Mathematical Physics Vol 1

3.2 Vector analysis

73

d ρ d t

d ρ d t

d r d t

d r c d t

r = r c + ρ ⇒ v =

=

+

=

.

We used here the fact that the vector r C is constant (the point C does not change its position), and thus its derivative is equal to zero. Given that the body is rigid 2 , it follows that the magnitude of the vector | ρ | = const, and thus, according to the previous example

i j k ω x ω y ω z x y z

d ρ d t

= ω × ρ =

= i ( ω y z − ω z y )+ j ( ω z x − ω x z )+ k ( ω x y − ω y x ) .

v =

We denoted here the angular velocity of the body by ω .

Problem31 Prove the Frenet-Serret 3 formulas

d t d s d n d s d b d s

1 ρ

n =( k 1 n ) ,

=

= − k 1 t + k 2 b = − 1 ρ

1 T

(3.5)

t +

b ,

1 T

= − k 2 n = −

n ,

where t , n and b are unit vectors of a natural trihedron, ρ is the radius of curvature, and T is the radius of torsion.

Solution Observe the natural trihedron formed by the unit vectors of the tangent t , main normal n and binormal b (see Figure 3.12). These three vectors form a right trihedron, and thus

t × n = b , n × b = t , b × t = n .

2 The term rigid body implies that the distance between any two points of the body, while moving, remains unchanged, that is, | −→ CM | = const . 3 Jean-Frédéric Frenet (1816-1900) and Joseph Alfred Serret (1819-1885), French mathematicians.