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1
Physics 435 : Grading
11 Home Work: 30%
Discussion Section: 5% attendance
3 Midterms : 30%
Final: 35%
Home Work due Friday in HW box
Friday < 5pm => 100% credit Monday < 5pm => 90% credit Thursday < 11 pm => 70% credit
Jim Wiss Office Hours
Thursdays 2:30pm to 4:30 pm in 413 LLP or by email
appointment. ([email protected])
Mid Terms (in class):
Fri Feb 22 Wed Mar 12 Wed Apr 23
I have heavily weighted the homework since it’s the only way of learning E&M at this level.
2
Welcome to Physics 435!
0
0 0
Gauss's Law :
Ampere-Maxwell
Faraday's Law:
ε
μ ε
∫ ⋅^ =
∫ ⋅^ =^ ⎜ +^ ∫ ⋅ ⎟
⎝ ⎣^ ⎦⎠
G G
G G G G
v i^ A
G G G (^) G v
E da Q encl d E d B da dt d B dl I E da dt
You saw 3 of 4 Maxwell Eqn in Physics 212
So why study E&M further?
Great applications of crucial mathematics: a) Vector Calculus b) Orthogonal functions c) Curvilinear coordinate systems
Physics beyond Phys 212 a) Differential form of Maxwell Eq, b) E&M in materials : Bound charge & currents c) Vector Potentials in Mech and Quant Mech d) Wave guides e) Radiation theory f) Relativistic transformations of E&M fields
E&M is the most practical branch of classical physics
It is correct as originally written by Maxwell in 1864:
a) Relativistically b) Quantum mechanically
E&M forms theoretical template for the successful gauge theory strong & weak interactions or the Standard Model.
Most of you have already been exposed to the basic physics of E&M by taking Physics 212 or its equivalent. Last semester we taught three of the four Maxwell Equations in integral form. You might recognize them. One of the important motivations for a second, two-semester course in E&M is that it gives us the excuse to introduce some really important and interesting mathematics that you will use in subsequent physics courses such as quantum mechanics. We will make extensive use of vector calculus such as divergence and curls. We will discuss the role of orthogonal functions in the solutions of partial differential equations. Fourier expansions is a possibly familiar use of orthogonal functions. Finally we will discuss the use of curvilinear coordinate systems such as spherical and cylindrical coordinate systems. These are generally the most elegant ways of handling spherical or cylindrically symmetric systems. The first chapter of your textbook , Electrodynamics by David J. Griffiths gives a very insightful overview of much of this mathematics. Of course we will also introduce some important physics beyond Physics 212. This includes obtaining the differential forms of Maxwell’s Eqn which nicely complement the integral forms that you are familiar with. The integral forms are particularly useful in very quickly solving highly symmetric problems such as the E-fields from spherically symmetric charge distributions. The differential forms are particularly useful in developing the theory of electrodynamics and in solving electromagnetic wave problems. We also discuss some of the very interesting and elegant physics involving electromagnetic fields in materials. We will show that the field is much richer than just substituting epsilon for epsilon_0 and mu for mu_0. The more complete theory makes extensive use of the concepts of bound charges and bound currents. I learned a lot about bound charges and currents while preparing to teach this course. We will also discuss the use of magnetic potentials. The most interesting magnetic potential is actual a vector potential with three components. The magnetic field can be written as derivatives of the vector potential much like the electric field can be written as the derivative of the scalar potential or voltage. It is fair to say that much of the modern physics created after 1960 that describes the subatomic world was inspired by the use of the vector potential to inject magnetic forces into Lagrangian dynamics and quantum mechanics. We will also discuss the radiation of electromagnetic waves, the theory of wave guides, and the role of relativity in understanding electrodynamics. The theory of relativity was invented by Einstein in the early1900’s who was exploring basic symmetry of electrodynamics.
Not only is E&M the most practical branch of classical physics but it is the only classical physics which is correct as far as we know. It requires no relativistic modifications although relativity gives considerable insight into the relationship between E and B fields. It carries over to quantum mechanics rather seamlessly as well and formed the basic template for modern field theories. Historically a good rule of thumb seems to be that the closer a new theory is to E&M the more likely the new theory is true.
4
Electrostatics: Coulomb & Gauss’s law
( )
( )
( )
3
3
0
0
∫
q r r E r r r
r r r E r d r r
G G
G G
G G
G G G
G G
G G
ε (^0) ∫ E da = Q inc
G G
i
ε 0 ε 0 ρ ( )
⎛ ∂ ∂ ∂ ⎞ ∇ = (^) ⎜ + + (^) ⎟= ⎝ ∂^ ∂^ ∂ ⎠
E Ex^ E^ y Ez r x y z
G G G i
E
G
λ
s
E
G E
G
σ
R
Q
volume
Divergence Theorem
surface
∫ ∇^ V d^^ τ= ∫ V da
G G^ G G
i i
( ) ( )
2 2
1 1
1
sin sin
sin
θ
θ
θ θ θ ρ θ θ ε (^0)
∂ ∂ ∇ Ε = + ∂ ∂ ∂
r E r E r r r E r
G G i
Coulomb Law for continuous distributions
Gauss’s Law in integral form
Gauss’s Law in differential form
Cylindrical and spherical coordinates
φ
y
x
z
r
r sin θ
r cos θ θ φ
y
x
z
s^ φˆ
ˆ z
r G
ˆ s
Here is what we will discuss in this first set of lectures. Some of this material may be
familiar from Physics 212. We will discuss computing the E-field from Coulomb’s
law both for point charges and continuous distributions. This will be followed by a
discussion of Gauss’s law and some of the classic Gauss’s law geometries. We
next write Gauss’s law in differential form which is better for some application. We
will typically work in Cartesian as well as curvilinear coordinate systems such as
spherical and cylindrical coordinates.
5
Coulomb's Law
12 2 1 2
2 2 9 12 2 2
1 12
2
Coulomb's Law
0
0
− 0
≈ × = ×
G G G
i
i
F r r r
r
r
q
N C
q
m
C N m
( )
( )
( ) ( ) ( )
1 2 1 2 (^12 ) 1 2
1 2 1 2 1 2 1 2 2 2 1 2 1 2 1 2 1 2
We can also write this as
where
G G
G
G G
G G
G G
q q r r
F
r r
r r x x y y z z
r r x x y y z z
12 ="force on 1 due to 2"
G
F
r 2
G
unit vector
r
r 1
G
q 1
q 2
12
r^ ˆ
O
Coulomb’s law is at the center of electrostatics and we will obtain many of the key
electrostatic results using this formulation. We will treat it as an experimental fact.
Coulomb says the force between two forces is proportional to the product of their
charges and inversely proportional to the square of the distance between them.
Coulomb’s constant will typically be written as 1/(4 pi epsilon_0) where epsilon_0 is
8.85 times 10^-12 in the MKS system which we will use throughout Physics 435
and 436. The force units are Newtons, the charge units are Coulomb and distance
is measured in meters. 1/(4 pi epsilon_0) implies an enormous force between
objects holding a Coulomb of charge – primarily since this is an astronomical
amount of charge. The charge of an electron or proton is 1.6 times 10^{-19}
Coulombs. The electrostatic force is a central force which means it points along the
line joining the two charges. We can thus write the force components between two
charges located with Cartesian coordates (x1,y1,z1) and (x2,y2,z2) in terms of the
indicated expression based on the difference of coordinates or in terms of the vector
difference between the displacement vectors r1 and r2. The form (r1-r2)/|r1 – r2|^
can be viewed as the unit vector r-hat = (r1-r2)/|r1 – r2| which sets the direction of
the force times an inverse r squared force or 1/|r1 – r2|^2. This is where the cubes
are coming from.
7
Continuous Distribution Example
(^2 2 2 ) 2 2 23 2 0
; 4 2 4
/ /
cos α cosα πε λ λ πε
0
0
= =
∫
A
y
y
dE dq^ h x h (^) x h dq dx E h^ dx x h
( )
( )
2
2 23 2 2
2 23 2 2
2 23 2
2 23 2 2 2 2
2
(^2 2 2 2 ) 2
(odd integra
1 4 0 4 4 2
4 2
nd)
integral table)
2
/ / / / / / / / / ' ' ' ' ' ' ' ' / /
(
λ πε
λ πε
λ πε
λ λ πε πε
0 −
0 −
0 −
(^0) − 0
−
− = =
=
=
⎛ ⎞ ⎛ ⎞ (^) ⎜ ⎟ = (^) ⎜ ⎟ = ⎝ + ⎠ ⎜⎜ ⎝ + ⎠
A A A A A A A A
A
A
x y
x
y
y
x h E E dx x h x E dx x h h E dx x h dx x
x a a^ x^ a
h x h E h x h (^) h h
2 2 4 2
ˆ λ πε (^0)
⎟⎟
⎛ ⎞ → = (^) ⎜ ⎟ ⎝ + ⎠
G (^) A
A
E y h (^) h
2 2
2 2
2 2
2 2
(^2 )
1 4
2 0 4 2
4 4
Limits of :
As
and (line charge)
As and
(point charge)
λ πε
λ πε
λ πε πε
0
0
0 0
⎛ ⎞ = (^) ⎜ ⎟ ⎝ + ⎠ → ∞ →
→
→ →
→ =
A A A A A
A A^ A A A
y
y
y
E h (^) h
h E h
h h q E h h
A
h
x^ ˆ
y^ ˆ
O
2 2
= ρ τ →λ = = − = −
− = +
G G
G G
G G
dq d dx r h r x r r x h
r r x h λ
( ) 3
∫
G G G
G G
G G
r r r E r d r r
(^2 2 2 )
2 2 23 2 0
4
2 4
/ /
cos ; cos
h
λ α α πε λ πε
0
0
= =
= ⊗ (^) ∫
A
y
y
dx h dE x h (^) x h dx E x h
Informal Method Add E in ±x pairs to cancel Ex and take y-component
h
x^ ˆ
α y ˆ
α
A
Here is an example of computing the field due to a finite, uniform line of charges. In order to simplify the problem, we put the observation point a distance Z above the center of the line of charge. The advantage of “centering” the observation point is that the component of the E-field parallel to the line of charge (i.e. the x-component) will cancel from +x and –x pairs leaving just the z-component. Our first step is to write an expression for the charge element dq. Since this is a 1 dimensional distribution the density expression rho d tau’ is replaced by lambda dx’ where lambda is in Coulombs/meter. We next write the observation point and source point vectors referenced to our origin (O) and coordinate system. We write just two components for simplicity and since nothing is happening in z. The observation point is at x=0 y=h. The source points are at x = x’ and y = 0. We take the difference of these two vectors to get r-r’ and use sqrt of sum of squares of components to construct |r – r’|. We next transcribe our expression for E from rho d tau’ to lambda dx’ The limits of integration are from -L/2 to L/2 since that is the only region with charge. We write the E and r-r’ vectors with explicit components. The Ex integral has an odd integrand over a symmetric domain and thus vanishes. For the Ey integral you will probably need an integral table. Fortunately a definite integral for this form exists. We use the integral table form with “a” =h and x = x’ and subtract the lower limit from the upper limit to get our final Ey expression. I hope you will find the forgoing straightforward although a bit tedious. We can cut through much of the tedium by an informal approach. We will integrate from 0 to L/2 and double our answer for Ey and realize that each +x charge will cancel the Ex field from each –x charge. We write the E- field element as the charge elements (Lambda dx) divided by the square of the observer to source distance which is the square of the hypotenuse of either right triangles times the constant. The E-field lies along the line from the source to the observation point and we want to take the Ey component by taking the cosine of angle with respect to the E-field direction and the y-axis which is marked alpha. This cosine is just the adjacent side (h) divided by the hypotenuse [sqrt(h^2+x^2)]. The sqrt(h^2+x^2) times r^2 gives us the (h^2+x^2)^(3/2) which appears in the denominator. We integrate from 0 to L/2 and double the result. We are left with the same integral as that in the formal treatment to the left. Frequently on exams or homework you will want to check your result by going to various limits where you know the answer. One limit is where the length of the charged region approaches infinity leaving us with an infinite line of charge. In that limit L/sqrt(L^2+4h^2) approaches 1 and we are left with the “line charge” expression that you might recognize from Physics 212. The other limit is where L approaches zero and the line charge approaches a point charge. This limit is more subtle. If we set L=0 we get zero field. This makes sense since if L=0 the total charge L times lambda vanishes and you get no E-field. A better approach is to start by multiplying Lambda L to get the total charge q and then set L=0 in the sqrt(L^2+4h^2) = 2h. Alternatively you can expand L/sqrt(L^2+4h^2) to lowest non-vanishing order in L. You can either do this with a formal Taylor expansion or write sqrt(L^2+4h^2) using the binomial expansion theorem: sqrt(L^2+4h^2)=(2h)[1+(L/2h)^2]^(1/2) approximately = (2h)[1+L^2/8h^2] which is second order in L.
Thus in the L approaches zero limit we have Coulomb’s constant times q/h^2 which is the correct field for a point charge.
8
Gauss’s Law
ε 0 ∫ = enc
G G
E da i Q
R
Q
Two Gaussian spheres for r > R and r < R
2
2
ε ε π
πε
0 0
0
∫ r
r
E da r E Q
Q
E r R
r
G G
i
3
3
3 3 2 3 3
3
inc ;^
0
0
⎣ ⎦^4
⎛ ⎞⎡^4 ⎤^ ⎛^ ⎞
r
r
r Q
Q
R
Q r r
r E Q
R R
Qr
E r R
R
Brute force superposition often involves an ugly
integral. In Physics 212 you learned a much
easier way for symmetric cases.
The forgoing superposition approach to computing the E-field even for one of the
simplest problems is a bit tedious and usually requires consulting integral tables.
Often direct superposition is the only way to compute the E-field as it was in our
finite charge line case. But if a problem has sufficient symmetry we can find the field
using Gauss’s Law. Although we will show that Gauss’s Law can be “derived” from
Coulomb’s law, we will think of it as a fact for a while. Gauss’s law references a
“bounding” surface around a volume of space and says that the surface integral of
the E- field times epsilon_0 is the total charge enclosed by the “bounding” surface.
Since the right-hand side of Gauss’s law is the scalar Q (ie no direction) and E is a
vector one needs to dot it into the surface vector on the left side of Gauss’s to get a
sensible relation. The surface vector element points normal to the surface and has a
“length” equal to the area of the element. To remind you how to use Gauss’s Law
we consider the E-field from a uniform (constant density) sphere of charge of radius
R and consider the E-field both inside and outside the sphere. The first step in
applying Gauss’s law is to find a “suitable” Gaussian surface. This is usually a
surface whose normal points in the direction of the E-field and a surface where the
magnitude of the E-field is constant. If no such surface can be found, Gauss’s Law
- while still true– is of little use and one must use superposition or some other trick
to find the E-field. In this case sphere’s centered on the charge center will work
since the E-field points radially outwards and is normal to the sphere surface and
the field only depends on radius and thus E is constant over the surface. This
means the surface integral is just the area of the surface times E (which is written
as E_r since E only has a radial component. Gauss (1777-1885) was the
mathematician who developed the divergence theorem which is the basis of
Gauss’s law along with a myriad of other contributions to physics and statistics.
10
Our 1-d : ( ) logically (eg isotropically) generalizes to:
written as w/
This is the differential form of Gauss's Law.
0
⎜ +^ +^ ⎟ =^ ∇ =⎜ ⎟
⎝ ∂^ ∂^ ∂^ ⎠ ⎝ ∂^ ∂^ ∂ ⎠
∇ =
G G G G i
x
x y z
E
x
x
E E E
r
x y z x y z
E
Searching for charge in 1-d
0 0 0
Δ = × = → = = =
E x x E x dq da dq
x
a
x x d
dq d
Can you tell if charges are present by looking at E- fields?
x
Ex
x
Ex
x 0 E x
∂ ≠ ∂
Δ x
σ 2 ε 0
σ 2 ε 0 E 0 (^) E 0
x ˆ
Imagine you are in a region between the plates of a parallel plate capacitor. As you
know from Physics 212 or can easily work out from Gauss’s Law, the field between
“infinite” capacitor plates is uniform and in the direction normal to the plates which
we will call the x-hat direction. So E_x will be constant and proportional to the
voltage between the plates. What if you thought you were this region and found that
E_x wasn’t constant but had a small discontinuity indicated by the red region? One
possible explanation was that you had encountered a thin slab of charge and were
observing the superimposed of the field from this slab of charge and the much
larger field from the capacitor. In the region before reaching the slab you see a
smaller field than that of the capacitor and beyond the slab you see a larger field
and (as you show in homework) the field has a constant slope while in the slab. The
thickness of the slab times the slope of the E-field will be twice the field of the
equivalent charge plane which as we saw is sigma/(2 epsilon_0). We can re-
arrange this expression to get epsilon_0 times the slope of the E-field is the charge
density divided by the thickness which is just the charge density rho. This is the one
dimensional version of the differential form of Gauss’s law. The “obvious” 3-
dimensional generalization of this example is to include derivatives in each
dimension. A neat notation for this 3 dimension derivative uses the inverted triangle
“del”. We think of del as a vector with the three indicated coordinates, We will see
how nifty this notation is and how frequently it generalizes. The divergence theorem
discovered by Gauss links the integral version of Gauss’s law (Physics 212) with
this differential version.
11
Spherical coordinates
r = ( r cos φ sin θ r sin φ sin θ r cosθ)
G
φ
y
x
z
r
r sin θ
r cos θ θ φ x
y
r sin θ
φ^ ˆ r ˆ
φ θ
y
x
z
θ
r ˆ
θ^ ˆ
0
ˆ (^) cos cos cos sin sin
ˆ (^) sin cos
r ˆ cos sin sin sin cos
θ θ φ θ φ θ
φ φ φ
φ θ φ θ θ
= −
= −
=
ˆ ˆ ˆ and ˆ ˆ ˆ
ˆ ˆ ˆ and ˆ ˆ ˆ
But unlike ˆ, ˆ, ,
ˆ ˆ, , ˆ depend on coords
x y z x y z
r r
x y
r
z
⊥ ⊥ × =
⊥ ⊥ × =
φ^ ˆ
Causes complications!
We will frequently use curvilinear coordinates in Physics 435 and 436. These are
coordinate systems with position dependent unit vectors. The big advantage of
Cartesian coordinates is the unit vectors are in the same direction at any location.
A good example is spherical coordinates which I illustrate here. A particle is located
by three coordinates, r , theta, and phi. The r coordinate is the distance from the
origin, the theta coordinate is the angle the particle makes with respect to the z axis.
The phi coordinate gives the angle between the particle and the x axis when one
“projects” the particle to the x-y plane. I give you the Cartesian coordinates of a
particle whose position is specified by r, theta, and phi. I hope you can understand
this expression through a combination of visualization and trigonometry. I also show
the three spherical unit vectors. You note they point in the direction of increasing r
(r-hat), increasing theta (theta-hat), and increasing phi (phi-hat). At any given point,
they are mutually perpendicular but there direction at a particle clearly depends on
the particle’s position. Using your right-hand you can easily see the various cross
products such as theta-hat cross phi-hat = r-hat and circular permutations such as
phi-hat cross r-hat = theta-hat. Using more visualization and trig you can confirm the
Cartesian components of the spherical unit vectors. Incidentally they are written on
the inside cover of Griffiths along with lots of other easy reference math.
13
Lets test out the spherical divergence
( ) (^ )^ (^ )
2 2
sin
sin sin
θ θ
0 0
E r E r E E
r r r r
G G G G
i i
3
r =^ <
Qr
E r R
R
( ) ( )
3
2 2
3 2 3 2 3 2 3 3
sin
sin sin
πε
ρ θ θ θ θ θ ε ρ π π ε
ρ ε πε ε π
0
0 0 0
0
0
∂ ⎛^ ⎞ ∂ ∂
+ ×
⎜ ⎟ +^ =
r
r r r r
Q Q Q
r r
R r r R r
r
Q
R R
Q
R
R
Q^3 / 3
ρ π
= 4
Q
R
It works!
First the r < R case
( ) ( )
3
2 2
3 2 3 2 3 2 3 3
sin
sin sin
πε
ρ θ θ θ θ θ ε ρ π π ε
ρ ε πε ε π
0
0 0 0
0
0
∂ ⎛^ ⎞ ∂ ∂
+ ×
⎜ ⎟ +^ =
r
r r r r
Q Q Q
r r
R r r R r
r
Q
R R
Q
R
( ) (^ )^ (^ )
2 2
sin
sin sin
θ θ
ρ ε ρ θ θ θ θ θ ε 0 0
E r E r E E
r r r r
G G G G
i i
Here is a quick sanity check of the differential Gaussian law in spherical
coordinates. Take the field inside of a uniform sphere of charge. The divergence of
this field (del dot E) should be the charge density divided by epsilon_0. We plug our
form into the spherical divergence formula. Since E only has an E_r component we
only use the first term. We only need to differentiate r^3 with respect to r which is
easy. We manage to get the right answer. A uniform ball of charge of radius R has
3Q/(4 pi R^3) as the volume density since (4 pi R^3/3) is the volume of a sphere.
14
Cylindrical Coordinates
2
ˆr (for r )
G Q
E R
r
2 2 2 2 4
This also checks since if r
0 0 ε^0
∂ ⎛^ ⎞^ ∂⎛^ ⎞
⎜ ⎟ =^ ⎜ ⎟=^ =^ →^ = 0
Q
r r
r r
Q
r r R
R
Q
( ) ( ) ( )
sE s E Ez
s s s z
φ
ρ
φ ε (^0)
G G
i
r = ( s cos φ s cosφ z )
G
φ
y
x
z
s φˆ
z ˆ
r
G
s ˆ
rest of the world Griffiths
ρ → s
Next the r > R case
Cylindrical coordinates
We next check the other case of the spherical ball, where we are outside of the ball
with r > R. In this case we have the usual 1/r^2 radial E-field. If we stick this field
into the radial part of the spherical divergence we get zero. It is interesting that a
non-zero, non-constant field can have a zero divergence. But of course this makes
sense since there is no charge once we are out of the ball of charge.
The other coordinate system that we will frequently use is the cylindrical coordinate
system. Here we specify a position using s, phi , and z. s is the radial coordinate
sqrt(x^2+y^2) , the same phi coordinate used in spherical coordinates, and the
Cartesian coordinate z. I show the Cartesian coordinates of a point specified by s,
phi, and z which you should try to get by visualization and trig. I also show the three
cylindrical unit vectors. When all the coordinate transformation tedium is performed
you get the indicated form for the cylindrical divergence on Griffiths’ cover.
16
Gauss’s Law and Divergence Theorem
volume
Divergence Theorem
surface
∇ V d τ = V da
∫ ∫
G G^ G G
i i
h
=^ ˆ
atop wd z
G
z^ ˆ
y^ ˆ
x^ ˆ w
d
a L
hd x
G
a R
hd x
G
abottom = − wd z^ ˆ
G
aF = − wh y^ ˆ
G
( )
2
1
2 1
2 1
surface
1-d Integral integral
Derivative divergence
3-d case of integral over derivative
Limits surface boun
dary
τ
∫
∫
x
x
df x dx f x f x dx
dx d
df x V r dx
f x f x V d
d
a
G G
i
G G
i
enc
volume surface
enc surface
enc
; d
d da
da
da
Integral Gauss Law Di
Hence:
fferential Gauss Law
0 0
0
0 0
∫ ∫
∫ ∫
∫
∫
E E d Q
E E
E Q
E Q E
G G^ G G
i i
G G G G
i i
G G
i
G G G G
i i
The mathematics that links the familiar Physics 212 form of Gauss’s Law to the
differential form is called the Divergence Thm obtained by Gauss. Griffiths makes
the very interesting observation that the Divergence Thm. Is really a three
dimensional version of the fundamental thm of calculus which is the integral of the
derivative is the function itself. Lets follow the interesting analogies. The one
dimensional version says the integral of the derivative of a function involves the
value of the functions at the two integration limits which we can think of as the
“boundaries” of the integration domain. In the Divergence Thm the one dimensional
integral becomes a 3-dimensional volume integral. The one dimensional derivative
becomes the divergence. The function evaluated at the “boundary” of the domain
becomes an integral of the function evaluated on the bounding surface The cube is
a reminder on how to evaluate a surface integral with an emphasis on the nature of
the surface integral. With the Divergence Theorem in hand, the bridge from the
differential law to the integral law is a painless sprint. One does a volume integral on
both sides of the divergence of E equals the charge density using an arbitrary
volume. The integral of the divergence is the same as the surface integral over the
“bounding” surface. The integral over the charge density is the charge “enclosed”
by the bounding surface. Hence the surface integral over the field is essentially the
charge enclosed by the surface. The integral and differential forms are equivalent.
We are now well set up to show that Gauss’s law follows from Coulomb’s Law.
17
A paradox?
2
2
2 2
2 2 2
inc sphe
all sphe
re
s
R
re
Hence ever
Consider E for point charge:
y
How can everywhere but
where?
But a
d
r
r r r r
q E
r
E Q
r
q r
q
r
q
R q R
0
0 0
0
⊆
0
0 0
0
∂ ⎛^ ⎞^ ∂⎛^ ⎞
⎜ ⎟ =^ ⎜ ⎟=^ →
×
∫ =
∫
G
i
G
G G
2
answer: is undefined
at not valid at. We are ok if ( ) and (
d?
r r r r
q
r
q
r
0 0
∂⎛^ ⎞
In 2-d this might look like this
x
y
ρ( x y , )
2 4 (^ )
The "fix" written as: ˆ πδ 3
⎛ ⎞ ∇ (^) ⎜ ⎟= ⎝ ⎠
r r r
G (^) G i
( )
( )
( )
( )
3 3
3
3
0 0
1
4
vol 0
( )
V
for r 0 and
Normalized such that
The shifted version is
( - ') '
The defining property of '
is ( ') ' ' ( )
δ δ
δ τ
πδ
δ
δ τ
3 3
3 ⊆
⊆
= ≠ = ∞
∫ =
⎛ ⎞ ∇ ⎜^ ⎟= − ⎜ ⎟ ⎝ ⎠
−
∫ −^ =
r
r
r
r d
r r r r r r
r r
f r r r d f r
G
G
G
G G G^ G
G
G G^ G G G i (^) G G
G G
G G G G
We start off with a mathematical paradox based on our two statements of Gauss’s
Law. Consider redoing the case of the E-field of a point charge. We form the
spherical divergence of this field and find that it vanishes. No surprise here! We
next “check” our result by the “equivalent” integral form of Gauss’s law. The surface
integral of the E-field centered around the point charge gives the q of the point
charge. No surprise here as well. What's the problem? How can the charge density
rho be zero everywhere and yet there is a point charge at the origin? I think the
problem must be that our spherical divergence does not really “work” at the origin
since it involves 1/r^2 terms that blow up at r=0. Perhaps rho is infinite at the
origin? Indeed this makes sense since the integral of rho is always equal to q no
matter how small in radius a sphere we pick. The only way this can happen is an
infinite charge at our origin and no charge anywhere else. The 2-dimensional spike
on the left side gives a useful way of visualizing the divergence of r-hat/r^2. At the
origin our divergence goes to infinity. But it does it so that the volume of the spike is
1. In three dimensions we say the divergence of r-hat/r^2 is 4 pi times a three
dimensional Dirac delta-function centered at the origin. We could shift the origin to a
source point r’ both on the left and right side of our divergence expression which is
just equivalent to substituting r for r-r’ The chief defining, property of delta-functions
is that the integral of a delta function times a function gives the value of the function
at where the argument of the delta-function is zero as long this zero point is in the
integration volume. This makes intuitive sense since (in our three-dimensional
example) the delta function only exists at the point r=r’ and thus f(r) will need to be
evaluated at r=r’ to return something proportional to f(r’). The delta-function has
been defined so that the proportional constant is 1. There are also 1 and 2
dimensions of the delta function which you may have seen in mechanics courses.
