Biomedical Engineering Reference
In-Depth Information
J = kn
ρ
(x)
x
β
x
2
x
1
x
u
x
0
FIGURE 3.3
The prescription for deriving Kramers rate of escape,
k
. The distribution of
particles in the well at
x
o
provide a source, while the location outside the barrier
x
u
acts as
a sink. Therefore, a steady flow of particles, or flux
J
, is passing over the barrier. The flux is
defined as the number of particles available
n
times rate of escape. Thus, estimating the flux
and integrating over the distribution of particles in the well provides the transition rate (see
discussion in text).
in the well provide a constant flux of particles over the barrier (steady-state condi-
tions). To the far left, the steeply rising energy is essentially a reflecting boundary.
To the right, after particles pass over the barrier and reach
x
u
, they are completely
unbound and carried away from the potential indefinitely. Therefore, we define
x
u
as
an absorbing boundary with ρ
0. To calculate the flux over the barrier, we look
to integrate over
x
o
to
x
u
. Multiplying both sides of Equation 3.9 by an integrating
factor exp
(
x
u
)
≡
(
U
(
x
)
/
k
B
T
)
,wehave
U
(
∂ρ
(
x
)
ρ
(
x
)
x
)
η
J
k
B
T
e
U
(
x
)
/
k
B
T
e
U
(
x
)
/
k
B
T
e
U
(
x
)
/
k
B
T
+
=
−
(3.10)
∂
x
k
B
T
ρ
e
U
(
x
)
/
k
B
T
d
d
x
η
J
k
B
T
e
U
(
x
)
/
k
B
T
(
x
)
=
−
(3.11)
Where we have exploited the fact that the left side of Equation 3.10 is just the chain
rule derivative of ρ
(
x
)
exp
(
U
(
x
)
/
k
B
T
)
. Upon integration, and noting that ρ
(
x
u
)=
0,
we have
e
U
(
x
0
)
/
k
B
T
(
)
k
B
T
η
ρ
x
0
J
=
(3.12)
x
u
x
0
e
U
(
x
)
/
k
B
T
d
x
Now the flux
J
is the steady flow of particles moving from a source, the well, over
the barrier and into a sink (or carried away). To determine the actual rate
k
of par-
ticles escaping per unit time, we recognize that the flux is the product of the rate
with the number of particles available to escape,
J
kn
. Therefore, we need to find
how the particle concentration changes along
x
.Define
n
as the number of particles
in the well. Near the bottom of the well, around
x
0
,theflux is nearly zero and we
=
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