29.1. DIVERGENCE AND CURL OF A VECTOR FIELD 569

This establishes the first formula. The second formula is done similarly. Now consider thethird.

(∇× (∇×f))i = ε i jk∂ j (∇×f)k = ε i jk∂ j (εkrs∂r fs)

=

=ε i jk︷︸︸︷εki j εkrs∂ j (∂r fs) = (δ irδ js −δ isδ jr)∂ j (∂r fs)

= ∂ j (∂i f j)−∂ j (∂ j fi) = ∂i (∂ j f j)−∂ j (∂ j fi)

=(

∇(∇ ·f)−∇2f)

i

This establishes the third identity.Consider the fourth identity.

∇ · (f ×g) = ∂i (f ×g)i = ∂iε i jk f jgk

= ε i jk (∂i f j)gk + ε i jk f j (∂igk)

=(εki j∂i f j

)gk −

(ε jik∂igk

)fk

= ∇×f ·g−∇×g ·f.

This proves the fourth identity.Consider the fifth.

(∇× (f ×g))i = ε i jk∂ j (f ×g)k = ε i jk∂ jεkrs frgs

= εki jεkrs∂ j ( frgs) = (δ irδ js −δ isδ jr)∂ j ( frgs)

= ∂ j ( fig j)−∂ j ( f jgi)

= (∂ jg j) fi +g j∂ j fi − (∂ j f j)gi − f j (∂ jgi)

= ((∇ ·g)f +(g ·∇)(f)− (∇ ·f)g− (f ·∇)(g))i

and this establishes the fifth identity.

29.1.2 Vector PotentialsOne of the above identities says ∇ ·(∇×f) = 0. Suppose now ∇ ·g= 0. Does it follow thatthere exists f such that g = ∇×f ? It turns out that this is usually the case and when suchan f exists, it is called a vector potential. Here is one way to do it, assuming everythingis defined so the following formulas make sense.

f (x,y,z)=(∫ z

0g2 (x,y, t) dt,−

∫ z

0g1 (x,y, t) dt +

∫ x

0g3 (t,y,0) dt,0

)T

. (29.1)

In verifying this you need to use the following manipulation which will generally holdunder reasonable conditions but which has not been carefully shown yet.

∂

∂x

∫ b

ah(x, t) dt =

∫ b

a

∂h∂x

(x, t) dt. (29.2)

The above formula seems plausible because the integral is a sort of a sum and the deriva-tive of a sum is the sum of the derivatives. However, this sort of sloppy reasoning willget you into all sorts of trouble. The formula involves the interchange of two limit opera-tions, the integral and the limit of a difference quotient. Such an interchange can only beaccomplished through a theorem. The following gives the necessary result.