Space-Antispace Transform Correspondence in Projective Geometric Algebra and Motor: Difference between pages

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'''By Eric Lengyel'''<br />
[[Image:proper_isom.svg|400px|thumb|right|'''Figure 1.''' A motor represents a proper Euclidean isometry, which can always be regarded as a rotation about a line $$\boldsymbol L$$ and a displacement along the same line.]]
May 20, 2022
A ''motor'' is an operator that performs a proper isometry in Euclidean space. Such isometries encompass all possible combinations of any number of [[rotations]] and [[translations]]. The name motor is a portmanteau of ''motion operator'' or ''moment vector''. Motors are equivalent to the set of ''dual quaternions'' used in conventional theories, and the functionality is properly generalized in rigid geometric algebra. Motors cannot perform improper isometries that include an odd number of [[reflections]]; those are instead performed by operators called [[flectors]].


[[Image:Antispace1.svg|480px|thumb|right|'''Figure 1.''' The coordinates $$(p_x, p_y, p_z)$$ can be interpreted as the one-dimensional span of a single vector representing a homogeneous point or as the $$(n - 1)$$-dimensional span of all orthogonal vectors representing a homogeneous hyperplane, which is a line when $$n = 3$$. Geometrically, these two interpretations are dual to each other, and their distances to the origin are reciprocals of each other.]]
In the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$, a motor $$\mathbf Q$$ has the general form
The concept of duality can be understood geometrically in an ''n''-dimensional projective setting by considering both the subspace that an object occupies and the complementary subspace that the object concurrently does not occupy. The dimensionalities of these two components always sum to ''n'', and they represent the ''space'' and ''antispace'' associated with the object. (Antispace is also known as negative space or counterspace.) The example shown in Figure 1 demonstrates the duality between homogeneous points and lines in a three-dimensional projective space. The triplet of coordinates $$(p_x, p_y, p_z)$$ can be interpreted as a vector pointing from the origin toward a specific location on the projection plane $$z = 1$$. This vector corresponds to the one-dimensional space of the point that it represents. The dual of a point materializes when we consider all of the directions of space that are orthogonal to the single direction $$(p_x, p_y, p_z)$$. As illustrated by the figure, these directions span an $$(n - 1)$$-dimensional subspace that intersects the projection plane at a line when $$n = 3$$. In this way, the coordinates $$(p_x, p_y, p_z)$$ can be interpreted as both a point and a line, and they are ''duals'' of each other.


When we express the coordinates $$(p_x, p_y, p_z)$$ on the vector basis as $$p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3$$, it explicitly states that we are working with a single spatial dimension representing a point, and the ambiguity is removed. Similarly, if we express the coordinates on the bivector basis as $$p_x \mathbf e_{23} + p_y \mathbf e_{31} + p_z \mathbf e_{12}$$, then we are working with the two orthogonal spatial dimensions representing a line. In each case, the subscripts of the basis elements tell us which basis vectors are present in the representation, and this defines the ''space'' of the object. The subscripts also tell us which basis vectors are absent in the representation, and this defines the ''antispace'' of the object. Acknowledging the existence of both the space and the antispace of any object and assigning equal meaningfulness to them allows us to explore the nature of duality to its fullest. A vector $$p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3$$ is never only a point, but both a point and a line simultaneously, where the point exists in space, and the line exists in antispace. Likewise, a bivector $$p_x \mathbf e_{23} + p_y \mathbf e_{31} + p_z \mathbf e_{12}$$ is never only a line, but both a line and a point simultaneously, where the line exists in space, and the point exists in antispace. If we study only the spatial facet of these objects and their higher-dimensional counterparts, then we are missing half of a bigger picture.
:$$\mathbf Q = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1d7d9}} + Q_{mx} \mathbf e_{23} + Q_{my} \mathbf e_{31} + Q_{mz} \mathbf e_{12} + Q_{mw} \mathbf 1$$ .


It is particularly interesting to consider the Euclidean isometries that map ''n''-dimensional space onto itself while preserving distances and angles. We know how each isometry transforms the space of a point, line, plane, etc., but for a complete understanding of the geometry, we must ask ourselves what happens to the antispace of those objects at the same time. Equivalently, when an object is transformed by an isometry, we would like to know how its dual is transformed. The answer requires that we first look at the invariants associated with each Euclidean isometry.
To possess the [[geometric property]], the components of $$\mathbf Q$$ must satisfy the equation


In the two-dimensional plane, the Euclidean isometries consist of a rotation about point, a translation in a specific direction, and a transflection with respect to a specific mirroring line. A reflection is a special case of transflection in which there is no motion parallel to the line, and translation is a special case of rotation in which the center lies in the horizon. Naturally, the invariant of a rotation is its center point, and the invariant of a transflection is its mirroring line. These objects are mapped onto themselves by their associated transforms, and this necessitates that their duals also be mapped onto themselves by whatever corresponding transforms occur in antispace. Because a rotation fixes its center point, the corresponding transform in antispace must fix the line that is dual to that point. And because a transflection fixes its mirroring line, the corresponding transform in antispace must fix the point that is dual to that line. These transforms occurring in antispace are dual analogs of the transforms occurring in space. There is a direct correspondence between the two transforms, and they are inextricably linked. Whenever one transform is applied in space, the other is applied in antispace, and vice-versa.
:$$Q_{vx} Q_{mx} + Q_{vy} Q_{my} + Q_{vz} Q_{mz} + Q_{vw} Q_{mw} = 0$$ .


[[Image:Antispace2.svg|640px|thumb|right|'''Figure 2.''' (Left) A regular rotation fixes the green center point at $$(1, 0)$$ and the horizon. (Right) The corresponding dual rotation fixes the green line at $$x = -1$$ dual to the center point and the origin.]]
Motors are capable of representing any general screw motion consisting of a rotation about a line combined with a displacement along the same line. By Chasles' theorem, all proper rigid motions in 3D space are screw motions.
Figure 2 shows a two-dimensional rotation transform and its dual analog. The green point represents the center of rotation, and the green line is the dual of that point. Under the regular rotation, the center point is fixed, and under the dual rotation, the line dual to the center point is fixed. These are not, however, the only fixed geometries. A regular rotation also fixes the horizon line, and thus its dual, the origin, must be fixed in the dual rotation. This is illustrated by the red point in the figure, which is the focus of the various conic-section orbits. Here, the green line is the directrix.


[[Image:Antispace3.svg|640px|thumb|right|'''Figure 3.''' (Left) A regular reflection fixes the green mirroring line at $$x = -1/2$$ and the point in the horizon in the perpendicular direction. (Right) The corresponding dual reflection fixes the green point $$(2, 0)$$ dual to the mirroring line and the line through the origin parallel to the mirroring line.]]
An element $$\mathbf x$$ is transformed by a motor $$\mathbf Q$$ through the operation $$\mathbf x' = \mathbf Q \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}$$, where $$\unicode{x27C7}$$ is the [[geometric antiproduct]].
Figure 3 shows a two-dimensional reflection transform and its dual analog. The green line represents the mirroring plane of the reflection, and the green point is the dual of that line. Under the regular reflection, the mirroring line is fixed, and under the dual reflection, the point dual to the mirroring line is fixed. As with rotation, there are additional fixed geometries under these transforms. A regular reflection fixes the point in the horizon in the direction perpendicular to the mirroring plane. The dual of this point is a line parallel to the mirroring line and containing the origin that remains fixed by the dual reflection. This is illustrated by the red line in the figure, which is clearly a reflection boundary in a sense.


[[Image:Antispace4.svg|640px|thumb|right|'''Figure 4.''' (Left) A regular translation fixes the point in the horizon perpendicular to the direction of translation and every line parallel to the direction of translation. (Right) The corresponding dual translation (a perspective projection) fixes the line parallel to the direction of translation through the origin and every point in the line through the origin perpendicular to the direction of translation.]]
The set of all motors, sometimes denoted by $$\unicode{x1D544}$$, forms a subgroup of index 2 in a geometric algebra. Its coset is the set of [[flectors]].
Finally, Figure 4 shows a two-dimensional translation transform and its dual analog. As mentioned above, a translation is a special case of rotation in which the center lies in the horizon. As such, there is no finite fixed geometry that can be shown in the figure for a regular translation. However, the dual of the center in the horizon must be a line containing the origin that is fixed by the dual translation, and that is illustrated by the red line in the figure. Dual translation is especially important because it is the one to which we can most easily assign some practical meaning. It is a ''perspective projection'' onto the line through the origin perpendicular to the direction of translation.


In the three-dimensional projective geometric algebra $$\mathbb R_{2,0,1}$$, a homogeneous representation of two-dimensional space, a regular rotation about a center point $$\mathbf c$$ is given by
== Simple Motors ==


:$$\mathbf Q = c_x \mathbf e_1 + c_y \mathbf e_2 + c_z \mathbf e_3 + r {\large\unicode{x1D7D9}}$$,
If $$Q_{mw} = 0$$ (i.e., the scalar part $$Q_{\mathbf 1}$$ is zero), then the motor $$\mathbf Q$$ is called a ''simple motor''. Every simple motor represents either a pure [[translation]] or a pure [[rotation]] about a line without any displacement along that line.


and this becomes a translation when $$c_z = 0$$. A regular transflection across the line $$\mathbf h$$ is given by
In the case of a pure translation, the motor $$\mathbf T$$ is given by


:$$\mathbf F = s + h_x \mathbf e_{23} + h_y \mathbf e_{31} + h_z \mathbf e_{12}$$,
:$$\mathbf T = t_x \mathbf e_{23} + t_y \mathbf e_{31} + t_z \mathbf e_{12} + {\large\unicode{x1d7d9}}$$ ,


and this becomes a pure reflection when $$s = 0$$. Together, these operators include all possible Euclidean isometries in the two-dimensional plane. Under the geometric antiproduct $$\unicode{x27C7}$$, arbitrary products of these operators form the group $$\mathrm E(2)$$ with $${\large\unicode{x1D7D9}}$$ as the identity, and they covariantly transform any object $$a$$ in the algebra through the sandwich products
and this performs a translation by twice the displacement vector $$(t_x, t_y, t_z)$$.


:$$a' = \mathbf Q \mathbin{\unicode{x27C7}} a \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}$$
In the case of a pure rotation, the motor $$\mathbf R$$ is given by
 
:$$\mathbf R = \boldsymbol l\sin\phi + {\large\unicode{x1d7d9}}\cos\phi$$ ,
 
and this performs a rotation by twice the angle $$\phi$$ about the [[line]] $$\boldsymbol l$$.
 
== Motors Built from Geometry ==
 
Motors having specific geometric constructions can be built from [[Point | points]], [[Line | lines]], and [[Plane | planes]] as shown in the following table.
 
{| class="wikitable"
! Motor|| Description
|-
| style="padding: 12px;" | $$\begin{split}\mathbf h \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{g}}} =\, &(g_yh_z - g_zh_y)\mathbf e_{41} + (g_zh_x - g_xh_z)\mathbf e_{42} + (g_xh_y - g_yh_x)\mathbf e_{43} \\ +\, &(h_xg_w - g_xh_w)\mathbf e_{23} + (h_yg_w - g_yh_w)\mathbf e_{31} + (h_zg_w - g_zh_w)\mathbf e_{12} \\ +\, &(g_xh_x + g_yh_y + g_zh_z)\smash{\large\unicode{x1d7d9}}\end{split}$$
| style="padding: 12px;" | Rotation about the line where planes $$\mathbf g$$ and $$\mathbf h$$ intersect by twice the angle between them in the direction from $$\mathbf g$$ to $$\mathbf h$$.
 
$$\mathbf g = g_x \mathbf e_{234} + g_y \mathbf e_{314} + g_z \mathbf e_{124} + g_w \mathbf e_{321}$$
 
$$\mathbf h = h_x \mathbf e_{234} + h_y \mathbf e_{314} + h_z \mathbf e_{124} + h_w \mathbf e_{321}$$
|-
| style="padding: 12px;" | $$\begin{split}\boldsymbol l \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{k}}}
=\, &(l_{vz} k_{vy} - l_{vy} k_{vz})\mathbf e_{41} + (l_{vx} k_{vz} - l_{vz} k_{vx})\mathbf e_{42} + (l_{vy} k_{vx} - l_{vx} k_{vy})\mathbf e_{43} \\
+\, &(l_{vz} k_{my} - l_{vy} k_{mz} + l_{mz} k_{vy} - l_{my} k_{vz})\mathbf e_{23} + (l_{vx} k_{mz} - l_{vz} k_{mx} + l_{mx} k_{vz} - l_{mz} k_{vx})\mathbf e_{31} + (l_{vy} k_{mx} - l_{vx} k_{my} + l_{my} k_{vx} - l_{mx} k_{vy})\mathbf e_{12} \\
+\, &(l_{vx} k_{mx} + l_{vy} k_{my} + l_{vz} k_{mz} + l_{mx} k_{vx} + l_{my} k_{vy} + l_{mz} k_{vz}) \\
+\, &(l_{vx} k_{vx} + l_{vy} k_{vy} + l_{vz} k_{vz})\smash{\large\unicode{x1d7d9}}\end{split}$$
| style="padding: 12px;" | Rotation about the line containing the closest points on lines $$\mathbf k$$ and $$\boldsymbol l$$ by twice the angle between the directions $$(l_{vx}, l_{vy}, l_{vz})$$ and $$(k_{vx}, k_{vy}, k_{vz})$$, and translation by twice the distance between the lines in the direction from $$\mathbf k$$ to $$\boldsymbol l$$.
 
$$\mathbf k = k_{vx} \mathbf e_{41} + k_{vy} \mathbf e_{42} + k_{vz} \mathbf e_{43} + k_{mx} \mathbf e_{23} + k_{my} \mathbf e_{31} + k_{mz} \mathbf e_{12}$$
 
$$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$
|-
| style="padding: 12px;" | $$\mathbf q \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{p}}} = (q_xp_w - p_xq_w)\mathbf e_{23} + (q_yp_w - p_yq_w)\mathbf e_{31} + (q_zp_w - p_zq_w)\mathbf e_{12} + (p_wq_w)\smash{\large\unicode{x1d7d9}}$$
| style="padding: 12px;" | Translation by twice the distance between points $$\mathbf p$$ and $$\mathbf q$$ in the direction from $$\mathbf p$$ to $$\mathbf q$$.
 
$$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$
 
$$\mathbf q = q_x \mathbf e_1 + q_y \mathbf e_2 + q_z \mathbf e_3 + q_w \mathbf e_4$$
|}
 
== Norm ==
 
The [[bulk norm]] of a motor $$\mathbf Q$$ is given by
 
:$$\left\Vert\mathbf Q\right\Vert_\unicode{x25CF} = \sqrt{\mathbf Q \mathbin{\unicode{x25CF}} \mathbf{\tilde Q}} = \sqrt{Q_{mx}^2 + Q_{my}^2 + Q_{mz}^2 + Q_{mw}^2}$$ ,
 
and its [[weight norm]] is given by
 
:$$\left\Vert\mathbf Q\right\Vert_\unicode{x25CB} = \sqrt{\mathbf Q \mathbin{\unicode{x25CB}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}\vphantom{\mathbf{\tilde Q}}} = {\large\unicode{x1D7D9}}\sqrt{Q_{vx}^2 + Q_{vy}^2 + Q_{vz}^2 + Q_{vw}^2}$$ .
 
The [[geometric norm]] of a motor $$\mathbf Q$$ is thus
 
:$$\widehat{\left\Vert\mathbf Q\right\Vert} = \sqrt{\dfrac{Q_{mx}^2 + Q_{my}^2 + Q_{mz}^2 + Q_{mw}^2}{Q_{vx}^2 + Q_{vy}^2 + Q_{vz}^2 + Q_{vw}^2}}$$ ,
 
and this is equal to half the distance that the origin is moved by the operator.
 
A motor is [[unitized]] when $$Q_{vx}^2 + Q_{vy}^2 + Q_{vz}^2 + Q_{vw}^2 = 1$$.
 
== Exponential Form ==
 
A motor $$\mathbf Q$$ can be expressed as the exponential of a [[unitized]] [[line]] $$\boldsymbol l$$ multiplied by $$d + \phi{\large\unicode{x1D7D9}}$$, where $$\phi$$ is half the angle of rotation about the line $$\boldsymbol l$$, and $$d$$ is half the displacement distance along the line $$\boldsymbol l$$. The exponential form can be written as
 
:$$\mathbf Q = \exp_\unicode{x27C7}((d + \phi{\large\unicode{x1D7D9}}) \mathbin{\unicode{x27C7}} \boldsymbol l) = \cos_\unicode{x27C7}(d + \phi{\large\unicode{x1D7D9}}) + \sin_\unicode{x27C7}(d + \phi{\large\unicode{x1D7D9}}) \mathbin{\unicode{x27C7}} \boldsymbol l$$ .
 
This expands to
 
:$$\mathbf Q = \boldsymbol l\sin\phi + (d \mathbin{\unicode{x27C7}} \boldsymbol l)\cos\phi - d\sin\phi + {\large\unicode{x1D7D9}}\cos\phi$$ ,
 
and replacing the [[line]] $$\boldsymbol l$$ with its components gives us
 
:$$\mathbf Q = (l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12})\sin\phi + d(l_{vx} \mathbf e_{23} + l_{vy} \mathbf e_{31} + l_{vz} \mathbf e_{12})\cos\phi - d\sin\phi + {\large\unicode{x1D7D9}}\cos\phi$$ .
 
A motor in exponential form is always [[unitized]], and its [[geometric norm]] can be written as
 
:$$\widehat{\left\Vert\mathbf Q\right\Vert} = \sqrt{d^2 + (l_{mx}^2 + l_{my}^2 + l_{mz}^2)\sin^2\phi}$$ .
 
The quantity $$d + \phi{\large\unicode{x1D7D9}}$$, as a homogeneous magnitude, is the ''pitch'' of the screw transformation performed by the motor, which is the amount of translation along the screw axis per radian of rotation.
 
== Logarithm ==
 
Given an arbitrary [[unitized]] motor $$\mathbf Q$$ with $$|Q_{vw}| < 1$$, the values of $$\boldsymbol l$$, $$d$$, and $$\phi$$ can be determined from the components of $$\mathbf Q$$, essentially taking the logarithm of a motor. First, the [[scalar]] and [[antiscalar]] terms satisfy the equations
 
:$$Q_{mw} = -d\sin\phi$$
:$$Q_{vw} = \cos\phi$$ .
 
Assuming that $$\phi > 0$$, we define $$s = \sin\phi = \sqrt{1 - Q_{vw}^2}$$. Then,
 
:$$d = -\dfrac{Q_{mw}}{s}$$
:$$\phi = \tan^{-1}\left(\dfrac{s}{Q_{vw}}\right)$$ .
 
If $$Q_{vw} = 0$$, then we assign $$\phi = \pi/2$$. If $$Q_{vw} < 0$$, then we can add $$\pi$$ to $$\phi$$ to make the angle positive, but this is not required.
 
The components of $$\boldsymbol l$$ are then given by
 
:$$(l_{vx}, l_{vy}, l_{vz}) = \dfrac{1}{s}(Q_{vx}, Q_{vy}, Q_{vz})$$
 
:$$(l_{mx}, l_{my}, l_{mz}) = \dfrac{1}{s}\left[(Q_{mx}, Q_{my}, Q_{mz}) - \dfrac{dQ_{vw}}{s}(Q_{vx}, Q_{vy}, Q_{vz})\right] = \dfrac{1}{s}\left[(Q_{mx}, Q_{my}, Q_{mz}) + \dfrac{Q_{vw} Q_{mw}}{s^2}(Q_{vx}, Q_{vy}, Q_{vz})\right]$$ .
 
In the special case that $$Q_{vw} = \pm 1$$, the motor must be a pure [[translation]] with $$\phi = 0$$ and $$(Q_{vx}, Q_{vy}, Q_{vz}) = (0, 0, 0)$$.
 
== Square Root ==
 
The square root of a [[quaternion]] can be calculated by summing with the identity and renormalizing. If we attempt to do the same thing with a motor $$\mathbf Q$$, we find that it works only under certain conditions. However, such a calculation does give us a starting point from which we can make a correction for the general case.
 
Using the exponential form of $$\mathbf Q$$, we first examine the [[weight norm]] of $$\mathbf Q + {\large\unicode{x1D7D9}}$$ and find that
 
:$$\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB} = \sqrt{\sin^2\phi + (\cos\phi + 1)^2} = \sqrt{\vphantom{\sin^2\phi}2 + 2Q_\smash{\large\unicode{x1D7D9}}}$$ ,
 
where $$Q_{\large\unicode{x1D7D9}} = \cos\phi$$. Applying the trigonometric identity $$\cos^2(\phi/2) = (1 + \cos\phi)/2$$, we can rewrite this as
 
:$$\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB} = 2\cos(\phi/2)$$ .
 
Applying several more trigonometric identities, we now observe
 
:$$\dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB}} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{2\cos(\phi/2)} = \boldsymbol l\sin(\phi/2) + \left(\dfrac{d}{2} \mathbin{\unicode{x27C7}} \boldsymbol l\right)\cos(\phi/2) - \dfrac{d}{2}\sin(\phi/2) + {\large\unicode{x1D7D9}}\cos(\phi/2) - \left(\dfrac{d}{2} \mathbin{\unicode{x27C7}} \boldsymbol l\right)\sin(\phi/2)\tan(\phi/2) - \dfrac{d}{2}\sin(\phi/2)$$ .
 
The first four terms are exactly the square root of $$\mathbf Q$$ because the distance $$d$$ and angle $$\phi$$ have both been halved. But there are two additional terms that we need to eliminate. It just so happens that
 
:$$\dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB}} \mathbin{\unicode{x27C7}} \dfrac{d}{2}\tan(\phi/2) = \left(\dfrac{d}{2} \mathbin{\unicode{x27C7}} \boldsymbol l\right)\sin(\phi/2)\tan(\phi/2) + \dfrac{d}{2}\sin(\phi/2)$$ ,
 
which means that
 
:$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} \mathbin{\unicode{x27C7}} \left({\large\unicode{x1D7D9}} + \dfrac{d}{2}\tan(\phi/2)\right)$$ .
 
The scalar component of $$\mathbf Q$$ can be rewritten as $$Q_{\mathbf 1} = -d\sin\phi = -2d\sin(\phi/2)\cos(\phi/2) $$. Dividing this by $$2 + 2Q_{\large\unicode{x1D7D9}} = 4\cos^2(\phi/2)$$ yields
 
:$$\dfrac{Q_\mathbf 1}{2 + 2Q_{\large\unicode{x1D7D9}}} = -\dfrac{d}{2}\tan(\phi/2)$$ .
 
The square root of a general motor $$\mathbf Q$$ is thus given by
 
:$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} \mathbin{\unicode{x27C7}} \left({\large\unicode{x1D7D9}} - \dfrac{Q_\mathbf 1}{2 + 2Q_{\large\unicode{x1D7D9}}}\right)$$ .
 
If $$\mathbf Q$$ is a simple motor, then $$Q_{\mathbf 1} = 0$$, and this reduces to
 
:$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB}}$$ .
 
== Factorization ==
 
Any [[unitized]] motor $$\mathbf Q$$ can be factored into the product of a motor $$\mathbf R = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1D7D9}}$$ corresponding to a pure [[rotation]] about a line through the origin and a motor $$\mathbf T = {t_x \mathbf e_{23} + t_y \mathbf e_{31} + t_z \mathbf e_{12} + \large\unicode{x1d7d9}}$$ corresponding to a pure [[translation]] by calculating
 
:$$\mathbf T = \mathbf Q \mathbin{\unicode{x27C7}} \mathbf{\underset{\Large\unicode{x7E}}{R}}$$ .
 
The motor $$\mathbf T$$ is then given by
 
:$$\mathbf T = (Q_{vy} Q_{mz} - Q_{vz} Q_{my} + Q_{mx} Q_{vw} - Q_{vx} Q_{mw})\mathbf e_{23} + (Q_{vz} Q_{mx} - Q_{vx} Q_{mz} + Q_{my} Q_{vw} - Q_{vy} Q_{mw})\mathbf e_{31} + (Q_{vx} Q_{my} - Q_{vy} Q_{mx} + Q_{mz} Q_{vw} - Q_{vz} Q_{mw})\mathbf e_{12} + {\large\unicode{x1d7d9}}$$ .
 
The original motor $$\mathbf Q$$ can now be expressed as
 
:$$\mathbf Q = \mathbf T \mathbin{\unicode{x27C7}} \mathbf R$$ ,
 
where both $$\mathbf R$$ and $$\mathbf T$$ are [[unitized]] simple motors.
 
== Conversion from Motor to Matrix ==
 
Given a specific [[Unitization | unitized]] motor $$\mathbf Q$$, define the matrices
 
:$$\mathbf A = \begin{bmatrix}1 - 2(Q_{vy}^2 + Q_{vz}^2) & 2Q_{vx} Q_{vy} & 2Q_{vz} Q_{vx} & 2(Q_{vy} Q_{mz} - Q_{vz} Q_{my}) \\ 2Q_{vx} Q_{vy} & 1 - 2(Q_{vz}^2 + Q_{vx}^2) & 2Q_{vy} Q_{vz} & 2(Q_{vz} Q_{mx} - Q_{vx} Q_{mz}) \\ 2Q_{vz} Q_{vx} & 2Q_{vy} Q_{vz} & 1 - 2(Q_{vx}^2 + Q_{vy}^2) & 2(Q_{vx} Q_{my} - Q_{vy} Q_{mx}) \\ 0 & 0 & 0 & 1\end{bmatrix}$$


and
and


:$$a' = \mathbf F \mathbin{\unicode{x27C7}} a \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{F}}}$$.
:$$\mathbf B = \begin{bmatrix}0 & -2Q_{vz} Q_{vw} & 2Q_{vy} Q_{vw} & 2(Q_{vw} Q_{mx} - Q_{vx} Q_{mw}) \\ 2Q_{vz} Q_{vw} & 0 & -2Q_{vx} Q_{vw} & 2(Q_{vw} Q_{my} - Q_{vy} Q_{mw}) \\ -2Q_{vy} Q_{vw} & 2Q_{vx} Q_{vw} & 0 & 2(Q_{vw} Q_{mz} - Q_{vz} Q_{mw}) \\ 0 & 0 & 0 & 0\end{bmatrix}$$ .


Symmetrically, a dual rotation about the line $$\mathbf c$$ is given by
Then the corresponding 4&times;4 matrix $$\mathbf M$$ that transforms a [[point]] $$\mathbf p$$, regarded as a column matrix, as $$\mathbf p' = \mathbf{Mp}$$ is given by


:$$\mathcal Q = c_x \mathbf e_{23} + c_y \mathbf e_{31} + c_z \mathbf e_{12} - r$$,
:$$\mathbf M = \mathbf A + \mathbf B$$ .


and a dual transflection across the point $$\mathbf h$$ is given by
The inverse of $$\mathbf M$$, which transforms a [[plane]] $$\mathbf g$$, regarded as a row matrix, as $$\mathbf g' = \mathbf{gM^{-1}}$$ is given by


:$$\mathcal F = h_x \mathbf e_1 + h_y \mathbf e_2 + h_z \mathbf e_3 - s {\large\unicode{x1D7D9}}$$.
:$$\mathbf M^{-1} = \mathbf A - \mathbf B$$ .


These two operators generate a different group of transformations under the geometric product $$\unicode{x27D1}$$. Arbitrary products of these operators form the dual Euclidean group $$\mathrm{RE}(2)$$ with $$\mathbf 1$$ as the identity, and they covariantly transform any object $$a$$ in the algebra through the sandwich products
== Conversion from Matrix to Motor ==


:$$a' = \mathcal Q \mathbin{\unicode{x27D1}} a \mathbin{\unicode{x27D1}} \mathcal{\tilde Q}$$
Let $$\mathbf M$$ be an orthogonal 4&times;4 matrix with determinant +1 having the form


and
:$$\mathbf M = \begin{bmatrix} M_{00} & M_{01} & M_{02} & M_{03} \\ M_{10} & M_{11} & M_{12} & M_{13} \\ M_{20} & M_{21} & M_{22} & M_{23} \\ 0 & 0 & 0 & 1\end{bmatrix}$$ .
 
Then, by equating the entries of $$\mathbf M$$ to the entries of $$\mathbf A + \mathbf B$$ from above, we have the following four relationships based on the diagonal entries of $$\mathbf M$$:
 
:$$M_{00} - M_{11} - M_{22} + 1 = 4Q_{vx}^2$$
 
:$$M_{11} - M_{22} - M_{00} + 1 = 4Q_{vy}^2$$
 
:$$M_{22} - M_{00} - M_{11} + 1 = 4Q_{vz}^2$$
 
:$$M_{00} + M_{11} + M_{22} + 1 = 4(1 - Q_{vx}^2 - Q_{vy}^2 - Q_{vz}^2) = 4Q_{vw}^2$$
 
And we have the following six relationships based on the off-diagonal entries of $$\mathbf M$$:
 
:$$M_{21} + M_{12} = 4Q_{vy} Q_{vz}$$
 
:$$M_{02} + M_{20} = 4Q_{vz} Q_{vx}$$
 
:$$M_{10} + M_{01} = 4Q_{vx} Q_{vy}$$
 
:$$M_{21} - M_{12} = 4Q_{vx} Q_{vw}$$
 
:$$M_{02} - M_{20} = 4Q_{vy} Q_{vw}$$
 
:$$M_{10} - M_{01} = 4Q_{vz} Q_{vw}$$
 
If $$M_{00} + M_{11} + M_{22} \geq 0$$, then we calculate
 
:$$Q_{vw} = \pm \dfrac{1}{2}\sqrt{M_{00} + M_{11} + M_{22} + 1}$$ ,
 
where either sign can be chosen. In this case, we know $$|Q_{vw}|$$ is at least $$1/2$$, so we can safely divide by $$4Q_{vw}$$ in the last three off-diagonal relationships above to solve for $$Q_{vx}$$, $$Q_{vy}$$, and $$Q_{vz}$$. Otherwise, if $$M_{00} + M_{11} + M_{22} < 0$$, then we select one of the first three diagonal relationships based on the largest diagonal entry $$M_{00}$$, $$M_{11}$$, or $$M_{22}$$. After calculating $$Q_{vx}$$, $$Q_{vy}$$, or $$Q_{vz}$$, we plug its value into two of the first three off-diagonal relationships to solve for the other two values of $$Q_{vx}$$, $$Q_{vy}$$, and $$Q_{vz}$$. Finally, we plug it into one of the last three off-diagonal relationships to solve for $$Q_{vw}$$.
 
Setting $$t_x = M_{03}$$, $$t_y = M_{13}$$, and $$t_z = M_{23}$$, the values of $$Q_{mx}$$, $$Q_{my}$$, $$Q_{mz}$$, and $$Q_{mw}$$ are given by
 
$$Q_{mx} = \dfrac{1}{2}(Q_{vw} t_x + Q_{vz} t_y - Q_{vy} t_z)$$
 
$$Q_{my} = \dfrac{1}{2}(Q_{vw} t_y + Q_{vx} t_z - Q_{vz} t_x)$$
 
$$Q_{mz} = \dfrac{1}{2}(Q_{vw} t_z + Q_{vy} t_x - Q_{vx} t_y)$$


:$$a' = \mathcal F \mathbin{\unicode{x27D1}} a \mathbin{\unicode{x27D1}} \mathcal{\tilde F}$$
$$Q_{mw} = -\dfrac{1}{2}(Q_{vx} t_x + Q_{vy} t_y + Q_{vz} t_z)$$ .


The groups $$\mathrm E(2)$$ and $$\mathrm{RE}(2)$$ are isomorphic, and they each contain the orthogonal group $$\mathrm O(2)$$ as a common subgroup. The complement operation provides a two-way mapping between transforms associated with members of $$\mathrm E(2)$$ and $$\mathrm{RE}(2)$$.
== Motor Transformations ==


The invariant geometries of the four types of transforms described above are summarized in Table 1. The Euclidean isometries always fix a coinvariant contained in the horizon, and the corresponding dual transforms always fix a coinvariant containing the origin. In general, if $$x$$ is the primary invariant of a Euclidean isometry, then the complement of the weight of $$x$$ gives the coinvariant. Symmetrically, if $$x$$ is the primary invariant of a dual transform, then the complement of the bulk of $$x$$ gives the coinvariant. When the primary invariant of a Euclidean isometry contains the origin, there is a corresponding dual transform that performs the same operation. Symmetrically, when the primary invariant of a projective transform is contained in the horizon, there is a corresponding Euclidean isometry that performs the same operation. These are where $$\mathrm E(2)$$ and $$\mathrm{RE}(2)$$ intersect at $$\mathrm O(2)$$.
[[Point | Points]], [[Line | lines]], and [[Plane | planes]] are transformed by a [[unitized]] motor $$\mathbf Q$$ as shown in the following table.


{| class="wikitable"
{| class="wikitable"
|+ style="caption-side: bottom; padding-top: 0.5em; text-align: left; font-size: 92%; font-weight: normal;" | '''Table 1.''' These are the invariants of transforms occurring in the 3D projective geometric algebra representing the 2D plane. The primary invariant of any regular transform (a Euclidean isometry) or dual transform is given by the vector or bivector components of the operator itself. The coinvariant is given by the weight complement of the primary invariant in the case of regular transforms and by the bulk complement of the primary invariant in the case of dual transforms.
! Type || Transformation
|-
|-
! Transform !! Primary Invariant !! Coinvariant
| style="padding: 12px;" | [[Point]]
 
$$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$
| style="padding: 12px;" | $$\begin{split}\mathbf Q \mathbin{\unicode{x27C7}} \mathbf p \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}} =\, &\left[(1 - 2Q_{vy}^2 - 2Q_{vz}^2)p_x + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})p_y + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})p_z + 2(Q_{vy}Q_{mz} - Q_{vz}Q_{my} + Q_{vw}Q_{mx} - Q_{vx}Q_{mw})p_w\right]\mathbf e_1 \\ +\, &\left[(1 - 2Q_{vz}^2 -2Q_{vx}^2))p_y + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})p_z + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})p_x + 2(Q_{vz}Q_{mx} - Q_{vx}Q_{mz} + Q_{vw}Q_{my} - Q_{vy}Q_{mw})p_w\right]\mathbf e_2 \\ +\, &\left[(1 - 2Q_{vx}^2 - 2Q_{vy}^2))p_z + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})p_x + 2(Q_{vy}Q_{vz} + Q_{vx}Q_{vw})p_y + 2(Q_{vx}Q_{my} - Q_{vy}Q_{mx} + Q_{vw}Q_{mz} - Q_{vz}Q_{mw})p_w\right]\mathbf e_3 \\ +\, &p_w\mathbf e_4\end{split}$$
|-
|-
| style="padding-left: 1em; padding-right: 1em;" | Regular rotation<br />$$\mathbf Q = c_x \mathbf e_1 + c_y \mathbf e_2 + c_z \mathbf e_3 + r {\large\unicode{x1D7D9}}$$
| style="padding: 12px;" | [[Line]]
| style="padding-left: 1em; padding-right: 1em;" | Point $$c_x \mathbf e_1 + c_y \mathbf e_2 + c_z \mathbf e_3$$
 
| style="padding-left: 1em; padding-right: 1em;" | Horizon line $$\mathbf e_{12}$$
$$\begin{split}\boldsymbol l =\, &l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} \\ +\, &l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}\end{split}$$
| style="padding: 12px;" | $$\begin{split}\mathbf Q \mathbin{\unicode{x27C7}} \boldsymbol l \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}} =\, &\left[(1 - 2Q_{vy}^2 - 2Q_{vz}^2)l_{vx} + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})l_{vy} + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})l_{vz}\right]\mathbf e_{41} \\ +\, &\left[(1 - 2Q_{vz}^2 - 2Q_{vx}^2)l_{vy} + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})l_{vz} + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})l_{vx}\right]\mathbf e_{42} \\ +\, &\left[(1 - 2Q_{vx}^2 - 2Q_{vy}^2)l_{vz} + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})l_{vx} + 2(Q_{vy}Q_{vz} + Q_{vx}Q_{vw})l_{vy}\right]\mathbf e_{43} \\ +\, &\left[-4(Q_{vy}Q_{my} + Q_{vz}Q_{mz})l_{vx} + 2(Q_{vy}Q_{mx} + Q_{vx}Q_{my} - Q_{vz}Q_{mw} - Q_{vw}Q_{mz})l_{vy} + 2(Q_{vz}Q_{mx} + Q_{vx}Q_{mz} + Q_{vy}Q_{mw} + Q_{vw}Q_{my})l_{vz} + (1 - 2Q_{vy}^2 - 2Q_{vz}^2)l_{mx} + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})l_{my} + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})l_{mz}\right]\mathbf e_{23} \\ +\, &\left[-4(Q_{vz}Q_{mz} + Q_{vx}Q_{mx})l_{vy} + 2(Q_{vz}Q_{my} + Q_{vy}Q_{mz} - Q_{vx}Q_{mw} - Q_{vw}Q_{mx})l_{vz} + 2(Q_{vx}Q_{my} + Q_{vy}Q_{mx} + Q_{vz}Q_{mw} + Q_{vw}Q_{mz})l_{vx} + (1 - 2Q_{vz}^2 - 2Q_{vx}^2)l_{my} + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})l_{mz} + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})l_{mx}\right]\mathbf e_{31} \\ +\, &\left[-4(Q_{vx}Q_{mx} + Q_{vy}Q_{my})l_{vz} + 2(Q_{vx}Q_{mz} + Q_{vz}Q_{mx} - Q_{vy}Q_{mw} - Q_{vw}Q_{my})l_{vx} + 2(Q_{vy}Q_{mz} + Q_{vz}Q_{my} + Q_{vx}Q_{mw} + Q_{vw}Q_{mx})l_{vy} + (1 - 2Q_{vx}^2 - 2Q_{vy}^2)l_{mz} + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})l_{mx} + 2(Q_{vy}Q_{vz} + 2Q_{vx}Q_{vw})l_{my}\right]\mathbf e_{12}\end{split}$$
|-
|-
| style="padding-left: 1em; padding-right: 1em;" | Regular transflection<br />$$\mathbf F = s + h_x \mathbf e_{23} + h_y \mathbf e_{31} + h_z \mathbf e_{12}$$
| style="padding: 12px;" | [[Plane]]
| style="padding-left: 1em; padding-right: 1em;" | Line $$h_x \mathbf e_{23} + h_y \mathbf e_{31} + h_z \mathbf e_{12}$$
 
| style="padding-left: 1em; padding-right: 1em;" | Point in horizon $$h_x \mathbf e_1 + h_y \mathbf e_2$$
$$\mathbf g = g_x \mathbf e_{423} + g_y \mathbf e_{431} + g_z \mathbf e_{412} + g_w \mathbf e_{321}$$
|-
| style="padding: 12px;" | $$\begin{split}\mathbf Q \mathbin{\unicode{x27C7}} \mathbf g \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}} =\, &\left[(1 - 2Q_{vy}^2 - 2Q_{vz}^2)g_x + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})g_y + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})g_z\right]\mathbf e_{423} \\ +\, &\left[(1 - 2Q_{vz}^2 - 2Q_{vx}^2)g_y + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})g_z + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})g_x\right]\mathbf e_{431} \\ +\, &\left[(1 - 2Q_{vx}^2 - 2Q_{vy}^2)g_z + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})g_x + 2(Q_{vy}Q_{vz} + Q_{vx}Q_{vw})g_y\right]\mathbf e_{412} \\ +\, &\left[2(Q_{vy}Q_{mz} - Q_{vz}Q_{my} + Q_{vx}Q_{mw} - Q_{vw}Q_{mx})g_x + 2(Q_{vz}Q_{mx} - Q_{vx}Q_{mz} + Q_{vy}Q_{mw} - Q_{vw}Q_{my})g_y + 2(Q_{vx}Q_{my} - Q_{vy}Q_{mx} + Q_{vz}Q_{mw} - Q_{vw}Q_{mz})g_z + g_w\right]\mathbf e_{321}\end{split}$$
| style="padding-left: 1em; padding-right: 1em;" | Dual rotation<br />$$\mathcal Q = c_x \mathbf e_{23} + c_y \mathbf e_{31} + c_z \mathbf e_{12} - r$$
| style="padding-left: 1em; padding-right: 1em;" | Line $$c_x \mathbf e_{23} + c_y \mathbf e_{31} + c_z \mathbf e_{12}$$
| style="padding-left: 1em; padding-right: 1em;" | Origin point $$\mathbf e_3$$
|-
| style="padding-left: 1em; padding-right: 1em;" | Dual transflection<br />$$\mathcal F = h_x \mathbf e_1 + h_y \mathbf e_2 + h_z \mathbf e_3 - s {\large\unicode{x1D7D9}}$$
| style="padding-left: 1em; padding-right: 1em;" | Point $$h_x \mathbf e_1 + h_y \mathbf e_2 + h_z \mathbf e_3$$
| style="padding-left: 1em; padding-right: 1em;" | Line through origin $$h_x \mathbf e_{23} + h_y \mathbf e_{31}$$
|}
|}


In the four-dimensional projective geometric algebra $$\mathbb R_{3,0,1}$$ representing three-dimensional space, every Euclidean isometry is either a screw transform $$\mathbf Q$$ or a rotoreflection $$\mathbf F$$. (The only geometrical difference between these is that the displacement along the rotation axis in a screw transform is replaced by a reflection in a plane perpendicular to the rotation axis in a rotoreflection.) The primary invariant of a screw transform is its bivector components, which corresponds to the line about which a rotation is taking place. A rotoreflection can have two primary invariants, one associated with its vector components and a second associated with its trivector components. These invariants, the invariants of the corresponding projective transforms, and the coinvariants for each are summarized in Table 2.
== See Also ==


{| class="wikitable"
* [[Flector]]
|+ style="caption-side: bottom; padding-top: 0.5em; text-align: left; font-size: 92%; font-weight: normal;" | '''Table 2.''' These are the invariants of transforms occurring in the 4D projective geometric algebra representing 3D space. The primary invariant of any regular transform (a Euclidean isometry) or dual transform is given by the vector, bivector, or trivector components of the operator itself. The coinvariants are given by the weight complement of each primary invariant in the case of regular transforms and by the bulk complement of each primary invariant in the case of projective transforms.
* [[Translation]]
|-
* [[Rotation]]
! Transform !! Primary Invariants !! Coinvariants
|-
| style="padding-left: 1em; padding-right: 1em;" | Regular screw transform<br />$$\mathbf Q = r_x \mathbf e_{41} + r_y \mathbf e_{42} + r_z \mathbf e_{43} + r_w {\large\unicode{x1D7D9}} + u_x \mathbf e_{23} + u_y \mathbf e_{31} + u_z \mathbf e_{12} + u_w$$
| style="padding-left: 1em; padding-right: 1em;" | Line $$r_x \mathbf e_{41} + r_y \mathbf e_{42} + r_z \mathbf e_{43} + u_x \mathbf e_{23} + u_y \mathbf e_{31} + u_z \mathbf e_{12}$$
| style="padding-left: 1em; padding-right: 1em;" | Line in horizon $$r_x \mathbf e_{23} + r_y \mathbf e_{31} + r_z \mathbf e_{12}$$
|-
| style="padding-left: 1em; padding-right: 1em;" | Regular rotoreflection<br />$$\mathbf F = s_x \mathbf e_1 + s_y \mathbf e_2 + s_z \mathbf e_3 + s_w \mathbf e_4 + h_x \mathbf e_{423} + h_y \mathbf e_{431} + h_z \mathbf e_{412} + h_w \mathbf e_{321}$$
| style="padding-left: 1em; padding-right: 1em;" | Plane $$h_x \mathbf e_{423} + h_y \mathbf e_{431} + h_z \mathbf e_{412} + h_w \mathbf e_{321}$$,<br />Point $$s_x \mathbf e_1 + s_y \mathbf e_2 + s_z \mathbf e_3 + s_w \mathbf e_4$$
| style="padding-left: 1em; padding-right: 1em;" | Point in horizon $$h_x \mathbf e_1 + h_y \mathbf e_2 + h_z \mathbf e_3$$,<br />Horizon $$e_{321}$$
|-
| style="padding-left: 1em; padding-right: 1em;" | Dual screw transform<br />$$\mathbf Q = u_x \mathbf e_{41} + u_y \mathbf e_{42} + u_z \mathbf e_{43} - u_w {\large\unicode{x1D7D9}} + r_x \mathbf e_{23} + r_y \mathbf e_{31} + r_z \mathbf e_{12} - r_w$$
| style="padding-left: 1em; padding-right: 1em;" | Line $$u_x \mathbf e_{41} + u_y \mathbf e_{42} + u_z \mathbf e_{43} + r_x \mathbf e_{23} + r_y \mathbf e_{31} + r_z \mathbf e_{12}$$
| style="padding-left: 1em; padding-right: 1em;" | Line through origin $$r_x \mathbf e_{41} + r_y \mathbf e_{42} + r_z \mathbf e_{43}$$
|-
| style="padding-left: 1em; padding-right: 1em;" | Dual rotoreflection<br />$$\mathcal F = h_x \mathbf e_1 + h_y \mathbf e_2 + h_z \mathbf e_3 - s {\large\unicode{x1D7D9}}$$
| style="padding-left: 1em; padding-right: 1em;" | Point $$h_x \mathbf e_1 + h_y \mathbf e_2 + h_z \mathbf e_3 + h_w \mathbf e_4$$,<br />Plane $$s_x \mathbf e_{423} + s_y \mathbf e_{431} + s_z \mathbf e_{412} + s_w \mathbf e_{321}$$
| style="padding-left: 1em; padding-right: 1em;" | Plane through origin $$h_x \mathbf e_{423} + h_y \mathbf e_{431} + h_z \mathbf e_{412}$$,<br />Origin $$\mathbf e_4$$
|}

Revision as of 05:36, 6 August 2023

Figure 1. A motor represents a proper Euclidean isometry, which can always be regarded as a rotation about a line $$\boldsymbol L$$ and a displacement along the same line.

A motor is an operator that performs a proper isometry in Euclidean space. Such isometries encompass all possible combinations of any number of rotations and translations. The name motor is a portmanteau of motion operator or moment vector. Motors are equivalent to the set of dual quaternions used in conventional theories, and the functionality is properly generalized in rigid geometric algebra. Motors cannot perform improper isometries that include an odd number of reflections; those are instead performed by operators called flectors.

In the 4D rigid geometric algebra $$\mathcal G_{3,0,1}$$, a motor $$\mathbf Q$$ has the general form

$$\mathbf Q = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1d7d9}} + Q_{mx} \mathbf e_{23} + Q_{my} \mathbf e_{31} + Q_{mz} \mathbf e_{12} + Q_{mw} \mathbf 1$$ .

To possess the geometric property, the components of $$\mathbf Q$$ must satisfy the equation

$$Q_{vx} Q_{mx} + Q_{vy} Q_{my} + Q_{vz} Q_{mz} + Q_{vw} Q_{mw} = 0$$ .

Motors are capable of representing any general screw motion consisting of a rotation about a line combined with a displacement along the same line. By Chasles' theorem, all proper rigid motions in 3D space are screw motions.

An element $$\mathbf x$$ is transformed by a motor $$\mathbf Q$$ through the operation $$\mathbf x' = \mathbf Q \mathbin{\unicode{x27C7}} \mathbf x \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}$$, where $$\unicode{x27C7}$$ is the geometric antiproduct.

The set of all motors, sometimes denoted by $$\unicode{x1D544}$$, forms a subgroup of index 2 in a geometric algebra. Its coset is the set of flectors.

Simple Motors

If $$Q_{mw} = 0$$ (i.e., the scalar part $$Q_{\mathbf 1}$$ is zero), then the motor $$\mathbf Q$$ is called a simple motor. Every simple motor represents either a pure translation or a pure rotation about a line without any displacement along that line.

In the case of a pure translation, the motor $$\mathbf T$$ is given by

$$\mathbf T = t_x \mathbf e_{23} + t_y \mathbf e_{31} + t_z \mathbf e_{12} + {\large\unicode{x1d7d9}}$$ ,

and this performs a translation by twice the displacement vector $$(t_x, t_y, t_z)$$.

In the case of a pure rotation, the motor $$\mathbf R$$ is given by

$$\mathbf R = \boldsymbol l\sin\phi + {\large\unicode{x1d7d9}}\cos\phi$$ ,

and this performs a rotation by twice the angle $$\phi$$ about the line $$\boldsymbol l$$.

Motors Built from Geometry

Motors having specific geometric constructions can be built from points, lines, and planes as shown in the following table.

Motor Description
$$\begin{split}\mathbf h \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{g}}} =\, &(g_yh_z - g_zh_y)\mathbf e_{41} + (g_zh_x - g_xh_z)\mathbf e_{42} + (g_xh_y - g_yh_x)\mathbf e_{43} \\ +\, &(h_xg_w - g_xh_w)\mathbf e_{23} + (h_yg_w - g_yh_w)\mathbf e_{31} + (h_zg_w - g_zh_w)\mathbf e_{12} \\ +\, &(g_xh_x + g_yh_y + g_zh_z)\smash{\large\unicode{x1d7d9}}\end{split}$$ Rotation about the line where planes $$\mathbf g$$ and $$\mathbf h$$ intersect by twice the angle between them in the direction from $$\mathbf g$$ to $$\mathbf h$$.

$$\mathbf g = g_x \mathbf e_{234} + g_y \mathbf e_{314} + g_z \mathbf e_{124} + g_w \mathbf e_{321}$$

$$\mathbf h = h_x \mathbf e_{234} + h_y \mathbf e_{314} + h_z \mathbf e_{124} + h_w \mathbf e_{321}$$

$$\begin{split}\boldsymbol l \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{k}}} =\, &(l_{vz} k_{vy} - l_{vy} k_{vz})\mathbf e_{41} + (l_{vx} k_{vz} - l_{vz} k_{vx})\mathbf e_{42} + (l_{vy} k_{vx} - l_{vx} k_{vy})\mathbf e_{43} \\ +\, &(l_{vz} k_{my} - l_{vy} k_{mz} + l_{mz} k_{vy} - l_{my} k_{vz})\mathbf e_{23} + (l_{vx} k_{mz} - l_{vz} k_{mx} + l_{mx} k_{vz} - l_{mz} k_{vx})\mathbf e_{31} + (l_{vy} k_{mx} - l_{vx} k_{my} + l_{my} k_{vx} - l_{mx} k_{vy})\mathbf e_{12} \\ +\, &(l_{vx} k_{mx} + l_{vy} k_{my} + l_{vz} k_{mz} + l_{mx} k_{vx} + l_{my} k_{vy} + l_{mz} k_{vz}) \\ +\, &(l_{vx} k_{vx} + l_{vy} k_{vy} + l_{vz} k_{vz})\smash{\large\unicode{x1d7d9}}\end{split}$$ Rotation about the line containing the closest points on lines $$\mathbf k$$ and $$\boldsymbol l$$ by twice the angle between the directions $$(l_{vx}, l_{vy}, l_{vz})$$ and $$(k_{vx}, k_{vy}, k_{vz})$$, and translation by twice the distance between the lines in the direction from $$\mathbf k$$ to $$\boldsymbol l$$.

$$\mathbf k = k_{vx} \mathbf e_{41} + k_{vy} \mathbf e_{42} + k_{vz} \mathbf e_{43} + k_{mx} \mathbf e_{23} + k_{my} \mathbf e_{31} + k_{mz} \mathbf e_{12}$$

$$\boldsymbol l = l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}$$

$$\mathbf q \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{p}}} = (q_xp_w - p_xq_w)\mathbf e_{23} + (q_yp_w - p_yq_w)\mathbf e_{31} + (q_zp_w - p_zq_w)\mathbf e_{12} + (p_wq_w)\smash{\large\unicode{x1d7d9}}$$ Translation by twice the distance between points $$\mathbf p$$ and $$\mathbf q$$ in the direction from $$\mathbf p$$ to $$\mathbf q$$.

$$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$

$$\mathbf q = q_x \mathbf e_1 + q_y \mathbf e_2 + q_z \mathbf e_3 + q_w \mathbf e_4$$

Norm

The bulk norm of a motor $$\mathbf Q$$ is given by

$$\left\Vert\mathbf Q\right\Vert_\unicode{x25CF} = \sqrt{\mathbf Q \mathbin{\unicode{x25CF}} \mathbf{\tilde Q}} = \sqrt{Q_{mx}^2 + Q_{my}^2 + Q_{mz}^2 + Q_{mw}^2}$$ ,

and its weight norm is given by

$$\left\Vert\mathbf Q\right\Vert_\unicode{x25CB} = \sqrt{\mathbf Q \mathbin{\unicode{x25CB}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}}\vphantom{\mathbf{\tilde Q}}} = {\large\unicode{x1D7D9}}\sqrt{Q_{vx}^2 + Q_{vy}^2 + Q_{vz}^2 + Q_{vw}^2}$$ .

The geometric norm of a motor $$\mathbf Q$$ is thus

$$\widehat{\left\Vert\mathbf Q\right\Vert} = \sqrt{\dfrac{Q_{mx}^2 + Q_{my}^2 + Q_{mz}^2 + Q_{mw}^2}{Q_{vx}^2 + Q_{vy}^2 + Q_{vz}^2 + Q_{vw}^2}}$$ ,

and this is equal to half the distance that the origin is moved by the operator.

A motor is unitized when $$Q_{vx}^2 + Q_{vy}^2 + Q_{vz}^2 + Q_{vw}^2 = 1$$.

Exponential Form

A motor $$\mathbf Q$$ can be expressed as the exponential of a unitized line $$\boldsymbol l$$ multiplied by $$d + \phi{\large\unicode{x1D7D9}}$$, where $$\phi$$ is half the angle of rotation about the line $$\boldsymbol l$$, and $$d$$ is half the displacement distance along the line $$\boldsymbol l$$. The exponential form can be written as

$$\mathbf Q = \exp_\unicode{x27C7}((d + \phi{\large\unicode{x1D7D9}}) \mathbin{\unicode{x27C7}} \boldsymbol l) = \cos_\unicode{x27C7}(d + \phi{\large\unicode{x1D7D9}}) + \sin_\unicode{x27C7}(d + \phi{\large\unicode{x1D7D9}}) \mathbin{\unicode{x27C7}} \boldsymbol l$$ .

This expands to

$$\mathbf Q = \boldsymbol l\sin\phi + (d \mathbin{\unicode{x27C7}} \boldsymbol l)\cos\phi - d\sin\phi + {\large\unicode{x1D7D9}}\cos\phi$$ ,

and replacing the line $$\boldsymbol l$$ with its components gives us

$$\mathbf Q = (l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} + l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12})\sin\phi + d(l_{vx} \mathbf e_{23} + l_{vy} \mathbf e_{31} + l_{vz} \mathbf e_{12})\cos\phi - d\sin\phi + {\large\unicode{x1D7D9}}\cos\phi$$ .

A motor in exponential form is always unitized, and its geometric norm can be written as

$$\widehat{\left\Vert\mathbf Q\right\Vert} = \sqrt{d^2 + (l_{mx}^2 + l_{my}^2 + l_{mz}^2)\sin^2\phi}$$ .

The quantity $$d + \phi{\large\unicode{x1D7D9}}$$, as a homogeneous magnitude, is the pitch of the screw transformation performed by the motor, which is the amount of translation along the screw axis per radian of rotation.

Logarithm

Given an arbitrary unitized motor $$\mathbf Q$$ with $$|Q_{vw}| < 1$$, the values of $$\boldsymbol l$$, $$d$$, and $$\phi$$ can be determined from the components of $$\mathbf Q$$, essentially taking the logarithm of a motor. First, the scalar and antiscalar terms satisfy the equations

$$Q_{mw} = -d\sin\phi$$
$$Q_{vw} = \cos\phi$$ .

Assuming that $$\phi > 0$$, we define $$s = \sin\phi = \sqrt{1 - Q_{vw}^2}$$. Then,

$$d = -\dfrac{Q_{mw}}{s}$$
$$\phi = \tan^{-1}\left(\dfrac{s}{Q_{vw}}\right)$$ .

If $$Q_{vw} = 0$$, then we assign $$\phi = \pi/2$$. If $$Q_{vw} < 0$$, then we can add $$\pi$$ to $$\phi$$ to make the angle positive, but this is not required.

The components of $$\boldsymbol l$$ are then given by

$$(l_{vx}, l_{vy}, l_{vz}) = \dfrac{1}{s}(Q_{vx}, Q_{vy}, Q_{vz})$$
$$(l_{mx}, l_{my}, l_{mz}) = \dfrac{1}{s}\left[(Q_{mx}, Q_{my}, Q_{mz}) - \dfrac{dQ_{vw}}{s}(Q_{vx}, Q_{vy}, Q_{vz})\right] = \dfrac{1}{s}\left[(Q_{mx}, Q_{my}, Q_{mz}) + \dfrac{Q_{vw} Q_{mw}}{s^2}(Q_{vx}, Q_{vy}, Q_{vz})\right]$$ .

In the special case that $$Q_{vw} = \pm 1$$, the motor must be a pure translation with $$\phi = 0$$ and $$(Q_{vx}, Q_{vy}, Q_{vz}) = (0, 0, 0)$$.

Square Root

The square root of a quaternion can be calculated by summing with the identity and renormalizing. If we attempt to do the same thing with a motor $$\mathbf Q$$, we find that it works only under certain conditions. However, such a calculation does give us a starting point from which we can make a correction for the general case.

Using the exponential form of $$\mathbf Q$$, we first examine the weight norm of $$\mathbf Q + {\large\unicode{x1D7D9}}$$ and find that

$$\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB} = \sqrt{\sin^2\phi + (\cos\phi + 1)^2} = \sqrt{\vphantom{\sin^2\phi}2 + 2Q_\smash{\large\unicode{x1D7D9}}}$$ ,

where $$Q_{\large\unicode{x1D7D9}} = \cos\phi$$. Applying the trigonometric identity $$\cos^2(\phi/2) = (1 + \cos\phi)/2$$, we can rewrite this as

$$\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB} = 2\cos(\phi/2)$$ .

Applying several more trigonometric identities, we now observe

$$\dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB}} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{2\cos(\phi/2)} = \boldsymbol l\sin(\phi/2) + \left(\dfrac{d}{2} \mathbin{\unicode{x27C7}} \boldsymbol l\right)\cos(\phi/2) - \dfrac{d}{2}\sin(\phi/2) + {\large\unicode{x1D7D9}}\cos(\phi/2) - \left(\dfrac{d}{2} \mathbin{\unicode{x27C7}} \boldsymbol l\right)\sin(\phi/2)\tan(\phi/2) - \dfrac{d}{2}\sin(\phi/2)$$ .

The first four terms are exactly the square root of $$\mathbf Q$$ because the distance $$d$$ and angle $$\phi$$ have both been halved. But there are two additional terms that we need to eliminate. It just so happens that

$$\dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB}} \mathbin{\unicode{x27C7}} \dfrac{d}{2}\tan(\phi/2) = \left(\dfrac{d}{2} \mathbin{\unicode{x27C7}} \boldsymbol l\right)\sin(\phi/2)\tan(\phi/2) + \dfrac{d}{2}\sin(\phi/2)$$ ,

which means that

$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} \mathbin{\unicode{x27C7}} \left({\large\unicode{x1D7D9}} + \dfrac{d}{2}\tan(\phi/2)\right)$$ .

The scalar component of $$\mathbf Q$$ can be rewritten as $$Q_{\mathbf 1} = -d\sin\phi = -2d\sin(\phi/2)\cos(\phi/2) $$. Dividing this by $$2 + 2Q_{\large\unicode{x1D7D9}} = 4\cos^2(\phi/2)$$ yields

$$\dfrac{Q_\mathbf 1}{2 + 2Q_{\large\unicode{x1D7D9}}} = -\dfrac{d}{2}\tan(\phi/2)$$ .

The square root of a general motor $$\mathbf Q$$ is thus given by

$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} \mathbin{\unicode{x27C7}} \left({\large\unicode{x1D7D9}} - \dfrac{Q_\mathbf 1}{2 + 2Q_{\large\unicode{x1D7D9}}}\right)$$ .

If $$\mathbf Q$$ is a simple motor, then $$Q_{\mathbf 1} = 0$$, and this reduces to

$$\sqrt{\mathbf Q} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\sqrt{2 + 2Q_\smash{\large\unicode{x1D7D9}}}} = \dfrac{\mathbf Q + {\large\unicode{x1D7D9}}}{\left\Vert\mathbf Q + \smash{\large\unicode{x1D7D9}}\right\Vert_\unicode{x25CB}}$$ .

Factorization

Any unitized motor $$\mathbf Q$$ can be factored into the product of a motor $$\mathbf R = Q_{vx} \mathbf e_{41} + Q_{vy} \mathbf e_{42} + Q_{vz} \mathbf e_{43} + Q_{vw} {\large\unicode{x1D7D9}}$$ corresponding to a pure rotation about a line through the origin and a motor $$\mathbf T = {t_x \mathbf e_{23} + t_y \mathbf e_{31} + t_z \mathbf e_{12} + \large\unicode{x1d7d9}}$$ corresponding to a pure translation by calculating

$$\mathbf T = \mathbf Q \mathbin{\unicode{x27C7}} \mathbf{\underset{\Large\unicode{x7E}}{R}}$$ .

The motor $$\mathbf T$$ is then given by

$$\mathbf T = (Q_{vy} Q_{mz} - Q_{vz} Q_{my} + Q_{mx} Q_{vw} - Q_{vx} Q_{mw})\mathbf e_{23} + (Q_{vz} Q_{mx} - Q_{vx} Q_{mz} + Q_{my} Q_{vw} - Q_{vy} Q_{mw})\mathbf e_{31} + (Q_{vx} Q_{my} - Q_{vy} Q_{mx} + Q_{mz} Q_{vw} - Q_{vz} Q_{mw})\mathbf e_{12} + {\large\unicode{x1d7d9}}$$ .

The original motor $$\mathbf Q$$ can now be expressed as

$$\mathbf Q = \mathbf T \mathbin{\unicode{x27C7}} \mathbf R$$ ,

where both $$\mathbf R$$ and $$\mathbf T$$ are unitized simple motors.

Conversion from Motor to Matrix

Given a specific unitized motor $$\mathbf Q$$, define the matrices

$$\mathbf A = \begin{bmatrix}1 - 2(Q_{vy}^2 + Q_{vz}^2) & 2Q_{vx} Q_{vy} & 2Q_{vz} Q_{vx} & 2(Q_{vy} Q_{mz} - Q_{vz} Q_{my}) \\ 2Q_{vx} Q_{vy} & 1 - 2(Q_{vz}^2 + Q_{vx}^2) & 2Q_{vy} Q_{vz} & 2(Q_{vz} Q_{mx} - Q_{vx} Q_{mz}) \\ 2Q_{vz} Q_{vx} & 2Q_{vy} Q_{vz} & 1 - 2(Q_{vx}^2 + Q_{vy}^2) & 2(Q_{vx} Q_{my} - Q_{vy} Q_{mx}) \\ 0 & 0 & 0 & 1\end{bmatrix}$$

and

$$\mathbf B = \begin{bmatrix}0 & -2Q_{vz} Q_{vw} & 2Q_{vy} Q_{vw} & 2(Q_{vw} Q_{mx} - Q_{vx} Q_{mw}) \\ 2Q_{vz} Q_{vw} & 0 & -2Q_{vx} Q_{vw} & 2(Q_{vw} Q_{my} - Q_{vy} Q_{mw}) \\ -2Q_{vy} Q_{vw} & 2Q_{vx} Q_{vw} & 0 & 2(Q_{vw} Q_{mz} - Q_{vz} Q_{mw}) \\ 0 & 0 & 0 & 0\end{bmatrix}$$ .

Then the corresponding 4×4 matrix $$\mathbf M$$ that transforms a point $$\mathbf p$$, regarded as a column matrix, as $$\mathbf p' = \mathbf{Mp}$$ is given by

$$\mathbf M = \mathbf A + \mathbf B$$ .

The inverse of $$\mathbf M$$, which transforms a plane $$\mathbf g$$, regarded as a row matrix, as $$\mathbf g' = \mathbf{gM^{-1}}$$ is given by

$$\mathbf M^{-1} = \mathbf A - \mathbf B$$ .

Conversion from Matrix to Motor

Let $$\mathbf M$$ be an orthogonal 4×4 matrix with determinant +1 having the form

$$\mathbf M = \begin{bmatrix} M_{00} & M_{01} & M_{02} & M_{03} \\ M_{10} & M_{11} & M_{12} & M_{13} \\ M_{20} & M_{21} & M_{22} & M_{23} \\ 0 & 0 & 0 & 1\end{bmatrix}$$ .

Then, by equating the entries of $$\mathbf M$$ to the entries of $$\mathbf A + \mathbf B$$ from above, we have the following four relationships based on the diagonal entries of $$\mathbf M$$:

$$M_{00} - M_{11} - M_{22} + 1 = 4Q_{vx}^2$$
$$M_{11} - M_{22} - M_{00} + 1 = 4Q_{vy}^2$$
$$M_{22} - M_{00} - M_{11} + 1 = 4Q_{vz}^2$$
$$M_{00} + M_{11} + M_{22} + 1 = 4(1 - Q_{vx}^2 - Q_{vy}^2 - Q_{vz}^2) = 4Q_{vw}^2$$

And we have the following six relationships based on the off-diagonal entries of $$\mathbf M$$:

$$M_{21} + M_{12} = 4Q_{vy} Q_{vz}$$
$$M_{02} + M_{20} = 4Q_{vz} Q_{vx}$$
$$M_{10} + M_{01} = 4Q_{vx} Q_{vy}$$
$$M_{21} - M_{12} = 4Q_{vx} Q_{vw}$$
$$M_{02} - M_{20} = 4Q_{vy} Q_{vw}$$
$$M_{10} - M_{01} = 4Q_{vz} Q_{vw}$$

If $$M_{00} + M_{11} + M_{22} \geq 0$$, then we calculate

$$Q_{vw} = \pm \dfrac{1}{2}\sqrt{M_{00} + M_{11} + M_{22} + 1}$$ ,

where either sign can be chosen. In this case, we know $$|Q_{vw}|$$ is at least $$1/2$$, so we can safely divide by $$4Q_{vw}$$ in the last three off-diagonal relationships above to solve for $$Q_{vx}$$, $$Q_{vy}$$, and $$Q_{vz}$$. Otherwise, if $$M_{00} + M_{11} + M_{22} < 0$$, then we select one of the first three diagonal relationships based on the largest diagonal entry $$M_{00}$$, $$M_{11}$$, or $$M_{22}$$. After calculating $$Q_{vx}$$, $$Q_{vy}$$, or $$Q_{vz}$$, we plug its value into two of the first three off-diagonal relationships to solve for the other two values of $$Q_{vx}$$, $$Q_{vy}$$, and $$Q_{vz}$$. Finally, we plug it into one of the last three off-diagonal relationships to solve for $$Q_{vw}$$.

Setting $$t_x = M_{03}$$, $$t_y = M_{13}$$, and $$t_z = M_{23}$$, the values of $$Q_{mx}$$, $$Q_{my}$$, $$Q_{mz}$$, and $$Q_{mw}$$ are given by

$$Q_{mx} = \dfrac{1}{2}(Q_{vw} t_x + Q_{vz} t_y - Q_{vy} t_z)$$

$$Q_{my} = \dfrac{1}{2}(Q_{vw} t_y + Q_{vx} t_z - Q_{vz} t_x)$$

$$Q_{mz} = \dfrac{1}{2}(Q_{vw} t_z + Q_{vy} t_x - Q_{vx} t_y)$$

$$Q_{mw} = -\dfrac{1}{2}(Q_{vx} t_x + Q_{vy} t_y + Q_{vz} t_z)$$ .

Motor Transformations

Points, lines, and planes are transformed by a unitized motor $$\mathbf Q$$ as shown in the following table.

Type Transformation
Point

$$\mathbf p = p_x \mathbf e_1 + p_y \mathbf e_2 + p_z \mathbf e_3 + p_w \mathbf e_4$$

$$\begin{split}\mathbf Q \mathbin{\unicode{x27C7}} \mathbf p \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}} =\, &\left[(1 - 2Q_{vy}^2 - 2Q_{vz}^2)p_x + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})p_y + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})p_z + 2(Q_{vy}Q_{mz} - Q_{vz}Q_{my} + Q_{vw}Q_{mx} - Q_{vx}Q_{mw})p_w\right]\mathbf e_1 \\ +\, &\left[(1 - 2Q_{vz}^2 -2Q_{vx}^2))p_y + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})p_z + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})p_x + 2(Q_{vz}Q_{mx} - Q_{vx}Q_{mz} + Q_{vw}Q_{my} - Q_{vy}Q_{mw})p_w\right]\mathbf e_2 \\ +\, &\left[(1 - 2Q_{vx}^2 - 2Q_{vy}^2))p_z + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})p_x + 2(Q_{vy}Q_{vz} + Q_{vx}Q_{vw})p_y + 2(Q_{vx}Q_{my} - Q_{vy}Q_{mx} + Q_{vw}Q_{mz} - Q_{vz}Q_{mw})p_w\right]\mathbf e_3 \\ +\, &p_w\mathbf e_4\end{split}$$
Line

$$\begin{split}\boldsymbol l =\, &l_{vx} \mathbf e_{41} + l_{vy} \mathbf e_{42} + l_{vz} \mathbf e_{43} \\ +\, &l_{mx} \mathbf e_{23} + l_{my} \mathbf e_{31} + l_{mz} \mathbf e_{12}\end{split}$$

$$\begin{split}\mathbf Q \mathbin{\unicode{x27C7}} \boldsymbol l \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}} =\, &\left[(1 - 2Q_{vy}^2 - 2Q_{vz}^2)l_{vx} + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})l_{vy} + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})l_{vz}\right]\mathbf e_{41} \\ +\, &\left[(1 - 2Q_{vz}^2 - 2Q_{vx}^2)l_{vy} + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})l_{vz} + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})l_{vx}\right]\mathbf e_{42} \\ +\, &\left[(1 - 2Q_{vx}^2 - 2Q_{vy}^2)l_{vz} + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})l_{vx} + 2(Q_{vy}Q_{vz} + Q_{vx}Q_{vw})l_{vy}\right]\mathbf e_{43} \\ +\, &\left[-4(Q_{vy}Q_{my} + Q_{vz}Q_{mz})l_{vx} + 2(Q_{vy}Q_{mx} + Q_{vx}Q_{my} - Q_{vz}Q_{mw} - Q_{vw}Q_{mz})l_{vy} + 2(Q_{vz}Q_{mx} + Q_{vx}Q_{mz} + Q_{vy}Q_{mw} + Q_{vw}Q_{my})l_{vz} + (1 - 2Q_{vy}^2 - 2Q_{vz}^2)l_{mx} + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})l_{my} + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})l_{mz}\right]\mathbf e_{23} \\ +\, &\left[-4(Q_{vz}Q_{mz} + Q_{vx}Q_{mx})l_{vy} + 2(Q_{vz}Q_{my} + Q_{vy}Q_{mz} - Q_{vx}Q_{mw} - Q_{vw}Q_{mx})l_{vz} + 2(Q_{vx}Q_{my} + Q_{vy}Q_{mx} + Q_{vz}Q_{mw} + Q_{vw}Q_{mz})l_{vx} + (1 - 2Q_{vz}^2 - 2Q_{vx}^2)l_{my} + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})l_{mz} + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})l_{mx}\right]\mathbf e_{31} \\ +\, &\left[-4(Q_{vx}Q_{mx} + Q_{vy}Q_{my})l_{vz} + 2(Q_{vx}Q_{mz} + Q_{vz}Q_{mx} - Q_{vy}Q_{mw} - Q_{vw}Q_{my})l_{vx} + 2(Q_{vy}Q_{mz} + Q_{vz}Q_{my} + Q_{vx}Q_{mw} + Q_{vw}Q_{mx})l_{vy} + (1 - 2Q_{vx}^2 - 2Q_{vy}^2)l_{mz} + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})l_{mx} + 2(Q_{vy}Q_{vz} + 2Q_{vx}Q_{vw})l_{my}\right]\mathbf e_{12}\end{split}$$
Plane

$$\mathbf g = g_x \mathbf e_{423} + g_y \mathbf e_{431} + g_z \mathbf e_{412} + g_w \mathbf e_{321}$$

$$\begin{split}\mathbf Q \mathbin{\unicode{x27C7}} \mathbf g \mathbin{\unicode{x27C7}} \smash{\mathbf{\underset{\Large\unicode{x7E}}{Q}}} =\, &\left[(1 - 2Q_{vy}^2 - 2Q_{vz}^2)g_x + 2(Q_{vx}Q_{vy} - Q_{vz}Q_{vw})g_y + 2(Q_{vz}Q_{vx} + Q_{vy}Q_{vw})g_z\right]\mathbf e_{423} \\ +\, &\left[(1 - 2Q_{vz}^2 - 2Q_{vx}^2)g_y + 2(Q_{vy}Q_{vz} - Q_{vx}Q_{vw})g_z + 2(Q_{vx}Q_{vy} + Q_{vz}Q_{vw})g_x\right]\mathbf e_{431} \\ +\, &\left[(1 - 2Q_{vx}^2 - 2Q_{vy}^2)g_z + 2(Q_{vz}Q_{vx} - Q_{vy}Q_{vw})g_x + 2(Q_{vy}Q_{vz} + Q_{vx}Q_{vw})g_y\right]\mathbf e_{412} \\ +\, &\left[2(Q_{vy}Q_{mz} - Q_{vz}Q_{my} + Q_{vx}Q_{mw} - Q_{vw}Q_{mx})g_x + 2(Q_{vz}Q_{mx} - Q_{vx}Q_{mz} + Q_{vy}Q_{mw} - Q_{vw}Q_{my})g_y + 2(Q_{vx}Q_{my} - Q_{vy}Q_{mx} + Q_{vz}Q_{mw} - Q_{vw}Q_{mz})g_z + g_w\right]\mathbf e_{321}\end{split}$$

See Also

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