chapter 1 of “modern problems in classical electrodynamics ... 611 spring... · chapter 1 of...

1Prof. Sergio B. MendesSpring 2018

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

Relativistic Kinematics

2

• Covariant formalism of electrodynamics

• Lorentz transformations

Prof. Sergio B. MendesSpring 2018

Relativistic Formalism of Electrodynamics

• Electromagnetic field tensor

• Electromagnetic field of a charge moving at constant speed

• Special relativity


Experimental Inconsistencies


Around mid-to-late 1800' all waves were assumed to require a material medium to propagate (e.g. sound waves, ocean waves, vibration on a solid material, acoustic guitar, etc)

As a consequence, the wave speed depends on the properties of the material medium where it propagates (T, P, Y, 𝝆𝝆, etc)

Relative motion between the observer and the material medium carrying the wave affect the measured speed of a particular wave

Following on this tradition, a material medium named ether was assumed as the medium where electromagnetic radiation propagates

Background


Michelson-Morley Experiment

An attempt to detect the existence of the ether (luminiferousmedium, the light medium)

Their goal was to show that different types of motion with respect to the ether give different speeds of light propagation

Applied an interferometric technique (known today as Michelson interferometer) to detect small changes in the transit time along different paths

Used Earth orbital speed around the sun (30 Km/s) as a lower limit of the motion of the Earth through the absolute ether

6

Michelson Interferometer

𝐿𝐿1𝐿𝐿2


B

A C


𝑡𝑡𝑡𝑡 = 𝑡𝑡𝐴𝐴−𝐵𝐵 + 𝑡𝑡𝐵𝐵−𝐶𝐶 =𝐿𝐿

𝑐𝑐2 − 𝑣𝑣2+

𝐿𝐿𝑐𝑐2 − 𝑣𝑣2

=2 𝐿𝐿𝑐𝑐2 − 𝑣𝑣2

𝑡𝑡𝑙𝑙 = 𝑡𝑡𝐴𝐴−𝐶𝐶 + 𝑡𝑡𝐶𝐶−𝐴𝐴 =𝐿𝐿

𝑐𝑐 + 𝑣𝑣+

𝐿𝐿𝑐𝑐 − 𝑣𝑣

=2 𝐿𝐿 𝑐𝑐𝑐𝑐2 − 𝑣𝑣2

∆𝑡𝑡 = 𝑡𝑡𝑙𝑙 − 𝑡𝑡𝑡𝑡 =2 𝐿𝐿 𝑐𝑐𝑐𝑐2 − 𝑣𝑣2

−2 𝐿𝐿𝑐𝑐2 − 𝑣𝑣2

=2 𝐿𝐿𝑐𝑐

1 −𝑣𝑣𝑐𝑐

2 −1

− 1 −𝑣𝑣𝑐𝑐

2 −1/2


𝑣𝑣𝑐𝑐≅

30 × 103 𝑚𝑚/𝑠𝑠3 × 108 𝑚𝑚/𝑠𝑠

≅ 10−4

∆𝑡𝑡 =2 𝐿𝐿𝑐𝑐 1 −

𝑣𝑣𝑐𝑐

2 −1

− 1 −𝑣𝑣𝑐𝑐

2 −1/2

≅2 𝐿𝐿𝑐𝑐

1 +𝑣𝑣𝑐𝑐

2− 1 +

12𝑣𝑣𝑐𝑐

2

≅𝐿𝐿 𝑣𝑣2

𝑐𝑐3

∆𝜙𝜙 = 𝜔𝜔 ∆𝑡𝑡

≅ 2 𝜋𝜋𝑐𝑐𝜆𝜆𝐿𝐿 𝑣𝑣2

𝑐𝑐3 = 2 𝜋𝜋𝐿𝐿𝜆𝜆𝑣𝑣2

𝑐𝑐2 𝐿𝐿 ≅ 11 m

𝜆𝜆 ≅ 500 × 10−9 m≅ 2 𝜋𝜋𝐿𝐿𝜆𝜆 10−8


“The Experiments on the relative motion of the earth and ether

have been completed and the result decidedly negative. The

expected deviation of the interference fringes from the zero

should have been 0.40 of a fringe – the maximum displacement

was 0.02 and the average much less than 0.01 – and then not in

the right place. As displacement is proportional to squares of the

relative velocities it follows that if the ether does slip past the

relative velocity is less than one sixth of the earth’s velocity.”

Albert Abraham Michelson, 1887


Theoretical Inconsistencies


System of reference to describe an event:• position in space (coordinate system)• time (clocks)

An inertial system of reference:In this system, a particle with no force acting on it will remain at rest or move at constant speed.

Systems of reference:If two systems of reference move at constant speed with respect to each other, and one of them is an inertial system of reference, then the other one is also an inertial frame of reference.

Preliminary Concepts


Galilean Transformation for two inertial systems of reference

𝒓𝒓′ = 𝒓𝒓 − 𝒗𝒗𝑜𝑜 𝑡𝑡

𝑡𝑡′ = 𝑡𝑡

𝑦𝑦

𝑧𝑧𝑥𝑥𝑧𝑧𝑧

𝑦𝑦𝑧

𝑥𝑥𝑧𝒗𝒗𝑜𝑜

𝓚𝓚

𝓞𝓞𝓞𝓞𝑧

𝓚𝓚’


𝑑𝑑𝑥𝑥𝑧𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑦𝑦𝑧𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑧𝑧𝑧𝑑𝑑𝑡𝑡𝑧

=𝑣𝑣𝑧𝑥𝑥𝑣𝑣𝑧𝑦𝑦𝑣𝑣𝑧𝑧𝑧

=𝑣𝑣𝑥𝑥 − 𝑣𝑣𝑜𝑜𝑥𝑥𝑣𝑣𝑦𝑦 − 𝑣𝑣𝑜𝑜𝑦𝑦𝑣𝑣𝑧𝑧 − 𝑣𝑣𝑜𝑜𝑧𝑧

𝑑𝑑𝑣𝑣𝑧𝑥𝑥𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑣𝑣𝑧𝑦𝑦𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑣𝑣𝑧𝑧𝑧𝑑𝑑𝑡𝑡𝑧

=𝑎𝑎𝑧𝑥𝑥𝑎𝑎𝑧𝑦𝑦𝑎𝑎𝑧𝑧𝑧

=𝑎𝑎𝑥𝑥𝑎𝑎𝑦𝑦𝑎𝑎𝑧𝑧

Velocity and Acceleration

velocity

acceleration

Invariance of Newton’s Law


Galilean Transformation as a four-vector linear transformation

𝑡𝑡𝑧𝑥𝑥𝑧𝑦𝑦𝑧𝑧𝑧𝑧

=

𝑡𝑡𝑥𝑥 − 𝑣𝑣𝑜𝑜𝑥𝑥 𝑡𝑡𝑦𝑦 − 𝑣𝑣𝑜𝑜𝑦𝑦 𝑡𝑡𝑧𝑧 − 𝑣𝑣𝑜𝑜𝑧𝑧 𝑡𝑡

𝒓𝒓′ = 𝒓𝒓 − 𝒗𝒗𝑜𝑜 𝑡𝑡

𝑡𝑡′ = 𝑡𝑡

𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑥𝑥𝑧𝑑𝑑𝑦𝑦𝑧𝑑𝑑𝑧𝑧𝑧

=

𝑑𝑑𝑡𝑡𝑑𝑑𝑥𝑥 − 𝑣𝑣𝑜𝑜𝑥𝑥 𝑑𝑑𝑡𝑡𝑑𝑑𝑦𝑦 − 𝑣𝑣𝑜𝑜𝑦𝑦 𝑑𝑑𝑡𝑡𝑑𝑑𝑧𝑧 − 𝑣𝑣𝑜𝑜𝑧𝑧 𝑑𝑑𝑡𝑡

𝑦𝑦


𝑦𝑦𝑧

𝑥𝑥𝑧𝒗𝒗𝑜𝑜

=1

−𝑣𝑣𝑜𝑜𝑥𝑥−𝑣𝑣𝑜𝑜𝑦𝑦−𝑣𝑣𝑜𝑜𝑧𝑧

0100

0010

0001

𝑡𝑡𝑥𝑥𝑦𝑦𝑧𝑧

=1


0100

0010

0001

𝑑𝑑𝑡𝑡𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑧𝑧

𝓚𝓚


𝓚𝓚’

four-components vector =

four-vector



=1


0100

0010

0001


𝑑𝑑𝑥𝑥𝑧𝜇𝜇 = �𝜈𝜈=0

3𝜕𝜕𝑥𝑥𝑧𝜇𝜇

𝜕𝜕𝑥𝑥𝜈𝜈𝑑𝑑𝑥𝑥𝜈𝜈

𝐺𝐺𝜈𝜈𝜇𝜇 ≡

𝜕𝜕𝑥𝑥𝑧𝜇𝜇

𝜕𝜕𝑥𝑥𝜈𝜈

𝑑𝑑𝑥𝑥𝑧0𝑑𝑑𝑥𝑥𝑧1𝑑𝑑𝑥𝑥𝑧2𝑑𝑑𝑥𝑥𝑧3

= =1


0100

0010

0001

𝑑𝑑𝑥𝑥0𝑑𝑑𝑥𝑥1𝑑𝑑𝑥𝑥2𝑑𝑑𝑥𝑥3

= �𝜈𝜈=0

3

𝐺𝐺𝜈𝜈𝜇𝜇 𝑑𝑑𝑥𝑥𝜈𝜈

Four-Vector Position

𝜈𝜈: collumn

𝜇𝜇: row


𝑡𝑡𝑥𝑥𝑦𝑦𝑧𝑧

=

𝑡𝑡𝑧𝑥𝑥′ + 𝑣𝑣𝑜𝑜𝑥𝑥 𝑡𝑡𝑧𝑦𝑦′ + 𝑣𝑣𝑜𝑜𝑦𝑦 𝑡𝑡𝑧𝑧𝑧′ + 𝑣𝑣𝑜𝑜𝑧𝑧 𝑡𝑡𝑧

𝒓𝒓 = 𝒓𝒓′ + 𝒗𝒗𝑜𝑜𝑡𝑡𝑧

𝑡𝑡 = 𝑡𝑡𝑧


=

𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑥𝑥𝑧 + 𝑣𝑣𝑜𝑜𝑥𝑥 𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑦𝑦𝑧 + 𝑣𝑣𝑜𝑜𝑦𝑦 𝑑𝑑𝑡𝑡𝑧𝑑𝑑𝑧𝑧𝑧 + 𝑣𝑣𝑜𝑜𝑧𝑧 𝑑𝑑𝑡𝑡𝑧

Inverse Galilean Transformation𝑦𝑦


𝑦𝑦𝑧

𝑥𝑥𝑧−𝒗𝒗𝑜𝑜

𝓚𝓚


𝓚𝓚’

=1𝑣𝑣𝑜𝑜𝑥𝑥𝑣𝑣𝑜𝑜𝑦𝑦𝑣𝑣𝑜𝑜𝑧𝑧

0100

0010

0001

𝑡𝑡𝑧𝑥𝑥𝑧𝑦𝑦𝑧𝑧𝑧𝑧


0100

0010

0001





0100

0010

0001


𝑑𝑑𝑥𝑥𝛼𝛼 = �𝜇𝜇=0

3𝜕𝜕𝑥𝑥𝛼𝛼

𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝑑𝑑𝑥𝑥𝑧𝜇𝜇

= �𝜈𝜈=0

3

𝛿𝛿𝛼𝛼𝜈𝜈 𝑑𝑑𝑥𝑥𝜈𝜈 = 𝑑𝑑𝑥𝑥𝛼𝛼 𝐻𝐻𝜇𝜇𝛼𝛼 ≡𝜕𝜕𝑥𝑥𝛼𝛼


= �𝜇𝜇=0

3

𝐻𝐻𝜇𝜇𝛼𝛼 𝑑𝑑𝑥𝑥𝑧𝜇𝜇


= =1𝑣𝑣𝑜𝑜𝑥𝑥𝑣𝑣𝑜𝑜𝑦𝑦𝑣𝑣𝑜𝑜𝑧𝑧

0100

0010

0001


= �𝜇𝜇=0

3

𝐻𝐻𝜇𝜇𝛼𝛼�𝜈𝜈=0

3

𝐺𝐺𝜈𝜈𝜇𝜇𝑑𝑑𝑥𝑥𝜈𝜈 = �

𝜈𝜈=0

3

�𝜇𝜇=0


𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝜕𝜕𝑥𝑥𝑧𝜇𝜇


𝜇𝜇: collumn

𝛼𝛼: row


�𝜇𝜇=0

3

𝐻𝐻𝜇𝜇𝛼𝛼 𝐺𝐺𝜈𝜈𝜇𝜇 =

1𝑣𝑣𝑜𝑜𝑥𝑥𝑣𝑣𝑜𝑜𝑦𝑦𝑣𝑣𝑜𝑜𝑧𝑧

0100

0010

0001

1−𝑣𝑣𝑜𝑜𝑥𝑥−𝑣𝑣𝑜𝑜𝑦𝑦−𝑣𝑣𝑜𝑜𝑧𝑧

0100

0010

0001

=1000

0100

0010

0001

= 𝛿𝛿𝛼𝛼𝜈𝜈


𝜕𝜕𝜕𝜕𝑡𝑡𝑧𝜕𝜕𝜕𝜕𝑥𝑥𝑧𝜕𝜕𝜕𝜕𝑦𝑦𝑧𝜕𝜕𝜕𝜕𝑧𝑧𝑧

Four-Vector Gradient Operator

=

𝜕𝜕𝜕𝜕𝑥𝑥𝑧0𝜕𝜕𝜕𝜕𝑥𝑥𝑧1𝜕𝜕𝜕𝜕𝑥𝑥𝑧2𝜕𝜕𝜕𝜕𝑥𝑥𝑧3

=

�𝛼𝛼=0


𝜕𝜕𝑥𝑥𝑧0𝜕𝜕𝜕𝜕𝑥𝑥𝛼𝛼

�𝛼𝛼=0



�𝛼𝛼=0



�𝛼𝛼=0



𝜕𝜕𝜕𝜕𝑥𝑥𝑧𝜇𝜇

= �𝛼𝛼=0


𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝜕𝜕𝜕𝜕𝑥𝑥𝛼𝛼

= �𝛼𝛼=0

3

𝐻𝐻𝜇𝜇𝛼𝛼𝜕𝜕𝜕𝜕𝑥𝑥𝛼𝛼


Four-Vector Gradient Operator for the Galilean Transformation

=

𝜕𝜕𝜕𝜕𝑡𝑡

+ 𝒗𝒗𝑜𝑜 .𝛁𝛁

𝜕𝜕𝜕𝜕𝑥𝑥𝜕𝜕𝜕𝜕𝑦𝑦𝜕𝜕𝜕𝜕𝑧𝑧

𝜕𝜕𝜕𝜕𝑡𝑡𝑧𝜕𝜕𝜕𝜕𝑥𝑥𝑧𝜕𝜕𝜕𝜕𝑦𝑦𝑧𝜕𝜕𝜕𝜕𝑧𝑧𝑧

=

𝜕𝜕𝜕𝜕𝑥𝑥𝑧0𝜕𝜕𝜕𝜕𝑥𝑥𝑧1𝜕𝜕𝜕𝜕𝑥𝑥𝑧2𝜕𝜕𝜕𝜕𝑥𝑥𝑧3

=1000

𝑣𝑣𝑜𝑜𝑥𝑥100

𝑣𝑣𝑜𝑜𝑦𝑦010

𝑣𝑣𝑜𝑜𝑧𝑧001

𝜕𝜕𝜕𝜕𝑥𝑥0𝜕𝜕𝜕𝜕𝑥𝑥1𝜕𝜕𝜕𝜕𝑥𝑥2𝜕𝜕𝜕𝜕𝑥𝑥3


𝜕𝜕2

𝜕𝜕𝑡𝑡𝑧2𝜕𝜕2

𝜕𝜕𝑥𝑥𝑧2𝜕𝜕2

𝜕𝜕𝑦𝑦𝑧2𝜕𝜕2

𝜕𝜕𝑧𝑧𝑧2

=

𝜕𝜕2

𝜕𝜕𝑡𝑡2+ 2𝒗𝒗𝑜𝑜.𝜵𝜵

𝜕𝜕𝜕𝜕𝑡𝑡

+ 𝒗𝒗𝑜𝑜.𝜵𝜵 𝒗𝒗𝑜𝑜.𝜵𝜵

𝜕𝜕2

𝜕𝜕𝑥𝑥2𝜕𝜕2

𝜕𝜕𝑦𝑦2𝜕𝜕2

𝜕𝜕𝑧𝑧2

Second-Order Four-Vector Operator for the Galilean

Transformation


“On the Galilean non-invariance of classical

electromagnetism”Eur. J. Phys. 30 (2009) 381–391

Giovanni Preti, Fernando de Felice, and Luca Masiero


Special Relativity


1. Laws of physics are the same (invariant) in any inertial system of reference, there is no absolute rest.

2. Light propagates in vacuum at a constant speed

The Two Postulates of Special Relativity


• What affects the speed of a wave is the relative motion between the medium carrying the wave and the observer.

• If light propagates in vacuum (therefore, if there is no medium carrying the light wave), then the speed of light (in empty space) is the same for any observer even if they have a relative motion with respective to each other.

• For light propagating in vacuum, the speed of light is the same for any inertial system of reference.

Consequence of the Two Postulates of Special Relativity


𝓞𝓞𝑧

As seen by 𝓚𝓚

𝓞𝓞𝒗𝒗𝑜𝑜

Two inertial systems of reference 𝓚𝓚 and 𝓚𝓚’

1. The axes along (x,y,z) are parallel to the corresponding axes along (x’,y’,z’).

𝓞𝓞𝑧𝓞𝓞−𝒗𝒗𝑜𝑜

As seen by 𝓚𝓚’

2. The relative motion between the two inertial systems of reference is along the x-axis (and x’-axis).

3. Consider that the origins 𝓞𝓞 and 𝓞𝓞′ of the two systems coincide at 𝒕𝒕 = 𝒕𝒕𝒕 = 𝟎𝟎


Two events as described by each inertial system of reference

Event 1: When the origins of the two systems coincide, a pulse of light is emitted.

𝑡𝑡1𝑥𝑥1𝑦𝑦1𝑧𝑧1

=0000

𝑡𝑡𝑧1𝑥𝑥𝑧1𝑦𝑦𝑧1𝑧𝑧𝑧1

=0000

Event 2: The pulse of light reaches a detector at a certain point in space and at a certain time.

𝑡𝑡2𝑥𝑥2𝑦𝑦2𝑧𝑧2

𝑡𝑡𝑧2𝑥𝑥𝑧2𝑦𝑦𝑧2𝑧𝑧𝑧2

𝓚𝓚: 𝓚𝓚’:

𝓚𝓚: 𝓚𝓚’:


𝓚𝓚: 𝑥𝑥2 − 𝑥𝑥1 2 + 𝑦𝑦2 − 𝑦𝑦1 2 + 𝑧𝑧2 − 𝑧𝑧1 2 = 𝑐𝑐2 𝑡𝑡2 − 𝑡𝑡1 2

𝓚𝓚𝑧: 𝑥𝑥𝑧2 − 𝑥𝑥𝑧1 2 + 𝑦𝑦𝑧2 − 𝑦𝑦𝑧1 2 + 𝑧𝑧𝑧2 − 𝑧𝑧𝑧1 2 = 𝑐𝑐2 𝑡𝑡𝑧2 − 𝑡𝑡𝑧1 2

𝑑𝑑𝑥𝑥 2 + 𝑑𝑑𝑦𝑦 2 + 𝑑𝑑𝑧𝑧 2 = 𝑐𝑐2 𝑑𝑑𝑡𝑡 2

𝑑𝑑𝑥𝑥𝑧 2 + 𝑑𝑑𝑦𝑦𝑧 2 + 𝑑𝑑𝑧𝑧𝑧 2 = 𝑐𝑐2 𝑑𝑑𝑡𝑡𝑧 2

𝑑𝑑𝑥𝑥 2 + 𝑑𝑑𝑦𝑦 2 + 𝑑𝑑𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 = 0

𝑑𝑑𝑥𝑥𝑧 2 + 𝑑𝑑𝑦𝑦𝑧 2 + 𝑑𝑑𝑧𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡𝑧 2 = 0


𝑑𝑑𝑥𝑥 2 + 𝑑𝑑𝑦𝑦 2 + 𝑑𝑑𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 =

= 𝑑𝑑𝑥𝑥𝑧 2 + 𝑑𝑑𝑦𝑦𝑧 2 + 𝑑𝑑𝑧𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡𝑧 2


𝑑𝑑𝑦𝑦 = 𝑑𝑑𝑦𝑦𝑧

𝑑𝑑𝑧𝑧 = 𝑑𝑑𝑧𝑧𝑧

as in Galilean transformation

𝑑𝑑𝑥𝑥 2 + 𝑑𝑑𝑦𝑦 2 + 𝑑𝑑𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 =

= 𝑑𝑑𝑥𝑥𝑧 2 + 𝑑𝑑𝑦𝑦𝑧 2 + 𝑑𝑑𝑧𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡𝑧 2

𝑑𝑑𝑥𝑥 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 = 𝑑𝑑𝑥𝑥𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡𝑧 2


𝑑𝑑𝑥𝑥′ = 𝑑𝑑𝑥𝑥 − 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡

𝑑𝑑𝑥𝑥 = 𝑑𝑑𝑥𝑥𝑧 + 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡𝑧

𝛾𝛾

𝛾𝛾due to symmetry

linear modification

= 𝛾𝛾 𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡 + 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡𝑧

𝑑𝑑𝑡𝑡𝑧 = 1𝛾𝛾 𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑𝑡𝑡solve for:


𝑑𝑑𝑥𝑥 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 = 𝑑𝑑𝑥𝑥𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡𝑧 2

𝑑𝑑𝑥𝑥′ = 𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡


1 − 𝛾𝛾2 𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑𝑡𝑡

𝑑𝑑𝑥𝑥 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 = 𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡 2 − 𝑐𝑐21𝛾𝛾 𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑𝑡𝑡2


𝑑𝑑𝑥𝑥 2: 1 = γ2 −𝑐𝑐2

𝛾𝛾2 𝑣𝑣𝑜𝑜2 1 − 𝛾𝛾2 2 1

1 − 𝛾𝛾2 = −𝑐𝑐2

𝛾𝛾2 𝑣𝑣𝑜𝑜2

1𝛾𝛾2

− 1 = −𝑣𝑣𝑜𝑜2

𝑐𝑐2

𝛾𝛾2 =1

1 − 𝛽𝛽2

𝑑𝑑𝑡𝑡 2: −𝑐𝑐2 = 𝛾𝛾2 𝑣𝑣𝑜𝑜2 − 𝑐𝑐2 𝛾𝛾2

1𝛾𝛾2

= 1 −𝑣𝑣𝑜𝑜2

𝑐𝑐2

2 𝑑𝑑𝑥𝑥 𝑑𝑑𝑡𝑡: 0 = −𝛾𝛾2 𝑣𝑣𝑜𝑜 − 𝑐𝑐21𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 𝛾𝛾2 𝑣𝑣𝑜𝑜 = −𝑐𝑐21𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 −𝑣𝑣𝑜𝑜2

𝑐𝑐2 =1𝛾𝛾2

− 1

𝛽𝛽 ≡𝑣𝑣𝑜𝑜𝑐𝑐


𝑑𝑑𝑥𝑥′ = 𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡


1 − 𝛾𝛾2 𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑𝑡𝑡

1𝛾𝛾 𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 =1𝛾𝛾 𝑣𝑣𝑜𝑜

−𝑣𝑣𝑜𝑜2

𝑐𝑐2 𝛾𝛾2 = −

𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2

𝑑𝑑𝑡𝑡′ = −𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2

𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑𝑡𝑡

𝑑𝑑 𝑐𝑐 𝑡𝑡𝑧 = −𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐

𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑 𝑐𝑐 𝑡𝑡

= 𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝛾𝛾 𝛽𝛽 𝑑𝑑 𝑐𝑐 𝑡𝑡

= −𝛾𝛾 𝛽𝛽 𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑 𝑐𝑐 𝑡𝑡


𝑑𝑑 𝑐𝑐 𝑡𝑡𝑧𝑑𝑑𝑥𝑥𝑧𝑑𝑑𝑦𝑦𝑧𝑑𝑑𝑧𝑧𝑧

=

−𝛾𝛾 𝛽𝛽 𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑 𝑐𝑐 𝑡𝑡𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝛾𝛾 𝛽𝛽 𝑑𝑑 𝑐𝑐 𝑡𝑡

𝑑𝑑𝑦𝑦𝑑𝑑𝑧𝑧

=

𝛾𝛾− 𝛾𝛾 𝛽𝛽

00

− 𝛾𝛾 𝛽𝛽𝛾𝛾00

0010

0001

𝑑𝑑 𝑐𝑐 𝑡𝑡𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑧𝑧

=


00


0010

0001



=



=


00


0010

0001


𝑥𝑥0𝑥𝑥1𝑥𝑥2𝑥𝑥3

≡𝑐𝑐 𝑡𝑡𝑥𝑥𝑦𝑦𝑧𝑧

𝑐𝑐 𝑡𝑡𝑧𝑥𝑥𝑧𝑦𝑦𝑧𝑧𝑧𝑧

≡𝑥𝑥𝑧0𝑥𝑥𝑧1𝑥𝑥𝑧2𝑥𝑥𝑧3

Lorentz Transformation

𝓞𝓞𝑧𝓞𝓞𝒗𝒗𝑜𝑜


𝛾𝛾 =1

1 − 𝛽𝛽2


𝑥𝑥

𝑐𝑐𝑡𝑡 𝑐𝑐𝑡𝑡’

𝑥𝑥𝑧

𝜃𝜃

𝜃𝜃

𝑡𝑡𝑎𝑎𝑡𝑡 𝜃𝜃 = 𝛽𝛽

𝛽𝛽 = 1

𝑑𝑑𝑥𝑥𝑧0 = 0 𝛾𝛾𝑑𝑑𝑥𝑥0 − 𝛾𝛾 𝛽𝛽 𝑑𝑑𝑥𝑥1 = 0 𝑑𝑑𝑥𝑥0

𝑑𝑑𝑥𝑥1= 𝛽𝛽

𝑑𝑑𝑥𝑥𝑧1 = 0 − 𝛾𝛾 𝛽𝛽 𝑑𝑑𝑥𝑥0 + 𝛾𝛾 𝑑𝑑𝑥𝑥1 = 0𝑑𝑑𝑥𝑥0

𝑑𝑑𝑥𝑥1= 1

𝛽𝛽



=

𝛾𝛾+ 𝛾𝛾 𝛽𝛽

00

+ 𝛾𝛾 𝛽𝛽𝛾𝛾00

0010

0001


𝑐𝑐 𝑡𝑡𝑥𝑥𝑦𝑦𝑧𝑧

≡𝑥𝑥0𝑥𝑥1𝑥𝑥2𝑥𝑥3

𝑥𝑥𝑧0𝑥𝑥𝑧1𝑥𝑥𝑧2𝑥𝑥𝑧3

≡

𝑐𝑐 𝑡𝑡𝑧𝑥𝑥𝑧𝑦𝑦𝑧𝑧𝑧𝑧

Inverse Lorentz Transformation

𝓞𝓞𝑧𝓞𝓞−𝒗𝒗𝑜𝑜


𝛾𝛾 =1

1 − 𝛽𝛽2


A Few Remarks

• 𝛽𝛽 = 0 then 𝛾𝛾 = 1𝑑𝑑𝑥𝑥𝑧 = 𝑑𝑑𝑥𝑥

𝑑𝑑𝑡𝑡𝑧 = 𝑑𝑑𝑡𝑡

• Although in general 𝑑𝑑𝑥𝑥𝑧 ≠ 𝑑𝑑𝑥𝑥 and 𝑑𝑑𝑡𝑡𝑧 = 𝑑𝑑𝑡𝑡, we always have:

𝑑𝑑𝑥𝑥 2 + 𝑑𝑑𝑦𝑦 2 + 𝑑𝑑𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2

= 𝑑𝑑𝑥𝑥𝑧 2 + 𝑑𝑑𝑦𝑦𝑧 2 + 𝑑𝑑𝑧𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡𝑧 2 = 𝑑𝑑𝑠𝑠 2

𝒅𝒅𝒅𝒅: space-time distance



=


00


0010

0001


Lorentz Transformation of the four-vector coordinates

𝑑𝑑𝑥𝑥𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑑𝑑𝑥𝑥𝛼𝛼 = �

𝛼𝛼=0


𝜕𝜕𝑥𝑥𝛼𝛼𝑑𝑑𝑥𝑥𝛼𝛼

𝓞𝓞𝑧𝓞𝓞𝒗𝒗𝑜𝑜

𝐿𝐿𝛼𝛼𝜇𝜇 =


𝜕𝜕𝑥𝑥𝛼𝛼


𝑑𝑑𝑥𝑥𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑑𝑑𝑥𝑥𝛼𝛼

𝑥𝑥𝑧𝜇𝜇 = 𝑥𝑥𝑧𝜇𝜇 𝑥𝑥0, 𝑥𝑥1, 𝑥𝑥2, 𝑥𝑥3 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑥𝑥𝛼𝛼

Lorentz Transformation for other four-vectors

where

𝑉𝑉𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑉𝑉𝛼𝛼

contravarianttensor of rank 1= four-vector

𝐿𝐿𝛼𝛼𝜇𝜇 =


𝜕𝜕𝑥𝑥𝛼𝛼


The phase of a wave should be invariant 𝜑𝜑′ = 𝜑𝜑 = constant:

𝑘𝑘′𝑥𝑥′ −𝜔𝜔′

𝑐𝑐𝑐𝑐 𝑡𝑡𝑧 = 𝑘𝑘 𝑥𝑥 −

𝜔𝜔𝑐𝑐𝑐𝑐 𝑡𝑡

= 𝑘𝑘 𝛾𝛾 𝑥𝑥𝑧 + 𝛾𝛾 𝛽𝛽 𝑐𝑐 𝑡𝑡𝑧 −𝜔𝜔𝑐𝑐𝛾𝛾 𝑐𝑐 𝑡𝑡𝑧 + 𝛾𝛾 𝛽𝛽 𝑥𝑥𝑧

= 𝛾𝛾 𝑘𝑘 − 𝛾𝛾 𝛽𝛽𝜔𝜔𝑐𝑐

𝑥𝑥′ − 𝛾𝛾𝜔𝜔𝑐𝑐− 𝛾𝛾 𝛽𝛽 𝑘𝑘 𝑐𝑐 𝑡𝑡′

𝜔𝜔′

𝑐𝑐= 𝛾𝛾

𝜔𝜔𝑐𝑐− 𝛾𝛾 𝛽𝛽 𝑘𝑘

𝑘𝑘′ = 𝛾𝛾 𝑘𝑘 − 𝛾𝛾 𝛽𝛽𝜔𝜔𝑐𝑐

�𝜔𝜔𝑧 𝑐𝑐𝑘𝑘𝑧𝑥𝑥𝑘𝑘𝑧𝑦𝑦𝑘𝑘𝑧𝑧𝑧

=


00


0010

0001

⁄𝜔𝜔 𝑐𝑐𝑘𝑘𝑥𝑥𝑘𝑘𝑦𝑦𝑘𝑘𝑧𝑧

Another Contravariant Vector

𝒌𝒌 = (𝑘𝑘, 0,0)


𝑘𝑘𝜇𝜇 =


Four-Vector “K”

𝑘𝑘𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑘𝑘𝛼𝛼

contravariant tensor of rank 1


Proper Time

𝑐𝑐2 𝑑𝑑𝑡𝑡 2 − 𝑑𝑑𝑥𝑥 2 − 𝑑𝑑𝑦𝑦 2 − 𝑑𝑑𝑧𝑧 2 = 𝑑𝑑𝑠𝑠 2 = 𝑐𝑐2 𝑑𝑑𝜏𝜏 2

𝑐𝑐2 − 𝑣𝑣𝑥𝑥 2 − 𝑣𝑣𝑦𝑦2 − 𝑣𝑣𝑧𝑧 2 = 𝑐𝑐2

𝑑𝑑𝜏𝜏𝑑𝑑𝑡𝑡

2

1 − 𝛽𝛽2 =𝑑𝑑𝜏𝜏𝑑𝑑𝑡𝑡

2

𝑑𝑑𝜏𝜏 =𝑑𝑑𝑡𝑡𝛾𝛾


=1𝑑𝑑𝜏𝜏

𝑐𝑐 𝑑𝑑𝑡𝑡𝑑𝑑𝑥𝑥𝑑𝑑𝑦𝑦𝑑𝑑𝑧𝑧

Four-Vector Velocity

𝑣𝑣𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑣𝑣𝛼𝛼


=𝛾𝛾𝑑𝑑𝑡𝑡


=

𝛾𝛾 𝑐𝑐𝛾𝛾 �𝑑𝑑𝑥𝑥

𝑑𝑑𝑡𝑡𝛾𝛾 �𝑑𝑑𝑦𝑦

𝑑𝑑𝑡𝑡𝛾𝛾 �𝑑𝑑𝑧𝑧

𝑑𝑑𝑡𝑡

𝑣𝑣𝜇𝜇 ≡𝑑𝑑𝑥𝑥𝜇𝜇

𝑑𝑑𝜏𝜏


=𝑚𝑚0

𝑑𝑑𝜏𝜏


Four-Vector Momentum

𝑝𝑝𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑝𝑝𝛼𝛼


= 𝑚𝑚0𝛾𝛾𝑑𝑑𝑡𝑡


=

𝛾𝛾 𝑚𝑚0 𝑐𝑐𝛾𝛾 𝑚𝑚0 �𝑑𝑑𝑥𝑥

𝑑𝑑𝑡𝑡𝛾𝛾 𝑚𝑚0 �𝑑𝑑𝑦𝑦

𝑑𝑑𝑡𝑡𝛾𝛾 𝑚𝑚0 �𝑑𝑑𝑧𝑧

𝑑𝑑𝑡𝑡

𝑝𝑝𝜇𝜇 ≡ 𝑚𝑚0𝑑𝑑𝑥𝑥𝜇𝜇

𝑑𝑑𝜏𝜏=

�𝐸𝐸 𝑐𝑐𝑝𝑝𝑥𝑥𝑝𝑝𝑦𝑦𝑝𝑝𝑧𝑧

𝐸𝐸2 − 𝑐𝑐 𝑝𝑝𝑥𝑥 2 − 𝑐𝑐 𝑝𝑝𝑦𝑦2 − 𝑐𝑐 𝑝𝑝𝑧𝑧 2 = 𝑚𝑚0

2𝑐𝑐4

𝑝𝑝𝑖𝑖 ≡ 𝛾𝛾 𝑚𝑚0𝑑𝑑𝑥𝑥𝑖𝑖𝑑𝑑𝑡𝑡

𝐸𝐸 ≡ 𝛾𝛾 𝑚𝑚0 𝑐𝑐2


𝐸𝐸2 − 𝑐𝑐 𝑝𝑝𝑥𝑥 2 − 𝑐𝑐 𝑝𝑝𝑦𝑦2 − 𝑐𝑐 𝑝𝑝𝑧𝑧 2 = 𝑚𝑚0

2𝑐𝑐4

𝑝𝑝𝑖𝑖 ≡ 𝛾𝛾 𝑚𝑚0 𝑣𝑣𝑖𝑖

𝐸𝐸2 − 𝑐𝑐 𝛾𝛾 𝑚𝑚0 𝑣𝑣𝑥𝑥 2 − 𝑐𝑐 𝛾𝛾 𝑚𝑚0 𝑣𝑣𝑦𝑦2 − 𝑐𝑐 𝛾𝛾 𝑚𝑚0 𝑣𝑣𝑧𝑧 2 = 𝑚𝑚0

2 𝑐𝑐4

𝐸𝐸2 = 𝑚𝑚02 𝑐𝑐4 1 +

𝛽𝛽2

1 − 𝛽𝛽2

𝐸𝐸 =𝑚𝑚0 𝑐𝑐2

1 − 𝛽𝛽2


Four-Vector Force contravariant tensor of rank 1

= 𝛾𝛾𝑑𝑑𝑑𝑑𝑡𝑡

�𝐸𝐸 𝑐𝑐𝛾𝛾 𝑚𝑚0 �𝑑𝑑𝑥𝑥

𝑑𝑑𝑡𝑡𝛾𝛾 𝑚𝑚0 �𝑑𝑑𝑥𝑥

𝑑𝑑𝑡𝑡𝛾𝛾 𝑚𝑚0 �𝑑𝑑𝑥𝑥

𝑑𝑑𝑡𝑡

𝐹𝐹𝜇𝜇 ≡𝑑𝑑𝑝𝑝𝜇𝜇

𝑑𝑑𝜏𝜏=

𝛾𝛾𝑐𝑐𝑃𝑃𝐹𝐹𝑥𝑥𝐹𝐹𝑦𝑦𝐹𝐹𝑧𝑧

=

𝛾𝛾𝑐𝑐𝑑𝑑𝐸𝐸𝑑𝑑𝑡𝑡

𝛾𝛾 𝑚𝑚0𝑑𝑑𝑑𝑑𝑡𝑡

𝛾𝛾𝑑𝑑𝑥𝑥𝑑𝑑𝑡𝑡


𝛾𝛾𝑑𝑑𝑦𝑦𝑑𝑑𝑡𝑡


𝛾𝛾𝑑𝑑𝑧𝑧𝑑𝑑𝑡𝑡


𝑑𝑑𝑥𝑥𝑧𝜇𝜇 = �𝜈𝜈=0



𝑥𝑥𝑧𝜇𝜇 = 𝑥𝑥𝑧𝜇𝜇 𝑥𝑥0, 𝑥𝑥1, 𝑥𝑥2, 𝑥𝑥3 = �𝜈𝜈=0

3

𝐴𝐴𝜈𝜈𝜇𝜇 𝑥𝑥𝜈𝜈 𝐴𝐴𝜈𝜈

𝜇𝜇 ≡𝜕𝜕𝑥𝑥𝑧𝜇𝜇

𝜕𝜕𝑥𝑥𝜈𝜈

Linear Transformations:

𝜕𝜕𝜕𝜕𝑥𝑥𝑧𝜇𝜇

= �𝜈𝜈=0

3𝜕𝜕𝑥𝑥𝜈𝜈

𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝜕𝜕𝜕𝜕𝑥𝑥𝜈𝜈

where

𝑉𝑉𝑧𝜇𝜇 = �𝜈𝜈=0


𝜕𝜕𝑥𝑥𝜈𝜈𝑉𝑉𝜈𝜈

𝑊𝑊𝑧𝜇𝜇 = �𝜈𝜈=0

3𝜕𝜕𝑥𝑥𝜈𝜈

𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝑊𝑊𝜈𝜈

contravarianttensor of rank 1= four-vector

covarianttensor of rank 1= four-vector


Inner Product of a Contravariant and a Covariant

tensor of rank 1

𝑉𝑉𝑧𝜇𝜇 = �𝜈𝜈=0


𝜕𝜕𝑥𝑥𝜈𝜈𝑉𝑉𝜈𝜈 𝑊𝑊𝑧𝜇𝜇 = �

𝛼𝛼=0


𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝑊𝑊𝛼𝛼

�𝜇𝜇=0

3

𝑉𝑉𝑧𝜇𝜇 𝑊𝑊𝑧𝜇𝜇

= �𝜇𝜇=0

3

�𝜈𝜈=0

3

�𝛼𝛼=0


𝜕𝜕𝑥𝑥𝜈𝜈𝜕𝜕𝑥𝑥𝛼𝛼

𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝑉𝑉𝜈𝜈 𝑊𝑊𝛼𝛼 = �

𝜈𝜈=0

3

�𝛼𝛼=0

3

𝛿𝛿𝜈𝜈𝛼𝛼 𝑉𝑉𝜈𝜈 𝑊𝑊𝛼𝛼 = �𝜈𝜈=0

3

𝑉𝑉𝜈𝜈 𝑊𝑊𝜈𝜈

= �𝜇𝜇=0

3

�𝜈𝜈=0


𝜕𝜕𝑥𝑥𝜈𝜈𝑉𝑉𝜈𝜈 �

𝛼𝛼=0


𝜕𝜕𝑥𝑥𝑧𝜇𝜇𝑊𝑊𝛼𝛼

invariant !!


Example 1: Charge Conservation

0 = 𝛁𝛁. 𝑱𝑱 +𝜕𝜕𝜌𝜌𝜕𝜕𝑡𝑡 = 𝛁𝛁. 𝑱𝑱 +

𝜕𝜕 𝑐𝑐 𝜌𝜌𝜕𝜕 𝑐𝑐 𝑡𝑡

=𝜕𝜕 𝑐𝑐 𝜌𝜌𝜕𝜕 𝑐𝑐 𝑡𝑡

+𝜕𝜕𝐽𝐽𝑥𝑥𝜕𝜕𝑥𝑥

+𝜕𝜕𝐽𝐽𝑦𝑦𝜕𝜕𝑦𝑦

+𝜕𝜕𝐽𝐽𝑧𝑧𝜕𝜕𝑧𝑧

=𝜕𝜕 𝑐𝑐 𝜌𝜌𝜕𝜕𝑥𝑥0

+𝜕𝜕𝐽𝐽𝑥𝑥𝜕𝜕𝑥𝑥1

+𝜕𝜕𝐽𝐽𝑦𝑦𝜕𝜕𝑥𝑥2

+𝜕𝜕𝐽𝐽𝑧𝑧𝜕𝜕𝑥𝑥3

𝐽𝐽𝜇𝜇 =

𝑐𝑐 𝜌𝜌𝐽𝐽𝑥𝑥𝐽𝐽𝑦𝑦𝐽𝐽𝑧𝑧

= �𝜇𝜇=0

3𝜕𝜕𝜕𝜕𝑥𝑥𝜇𝜇

𝐽𝐽𝜇𝜇contravariant

covariant


Four-Vector Current Density “J”

𝐽𝐽𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝐽𝐽𝛼𝛼

𝐽𝐽𝜇𝜇 =




Consider a charge 𝑞𝑞 at rest at a particular point 𝒓𝒓𝟎𝟎 in space.

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

reference moving along x-axis at a constant speed 𝑣𝑣0 = 𝛽𝛽 𝑐𝑐 ?


𝑱𝑱𝜇𝜇 𝒓𝒓, 𝑡𝑡 =


𝑱𝑱 𝒓𝒓, 𝑡𝑡 = (0,0,0)𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎

𝐽𝐽𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝐽𝐽𝛼𝛼 =


00


0010

0001

𝑐𝑐 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎000

=

𝛾𝛾 𝑐𝑐 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎− 𝛾𝛾 𝛽𝛽 𝑐𝑐 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎

00

=𝑐𝑐 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎

000

&


𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎

𝑐𝑐 𝑡𝑡 − 𝑡𝑡0𝑥𝑥 − 𝑥𝑥0𝑦𝑦 − 𝑦𝑦0𝑧𝑧 − 𝑧𝑧0

= �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝐼𝐼, 𝜇𝜇 𝑥𝑥𝑧𝛼𝛼 − 𝑥𝑥0𝑧𝛼𝛼 =


00


0010

0001

𝑐𝑐 𝑡𝑡′ − 𝑡𝑡𝑧0𝑥𝑥𝑧 − 𝑥𝑥𝑧0𝑦𝑦𝑧 − 𝑦𝑦𝑧0𝑧𝑧𝑧 − 𝑧𝑧𝑧0

=

𝛾𝛾 𝑐𝑐 𝑡𝑡′ − 𝑡𝑡𝑧0 + 𝛾𝛾 𝛽𝛽 𝑥𝑥𝑧 − 𝑥𝑥𝑧0𝛾𝛾 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝛾𝛾 𝛽𝛽 𝑐𝑐 𝑡𝑡′ − 𝑡𝑡𝑧0

𝑦𝑦𝑧 − 𝑦𝑦𝑧0𝑧𝑧𝑧 − 𝑧𝑧𝑧0

= 𝛿𝛿 𝛾𝛾 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝛾𝛾 𝛽𝛽 𝑐𝑐 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0

𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎 = 𝛿𝛿 𝑥𝑥 − 𝑥𝑥0 𝛿𝛿 𝑦𝑦 − 𝑦𝑦0 𝛿𝛿 𝑧𝑧 − 𝑧𝑧0

= 1𝛾𝛾𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0


𝐽𝐽𝑧𝜇𝜇 =

𝛾𝛾 𝑐𝑐 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎− 𝛾𝛾 𝛽𝛽 𝑐𝑐 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎

00

𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎 =1𝛾𝛾𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0

𝐽𝐽𝑧𝜇𝜇 =

𝑐𝑐 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0− 𝛽𝛽 𝑐𝑐 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0

00

𝜌𝜌′ = 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0

𝐽𝐽𝑧𝑥𝑥 = −𝑣𝑣0 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0


𝜌𝜌𝑧 𝒓𝒓𝑧, 𝑡𝑡𝑧 = 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0

𝐽𝐽𝑧𝑥𝑥 𝒓𝒓𝑧, 𝑡𝑡𝑧𝐽𝐽𝑧𝑦𝑦 𝒓𝒓𝑧, 𝑡𝑡𝑧𝐽𝐽𝑧𝑧𝑧 𝒓𝒓𝑧, 𝑡𝑡𝑧

=−𝑣𝑣0 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 + 𝑣𝑣0 𝑡𝑡′ − 𝑡𝑡𝑧0 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0

00

𝑡𝑡𝑧0 = 0 𝑥𝑥𝑧0 = 0

𝜌𝜌𝑧 𝒓𝒓𝑧, 𝑡𝑡𝑧 = 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 + 𝑣𝑣0 𝑡𝑡′ 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0 = 𝑞𝑞 𝛿𝛿3 𝒓𝒓𝑧 − 𝒓𝒓𝑧𝟎𝟎 𝑡𝑡′

𝑱𝑱𝑧 𝒓𝒓𝑧, 𝑡𝑡𝑧 =−𝑣𝑣0 𝑞𝑞 𝛿𝛿 𝑥𝑥𝑧 + 𝑣𝑣0 𝑡𝑡′ 𝛿𝛿 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 𝛿𝛿 𝑧𝑧𝑧 − 𝑧𝑧𝑧0

00

=−𝑣𝑣0 𝑞𝑞 𝛿𝛿3 𝒓𝒓𝑧 − 𝒓𝒓𝑧𝟎𝟎 𝑡𝑡′

00

𝒓𝒓𝑧𝟎𝟎 𝑡𝑡′ =−𝑣𝑣0 𝑡𝑡′𝑦𝑦𝑧0𝑧𝑧𝑧0


Example 2: Lorenz Gauge

0 = 𝛻𝛻.𝑨𝑨 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕𝜕𝜕𝑡𝑡Φ = 𝛁𝛁.𝑨𝑨 +

𝜕𝜕 Φ/𝑐𝑐𝜕𝜕 𝑐𝑐 𝑡𝑡

=𝜕𝜕 Φ/𝑐𝑐𝜕𝜕 𝑐𝑐 𝑡𝑡

+𝜕𝜕𝐴𝐴𝑥𝑥𝜕𝜕𝑥𝑥

+𝜕𝜕𝐴𝐴𝑦𝑦𝜕𝜕𝑦𝑦

+𝜕𝜕𝐴𝐴𝜕𝜕𝑧𝑧

=𝜕𝜕 Φ/𝑐𝑐𝜕𝜕𝑥𝑥0

+𝜕𝜕𝐴𝐴𝑥𝑥𝜕𝜕𝑥𝑥1

+𝜕𝜕𝐴𝐴𝑦𝑦𝜕𝜕𝑥𝑥2

+𝜕𝜕𝐴𝐴𝑧𝑧𝜕𝜕𝑥𝑥3

= �𝜇𝜇=0

3𝜕𝜕𝜕𝜕𝑥𝑥𝜇𝜇

𝐴𝐴𝜇𝜇

covariant

contravariant𝐴𝐴𝜇𝜇 =

Φ/𝑐𝑐𝐴𝐴𝑥𝑥𝐴𝐴𝑦𝑦𝐴𝐴𝑧𝑧


Four-Vector Potential “A”

𝐴𝐴𝜇𝜇 =


𝐴𝐴𝑧𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝐴𝐴𝛼𝛼



Consider a charge 𝑞𝑞 moving at a constant speed 𝑣𝑣0 = 𝛽𝛽 𝑐𝑐 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?


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

= 𝜇𝜇𝑜𝑜 𝑱𝑱 𝒓𝒓, 𝑡𝑡

−𝛻𝛻2Φ 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕2Φ 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡2=𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

𝜌𝜌 𝒓𝒓, 𝑡𝑡 = 𝑞𝑞 𝛿𝛿 𝑥𝑥 + 𝑣𝑣0 𝑡𝑡 𝛿𝛿 𝑦𝑦 − 𝑦𝑦0 𝛿𝛿 𝑧𝑧 − 𝑧𝑧0 = 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎 𝑡𝑡

𝑱𝑱 𝒓𝒓, 𝑡𝑡 =𝑣𝑣0 𝑞𝑞 𝛿𝛿 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡 𝛿𝛿 𝑦𝑦 − 𝑦𝑦0 𝛿𝛿 𝑧𝑧 − 𝑧𝑧0

00

=𝑣𝑣0 𝑞𝑞 𝛿𝛿3 𝒓𝒓 − 𝒓𝒓𝟎𝟎 𝑡𝑡

00

Hard Way


Easier Way:Start with an inertial frame of reference K’

moving in the same way as the charge

=1

4 𝜋𝜋 𝜖𝜖0𝑞𝑞

𝒓𝒓𝑧 − 𝒓𝒓𝑧0

𝜌𝜌𝑧 𝒓𝒓′ = 𝑞𝑞 𝛿𝛿3 𝒓𝒓𝑧 − 𝒓𝒓𝑧𝟎𝟎

Φ𝑧 𝒓𝒓𝑧 =1

4 𝜋𝜋 𝜖𝜖0�−∞

+∞𝜌𝜌𝑧 𝒓𝒓𝑧𝑧𝒓𝒓𝑧 − 𝒓𝒓𝑧𝑧

𝑑𝑑𝑉𝑉𝑧𝑧

electrostatic problem


𝑱𝑱𝑧 𝒓𝒓𝑧 =000

𝑨𝑨𝑧 𝒓𝒓𝑧 =𝜇𝜇𝑜𝑜

4 𝜋𝜋�−∞

+∞𝑱𝑱𝑧 𝒓𝒓𝑧𝑧𝒓𝒓 − 𝒓𝒓′′

𝑑𝑑𝑉𝑉𝑧𝑧

=000

magnetostaticproblem


𝐴𝐴𝜇𝜇 = �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝐼𝐼, 𝜇𝜇 𝐴𝐴𝑧𝛼𝛼

𝐴𝐴𝑧𝛼𝛼 =

Φ𝑧/𝑐𝑐𝐴𝐴𝑧𝑥𝑥𝐴𝐴𝑧𝑦𝑦𝐴𝐴𝑧𝑧𝑧

𝐿𝐿𝛼𝛼𝐼𝐼, 𝜇𝜇 =


00


0010

0001

=

Φ𝑧/𝑐𝑐000

𝐴𝐴𝜇𝜇 =

𝛾𝛾 Φ𝑧/𝑐𝑐𝛾𝛾 𝛽𝛽 Φ𝑧/𝑐𝑐

00 𝑨𝑨 𝒓𝒓, 𝑡𝑡 =

𝛾𝛾 𝛽𝛽 Φ𝑧/𝑐𝑐00

Φ 𝒓𝒓, 𝑡𝑡 = 𝛾𝛾 Φ𝑧 =𝛾𝛾 𝑞𝑞

4 𝜋𝜋 𝜖𝜖0 𝒓𝒓𝑧 − 𝒓𝒓𝑧0

=

𝛾𝛾 𝑞𝑞 𝑣𝑣04 𝜋𝜋 𝜖𝜖0 𝑐𝑐2 𝒓𝒓𝑧 − 𝒓𝒓𝑧0

00


𝒓𝒓𝑧 − 𝒓𝒓𝑧0


= �𝛼𝛼=0

3

𝐿𝐿𝛼𝛼𝜇𝜇 𝑥𝑥𝛼𝛼 − 𝑥𝑥0𝛼𝛼 =


00


0010

0001


=

𝛾𝛾 𝑐𝑐 𝑡𝑡 − 𝑡𝑡0 − 𝛾𝛾 𝛽𝛽 𝑥𝑥 − 𝑥𝑥0𝛾𝛾 𝑥𝑥 − 𝑥𝑥0 − 𝛾𝛾 𝛽𝛽 𝑐𝑐 𝑡𝑡 − 𝑡𝑡0

𝑦𝑦 − 𝑦𝑦0𝑧𝑧 − 𝑧𝑧0

𝒓𝒓𝑧 − 𝒓𝒓𝑧0 = 𝑥𝑥𝑧 − 𝑥𝑥𝑧0 2 + 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 2 + 𝑧𝑧𝑧 − 𝑧𝑧𝑧0 2

= 𝛾𝛾 𝑥𝑥 − 𝑥𝑥0 − 𝛾𝛾 𝛽𝛽 𝑐𝑐 𝑡𝑡 − 𝑡𝑡0 2 + 𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2


𝑡𝑡0 = 0 𝑥𝑥0 = 0

Φ 𝒓𝒓, 𝑡𝑡 =𝛾𝛾 𝑞𝑞

4 𝜋𝜋 𝜖𝜖0 𝒓𝒓𝑧 − 𝒓𝒓𝑧0

=𝛾𝛾 𝑞𝑞

4 𝜋𝜋 𝜖𝜖0 𝛾𝛾2 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡 2 + 𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2

𝑨𝑨 𝒓𝒓, 𝑡𝑡 =

𝛾𝛾 𝑞𝑞 𝑣𝑣04 𝜋𝜋 𝜖𝜖0 𝑐𝑐2 𝒓𝒓𝑧 − 𝒓𝒓𝑧0

00

=

𝛾𝛾 𝑞𝑞 𝑣𝑣04 𝜋𝜋 𝜖𝜖0 𝑐𝑐2 𝛾𝛾2 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡 2 + 𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2

00


𝜵𝜵.𝑨𝑨 𝒓𝒓, 𝑡𝑡 + 𝜇𝜇𝑜𝑜 𝜖𝜖0𝜕𝜕Φ 𝒓𝒓, 𝑡𝑡

𝜕𝜕𝑡𝑡= 0

HW: From the previous expressions for the vector and scalar potentials, prove:

HW: Calculate the electric field of a charge moving at constant speed

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

HW: Calculate the magnetic field of a charge moving at constant speed

𝑩𝑩 𝒓𝒓, 𝒕𝒕 = 𝛻𝛻 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡


Metric Tensor and Invariants𝑑𝑑𝑠𝑠 2 = 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 − 𝑑𝑑𝑥𝑥 2 − 𝑑𝑑𝑦𝑦 2 − 𝑑𝑑𝑧𝑧 2

= 𝑑𝑑𝑥𝑥0 2 − 𝑑𝑑𝑥𝑥1 2 − 𝑑𝑑𝑥𝑥2 2 − 𝑑𝑑𝑥𝑥3 2

𝑑𝑑𝑥𝑥0 = 𝑑𝑑𝑥𝑥0𝑑𝑑𝑥𝑥1 = −𝑑𝑑𝑥𝑥1𝑑𝑑𝑥𝑥2 = −𝑑𝑑𝑥𝑥2𝑑𝑑𝑥𝑥3 = −𝑑𝑑𝑥𝑥3

𝑑𝑑𝑠𝑠 2 = �𝜈𝜈=0

3

𝑑𝑑𝑥𝑥𝜈𝜈 𝑑𝑑𝑥𝑥𝜈𝜈

𝑑𝑑𝑥𝑥𝜇𝜇 = �𝜈𝜈=0

3

𝑔𝑔𝜇𝜇𝜈𝜈 𝑑𝑑𝑥𝑥𝜈𝜈

𝑔𝑔𝜇𝜇𝜈𝜈 =1000

0−100

00−10

000−1

Metric Tensor


𝑘𝑘𝜇𝜇 =


𝑘𝑘𝜇𝜇 = �𝜈𝜈=0

3

𝑔𝑔𝜇𝜇𝜈𝜈 𝑘𝑘𝜈𝜈 =

⁄𝜔𝜔 𝑐𝑐−𝑘𝑘𝑥𝑥−𝑘𝑘𝑦𝑦−𝑘𝑘𝑧𝑧

�𝜇𝜇=0

3

𝑘𝑘𝜇𝜇 𝑘𝑘𝜇𝜇 = ⁄𝜔𝜔 𝑐𝑐 2 − 𝑘𝑘𝑥𝑥 2 − 𝑘𝑘𝑦𝑦2 − 𝑘𝑘𝑧𝑧 2 = 𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠𝑡𝑡𝑎𝑎𝑡𝑡𝑡𝑡


𝐽𝐽𝜇𝜇 =


𝐽𝐽𝜇𝜇 = �𝜈𝜈=0

3

𝑔𝑔𝜇𝜇𝜈𝜈 𝐽𝐽𝜈𝜈 =

𝑐𝑐 𝜌𝜌−𝐽𝐽𝑥𝑥−𝐽𝐽𝑦𝑦−𝐽𝐽𝑧𝑧

�𝜇𝜇=0

3

𝐽𝐽𝜇𝜇 𝐽𝐽𝜇𝜇 = 𝑐𝑐 𝜌𝜌 2 − 𝐽𝐽𝑥𝑥 2 − 𝐽𝐽𝑦𝑦2 − 𝐽𝐽𝑧𝑧 2 = 𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠𝑡𝑡𝑎𝑎𝑡𝑡𝑡𝑡


𝐴𝐴𝜇𝜇 =


𝐴𝐴𝜇𝜇 = �𝜈𝜈=0

3

𝑔𝑔𝜇𝜇𝜈𝜈 𝐴𝐴𝜈𝜈 =

Φ/𝑐𝑐−𝐴𝐴𝑥𝑥−𝐴𝐴𝑦𝑦−𝐴𝐴𝑧𝑧

�𝜇𝜇=0

3

𝐴𝐴𝜇𝜇 𝐴𝐴𝜇𝜇 = �Φ 𝑐𝑐2− 𝐴𝐴𝑥𝑥 2 − 𝐴𝐴𝑦𝑦

2 − 𝐴𝐴𝑧𝑧 2 = 𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠𝑡𝑡𝑎𝑎𝑡𝑡𝑡𝑡


�𝜇𝜇=0

3

𝑝𝑝𝜇𝜇 𝑝𝑝𝜇𝜇 = 𝐸𝐸 2 − 𝑐𝑐 𝑝𝑝𝑥𝑥 2 − 𝑐𝑐 𝑝𝑝𝑥𝑥 2 − 𝑐𝑐 𝑝𝑝𝑧𝑧 2 = 𝑐𝑐𝑐𝑐𝑡𝑡𝑠𝑠𝑡𝑡𝑎𝑎𝑡𝑡𝑡𝑡

𝑝𝑝𝜇𝜇 =𝐸𝐸𝑐𝑐 𝑝𝑝𝑥𝑥𝑐𝑐 𝑝𝑝𝑦𝑦𝑐𝑐 𝑝𝑝𝑧𝑧

𝑝𝑝𝜇𝜇 = �𝜈𝜈=0

3

𝑔𝑔𝜇𝜇𝜈𝜈 𝑝𝑝𝜈𝜈 =𝐸𝐸

−𝑐𝑐 𝑝𝑝𝑥𝑥−𝑐𝑐 𝑝𝑝𝑦𝑦−𝑐𝑐 𝑝𝑝𝑧𝑧


𝑑𝑑𝑠𝑠 2 = �𝜈𝜈=0

3

𝑑𝑑𝑥𝑥𝜈𝜈 𝑑𝑑𝑥𝑥𝜈𝜈 = �𝜈𝜈=0

3

�𝜇𝜇=0

3

𝑔𝑔𝜈𝜈𝜇𝜇 𝑑𝑑𝑥𝑥𝜇𝜇 𝑑𝑑𝑥𝑥𝜈𝜈

= �𝜈𝜈=0

3

�𝜇𝜇=0

3

𝑔𝑔𝜈𝜈𝜇𝜇𝑑𝑑𝑥𝑥𝜇𝜇 𝑑𝑑𝑥𝑥𝜈𝜈

𝑔𝑔𝜈𝜈𝜇𝜇 = 𝑔𝑔𝜈𝜈𝜇𝜇 =1000

0−100

00−10

000−1

𝑑𝑑𝑥𝑥𝜈𝜈 = �𝜇𝜇=0

3

𝑔𝑔𝜈𝜈𝜇𝜇 𝑑𝑑𝑥𝑥𝜇𝜇 𝑑𝑑𝑥𝑥𝜈𝜈 = �𝜇𝜇=0

3

𝑔𝑔𝜈𝜈𝜇𝜇𝑑𝑑𝑥𝑥𝜇𝜇


Tensors of Rank 1

𝑑𝑑𝑥𝑥𝜇𝜇𝑑𝑑𝑥𝑥𝜇𝜇

𝜕𝜕𝜕𝜕𝑥𝑥𝜇𝜇

≡ 𝜕𝜕𝜇𝜇𝜕𝜕𝜕𝜕𝑥𝑥𝜇𝜇

≡ 𝜕𝜕𝜇𝜇

covariantcontravariant

𝑔𝑔𝜈𝜈𝜇𝜇 ,𝑔𝑔𝜈𝜈𝜇𝜇

𝑔𝑔𝜈𝜈𝜇𝜇 ,𝑔𝑔𝜈𝜈𝜇𝜇


Contravariant Tensor of Rank 2

𝑉𝑉𝑧𝛼𝛼 = �𝜇𝜇=0

3𝜕𝜕𝑥𝑥𝑧𝛼𝛼

𝜕𝜕𝑥𝑥𝜇𝜇𝑉𝑉𝜇𝜇 = �

𝜇𝜇=0

3

𝐿𝐿𝜇𝜇𝛼𝛼 𝑉𝑉𝜇𝜇 𝑊𝑊𝑧𝛽𝛽 = �𝜈𝜈=0

3𝜕𝜕𝑥𝑥𝑧𝛽𝛽

𝜕𝜕𝑥𝑥𝜈𝜈𝑊𝑊𝜈𝜈 = �

𝜈𝜈=0

3

𝐿𝐿𝜈𝜈𝛽𝛽 𝑊𝑊𝜈𝜈

𝑇𝑇𝛼𝛼𝛽𝛽 ≡ 𝑉𝑉𝛼𝛼 𝑊𝑊𝛽𝛽

𝑇𝑇𝑧𝛼𝛼𝛽𝛽 = 𝑉𝑉𝑧𝛼𝛼 𝑊𝑊𝑧𝛽𝛽 = �𝜇𝜇=0

3𝜕𝜕𝑥𝑥𝑧𝛼𝛼

𝜕𝜕𝑥𝑥𝜇𝜇𝑉𝑉𝜇𝜇�

𝜈𝜈=0

3𝜕𝜕𝑥𝑥𝑧𝛽𝛽

𝜕𝜕𝑥𝑥𝜈𝜈𝑊𝑊𝜈𝜈 = �

𝜇𝜇=0

3

𝐿𝐿𝜇𝜇𝛼𝛼 𝑉𝑉𝜇𝜇�𝜈𝜈=0

3

𝐿𝐿𝜈𝜈𝛽𝛽 𝑊𝑊𝜈𝜈

= �𝜇𝜇=0

3

𝐿𝐿𝜇𝜇𝛼𝛼 �𝜈𝜈=0

3

𝐿𝐿𝜈𝜈𝛽𝛽 𝑇𝑇𝜇𝜇𝜈𝜈 = 𝐿𝐿 𝑇𝑇 𝐿𝐿𝑡𝑡 𝛼𝛼𝛽𝛽


𝐴𝐴𝛽𝛽 =


𝜕𝜕𝜕𝜕𝑥𝑥𝛼𝛼

= 𝜕𝜕𝛼𝛼

𝐹𝐹𝛼𝛼𝛽𝛽 ≡ 𝜕𝜕𝛼𝛼 𝐴𝐴𝛽𝛽 − 𝜕𝜕𝛽𝛽 𝐴𝐴𝛼𝛼

𝐹𝐹𝛼𝛼𝛽𝛽 = −𝐹𝐹𝛽𝛽𝛼𝛼

𝐹𝐹𝛼𝛼𝛼𝛼 = 0

Electromagnetic Field Tensor


Electric Field as a component of the electromagnetic field tensor

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

𝐴𝐴𝜇𝜇 =


𝐸𝐸𝑖𝑖 = −𝜕𝜕Φ𝜕𝜕𝑥𝑥𝑖𝑖

−𝜕𝜕𝐴𝐴𝑖𝑖

𝜕𝜕𝑡𝑡= −

𝜕𝜕 𝑐𝑐 𝐴𝐴0

𝜕𝜕𝑥𝑥𝑖𝑖− 𝑐𝑐

𝜕𝜕𝐴𝐴𝑖𝑖

𝜕𝜕𝑥𝑥0= 𝑐𝑐 −

𝜕𝜕𝐴𝐴0

𝜕𝜕𝑥𝑥𝑖𝑖−𝜕𝜕𝐴𝐴𝑖𝑖

𝜕𝜕𝑥𝑥0

𝐸𝐸𝑖𝑖

𝑐𝑐= −

𝜕𝜕𝐴𝐴0

𝜕𝜕𝑥𝑥𝑖𝑖−𝜕𝜕𝐴𝐴𝑖𝑖

𝜕𝜕𝑥𝑥0

𝑖𝑖 ≡ 1, 2, 3

= 𝜕𝜕𝑖𝑖𝐴𝐴0 − 𝜕𝜕0𝐴𝐴𝑖𝑖 = 𝐹𝐹𝑖𝑖0


𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 𝛁𝛁 × 𝑨𝑨 𝒓𝒓, 𝑡𝑡 𝐴𝐴𝜇𝜇 =


𝐵𝐵𝑖𝑖 = �𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘𝜕𝜕𝐴𝐴𝑘𝑘

𝜕𝜕𝑥𝑥𝑗𝑗

𝑖𝑖, 𝑗𝑗, 𝑘𝑘 ≡ 1, 2, 3

= −�𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘 𝜕𝜕𝑗𝑗𝐴𝐴𝑘𝑘

𝐵𝐵1 = − 𝜕𝜕2𝐴𝐴3 − 𝜕𝜕3𝐴𝐴2 = −𝐹𝐹23

𝐵𝐵2 = − 𝜕𝜕3𝐴𝐴1 − 𝜕𝜕1𝐴𝐴3 = −𝐹𝐹31

𝐵𝐵3 = − 𝜕𝜕1𝐴𝐴2 − 𝜕𝜕2𝐴𝐴1 = −𝐹𝐹12

Magnetic Field as a component of the electromagnetic field tensor


𝐹𝐹𝛼𝛼𝛽𝛽 ≡ 𝜕𝜕𝛼𝛼𝐴𝐴𝛽𝛽 − 𝜕𝜕𝛽𝛽𝐴𝐴𝛼𝛼 =1𝑐𝑐

0 −𝐸𝐸1𝐸𝐸1 0

−𝐸𝐸2 −𝐸𝐸3−𝑐𝑐𝐵𝐵3 𝑐𝑐𝐵𝐵2

𝐸𝐸2 𝑐𝑐𝐵𝐵3𝐸𝐸3 −𝑐𝑐𝐵𝐵2

0 −𝑐𝑐𝐵𝐵1𝑐𝑐𝐵𝐵1 0

𝐸𝐸𝑖𝑖

𝑐𝑐= 𝐹𝐹𝑖𝑖0

𝐵𝐵1 = − 𝜕𝜕2𝐴𝐴3 − 𝜕𝜕3𝐴𝐴2 = −𝐹𝐹23

𝐵𝐵2 = − 𝜕𝜕3𝐴𝐴1 − 𝜕𝜕1𝐴𝐴3 = −𝐹𝐹31

𝐵𝐵3 = − 𝜕𝜕1𝐴𝐴2 − 𝜕𝜕2𝐴𝐴3 = −𝐹𝐹12

Components of the Electromagnetic Field Tensor


𝐹𝐹𝑧𝛼𝛼𝛽𝛽 = 𝐿𝐿 𝐹𝐹 𝐿𝐿𝑡𝑡 𝛼𝛼𝛽𝛽

How the Electromagnetic Field Tensor changes under a Lorentz

transformation:

𝐹𝐹 =1𝑐𝑐

0 −𝐸𝐸1𝐸𝐸1 0




𝐿𝐿 =


00


0010

0001

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

constant speed 𝑣𝑣0 = 𝛽𝛽 𝑐𝑐


𝐹𝐹′ =1𝑐𝑐

0 −𝐸𝐸′1

𝐸𝐸′1 0−𝐸𝐸′2 −𝐸𝐸′3

−𝑐𝑐𝐵𝐵′3 𝑐𝑐𝐵𝐵′2

𝐸𝐸′2 𝑐𝑐𝐵𝐵′3

𝐸𝐸′3 −𝑐𝑐𝐵𝐵′20 −𝑐𝑐𝐵𝐵′1

𝑐𝑐𝐵𝐵′1 0

= 𝐿𝐿 𝐹𝐹 𝐿𝐿𝑡𝑡

𝐸𝐸𝑧1 = 𝐸𝐸1

𝐸𝐸𝑧2 = 𝛾𝛾 𝐸𝐸2 − 𝛽𝛽 𝑐𝑐 𝐵𝐵3

𝐸𝐸𝑧3 = 𝛾𝛾 𝐸𝐸3 + 𝛽𝛽 𝑐𝑐 𝐵𝐵2

𝐵𝐵𝑧1 = 𝐵𝐵1

𝐵𝐵𝑧2 = 𝛾𝛾 𝐵𝐵2 +𝛽𝛽𝑐𝑐𝐸𝐸3

𝐵𝐵𝑧3 = 𝛾𝛾 𝐵𝐵3 −𝛽𝛽𝑐𝑐𝐸𝐸2


Note: Galilean transformation leads to different and incorrect relations:

𝑭𝑭 = 𝑞𝑞 𝑬𝑬𝑧 + 𝑞𝑞 𝒗𝒗𝑧 × 𝑩𝑩𝑧

𝒗𝒗 = 𝒗𝒗′ + 𝒗𝒗0

= 𝑞𝑞 𝑬𝑬 + 𝑞𝑞 𝒗𝒗′ + 𝒗𝒗0 × 𝑩𝑩

𝑬𝑬′ = 𝑬𝑬 + 𝒗𝒗0 × 𝑩𝑩 𝑩𝑩′ = 𝑩𝑩

= 𝑞𝑞 𝑬𝑬 + 𝑞𝑞 𝒗𝒗 × 𝑩𝑩

𝒗𝒗′: particle velocity with respect to K’

= 𝑞𝑞 𝑬𝑬 + 𝒗𝒗0 × 𝑩𝑩 + 𝑞𝑞 𝒗𝒗′ × 𝑩𝑩


Electric and Magnetic Fields of a Moving Charge


Consider a charge 𝑞𝑞 moving at a constant speed 𝑣𝑣0 = 𝛽𝛽 𝑐𝑐 along a straight

line parallel to the x-axis.

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


𝐸𝐸𝑧1 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝑥𝑥𝑧 − 𝑥𝑥𝑧0

𝑥𝑥𝑧 − 𝑥𝑥𝑧0 2 + 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 2 + 𝑧𝑧𝑧 − 𝑧𝑧𝑧0 2 3/2


4 𝜋𝜋 𝜖𝜖0𝑦𝑦𝑧 − 𝑦𝑦𝑧0



4 𝜋𝜋 𝜖𝜖0𝑧𝑧𝑧 − 𝑧𝑧𝑧0


𝐵𝐵𝑧1 = 0

𝐵𝐵𝑧2 = 0

𝐵𝐵𝑧3 = 0

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


𝐸𝐸𝑧1 = 𝐸𝐸1

𝐸𝐸𝑧2 = 𝛾𝛾 𝐸𝐸2 − 𝛽𝛽 𝑐𝑐 𝐵𝐵3

𝐸𝐸𝑧3 = 𝛾𝛾 𝐸𝐸3 + 𝛽𝛽 𝑐𝑐 𝐵𝐵2

𝐵𝐵𝑧1 = 𝐵𝐵1

𝐵𝐵𝑧2 = 𝛾𝛾 𝐵𝐵2 +𝛽𝛽𝑐𝑐𝐸𝐸3

𝐵𝐵𝑧3 = 𝛾𝛾 𝐵𝐵3 −𝛽𝛽𝑐𝑐𝐸𝐸2

𝐸𝐸1 = 𝐸𝐸𝑧1

𝐸𝐸2 = 𝛾𝛾 𝐸𝐸𝑧2 + 𝛽𝛽 𝑐𝑐 𝐵𝐵𝑧3

𝐸𝐸3 = 𝛾𝛾 𝐸𝐸𝑧3 − 𝛽𝛽 𝑐𝑐 𝐵𝐵𝑧2

𝐵𝐵1 = 𝐵𝐵𝑧1

𝐵𝐵2 = 𝛾𝛾 𝐵𝐵𝑧2 −𝛽𝛽𝑐𝑐𝐸𝐸𝑧3

𝐵𝐵3 = 𝛾𝛾 𝐵𝐵𝑧3 +𝛽𝛽𝑐𝑐𝐸𝐸𝑧2



𝐸𝐸2 = 𝛾𝛾 𝐸𝐸𝑧2

𝐸𝐸3 = 𝛾𝛾 𝐸𝐸𝑧3

𝐵𝐵1 = 0

𝐵𝐵2 = −𝛾𝛾 𝛽𝛽𝑐𝑐𝐸𝐸𝑧3

𝐵𝐵3 =𝛾𝛾 𝛽𝛽𝑐𝑐𝐸𝐸𝑧2

= −𝛽𝛽𝑐𝑐𝐸𝐸3

=𝛽𝛽𝑐𝑐𝐸𝐸2


𝐸𝐸2 = 𝛾𝛾 𝐸𝐸𝑧2 + 𝛽𝛽 𝑐𝑐 𝐵𝐵𝑧3

𝐸𝐸3 = 𝛾𝛾 𝐸𝐸𝑧3 − 𝛽𝛽 𝑐𝑐 𝐵𝐵𝑧2

𝐵𝐵1 = 𝐵𝐵𝑧1

𝐵𝐵2 = 𝛾𝛾 𝐵𝐵𝑧2 −𝛽𝛽𝑐𝑐𝐸𝐸𝑧3

𝐵𝐵3 = 𝛾𝛾 𝐵𝐵𝑧3 +𝛽𝛽𝑐𝑐𝐸𝐸𝑧2


𝐸𝐸1 = 𝐸𝐸𝑧1 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝑥𝑥𝑧 − 𝑥𝑥𝑧0


𝐸𝐸2 = 𝛾𝛾 𝐸𝐸𝑧2 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝛾𝛾 𝑦𝑦𝑧 − 𝑦𝑦𝑧0


𝐸𝐸3 = 𝛾𝛾 𝐸𝐸𝑧3 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝛾𝛾 𝑧𝑧𝑧 − 𝑧𝑧𝑧0



𝐵𝐵1 = 0

𝐵𝐵2 = −𝛾𝛾 𝛽𝛽𝑐𝑐 𝐸𝐸′3 = −

𝛽𝛽𝑐𝑐

𝑞𝑞4 𝜋𝜋 𝜖𝜖0

𝛾𝛾 𝑧𝑧𝑧 − 𝑧𝑧𝑧0𝑥𝑥𝑧 − 𝑥𝑥𝑧0 2 + 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 2 + 𝑧𝑧𝑧 − 𝑧𝑧𝑧0 2 3/2

𝐵𝐵3 =𝛾𝛾 𝛽𝛽𝑐𝑐𝐸𝐸𝑧2 =

𝛽𝛽𝑐𝑐

𝑞𝑞4 𝜋𝜋 𝜖𝜖0

𝛾𝛾 𝑦𝑦𝑧 − 𝑦𝑦𝑧0𝑥𝑥𝑧 − 𝑥𝑥𝑧0 2 + 𝑦𝑦𝑧 − 𝑦𝑦𝑧0 2 + 𝑧𝑧𝑧 − 𝑧𝑧𝑧0 2 3/2



=


00


0010

0001


𝑡𝑡0 = 𝑡𝑡𝑧0 = 0 𝑥𝑥0 = 𝑥𝑥𝑧0 = 0


=

𝛾𝛾 𝑐𝑐 𝑡𝑡 − 𝛽𝛽 𝑥𝑥𝛾𝛾 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡𝑦𝑦 − 𝑦𝑦0𝑧𝑧 − 𝑧𝑧0


𝐸𝐸1 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝛾𝛾 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡

𝛾𝛾2 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡 2 + 𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2 3/2

𝐸𝐸2 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝛾𝛾 𝑦𝑦 − 𝑦𝑦0


𝐸𝐸3 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝛾𝛾 𝑧𝑧 − 𝑧𝑧0



𝛾𝛾2 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡 2 + 𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2

𝑥𝑥, 𝑦𝑦, 𝑧𝑧

𝑣𝑣0 𝑡𝑡,𝑦𝑦0, 𝑧𝑧0

=1

1 − 𝛽𝛽2𝑥𝑥 − 𝑣𝑣0𝑡𝑡 2 + 𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2 1 − 𝛽𝛽2 𝑠𝑠𝑖𝑖𝑡𝑡 𝜃𝜃 2

𝜃𝜃 𝑥𝑥

𝑠𝑠𝑖𝑖𝑡𝑡2 𝜃𝜃 ≡𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2

𝑥𝑥 − 𝑣𝑣0 𝑡𝑡 2 + 𝑦𝑦 − 𝑦𝑦0 2 + 𝑧𝑧 − 𝑧𝑧0 2

𝒗𝒗𝟎𝟎

𝒅𝒅

𝒅𝒅 ≡ 𝑥𝑥 − 𝑣𝑣0 𝑡𝑡 , 𝑦𝑦 − 𝑦𝑦0 , 𝑧𝑧 − 𝑧𝑧0


𝑬𝑬 𝒓𝒓, 𝑡𝑡 =𝑞𝑞

4 𝜋𝜋 𝜖𝜖0𝒅𝒅𝑑𝑑3

1 − 𝛽𝛽2

1 − 𝛽𝛽2 𝑠𝑠𝑖𝑖𝑡𝑡2 𝜃𝜃 3/2

classical Coulomb term

relativistic correction

𝛽𝛽 = 0.00 & 0.20

𝛽𝛽 = 0.70

𝛽𝛽 = 0.90

𝛽𝛽 = 0.99

𝒗𝒗𝟎𝟎

• The electric field points away from the charge at present time (t).

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

• However, it is not isotropic.

• The electric field amplitude depends on the direction away from the charge. 1 − 𝛽𝛽2

𝜃𝜃 ≅ 0

𝜃𝜃 ≅ 90°1

1 − 𝛽𝛽2


𝐵𝐵1 = 0

𝐵𝐵2 = −𝛽𝛽𝑐𝑐𝐸𝐸3

𝐵𝐵3 =𝛽𝛽𝑐𝑐𝐸𝐸2

𝑩𝑩 = 𝜷𝜷 ×𝑬𝑬𝑐𝑐

𝑩𝑩 𝒓𝒓, 𝑡𝑡 =𝜇𝜇0

4 𝜋𝜋𝑞𝑞 𝒗𝒗𝟎𝟎 ×

𝒅𝒅𝑑𝑑3

1 − 𝛽𝛽2

1 − 𝛽𝛽2 𝑠𝑠𝑖𝑖𝑡𝑡2 𝜃𝜃 3/2

classical Biot-Savart

term

relativistic correction


Maxwell’s Equations in terms of the Electromagnetic Field Tensor


0 −𝐸𝐸1𝐸𝐸1 0




𝐽𝐽𝜇𝜇 =


sources

fields


The Four Inhomogeneous Maxwell’s Equations

Gauss’s Law

Ampere’s Law

𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

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



𝛁𝛁.𝑬𝑬 𝒓𝒓, 𝑡𝑡 =𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

𝐸𝐸𝑖𝑖 = 𝑐𝑐 𝐹𝐹𝑖𝑖0

𝜌𝜌 𝒓𝒓, 𝑡𝑡 =𝐽𝐽0

𝑐𝑐

�𝑖𝑖=1

3𝜕𝜕𝜕𝜕𝑥𝑥𝑖𝑖

𝐸𝐸𝑖𝑖 = �𝑖𝑖=1


𝑐𝑐 𝐹𝐹𝑖𝑖0 = 𝑐𝑐�𝑖𝑖=1

3

𝜕𝜕𝑖𝑖 𝐹𝐹𝑖𝑖0

= 𝑐𝑐�𝛼𝛼=0

3

𝜕𝜕𝛼𝛼 𝐹𝐹𝛼𝛼0

=𝜌𝜌 𝒓𝒓, 𝑡𝑡𝜖𝜖0

=𝐽𝐽0

𝜖𝜖0 𝑐𝑐

�𝛼𝛼=0

3

𝜕𝜕𝛼𝛼 𝐹𝐹𝛼𝛼0 = 𝜇𝜇0 𝐽𝐽0

+ 𝑐𝑐 𝜕𝜕0𝐹𝐹00


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


�𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘𝜕𝜕𝐵𝐵𝑘𝑘

𝜕𝜕𝑥𝑥𝑗𝑗− 𝜇𝜇𝑜𝑜 𝜖𝜖0

𝜕𝜕𝐸𝐸𝑖𝑖

𝜕𝜕𝑡𝑡

= �𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘 𝜕𝜕𝑗𝑗𝐵𝐵𝑘𝑘 −𝜕𝜕 �𝐸𝐸𝑖𝑖 𝑐𝑐𝜕𝜕 𝑐𝑐 𝑡𝑡

= �𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘 𝜕𝜕𝑗𝑗𝐵𝐵𝑘𝑘 + 𝜕𝜕0 �−𝐸𝐸𝑖𝑖 𝑐𝑐

= 𝜇𝜇𝑜𝑜 𝐽𝐽𝑖𝑖


𝐵𝐵1 = −𝐹𝐹23 𝐵𝐵2 = −𝐹𝐹31 𝐵𝐵3 = −𝐹𝐹12

�𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘 𝜕𝜕𝑗𝑗𝐵𝐵𝑘𝑘 + 𝜕𝜕0−𝐸𝐸𝑖𝑖

𝑐𝑐= 𝜇𝜇𝑜𝑜 𝐽𝐽𝑖𝑖

−𝐸𝐸𝑖𝑖

𝑐𝑐= 𝐹𝐹0𝑖𝑖

𝑖𝑖 = 1 𝜕𝜕2𝐵𝐵3 − 𝜕𝜕3𝐵𝐵2 + 𝜕𝜕0𝐹𝐹01

= 𝜕𝜕2𝐹𝐹21 + 𝜕𝜕3𝐹𝐹31 + 𝜕𝜕0𝐹𝐹01

= �𝛼𝛼=0

3

𝜕𝜕𝛼𝛼𝐹𝐹𝛼𝛼1 = 𝜇𝜇𝑜𝑜 𝐽𝐽1

�𝛼𝛼=0

3


+𝜕𝜕1𝐹𝐹11


�𝑗𝑗=1

3

�𝑘𝑘=1

3



𝑖𝑖 = 2 𝜕𝜕3𝐵𝐵1 − 𝜕𝜕1𝐵𝐵3 + 𝜕𝜕0𝐹𝐹02

= 𝜕𝜕3𝐹𝐹32 + 𝜕𝜕1𝐹𝐹12 + 𝜕𝜕0𝐹𝐹02 +𝜕𝜕2𝐹𝐹22

= �𝛼𝛼=0

3


�𝛼𝛼=0

3


𝐵𝐵1 = −𝐹𝐹23 𝐵𝐵2 = −𝐹𝐹31 𝐵𝐵3 = −𝐹𝐹12 −𝐸𝐸𝑖𝑖



�𝑗𝑗=1

3

�𝑘𝑘=1

3



𝑖𝑖 = 3 𝜕𝜕1𝐵𝐵2 − 𝜕𝜕2𝐵𝐵1 + 𝜕𝜕0𝐹𝐹03

= 𝜕𝜕1𝐹𝐹13 + 𝜕𝜕2𝐹𝐹23 + 𝜕𝜕0𝐹𝐹03 +𝜕𝜕3𝐹𝐹33

= �𝛼𝛼=0

3


�𝛼𝛼=0

3


𝐵𝐵1 = −𝐹𝐹23 𝐵𝐵2 = −𝐹𝐹31 𝐵𝐵3 = −𝐹𝐹12 −𝐸𝐸𝑖𝑖



The Four Inhomogeneous Maxwell’s Equations

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

�𝛼𝛼=0

3

𝜕𝜕𝛼𝛼𝐹𝐹𝛼𝛼𝛽𝛽 = 𝜇𝜇0 𝐽𝐽𝛽𝛽

sourcesfields


𝜕𝜕𝛼𝛼𝐹𝐹𝛼𝛼𝛽𝛽 = 𝜇𝜇0 𝐽𝐽𝛽𝛽 𝐹𝐹𝛼𝛼𝛽𝛽 ≡ 𝜕𝜕𝛼𝛼𝐴𝐴𝛽𝛽 − 𝜕𝜕𝛽𝛽𝐴𝐴𝛼𝛼

= 𝜕𝜕𝛼𝛼 𝜕𝜕𝛼𝛼𝐴𝐴𝛽𝛽 − 𝜕𝜕𝛽𝛽𝐴𝐴𝛼𝛼

= 𝜇𝜇0 𝐽𝐽𝛽𝛽

= 𝜕𝜕𝛼𝛼𝜕𝜕𝛼𝛼𝐴𝐴𝛽𝛽 − 𝜕𝜕𝛽𝛽 𝜕𝜕𝛼𝛼𝐴𝐴𝛼𝛼𝜕𝜕𝛼𝛼 𝐹𝐹𝛼𝛼𝛽𝛽 = 𝜕𝜕𝛼𝛼𝜕𝜕𝛼𝛼𝐴𝐴𝛽𝛽

0

𝜕𝜕𝛼𝛼𝜕𝜕𝛼𝛼𝐴𝐴𝛽𝛽 = 𝜇𝜇0 𝐽𝐽𝛽𝛽

Lorenz gauge

𝜕𝜕𝛼𝛼𝜕𝜕𝛼𝛼 ≡ □2 d’Alembertian, which is an invariant


The Four Homogeneous Maxwell’s Equations

Gauss’s Law of Magnetism

Faraday’s Law

𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0

𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0


𝛁𝛁.𝑩𝑩 𝒓𝒓, 𝑡𝑡 = 0

𝐵𝐵1 = −𝐹𝐹23 𝐵𝐵2 = −𝐹𝐹31 𝐵𝐵3 = −𝐹𝐹12

�𝑖𝑖=1


𝐵𝐵𝑖𝑖 =𝜕𝜕𝐵𝐵1

𝜕𝜕𝑥𝑥1+𝜕𝜕𝐵𝐵2

𝜕𝜕𝑥𝑥2+𝜕𝜕𝐵𝐵3

𝜕𝜕𝑥𝑥3

= −𝜕𝜕𝐹𝐹23

𝜕𝜕𝑥𝑥1−𝜕𝜕𝐹𝐹31

𝜕𝜕𝑥𝑥2−𝜕𝜕𝐹𝐹12

𝜕𝜕𝑥𝑥3

= 𝜕𝜕1𝐹𝐹23 + 𝜕𝜕2𝐹𝐹31 + 𝜕𝜕3𝐹𝐹12 = 0

𝜕𝜕1𝐹𝐹23 + 𝜕𝜕2𝐹𝐹31 + 𝜕𝜕3𝐹𝐹12 = 0


𝛁𝛁 × 𝑬𝑬 𝒓𝒓, 𝑡𝑡 +𝜕𝜕𝑩𝑩 𝒓𝒓, 𝑡𝑡𝜕𝜕𝑡𝑡

= 0

�𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘𝜕𝜕 𝐸𝐸𝑘𝑘

𝑐𝑐 𝜕𝜕𝑥𝑥𝑗𝑗+𝜕𝜕𝐵𝐵𝑖𝑖

𝑐𝑐 𝜕𝜕𝑡𝑡= 0

�𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘 𝜕𝜕𝑗𝑗 𝐸𝐸𝑘𝑘/𝑐𝑐 − 𝜕𝜕0𝐵𝐵𝑖𝑖 = 0

�𝑗𝑗=1

3

�𝑘𝑘=1

3

𝜖𝜖𝑖𝑖𝑗𝑗𝑘𝑘 𝜕𝜕𝑗𝑗 𝐸𝐸𝑘𝑘/𝑐𝑐 + 𝜕𝜕0𝐵𝐵𝑖𝑖 = 0


�𝑗𝑗=1

3

�𝑘𝑘=1

3


𝐵𝐵1 = −𝐹𝐹23 𝐵𝐵2 = −𝐹𝐹31 𝐵𝐵3 = −𝐹𝐹12 𝐸𝐸𝑖𝑖


𝑖𝑖 = 1

𝜕𝜕2 𝐸𝐸3/𝑐𝑐 − 𝜕𝜕3 𝐸𝐸2/𝑐𝑐 − 𝜕𝜕0𝐵𝐵1 = 0

𝜕𝜕2𝐹𝐹30 + 𝜕𝜕3𝐹𝐹02 + 𝜕𝜕0𝐹𝐹23 = 0




𝑖𝑖 = 2

𝜕𝜕3 𝐸𝐸1/𝑐𝑐 − 𝜕𝜕1 𝐸𝐸3/𝑐𝑐 − 𝜕𝜕0𝐵𝐵2 = 0

𝜕𝜕3𝐹𝐹10 + 𝜕𝜕1𝐹𝐹03 + 𝜕𝜕0𝐹𝐹31 = 0

�𝑗𝑗=1

3

�𝑘𝑘=1

3



𝑖𝑖 = 3

𝜕𝜕1 𝐸𝐸2/𝑐𝑐 − 𝜕𝜕2 𝐸𝐸1/𝑐𝑐 − 𝜕𝜕0𝐵𝐵3 = 0

𝜕𝜕1𝐹𝐹20 + 𝜕𝜕2𝐹𝐹01 + 𝜕𝜕0𝐹𝐹12 = 0

�𝑗𝑗=1

3

�𝑘𝑘=1

3





𝜕𝜕1𝐹𝐹23 + 𝜕𝜕2𝐹𝐹31 + 𝜕𝜕3𝐹𝐹12 = 0

𝜕𝜕3𝐹𝐹01 + 𝜕𝜕0𝐹𝐹13 + 𝜕𝜕1𝐹𝐹30 = 0

𝜕𝜕0𝐹𝐹12 + 𝜕𝜕1𝐹𝐹20 + 𝜕𝜕2𝐹𝐹01 = 0

𝑖𝑖 = 1

𝑖𝑖 = 2

𝑖𝑖 = 3

𝑖𝑖 = 00

23

0 1

23

0 1

23

1

𝜕𝜕2𝐹𝐹30 + 𝜕𝜕3𝐹𝐹02 + 𝜕𝜕0𝐹𝐹23 = 0

0 1

23


𝜕𝜕𝛼𝛼𝐹𝐹𝛽𝛽𝛾𝛾 + 𝜕𝜕𝛽𝛽𝐹𝐹𝛾𝛾𝛼𝛼 + 𝜕𝜕𝛾𝛾𝐹𝐹𝛼𝛼𝛽𝛽 = 0

The Four Homogeneous Maxwell’s Equations

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

can be written as:

0 1

23𝜀𝜀𝛿𝛿𝛼𝛼𝛽𝛽𝛾𝛾 𝜕𝜕𝛼𝛼𝐹𝐹𝛽𝛽𝛾𝛾 = 0

or 𝜀𝜀𝛿𝛿𝛼𝛼𝛽𝛽𝛾𝛾


𝜕𝜕𝛼𝛼𝐹𝐹𝛼𝛼𝛽𝛽 = 𝜇𝜇0 𝐽𝐽𝛽𝛽

𝜕𝜕𝛼𝛼𝐹𝐹𝛽𝛽𝛾𝛾 + 𝜕𝜕𝛽𝛽𝐹𝐹𝛾𝛾𝛼𝛼 + 𝜕𝜕𝛾𝛾𝐹𝐹𝛼𝛼𝛽𝛽 = 0


0 −𝐸𝐸1𝐸𝐸1 0




𝐽𝐽𝜇𝜇 =


𝐴𝐴𝜇𝜇 =


Electromagnetic Theory


A Few Notes on Relativistic Mechanics

and Field TheoryChapter 2 of “Modern Problems in Classical

Electrodynamics” by Charles Brau


𝒮𝒮 = �𝑡𝑡1

𝑡𝑡2𝐿𝐿 𝑞𝑞𝑖𝑖 , 𝑞𝑞𝑖𝑖 𝑑𝑑𝑡𝑡

𝐿𝐿 = �𝑖𝑖

𝐿𝐿𝑖𝑖 𝑞𝑞𝑖𝑖 , 𝑞𝑞𝑖𝑖

𝑑𝑑𝑑𝑑𝑡𝑡

𝜕𝜕𝐿𝐿𝜕𝜕 𝑞𝑞𝑖𝑖

=𝜕𝜕𝐿𝐿𝜕𝜕𝑞𝑞𝑖𝑖

Lagrangian of Discrete Particles



𝑡𝑡2𝐿𝐿 𝑞𝑞𝑖𝑖 , 𝑞𝑞𝑖𝑖 𝑑𝑑𝑡𝑡

𝐿𝐿 = �𝑖𝑖

𝐿𝐿𝑖𝑖 𝑞𝑞𝑖𝑖 , 𝑞𝑞𝑖𝑖 𝐿𝐿 = �ℒ 𝜙𝜙𝑘𝑘,𝜕𝜕𝛽𝛽𝜙𝜙𝑘𝑘 𝑑𝑑𝑉𝑉

𝑑𝑑𝑑𝑑𝑡𝑡


=𝜕𝜕𝐿𝐿𝜕𝜕𝑞𝑞𝑖𝑖

𝜕𝜕𝛽𝛽𝜕𝜕ℒ

𝜕𝜕 𝜕𝜕𝛽𝛽𝜙𝜙𝑘𝑘=

𝜕𝜕ℒ𝜕𝜕𝜙𝜙𝑘𝑘


𝑡𝑡2𝑑𝑑𝑡𝑡 �ℒ 𝜙𝜙𝑘𝑘 ,𝜕𝜕𝛽𝛽𝜙𝜙𝑘𝑘 𝑑𝑑𝑉𝑉

=1𝑐𝑐�𝑡𝑡1

𝑡𝑡2ℒ 𝜙𝜙𝑘𝑘,𝜕𝜕𝛽𝛽𝜙𝜙𝑘𝑘 𝑑𝑑4𝑥𝑥

Lagrangian of Continuous Fields and Particles: 𝜙𝜙𝑘𝑘: 𝐴𝐴𝜇𝜇 & 𝐽𝐽𝜇𝜇


ℒ = −1

4 𝜇𝜇0𝐹𝐹𝜇𝜇𝜈𝜈 𝐹𝐹𝜇𝜇𝜈𝜈 − 𝐽𝐽𝜇𝜇 𝐴𝐴𝜇𝜇

𝜕𝜕ℒ𝜕𝜕𝐴𝐴𝛼𝛼

= −𝐽𝐽𝛼𝛼

𝜕𝜕ℒ𝜕𝜕 𝜕𝜕𝛽𝛽𝐴𝐴𝛼𝛼

= −1𝜇𝜇0𝐹𝐹𝛽𝛽𝛼𝛼

𝜕𝜕𝛽𝛽𝜕𝜕ℒ𝜕𝜕𝛽𝛽𝐴𝐴𝛼𝛼

= −1𝜇𝜇0𝜕𝜕𝛽𝛽𝐹𝐹𝛽𝛽𝛼𝛼

𝜕𝜕𝛽𝛽𝜕𝜕ℒ

𝜕𝜕 𝜕𝜕𝛽𝛽𝐴𝐴𝛼𝛼=

𝜕𝜕ℒ𝜕𝜕𝐴𝐴𝛼𝛼

𝜕𝜕𝛽𝛽𝐹𝐹𝛽𝛽𝛼𝛼 = 𝜇𝜇0 𝐽𝐽𝛼𝛼

HW:

Relativistic Lagrangianof Fields and Particles


𝐻𝐻 = �𝑖𝑖

𝑝𝑝𝑖𝑖 𝑞𝑞𝑖𝑖 − 𝐿𝐿𝑖𝑖 𝑞𝑞𝑖𝑖 , 𝑞𝑞𝑖𝑖

𝑝𝑝𝑖𝑖 ≡𝜕𝜕𝐿𝐿𝜕𝜕 𝑞𝑞𝑖𝑖

Hamiltonian of Particles


Relativistic Hamiltonian of Fields and Particles

𝐻𝐻 = �𝑖𝑖


𝑞𝑞𝑖𝑖 − 𝐿𝐿𝑖𝑖 𝑞𝑞𝑖𝑖 , 𝑞𝑞𝑖𝑖

ℋ𝛼𝛼𝛽𝛽 =

𝜕𝜕ℒ𝜕𝜕 𝜕𝜕𝛼𝛼𝐴𝐴𝛾𝛾

𝜕𝜕𝛽𝛽𝐴𝐴𝛾𝛾 − 𝑔𝑔𝛼𝛼𝛽𝛽 ℒ

= −1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝜕𝜕𝛽𝛽𝐴𝐴𝛾𝛾 − 𝑔𝑔𝛼𝛼

𝛽𝛽 −1



= −1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝐹𝐹𝛽𝛽𝛾𝛾 + 𝜕𝜕𝛾𝛾𝐴𝐴𝛽𝛽 +

14 𝜇𝜇0

𝐹𝐹𝜇𝜇𝜈𝜈 𝐹𝐹𝜇𝜇𝜈𝜈𝑔𝑔𝛼𝛼𝛽𝛽 + 𝐽𝐽𝜇𝜇 𝐴𝐴𝜇𝜇𝑔𝑔𝛼𝛼

𝛽𝛽

ℋ𝛼𝛼𝛽𝛽 = −

1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝜕𝜕𝛽𝛽𝐴𝐴𝛾𝛾 − 𝑔𝑔𝛼𝛼

𝛽𝛽 −1


= −1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝐹𝐹𝛽𝛽𝛾𝛾 +

14 𝜇𝜇0

𝐹𝐹𝜇𝜇𝜈𝜈 𝐹𝐹𝜇𝜇𝜈𝜈𝑔𝑔𝛼𝛼𝛽𝛽 −

1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝜕𝜕𝛾𝛾𝐴𝐴𝛽𝛽 + 𝐽𝐽𝜇𝜇 𝐴𝐴𝜇𝜇𝑔𝑔𝛼𝛼

𝛽𝛽

= 𝑇𝑇𝛼𝛼𝛽𝛽 −

1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝜕𝜕𝛾𝛾𝐴𝐴𝛽𝛽 + 𝐽𝐽𝜇𝜇 𝐴𝐴𝜇𝜇𝑔𝑔𝛼𝛼

𝛽𝛽

𝑇𝑇𝛼𝛼𝛽𝛽 ≡ −

1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝐹𝐹𝛽𝛽𝛾𝛾 +

14 𝜇𝜇0

𝐹𝐹𝜇𝜇𝜈𝜈 𝐹𝐹𝜇𝜇𝜈𝜈𝑔𝑔𝛼𝛼𝛽𝛽


𝑇𝑇𝛼𝛼𝛽𝛽 ≡ −

1𝜇𝜇0𝐹𝐹𝛼𝛼𝛾𝛾 𝐹𝐹𝛽𝛽𝛾𝛾 +

14 𝜇𝜇0

𝐹𝐹𝜇𝜇𝜈𝜈 𝐹𝐹𝜇𝜇𝜈𝜈𝑔𝑔𝛼𝛼𝛽𝛽

𝑇𝑇𝛼𝛼𝛽𝛽 =

𝑢𝑢𝑐𝑐𝑔𝑔𝑥𝑥𝑐𝑐𝑔𝑔𝑦𝑦𝑐𝑐𝑔𝑔𝑧𝑧

𝑐𝑐𝑔𝑔𝑥𝑥𝑇𝑇𝑥𝑥𝑥𝑥𝑇𝑇𝑦𝑦𝑥𝑥𝑇𝑇𝑧𝑧𝑥𝑥

𝑐𝑐𝑔𝑔𝑦𝑦𝑇𝑇𝑥𝑥𝑦𝑦𝑇𝑇𝑦𝑦𝑦𝑦𝑇𝑇𝑧𝑧𝑦𝑦

𝑐𝑐𝑔𝑔𝑧𝑧𝑇𝑇𝑥𝑥𝑧𝑧𝑇𝑇𝑦𝑦𝑧𝑧𝑇𝑇𝑧𝑧𝑧𝑧

𝛽𝛽 = 0 𝜕𝜕𝛼𝛼𝑇𝑇𝛼𝛼0 = 𝜕𝜕0 𝑢𝑢+𝜕𝜕1 𝑐𝑐𝑔𝑔𝑥𝑥 +𝜕𝜕2 𝑐𝑐𝑔𝑔𝑦𝑦 +𝜕𝜕3 𝑐𝑐𝑔𝑔𝑧𝑧 =1𝑐𝑐𝜕𝜕𝑢𝑢𝜕𝜕𝑡𝑡

+𝛁𝛁.𝑺𝑺

𝛽𝛽 = 𝑖𝑖 𝜕𝜕𝛼𝛼𝑇𝑇𝛼𝛼𝑖𝑖 = 𝜕𝜕0 𝑐𝑐𝑔𝑔𝑖𝑖 + 𝜕𝜕𝑗𝑗𝑇𝑇𝑗𝑗𝑖𝑖

chapter 1 of “modern problems in classical electrodynamics ... 611 spring... · chapter 1 of...

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