Triple Scalar Product
Sections 1.3 and 1.4 cover the two types of multiplication of interest here. However, there are combinations of three vectors, Aã(BìC)and Aì(BìC), that occur with sufficient frequency to deserve further attention. The combination
Aã(BìC)
is known as the triple scalar product. B×C yields a vector that, dotted into A, gives a scalar. We note that(AãB)ìC represents a scalar crossed into a vector, an operation that is not defined. Hence, if we agree to exclude this undefined interpretation, the parentheses may be omitted and the triple scalar product written AãBìC.
Using Eqs. (1.38) for the cross product and Eq. (1.24) for the dot product, we obtain AãBìC=Ax(ByCz−BzCy)+Ay(BzCx−BxCz)+Az(BxCy−ByCx)
=BãCìA=CãAìB
= −AãCìB= −CãBìA= −BãAìC, and so on. (1.48) There is a high degree of symmetry in the component expansion. Every term contains the factorsAi,Bj, andCk. Ifi,j, andkare in cyclic order(x, y, z), the sign is positive. If the order is anticyclic, the sign is negative. Further, the dot and the cross may be interchanged,
AãBìC=AìBãC. (1.49)
FIGURE1.16 Parallelepiped representation of triple scalar product.
A convenient representation of the component expansion of Eq. (1.48) is provided by the determinant
AãBìC=
Ax Ay Az Bx By Bz Cx Cy Cz
. (1.50)
The rules for interchanging rows and columns of a determinant12 provide an immediate verification of the permutations listed in Eq. (1.48), whereas the symmetry of A, B, and C in the determinant form suggests the relation given in Eq. (1.49). The triple products encountered in Section 1.4, which showed that A×B was perpendicular to both A and B, were special cases of the general result (Eq. (1.48)).
The triple scalar product has a direct geometrical interpretation. The three vectors A, B, and C may be interpreted as defining a parallelepiped (Fig. 1.16):
|B×C| =BCsinθ
=area of parallelogram base. (1.51) The direction, of course, is normal to the base. Dotting A into this means multiplying the base area by the projection of A onto the normal, or base times height. Therefore
AãBìC=volume of parallelepiped defined by A,B,and C.
The triple scalar product finds an interesting and important application in the construc- tion of a reciprocal crystal lattice. Let a, b, and c (not necessarily mutually perpendicular)
12See Section 3.1 for a summary of the properties of determinants.
represent the vectors that define a crystal lattice. The displacement from one lattice point to another may then be written
r=naa+nbb+ncc, (1.52)
withna, nb, andnctaking on integral values. With these vectors we may form a′= b×c
aãbìc, b′= cìa
aãbìc, c′= aìb
aãbìc. (1.53a) We see that a′is perpendicular to the plane containing b and c, and we can readily show that
a′ãa=b′ãb=c′ãc=1, (1.53b) whereas
a′ãb=a′ãc=b′ãa=b′ãc=c′ãa=c′ãb=0. (1.53c) It is from Eqs. (1.53b) and (1.53c) that the name reciprocal lattice is associated with the points r′=n′aa′+n′bb′+n′cc′. The mathematical space in which this reciprocal lattice ex- ists is sometimes called a Fourier space, on the basis of relations to the Fourier analysis of Chapters 14 and 15. This reciprocal lattice is useful in problems involving the scattering of waves from the various planes in a crystal. Further details may be found in R. B. Leighton’s Principles of Modern Physics, pp. 440–448 [New York: McGraw-Hill (1959)].
Triple Vector Product
The second triple product of interest is A×(B×C), which is a vector. Here the parentheses must be retained, as may be seen from a special case(xˆ× ˆx)× ˆy=0, whilexˆ×(xˆ× ˆy)= ˆ
x× ˆz= −ˆy.
Example 1.5.1 A TRIPLEVECTORPRODUCT
For the vectors
A= ˆx+2yˆ− ˆz=(1,2,−1), B= ˆy+ ˆz=(0,1,1), C= ˆx− ˆy=(0,1,1), B×C=
ˆ
x yˆ zˆ
0 1 1
1 −1 0
= ˆx+ ˆy− ˆz, and
A×(B×C)= ˆ
x yˆ zˆ 1 2 −1 1 1 −1
= −ˆx− ˆz= −(yˆ+ ˆz)−(xˆ− ˆy)
= −B−C.
By rewriting the result in the last line of Example 1.5.1 as a linear combination of B and C, we notice that, taking a geometric approach, the triple vector product is perpendicular
FIGURE1.17 B and C are in thexy-plane.
B×C is perpendicular to thexy-plane and is shown here along thez-axis. Then A×(B×C)is perpendicular to thez-axis
and therefore is back in thexy-plane.
to A and to B×C.The plane defined by B and C is perpendicular to B×C, and so the triple product lies in this plane (see Fig. 1.17):
A×(B×C)=uB+vC. (1.54)
Taking the scalar product of Eq. (1.54) with A gives zero for the left-hand side, so uAãB+vAãC=0. Henceu=wAãC andv= −wAãB for a suitablew. Substitut- ing these values into Eq. (1.54) gives
A×(B×C)=w
B(AãC)−C(AãB); (1.55)
we want to show that
w=1
in Eq. (1.55), an important relation sometimes known as the BAC–CAB rule. Since Eq. (1.55) is linear in A,B, andC,w is independent of these magnitudes. That is, we only need to show that w=1 for unit vectors A,ˆ B,ˆ C. Let us denoteˆ Bˆ ã ˆC=cosα,
ˆ
Cã ˆA=cosβ,Aˆ ã ˆB=cosγ, and square Eq. (1.55) to obtain Aˆ ì(Bˆì ˆC)2= ˆA2(Bˆì ˆC)2−Aˆ ã(Bˆì ˆC) 2
=1−cos2α−Aˆ ã(Bˆ ì ˆC)2
=w2
(Aˆã ˆC)2+(Aˆ ã ˆB)2−2(Aˆã ˆB)(Aˆ ã ˆC)(Bˆã ˆC)
=w2
cos2β+cos2γ−2 cosαcosβcosγ
, (1.56)
using(Aˆ ì ˆB)2= ˆA2Bˆ2−(Aˆ ã ˆB)2repeatedly (see Eq. (1.43) for a proof). Consequently, the (squared) volume spanned byA,ˆ B,ˆ C that occurs in Eq. (1.56) can be written asˆ
Aˆ ã(Bˆì ˆC) 2=1−cos2α−w2
cos2β+cos2γ−2 cosαcosβcosγ . Herew2=1, since this volume is symmetric inα, β, γ. That is,w= ±1 and is inde- pendent ofA,ˆ B,ˆ C. Using again the special caseˆ xˆ×(xˆ× ˆy)= −ˆy in Eq. (1.55) finally givesw=1. (An alternate derivation using the Levi-Civita symbolεij k of Chapter 2 is the topic of Exercise 2.9.8.)
It might be noted here that just as vectors are independent of the coordinates, so a vector equation is independent of the particular coordinate system. The coordinate system only determines the components. If the vector equation can be established in Cartesian coor- dinates, it is established and valid in any of the coordinate systems to be introduced in Chapter 2. Thus, Eq. (1.55) may be verified by a direct though not very elegant method of expanding into Cartesian components (see Exercise 1.5.2).
Exercises
1.5.1 One vertex of a glass parallelepiped is at the origin (Fig. 1.18). The three adjacent vertices are at(3,0,0),(0,0,2), and(0,3,1). All lengths are in centimeters. Calculate the number of cubic centimeters of glass in the parallelepiped using the triple scalar product.
1.5.2 Verify the expansion of the triple vector product
Aì(BìC)=B(AãC)−C(AãB)
FIGURE1.18 Parallelepiped: triple scalar product.
by direct expansion in Cartesian coordinates.
1.5.3 Show that the first step in Eq. (1.43), which is
(AìB)ã(AìB)=A2B2−(AãB)2, is consistent with the BAC–CAB rule for a triple vector product.
1.5.4 You are given the three vectors A, B, and C, A= ˆx+ ˆy, B= ˆy+ ˆz, C= ˆx− ˆz.
(a) Compute the triple scalar product, AãBìC. Noting that A=B+C, give a geo- metric interpretation of your result for the triple scalar product.
(b) Compute A×(B×C).
1.5.5 The orbital angular momentum L of a particle is given by L=r×p=mr×v, where p is the linear momentum. With linear and angular velocity related by v=ω×r, show that
L=mr2
ω− ˆr(rˆãω).
Herer is a unit vector in the r-direction. For rˆ ãω=0 this reduces to L=Iω, with the moment of inertiaI given bymr2. In Section 3.5 this result is generalized to form an inertia tensor.
1.5.6 The kinetic energy of a single particle is given byT =12mv2. For rotational motion this becomes 12m(ω×r)2. Show that
T =1 2m
r2ω2−(rãω)2 .
For rãω=0 this reduces toT =12I ω2, with the moment of inertiaI given bymr2. 1.5.7 Show that13
a×(b×c)+b×(c×a)+c×(a×b)=0.
1.5.8 A vector A is decomposed into a radial vector Ar and a tangential vector At. Ifˆr is a unit vector in the radial direction, show that
(a) Ar = ˆr(Aã ˆr)and (b) At= −ˆr×(rˆ×A).
1.5.9 Prove that a necessary and sufficient condition for the three (nonvanishing) vectors A, B, and C to be coplanar is the vanishing of the triple scalar product
AãBìC=0.
13This is Jacobi’s identity for vector products; for commutators it is important in the context of Lie algebras (see Eq. (4.16) in Section 4.2).
1.5.10 Three vectors A, B, and C are given by
A=3xˆ−2yˆ+2z,ˆ B=6xˆ+4yˆ−2z,ˆ C= −3xˆ−2yˆ−4z.ˆ
Compute the values of AãBìC and Aì(BìC),Cì(AìB)and Bì(CìA).
1.5.11 Vector D is a linear combination of three noncoplanar (and nonorthogonal) vectors:
D=aA+bB+cC.
Show that the coefficients are given by a ratio of triple scalar products, a=DãBìC
AãBìC, and so on.
1.5.12 Show that
(AìB)ã(CìD)=(AãC)(BãD)−(AãD)(BãC).
1.5.13 Show that
(AìB)ì(CìD)=(AãBìD)C−(AãBìC)D.
1.5.14 For a spherical triangle such as pictured in Fig. 1.14 show that sinA
sinBC= sinB
sinCA = sinC sinAB.
Here sinAis the sine of the included angle atA, whileBC is the side opposite (in radians).
1.5.15 Given
a′= b×c
aãbìc, b′= cìa
aãbìc, c′= aìb aãbìc, and aãbìc=0, show that
(a) xãy′=δxy, (x,y=a,b,c), (b) a′ãb′ìc′=(aãbìc)−1, (c) a= b′×c′
a′ãb′ìc′.
1.5.16 If xãy′=δxy, (x,y=a,b,c), prove that a′= b×c
aãbìc. (This is the converse of Problem 1.5.15.)
1.5.17 Show that any vector V may be expressed in terms of the reciprocal vectors a′, b′, c′(of Problem 1.5.15) by
V=(Vãa)a′+(Vãb)b′+(Vãc)c′.
1.5.18 An electric chargeq1moving with velocity v1produces a magnetic induction B given by
B=à0
4πq1v1× ˆr
r2 (mks units),
wherer points fromˆ q1to the point at which B is measured (Biot and Savart law).
(a) Show that the magnetic force on a second chargeq2, velocity v2, is given by the triple vector product
F2=à0 4π
q1q2
r2 v2×(v1× ˆr).
(b) Write out the corresponding magnetic force F1thatq2exerts onq1. Define your unit radial vector. How do F1and F2compare?
(c) Calculate F1and F2for the case ofq1andq2moving along parallel trajectories side by side.
ANS.
(b) F1= −à0
4π q1q2
r2 v1×(v2× ˆr).
In general, there is no simple relation between F1and F2. Specifically, Newton’s third law, F1= −F2, does not hold.
(c) F1=à0
4π q1q2
r2 v2ˆr= −F2. Mutual attraction.