# On the instantaneous screw axes of two parameter motions in Lorentzian space.

[section]1. Introduction and preliminaries

Let [IR.sup.n] = {([r.sub.1],[r.sub.2],... ,[r.sub.n]) | [r.sub.1],[r.sub.1],... ,[r.sub.n] [member of] IR} be a n-dimensional vector space, r = ([r.sub.1],[r.sub.2],... ,[r.sub.n]) and s = ([s.sub.1],[s.sub.2],... ,[s.sub.n]) be two vectors in [IR.sup.n], the Lorentz scalar product of r and s is defined by

[<r,s>.sub.L] = [-r.sub.1][s.sub.1] + [r.sub.2][s.sub.2] + ... + [r.sub.n][s.sub.n].

[L.sup.n] = ([IR.sup.n], [<,>.sub.L]) is called n-dimensional Lorentz space, or Minkowski n-space. We denote [L.sup.] as ([IR.sup.n], [<,>.sub.L]). For any r = ([r.sub.1], [r.sub.2], [r.sub.3]), s = ([s.sub.1],[s.sub.2], [s.sub.3]) [member of] [L.sup.3], in the meaning Lorentz vector product of r and s is defined by

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

where [e.sub.1] [[conjunction].sub.L] [e.sub.2] = [-e.sub.3], [e.sub.2] [[conjunction].sub.L] [e.sub.3] = [e.sub.1] vector, a lightlike vector or a timelike vector if [<r,r>.sub.L] > 0, [<r,r>.sub.L] = 0 or [<r,r>.sub.L] < 0 respectively. For r [member of] [L.sup.n], the norm of r defined by [[parallel]r[parallel].sub.L] [square root |[<r,r>.sub.L]|], and r is called a unit vector if [[parallel]r[parallel].sub.L] = 1 [5]. In the Minkowski n-space, the two parameter motion of a rigid body is defined by

(1.1) Y([lambda],[mu]) = A([lambda],[mu])X + C([lambda],[mu]),

where A [member of] SO(n,1) is a positive semi orthogonal matrix, C [member of] [IR.sup.n.sub.1] is a column matrix, Y and X are position vectories of the same point B respectively, for the fixed and moving space with respect to semi orthonormal coordinate systems. The two parameter motion is given by (1.1), for ([lambda],[mu]) = (0,0), we have

A(0,0) = [A.sup.-1](0,0) = [A.sup.T] (0,0) = I

and

C(0,0) = 0.

Then fixed and moving space is coincided. If [lambda] = [lambda](t),[mu] = [mu](t), then one parameter motion is obtained from two parameter motion. Since A [member of] SO(n,1), we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

For the sake of the short we shall take as [A.sup.T]([lambda],[mu]) = [A.sup.T] and A([lambda],[mu]) = A.

Definition 1. Taking the derivation with respect to tin equation Y([lambda],[mu]) = A([lambda],[mu])X + C([lambda],[mu]) where let [lambda] = [lambda](t) and [mu] = [mu](t), then it follows that

[??] = [Y.sub.[lambda]][??] + [Y.sub.[mu]][??],

[??] = [A.sub.[lambda]][??] + [A.sub.[mu]][??],

[??] = [C.sub.[lambda]][??] + [C.sub.[mu]][??],

[??] = [??]X + [??] + A[??].

So [??],[??]X +[??], A[??] are called absolutely sliding and relative velocities of the point B has position vectories [??] , respectively. Let X be solution of system of [[??].sub.f] = [??]X + [??] = 0 and the solution is constant on the fixed and moving space at position t. These points X is called instantaneous pole points at every position t.

Definition 2. If rank [??] = n - 1 = r be an even number on the two parameter motion given by equation Y([lambda],[mu]) = A([lambda],[mu])X +C([lambda],[mu]), then at any position of points the locus having a velocity vector with stationary norm is a line. The line is called instantaneous screw axis and denoted by I.S.A. [2]. Furthermore the moving space screw axis is defined by X = P + [sigma] E where P is a particular solution of equation [??]X + [??] = 0 and E represent a bases of solution space of homogeneous equation [??]X = 0.

[section]2. The instantaneous screw axes of two parameter motions

Theorem 1. Let A [member of] SO(n,1) and let n be an odd number. Then the rank of [A.sub.[lambda]] and [A.sub.[mu]] are even.

Proof. Since

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and A(0,0) = [A.sup.T](0,0) = I, then

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[A.sub.[lambda]] + [epsilon][A.sup.T.sub.[lambda]][epsilon = 0

and

[A.sub.[mu]][epsilon][A.sup.T.][epsilon] + A[epsilon][A.sup.T.sub.[mu]][epsilon] = 0

[A.sub.[mu]] + [epsilon][A.sup.T.sub.[mu]][epsilon] = 0.

Thus [A.sub.[lambda]] and [A.sub.[mu]] are semi skew-symmetric matrices. Since n is an odd number it follows that

det [A.sub.[lambda]] = 0,

det [A.sub.[mu]] = 0.

Thus it must be rank ([A.sub.[lambda]]) = r (even) and rank ([A.sub.[mu]]) = r (even).

Theorem 2. Let A [member of] SO(n, 1). Then

rank [A.sub.[lambda][lambda]] = 0 [??] rank [A.sub.[lambda]] = 0

and

rank [A.sub.[mu][mu]] = 0 [??] rank [A.sub.[mu]] = 0.

Proof. Since

(2.1) A([lambda],[mu][epsilon][A.sup.T] ([lambda],[mu][epsilon] = [I.sub.n]

it takes derivation with respect to ~[lambda], it follows that

[A.sub.[lambda]][epsilon][A.sup.T][epsilon] + A[epsilon][A.sup.T.sub.[lambda]][epsilon] = 0

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Since rank [A.sub.[lambda][lambda]] = 0, we get [A.sub.[lambda][lambda]] = 0 and [A.sup.T.sub.[lambda][lambda]] = 0. We have following that (2.1)

[A.sub.[lambda]][epsilon][A.sup.T.sub.[lambda]][epsilon] = 0 [??] [A.sub.[lambda]][epsilon][A.sup.T.sub.[lambda]] = 0.

For every x [member of] [IR.sup.1.sub.n], we have

([A.sub.[lambda]][epsilon][A.sup.T.sub.[lambda]]) [x.sup.T] = (0) [x.sup.T]

([A.sub.[lambda]][epsilon][A.sup.T.sub.[lambda]]) [x.sup.T] = 0

x ([A.sub.[lambda]][epsilon][A.sup.T.sub.[lambda]]) [x.sup.T] = 0

([xA.sub.[lambda]] [epsilon] [(xA[mu]).sup.T] = 0

So that [<x[A.sub.[lambda]], x[A.sub.[lambda]]>.sub.L] = 0 (from the non-degenere property), x[A.sub.[lambda]] = 0. Since it is true for every x [member of] [IR.sup.1.sub.n], so we get [A.sub.[lambda]] = 0 and rank ([A.sub.[lambda]]) = 0. It takes derivation with respect to [mu] in the equation (2.1) similarly it follows that

[A.sub.[mu]][epsilon][A.sup.T][epsilon] + A[epsilon][A.sup.T.sub.[mu]][epsilon] = 0,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Since rank ([A.sub.[mu][mu]]) = 0, we get [A.sub.[mu][mu]] = 0 and [A.sup.T.sub.[mu][mu]] = 0, thus we have

[A.sub.[mu]][epsilon][A.sub.T.sub.[mu]][epsilon = 0 [??] [A.sub.[mu]][epsilon][A.sub.T.sub.[mu] = 0

Since it is true for every x [member of] [IR.sup.1.sub.n], we have the following

([A.sub.[mu]][epsilon][A.sub.T.sub.[mu]]) [x.sup.T]= (0) [x.sup.T]

([A.sub.[mu]][epsilon][A.sup.T.sub.[mu]]) [x.sup.T] = 0

x ([A.sub.[mu]][epsilon][A.sup.T.sub.[mu]]) [x.sup.T] = 0

([xA.sub.[mu][lambda]] [epsilon] [([xA.sub.[mu]]).sup.T.] = 0

Hence

[<[xA.sub.[mu]],[xA.sub.[mu]]>.sub.L] = 0

[xA.sub.[mu]] = 0.

For every x [member of] [IR.sup.1.sub.n] it is true, we get [A.sub.[mu]] = 0 and rank ([A.sub.[mu]]) = 0. Conversely it is obviously to see.

[section]3. Special case n = 3

Since [A.sub.[lambda]] and [A.sub.[mu]] are semi skew-symmetric matrices especially

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Let [A.sub.[lambda]] = -[epsilon][A.sup.T.sub.[lambda]][epsilon],[A.sub.[mu]] = -[epsilon][A.sup.T.sub.[mu]][epsilon]. The equation X = [A.sup.-1]([lambda],[mu])Y([lambda],[mu]) is obtained from the equation of Y([lambda],[mu]) = A([lambda],[mu])X. By differentiating the equation Y([lambda],[mu]) = A([lambda],[mu])X with respect to t, we have

[Y.sub.[lambda]][??] + [Y.sub.[mu]][??] = ([A.sub.[lambda]][??] + [A.sub.[mu]][??]) X

= ([Y.sub.[lambda]][??] + [Y.sub.[mu]][??]) [A.sup.-1]([lambda],[mu])Y([lambda],[mu]).

In the position ([lambda],[mu]) (0, 0), we have

[Y.sub.[lambda]][??] + [Y.sub.[mu]][??] = ([A.sub.[lambda]][??] + [A.sub.[mu]][??]) Y([lambda],[mu])

= [Omega]Y[[lambda],[mu]]).

Since [A.sub.[lambda]] and [A.sub.[mu]] are semi skew-symmetric matrices, we get

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and also angular velocity matrix is an semi skew-symmetric. Since A matrix is semi orthogonal, we have the following equation

A([lambda].[mu])[epsilon][A.sup.T] ([lambda],[mu])[epsilon] = I.

Now differentiating with respect to t, it follows that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

For ([lambda],[mu]) = (0, 0), we obtain

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Since [Omega] = [A.sub.[lambda]][lambda] + [A.sub.[mu]][mu], it follows that

[Omega] + [epsilon][[Omega].sup.T][epsilon] = 0,

where [Omega] is semi skew-symmetric matrix. Since pole points which are the points of sliding velocity is zero given by

[[??].sub.f = ([A.sub.[lambda]][??] + [A.sub.[mu]][??]) X + ([C.sub.[lambda]][??] + [C.sub.[mu]][??])

is pole points of two parameter motion by

Y([lambda],[mu]) = A([lambda],[mu]) X + C([lambda],[mu]).

The equation

(3.1) ([A.sub.[lambda]][??] + [A.sub.[mu]][??]) X + ([C.sub.[lambda]][??] + [C.sub.[mu]][??]) = 0

can be solution it must be rank ([A.sub.[lambda]][??] + [A.sub.[mu]][??]) = rank [Omega] = 2. It follows that

Y([lambda],[mu]) = A([lambda],[mu])X + C([lambda],[mu])

A([lambda],[mu])X = Y([lambda],[mu]) - C([lambda],[mu])

[A.sup.-1]([lambda],[mu])A([lambda],[mu])X = [A.sup.-1]([lambda],[mu])(Y([lambda],[mu]) - C([lambda],[mu]))

X = [A.sup.-1]([lambda],[mu])(Y([lambda],[mu]) - C([lambda],[mu])).

If we get write this value of X in the equation (3.1), we have

(3.2) [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

In the position ([lambda],[mu]) = (0, 0), we have

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

And if we say

Y([lambda],[mu]) - C([lambda],[mu]) = Y*([lambda],[mu]),

then the equation (3.2) form

(3.3) [Omega] [[conjunction].sub.L] Y*([lambda],[mu]) + ([C.sub.[lambda]][??] + [C.sub.[mu]][??]) = 0.

The solution of equation (3.3) gives fixed pole points of the motion. For the solution of equation (3.3) it must be verified the condition

<[Omega], [<[C.sub.[lambda]][??] + [C.sub.[mu]][??]>.sub.L] = 0.

In generally, this condition can't verify. But we can separate two composite velocities [C.sub.[lambda]][lambda] + [C.sub.[mu]][mu]either orthogonal [??] or parallel. If W is angular velocity matrix of moving space, then it can find W = [A.sup.-1]([lambda],[mu])[Omega]. Since [A.sup.-1](0,0) = I, we have W = [Omega] in the position ([lambda],[mu]) = (0,0).

Let

E = [Omega]/[[parallel][Omega][parallel].sub.L]

and

E* = W/[[parallel]W[parallel].sub.L].

Then we have E* = [n]AE, where n [member of] IR. For ([lambda],[mu]) = (0, 0) and n = 1, it follows that

E* = E.

We can separate two components of the velocity [C.sub.[lambda]][??] + [C.sub.[mu]][??] which form

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and

V = [<[Omega],[C.sub.[lambda]][??] + [C.sub.[mu]][??]>.sub.L/[<[Omega],[Omega].sub.L] [Omega],

one is orthogonal to vector [??] , another is parallel to vector [??] respectively, in which

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

That is

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

and

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

If we write velocity U replacing [C.sub.[lambda]][??] + [C.sub.[mu]][??] in equation (3.3), we have this condition [<[Omega, U>.sub.L] = 0 is verified in this equation. Let rank [Omega] = 2 ,then the system can be solution. Now we get the solution of the equation

[Omega] [[conjunction].sub.L] Y*([lambda],[mu]) + U = 0.

Here [??][[conjunction].sub.L] ([??][[conjunction].sub.L][??]) = [<[??],[??]>.sub.L][??] = [<[??],[??]>.sub.L][??]. It follows that

[Omega] [conjunction] [L.sub.Y]*([lambda],[mu]) + U = 0,

[Omega] [[conjunction].sub.L] ([Omega] [[conjunction].sub.L] Y*([lambda],[mu]) + U = 0,

[Omega] [[conjunction].sub.L] ([Omega] [[conjunction].sub.L] Y*([lambda],[mu])) + [Omega] [[conjunction].sub.L] U = 0,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

Y*([lambda],[mu]) = -[Omega] [[conjunction].sub.L] U/[<[Omega],[Omega]>.sub.L] + [sigma][Omega].

If we write the last equation replacing by the equation

Y*([lambda],[mu]) = Y([lambda],[mu]) - C([lambda],[mu]),

so we have

Y([lambda],[mu]) = -[Omega] [[conjunction].sub.L] U/[<[Omega],[Omega]>.sub.L] + C([lambda],[mu]) + [sigma][Omega].

Hence we get

(3.4) Y([lambda],[mu]) = Q + [sigma][Omega], [sigma] [member of] IR.

This means that it is a line which is passes though point Q and straight [Omega]. The line is called fixed pole axis in the fixed space, the expression of the fixed pole axis in the moving space find to write instead of value Y([lambda],[mu]) in the equation (1.1), it follows that

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

-[Omega] [[conjunction].sub.L] U/[<[Omega],[Omega]>].sub.L] + [sigma][Omega] = A([lambda],[mu])X,

[MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII]

In the position ([lambda],[mu]) = (0, 0), it follows that

X = -[Omega] [[conjunction].sub.L] U/[<[Omega],[Omega]>].sub.L] + [sigma][Omega],

X = P + [sigma][Omega].

If the pole axis of fixed and moving space coincide in the position ([lambda],[mu]) (0, 0), we have P = Q and C(0, 0) = 0 . Thus lines passes though points Q and P is straight [??], which are the pole axis in fixed and moving space. In the position ([lambda],[mu]) of motion the lines have a velocity vector with stationary norm, locus of the lines which called instantaneous screw axis. The equations

Y = Q + [sigma][Omega], [sigma] [member of] IR

Y = P + [sigma][Omega], [sigma] [member of] IR

depend only on parameters [?? and [??. Thus there is [[infinity].sub.2] the one parameter motion. There are [infinity] instantaneous screw axis since the parameters [?? and [?? depend only on t [1]. The locus of this screw axis is a ruled surface. Indeed the following equations determine a ruled surface,

Y(t,[sigma]) = Q(t) + [sigma][Omega](t), [sigma] [member of] IR

X(t,[sigma]) = P(t) + [sigma][Omega](t), [sigma] [member of] IR.

References

[1] Bottema O., Roth B., Theotical Kinematics, North Holland Publ. Com., 1979.

[2] Hacisalihoglu H.H., On The Geometry of Motion In The Euclidean n-Space, Communications. Ank., Univ. Seri 23 A., 1974, 95-108.

[3] Karacan M. K., Kinematic Applications of Two-Parameter Motion, Ankara University, Graduate School and Natural Sciences, Ph.D Thesis, 2004.

[4] Karacan M.K., Yayli Y., On The Instantaneous Screw Axes of Two Parameter Motions, Facta Universitatis, Series, Mechanics, Automatic, Control and Robotics, 6(2007), No.1, 81-88.

[5] O'Neill B., Semi-Riemannian Geometry with Applications to Relativity, Academic Press, London, 1983.

Murat Kemal Karacan ([dagger]) and Levent Kula ([double dagger])

([dagger]) Usak University, Faculty of Sciences and Arts, Department of Mathematics, Usak, Turkey

([double dagger]) Ahi Evran University, Faculty of Sciences and Arts, Department of Mathematics, Kirsehir, Turkey

murat.karacan@usak.edu.tr lkula@ahievran.edu.tr