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1 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 = E 2 8π φ 21 = E d s P 1 P 2 = φ(P 2 ) φ(P 1 ) unit: erg/esu = statvolt E = −∇φ φ = q j r j j =1 N φ = dq r U = 1 2 ρφ dv

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Page 1: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

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∫

Page 2: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

Page 3: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

Page 4: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

Page 5: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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∫

Page 6: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

Page 7: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

q

Page 8: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

Page 9: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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

Page 10: Electromagnetism - Harvard Universityusers.physics.harvard.edu/~morii/phys15b/lectures/Lecture4.pdf · 1 Electromagnetism Physics 15b Lecture #4 Divergence and Laplacian Purcell 2.7–2.12

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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