equation of their equilibrium will be

$\Sigma PD\sin\nu=0.$ (237)

Here each term $PD\sin\nu$ expresses the static moment [[torque]] of the force
$P$ about the fixed axis; it is equal to the product of the distance from the
point of application of that force to the fixed axis with the component of
that force normal to that axis. The author here combines $P$ and $\nu$ into a
single quantity, a component of the force. He should say also, normal to the
distance $D$.

Repetitive paragraph omitted.

The conditions of equilibrium of forces applied to a solid system, which the
secondary but admirably ingenious consideration of couples reduces in finite
terms to two conditions, are therefore comprised in a single law, similarly
expressed in finite terms, that the sum of the moments of the forces be null
about an arbitrary axis. This law is general and applies, as does that of the
principle of virtual speeds - equivalent to an infinitesimal transformation -
to the equilibrium of any system rigid or not, provided that the conditions of
the connections among points of the system be replaced by the introduction of
forces that make it possible to regard these points as entirely free. Here the
author definitely is speaking of Mechanics; his idea is that even if the
system is not solid but consists of independently mobile points, it can be
made to act like a solid if the forces acting on it are such as to leave
invariant all the distances $|\vec{r}_{i}-\vec{r}_{j}|$. These forces are
determined by analysis and eliminated in accordance with the equations that
give the equilibrium of each point.

By this means the law of equilibrium of a point immediately implies that of a
solid system. We shall not linger over this. We merely remark that in the
particular case of the forces introduced being equal and opposite in pairs,
the sum of the moments of all the forces applied to all the points, which must
be null for equilibrium, contains only the sum of the given forces, and thus
expresses, in identical form, the law of equilibrium of one or many points
entirely free and that of a rigid system.

###  33.1 On the equation of virtual speeds.

Consider an infinitesimal displacement of the system, producing changes
characterized by the symbol $\delta$ in the coordinates. If the force $P$
tends to change the coordinate $p$, (we claim that) the product $P\delta p$
will be equal to $PD\theta\sin\nu$, where $D\theta\sin\nu$ is the moment of
the virtual rotation $\theta$.

In fact, since this infinitesimal displacement must reduce to either a single
or to two successive rotations, it need only be considered in its unique
rotation or in the first of the two. And so the infinitesimal arc described by
the point at which $P$ is applied, projected along the direction of that
force, will be equal to the infinitesimal change in the coordinate $p$ on
which the force acts. This (indeed) gives $\delta p=\theta D\sin\nu$, or

$P\delta p=\theta PD\sin\nu.$ (238)

Hence the general equation (237) of equilibrium of forces

transforms into

$\Sigma P\delta p=0,$ (239)

which says that if a solid system is in equilibrium under a set of forces, and
by any cause this system is infinitesimally dislodged from its present
position, the sum of the products of each force with the infinitesimal
distance traversed by the point (of application of the force) of the system
along the direction of that force must be zero, and conversely; this is the
principle of virtual speeds.As we would say nowadays, the solid is in
equilbrium under a set of forces applied at certain points in a certain manner
if and only if the virtual work that would be done by these forces in any
hypothetical infinitesimal displacement obeying the geometrical constraints of
the solid is zero.

Equation (239), although it is certainly superior, algebraically speaking, to
(237), is no more general at bottom; but it expresses in the simplest possible
way the law of equilibrium of any system in which the conditions binding the
parts together can be transformed into a set of linear equations among the
changes of coordinates of the various points of the system.

The remainder of this section, and hence of the whole treatise, is devoted to
elucidating the above remark.

In fact, (suppose that) these conditions are expressed by equations such as
$\delta L=0,\delta L^{\prime}=0,\delta L^{\prime\prime}=0,$ etc. or more
generally by a single equation such as

$\lambda\delta L+\lambda^{\prime}\delta
L^{\prime}+\lambda^{\prime\prime}\delta L^{\prime\prime}+...=0,$ (240)

in which $\lambda,\lambda^{\prime},\lambda^{\prime\prime},...$ are arbitrary
multipliers (what we nowadays call Lagrange multipliers). Let us denote by
$\bar{x},\bar{y},\bar{z}$ the coefficients of the changes $\delta x,\delta
y,\delta z$ (hidden within) this equation; by $r$ a linear distance in the
direction of $(\bar{x},\bar{y},\bar{z})$; and by $R$ a force equal to
$\sqrt{\bar{x}^{2}+\bar{y}^{2}+\bar{z}^{2}}$, applied at the point $(x,y,z)$
in the direction of the line that it tends to change, Then we have

$\delta r=\frac{\bar{x}\delta x+\bar{y}\delta y+\bar{z}\delta
z}{\sqrt{\bar{x}^{2}+\bar{y}^{2}+\bar{z}^{2}}}$ (241)

and similarly for $\delta r^{\prime}$,$\delta r^{\prime\prime}$, etc. Equation
(240) will now take the form $R\delta r+R^{\prime}\delta
r^{\prime}+R^{\prime\prime}\delta r^{\prime\prime}+...=0$, and the equations
(239) and

$\Sigma P\delta p+R\delta r+R^{\prime}\delta r^{\prime}+R^{\prime\prime}\delta
r^{\prime\prime}+...=0$ (242)

will have equal generality, the changes $\delta p$ being _limited_ in (239) by
the equation of conditions (240), and in (242) being completely independent.
Now, in the second case, (242) expresses the condition of equilibrium of all
the points of the system, independent of all binding, but implied by the
external forces $P,P^{\prime}P^{\prime\prime},...$ and by other forces
$R,R^{\prime},R^{\prime\prime}...$ which _statically_ have replaced the
assumed conditions of binding.

The relation between (240) and (242) is the familiar one that arises in the
Lagrange multiplier method: one relaxes the rigidity conditions requiring
$\delta L$, etc. to vanish, at the cost of introducing the extra terms
$R\delta r$, etc. into the equation to be solved. It is interesting that the
whole procedure, nowadays presented as a purely mathematical manipulation in
the spirit of [4], is here explicated in a totally physical way.

By eliminating these forces $R,R^{\prime},R^{\prime\prime}...$ from (242) one
can obtain the definitive equations of equilibrium of the external forces
subject to the binding of the system. And conversely, if these equations hold,
there will be equilibrium, since the external forces are then determined, and
the equations (242) on independent values of $\delta r$, etc. establish the
immobility of all the points of the system resulting from the action of the
external forces plus that of the others that are statically equivalent to the
given binding among the various points of the system.

When the system under consideration is continuous, the equations of binding
contain definite integrals which represent in some way an infinite number of
linear conditions among the changes of coordinates of the system. The
arbitrary multipliers may be moved inside the integral sign, and it then
remains to solve for the changes by a method entirely analytic and independent
of any static consideration. One easily arrives at the following general
formula:

$\delta
S^{n}Udx_{1}dx_{2}dx_{3}...dx_{n}=S^{n}dx_{1}dx_{2}dx_{3}...dx_{n}[\delta
U+U(\frac{\partial\delta x_{1}}{\partial x_{1}}+\frac{\partial\delta
x_{2}}{\partial x_{2}}+...+\frac{\partial\delta x_{n}}{\partial x_{n}})],$
(243)

where $U$ is an arbitrary function of the independent variables
$x_{1},x_{2},...,x_{n}$, and the symbol $S^{n}$ denotes a multiple definite
integral of order $n$.

In today’s notation (243) could be written as

$\delta\int dx_{1}...\int dx_{n}U(x_{1},...x_{n})=\int dx_{1}...\int
dx_{n}[\delta U(x_{1},...x_{n})+U(x_{1},...x_{n})(\frac{\partial\delta
x_{1}}{\partial x_{1}}+...+\frac{\partial\delta x_{n}}{\partial x_{n}})].$
(244)

The point of interest is that $U$ is an arbitrary function, not necessarily
linear in the $x_{i}$. This is why the term $\delta U$ enters the right side
of the equation.

_December 5, 1840._

##  34 Acknowledgment

This translation has been truly a labor of love. I could never have dared to
undertake it, much less carry it through to completion, without the unflagging
interest and support of my associate Dr. Johannes Familton, with whom I
discussed every part of the translation as it took shape. Dr. Familton also
gave me indispensable help in designing and implementing the figures.

## References

  * [1] L. Euler, Formulae generales pro translatione quacunque corporum rigidorum. Novi Commentarii Academiae Scientiarum Petropolitanae 20 (1776), 189207.
  * [2] M. Chasles, Bull. universel des Sciences, 14 (1830).
  * [3] L. Euler, Problema algebraicum ob affectiones prorsus singulares memorabile (1771). Euler Archive - All Works. 407
  * [4] J. L. Lagrange, Mécanique Analytique, Tome 1. Courcier, Paris, 1776.

Figures

1) Section 3 (Euler)

2) Section 5 (Central axis)

3) Section 6 (spherical geodesic)

4) Section 10 (parallel axes)