4.2 Multiple Vector Products 51
a×(b×c)=a·cb−a·bc. (4.17)Noticethat thepositionof theparenthesis(.. .)is essential inthis case. Theexpression
a×b×cis not well defined sincea×(b×c)=(a×b)×c, in general. The latter
double cross product yields a vector with component parallel toaandb.
The proof of (4.17), is also based on the relation (4.10). In component notation,
one has
[a×(b×c)]μ=εμνλaν(b×c)λ=εμνλελμ′ν′aνbμ′cν′.Now use of (4.10), together with the symmetry properties of the epsilon-tensor yields
[a×(b×c)]μ=bμaνcν−cμaνbν, (4.18)which is equivalent to (4.17).
For the special casea=c=e, whereeis a unit vector, (4.17) reduces to
e×(b×e)=b−e·be=b−b‖:=b⊥. (4.19)Hereb‖:=e·beandb⊥are the parts ofbwhich are parallel and perpendicular,
respectively, toe. Clearly, one hasb‖+b⊥=b.
4.3 Applications.
4.3.1 Angular Momentum for the Motion on a Circle
The linear momentump=mvof a particle with massmand velocityv,movingon
a circle, with the angular velocityw,cf.(3.46), is given by
p=mw×r.Herer=0 is a point on the axis of rotation which is parallel tow. With the help of
(4.18), the resulting orbital angular momentumL=r×pis found to be
L=mr×(w×r)=m(r^2 w−r·wr). (4.20)In component notation, this equation linking the angular momentum with the angular
velocity, is equivalent to
Lμ=m(r^2 wμ−rνwνrμ)=m(r^2 δμν−rμrν)wν. (4.21)