Foundations of Electromagnetic Theory - … 611 spring 18/classnotes... · Foundations of...

1Prof. Sergio B. MendesSpring 2018

Prologue of “Modern Problems in Classical Electrodynamics” by Charles Brau

Foundations of Electromagnetic Theory

2

• Topics to be covered in this chapter (prologue): electrostatics, magnetostatics, electrodynamics, electromagnetic waves, conservation laws, and Maxwell’s stress tensor.

• Almost all forces perceived in Nature (except for gravity) are electromagnetic forces.

• For most of the topics listed above, this is intended to be a review, not an introduction.

Prof. Sergio B. MendesSpring 2018

Foundations of Electromagnetic Theory

Electrostatic Theory

3

charges are not moving, they are fixed in space

charge density: 𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 𝜌𝜌 𝒓𝒓


4

Milestones in Electrostatics• 600 B.C.: Thales of Miletus, amber rubbed with fur attracts other

object, electron = amber in greek

• 1600 D.C.: Gilbert, “amberness” phenomena is also displayed by other materials

• 1733: du Fay, repulsion is also possible (in addition to attraction), two flavors of electric charge

• 1834: Faraday, electric charge comes in discrete amounts (is quantized in modern terminology)

• 1909: Millikan, 𝑒𝑒 ≅ −1.62 × 10−19 𝐶𝐶

• 1746: Watson, 1747: Franklin, electricity as a fluid that moves from one body to another, conservation of electric charge

• 1785: Coulomb, force between small electrically-charged objects


5

𝑭𝑭𝒒𝒒 𝒓𝒓 =𝑞𝑞 𝑞𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓𝑞 3

Coulomb Law

𝑞𝑞𝑞𝑞𝑞

𝒓𝒓𝒓𝒓𝑞

𝒓𝒓 − 𝒓𝒓𝑞𝑭𝑭

𝒪𝒪

(force on 𝑞𝑞 located at 𝒓𝒓due to 𝑞𝑞𝑞 located at 𝒓𝒓𝑞)

𝑭𝑭𝒒𝒒 𝒓𝒓

∝ 𝑞𝑞 𝑞𝑞𝑞

𝜖𝜖0 ≅ 8.854187817 × 10−12𝐶𝐶2

𝑁𝑁 𝑚𝑚2

(permittivity of free space)


∝𝟏𝟏


6

𝑭𝑭𝒒𝒒 𝒓𝒓 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0�𝑖𝑖 = 𝑎𝑎𝑎𝑎𝑎𝑎

𝑞𝑞𝑖𝑖𝒓𝒓 − 𝒓𝒓𝑖𝑖𝒓𝒓 − 𝒓𝒓𝑖𝑖 3

𝑭𝑭𝒒𝒒 𝒓𝒓 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓′ 3𝑑𝑑𝑑𝑑𝑞

𝑑𝑑𝑞𝑞 𝒓𝒓𝑞 = 𝜌𝜌 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑𝑞

• Continuous distribution of charges 𝜌𝜌 𝒓𝒓𝑞 :

• Discrete charges 𝑞𝑞𝑖𝑖 located at 𝒓𝒓𝑖𝑖:

Net Force from Multiple ChargesPrinciple of Superposition: forces are added (vectorially)


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The force 𝑭𝑭𝒒𝒒 𝒓𝒓 on 𝑞𝑞 is linearly proportional to 𝑞𝑞.

The proportionality constant is called the electric field 𝑬𝑬 𝒓𝒓 :

𝑬𝑬 𝒓𝒓 ≡𝑭𝑭𝒒𝒒 𝒓𝒓𝒒𝒒


An Important Consequence

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𝑬𝑬 𝒓𝒓 =1




4 𝜋𝜋 𝜖𝜖0�−∞

+∞



• Multiple discrete charges 𝑞𝑞𝑖𝑖 located at 𝒓𝒓𝑖𝑖:

• Continuous distribution of charges 𝜌𝜌 𝒓𝒓𝑞 :

• Single charge 𝑞𝑞𝑞 located at 𝒓𝒓𝑞:

𝑬𝑬 𝒓𝒓 =𝑞𝑞𝑞



Electric Field


𝒓𝒓

𝒪𝒪

𝒓𝒓

𝒪𝒪

𝒪𝒪

𝒓𝒓𝑞 𝑞𝑞𝑞 𝒓𝒓

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Flux of the Electric Field on a Arbitrary Surface

Φ𝑬𝑬 ≡�𝑆𝑆𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅

𝑬𝑬 𝒓𝒓

𝒅𝒅𝒅𝒅𝑆𝑆

𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = 𝑬𝑬 𝒓𝒓 cos 𝜃𝜃 𝒓𝒓 𝑑𝑑𝑆𝑆

𝜃𝜃 𝒓𝒓

𝒪𝒪

𝒓𝒓

vector vectordot

product

𝒅𝒅𝒅𝒅 ≡ �𝒏𝒏 𝒓𝒓 𝑑𝑑𝑆𝑆

𝑑𝑑𝑆𝑆

vector unit normal vector

differential area


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Divergence Theorem

�𝑆𝑆𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑸𝑸 𝒓𝒓 𝑑𝑑𝑑𝑑

for any continuously differentiable vector field 𝑸𝑸 𝒓𝒓 and any closed surface S:

closed surface S surrounds volume V

also known as Gauss's theorem or

Ostrogradsky's theorem


S𝑸𝑸 𝒓𝒓V

S

V

11

Electric Flux on a closed surface:

Φ𝑬𝑬, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒅𝒅 = �𝑆𝑆𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅

Φ𝑬𝑬,𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒅𝒅 = �𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑

Divergence Theorem


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Single charge 𝑞𝑞𝑞 located at 𝒓𝒓𝑞:

𝑬𝑬 𝒓𝒓 =𝑞𝑞𝑞



𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

=𝒓𝒓 − 𝒓𝒓′


�𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑 = �

𝑉𝑉

𝑞𝑞𝑞4 𝜋𝜋 𝜖𝜖0

𝛻𝛻. 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑

HW:

HW:

=𝑞𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝛻𝛻

−1𝒓𝒓 − 𝒓𝒓𝑞

𝛻𝛻. 𝛻𝛻𝜓𝜓 𝒓𝒓 = 𝛻𝛻2 𝜓𝜓 𝒓𝒓

= �𝑉𝑉


𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑


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𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞

= 4 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑞HW:

𝑞𝑞′

𝜖𝜖0

0

if q’ is inside V

if q’ is outside V

= �𝑉𝑉

𝑞𝑞′

4 𝜋𝜋 𝜖𝜖04 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑


𝑉𝑉


𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑 =


{=

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Multiple discrete charges 𝑞𝑞𝑖𝑖 located at 𝒓𝒓𝑖𝑖:




=1


𝑞𝑞𝑖𝑖 𝛻𝛻−1

𝒓𝒓 − 𝒓𝒓𝑖𝑖

= �𝑉𝑉

14 𝜋𝜋 𝜖𝜖0

�𝑖𝑖 = 𝑎𝑎𝑎𝑎𝑎𝑎

𝑞𝑞𝑖𝑖 𝛻𝛻2−1

𝒓𝒓 − 𝒓𝒓𝑖𝑖𝑑𝑑𝑑𝑑

= �𝑉𝑉

14 𝜋𝜋 𝜖𝜖0

�𝑖𝑖 = 𝑎𝑎𝑎𝑎𝑎𝑎

𝑞𝑞𝑖𝑖 4 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑖𝑖 𝑑𝑑𝑑𝑑

= �𝑖𝑖= 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖

𝑞𝑞𝑖𝑖𝜖𝜖0

= �𝑖𝑖=𝑎𝑎𝑎𝑎𝑎𝑎


�𝑉𝑉𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑖𝑖 𝑑𝑑𝑑𝑑


𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑


15


𝑖𝑖= 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖


=𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝜖𝜖0

independent of charge location inside S

independent of shape of surface S


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4 𝜋𝜋 𝜖𝜖0�−∞

+∞



Continuous distribution of charges 𝜌𝜌 𝒓𝒓𝑞 :

=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

𝑑𝑑𝑑𝑑𝑞

𝛻𝛻.𝑬𝑬 𝒓𝒓

=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻2−1𝒓𝒓 − 𝒓𝒓𝑞


=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 4 𝜋𝜋 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝑞 𝑑𝑑𝑑𝑑𝑞 =𝜌𝜌 𝒓𝒓𝜖𝜖0

= 𝛻𝛻.1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞



17


𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓 𝑑𝑑𝑑𝑑 = �

𝑉𝑉

𝜌𝜌 𝒓𝒓𝜖𝜖0

𝑑𝑑𝑑𝑑

𝛻𝛻.𝑬𝑬 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Gauss’s Law (differential form)

=𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝜖𝜖0

�𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒅𝒅 =

𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝜖𝜖0

Gauss’s Law (integral form)


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Back to the electric field:

= −𝛻𝛻Φ 𝒓𝒓

Φ 𝒓𝒓 ≡1


𝑞𝑞𝑖𝑖𝒓𝒓 − 𝒓𝒓𝑖𝑖

Electric Field 𝑬𝑬 𝒓𝒓is a vector field

Electric Potential Φ 𝒓𝒓is a scalar function

(discrete charges)

=1


𝑞𝑞𝑖𝑖 𝛻𝛻−1

𝒓𝒓 − 𝒓𝒓𝑖𝑖


19

Back to the electric field:(continuous charge distribution)

= −𝛻𝛻Φ 𝒓𝒓

Φ 𝒓𝒓 ≡1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′


=1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞

𝜌𝜌 𝒓𝒓𝑞 𝛻𝛻−1

𝒓𝒓 − 𝒓𝒓′𝑑𝑑𝑑𝑑𝑞


4 𝜋𝜋 𝜖𝜖0�−∞

+∞




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𝑬𝑬 𝒓𝒓 = −𝛻𝛻Φ 𝒓𝒓

Φ𝑞 𝒓𝒓 = Φ 𝒓𝒓 + Λ

𝛻𝛻.𝑬𝑬 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

A couple of observations:

(1) Adding a constant Λ to the electric potential Φ 𝒓𝒓 has no impact on the field:

(2) We can solve the electric potential first and then use it to determine the electric field:

𝛻𝛻. −𝛻𝛻Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

−𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0

Poisson’s equation


21

𝛻𝛻 × 𝛻𝛻𝑓𝑓 𝒓𝒓 = 0HW:


𝛻𝛻 × 𝑬𝑬 𝒓𝒓 = 𝛻𝛻 × −𝛻𝛻Φ 𝒓𝒓 = −𝛻𝛻 × 𝛻𝛻Φ 𝒓𝒓 = 0

𝛻𝛻 × 𝑬𝑬 𝒓𝒓 = 0


22

Stokes’ Theorem

�𝐶𝐶𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒓𝒓

closed loop C surrounds surface S


for any continuously differentiable vector field 𝑸𝑸 𝒓𝒓 and any closed loop C:

CS

𝛻𝛻 × 𝑸𝑸 𝒓𝒓

𝑸𝑸 𝒓𝒓

𝐶𝐶 𝐶𝐶

𝑆𝑆𝑆𝑆𝑞

= �𝑆𝑆𝛻𝛻 × 𝑸𝑸 𝒓𝒓 . 𝒅𝒅𝒅𝒅

23

�𝑎𝑎, 𝐶𝐶1

𝑏𝑏𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = �

𝐶𝐶𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓+�

𝑏𝑏, 𝐶𝐶2

𝑎𝑎𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= �𝑆𝑆𝛻𝛻 × 𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒅𝒅

= �𝑆𝑆

0 . 𝒅𝒅𝒅𝒅

= 0

�𝑎𝑎, 𝐶𝐶1

𝑏𝑏𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = −�

𝑏𝑏, 𝐶𝐶2

𝑎𝑎𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= �𝑎𝑎, 𝐶𝐶2

𝑏𝑏𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = constant, regardless of the path

𝒂𝒂

𝒃𝒃

𝐶𝐶1 𝐶𝐶2


𝑬𝑬 𝒓𝒓

𝒅𝒅𝒓𝒓

24

Φ 𝒓𝒓 ≡ −�−∞

𝒓𝒓𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

�𝒂𝒂

𝒃𝒃𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = �

𝒂𝒂

−∞𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 + �

−∞

𝒃𝒃𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓

= Φ 𝒂𝒂 −Φ 𝒃𝒃

regardless of the path 𝐶𝐶𝒂𝒂

𝒃𝒃

𝐶𝐶



25

Workdone by the electrostatic force 𝑭𝑭𝒒𝒒 𝒓𝒓 when the charge q moves from a to b

𝑊𝑊 = �𝒂𝒂

𝒃𝒃𝑭𝑭𝒒𝒒 𝒓𝒓 . 𝒅𝒅𝒓𝒓

𝑭𝑭𝒒𝒒 𝒓𝒓 = 𝑞𝑞 𝑬𝑬 𝒓𝒓

𝑊𝑊 = �𝒂𝒂

𝒃𝒃𝑞𝑞 𝑬𝑬 𝒓𝒓 . 𝒅𝒅𝒓𝒓 = 𝑞𝑞 Φ 𝒂𝒂 − Φ 𝒃𝒃

𝒂𝒂

𝒃𝒃

𝐶𝐶

regardless of the path


26

External Workdone by an external force to perfectly balance the electrostatic force

𝑊𝑊𝑖𝑖𝑒𝑒𝑒𝑒 = 𝑞𝑞 Φ 𝒃𝒃 − Φ 𝒂𝒂𝒂𝒂

𝒃𝒃

𝑊𝑊𝑖𝑖𝑒𝑒𝑒𝑒 = 𝑞𝑞 Φ 𝒓𝒓 − Φ −∞

−∞

𝒓𝒓

= 𝑞𝑞 Φ 𝒓𝒓

𝑭𝑭𝒄𝒄𝒆𝒆𝒆𝒆 𝒓𝒓 = − 𝑭𝑭𝒒𝒒 𝒓𝒓

stored energy !!Prof. Sergio B. MendesSpring 2018

𝐶𝐶

𝐶𝐶

𝒂𝒂 → ∞0

27

𝑊𝑊1 & 2 = 𝑞𝑞1 Φ2 𝒓𝒓1

= 𝑞𝑞2 Φ1 𝒓𝒓2

=12𝑞𝑞1 Φ2 𝒓𝒓1 +

12𝑞𝑞2 Φ1 𝒓𝒓2

𝑞𝑞1 𝑞𝑞2

𝒪𝒪

𝒓𝒓1 𝒓𝒓2

External work done to assemble charges = Stored Energy

two point charges


𝑞𝑞2

28

𝑞𝑞1 𝑞𝑞2

𝒪𝒪

𝒓𝒓1 𝒓𝒓2

Additional work for bringing a third charge

𝑞𝑞3∆𝑊𝑊 = 𝑞𝑞3 Φ1 𝒓𝒓3 + 𝑞𝑞3 Φ2 𝒓𝒓3

= 𝑞𝑞1 Φ3 𝒓𝒓1 + 𝑞𝑞2 Φ3 𝒓𝒓2

=12𝑞𝑞3 Φ1 𝒓𝒓3 +

12𝑞𝑞1 Φ3 𝒓𝒓1 +

12𝑞𝑞3 Φ2 𝒓𝒓3 +



𝒓𝒓3

𝑞𝑞1 𝑞𝑞2

29

𝑊𝑊1 & 2 & 3 =12𝑞𝑞1 Φ2 𝒓𝒓1 +


+12𝑞𝑞3 Φ1 𝒓𝒓3 +


+12𝑞𝑞3 Φ2 𝒓𝒓3 +

12𝑞𝑞2 Φ3 𝒓𝒓2 =

=12𝑞𝑞1 Φ2 𝒓𝒓1 + Φ3 𝒓𝒓1

+12𝑞𝑞2 Φ3 𝒓𝒓2 + Φ1 𝒓𝒓2

+12𝑞𝑞3 Φ2 𝒓𝒓3 + Φ3 𝒓𝒓3

Work needed to assemble three charges(stored energy)


30

𝑑𝑑𝑊𝑊 =12𝑑𝑑𝑞𝑞 𝒓𝒓 Φ 𝒓𝒓

𝑑𝑑𝑞𝑞 𝒓𝒓 = 𝜌𝜌 𝒓𝒓 𝑑𝑑𝑑𝑑

𝜌𝜌 𝒓𝒓 = −𝜖𝜖0 𝛻𝛻2Φ 𝒓𝒓

𝑑𝑑𝑊𝑊 = −12𝜖𝜖0 Φ 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

Energy stored in a continuous distribution of charges

−Φ 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 = − 𝛻𝛻. Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 + 𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓HW:


31

=12𝜖𝜖0 − 𝛻𝛻. Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 + 𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

𝑊𝑊 =12𝜖𝜖0�

−∞

+∞

− 𝛻𝛻. Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 + 𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

=12𝜖𝜖0 − �

−∞

+∞

Φ 𝒓𝒓 𝛻𝛻Φ 𝒓𝒓 . 𝒅𝒅𝒅𝒅 + �−∞

+∞

𝛻𝛻Φ 𝒓𝒓 . 𝛻𝛻Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

=12𝜖𝜖0�

−∞

+∞

𝛻𝛻Φ 𝒓𝒓 2 𝑑𝑑𝑑𝑑


𝑑𝑑𝑊𝑊 = −12𝜖𝜖0 Φ 𝒓𝒓 𝛻𝛻2Φ 𝒓𝒓 𝑑𝑑𝑑𝑑

0

=12𝜖𝜖0�

−∞

+∞

𝑬𝑬 𝒓𝒓 2 𝑑𝑑𝑑𝑑

32

𝑢𝑢𝐸𝐸 𝒓𝒓 ≡𝑑𝑑𝑊𝑊𝑑𝑑𝑑𝑑

=12𝜖𝜖0 𝑬𝑬 𝒓𝒓 2

(Electric) Energy Density


Magnetostatic Theory

33

steady (constant) current

current density: 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 𝑱𝑱 𝒓𝒓

𝑱𝑱 ≡𝑑𝑑𝑑𝑑𝑑𝑑𝑆𝑆

�𝒏𝒏


34

Milestones in Magnetostatics• 2637 B.C.: reports of magnets by Chinese civilization

• 3rd century A.D.: Chinese ships used compass

• 1600: Gilbert, (De Magnete) described Earth as a magnet

• 800 A.D.: Greek reports on lodestones (magnetite Fe2O3)

• 1820: Oersted, an electric current deflects a compass

• 1830’s: Ampere, nearby currents create a force on each other


35

Conservation of Electric Charge

�s𝑱𝑱 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅 = −

𝑑𝑑𝑄𝑄𝑉𝑉𝑑𝑑𝑡𝑡

= −𝑑𝑑𝑑𝑑𝑡𝑡�

𝑉𝑉𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑

�𝑉𝑉𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑 = −�

𝑉𝑉

𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝑑𝑑𝑑𝑑

𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 𝜌𝜌 𝒓𝒓 𝜕𝜕𝜌𝜌 𝒓𝒓𝜕𝜕𝑡𝑡

= 0 𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 𝛁𝛁. 𝑱𝑱 𝒓𝒓 = 0if then and

always valid due to conservation of charge


36

𝛁𝛁. 𝑱𝑱 𝒓𝒓 = 0

Satisfies × Violates


�s𝑱𝑱 𝒓𝒓 . 𝒅𝒅𝒅𝒅 = 0

37

Magnetic Force• on a moving electric charge due to a magnetic field 𝑩𝑩

𝑑𝑑𝑭𝑭 = 𝑑𝑑𝑞𝑞 𝒗𝒗 × 𝑩𝑩 = I dr × 𝑩𝑩 = dV 𝑱𝑱 × 𝑩𝑩

𝑭𝑭 = 𝑞𝑞 𝒗𝒗 × 𝑩𝑩

• For a non-zero magnetic force, charges must be moving and in a path not collinear with the magnetic field

• Static magnetic force is always orthogonal to the direction of charge motion

• Work: 𝑑𝑑𝑊𝑊 = 𝑭𝑭 . 𝒅𝒅𝒓𝒓 = 𝑞𝑞 𝒗𝒗 × 𝑩𝑩 . 𝒗𝒗 𝑑𝑑𝑡𝑡 = 0

• [B] = Tesla in S.I. units

Lorentz force


• on an electric current due to a magnetic field 𝑩𝑩

38

Magnetic Fieldis created by electric charges in motion (electric current)

𝑩𝑩 𝒓𝒓 =𝜇𝜇𝑜𝑜4 𝜋𝜋

�𝐶𝐶

I 𝒓𝒓′ d 𝒓𝒓′ ×𝒓𝒓 − 𝒓𝒓′


=𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

𝑱𝑱 𝒓𝒓′ dV𝑞 ×𝒓𝒓 − 𝒓𝒓′


Biot-Savart law

𝜇𝜇𝑜𝑜 = 4 π × 10−7𝑁𝑁𝐴𝐴2

(permeability of free space)


39

𝛻𝛻.𝑩𝑩 𝒓𝒓 = 𝛻𝛻.𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞 𝑱𝑱 𝒓𝒓′ ×𝒓𝒓 − 𝒓𝒓′


𝛻𝛻. 𝒂𝒂 × 𝒃𝒃 = 𝒃𝒃 .𝛻𝛻 × 𝒂𝒂 − 𝒂𝒂 . 𝛻𝛻 × 𝒃𝒃HW:

=𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓𝑞 3.𝛻𝛻 × 𝑱𝑱 𝒓𝒓′ − 𝑱𝑱 𝒓𝒓′ .𝛻𝛻 ×

𝒓𝒓 − 𝒓𝒓′


= −𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞 𝑱𝑱 𝒓𝒓′ .𝛻𝛻 × 𝛻𝛻−1𝒓𝒓 − 𝒓𝒓𝑞

𝛻𝛻 × 𝛻𝛻𝑓𝑓 𝒓𝒓 = 𝟎𝟎 𝛻𝛻.𝑩𝑩 𝒓𝒓 = 𝟎𝟎because: = 0,


0

40

𝛻𝛻 × 𝑩𝑩 𝒓𝒓 = 𝛻𝛻 ×𝜇𝜇𝑜𝑜4 𝜋𝜋

�−∞

+∞

dV𝑞 J 𝒓𝒓′ ×𝒓𝒓 − 𝒓𝒓′

𝒓𝒓 − 𝒓𝒓𝑞 3 = 𝜇𝜇𝑜𝑜 J 𝒓𝒓HW:

𝛻𝛻 × 𝑩𝑩 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

Ampère’s law(differential form)

�𝑆𝑆𝛻𝛻 × 𝑩𝑩 𝒓𝒓 .𝒅𝒅𝒅𝒅 = 𝜇𝜇𝑜𝑜�

𝑆𝑆𝑱𝑱 𝒓𝒓 .𝒅𝒅𝒅𝒅

�𝐶𝐶𝑩𝑩 𝒓𝒓 .𝒅𝒅𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑑𝑑

Ampère’s law(integral form)


41

Vector Potential: 𝑨𝑨 𝒓𝒓

𝛻𝛻.𝑩𝑩 𝒓𝒓 = 0

𝑩𝑩 𝒓𝒓 ≡ 𝛻𝛻 × 𝑨𝑨 𝒓𝒓

HW: 𝛻𝛻. 𝛻𝛻 × 𝒂𝒂 = 0

𝛻𝛻.𝑩𝑩 𝒓𝒓 = 𝛻𝛻. 𝛻𝛻 × 𝑨𝑨 𝒓𝒓 = 0


42

𝛻𝛻 × 𝛻𝛻 × 𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

𝛻𝛻 × 𝛻𝛻 × 𝒂𝒂 = 𝛻𝛻 𝛻𝛻.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂HW:


𝛻𝛻 × 𝑩𝑩 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

𝜵𝜵 × 𝜵𝜵 × 𝑨𝑨 𝒓𝒓 = 𝜵𝜵 𝜵𝜵.𝑨𝑨 𝒓𝒓 − 𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

𝑩𝑩 𝒓𝒓 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓


𝑨𝑨𝑞 𝒓𝒓 = 𝑨𝑨 𝒓𝒓 + 𝜵𝜵𝛬𝛬 𝒓𝒓

𝑩𝑩 𝒓𝒓 = 𝜵𝜵 × 𝑨𝑨′ 𝒓𝒓 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓

𝛁𝛁 × 𝛁𝛁𝛬𝛬 𝒓𝒓 = 0

because

and

Certain freedom to choose the Vector Potential:


𝑨𝑨𝑞 𝒓𝒓 = 𝑨𝑨 𝒓𝒓 + 𝜵𝜵𝛬𝛬 𝒓𝒓

𝛁𝛁.𝑨𝑨𝑞 𝒓𝒓 = 𝛁𝛁.𝑨𝑨 𝒓𝒓 + 𝛁𝛁.𝛻𝛻𝛬𝛬 𝒓𝒓

= 𝛁𝛁.𝑨𝑨 𝒓𝒓 + 𝛻𝛻2𝛬𝛬 𝒓𝒓

Now, we will find 𝛬𝛬 𝒓𝒓 such that 𝛻𝛻2𝛬𝛬 𝒓𝒓 = −𝛁𝛁.𝑨𝑨 𝒓𝒓 , then:

𝛁𝛁.𝑨𝑨𝑞 𝒓𝒓 = 0

𝜵𝜵 × 𝜵𝜵 × 𝑨𝑨𝑞 𝒓𝒓 = 𝜵𝜵 𝜵𝜵.𝑨𝑨𝑞 𝒓𝒓 − 𝛻𝛻𝟐𝟐𝑨𝑨𝑞 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

−𝛻𝛻𝟐𝟐𝑨𝑨𝑞 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓

Coulomb gauge

How can we benefit from this freedom?

0


Big picture in

Electrostatics and Magnetostatics:


𝑩𝑩 𝒓𝒓 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓−𝛻𝛻2Φ 𝒓𝒓 =𝜌𝜌 𝒓𝒓𝜖𝜖0


Φ 𝒓𝒓 ≡1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′

𝑑𝑑𝑑𝑑𝑞 𝑨𝑨 𝒓𝒓 ≡𝜇𝜇𝑜𝑜

4 𝜋𝜋�−∞

+∞𝑱𝑱 𝒓𝒓𝑞𝒓𝒓 − 𝒓𝒓′


Electrostatics Magnetostatics

𝜌𝜌 𝒓𝒓𝑱𝑱 𝒓𝒓

𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0−𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡 = 0

charge conservation

𝑰𝑰

Electrodynamic Theory

47

𝜌𝜌 𝒓𝒓, 𝑡𝑡


𝑱𝑱 𝒓𝒓, 𝑡𝑡−𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡

time-dependent theory for electric and magnetic fields

}


What remains valid ?


Φ𝑬𝑬, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒅𝒅 𝑡𝑡 = �𝑆𝑆𝑬𝑬 𝒓𝒓, 𝑡𝑡 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑 = �

𝑉𝑉

𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

𝑑𝑑𝑑𝑑

𝛻𝛻.𝑬𝑬 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0


Φ𝑩𝑩, 𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒄𝒅𝒅 𝑡𝑡 = �𝑆𝑆𝑩𝑩 𝒓𝒓, 𝑡𝑡 .𝒅𝒅𝒅𝒅 = �

𝑉𝑉𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑑𝑑𝑑𝑑 = 0

𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0


What is new ?


Φ𝑩𝑩 𝑡𝑡 = �𝑆𝑆𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅

−𝑑𝑑𝑑𝑑𝑡𝑡Φ𝑩𝑩 𝑡𝑡 = �

𝐶𝐶𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒓𝒓

Faraday’s Law

−𝑑𝑑𝑑𝑑𝑡𝑡�𝑆𝑆𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅 = �

𝑆𝑆𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝒅𝒅𝒅𝒅

𝛻𝛻 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡


What needs to be modified ?


𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0


𝜕𝜕𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜖𝜖0𝜕𝜕 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 + 𝜖𝜖0𝜕𝜕 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡= 0

𝛁𝛁. 𝑱𝑱 𝒓𝒓, 𝑡𝑡 + 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0

= 𝛁𝛁. 𝛁𝛁 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝛁𝛁. 𝛁𝛁 × 𝑸𝑸 𝒓𝒓, 𝑡𝑡


In summary, Maxwell’s equations:

𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0

𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡


1.

2.

3.

4.

Gauss’s law

Faraday’s law

GeneralizedAmpère’s law

Gauss’s law of magnetism

56

Vector Potential: 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝒆𝒆 = 0

𝑩𝑩 𝒓𝒓, 𝒆𝒆 ≡ 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝛻𝛻. 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 = 0



Scalar Potential: Φ 𝒓𝒓, 𝑡𝑡


𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝛻𝛻 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

𝛻𝛻 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝟎𝟎

𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

≡ −𝛻𝛻Φ 𝒓𝒓, 𝑡𝑡

𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝛻𝛻Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

&

≡ 𝛻𝛻 × −𝛻𝛻Φ 𝒓𝒓, 𝑡𝑡


𝑩𝑩 𝒓𝒓, 𝒆𝒆 = 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

Fields: 𝑬𝑬 𝒓𝒓, 𝑡𝑡 & 𝑩𝑩 𝒓𝒓, 𝑡𝑡

and

Potentials: Φ 𝒓𝒓, 𝑡𝑡 & 𝑨𝑨 𝒓𝒓, 𝑡𝑡



How to determine

the scalar Φ 𝒓𝒓, 𝑡𝑡 and vector 𝑨𝑨 𝒓𝒓, 𝑡𝑡 potentials

directly from

the charge 𝜌𝜌 𝒓𝒓, 𝑡𝑡 and current 𝑱𝑱 𝒓𝒓, 𝑡𝑡 densities ?




𝛻𝛻. −𝛻𝛻Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

=𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡

𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

Gauss’s law



𝛁𝛁 × 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡

−𝛻𝛻Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡


𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡


𝛻𝛻 × 𝛻𝛻 × 𝒂𝒂 = 𝛻𝛻 𝛻𝛻.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂

𝛻𝛻 𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 − 𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝛻𝛻𝜕𝜕𝜕𝜕𝑡𝑡 Φ 𝒓𝒓, 𝑡𝑡 +

𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Remember HW 9:

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 + 𝛻𝛻 𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕𝜕𝜕𝑡𝑡 Φ 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Ampère’s law


−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡




The scalar Φ 𝒓𝒓, 𝑡𝑡 and vector 𝑨𝑨 𝒓𝒓, 𝑡𝑡potentials determined from the charge 𝜌𝜌 𝒓𝒓, 𝑡𝑡 and current 𝑱𝑱 𝒓𝒓, 𝑡𝑡 densities


How can we simplify those equations ?


We have certain freedom to choose the vector and scalar potentials:

𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 ≡ 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

Φ′ 𝒓𝒓, 𝑡𝑡 ≡ Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝜵𝜵 × 𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

−𝛻𝛻Φ𝑞 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

= −𝛻𝛻Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡𝑨𝑨 𝒓𝒓, 𝑡𝑡

= −𝛻𝛻 Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝑬𝑬 𝒓𝒓, 𝑡𝑡

= 𝑩𝑩 𝒓𝒓, 𝑡𝑡

Conclusion: if 𝑨𝑨 𝒓𝒓, 𝑡𝑡 & Φ 𝒓𝒓, 𝑡𝑡 is a solution then 𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 & Φ′ 𝒓𝒓, 𝑡𝑡 (as defined above) is also a

solution, and vice-versa.

= 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡 = 𝜵𝜵 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡

−𝜕𝜕𝜕𝜕𝑡𝑡 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡


𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 = 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

Φ′ 𝒓𝒓, 𝑡𝑡 = Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

How can we use the freedom in the choice of the potentials

−𝛻𝛻2Φ𝑞 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡

𝛻𝛻.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

−𝛻𝛻𝟐𝟐𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡2+ 𝛻𝛻 𝛻𝛻.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕𝜕𝜕𝑡𝑡Φ𝑞 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

to simplify the following equations:


𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 = 𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

Φ′ 𝒓𝒓, 𝑡𝑡 = Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝛻𝛻.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ𝑞 𝒓𝒓, 𝑡𝑡 = 𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝛻𝛻.𝜵𝜵𝛬𝛬 𝒓𝒓, 𝑡𝑡

+ 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕𝜕𝜕𝑡𝑡

−𝜕𝜕𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝛻𝛻𝟐𝟐𝛬𝛬 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝛬𝛬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

= − 𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡

We now solve for 𝛬𝛬 𝒓𝒓, 𝑡𝑡 such that 𝛻𝛻.𝑨𝑨′ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑒𝑒Φ′ 𝒓𝒓, 𝑡𝑡 = 𝟎𝟎 :

= 𝑓𝑓 𝒓𝒓, 𝑡𝑡





𝛻𝛻.𝑨𝑨′ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ′ 𝒓𝒓, 𝑡𝑡 = 0

−𝛻𝛻2Φ𝑞 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡

𝛻𝛻.𝑨𝑨𝑞 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0





𝜕𝜕𝑡𝑡2= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

−𝛻𝛻2Φ′ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2Φ𝑞 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡2=𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

Lorenz gauge


𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 = 0

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2


−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2Φ 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡2=𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

In the Lorenz gauge:

𝜌𝜌 𝒓𝒓, 𝑡𝑡 determines Φ 𝒓𝒓, 𝑡𝑡

𝑱𝑱 𝒓𝒓, 𝑡𝑡 determines 𝑨𝑨 𝒓𝒓, 𝑡𝑡


In addition to the Lorenz gauge, there are other possible choices to

simplify the equations below !!

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡


−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

+ 𝛻𝛻 𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡


The Coulomb gauge

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 −𝜕𝜕𝜕𝜕𝑡𝑡




𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 = 0

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 + 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝛻𝛻


or radiation gauge or transverse gauge


Φ 𝒓𝒓, 𝑡𝑡 =1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓′, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′


−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0


−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜖𝜖0 𝛻𝛻

𝜕𝜕𝜕𝜕𝑡𝑡 Φ 𝒓𝒓, 𝑡𝑡

= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 −1

4 𝜋𝜋 𝛻𝛻𝜕𝜕𝜕𝜕𝑡𝑡

�−∞

+∞𝜌𝜌 𝒓𝒓′, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′


= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 +1

4 𝜋𝜋 𝛻𝛻 �−∞

+∞𝛻𝛻𝑞. 𝑱𝑱 𝒓𝒓𝑞, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′


−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 𝜇𝜇𝑜𝑜 𝑱𝑱𝑻𝑻 𝒓𝒓, 𝑡𝑡

𝛻𝛻. 𝑱𝑱𝑻𝑻 𝒓𝒓, 𝑡𝑡 = 0𝑱𝑱𝑻𝑻 𝒓𝒓, 𝑡𝑡 ≡ 𝑱𝑱 𝒓𝒓, 𝑡𝑡 +1

4 𝜋𝜋 𝛻𝛻 �−∞

+∞𝛻𝛻𝑞. 𝑱𝑱 𝒓𝒓𝑞, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′ 𝑑𝑑𝑑𝑑𝑞

Φ 𝒓𝒓, 𝑡𝑡 =1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌 𝒓𝒓′, 𝑡𝑡𝒓𝒓 − 𝒓𝒓′ 𝑑𝑑𝑑𝑑𝑞



𝑬𝑬 𝒓𝒓, 𝑡𝑡 = −𝜕𝜕𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 0 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0

Φ 𝒓𝒓, 𝑡𝑡 = 0

−𝛻𝛻𝟐𝟐𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑨𝑨 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2 = 0

The Coulomb gauge is particularly useful in the absence of charges and currents

𝛻𝛻.𝑨𝑨 𝒓𝒓, 𝑡𝑡 = 0

&


Energy and Power from E & B fields



𝑩𝑩 𝒓𝒓, 𝑡𝑡 .

𝑬𝑬 𝒓𝒓, 𝑡𝑡 .

𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 =

= −𝑩𝑩 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

−𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡


𝛻𝛻. 𝒂𝒂 × 𝒃𝒃 = 𝒃𝒃 .𝛻𝛻 × 𝒂𝒂 − 𝒂𝒂 . 𝛻𝛻 × 𝒃𝒃HW:

= 𝑩𝑩 𝒓𝒓, 𝑡𝑡 . 𝛻𝛻 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡

= −12𝜕𝜕 𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡− 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0

12𝜕𝜕 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡

= −𝑩𝑩 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

− 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 .𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

𝛻𝛻. 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

+𝜖𝜖02𝜕𝜕 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡+

12 𝜇𝜇𝑜𝑜

𝜕𝜕 𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡+ 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0

𝛻𝛻. 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡


𝛻𝛻. 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

+𝜖𝜖02𝜕𝜕 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡 +1

2 𝜇𝜇𝑜𝑜𝜕𝜕 𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝑡𝑡 + 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0

�𝑉𝑉

𝛻𝛻.𝓢𝓢 +𝜕𝜕𝑢𝑢𝐸𝐸𝜕𝜕𝑡𝑡 +

𝜕𝜕𝑢𝑢𝐵𝐵𝜕𝜕𝑡𝑡 +

𝜕𝜕𝜕𝜕𝜕𝜕𝑡𝑡 𝑑𝑑𝑑𝑑 = 0

�𝑆𝑆𝓢𝓢 .𝒅𝒅𝒅𝒅 +

𝑑𝑑𝑈𝑈𝐸𝐸𝑑𝑑𝑡𝑡 +

𝑑𝑑𝑈𝑈𝐵𝐵𝑑𝑑𝑡𝑡 +

𝑑𝑑𝑊𝑊𝑑𝑑𝑡𝑡 = 0

𝛻𝛻.𝓢𝓢 +𝜕𝜕𝑢𝑢𝐸𝐸𝜕𝜕𝑡𝑡

+𝜕𝜕𝑢𝑢𝐵𝐵𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕𝜕𝜕𝑡𝑡

= 0

�𝑆𝑆𝓢𝓢 .𝒅𝒅𝒅𝒅 +

𝑑𝑑𝑑𝑑𝑡𝑡�𝑉𝑉

𝑢𝑢𝐸𝐸 + 𝑢𝑢𝐵𝐵 + 𝜕𝜕 𝑑𝑑𝑑𝑑 = 0

𝓢𝓢 ≡ 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

𝑢𝑢𝐸𝐸 =𝜖𝜖02 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2 𝑢𝑢𝐵𝐵 =

12 𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕𝜕𝜕𝜕𝜕𝑡𝑡

= 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Differential form of conservation of energy

(per-unit-time)

Integral form of conservation of

energy (per-unit-time)

Poynting vector


HW: From Maxwell’s equations, prove the following relation:

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡

+ 𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡

+𝜕𝜕𝜕𝜕𝑡𝑡

𝜖𝜖0 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

= 0


HW: Show that

= �𝑗𝑗=1

3𝜕𝜕𝜕𝜕𝑥𝑥𝑗𝑗

𝜖𝜖012𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗 +

1𝜇𝜇𝑜𝑜

12𝑩𝑩 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑖𝑖 +

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑖𝑖 =


Linear Momentum and Forcefrom E & B fields

𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0

𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡 − 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 01𝜇𝜇𝑜𝑜𝑩𝑩 𝒓𝒓, 𝑡𝑡 ×

− 𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 −𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

= 0

𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0−1𝜇𝜇𝑜𝑜𝑩𝑩 𝒓𝒓, 𝑡𝑡

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×



+1𝜇𝜇𝑜𝑜



+𝜕𝜕𝜕𝜕𝑡𝑡


= 0


𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡𝑑𝑑𝑑𝑑 = 𝑑𝑑𝑭𝑭𝐿𝐿 𝒓𝒓, 𝑡𝑡

𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 =𝑑𝑑𝑭𝑭𝐿𝐿 𝒓𝒓, 𝑡𝑡

𝑑𝑑𝑑𝑑≡ 𝒇𝒇𝐿𝐿 𝒓𝒓, 𝑡𝑡

the Lorentz force (per-unit-volume)due to the fields acting on the

charged particles (charges and currents)

The rate of change of the

linear momentum (per-unit-volume) of the charged

particles

=𝜕𝜕𝒑𝒑 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

82Prof. Sergio B. MendesSpring 2018+𝜕𝜕𝒑𝒑 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

+𝜕𝜕𝜕𝜕𝑡𝑡


= 0


+1𝜇𝜇𝑜𝑜



+ 𝒇𝒇𝐿𝐿 𝒓𝒓, 𝑡𝑡


𝒈𝒈 𝒓𝒓, 𝑡𝑡 ≡ 𝜖𝜖0 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

= 𝜖𝜖0 𝜇𝜇𝑜𝑜 𝓢𝓢

Linear momentum (per-unit-volume) associated withthe E & B fields

𝜕𝜕𝜕𝜕𝑡𝑡


=𝜕𝜕𝜕𝜕𝑡𝑡𝒈𝒈 𝒓𝒓, 𝑡𝑡


+𝜕𝜕𝒑𝒑 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

+𝜕𝜕𝒈𝒈 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0


+1𝜇𝜇𝑜𝑜


+𝜕𝜕𝜕𝜕𝑡𝑡



+𝜕𝜕𝑝𝑝𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

+𝜕𝜕𝑔𝑔𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

= 0

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑖𝑖

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑖𝑖

Let’s calculate one Cartesian component:


𝑖𝑖

𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖

= 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝑥𝑥𝑎𝑎

− 𝐸𝐸𝑖𝑖𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑗𝑗

𝑃𝑃𝑖𝑖 = 𝛁𝛁 × 𝑬𝑬 𝑖𝑖

𝑬𝑬 × 𝑷𝑷 𝑖𝑖

𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡

𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖

= 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝛁𝛁 × 𝑬𝑬 𝑖𝑖 − 𝐸𝐸𝑖𝑖𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑗𝑗

= 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝑃𝑃𝑖𝑖

= 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝑥𝑥𝑎𝑎

= 𝐸𝐸𝑖𝑖𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑗𝑗

𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖

otherwise

(1,2,3) or (2,3,1) or (3,1,2)

(1,3,2) or (3,2,1) or (2,1,3)

= 0

= + 1

= - 1

= �𝑗𝑗=1

3

�𝑖𝑖=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝐸𝐸𝑗𝑗 𝑃𝑃𝑖𝑖

= �𝑘𝑘=1

3

�𝑎𝑎=1

3

𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝑥𝑥𝑎𝑎

= 𝐸𝐸𝑖𝑖�𝑗𝑗=1

3𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑗𝑗


𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘

= 𝛿𝛿𝑖𝑖𝑎𝑎 𝛿𝛿𝑗𝑗𝑘𝑘

𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖 = 𝜖𝜖𝑖𝑖𝑗𝑗𝑖𝑖 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝑥𝑥𝑎𝑎


= 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑖𝑖

−𝜕𝜕𝐸𝐸𝑖𝑖𝜕𝜕𝑥𝑥𝑗𝑗


= 𝛿𝛿𝑖𝑖𝑎𝑎 𝛿𝛿𝑗𝑗𝑘𝑘 − 𝛿𝛿𝑖𝑖𝑘𝑘 𝛿𝛿𝑗𝑗𝑎𝑎 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑘𝑘𝜕𝜕𝑥𝑥𝑎𝑎


− 𝛿𝛿𝑖𝑖𝑘𝑘 𝛿𝛿𝑗𝑗𝑎𝑎

= 𝜖𝜖𝑖𝑖𝑖𝑖𝑗𝑗 𝜖𝜖𝑖𝑖𝑎𝑎𝑘𝑘


𝑬𝑬 × 𝛁𝛁 × 𝑬𝑬 − 𝑬𝑬 𝛁𝛁.𝑬𝑬 𝑖𝑖 = 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑖𝑖

−𝜕𝜕𝐸𝐸𝑖𝑖𝜕𝜕𝑥𝑥𝑗𝑗


= 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑖𝑖

− 𝐸𝐸𝑗𝑗𝜕𝜕𝐸𝐸𝑖𝑖𝜕𝜕𝑥𝑥𝑗𝑗


=12𝜕𝜕 𝐸𝐸𝑗𝑗 𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑖𝑖

−𝜕𝜕 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑗𝑗

=12𝜕𝜕 𝑬𝑬 2

𝜕𝜕𝑥𝑥𝑗𝑗𝛿𝛿𝑖𝑖𝑗𝑗 −

𝜕𝜕 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗𝜕𝜕𝑥𝑥𝑗𝑗

=𝜕𝜕𝜕𝜕𝑥𝑥𝑗𝑗

12𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗


+𝜕𝜕𝑝𝑝𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

+𝜕𝜕𝑔𝑔𝑖𝑖 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡

= 0

𝜖𝜖0 𝑬𝑬 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 𝑖𝑖

+1𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 𝑖𝑖


+𝜕𝜕𝑝𝑝𝑖𝑖𝜕𝜕𝑡𝑡

+𝜕𝜕𝑔𝑔𝑖𝑖𝜕𝜕𝑡𝑡

= 0

𝜕𝜕𝜕𝜕𝑥𝑥𝑗𝑗

𝜖𝜖012𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗

+𝜕𝜕𝜕𝜕𝑥𝑥𝑗𝑗

1𝜇𝜇𝑜𝑜

12𝑩𝑩 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

�𝑗𝑗=1

3𝜕𝜕𝑇𝑇𝑖𝑖𝑗𝑗𝜕𝜕𝑥𝑥𝑗𝑗


𝑇𝑇𝑖𝑖𝑗𝑗 ≡ 𝜖𝜖012𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗 +

1𝜇𝜇𝑜𝑜

12𝑩𝑩 2 𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

�𝑗𝑗=1


+𝜕𝜕𝑔𝑔𝑖𝑖𝜕𝜕𝑡𝑡


= 0

�𝑉𝑉�𝑗𝑗=1


𝑑𝑑𝑑𝑑 + �𝑉𝑉

𝜕𝜕𝑔𝑔𝑖𝑖𝜕𝜕𝑡𝑡


𝑑𝑑𝑑𝑑 = 0

�𝑆𝑆�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 𝑑𝑑𝑆𝑆𝑗𝑗 +𝑑𝑑𝑑𝑑𝑡𝑡�

𝑉𝑉𝑔𝑔𝑖𝑖 + 𝑝𝑝𝑖𝑖 𝑑𝑑𝑑𝑑 = 0

Maxwell’s stress tensor

−𝐹𝐹𝑖𝑖


𝐹𝐹𝑖𝑖 = −�𝑆𝑆�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 𝑑𝑑𝑆𝑆𝑗𝑗

𝑭𝑭 = −�𝑆𝑆𝑻𝑻 . �𝒏𝒏 𝑑𝑑𝑆𝑆

𝑑𝑑𝑭𝑭𝑑𝑑𝑆𝑆

= − 𝑻𝑻 . �𝒏𝒏

= −�𝑆𝑆�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 �𝑛𝑛𝑗𝑗 𝑑𝑑𝑆𝑆�𝒏𝒏

unit vector normal to

the surface

𝑑𝑑𝐹𝐹𝑖𝑖𝑑𝑑𝑆𝑆

= −�𝑗𝑗=1

3

𝑇𝑇𝑖𝑖𝑗𝑗 �𝑛𝑛𝑗𝑗


𝑇𝑇𝑖𝑖𝑗𝑗 ≡ 𝜖𝜖012𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗 +

1𝜇𝜇𝑜𝑜

12𝑩𝑩 2 𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

= 𝜖𝜖0 −12𝑬𝑬 2 �𝒏𝒏 + 𝑬𝑬 𝑬𝑬. �𝒏𝒏 +

1𝜇𝜇𝑜𝑜

−12𝑩𝑩 2 �𝒏𝒏 + 𝑩𝑩 𝑩𝑩. �𝒏𝒏


= − 𝑻𝑻 . �𝒏𝒏


= −𝜖𝜖02𝑬𝑬 2 +

12 𝜇𝜇𝑜𝑜

𝑩𝑩 2 �𝒏𝒏 + 𝜖𝜖0 𝑬𝑬 𝑬𝑬. �𝒏𝒏 +1𝜇𝜇𝑜𝑜𝑩𝑩 𝑩𝑩. �𝒏𝒏

along − �𝒏𝒏,

compressing the surface inwards

along + �𝒏𝒏,

tensioning the surface outwards


𝑬𝑬 ⊥ �𝒏𝒏


= −𝜖𝜖02𝑬𝑬 2 �𝒏𝒏

𝑩𝑩 ⊥ �𝒏𝒏


= −1

2 𝜇𝜇𝑜𝑜𝑩𝑩 2�𝒏𝒏

𝑩𝑩 = 𝟎𝟎 𝑬𝑬 = 0

𝑑𝑑𝑭𝑭�𝒏𝒏 𝑑𝑑𝑭𝑭

�𝒏𝒏


𝑑𝑑𝑭𝑭�𝒏𝒏

�𝒏𝒏𝑑𝑑𝑭𝑭

crashing can

https://www.youtube.com/watch?v=YFTH8ywXaZs

https://www.youtube.com/watch?v=YFTH8ywXaZs



𝑬𝑬 ∥ �𝒏𝒏 or −�𝒏𝒏𝑩𝑩 = 𝟎𝟎


= +𝜖𝜖02𝑬𝑬 2 �𝒏𝒏


= +1

2 𝜇𝜇𝑜𝑜𝑩𝑩 2�𝒏𝒏

𝑩𝑩 ∥ �𝒏𝒏 or − �𝒏𝒏𝑬𝑬 = 𝟎𝟎






𝑑𝑑𝑭𝑭

�𝒏𝒏

�𝒏𝒏

𝑑𝑑𝑭𝑭

𝑑𝑑𝑭𝑭

�𝒏𝒏

�𝒏𝒏

𝑑𝑑𝑭𝑭


Conservation Laws


𝛁𝛁. 𝑱𝑱 +𝜕𝜕𝜌𝜌𝜕𝜕𝑡𝑡

= 0

Conservation of Charge


𝛁𝛁.𝓢𝓢 +𝜕𝜕𝜕𝜕𝑡𝑡

𝑢𝑢𝐸𝐸 + 𝑢𝑢𝐵𝐵 + 𝜕𝜕 = 0

𝓢𝓢 = 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

𝑢𝑢𝐸𝐸 =𝜖𝜖02 𝑬𝑬 𝒓𝒓, 𝑡𝑡 2 𝑢𝑢𝐵𝐵 =

12 𝜇𝜇𝑜𝑜

𝑩𝑩 𝒓𝒓, 𝑡𝑡 2

𝜕𝜕 = 𝑬𝑬 𝒓𝒓, 𝑡𝑡 . 𝑱𝑱 𝒓𝒓, 𝑡𝑡

Conservation of Energy


𝛁𝛁.𝑻𝑻 +𝜕𝜕𝜕𝜕𝑡𝑡

𝒑𝒑 + 𝒈𝒈 = 0

𝒈𝒈 𝒓𝒓, 𝑡𝑡 = 𝜖𝜖0 𝜇𝜇𝑜𝑜 𝑬𝑬 𝒓𝒓, 𝑡𝑡 ×𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜇𝜇𝑜𝑜

= 𝜖𝜖0 𝜇𝜇𝑜𝑜 𝓢𝓢

𝑇𝑇𝑖𝑖𝑗𝑗 = 𝜖𝜖012 𝑬𝑬 2𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐸𝐸𝑖𝑖 𝐸𝐸𝑗𝑗 +

1𝜇𝜇𝑜𝑜

12 𝑩𝑩 2 𝛿𝛿𝑖𝑖𝑗𝑗 − 𝐵𝐵𝑖𝑖 𝐵𝐵𝑗𝑗

𝜕𝜕𝜕𝜕𝑡𝑡 𝒑𝒑 = 𝒇𝒇𝐿𝐿 𝒓𝒓, 𝑡𝑡 = 𝜌𝜌 𝒓𝒓, 𝑡𝑡 𝑬𝑬 𝒓𝒓, 𝑡𝑡 + 𝑱𝑱 𝒓𝒓, 𝑡𝑡 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡

Conservation of Linear Momentum


𝛻𝛻.𝑬𝑬 𝒓𝒓, 𝑡𝑡 =

𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0


𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

Electromagnetic Waves𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 0 𝑱𝑱 𝒓𝒓, 𝑡𝑡 = 0

+ 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

Consider: &

0



𝜵𝜵 ×

𝛻𝛻 × 𝛻𝛻 × 𝒂𝒂 = 𝛻𝛻 𝛻𝛻.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂HW:

𝜵𝜵 × 𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝛁𝛁 𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 − 𝛁𝛁𝟐𝟐𝑬𝑬 𝒓𝒓, 𝑡𝑡


𝛻𝛻.𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝟎𝟎

𝛻𝛻𝟐𝟐𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

= −𝛁𝛁 ×𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= −𝜕𝜕𝜕𝜕𝑡𝑡

𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡



𝜵𝜵 ×

𝛻𝛻 × 𝛻𝛻 × 𝒂𝒂 = 𝛻𝛻 𝛻𝛻.𝒂𝒂 − 𝛻𝛻𝟐𝟐𝒂𝒂HW:

𝜵𝜵 × 𝛁𝛁 × 𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝛁𝛁 𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 − 𝛁𝛁𝟐𝟐𝑩𝑩 𝒓𝒓, 𝑡𝑡


𝛻𝛻.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝟎𝟎

𝛻𝛻𝟐𝟐𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

= 𝜇𝜇𝑜𝑜 𝜖𝜖0 𝛁𝛁 ×𝜕𝜕𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡

𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡


𝛻𝛻𝟐𝟐𝑬𝑬 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑬𝑬 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

𝛻𝛻𝟐𝟐𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡2

𝑣𝑣 =1𝜇𝜇𝑜𝑜 𝜖𝜖0

≅ 2.99792458 × 108 m/s

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