Page 263 - Electromagnetics
P. 263
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and a frequency-domain magnetic source current J m . The fields (E 1 ,D 1 ,B 1 ,H 1 ) within
the region (which may also contain arbitrary media) are described by
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∇× E 1 =−J m − jωB 1 , (4.181)
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∇× H 1 = J + jωD 1 , (4.182)
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∇· D 1 = ˜ρ, (4.183)
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∇· B 1 = ˜ρ m . (4.184)
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Suppose we have been given a mathematical description of the sources (J, J m ) and have
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solved for the field vectors (E 1 , D 1 , B 1 , H 1 ). Of course, we must also have been supplied
with a set of boundary values and constitutive relations in order to make the solution
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unique. We note that if we replace the formula for J with the formula for J m in (4.182)
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(and ˜ρ with ˜ρ m in (4.183)) and also replace J m with −J in (4.181) (and ˜ρ m with −˜ρ in
(4.184)) we get a new problem. However, the symmetry of the equations allows us to
specify the solution immediately. The new set of curl equations requires
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∇× E 2 = J − jωB 2 , (4.185)
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∇× H 2 = J m + jωD 2 . (4.186)
If we can resolve the question of how the constitutive parameters must be altered to
reflect these replacements, then we can conclude by comparing (4.185) with (4.182) and
(4.186) with (4.181) that
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E 2 = H 1 , B 2 =−D 1 , D 2 = B 1 , H 2 =−E 1 .
The discussion regarding units in § 2.9.2 carries over to the present case. Multiplying
Ampere’s law by η 0 = (µ 0 / 0 ) 1/2 , we have
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∇× E =−J m − jωB, ∇× (η 0 H) = (η 0 J) + jω(η 0 D).
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Thus if the original problem has solution (E 1 ,η 0 D 1 , B 1 ,η 0 H 1 ), then the dual problem
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with J replaced by J m /η 0 and J m replaced by −η 0 J has solution
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E 2 = η 0 H 1 , (4.187)
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B 2 =−η 0 D 1 , (4.188)
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η 0 D 2 = B 1 , (4.189)
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η 0 H 2 =−E 1 . (4.190)
As with duality in the time domain, the constitutive parameters for the dual problem
must be altered from those of the original problem. For linear anisotropic media we have
from (4.13) and (4.14) the constitutive relationships
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D 1 = ˜ ¯ 1 · E 1 , (4.191)
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B 1 = ˜ ¯µ · H 1 , (4.192)
1
for the original problem, and
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D 2 = ˜ ¯ 2 · E 2 , (4.193)
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B 2 = ˜ ¯µ · H 2 , (4.194)
2
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