Fill in the blanks (complete proof is required!). required!). Preliminaries: we have the notation, ( v) k [ve [vector] vk ; s [scalar]; r]; (p) k [ps [pseudovector] v1iv 2j ijk [cr [c ross-product of vectors]; s];
(1.1)
a [ps [pseudosca scalar] v1i v2j ijk v3k [tr [tripl iple-product of vectors]; s];
Scalars, pseudoscalars, vectors, and pseudovectors are defined b y the following behaviours under the parity1 operation, which transforms scalars and vectors, and distributes across multiplication and addition, (1.2) P P ; P(s) s; P(vi ) vi ; P(s1v1 s2 v2 ) s1P (v1 ) s2 P (v 2 ); Thus, pseudoscalars and pseudovectors transform as, P(p) P( v1 v2 ) P( v1 ) P ( v2 ) ( v1) ( v 2 )
v1 v2
p;
(1.3)
P(a) P( v1 v 2 v3 ) P( v1 v2 ) P( v3 ) v1 v 2 ( v3) a; Properties of multiplication: Summarily, we have the following transform-laws, P (s ) s; P (a ) a ; P ( v ) v; P (p ) p; P (s v ) P (s )P (v ) (s )( v) s v ;
(1.4)
P(av) av av; P (sa ) P (s )P (a ) sa ; P (a p ) P (a )P (p ) a p;
Formula to work with vector-products: We shall often refer to the following frequentl y-used formula, ab A B aA bB aB Ab ; ijp ijk 2 pk ;
(1.5)
Then, we can look at each and write, (a)
( v p) k vi p j ijk vi (vi vj i j j )ijk vi vi vj i j j ikj vi vi vj i i j k i k j i
(p p) k pi p j ijk [( v1 v2 ) ( v3 v 4 )]k ( v1i v2j i j i )( v3i v4j i j j ) ijk v1i v2j v3i v4j i j i i k (b)
(1.6)
vi vivk vi vivk ( v v)vk ( v v)vk linear combination of vectors [ vector ] v
v1iv2jv3k v4i ij i v1i v2j v3i v4k ij i [( v1 v2 ) v 4 ]v3k [( v1 v2 ) v3 ]v4k
j i
i i kj
(1.7)
(p p)k a124v3k a123v4k vectors with pseudoscalar coefficients pseudovector pk v v v
(c) p p p
(d)
sum of scal scalar arss scal scalar ar v sum
p ( v1 v 2 ) ( v3 v 4 ) v1iv2j ij v3i v4j i j v1i v 2j v3i v4j
ii
jj
(1.8)
ij i j v 1i v 2j vi3v4j v1i v 2j v3j vi4
( v1 v 3 )( v 2 v 4 ) ( v1 v 4 )( v 2 v3 ) scalar minus scalar scalar s( v p) s( v ( v1 v2 )) sa [scalar] [pseudoscalar] [pseudoscalar]
(e)
(1.9)
(1.10)
To find out what ap is, dot it with the vector u , which could be either vector or pseudovector, (f)
1
ap u apk uk [ v1 ( v2 v3 )] (v 4 v5 )k uk v1 v 2i v3j ijij v4i v5j i j k u k v1 v2i v3j v4i v5j uk
The proof is simple:
P 1 ABP ABP P 1 AP 1PBP PBP , due to P being its own inverse.
ij
i j k
(1.11)
If the vector u is a pseudovector, then ap u
u p v 6 v7
i
j
i
u
v6 v 7 . If it is a vector, then
j i
j
v1 v2v3 v4 v5 v6 v 7
ij k
ij
i j k
u
v v v v i j 4 5
i 6
j 7
ii
v. j j
Looking at the cases,
i j i j v1vi2v 3j ij
(f) (cont’d)
v 4 v6 v5 v7 v1 ( v 2 v3 ) scalar pseudoscalar pseudoscalar v v v v 4 7 5 6
(f) (cont’d)
ap u [ v1 (v 2 v3 )] [( v 4 v5 ) v] pseudoscalar pseudoscalar scalar
(1.12)
u v
(1.13)
Thus, ap p [pseudoscalar] , and ap v [scalar] . That must mean ap [vector] . Finally, we consider, v (p p)
(g)
v ( p1i p2 j ij ) v v1iv2jv3iv4j ( i j i i j j ij ) v v1i v2j v3i v4j ( i j
j
i jj ) i j j
v v2 v1 j v v1 v2j v3iv4j ij j ( v v 2 ) ( v1 v3 v4 ) ( v v1) ( v2 v3 v4 ) sa sa
Under parity, we have sa sa s(a) s(a) sa s a a ; that is, we have a pseudoscalar.
(1.14)
