special relativity and electrodynamics 611 spring 20/classnotes... · special relativity and...

Special Relativityand

Electrodynamics

Spring 2020 Prof. Sergio B. Mendes 1

• Chapter 1 of “Modern Problems in Classical Electrodynamics” by Brau

• Chapter 11 “Classical Electrodynamics” by Jackson, 3rd ed.

• Chapter 1-4 “The Classical Theory of Fields” by Landau and Lifshitz, 4th ed.

• Chapter 5 “Electrodynamics” by Melia

• Special Relativity and Electrodynamics, by Prof. Leonard Susskind, part of the Theoretical Minimum Lectures from Stanford University

https://theoreticalminimum.com/courses/special-relativity-and-electrodynamics/2012/spring

2Prof. Sergio B. MendesSpring 2018

1. Laws of physics are the same (invariant) in any inertial frame of reference.

2. Light propagates in vacuum at constant speed.

Einstein’s Two Postulates of Special Relativity

Outcome of the Two Postulates:

Spring 2020 Prof. Sergio B. Mendes 3

𝑐 ∆𝑡 2 − ∆𝑥 2 = 𝑐 ∆𝑡′ 2 − ∆𝑥′ 2

Relating the Coordinates of the Two Frames of Reference

Fall 2018 Prof. Sergio B. Mendes 4

𝑥′ = 𝑥 − 𝑣𝑜 𝑡

𝑥 = 𝑥′ + 𝑣𝑜 𝑡′

𝛾

𝛾due to symmetry

linear modification

= 𝛾 𝛾 𝑥 − 𝑣𝑜 𝑡 + 𝑣𝑜 𝑡′

𝑡′ =1

𝛾 𝑣𝑜1 − 𝛾2 𝑥 + 𝛾 𝑡solving for 𝑡′:

in Galilean transformation

Inserting the Results into the Postulates


∆𝑥′ = 𝛾 ∆𝑥 − 𝑣𝑜 ∆𝑡∆𝑡′ =1

𝛾 𝑣𝑜1 − 𝛾2 ∆𝑥 + 𝛾 ∆𝑡

𝑐 ∆𝑡 2 − ∆𝑥 2 = 𝑐 ∆𝑡′ 2 − ∆𝑥′ 2

𝑥′ = 𝛾 𝑥 − 𝑣𝑜 𝑡𝑡′ =1

𝛾 𝑣𝑜1 − 𝛾2 𝑥 + 𝛾 𝑡

Finding 𝛾:


𝛾 =1

1 − 𝛽2𝛽 ≡

𝑣𝑜𝑐

∆𝑥 2: 1 = γ2 −𝑐2

𝛾2 𝑣𝑜2 1 − 𝛾2 2 1

1 − 𝛾2= −

𝑐2

𝛾2 𝑣𝑜2

1

𝛾2− 1 = −

𝑣𝑜2

𝑐2

∆𝑡 2: −𝑐2 = 𝛾2 𝑣𝑜2− 𝑐2 𝛾2

1

𝛾2= 1 −

𝑣𝑜2

𝑐2

2 ∆𝑥 ∆𝑡: 0 = −𝛾2 𝑣𝑜 − 𝑐21

𝑣𝑜1 − 𝛾2 𝛾2 𝑣𝑜 = −𝑐2

1

𝑣𝑜1 − 𝛾2 −

𝑣𝑜2

𝑐2=

1

𝛾2− 1


𝑦′ = 𝑦

𝑧′ = 𝑧

𝑥′ = 𝛾 𝑥 − 𝑣𝑜 𝑡

𝛾 =1

1 − 𝛽2

𝑡′ =1

𝛾 𝑣𝑜1 − 𝛾2 𝑥 + 𝛾 𝑡

Lorentz Transformations

= −𝛾 𝛽 𝑐 𝑡 + 𝛾 𝑥

𝛽 ≡𝑣𝑜𝑐

𝑐 𝑡′ = 𝛾 𝑐 𝑡 − 𝛾 𝛽 𝑥


In general 𝑑𝑥′ ≠ 𝑑𝑥 and 𝑑𝑡′ = 𝑑𝑡,

𝑐2 𝑑𝑡 2 − 𝑑𝑥 2 − 𝑑𝑦 2 − 𝑑𝑧 2 = 𝑐2 𝑑𝑡′ 2 − 𝑑𝑥′ 2− 𝑑𝑦′ 2 − 𝑑𝑧′ 2

≡ 𝑑𝑠 2

𝑑𝑠 ≡ space-time distance

An Invariant

≡ 𝑐2 𝑑𝜏 2

𝑑𝜏 ≡ proper time

however we always have:


Proper Time

𝑐2 𝑑𝑡 2 − 𝑑𝑥 2 − 𝑑𝑦 2 − 𝑑𝑧 2 = 𝑐2 𝑑𝜏 2

𝑐2 −𝑑𝑥

𝑑𝑡

2

−𝑑𝑦

𝑑𝑡

2

−𝑑𝑧

𝑑𝑡

2

= 𝑐2𝑑𝜏

𝑑𝑡

2

1 − 𝛽2 =𝑑𝜏

𝑑𝑡

2

𝑑𝜏 =𝑑𝑡

𝛾

÷ 𝑑𝑡 2

÷ 𝑐2

1

𝛾2=


𝑐 𝑡 ≡ 𝑥0

𝑥 ≡ 𝑥1

𝑦 ≡ 𝑥2

𝑧 ≡ 𝑥3

Let’s adopt the following notation:

𝑐 𝑡′ ≡ 𝑥′0

𝑥 ≡ 𝑥′1

𝑦 ≡ 𝑥′2

𝑧 ≡ 𝑥′3


𝑥′1 ≡ 𝑥′1 𝑥0, 𝑥1, 𝑥2, 𝑥3 = −𝛾 𝛽 𝑥0 + 𝛾 𝑥1

𝑥′0 ≡ 𝑥′0 𝑥0, 𝑥1, 𝑥2, 𝑥3 = 𝛾 𝑥0 − 𝛾 𝛽 𝑥1

𝑥′2 ≡ 𝑥′2 𝑥0, 𝑥1, 𝑥2, 𝑥3 = 𝑥2

𝑥′3 ≡ 𝑥′3 𝑥0, 𝑥1, 𝑥2, 𝑥3 = 𝑥3


𝑥′0

𝑥′1

𝑥′2

𝑥′3

=

𝛾−𝛾𝛽00

−𝛾𝛽𝛾00

0010

0001

𝑥0

𝑥1

𝑥2

𝑥3


Lorentz Transformationsfor Differentials

𝑑𝑥′0

𝑑𝑥′1

𝑑𝑥′2

𝑑𝑥′3

=

𝜕𝑥′0

𝜕𝑥0

𝜕𝑥′1

𝜕𝑥0

𝜕𝑥′2

𝜕𝑥0

𝜕𝑥′3

𝜕𝑥0

𝜕𝑥′0

𝜕𝑥1

𝜕𝑥′1

𝜕𝑥1

𝜕𝑥′2

𝜕𝑥1

𝜕𝑥′3

𝜕𝑥1

𝜕𝑥′0

𝜕𝑥2

𝜕𝑥′1

𝜕𝑥2

𝜕𝑥′2

𝜕𝑥2

𝜕𝑥′3

𝜕𝑥2

𝜕𝑥′0

𝜕𝑥3

𝜕𝑥′1

𝜕𝑥3

𝜕𝑥′2

𝜕𝑥3

𝜕𝑥′3

𝜕𝑥3

𝑑𝑥0

𝑑𝑥1

𝑑𝑥2

𝑑𝑥3


𝑑𝑥′𝜇 =

𝛼=0

3𝜕𝑥′𝜇

𝜕𝑥𝛼𝑑𝑥𝛼 =

𝛼=0

3

𝐿𝛼𝜇𝑑𝑥𝛼 = 𝐿𝛼

𝜇𝑑𝑥𝛼

𝐿𝛼𝜇≡𝜕𝑥′𝜇

𝜕𝑥𝛼

𝐿𝛼𝜇≡𝜕𝑥′𝜇

𝜕𝑥𝛼=

𝛾−𝛾𝛽00

−𝛾𝛽𝛾00

0010

0001


𝑥1 ≡ 𝑥1 𝑥′0, 𝑥′

1, 𝑥′

2, 𝑥′

3= 𝛾 𝛽 𝑥′

0+ 𝛾 𝑥′1

𝑥0 ≡ 𝑥0 𝑥′0, 𝑥′

1, 𝑥′

2, 𝑥′

3= 𝛾 𝑥′

0+ 𝛾 𝛽 𝑥′1

𝑥2 ≡ 𝑥2 𝑥′0, 𝑥′

1, 𝑥′

2, 𝑥′

3= 𝑥′

2

𝑥3 ≡ 𝑥3 𝑥′0, 𝑥′

1, 𝑥′

2, 𝑥′

3= 𝑥′

3

Inverse Transformations

𝑥0

𝑥1

𝑥2

𝑥3

=

𝛾+𝛾𝛽00

+𝛾𝛽𝛾00

0010

0001

𝑥′0

𝑥′1

𝑥′2

𝑥′3


𝑑𝑥0

𝑑𝑥1

𝑑𝑥2

𝑑𝑥3

=

𝜕𝑥0

𝜕𝑥′0

𝜕𝑥1

𝜕𝑥′0

𝜕𝑥2

𝜕𝑥′0

𝜕𝑥3

𝜕𝑥′0

𝜕𝑥0

𝜕𝑥′1

𝜕𝑥1

𝜕𝑥′1

𝜕𝑥2

𝜕𝑥′1

𝜕𝑥3

𝜕𝑥′1

𝜕𝑥0

𝜕𝑥′2

𝜕𝑥1

𝜕𝑥′2

𝜕𝑥2

𝜕𝑥′2

𝜕𝑥3

𝜕𝑥′2

𝜕𝑥0

𝜕𝑥′3

𝜕𝑥1

𝜕𝑥′3

𝜕𝑥2

𝜕𝑥′3

𝜕𝑥3

𝜕𝑥′3

𝑑𝑥′0

𝑑𝑥′1

𝑑𝑥′2

𝑑𝑥′3

Inverse Lorentz Transformationsfor Differentials


𝐼𝐿𝜇𝛼 ≡

𝜕𝑥𝛼

𝜕𝑥′𝜇=

𝛾+𝛾𝛽00

+𝛾𝛽𝛾00

0010

0001

𝑑𝑥𝛼 =

𝛼=0

3𝜕𝑥𝛼

𝜕𝑥′𝜇𝑑𝑥′𝜇 =

𝛼=0

3

𝐼𝐿𝜇𝛼 𝑑𝑥′𝜇 = 𝐼𝐿𝜇

𝛼 𝑑𝑥′𝜇


𝜕𝑥𝛼

𝜕𝑥′𝜇


𝑑𝑥′𝜇 =𝜕𝑥′𝜇

𝜕𝑥𝛼𝑑𝑥𝛼 𝑑𝑥𝛼 =

𝜕𝑥𝛼

𝜕𝑥′𝛽𝑑𝑥′𝛽

𝑑𝑥′𝜇 =𝜕𝑥′𝜇

𝜕𝑥𝛼𝜕𝑥𝛼

𝜕𝑥′𝛽𝑑𝑥′𝛽 =

𝜕𝑥′𝜇

𝜕𝑥′𝛽𝑑𝑥′𝛽 = 𝛿𝛽

𝜇

𝜕𝑥′𝜇

𝜕𝑥𝛼𝜕𝑥𝛼

𝜕𝑥′𝛽= 𝐿𝛼

𝜇𝐼𝐿𝛽𝛼 = 𝛿𝛽

𝜇

Yes, it’s consistent !!



contravarianttensor of rank 1

𝐿𝛼𝜇=𝜕𝑥′𝜇

𝜕𝑥𝛼

and Contravariant Four-Vectors

𝑑𝑥′𝜇 =

𝛼=0

3𝜕𝑥′𝜇

𝜕𝑥𝛼𝑑𝑥𝛼 =

𝛼=0

3

𝐿𝛼𝜇𝑑𝑥𝛼 = 𝐿𝛼

𝜇𝑑𝑥𝛼

𝑉′𝜇 =

𝛼=0

3𝜕𝑥′𝜇

𝜕𝑥𝛼𝑉𝛼 =

𝛼=0

3

𝐿𝛼𝜇𝑉𝛼 = 𝐿𝛼

𝜇𝑉𝛼


Examples of Contravariant Four-Vectors


𝑑𝑥𝜇 ≡

𝑐 𝑑𝑡𝑑𝑥𝑑𝑦𝑑𝑧

Four-Vector Distance

𝑑𝑥′𝜇 =

𝛼=0

3

𝐿𝛼𝜇𝑑𝑥𝛼

contravariant tensor of rank 1

𝑑𝑥1 ≡ 𝑑𝑥

𝑑𝑥0 ≡ 𝑐 𝑑𝑡

𝑑𝑥2 ≡ 𝑑𝑦

𝑑𝑥3 ≡ 𝑑𝑧


=1

𝑑𝜏


Four-Vector Velocity

𝑣′𝜇 =

𝛼=0

3

𝐿𝛼𝜇𝑣𝛼


=𝛾

𝑑𝑡


=

𝛾 𝑐

𝛾 ൗ𝑑𝑥𝑑𝑡

𝛾 ൗ𝑑𝑦𝑑𝑡

𝛾 ൗ𝑑𝑧𝑑𝑡

𝑣𝜇 ≡𝑑𝑥𝜇

𝑑𝜏

𝑣1 ≡ 𝛾𝑑𝑥

𝑑𝑡≡ 𝛾 𝑣𝑥

𝑣0 ≡ 𝛾 𝑐

=

𝛾 𝑐𝛾 𝑣𝑥𝛾 𝑣𝑦𝛾 𝑣𝑧

𝑣2 ≡ 𝛾𝑑𝑦

𝑑𝑡≡ 𝛾 𝑣𝑦

𝑣3 ≡ 𝛾𝑑𝑧

𝑑𝑡≡ 𝛾 𝑣𝑧


Four-Vector Linear Momentum

𝑝′𝜇 =

𝛼=0

3

𝐿𝛼𝜇𝑝𝛼

= 𝑚𝛾

𝑑𝑡


=

𝛾 𝑚 𝑐

𝛾 𝑚 ൗ𝑑𝑥𝑑𝑡

𝛾 𝑚 ൗ𝑑𝑦𝑑𝑡

𝛾 𝑚 ൗ𝑑𝑧𝑑𝑡

𝑝𝜇 ≡ 𝑚𝑑𝑥𝜇

𝑑𝜏=

𝛾 𝑚 𝑐𝑝𝑥𝑝𝑦𝑝𝑧

𝑝1 ≡ 𝑝𝑥 ≡ 𝛾 𝑚𝑑𝑥

𝑑𝑡= 𝛾 𝑚 𝑣𝑥

𝑝0 ≡ 𝛾 𝑚 𝑐


𝑝2 ≡ 𝑝𝑦 ≡ 𝛾 𝑚𝑑𝑦

𝑑𝑡= 𝛾 𝑚 𝑣𝑦

𝑝1 ≡ 𝑝𝑧 ≡ 𝛾 𝑚𝑑𝑧

𝑑𝑡= 𝛾 𝑚 𝑣𝑧


Four-Vector Force

= 𝛾𝑑

𝑑𝑡

𝛾 𝑚 𝑐

𝛾 𝑚 ൗ𝑑𝑥𝑑𝑡

𝛾 𝑚 ൗ𝑑𝑦𝑑𝑡

𝛾 𝑚 ൗ𝑑𝑧𝑑𝑡

𝐹𝜇 ≡𝑑𝑝𝜇

𝑑𝜏=

𝛾𝑑

𝑑𝑡𝛾 𝑚 𝑐

𝛾 𝑚𝑑

𝑑𝑡𝛾𝑑𝑥

𝑑𝑡

𝛾 𝑚𝑑

𝑑𝑡𝛾𝑑𝑦

𝑑𝑡

𝛾 𝑚𝑑

𝑑𝑡𝛾𝑑𝑧

𝑑𝑡


𝐹𝑖 ≡ 𝛾 𝑚𝑑

𝑑𝑡𝛾𝑑𝑥𝑖

𝑑𝑡

𝐹0 ≡ 𝛾𝑑

𝑑𝑡𝛾 𝑚 𝑐

𝐹′𝜇 =

𝛼=0

3

𝐿𝛼𝜇𝐹𝛼


A Different Kind of Four-Vector:

the Four-Vector Gradient, which transforms through the Inverse Lorentz Transformations


Four-Vector Gradient, Inverse Lorentz Transformations,

covarianttensor of rank 1

and Covariant Four-Vectors

𝑊′𝜇 =

𝛼=0

3𝜕𝑥𝛼

𝜕𝑥′𝜇𝑊𝛼 =

𝛼=0

3

𝐼𝐿𝜇𝛼 𝑊𝛼 = 𝐼𝐿𝜇

𝛼 𝑊𝛼

𝜕

𝜕𝑥′𝜇=

𝜈=0

3𝜕𝑥𝛼

𝜕𝑥′𝜇𝜕

𝜕𝑥𝛼=

𝛼=0

3

𝐼𝐿𝜇𝛼

𝜕

𝜕𝑥𝛼= 𝐼𝐿𝜇

𝛼𝜕

𝜕𝑥𝛼


𝜕𝑥𝛼

𝜕𝑥′𝜇


Inner Product of Contravariant and Covariant Tensors of Rank 1

𝑉′𝜇 =

𝜈=0

3𝜕𝑥′𝜇

𝜕𝑥𝜈𝑉𝜈 𝑊′𝜇 =

𝛼=0

3𝜕𝑥𝛼

𝜕𝑥′𝜇𝑊𝛼

𝜇=0

3

𝑉′𝜇 𝑊′𝜇

=

𝜇=0

3

𝜈=0

3

𝛼=0

3𝜕𝑥′𝜇

𝜕𝑥𝜈𝜕𝑥𝛼

𝜕𝑥′𝜇𝑉𝜈 𝑊𝛼 =

𝜈=0

3

𝛼=0

3

𝛿𝜈𝛼 𝑉𝜈 𝑊𝛼 =

𝜈=0

3

𝑉𝜈 𝑊𝜈

=

𝜇=0

3

𝜈=0

3𝜕𝑥′𝜇

𝜕𝑥𝜈𝑉𝜈

𝛼=0

3𝜕𝑥𝛼

𝜕𝑥′𝜇𝑊𝛼

invariant !!


𝜇=0

3

𝑉′𝜇 𝑊′𝜇 =

𝜈=0

3

𝑉𝜈 𝑊𝜈

Inner Product and Invariants


Relation between Contravariant and Covariant Tensors

𝑐2 𝑑𝜏 2 = 𝑐2 𝑑𝑡 2 − 𝑑𝑥 2 − 𝑑𝑦 2 − 𝑑𝑧 2

= 𝑑𝑥0 2 − 𝑑𝑥1 2 − 𝑑𝑥2 2 − 𝑑𝑥3 2 =

𝜈=0

3

𝑑𝑥𝜇 𝑑𝑥𝜇

𝑑𝑥0 ≡ 𝑑𝑥0

𝑑𝑥1 ≡ − 𝑑𝑥1𝑑𝑥𝜇 ≡

𝑑𝑥0𝑑𝑥1𝑑𝑥2𝑑𝑥3

=

𝑑𝑥0

−𝑑𝑥1

−𝑑𝑥2

−𝑑𝑥3𝑑𝑥2 ≡ − 𝑑𝑥2

𝑑𝑥3 ≡ − 𝑑𝑥3


The Metric Tensor

𝑑𝑥𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝑑𝑥𝜈 = 𝑔𝜇𝜈 𝑑𝑥

𝜈

𝑔𝜇𝜈 =

1000

0−100

00−10

000−1

𝑑𝑥𝜇 =

𝑑𝑥0𝑑𝑥1𝑑𝑥2𝑑𝑥3

=

𝑑𝑥0

−𝑑𝑥1

−𝑑𝑥2

−𝑑𝑥3

=

1000

0−100

00−10

000−1

𝑑𝑥0

𝑑𝑥1

𝑑𝑥2

𝑑𝑥3


𝑔𝜇𝜈 = 𝑔𝜇𝜈

𝑑𝑥𝜇 =

𝜈=0

3

𝑔𝜇𝜈𝑑𝑥𝜈

𝑑𝑥𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝑑𝑥𝜈


𝑐 𝑑𝜏 2 =

𝜇=0

3

𝑑𝑥𝜇 𝑑𝑥𝜇

=

𝜇=0

3

𝑑𝑥𝜇

𝜈=0

3

𝑔𝜇𝜈 𝑑𝑥𝜈

=

𝜇=0

3

𝜈=0

3

𝑔𝜇𝜈𝑑𝑥𝜈 𝑑𝑥𝜇

𝑑𝑥𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝑑𝑥𝜈 𝑑𝑥𝜇 =

𝜈=0

3

𝑔𝜇𝜈𝑑𝑥𝜈

=

𝜇=0

3

𝜈=0

3

𝑔𝜇𝜈 𝑑𝑥𝜈 𝑑𝑥𝜇

Proof:


𝜇=0

3

𝑑𝑥𝜇 𝑑𝑥𝜇 = 𝑐 𝑑𝑡 2 − 𝑑𝑥 2 − 𝑑𝑦 2 − 𝑑𝑧 2 = 𝑐2 𝑑𝜏2

𝑑𝑥𝜇 =


𝑑𝑥𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝑑𝑥𝜈 =

𝑐 𝑑𝑡−𝑑𝑥−𝑑𝑦−𝑑𝑧

Four-Vector Distance Invariant


𝜇=0

3

𝑣𝜇 𝑣𝜇 = 𝛾 𝑐 2 − 𝛾 𝑣𝑥2 − 𝛾 𝑣𝑦

2− 𝛾 𝑣𝑧

2

𝑣𝜇 =

𝛾 𝑐𝛾 𝑣𝑥𝛾 𝑣𝑦𝛾 𝑣𝑧

𝑣𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝑣𝜈 =

𝛾 𝑐−𝛾 𝑣𝑥−𝛾 𝑣𝑦−𝛾 𝑣𝑧

= 𝛾 𝑐 2 1 −1

𝑐2𝑣𝑥

2 + 𝑣𝑦2 + 𝑣𝑧

2

= 𝑐2

Four-Vector Velocity Invariant


𝜇=0

3

𝑝𝜇 𝑝𝜇 = 𝛾 𝑚 𝑐 2 − 𝛾 𝑚 𝑣𝑥2 − 𝛾 𝑚 𝑣𝑦

2− 𝛾 𝑚 𝑣𝑧

2

𝑝𝜇 =

𝛾 𝑚 𝑐𝛾 𝑚 𝑣𝑥𝛾 𝑚 𝑣𝑦𝛾 𝑚 𝑣𝑧

𝑝𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝑝𝜈 =

𝛾 𝑚 𝑐−𝛾 𝑚 𝑣𝑥−𝛾 𝑚 𝑣𝑦−𝛾 𝑚 𝑣𝑧

= 𝛾 𝑚 𝑐 2 1 −1

𝑐2𝑣𝑥

2 + 𝑣𝑦2+ 𝑣𝑧

2

Four-Vector Linear Momentum Invariant

= 𝑚2 𝑐2


Inner Products and Invariants

𝑑𝑥𝜈 𝑑𝑥𝜈 = 𝑐2 𝑑𝜏2

𝑣𝜈 𝑣𝜈 = 𝑐2

𝑝𝜈 𝑝𝜈 = 𝑚2 𝑐2


Example 1: Charge Conservation

0 = 𝛁. 𝑱 +𝜕𝜌

𝜕𝑡= 𝛁. 𝑱 +

𝜕 𝑐 𝜌

𝜕 𝑐 𝑡

=𝜕 𝑐 𝜌

𝜕 𝑐 𝑡+𝜕𝐽𝑥𝜕𝑥

+𝜕𝐽𝑦𝜕𝑦

+𝜕𝐽𝑧𝜕𝑧

=𝜕 𝑐 𝜌

𝜕𝑥0+𝜕𝐽𝑥𝜕𝑥1

+𝜕𝐽𝑦𝜕𝑥2

+𝜕𝐽𝑧𝜕𝑥3

𝐽𝜇 =

𝑐 𝜌𝐽𝑥𝐽𝑦𝐽𝑧

=

𝜇=0

3𝜕

𝜕𝑥𝜇𝐽𝜇

contravariant

covariant


Four-Vector Current Density “J”

𝐽′𝜇 =

𝛼=0

3

𝐿𝛼𝜇𝐽𝛼

𝐽𝜇 =




𝐽𝜇 =


𝐽𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝐽𝜈 =

𝑐 𝜌−𝐽𝑥−𝐽𝑦−𝐽𝑧

𝜇=0

3

𝐽𝜇 𝐽𝜇 = 𝑐 𝜌 2 − 𝐽𝑥2 − 𝐽𝑦

2− 𝐽𝑧

2 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

Contravariant Transformations

Covariant Transformations

𝑉′𝜇 =

𝜈=0

3𝜕𝑥′𝜇

𝜕𝑥𝜈𝑉𝜈 𝑊′𝜇 =

𝜈=0

3𝜕𝑥𝜈

𝜕𝑥′𝜇𝑊𝜈

𝑥′0

𝑥′1

𝑥′2

𝑥′3

=

𝛾−𝛾𝛽00

−𝛾𝛽𝛾00

0010

0001

𝑥0

𝑥1

𝑥2

𝑥3

𝑥0

𝑥1

𝑥2

𝑥3

=

𝛾𝛾𝛽00

𝛾𝛽𝛾00

0010

0001

𝑥′0

𝑥′1

𝑥′2

𝑥′3

𝜕𝑥′𝜇

𝜕𝑥𝜈= =

𝛾−𝛾𝛽00

−𝛾𝛽𝛾00

0010

0001

𝜕𝑥𝜈

𝜕𝑥′𝜇=

𝛾𝛾𝛽00

𝛾𝛽𝛾00

0010

0001

𝑑𝑥′𝜇 =

𝜈=0

3𝜕𝑥′𝜇

𝜕𝑥𝜈𝑑𝑥𝜈

𝜕

𝜕𝑥′𝜇=

𝜈=0

3𝜕𝑥𝜈

𝜕𝑥′𝜇𝜕

𝜕𝑥𝜈39Prof. Sergio B. MendesSpring 2018

Contravariant Four Vectors Covariant Four Vector


𝑘𝜇 =

Τ𝜔 𝑐𝑘𝑥𝑘𝑦𝑘𝑧

𝑑𝑥𝜇 =


𝑝𝜇 ≡ 𝑚0 𝑣𝜇𝑣𝜇 ≡

𝑑𝑥𝜇

𝑑𝜏𝐹𝜇 ≡

𝑑𝑝𝜇

𝑑𝜏

𝜕

𝜕𝑥𝜇=

𝜕

𝑐 𝜕𝑡𝜕

𝜕𝑥𝜕

𝜕𝑦𝜕

𝜕𝑧

𝐽𝜇 =


𝐴𝜇 =

Φ/𝑐𝐴𝑥𝐴𝑦𝐴𝑧


What are the charge density and current density for a frame of

reference moving along x-axis at a constant speed 𝑣0 = 𝛽 𝑐 ?

Consider a charge 𝑞 at rest at a particular point 𝒓𝟎 in space.


𝑱𝜇 𝒓, 𝑡 =


𝑱 𝒓, 𝑡 = (0,0,0)𝜌 𝒓, 𝑡 = 𝑞 𝛿3 𝒓 − 𝒓𝟎

𝐽′𝜇 =

𝛼=0

3

𝐿𝛼𝜇𝐽𝛼 =

𝛾−𝛾𝛽00

−𝛾𝛽𝛾00

0010

0001

𝑐 𝑞 𝛿3 𝒓 − 𝒓𝟎000

=

𝛾 𝑐 𝑞 𝛿3 𝒓 − 𝒓𝟎− 𝛾 𝛽 𝑐 𝑞 𝛿3 𝒓 − 𝒓𝟎

00

=

𝑐 𝑞 𝛿3 𝒓 − 𝒓𝟎000

&


𝛿3 𝒓 − 𝒓𝟎

𝑐 𝑡 − 𝑡0𝑥 − 𝑥0𝑦 − 𝑦0𝑧 − 𝑧0

=

𝛼=0

3

𝐼𝐿𝛼𝜇

𝑥′𝛼 − 𝑥0′𝛼 =

𝛾𝛾𝛽00

𝛾𝛽𝛾00

0010

0001

𝑐 𝑡′ − 𝑡′0𝑥′ − 𝑥′0𝑦′ − 𝑦′0𝑧′ − 𝑧′0

=

𝛾 𝑐 𝑡′ − 𝑡′0 + 𝛾 𝛽 𝑥′ − 𝑥′0𝛾 𝑥′ − 𝑥′0 + 𝛾 𝛽 𝑐 𝑡′ − 𝑡′0

𝑦′ − 𝑦′0𝑧′ − 𝑧′0

= 𝛿 𝛾 𝑥′ − 𝑥′0 + 𝛾 𝛽 𝑐 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0

𝛿3 𝒓 − 𝒓𝟎 = 𝛿 𝑥 − 𝑥0 𝛿 𝑦 − 𝑦0 𝛿 𝑧 − 𝑧0

=1

𝛾𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0


𝐽′𝜇 =

𝛾 𝑐 𝑞 𝛿3 𝒓 − 𝒓𝟎− 𝛾 𝛽 𝑐 𝑞 𝛿3 𝒓 − 𝒓𝟎

00

𝛿3 𝒓 − 𝒓𝟎 =1

𝛾𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0

𝐽′𝜇 =

𝑐 𝑞 𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0− 𝛽 𝑐 𝑞 𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0

00

𝜌′ = 𝑞 𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0

𝐽′𝑥 = −𝑣0 𝑞 𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0


𝜌′ 𝒓′, 𝑡′ = 𝑞 𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0

𝐽′𝑥 𝒓′, 𝑡′

𝐽′𝑦 𝒓′, 𝑡′

𝐽′𝑧 𝒓′, 𝑡′

=−𝑣0 𝑞 𝛿 𝑥′ − 𝑥′0 + 𝑣0 𝑡′ − 𝑡′0 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0

00

𝑡′0 = 0 𝑥′0 = 0

= 𝑞 𝛿3 𝒓′ − 𝒓′𝟎 𝑡′

𝑱′ 𝒓′, 𝑡′ =−𝑣0 𝑞 𝛿 𝑥′ + 𝑣0 𝑡

′ 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′000

=−𝑣0 𝑞 𝛿

3 𝒓′ − 𝒓′𝟎 𝑡′

00

𝒓′𝟎 𝑡′ =

−𝑣0 𝑡′

𝑦′0𝑧′0

𝜌′ 𝒓′, 𝑡′ = 𝑞 𝛿 𝑥′ + 𝑣0 𝑡′ 𝛿 𝑦′ − 𝑦′0 𝛿 𝑧′ − 𝑧′0


Example 2: Lorenz Gauge

0 = 𝛻.𝑨 + 𝜇𝑜 𝜖0𝜕

𝜕𝑡Φ = 𝛁. 𝑨 +

𝜕 Φ/𝑐

𝜕 𝑐 𝑡

=𝜕 Φ/𝑐

𝜕 𝑐 𝑡+𝜕𝐴𝑥𝜕𝑥

+𝜕𝐴𝑦𝜕𝑦

+𝜕𝐴

𝜕𝑧

=𝜕 Φ/𝑐

𝜕𝑥0+𝜕𝐴𝑥𝜕𝑥1

+𝜕𝐴𝑦𝜕𝑥2

+𝜕𝐴𝑧𝜕𝑥3

=

𝜇=0

3𝜕

𝜕𝑥𝜇𝐴𝜇

covariant

contravariant𝐴𝜇 =



Four-Vector Potential “A”

𝐴𝜇 =


𝐴′𝜇 =

𝛼=0

3

𝐿𝛼𝜇𝐴𝛼



𝐴𝜇 =


𝐴𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝐴𝜈 =

Φ/𝑐−𝐴𝑥−𝐴𝑦−𝐴𝑧

𝜇=0

3

𝐴𝜇 𝐴𝜇 = ൗΦ 𝑐2− 𝐴𝑥

2 − 𝐴𝑦2− 𝐴𝑧



Consider a charge 𝑞 moving at a constant speed 𝑣0 = 𝛽 𝑐 along a straight

line parallel to the x-axis.

1. What are the scalar potential and vector potential created by this moving charge ?

2. What are the electric field and magnetic field created by this moving charge?


−𝛻𝟐𝑨 𝒓, 𝑡 + 𝜇𝑜 𝜖0𝜕2𝑨 𝒓, 𝑡

𝜕𝑡2= 𝜇𝑜 𝑱 𝒓, 𝑡

−𝛻2Φ 𝒓, 𝑡 + 𝜇𝑜 𝜖0𝜕2Φ 𝒓, 𝑡

𝜕𝑡2=𝜌 𝒓, 𝑡

𝜖0

𝜌 𝒓, 𝑡 = 𝑞 𝛿 𝑥 − 𝑣0 𝑡 𝛿 𝑦 − 𝑦0 𝛿 𝑧 − 𝑧0 = 𝑞 𝛿3 𝒓 − 𝒓𝟎 𝑡

𝑱 𝒓, 𝑡 =𝑣0 𝑞 𝛿 𝑥 − 𝑣0 𝑡 𝛿 𝑦 − 𝑦0 𝛿 𝑧 − 𝑧0

00

=𝑣0 𝑞 𝛿

3 𝒓 − 𝒓𝟎 𝑡

00

Hard Way


Easier Way:

Start with an inertial frame of reference K’ moving the same way as the charge

=1

4 𝜋 𝜖0

𝑞

𝒓′ − 𝒓′0

𝜌′ 𝒓′ = 𝑞 𝛿3 𝒓′ − 𝒓′𝟎

Φ′ 𝒓′ =1

4 𝜋 𝜖0ම

−∞

+∞𝜌′ 𝒓′′

𝒓′ − 𝒓′′𝑑𝑉′′

electrostatic problem


𝑱′ 𝒓′ =000

𝑨′ 𝒓′ =𝜇𝑜4 𝜋

ම

−∞

+∞𝑱′ 𝒓′′

𝒓 − 𝒓′′𝑑𝑉′′

=000

magnetostaticproblem


𝐴𝜇 =

𝛼=0

3

𝐼𝐿𝛼𝜇𝐴′𝛼

𝐴′𝛼 =

Φ′/𝑐

𝐴′𝑥𝐴′𝑦𝐴′𝑧

𝐼𝐿𝛼𝜇=

𝛾

+ 𝛾 𝛽00

+ 𝛾 𝛽𝛾00

0010

0001

=

Φ′/𝑐000

𝐴𝜇 =

𝛾 Φ′/𝑐

𝛾 𝛽 Φ′/𝑐00

𝑨 𝒓, 𝑡 =𝛾 𝛽 Φ′/𝑐

00

Φ 𝒓, 𝑡 = 𝛾 Φ′ =𝛾 𝑞

4 𝜋 𝜖0 𝒓′ − 𝒓′0

=

𝛾 𝑞 𝑣04 𝜋 𝜖0 𝑐

2 𝒓′ − 𝒓′000


𝒓′ − 𝒓′0

𝑐 𝑡′ − 𝑡′0𝑥′ − 𝑥′0𝑦′ − 𝑦′0𝑧′ − 𝑧′0

=

𝛼=0

3

𝐿𝛼𝜇

𝑥𝛼 − 𝑥0𝛼 =

𝛾

− 𝛾 𝛽00

− 𝛾 𝛽𝛾00

0010

0001

𝑐 𝑡 − 𝑡0𝑥 − 𝑥0𝑦 − 𝑦0𝑧 − 𝑧0

=

𝛾 𝑐 𝑡 − 𝑡0 − 𝛾 𝛽 𝑥 − 𝑥0𝛾 𝑥 − 𝑥0 − 𝛾 𝛽 𝑐 𝑡 − 𝑡0

𝑦 − 𝑦0𝑧 − 𝑧0

𝒓′ − 𝒓′0 = 𝑥′ − 𝑥′02 + 𝑦′ − 𝑦′0

2 + 𝑧′ − 𝑧′02

= 𝛾 𝑥 − 𝑥0 − 𝛾 𝛽 𝑐 𝑡 − 𝑡02 + 𝑦 − 𝑦0

2 + 𝑧 − 𝑧02


𝑡0 = 0 𝑥0 = 0

Φ 𝒓, 𝑡 =𝛾 𝑞

4 𝜋 𝜖0 𝒓′ − 𝒓′0

=𝛾 𝑞

4 𝜋 𝜖0 𝛾2 𝑥 − 𝑣0 𝑡2 + 𝑦 − 𝑦0

2 + 𝑧 − 𝑧02

𝑨 𝒓, 𝑡 =

𝛾 𝑞 𝑣04 𝜋 𝜖0 𝑐

2 𝒓′ − 𝒓′000

=

𝛾 𝑞 𝑣0

4 𝜋 𝜖0 𝑐2 𝛾2 𝑥 − 𝑣0 𝑡

2 + 𝑦 − 𝑦02 + 𝑧 − 𝑧0

2

00


𝑘𝜇 =

Τ𝜔 𝑐𝑘𝑥𝑘𝑦𝑘𝑧

𝑘𝜇 =

𝜈=0

3

𝑔𝜇𝜈 𝑘𝜈 =

Τ𝜔 𝑐−𝑘𝑥−𝑘𝑦−𝑘𝑧

𝜇=0

3

𝑘𝜇 𝑘𝜇 = Τ𝜔 𝑐

2 − 𝑘𝑥2 − 𝑘𝑦

2− 𝑘𝑧



Tensors of Rank 1

𝑑𝑥𝜈𝑑𝑥𝜇

𝜕

𝜕𝑥𝜈≡ 𝜕𝜈

𝜕

𝜕𝑥𝜇≡ 𝜕𝜇

covariantcontravariant

𝜕𝜇 = 𝑔𝜇𝜈 𝜕𝜈

𝜕𝜈 = 𝑔𝜈𝜇 𝜕𝜇

𝑑𝑥𝜇 = 𝑔𝜇𝜈 𝑑𝑥𝜈

𝑑𝑥𝜈 = 𝑔𝜈𝜇 𝑑𝑥𝜇


Contravariant Tensor of Rank 2

𝑉′𝛼 =

𝜇=0

3𝜕𝑥′𝛼

𝜕𝑥𝜇𝑉𝜇 =

𝜇=0

3

𝐿𝜇𝛼 𝑉𝜇 𝑊′𝛽 =

𝜈=0

3𝜕𝑥′𝛽

𝜕𝑥𝜈𝑊𝜈 =

𝜈=0

3

𝐿𝜈𝛽𝑊𝜈

𝑇𝛼𝛽 ≡ 𝑉𝛼 𝑊𝛽

𝑇′𝛼𝛽 = 𝑉′𝛼 𝑊′𝛽 =

𝜇=0

3𝜕𝑥′𝛼

𝜕𝑥𝜇𝑉𝜇

𝜈=0

3𝜕𝑥′𝛽

𝜕𝑥𝜈𝑊𝜈 =

𝜇=0

3

𝐿𝜇𝛼 𝑉𝜇

𝜈=0

3

𝐿𝜈𝛽𝑊𝜈

=

𝜇=0

3

𝐿𝜇𝛼

𝜈=0

3

𝐿𝜈𝛽𝑇𝜇𝜈 = 𝐿 𝑇 𝐿𝑡 𝛼𝛽


𝐴𝛽 =


𝜕

𝜕𝑥𝛼= 𝜕𝛼

𝐹𝛼𝛽 ≡ 𝜕𝛼 𝐴𝛽 − 𝜕𝛽 𝐴𝛼

𝐹𝛼𝛽 = −𝐹𝛽𝛼

𝐹𝛼𝛼 = 0

Electromagnetic Field Tensor


Electric Field as a component of the electromagnetic field tensor

𝑬 𝒓, 𝑡 = −𝛻Φ 𝒓, 𝑡 −𝜕𝑨 𝒓, 𝑡

𝜕𝑡𝐴𝜇 =


𝐸𝑖 = −𝜕Φ

𝜕𝑥𝑖−𝜕𝐴𝑖

𝜕𝑡= −

𝜕 𝑐 𝐴0

𝜕𝑥𝑖− 𝑐

𝜕𝐴𝑖

𝜕𝑥0= 𝑐 −

𝜕𝐴0


𝜕𝑥0

𝐸𝑖

𝑐= −

𝜕𝐴0


𝜕𝑥0

𝑖 ≡ 1, 2, 3

= 𝜕𝑖𝐴0 − 𝜕0𝐴𝑖 = 𝐹𝑖0


𝑩 𝒓, 𝑡 = 𝛁 × 𝑨 𝒓, 𝑡𝐴𝜇 =


𝐵𝑖 =

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘𝜕𝐴𝑘

𝜕𝑥𝑗

𝑖, 𝑗, 𝑘 ≡ 1, 2, 3

= −

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘 𝜕𝑗𝐴𝑘

𝐵1 = − 𝜕2𝐴3 − 𝜕3𝐴2 = −𝐹23

𝐵2 = − 𝜕3𝐴1 − 𝜕1𝐴3 = −𝐹31

𝐵3 = − 𝜕1𝐴2 − 𝜕2𝐴1 = −𝐹12

Magnetic Field as a component of the electromagnetic field tensor


𝐹𝛼𝛽 ≡ 𝜕𝛼𝐴𝛽 − 𝜕𝛽𝐴𝛼 =1

𝑐

0 −𝐸1

𝐸1 0−𝐸2 −𝐸3

−𝑐𝐵3 𝑐𝐵2

𝐸2 𝑐𝐵3

𝐸3 −𝑐𝐵20 −𝑐𝐵1

𝑐𝐵1 0

𝐸𝑖

𝑐= 𝐹𝑖0

𝐵1 = − 𝜕2𝐴3 − 𝜕3𝐴2 = −𝐹23

𝐵2 = − 𝜕3𝐴1 − 𝜕1𝐴3 = −𝐹31

𝐵3 = − 𝜕1𝐴2 − 𝜕2𝐴3 = −𝐹12

Components of the Electromagnetic Field Tensor


𝐹′𝛼𝛽 = 𝐿 𝐹 𝐿𝑡 𝛼𝛽

How the Electromagnetic Field Tensor changes under a Lorentz

transformation:

𝐹 =1

𝑐

0 −𝐸1

𝐸1 0−𝐸2 −𝐸3


𝐸2 𝑐𝐵3

𝐸3 −𝑐𝐵20 −𝑐𝐵1

𝑐𝐵1 0

𝐿 =

𝛾

− 𝛾 𝛽00

− 𝛾 𝛽𝛾00

0010

0001

For a frame of reference K’ moving along x-axis at a

constant speed 𝑣0 = 𝛽 𝑐


𝐹′ =1

𝑐

0 −𝐸′1

𝐸′1

0

−𝐸′2

−𝐸′3

−𝑐𝐵′3

𝑐𝐵′2

𝐸′2

𝑐𝐵′3

𝐸′3

−𝑐𝐵′2

0 −𝑐𝐵′1

𝑐𝐵′1

0

= 𝐿 𝐹 𝐿𝑡

𝐸′1 = 𝐸1

𝐸′2 = 𝛾 𝐸2 − 𝛽 𝑐 𝐵3

𝐸′3 = 𝛾 𝐸3 + 𝛽 𝑐 𝐵2

𝐵′1 = 𝐵1

𝐵′2 = 𝛾 𝐵2 +𝛽

𝑐𝐸3

𝐵′3 = 𝛾 𝐵3 −𝛽

𝑐𝐸2


Note: Galilean Transformations lead to different (and incorrect) relations:

𝒗′: particle velocity with respect to K’

𝑭′ = 𝑞′ 𝑬′ + 𝑞′ 𝒗′ × 𝑩′

𝑩 = 𝑩′𝑬 = 𝑬′ − 𝒗𝑜 × 𝑩′

= 𝑞 𝑬′ + 𝑞 𝒗 − 𝒗𝑜 × 𝑩′

= 𝑞 𝑬′ − 𝒗𝑜 × 𝑩′ + 𝑞 𝒗 × 𝑩′

= 𝑞 𝑬 + 𝑞 𝒗 × 𝑩 = 𝑭

𝒗 = 𝒗′ + 𝒗0𝑣: particle velocity with respect to K

according to (incorrect) Galilean Transformations


Electric and Magnetic Fields of a Moving Charge


Consider a charge 𝑞 moving at a constant speed 𝑣0 = 𝛽 𝑐 along a straight

line parallel to the x-axis.

What are the electric and magnetic fields created by this moving charge?


𝐸′1 =𝑞

4 𝜋 𝜖0

𝑥′ − 𝑥′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐸′2 =𝑞

4 𝜋 𝜖0

𝑦′ − 𝑦′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐸′3 =𝑞

4 𝜋 𝜖0

𝑧′ − 𝑧′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐵′1 = 0

𝐵′2 = 0

𝐵′3 = 0

Consider that the charge is at rest in the reference frame K’


𝐸′1 = 𝐸1

𝐸′2 = 𝛾 𝐸2 − 𝛽 𝑐 𝐵3

𝐸′3 = 𝛾 𝐸3 + 𝛽 𝑐 𝐵2

𝐵′1 = 𝐵1

𝐵′2 = 𝛾 𝐵2 +𝛽

𝑐𝐸3

𝐵′3 = 𝛾 𝐵3 −𝛽

𝑐𝐸2

𝐸1 = 𝐸′1

𝐸2 = 𝛾 𝐸′2 + 𝛽 𝑐 𝐵′3

𝐸3 = 𝛾 𝐸′3 − 𝛽 𝑐 𝐵′2

𝐵1 = 𝐵′1

𝐵2 = 𝛾 𝐵′2 −𝛽

𝑐𝐸′3

𝐵3 = 𝛾 𝐵′3 +𝛽

𝑐𝐸′2


𝐸1 = 𝐸′1

𝐸2 = 𝛾 𝐸′2

𝐸3 = 𝛾 𝐸′3

𝐵1 = 0

𝐵2 = −𝛾 𝛽

𝑐𝐸′3

𝐵3 =𝛾 𝛽

𝑐𝐸′2

= −𝛽

𝑐𝐸3

=𝛽

𝑐𝐸2

𝐸1 = 𝐸′1

𝐸2 = 𝛾 𝐸′2 + 𝛽 𝑐 𝐵′3

𝐸3 = 𝛾 𝐸′3 − 𝛽 𝑐 𝐵′2

𝐵1 = 𝐵′1

𝐵2 = 𝛾 𝐵′2 −𝛽

𝑐𝐸′3

𝐵3 = 𝛾 𝐵′3 +𝛽

𝑐𝐸′2


𝐸1 = 𝐸′1 =𝑞

4 𝜋 𝜖0

𝑥′ − 𝑥′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐸2 = 𝛾 𝐸′2 =𝑞

4 𝜋 𝜖0

𝛾 𝑦′ − 𝑦′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐸3 = 𝛾 𝐸′3 =𝑞

4 𝜋 𝜖0

𝛾 𝑧′ − 𝑧′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2


𝐵1 = 0

𝐵2 = −𝛾 𝛽

𝑐𝐸′

3= −

𝛽

𝑐

𝑞

4 𝜋 𝜖0

𝛾 𝑧′ − 𝑧′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐵3 =𝛾 𝛽

𝑐𝐸′2 =

𝛽

𝑐

𝑞

4 𝜋 𝜖0

𝛾 𝑦′ − 𝑦′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2


𝑐 𝑡′ − 𝑡′0𝑥′ − 𝑥′0𝑦′ − 𝑦′0𝑧′ − 𝑧′0

=

𝛾

− 𝛾 𝛽00

− 𝛾 𝛽𝛾00

0010

0001

𝑐 𝑡 − 𝑡0𝑥 − 𝑥0𝑦 − 𝑦0𝑧 − 𝑧0

𝑡0 = 𝑡′0 = 0 𝑥0 = 𝑥′0 = 0

𝑐 𝑡′ − 𝑡′0𝑥′ − 𝑥′0𝑦′ − 𝑦′0𝑧′ − 𝑧′0

=

𝛾 𝑐 𝑡 − 𝛽 𝑥

𝛾 𝑥 − 𝑣0 𝑡𝑦 − 𝑦0𝑧 − 𝑧0


=𝑞

4 𝜋 𝜖0

𝛾 𝑥 − 𝑣0 𝑡

𝛾2 𝑥 − 𝑣0 𝑡2 + 𝑦 − 𝑦0

2 + 𝑧 − 𝑧02 3/2

=𝑞

4 𝜋 𝜖0

𝛾 𝑦 − 𝑦0𝛾2 𝑥 − 𝑣0 𝑡

2 + 𝑦 − 𝑦02 + 𝑧 − 𝑧0

2 3/2

=𝑞

4 𝜋 𝜖0

𝛾 𝑧 − 𝑧0𝛾2 𝑥 − 𝑣0 𝑡

2 + 𝑦 − 𝑦02 + 𝑧 − 𝑧0

2 3/2

𝐸1 = 𝐸′1 =𝑞

4 𝜋 𝜖0

𝑥′ − 𝑥′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐸2 = 𝛾 𝐸′2 =𝑞

4 𝜋 𝜖0

𝛾 𝑦′ − 𝑦′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2

𝐸3 = 𝛾 𝐸′3 =𝑞

4 𝜋 𝜖0

𝛾 𝑧′ − 𝑧′0𝑥′ − 𝑥′0

2 + 𝑦′ − 𝑦′02 + 𝑧′ − 𝑧′0

2 3/2


𝛾2 𝑥 − 𝑣0 𝑡2 + 𝑦 − 𝑦0

2 + 𝑧 − 𝑧02

𝑥, 𝑦, 𝑧

𝑣0 𝑡, 𝑦0, 𝑧0

= 𝛾2 𝑑2 1 − 𝛽2 𝑠𝑖𝑛2 𝜃

𝜃 𝑥

𝑠𝑖𝑛2 𝜃 ≡𝑦 − 𝑦0

2 + 𝑧 − 𝑧02

𝑑2

𝒗𝟎

𝒅

𝒅 ≡ 𝑥 − 𝑣0 𝑡 , 𝑦 − 𝑦0 , 𝑧 − 𝑧0

particle position at

time 𝑡

point to determine the field at time 𝑡

=𝑦 − 𝑦0

2 + 𝑧 − 𝑧02

𝑥 − 𝑣0 𝑡2 + 𝑦 − 𝑦0

2 + 𝑧 − 𝑧02


𝑬 𝒓, 𝑡 =𝑞

4 𝜋 𝜖0

𝒅

𝑑31 − 𝛽2

1 − 𝛽2 𝑠𝑖𝑛2 𝜃 3/2

classical Coulomb term

relativistic correction

𝛽 = 0.00 & 0.20

𝛽 = 0.70

𝛽 = 0.90

𝛽 = 0.99

𝒗𝟎

• The electric field 𝑬 points away from the position of the charge at the time of observation (t).

• The amplitude shows higher strength for directions perpendicular to the direction of propagation.

• However, 𝑬 is not isotropic.

• The electric field amplitude 𝑬depends on the direction away from the charge. 1 − 𝛽2

𝜃 ≅ 0

𝜃 ≅ 90°1

1 − 𝛽2


𝐵1 = 0

𝐵2 = −𝛽

𝑐𝐸3

𝐵3 =𝛽

𝑐𝐸2

𝑩 = 𝜷 ×𝑬

𝑐

𝑩 𝒓, 𝑡 =𝜇04 𝜋

𝑞 𝒗𝟎 ×𝒅

𝑑31 − 𝛽2

1 − 𝛽2 𝑠𝑖𝑛2 𝜃 3/2

classical Biot-Savart

term

relativistic correction


Maxwell’s Equations in terms of the Electromagnetic Field Tensor


𝑐

0 −𝐸1

𝐸1 0−𝐸2 −𝐸3


𝐸2 𝑐𝐵3

𝐸3 −𝑐𝐵20 −𝑐𝐵1

𝑐𝐵1 0

𝐽𝜇 =


sources

fields


The Four Inhomogeneous Maxwell’s Equations

Gauss’s Law

Ampere’s Law

𝛁. 𝑬 𝒓, 𝑡 =𝜌 𝒓, 𝑡

𝜖0

𝛁 × 𝑩 𝒓, 𝑡 − 𝜇𝑜 𝜖0𝜕𝑬 𝒓, 𝑡

𝜕𝑡= 𝜇𝑜 𝑱 𝒓, 𝑡


𝛁. 𝑬 𝒓, 𝑡 =𝜌 𝒓, 𝑡

𝜖0𝐸𝑖 = 𝑐 𝐹𝑖0

𝜌 𝒓, 𝑡 =𝐽0

𝑐

𝑖=1

3𝜕

𝜕𝑥𝑖𝐸𝑖 =

𝑖=1

3𝜕

𝜕𝑥𝑖𝑐 𝐹𝑖0 = 𝑐

𝑖=1

3

𝜕𝑖 𝐹𝑖0

= 𝑐

𝛼=0

3

𝜕𝛼 𝐹𝛼0

=𝜌 𝒓, 𝑡

𝜖0=

𝐽0

𝜖0 𝑐

𝛼=0

3

𝜕𝛼 𝐹𝛼0 = 𝜇0 𝐽

0

+ 𝑐 𝜕0𝐹00


𝛁 × 𝑩 𝒓, 𝑡 − 𝜇𝑜 𝜖0𝜕𝑬 𝒓, 𝑡

𝜕𝑡= 𝜇𝑜 𝑱 𝒓, 𝑡

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘𝜕𝐵𝑘

𝜕𝑥𝑗− 𝜇𝑜 𝜖0

𝜕𝐸𝑖

𝜕𝑡

=

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘 𝜕𝑗𝐵𝑘 −

𝜕 ൗ𝐸𝑖𝑐

𝜕 𝑐 𝑡

=

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘 𝜕𝑗𝐵𝑘 + 𝜕0 ൗ−𝐸𝑖

𝑐

= 𝜇𝑜 𝐽𝑖


𝐵1 = −𝐹23 𝐵2 = −𝐹31 𝐵3 = −𝐹12

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘 𝜕𝑗𝐵𝑘 + 𝜕0

−𝐸𝑖

𝑐= 𝜇𝑜 𝐽

𝑖

−𝐸𝑖

𝑐= 𝐹0𝑖

𝑖 = 1 𝜕2𝐵3 − 𝜕3𝐵

2 + 𝜕0𝐹01

= 𝜕2𝐹21 + 𝜕3𝐹

31 + 𝜕0𝐹01

=

𝛼=0

3

𝜕𝛼𝐹𝛼1 = 𝜇𝑜 𝐽

1

𝛼=0

3

𝜕𝛼 𝐹𝛼1 = 𝜇0 𝐽

1

+ 𝜕1𝐹11


𝑗=1

3

𝑘=1

3


−𝐸𝑖

𝑐= 𝜇𝑜 𝐽

𝑖

𝑖 = 2 𝜕3𝐵1 − 𝜕1𝐵

3 + 𝜕0𝐹02

= 𝜕3𝐹32 + 𝜕1𝐹

12 + 𝜕0𝐹02 + 𝜕2𝐹

22

=

𝛼=0

3


2

𝛼=0

3

𝜕𝛼 𝐹𝛼2 = 𝜇0 𝐽

2

𝐵1 = −𝐹23 𝐵2 = −𝐹31 𝐵3 = −𝐹12−𝐸𝑖

𝑐= 𝐹0𝑖


𝑗=1

3

𝑘=1

3


−𝐸𝑖

𝑐= 𝜇𝑜 𝐽

𝑖

𝑖 = 3 𝜕1𝐵2 − 𝜕2𝐵

1 + 𝜕0𝐹03

= 𝜕1𝐹13 + 𝜕2𝐹

23 + 𝜕0𝐹03 + 𝜕3𝐹

33

=

𝛼=0

3


3

𝛼=0

3

𝜕𝛼 𝐹𝛼3 = 𝜇0 𝐽

3

𝐵1 = −𝐹23 𝐵2 = −𝐹31 𝐵3 = −𝐹12−𝐸𝑖

𝑐= 𝐹0𝑖


The Four Inhomogeneous Maxwell’s Equations

(Gauss’s and Ampere’s Laws) can be written as:

𝛼=0

3

𝜕𝛼𝐹𝛼𝛽 = 𝜇0 𝐽

𝛽

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𝛽 𝐹𝛼𝛽 ≡ 𝜕𝛼𝐴𝛽 − 𝜕𝛽𝐴𝛼

= 𝜕𝛼 𝜕𝛼𝐴𝛽 − 𝜕𝛽𝐴𝛼

= 𝜇0 𝐽𝛽

= 𝜕𝛼𝜕𝛼𝐴𝛽 − 𝜕𝛽 𝜕𝛼𝐴

𝛼𝜕𝛼 𝐹𝛼𝛽 = 𝜕𝛼𝜕𝛼𝐴𝛽

0

𝜕𝛼𝜕𝛼𝐴𝛽 = 𝜇0 𝐽

𝛽

Lorenz gauge

𝜕𝛼𝜕𝛼 ≡ □2

d’Alembertian, which is an invariant


The Four Homogeneous Maxwell’s Equations

Gauss’s Law of Magnetism

Faraday’s Law

𝛁.𝑩 𝒓, 𝑡 = 0

𝛁 × 𝑬 𝒓, 𝑡 +𝜕𝑩 𝒓, 𝑡

𝜕𝑡= 0


𝛁.𝑩 𝒓, 𝑡 = 0

𝐵1 = −𝐹23 𝐵2 = −𝐹31 𝐵3 = −𝐹12

𝑖=1

3𝜕

𝜕𝑥𝑖𝐵𝑖 =

𝜕𝐵1

𝜕𝑥1+𝜕𝐵2

𝜕𝑥2+𝜕𝐵3

𝜕𝑥3

= −𝜕𝐹23

𝜕𝑥1−𝜕𝐹31

𝜕𝑥2−𝜕𝐹12

𝜕𝑥3

= 𝜕1𝐹23 + 𝜕2𝐹31 + 𝜕3𝐹12 = 0

𝜕1𝐹23 + 𝜕2𝐹31 + 𝜕3𝐹12 = 0


𝛁 × 𝑬 𝒓, 𝑡 +𝜕𝑩 𝒓, 𝑡

𝜕𝑡= 0

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘𝜕 𝐸𝑘

𝑐 𝜕𝑥𝑗+𝜕𝐵𝑖

𝑐 𝜕𝑡= 0

−

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘 𝜕𝑗𝐸𝑘

𝑐+ 𝜕0𝐵𝑖 = 0

𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘 𝜕𝑗 𝐸𝑘/𝑐 + 𝜕0𝐵𝑖 = 0


𝑗=1

3

𝑘=1

3

𝜖𝑖𝑗𝑘 𝜕𝑗 𝐸𝑘/𝑐 − 𝜕0𝐵𝑖 = 0

𝐵1 = −𝐹23 𝐵2 = −𝐹31 𝐵3 = −𝐹12𝐸𝑖

𝑐= 𝐹𝑖0

𝑖 = 1

𝜕2 𝐸3/𝑐 − 𝜕3 𝐸2/𝑐 − 𝜕0𝐵1 = 0

𝜕2𝐹30 + 𝜕3𝐹02 + 𝜕0𝐹23 = 0


𝐵1 = −𝐹23 𝐵2 = −𝐹31 𝐵3 = −𝐹12𝐸𝑖

𝑐= 𝐹𝑖0

𝑖 = 2

𝜕3 𝐸1/𝑐 − 𝜕1 𝐸3/𝑐 − 𝜕0𝐵2 = 0

𝜕3𝐹10 + 𝜕1𝐹03 + 𝜕0𝐹31 = 0

𝑗=1

3

𝑘=1

3



𝑖 = 3

𝜕1 𝐸2/𝑐 − 𝜕2 𝐸1/𝑐 − 𝜕0𝐵3 = 0

𝜕1𝐹20 + 𝜕2𝐹01 + 𝜕0𝐹12 = 0

𝑗=1

3

𝑘=1

3


𝐵1 = −𝐹23 𝐵2 = −𝐹31 𝐵3 = −𝐹12𝐸𝑖

𝑐= 𝐹𝑖0


𝜕1𝐹23 + 𝜕2𝐹31 + 𝜕3𝐹12 = 0

𝜕3𝐹01 + 𝜕0𝐹13 + 𝜕1𝐹30 = 0

𝜕0𝐹12 + 𝜕1𝐹20 + 𝜕2𝐹01 = 0

𝑖 = 1

𝑖 = 2

𝑖 = 3

𝑖 = 00

23

0 1

23

0 1

23

1

𝜕2𝐹30 + 𝜕3𝐹02 + 𝜕0𝐹23 = 0

0 1

23


The Four Homogeneous Maxwell’s Equations

(Gauss’s Law of Magnetism and Faraday’s Law)

can be written as:

0 1

23

휀𝛿𝛼𝛽𝛾 𝜕𝛼𝐹𝛽𝛾 = 0 휀𝛿𝛼𝛽𝛾



𝛽


𝑐

0 −𝐸1

𝐸1 0−𝐸2 −𝐸3


𝐸2 𝑐𝐵3

𝐸3 −𝑐𝐵20 −𝑐𝐵1

𝑐𝐵1 0

𝐽𝜇 =


𝐴𝜇 =


Electromagnetic Theory

휀𝛿𝛼𝛽𝛾 𝜕𝛼𝐹𝛽𝛾 = 0

special relativity and electrodynamics 611 spring 20/classnotes... · special relativity and...

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