Mathematical Physics Vol 1

Chapter 3. Examples

74

Figure 3.12: Natural trihedron.

The space curve can be represented by the relation

r = r ( s )= x ( s ) i + y ( s ) j + z ( s ) k ,

(3.6)

where r is the position vector of an arbitrary point of the curve, s an arc of the curve, and i , j and k are unit vectors of the axes x , y and z , respectively. It follows that

d r d s

= t ,

d 2 r d s 2

d t d s

1 ρ

n ,

(3.7)

=

=

where 1 / ρ = k 1 is the first curvature (flexion). Given that b = t × n ⇒ d b d s = d t d s ×

d n d s ⇒

n + t ×

d b d s

d n d s

= t ×

(3.8)

,

which follows from

d t d s

1 ρ

n ,

=

d t d s and n are collinear, and their vector product is equal to zero. Given that b · b = 1 , we obtain d b d s · b = 0 .

vectors

(3.9)

d b d s (torsion vector) is orthogonal to both t and b ,

From (3.8) and (3.9) it follows that

and thus this vector is collinear with the main normal n , that is, d b d s = − k 2 n .

(3.10)

Note that k 2 < 0, if vectors k 2 and n have the same direction, when d s > 0, while k 2 > 0, if these vectors have opposite directions. The variable k 2 is called the curve torsion(3.6).