Handbook of Solvents - George Wypych - ChemTech - Ventech!

134 Abraham Nitzan
t
Dt ()= ∫ dt′ ε ( t− t′ ) E( t′
)
[4.3.2]
−∞
−iωt
and their Fourier transforms (e.g. E( ω)
= dte E( t ) satisfy
where
∞
∫
−∞
D( ω) = ε( ω) E(
ω)
[4.3.3]
∞
−iωt
() ≡∫dte
() t
εω ε
rewriting ε(t) in the form
we get
0
() t = () t + () t
[4.3.4]
ε 2 ε δ ~ ε
[4.3.5]
e
t
() = e () + ∫ ′ ( − ′ ) ( ′ )
Dt ε Et dtε t t Et ~ [4.3.6]
−∞
() ω ε () ω ε()() ω ω
D = E + E ~ [4.3.7]
e
In Eq. [4.3.5] ε e is the “instantaneous” part of the solvent response, associated with its
electronic polarizability. For simplicity we limit ourselves to the Debye model for dielectric
relaxation in which the kernel ~ ε in [4.3.5] takes the form
~ εs εe
ε()
t = e
τ
− −t/
τD
D
[4.3.8]
This function is characterized by three parameters: the electronic ε e and static ε s response
constants, and the Debye relaxation time, τ D. In this case
leads to
()
εω = ε +
e
∞
∫
0
εs −εe
/
dt e e
τ
D
−t τ −iωt
ε −ε
= εe
+
1+
iωτ
s e
In this model a step function change in the electrostatic field
() ()
D
[4.3.9]
Et = 0, t< 0; Et = Et , ≥0
[4.3.10]
t
εs −εe
−( t− t′
)/ τD −t/
τD
() = εe()
+
() s(
1 )
∫
Dt Et e Et′ dt′ = ε − e + εee
τ
0
D
[ ]
−t/τD E
[4.3.11]