25.9. CURL AND CROSS PRODUCTS 483

= εrst det

(∂xp

∂yq

)det(

∂ zi

∂xk

)= ε

rst det(MN)

where N is the matrix whose pqth entry is ∂xp

∂yq and M is the matrix whose ikth entry is ∂ zi

∂xk .Therefore, from the definition of matrix multiplication and the chain rule, this equals

= εrst det

(∂ zi

∂yp

)≡ ε

rst (z)

from the above discussion.Now ε i jk (y) = ε i jk and for a vector field, F,

curl(F )≡ εi jk (y)Fk, j (y)ei (y) .

Therefore, since we know how everything transforms assuming 25.45, it is routine to writethis in terms of x.

curl(F ) = εrst (x)

∂yi

∂xr∂y j

∂xs∂yk

∂xt Fp,q (x)∂xp

∂yk∂xq

∂y j em (x)∂xm

∂yi

= εrst (x)δ

mr δ

qs δ

pt Fp,q (x)em (x) = ε

mqp (x)Fp,q (x)em (x) . (25.48)

More simplification is possible. Recalling the definition of Fp,q (x) ,

∂F

∂xq ≡ Fp,q (x)ep (x) =

∂xq [Fp (x)ep (x)]

=∂Fp (x)

∂xq ep (x)+Fp (x)∂ep

∂xq =∂Fp (x)

∂xq ep (x)−Fr (x)

{r

pq

}ep (x)

by Theorem 25.6.2. Therefore,

Fp,q (x) =∂Fp (x)

∂xq −Fr (x)

{r

pq

}and so

curl(F ) = εmqp (x)

∂Fp (x)

∂xq em (x)− εmqp (x)Fr (x)

{r

pq

}em (x) .

However, because

{r

pq

}=

{r

qp

}, the second term in this expression equals 0. To

see this,

εmqp (x)

{r

pq

}= ε

mpq (x)

{r

qp

}=−ε

mqp (x)

{r

pq

}.

Therefore, by 25.48,

curl(F ) = εmqp (x)

∂Fp (x)

∂xq em (x) . (25.49)

What about the cross product of two vector fields? Let F and G be two vector fields.Then in terms of standard coordinates y,

F ×G= εi jk (y)Fj (y)Gk (y)ei (y)

25.9. CURL AND CROSS PRODUCTS 483P i= e'™ det (55) det (53) = e™ det (MN)where N is the matrix whose pq" entry is x and M is the matrix whose ik” entry is 2,Therefore, from the definition of matrix multiplication and the chain rule, this equalsi= e’™ det (55) =e (z)from the above discussion.Now e¥! (y) = e* and for a vector field, F,curl (F) = e* (y) Fj (y) ei (y)-Therefore, since we know how everything transforms assuming 25.45, it is routine to writethis in terms of a.dy! dy! ay* AxP Ax4 ox"Dat aw ax Pd") Tye Dyin ®) Bei=e! (x) 67 575? Fy g (x) Em (a) = €""4? (x) Fog (@) em (x). (25.48)More simplification is possible. Recalling the definition of Fy, (a),oF = Fpq(x)e? (x) = 2 [Fp (aw) e? (x)= MOL) or (a) 4 Fy (a) 2 = OP) 6) F, | ’ he (x)curl (F’) = e’” (x)x4 Oxd Oxdby Theorem 25.6.2. Therefore,OF, (x rFrat) = EO) rf ,and soOF (x)x4curl (EF) = e”!? (x)€m (x) — €™? (ax) F, (2) " Cm (@).Pqr rHowever, because = , the second term in this expression equals 0. ToPq qPemp | 0g \ em) op \ =e | 0g \.Therefore, by 25.48,see this,OF, (a)Ox4What about the cross product of two vector fields? Let F’ and G be two vector fields.Then in terms of standard coordinates y,Fx G=e'* (y) F;(y) Gx (y) ei(y)curl (F’) = e"!? (a) Cm (x). (25.49)