# 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 . 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. . 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

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

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[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]

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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]

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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

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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

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

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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

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That is

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and

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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 . 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

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

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

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

 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.

 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
Author: Printer friendly Cite/link Email Feedback Karacan, Murat Kemal; Kula, Levent Scientia Magna Report 7TURK Jan 1, 2009 2527 Browder's theorem and generalized Weyl's theorem. Cyclic dualizing elements in Girard quantales. Matrices Matrices (Mathematics) Rotation (Motion) Rotational motion Topological spaces