Handbook of Solvents - George Wypych - ChemTech - Ventech!

4.3 Polar solvation dynamics 135
For t→ 0, D(t) becomes ε eE, and for t→∞it is D = ε sE. The relaxation process which
carries the initial response to its final value is exponential, with the characteristic relaxation
time, τ D.
The result [4.3.11] is relevant for an experiment in which a potential difference is suddenly
switched on and held constant between two electrodes separated by a dielectric
spacer. This means that the electrostatic field is held constant as the solvent polarization relaxes.
For this to happen the surface charge density on the electrodes, i.e. the dielectric displacement
D, has to change under the voltage source so as to keep the field constant.
The solvation dynamics experiment of interest here is different: Here at timet=0the
charge distribution ρ(r) is switched on and is kept constant as the solvent relaxes. In other
words, the dielectric displacement D, the solution of the Poisson equation ∇D =4πρ that
corresponds to the given ρ is kept constant while the solvent polarization and the electrostatic
field relax to equilibrium. To see how the relaxation proceeds in this case we start
again from
t
() = e () + ∫ ′ ( − ′ ) ( ′ )
Dt ε Et dtε t t Et ~ [4.3.12]
−∞
take the time derivative of both sides with respect to t
dD
dt
dE
Et () ( ) dt ()
dt
d
t
= e + + ′ Et
dt
⎛ ~ ⎞
ε ~
ε
ε 0 ∫ ⎜ ⎟ ′
[4.3.13]
⎝ ⎠
use the relations ~ ε(0) = ( )
t
dt d
′
dt
⎛ ~ ε⎞
⎝ ⎠
−∞ t− t′
ε − ε / τ and
s e D
⎜ ⎟ Et () ′ =− ~
∫ ∫ ε(
t− t′ ) Et ( ′ ) =− ( Dt () −εeEt () )
−∞ t− t′ D −∞
(cf Eq. [4.3.8]), to get
d
dt
( D − ε E) =− ( D −ε
E)
e
1
τ
D
t
1 1
τ
s
τ
D
[4.3.14]
When D evolves under a constant E, Eq. [4.3.14] implies that (d/dt)D = (-1/τ D)D + constant,
so that D relaxes exponentially with the time constant τ D, as before. However if E relaxes
under a constant D, the time evolution of E is given by
i.e.
d
dt
εs
⎛ 1 ⎞
E =− ⎜
⎜E
− D⎟
εeτ ⎟
D ⎝ εs
⎠
1
Et () = D+ Ae
ε
s
−t/
τL
[4.3.15]
[4.3.16]