Key Concepts for this section
1: Lorentz force law, Field, Maxwell’s equation
2: Ion Transport, Nernst-Planck equation
3: (Quasi)electrostatics, potential function,
4: Laplace’s equation, Uniqueness
5: Debye layer, electroneutrality
Goals of Part II:
(1) Understand when and why electromagnetic (E and B)
interaction is relevant (or not relevant) in biological
systems.
(2) Be able to analyze quasistatic electric fields in 2D
and 3D.
Electrostatics
Steady Diffusion
Φ=V0
Φ=0
Gel or tissue
(σ,ε)
Φ=0
2
∇ Φ=0
r
J e = −σ∇Φ
C=C0
Φ=0
C=0
Gel or tissue
(D)
C=0
∇ 2C = 0
r
J i = − Di ∇ci
C=0
2
2
2
∂
Φ
∂
Φ
∂
Φ
∇ 2Φ = 2 + 2 + 2 = 0
∂x
∂y
∂z
Assume Φ ( x, y, z ) = Χ( x) ϒ( y ) Ζ( z )
2
2
2
∂
Χ
∂
ϒ
∂
Ζ
∇ 2 Φ = ϒΖ 2 + ΧΖ 2 + Χϒ 2 = 0
∂x
∂y
∂z
1 ∂2Χ 1 ∂2ϒ 1 ∂2Ζ
+
+
=0
2
2
2
Χ ∂x
ϒ ∂y
Ζ ∂z
123
123 123
function
of x
function
of y
function
of z
Three possibilities
1 ∂2Χ
+ kx x
− kx x
2
=
⇒
Χ
(
)
=
,
k
x
e
e
x
Χ ∂x 2
or = − k x 2 ⇒ Χ( x) = sin(k x x), cos(k x x)
or = 0 ⇒ Χ( x) = ax + b (a, b : constants )
∇ 2 Φ = 0, Φ ( x, y ) = Χ( x) ϒ( y )
2
Φ=V0
2
1 ∂ Χ 1∂ ϒ
+
=0
Χ ∂x 2 ϒ ∂y 2
1 ∂2Χ
2
=
−
k
X ( x) ~ sin(kx)
2
Χ ∂x
sin(kL) = 0 ⇒ kL = nπ (n : integer)
nπ
Eigenvalue : kn =
L
Φ=0
expand Χ(x) using Fourier sine series
Gel or tissue
(σ,ε)
Φ=0
⎛ nπ x ⎞
Χ( x) = ∑ An sin ⎜
⎟ (This satisfies B. C. at x=0, L)
L
⎝
⎠
n
∂ 2 ϒ( y )
⎛ nπ y ⎞
⎛ nπ y ⎞
2
−
ϒ
=
⇒
ϒ
k
y
y
or
then,
(
)
0
(
)
~
sinh
cosh
n
⎜
⎟
⎜
⎟
∂y 2
⎝ L ⎠
⎝ L ⎠
⎛ nπ y ⎞
⎛ nπ x ⎞
⎛ nπ y ⎞
x
x
y
A
sinh
since
(
,
0)
0
(
,
)
sin
ϒ( y ) = sinh ⎜
Φ
=
∴
Φ
=
∑
n
⎜
⎟
⎜
⎟
⎟
⎝ L ⎠
⎝ L ⎠
⎝ L ⎠
n
Determining A n : use boundary condition
⎛ nπ x ⎞
Φ ( x, L) = V0 = ∑ An sin ⎜
⎟ sinh ( nπ )
L
⎝
⎠
n
L
2V (1 − cos(nπ ))
⎛ mπ x ⎞
operate ∫ sin ⎜
on both sides ⇒ An = 0
⎟
0
nπ sinh(nπ )
⎝ L ⎠
Φ=0
Solving Laplace’s Equation (Numerically)
2
d Φ
= 0 → Φ ( x) = ax + b
1D case:
2
dx
Φ (n − 1, m)
2
2
∂
Φ
∂
Φ
2D case:
+
=0
2
2
∂x
∂y
∂Φ
1
(n + , m) = Φ (n + 1, m) − Φ (n, m)
∂x
2
Φ (n, m + 1)
Φ ( n, m )
Φ (n + 1, m)
Φ (n, m − 1)
y (m)
x (n)
∂Φ
1
(n − , m) = Φ (n, m) − Φ (n − 1, m)
∂x
2
1
1
∂ 2Φ
∂Φ
∂Φ
(
n
,
m
)
(
n
,
m
)
(
n
, m) = Φ (n + 1, m) + Φ (n − 1, m) − 2Φ (n, m)
=
+
−
−
2
2
2
∂x
∂x
∂x
Φ (n, m + 1)
Laplace’s equation
In discretized form
Φ (n − 1, m)
Φ ( n, m )
Φ (n + 1, m)
Φ (n, m − 1)
y (m)
∂ 2Φ
∂ 2Φ
( n, m ) + 2 ( n , m ) =
2
∂x
∂y
Φ (n + 1, m) + Φ (n − 1, m) + Φ (n, m + 1) + Φ (n, m − 1) − 4Φ (n, m) = 0
x (n)
Φ (n + 1, m) + Φ (n − 1, m) + Φ (n, m + 1) + Φ (n, m − 1)
Φ ( n, m ) =
4
Value in the middle = average of surrounding values
Finite Element Method
Known Solutions for Laplace equations
Cylindrical Coordinates
∂ 2 Φ 1 ∂Φ 1 ∂ 2 Φ ∂ 2 Φ
∇ Φ( ρ , ϕ , z ) = 0 ⇒
+
+ 2
+ 2 =0
2
2
ρ ∂ρ ρ ∂ϕ
∂ρ
∂z
Φ ( ρ , ϕ , z ) = R( ρ )Ψ (ϕ ) Ζ( z )
R( ρ ) ⇒ Bessel Functions ( J n , N n , I n , K n )
2
Ψ (ϕ ) ⇒ Trigonometric (sin, cos,sinh, cosh)
Ζ( z ) ⇒ Trigonometric (sin, cos,sinh, cosh)
Spherical Coordinates
∇ Φ(r ,θ , ϕ ) = 0 ⇒
2
1 ∂ ⎛ 2 ∂Φ ⎞
1
∂ ⎛
∂Φ ⎞
1
∂ 2Φ
=0
⎜r
⎟+ 2
⎜ sin θ
⎟+ 2 2
2
2
r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝
∂θ ⎠ r sin θ ∂ϕ
Φ (r ,θ , ϕ ) = R(r )Θ(θ )Ψ (ϕ )
R(r ) ⇒ Spherical Bessel Functions
Θ(θ ) ⇒ Legendre Functions ( Pn (cos θ ))
Ψ (ϕ ) ⇒ Trigonometric (sin ϕ , cos ϕ )
Cell in a field
σ0, ε (medium)
Eext
R
σi , ε
ẑ
Equation to solve :
r
r
∇ ⋅ J e = ∇ ⋅ (σ E ) = −∇ ⋅ (σ∇Φ) = 0 ∴∇ 2 Φ = 0 ( Laplace ' s Equation)
1 ∂ ⎛ 2 ∂Φ ⎞
1
∂ ⎛
∂Φ ⎞
1
∂ 2Φ
∇ Φ (r ,θ , ϕ ) = 0 ⇒
=0
⎜r
⎟+
⎜ sin θ
⎟+
r 2 ∂r ⎝ ∂r ⎠ r 2 sin θ ∂θ ⎝
∂θ ⎠ r 2 sin 2 θ ∂ϕ 2
Φ (r , θ , ϕ ) = R(r )Θ(θ )
separate and solve,
1
R(r ) ⇒ Ar n + B n +1
r
Θ(θ ) ⇒ Legendre Functions ( Pn (cos θ ))
2
0
Guessing the solution
r
E → Eext zˆ
as
r →∞
Pn(cosθ) ~ cos nθ
Only n =1 term contributes
(should be “dipole” field)
Φ = − Eext z = − Eext r cos θ
Eext
Trial Solution:
--
r →∞
+
+
θ -z
+
σ
,
ε
++ i
+
σ0, ε (medium)
1
cos θ (for r ≥ R)
2
r
1
Φ i = Cr cos θ + D 2 cos θ (for r ≤ R)
r
D = 0 (Φ i finite at r=0)
Φ o = Ar cos θ + B
A = − Eext
+
as
(Φ o → − Eext r cos θ when r → ∞)
Boundary Conditions (For EQS approximation)
ur
∇ ⋅ (ε E ) = ρe
r
∇× E = 0
uur
∂ρ
∇ ⋅ Je = −
∂t
uur
uur
nˆ ⋅ (ε1 E1 − ε 2 E2 ) = σ s
uur
uur
r
nˆ × E1 = nˆ × E2 ( E1
tangential
r
= E2
tangential
)
uur
uur
∂σ
nˆ ⋅ (σ 1 E1 − σ 2 E2 ) = − s
∂t
Figure 5.3.1 (a) Differential contour intersecting surface supporting surface charge density. (b) Differential
volume enclosing surface charge on surface having normal n.
Courtesy of Herman Haus and James Melcher. Used with permission.
Source: http://web.mit.edu/6.013_book/www/
Some plots for the solution
Cell is less conductive than media
Insulating Cell
σ < σ0
Cell is more conductive than media
σ > σ0
σ=0
Perfectly conducting Cell
σ=∞
Figure by MIT OCW.