Advanced Calculus fi..

Chapter 5 Vector Integral Calculus 341In the absence of conductors, E and H satisfy the Marwell equations:a) div E = 47rp, b) div H = 0,c) curlE =1 aH1 aEd) curl H = --c at'(5.120)c ar 'where p is the charge density and c is a universal constant.We now apply (a), ...,(d) of (5.120) and our rules of vector calculus toshow that E and H are expressible in terms of suitable potentials, which in turnsatisfy certain partial differential equations. We assume that all derivativesappearing here are continuous, so that the order of differentiation can bechanged and that the discussion is confined to a simply connected domain towhich Theorem IV of Section 5.13 applies, so that each irrotational field isa gradient and each solenoidal field is a curl.From (b) of (5.120), we can choose a vector field A (the vectorpotential)such thatH = curl A. (5.121)Here both H and A depend on x , y , z and t, as do all functions appearing inthis analysis. As in Problem 5 following Section 5.13, A is not unique; to anychoice of A one can add grad 4, where 4 is an arbitrary function of x, y, zand t. From (c) and (5.121) we deduce that.[SE 1 .?(5.122)and hence that Q (the scalar potential) can be chosen so thatIf we take the divergence of both sides and use (a), we obtainFrom (d) and (5.121) we obtain:T1 aEcurl curl A --- = 0.* ac at(5.125)From (5.123), by differentiation,Addition of the last two equations and application of the identity (3.40) ofSection 3.6 for curl curl A givesWe now use the freedom in the choice of A to achieve the Lorentz condition: