On The Volume Of The Trajectory Surface Under The Galilean Motions In The Galilean Space

On The Volume Of The Trajectory Surface Under The Galilean Motions In The Galilean Space

Mücahit Akb¬y¬k

, Salim Yüce

Received 15 Feb 2017

Abstract

In this work, the volumes of the trajectory surfaces which are traced by …xed points during 3-parameter Galilean space motions are studied. Also, the well- known classical Holditch theorem [3] is generalized for the volumes of the trajec- tory surfaces in the Galilean space.

1 Introduction

In 1958, H. Holditch, [3], came up with the following outstanding classical theorem: If the endpointsA andB of a …xed line segmentAB with lengtha+bare rotated once along an ovalkin the Euclidean planeE²;then a given …xed pointX(AX=a; XB=b) ofABdescribes a closed not necessarily convex curvekX:The areaF of the Holditch- Ring bounded by the curves kandk_X isF = ab:

Later, this theorem was studied by di¤erent methods [1,2,6–10] and in di¤erent spaces [13–15]. One of the generalizations of this theorem is on the volumes of the surfaces of 3-dimensional Euclidean space which are traced by …xed points during 3- parameter motions are given by H. R. Müller [6–8] and W. Blashke [1].

In this paper, the volumes of the trajectory surfaces of …xed points under 3- parameter Galilean space motions are calculated. Also, by the help of a special distance that we have de…ned, we generalize the well-known classical Holditch theorem for the volumes of the trajectory surfaces of …xed points under 3-parameter Galilean space motions.

2 Preliminaries

Galilean geometry G³ can be described as the study of properties of 3-dimensional space with coordinates that are invariant under Galilean transformations

8<

:

x⁰ =x+a;

y⁰= (vcos )x+ (cos')y+ (sin')z+b;

z⁰ = (vsin )x+ ( sin')y+ (cos')z+d:

Mathematics Sub ject Classi…cations: 53A17, 53A35.

yDepartment of Mathematics, Yildiz Technical University, Esenler, Istanbul 34220, Turkey

zDepartment of Mathematics, Yildiz Technical University, Esenler, Istanbul 34220, Turkey

297

The Galilean transformations consist of translation, rotation and shear motions, and are described by I. M. Yaglom in [11]. In the literature, the basic information about Galilean Geometry is …rstly given by I. M. Yaglom. Then, the di¤erential geometry of curves and surfaces in the Galilean space G³ is worked in detail by O. Röshcel in [9].

Also, the quadrics in the Galilean space are examined by Kamenarovic in [4]. Now, let’s give some basic information about the Galilean space G³. Leta = (x; y; z) and b= (x1; y1; z1)be two vectors in the Galilean space. The scalar product ofa andbis de…ned by

<a;b>_G=xx1:

The vectors in the Galilean space are divided into two classes as non-isotropic vectors and isotropic vectors which are of the form a = (x; y; z); x 6= 0 and p = (0; y; z);

respectively:Moreover, the special scalar product of isotropic vectorsp= (0; y; z)and q= (0; y1; z1)is de…ned by

<p;q> =yy1+zz1:

Ifa= (x; y; z)andb= (x1; y1; z1)are vectors in Galilean space, the vector product of a andbis de…ned as the following in [5]:

a b=

0 e2 e3

x y z

x1 y1 z1

:

Letg₁be a nonisotropic vector,g₂and g₃ be isotropic vectors in the Galilean space.

If the vectors g₁;g₂; and g₃ satisfy that <g₁;g₁ >_G=<g₂;g₂ > =<g₃;g₃ > = 1 and <g₁;g₂>_G=<g₁;g₃>_G=<g₂;g₃> = 0;then the vector systemfg₁;g₂;g₃g is called an orthonormal frame of Galilean space. More information about the Galilean geometry can be found in [9, 11].

3 One Parameter Galilean Space Motion

LetR andR⁰ be two3 dimensional Galilean spaces. Let fO;g1;g2;g3g and fO⁰;g₁⁰;g⁰₂;g⁰₃g

are orthonormal frames of spaces R and R⁰, respectively. Assume that the frame fO;g₁;g₂;g₃gmoves with respect to frame fO⁰;g⁰₁;g⁰₂;g₃⁰g: Then, it is accepted that the space R moves according to the space R⁰. The spaces R and R⁰ are called moving space and …xed space, respectively. Moreover, the frames fO;g1;g2;g3g and fO⁰;g⁰₁;g⁰₂;g⁰₃g are calledmoving frame and …xed frame, respectively. This motion is called one parameter Galilean space motion and is denoted byB=R=R⁰: Here, the spaces Rand R⁰ are orientated in the same direction. During the motionB =R=R⁰; it is clear that

x⁰ = u+x;

where x⁰ and x correspond to the position vectors of any pointX 2 R according to the rectangular coordinate systems ofR⁰,R, respectively, and

u=OO⁰=u1g1+u2g2+u3g3:

Here, the vectoruis calledtranslation vector. In the motionB=R=R⁰, the vectorsx⁰; x and uare continuously di¤erentiable functions of a real parameter t:If the frames fg1;g2;g3gand fg⁰₁;g⁰₂;g₃⁰g are written as

G= 2 4 g1

g2

g₃ 3

5 andG⁰= 2 4 g⁰₁

g⁰₂ g⁰₃

3 5

in the matrix form, respectively, then

G=AG⁰ andG⁰=A ¹G (1)

are hold. Here,A is an invertible matrix. In this case, by di¤erentiating both sides of the equation (1) and considering that G⁰ is …xed frame, we get

dG= G; (2)

where =dAA ¹. If we calculate from above equation (2) and by the necessary operations with the basis vectorsg_i;1 i 3;we get

= 2

4 0 !3 !2

0 0 !1

0 !1 0 3

5; (3)

where !i;1 i 3 are the linear di¤erential forms with respect to t, that is, !i = fi(t)dt:

The vector

!=!1g1+!2g2+!3g3

which is de…ned by nonzero components of is called the instantaneousPfa¢ an vector of the motion B =R=R⁰.

Especially, if!₁= 0; then, the motion ofB consists of only translation and shear motions. The motion doesn’t contain the rotation. We will, therefore, accept!₁ 6= 0 in this work.

If it is used the equality (3) given for , we get

dg1= !3g2+!2g3; dg2= !1g3; dg3=!1g2: (4) Moreover, if we calculate the exterior derivation of equation (4), by considering that basis vectors gi; i= 1;2;3, are linearly independent, we obtain

d!₁= 0; d!₂= !₃^!₁; d!₃=!₂^!₁;

where "^" is the wedge product of the di¤erential forms. Hence, the conditions of integration for components of the pfa¢ an vector of the motionR=R⁰ are found as

d!1= 0; d!2= !3^!1; d!3=!2^!1: (5) On the other hand, by di¤erentiating of translation vector

O⁰O= u= u1g1 u2 g₂ u3g3

and by the aid of using the equation (4), we have

0 = du= ₁g1+ ₂g2+ ₃g3;

where

0 = du

and

1= du1; 2= du2+u1!3 u3!1; 3= du3 u1!2+u2!1:

The equations 8

>>

<

>>

:

dg1=!2g3 !3g2; dg2= !1g3; dg3=!1g2;

0= du= ₁g1+ ₂g2+ ₃g3;

(6)

are called derivative equations of motionR=R⁰. Furthermore,

0=d ⁰=d 1g1+ (d 2 1^!3+ 3^!1)g2+ (d 3+ 1^!2 2^!1)g3

and because of the fact thatfg₁;g₂; g₃g are linearly independent, we …nd

d 1= 0; d 2= 1^!3 3^!1; d 3= 1^!2+ 2^!1: (7) So, the conditions of integration obtained for the translation vector of the motionR=R⁰ are equations (7).

Now, let’s examine the velocity vectors of the pointX under the motionR=R⁰:Let X be any point inR. So, we can write

x= X3 i=1

x_ig_i

with respect to the moving frame fO;g1;g2;g3g. Since we have x⁰ = u+x

for position vector x⁰ of the pointX with respect to …xed framefO⁰;g⁰₁; g⁰₂;g⁰₃g, the di¤erential of position vectorx⁰ can be expressed as

dx⁰= du+dx:

By (6), it is calculated as

dx⁰ = ₁g1+ ( ₂ x₁!₃+x₃!₁)g2+ ( ₃+x₁!₂ x₂!₁)g3+dx₁g1+dx₂g2+dx₃g3: During the motion R=R⁰; the velocity vector of any point X 2R with respect to

…xed spaceR⁰ and moving space Ris calledabsolute velocity XAandrelative velocity XR, respectively. XA and XR are expressed by

XA=dx⁰

and

XR=dx1g1+dx2g2+dx3g3:

The di¤erence between the absolute and the relative velocities is calledsliding velocity X_F of the pointX and it is stated as

XF = 1g1+ ( 2 x1!3+x3!1)g2+ ( 3+x1!2 x2!1)g3:

In this way, the following theorem can be given:

THEOREM 1. LetX be a point inR and XA;XF; andXR be the absolute, the sliding and the relative velocity of the point X under the motion R=R⁰, respectively.

Then, the relation between the velocities is given by XA=XF +XR:

If any pointX in Ris …xed, then the relative velocity XR=0and XA=XF

under the motion R=R⁰: Furthermore, by considering derivative equations (6), the following equation holds:

dx= ( x1!3+x3!1)g2+ (x1!2 x2w1)g3

or

dx=x !

for any …xed pointX inR:So, for any …xed pointX during the motionR=R⁰, one can state

dx⁰ = ⁰+x !:

Also, one can rewrite

dx⁰= 1g1+ 2g2+ 3g3; where

1= 1; 2= 2+x3!1 !3x1; 3= 3+x1!2 !1x2: (8)

4 The Volume of the Trajectory Surface in G

I: Until now, we have considered that the translation vector u and basis vectors gi

for 1 i 3 of the motionB are functions of a real parameter t:From now on, we assume that the translation vector uand basis vectorsg_i for1 i 3 of the motion B are functions of real parameterst₁; t₂andt₃:And this motion is called3-parameter Galilean space motion and we shall denote this 3-parameter motion byB₃:During the motionB₃; !_iand _i are the linear di¤erential forms with respect tot₁; t₂andt₃:So, the equations (5), (7) and (8) are not changed for the motion B3.

Under the motion B₃; any …xed point X in the moving space R determines a volumetric trajectory surface inR⁰:The volume element of the trajectory surface ofX under the motionB3, is de…ned by

dJX = 1^ ²^ ³: (9)

Thus, the integration of the volume element over a region Gdetermined by the …xed pointXof the parameter space during the motionB₃yields the volume of the trajectory surface, i.e.,

JX = Z

G

dJX:

By putting equation (8) into (9) and after making necessary arrangement, the volume of the trajectory surface of …xed pointX inR during the motionB3 is calculated as

JX=JO+ax²₁+ X3 i=2

bixix1+ X3 i=1

cixi; (10)

where J_O= Z

G

1^ 2^ 3; a= Z

G

1^!₂^!₃; b₂= Z

G

( ₁^!₁^!₃); b₃= Z

G

( ₁^!₁^!₂);

c1= Z

G

1^ ²^!2 1^!3^ ³; c2= Z

G

( 1^ ²^!1); c3= Z

G

1^!1^ ³:

J_Ois the volume of trajectory surface of origin pointO:So, the volumeJ_Xof trajectory surface is a quadratic polynomial ofx_i:

THEOREM 2. All …xed points inR whose trajectory surfaces have equal volume under the motionB3lie on the same quadric.

II: LetX = (x_i)and Y = (y_i)be two …xed points in R and Z = (z_i)be another point on the line segment XY; that is, zi = xi + yi; + = 1 in barycentric coordinates. Then, the volume of the region in R⁰ determined by the …xed point Z under the motionB3, by using equation (10), is obtained as

JZ = ²JX+ 2 JXY + ²JY; where

JXY =JO+ax1y1+1 2

X3 i=2

bi(x1yi+y1xi) +1 2

X3 i=1

ci(xi+yi): (11) Also,J_XY is called themixture trajectory surface volume. It is clearly seen thatJ_XX = J_X andJ_XY =J_{Y X}: Since

J_X 2J_XY +J_Y =a(x₁ y₁)²+ X3 i=2

b_i(x₁ y₁) (x_i y_i); (12)

we can restate the volume trajectory surface of the …xed pointZ asJ_Z= J_X+ J_Y a(x1 y1)²+

P3 i=2

bi(x1 y1) (xi yi) :So, we may give following theorem:

THEOREM 3. LetX and Y be two di¤erent …xed points inR;andZ be another point on the segmentXY:During the motionB₃, the relation between the volumes of the trajectory surfaces of …xed pointsX; Y andZ is as follows:

JZ = JX+ JY a(x1 y1)²+ X3 i=2

bi(x1 y1) (xi yi)

!

: (13)

We will de…ne the distanceD(X; Y)between the …xed pointsX; Y 2R;by

D²(X; Y) =" a(x1 y1)²+ X3 i=2

bi(x1 y1) (xi yi)

!

; "= 1: (14)

Therefore, by the help of the distance (14), the equation (13) can be restated as follows:

JZ = JX+ JY " D²(X; Y): (15) SinceX; Y andZ are collinear, we can writeD(X; Z) +D(Z; Y) =D(X; Y):Hence, if we represent =^D(Z;Y_D(X;Y⁾₎; = ^D(X;Z)_D(X;Y);where D(X; Y)6= 0; i.e.,x16=y1, then from equation (15), we have

J_Z = 1

D(X; Y)[D(Z; Y)J_X+D(X; Z)J_Y] "D(Z; Y)D(X; Z): (16) Now, we consider that the …xed points X andY trace the same trajectory surface. In this case, we getJX =JY:Then, from the above equation (16), we get

JX JZ="D(Z; Y)D(X; Z):

So, we may give the following theorem:

THEOREM 4. Let X = (x₁; x₂; x₃) and Y = (y₁; y₂; y₃); where x₁ 6= y₁; be two di¤ erent …xed points in R and Z be another point on the segment XY: Let the

…xed points X and Y trace the same trajectory surface and the …xed pointZ trace the di¤erent trajectory surface during the motion B3. Then, the di¤erence between the volumes of these two trajectory surfaces depends on the distances of Z from the endpoints which are de…ned in (14) with respect to the motionB3:

In case of x1 = y1; then the equation (13) is arranged as JZ = JX+ JY: If the …xed pointsX andY trace the same trajectory surface during the motion B3, we obtain

JX JZ= 0:

COROLLARY. Let X = (x₁; x₂; x₃) and Y = (y₁; y₂; y₃); where x₁ = y₁ be two di¤ erent …xed points in R and Z be another point on the segment XY:If the …xed pointsXandY trace the same trajectory surface, the …xed pointZtrace the trajectory surface with same volume during the motionB3.

III:

THEOREM 5. Let us consider a triangle in R whose vertices are points X1 = (x1; x2; x3); X2 = (y1; y2; y3) and X3 = (z1; z2; z3); where x1 6= y1, x1 6= z1 and y16=z1. If the vertices of this triangle trace the same trajectory surface inR⁰;then a di¤erent point Qon the plane which is determined byX1; X2 and X3 traces another surface. The di¤erence between the volumes of these two trajectory surfaces depends on the distances D(Xk; Q); D(Xk; Qk); D(Qk; Xj) and D(Xi; Qk) which are measured with respect to the motionB₃. Here, the pointQ_iis a intersection point of the segments X_iQandX_jX_k; i; j; k= 1;2;3; 2;3;1; 3;1;2:

PROOF. LetX1= (xi); X2= (yi)and X3 = (zi); x1 6=y1,x1 6=z1 and y1 6=z1

be three non- collinear …xed points in R;andQ= (qi) be another …xed point on the plane which is determined byX1= (xi); X2= (yi)andX3= (zi):Then, for the point Q, we can write

qi = 1x1+ 2yi+ 3zi; 1+ 2+ 3= 1: (17) Under the motion B3, by the help of equations (10), (11) and (17), the volume of the region determined by the point Qcan be calculated as

JQ= ²₁JX1+ ²₂JX2+ ²₃JX3+ 2 1 2JX1X2+ 2 1 3JX1X3+ 2 2 3JX2X3: Also, by considering the equations (12) and (14), we get

JQ = 1JX1+ 2JX2+ 3JX3

"_{12 1 2}D²(X₁; X₂) +"_{13 1 3}D²(X₁; X₃) +"_{23 2 3}D²(X₂; X₃) : Let the pointsQ1= (q1i)are intersection point of the segmentsX1QandX2X3. Then, we can write

q1i = ₁yi+ ₂zi; qi= ₃xi+ ₄ai;

where ₁+ ₂= ₃+ ₄= 1:Hence, we have ₁= ₃; ₂= _{1 4}; ₂= _{2 4}i.e.,

1= D(Q; Q₁)

D(X1; Q1); ₂= D(Q₁; X₃) D(X2; X3)

D(X₁; Q)

D(X1; Q1); ₃= D(X₂; Q₁) D(X2; X3)

D(X₁; Q) D(X1; Q1): Similarly, for Q2andQ3;we calculate

i= D(Q; Q_i)

D(Xi; Qi) = D(X_j; Q) D(Xj; Qj)

D(X_k; Q_j)

D(Xk; Xi) = D(X_k; Q) D(Xk; Qk)

D(Q_k; X_j) D(Xi; Xj)

fori; j; k= 1;2;3; 2;3;1; 3;1;2. So, the volume of the trajectory surface determined by the pointQcan be rearranged as

J_Q= X3 i=1

D(Q; Q_i) D(X; Qi)J_Xi

X3 i=1

"_ij D(X_k; Q) D(Xk; Qk)

2

D(Q_k; X_j)D(X_i; Q_k): (18)

During the motionB₃;since the pointsX₁; X₂andX₃trace the same trajectory surface in R⁰; we can write JX1 = JX2 =JX3: So, from the equation 1+ 2+ 3 = 1 and equation (18), we can get

JXi JQ = X3 i=1

"ij

D(Xk; Q) D(X_k; Q_k)

2

D(Qk; Xj)D(Xi; Qk);

fori; j; k (cyclic):

Finally, the di¤erence between the volumesJX1 andJQ only depends on distances on triangleX₁X⁴₂X₃ de…ned in (14) according to the motionB₃:

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