J2 V 1 J2 1v J2

V • [$2 Ji ] = V $2 • Ji + $2 V • Ji where Ji = —oi V$i for i = 1, 2. Subtracting these equations and assuming that o1 = o2 = o, the first terms on the RHS of each equation cancel. By assumption V • J2 = 0 in V and J1 • n = 0 on S. Integrating over the volume V and using the divergence theorem to write the LHS as a surface integral over S leads to

Js JV

Fig. 4. Reciprocal source and measurement configurations for EEG reciprocity theorem

5 The derivations presented here assume isotropy for simplicity.

In configuration ti, let the current source and sink be located at r± = ri ^ d/2 and let the dipole strength be p = I1d, where d is the dipole separation. We have

where the sign convention is such that J1 = —a{V$1. In the notation of

0 0

Post a comment