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Electromagnetism Physics 15b
Lecture #4 Divergence and Laplacian
Purcell 2.7–2.12
What We Did Last Time Used Gauss’s Law on infinite sheet of charge Uniform electric field E = 2πσ above and below the sheet Electric field has energy with volume density given by
Defined electric potential by line integral
Electric field is negative gradient of electric potential
Potential due to charge distribution: or
Total energy of a charge distribution:
u =E2
8π
φ21 = − E ⋅ds
P1
P2∫ = φ(P2) −φ(P1) unit: erg/esu = statvolt
E = −∇φ
φ =
qj
rjj =1
N
∑ φ =
dqr∫
U =
12
ρφ dv∫
2
Today’s Goals Introduce divergence of vector field How much “flow” is coming out per unit volume
Translate Gauss’s Law into a differential (local) form Gauss’s Divergence Theorem connects the two forms
Look in the energy again
Equivalence of and
Define the Laplacian = divergence of gradient Re-express Gauss’s Law with a Laplacian
Study mathematical properties of Laplace’s equation Conclude with a Uniqueness Theorem
U =
12
ρφ dv∫ U =
E 2
8πdV∫
Shrinking Gauss’s Law Charge is distributed with a volume density ρ(r) Draw a surface S enclosing a volume V
Guass’s Law:
Now, make V so small that ρ is constant inside V
As we make V smaller, the total flux out of S scales with V
Therefore:
LHS is “how much E is flowing out per unit volume” Let’s call it the divergence of E
E ⋅da
S∫ = 4π ρdvV∫ Total charge in V
E ⋅da
S∫ = 4πρV for very small V
limV→0
E ⋅daS∫
V= 4πρ
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Divergence In the small-V limit, the integral depend on volume, but not on the shape We can use a rectangular box Consider the left (S1) and right (S2) walls
Add up all walls:
divE ≡ lim
V→0
E ⋅daS∫
V= 4πρ
dx
dy
dz
E(x + dx,y,z) E(x,y,z)S1 S2
E ⋅da
S1∫ = E(x,y,z) ⋅ (−x)dydz
E ⋅da
S2∫ = E(x + dx,y,z) ⋅ xdydz
Sum = Ex(x + dx) − Ex(x)( )dydz
=∂Ex
∂xdxdydz
E ⋅da
S∫ =∂Ex
∂x+∂Ey
∂y+∂Ez
∂z
⎛
⎝⎜
⎞
⎠⎟V = ∇ ⋅E( )V
div E
Gauss’s Law, Local Version We now have Gauss’s Law for a very small volume/surface
Connects local properties of E with the local charge density
Divergence has the easy form in Cartesian coordinates In cylindrical coordinates:
In spherical coordinates:
divE = 4πρ where divE ≡ ∇ ⋅E =
∂Ex
∂x+∂Ey
∂y+∂Ez
∂z
⎛
⎝⎜
⎞
⎠⎟
∇ ⋅F =
1r 2
∂(r 2Fr )∂r
+1
r sinθ∂(Fθ sinθ)
∂θ+
1r sinθ
∂Fφ
∂φ
∇ ⋅F =
1r∂(rFr )∂r
+1r∂Fφ
∂φ+∂Fz
∂z
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Coulomb Field Let’s calculate div E for
We can do this by expressing E in x-y-z :
Or we can use div in spherical coordinates
Since only Er is non-zero, we get
This is correct — we have no charge except at r = 0 At r = 0, 1/r2 gives us an infinity That’s OK because a “point” charge has an infinite density
E =
qr 2 r
E =
qx2 + y 2 + z2
xx + yy + zz
x2 + y 2 + z2
∇ ⋅F =
1r 2
∂(r 2Fr )∂r
+1
r sinθ∂(Fθ sinθ)
∂θ+
1r sinθ
∂Fφ
∂φ
∇ ⋅E =
1r 2
∂(r 2Er )∂r
=1r 2
∂q∂r
= 0
Spherical Charge Let’s give the “point” charge a small radius R We did this in Lecture 2, and the solution was
For r < R,
The charge density of the sphere is
It works everywhere (as long as ρ is finite)
E =
Qr2 r for r ≥ R
QrR3 r for r < R
⎧
⎨⎪⎪
⎩⎪⎪
This part is same as a point charge. We know div E = 0.
Let’s work on this part
∇ ⋅E =
1r 2
∂(r 2Er )∂r
=1r 2
∂∂r
Qr 3
R3
⎛
⎝⎜⎞
⎠⎟=
3QR3
ρ =
Q4π3 R3 ∇ ⋅E = 4πρ
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Divergence Theorem We got div E = 4πρ from the “original” Gauss’s Law by shrinking the volume/surface We should be able to go back by “integrating”
Start from a volume V and cut into sub-volumes V1 and V2 Surface integrals add up:
because the integrals on the boundary cancel Divide V1 and V2 into smaller volumes …
E ⋅da
S∫ = E ⋅daS1∫ + E ⋅da
S2∫ V1
V2
E ⋅da
S∫ = E ⋅daSj∫
j∑ where V = Vj
j∑
E ⋅da
S∫ = limVj →0
E ⋅daSj∫
j∑ = lim
Vj →0(∇ ⋅E)Vj
j∑ = ∇ ⋅E
V∫ dv
Divergence Theorem For any vector field F
This is Gauss’s Divergence Theorem This is a mathematical theorem — No physics in it
Two forms of Gauss’s Law (physics) are connected by the Divergence Theorem
As a math theorem, Divergence Theorem can be useful in other ways, too Question from Lecture 3: How can the electrostatic energy be
F ⋅da
S∫ = ∇ ⋅FV∫ dv
F ⋅da
S∫ = ∇ ⋅FV∫ dv
E ⋅da
S∫ = 4π ρdvV∫ ∇ ⋅E = 4πρ
U =
E 2
8πdV∫ and U =
12
ρφ dv∫
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Electrostatic Energy Consider the divergence of the product Eϕ
Integrate LHS over very large volume and use Divergence Theorem
Integral of RHS must be 0, too
∇ ⋅ (Eφ) = ∂∂x
(Exφ) + ∂∂y
(Eyφ) + ∂∂z
(Ezφ)
=∂Ex
∂xφ + Ex
∂φ∂x
+∂Ey
∂yφ + Ey
∂φ∂y
+∂Ez
∂zφ + Ez
∂φ∂z
= (∇ ⋅E)φ +E ⋅ (∇φ) = 4πρφ − E 2
∇ ⋅ (Eφ)dv
V∫ = (Eφ) ⋅daS∫ = 0 assuming Eφ → 0 at far away
4π ρφ dv∫ − E 2 dv∫ = 0
12
ρφ dv∫ =E 2
8πdv∫
Laplacian Now we know: Let’s combine them
Laplacian is defined as the “divergence of the gradient” It represents the local curvature of the function
Laplacian allows us to calculate ρ from ϕ
E = −∇φ ∇ ⋅E = 4πρ
∇ ⋅ (−∇φ) = −
∂2φ∂x2 +
∂2φ∂y 2 +
∂2φ∂z2
⎛
⎝⎜⎞
⎠⎟≡ −∇2φ = 4πρ
Laplacian
x
y
φ ∇
2φ < 0
x
y
φ
∇2φ > 0
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Charge, Field, and Potential
Electric FieldE
Chargedensity
ρ φ Electric
potential
4πρ = ∇ ⋅E
E =
ρr 2 r dv∫
4πρ = −∇2φ
φ =
ρr
dv∫
E = −∇φ
φ = − E ⋅ds∫
Laplace’s Equation Where there is no charge, electric potential satisfies
This applies to almost everywhere in any E&M problem
Solutions of Laplace’s Equation are not necessarily trivial
Solutions to Laplace’s Equation have interesting and useful mathematical properties
∇2φ = 0 Laplace’s Equation
+σ
−σ
q
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Average Theorem Theorem: If ϕ satisfies Laplace’s equation, the average value of ϕ over a surface of any sphere equals to the value of ϕ at the center of the sphere Consider two concentric spheres S and S’ with radii r and r + dr Average values of ϕ over S and S’ are
Consider the difference φ =
14π
φ(r)dΩ∫ , ′φ =1
4πφ(r + dr)dΩ∫
dr
r
′φ − φ =1
4πφ(r + dr) −φ(r)( )dΩ∫
=1
4π∇φ(r) ⋅dr dΩ∫
=dr
4πr 2 ∇φ daS∫
Average Theorem Apply the Divergence theorem
Thus, the average values over S and S’ are the same We can repeat this with successively smaller spheres, until we reach
the center of the sphere
Theorem: If ϕ satisfies Laplace’s equation in a given volume, it has no maxima or minima inside the volume Note: it may have maxima or minima at the border Suppose ϕ has a maximum at point P. One can draw a (small)
spherical surface S around P so that the value of ϕ on S is smaller than that at P. This contradicts the average theorem
dr4πr 2 ∇φ da
S∫ =dr
4πr 2 ∇2
V∫ φdv = 0 ∇2φ = 0
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Impossibility Theorem Theorem: No electrostatic field can hold a charged particle in a stable equilibrium in empty space Such a field must have a local minimum (or maximum), which is
impossible
This Impossibility Theorem implies that one cannot build a configuration of more than one electric charges that would remain stable with the electrostatic forces alone Same applies to gravity, too Then, how can anything stable exist at all?
Stable systems can exist either: because they are not static, or because of quantum mechanics
Uniqueness Theorem Theorem: The potential ϕ inside a volume is uniquely determined if the charge density ρ in the volume and the potential ϕ at the boundary are given Suppose there are two solutions, ϕ1 and ϕ2 that satisfy
Take the difference: φ = ϕ1 − ϕ2
i.e., φ is a solution to Laplace’s equation Because of the no-max/min theorem, φ = 0 everywhere in V
It means that our formulas are complete (as well as being consistent) for describing physical systems
−∇2φ1 = −∇2φ2 = 4πρ inside V and φ1 = φ2 on the bounrdary S
−∇2ϕ = −∇2(φ1 −φ2) = 0 in V and ϕ = φ1 −φ2 = 0 on S
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Summary Defined divergence:
Guass’s Law (local version): Linked to the integral version
by the Divergence Theorem:
Defined the Laplacian:
From Gauss’s Law:
Laplace’s equation: Average theorem, no-max/min theorem,
impossibility theorem, uniqueness theorem
divF = lim
V→0
F ⋅daS∫
V= ∇ ⋅F =
∂Fx
∂x+∂Fy
∂y+∂Fz
∂z
⎛
⎝⎜
⎞
⎠⎟
∇ ⋅E = 4πρ
F ⋅da
S∫ = ∇ ⋅FV∫ dv
∇2f = ∇ ⋅ ∇f( ) = ∂2f
∂x2 +∂2f∂y 2 +
∂2f∂z2
⎛
⎝⎜⎞
⎠⎟
4πρ = −∇2φ
Electric FieldE
Chargedensity
ρ φ Electric
potential
4πρ = ∇ ⋅E
E =
ρr 2 r dv∫
4πρ = −∇2φ
φ =
ρr
dv∫
E = −∇φ
φ = − E ⋅ds∫
∇2φ = 0