Page 57 - Electromagnetics

¯
It is somewhat laborious to obtain the constitutive matrix [C] for an arbitrary moving
medium. Detailed expressions for isotropic, bianisotropic, gyrotropic, and uniaxial media
are given by Kong [101]. The rather complicated expressions can be written in a more
compact form if we consider the expressions for B and D in terms of the pair (E, H).

For a linear isotropic material such that D =
E and B = µ H in the moving frame,

the relationships in the laboratory frame are [101]
¯
B = µ A · H − Ω × E, (2.107)
¯
D =
A · E + Ω × H, (2.108)
where
2
1 − β 2 n − 1
¯ ¯
A = I − ββ , (2.109)
2 2
1 − n β 1 − β 2
2
n − 1 β
Ω = , (2.110)
2 2
1 − n β c
1/2
and where n = c(µ
) is the optical index of the medium. A moving material that
is isotropic in its own moving reference frame is bianisotropic in the laboratory frame.
If, for instance, we tried to measure the relationship between the fields of a moving
isotropic fluid, but used instruments that were stationary in our laboratory (e.g., attached
to our measurement bench) we would find that D depends not only on E but also on
H, and that D aligns with neither E nor H. That a moving material isotropic in its
own frame of reference is bianisotropic in the laboratory frame was known long ago.
Roentgen showed experimentally in 1888 that a dielectric moving through an electric
field becomes magnetically polarized, while H.A. Wilson showed in 1905 that a dielectric
moving through a magnetic field becomes electrically polarized [139].
2
2
If v /c 1, we can consider the form of the constitutive equations for a first-order
2
2
Lorentz transformation. Ignoring terms to order v /c in (2.109) and (2.110), we obtain
¯
¯
2
2
A = I and Ω = v(n − 1)/c . Then, by (2.107) and (2.108),
v × E
2

B = µ H − (n − 1) , (2.111)
c 2
v × H
2

D =
E + (n − 1) . (2.112)
c 2
We can also derive these from the first-order field conversion equations (2.61)–(2.64).
From (2.61) and (2.62) we have
2

D = D + v × H/c =
E =
(E + v × B).
Eliminating B via (2.64), we have
2
2

D + v × H/c =
E +
v × (v × E/c ) +
v × B =
E +
v × B
2
2
where we have neglected terms of order v /c . Since B = µ H = µ (H − v × D),we

have
2

D + v × H/c =
E +
µ v × H −
µ v × v × D.
2
2
2
2

Using n = c µ
and neglecting the last term since it is of order v /c , we obtain
v × H
2

D =
E + (n − 1) ,
c 2
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