Physics 3 for Electrical Engineering Ben Gurion University of the Negev Week 3. Special relativity – transformation of the E and B fields the field tensor F  electromagnetic field of a charge moving at constant velocity Sources: Feynman Lectures II, Chap. 13 Sect. 6, Chap. 25, and Chap. 26, Sects. 2-3; E. M. Purcell, Electricity and Magnetism, Berkeley Physics Course Vol. 2, Second Edition, 1985, Sect. 6.7 Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin

Transformation of the E and B fields Now that we have derived the relativistic transformation laws for space, time, velocity, momentum and energy, let ’ s look for the transformation laws for E and B. These fields are determined by Maxwell ’ s equations, and Maxwell ’ s equations already obey Einstein ’ s postulate that the speed of light is a universal constant. Hence the relativistic transformation laws for E and B should not present any problem. But how do we discover them?

Transformation of the E and B fields Now that we have derived the relativistic transformation laws for space, time, velocity, momentum and energy, let ’ s look for the transformation laws for E and B. These fields are determined by Maxwell ’ s equations, and Maxwell ’ s equations already obey Einstein ’ s postulate that the speed of light is a universal constant. Hence the relativistic transformation laws for E and B should not present any problem. But how do we discover them? No progress without a paradox! – John A. Wheeler

A paradox: A current I = ρ – A v runs in a straight wire, where A is the cross-sectional area of the wire and ρ – < 0 is the (negative) charge density of the electrons. A particle of negative charge q moves parallel to the wire, with the same velocity v. Since the charged particle moves in the magnetic field of the current, it feels the Lorentz force F L = q v × B. FLFL B I → v

B A paradox: But in the rest frame of the particle, it feels no Lorentz force! I →

A paradox: Let ’ s go back to the rest frame of the wire. Let ρ + > 0 denote the (positive) charge density of the protons. We will assume |ρ – | = ρ + and therefore there is no Coulomb force F C in this frame. F C = 0 B I → v

A paradox: Calculation: in the rest frame of the wire, the field is B = I / 2π rc 2 ε 0 = ρ + A v / 2π rc 2 ε 0, where r is the distance from the charged particle to the axis of the wire. Thus the Lorentz force is F L = qvB = qρ + A v 2 / 2π rc 2 ε 0. F B I → FLFL v

B A paradox: In the rest frame of the particle, there is no Lorentz force, but there is instead a Coulomb force due to unbalanced charge concentrations ρ + ' > 0 and ρ – ' < 0: I → FC'FC'

B A paradox: In the rest frame of the particle, there is no Lorentz force, but there is instead a Coulomb force of magnitude I → FC'FC'

B A paradox: Actually, what has to be equal is not F L and F C ' but the change in transverse momentum F L Δt and F C 'Δt'. Since Δt = γΔt ‘ we indeed find I → FC'FC'

We have learned an important lesson: If we want to understand how E and B transform from one inertial reference frame to another, we should look at how the charge and current densities transform.

K So let ’ s consider two infinite, parallel sheets of uniform surface charge densities σ and –σ (as measured in the inertial frame K), moving with speed v 0 in the positive x-direction. We have E x = 0 = E z while E y = σ/ε 0 and B x = 0 = B y while B z = σ v 0 /c 2 ε 0 (between the parallel sheets).

Now in an inertial frame K ′, moving with speed v in the positive x-direction relative to K, the speed of the sheets v 0 ′ is v K K′K′

In K′ the magnitude of the surface charge density is σ′ = σ γ 0 ′/γ 0 = σ γ(1 – v 0 v/c 2 ), where v K K′K′

Some algebra:

In K′ the surface charge density is σ ′ = σ γ(1 – v 0 v/c 2 ) and the surface current density is σ′v 0 ′ which is v K K′K′

We now use the transformed surface charge density and surface current density to calculate the transformed fields: By rotating these equations –π/2 around the x-axis, we get two more relations:

Finally, the components of E and B parallel to the boost axis do not change: How do we know? A boost perpendicular to the plates only changes the separation of the plates – that does not change the electric field. A boost along the axis of a solenoid multiplies the density of the wires around the solenoid by a factor γ but reduces the current in each by a factor of γ (time dilation) so it does not change the magnetic field.

Summary: E ║ ′ = E ║ E ┴ ′ = γ (E ┴ + v × B ┴ ) B ║ ′ = B ║ B ┴ ′ = γ (B ┴ – v × E ┴ /c 2 )

The field tensor F  (Here we take c = 1.) We can write the electromagnetic field as a tensor F: If L is any Lorentz transformation in matrix form, then F ’ = LFL T.

For example, if L is then F ’ is

Electromagnetic field of a charge moving at constant velocity The field of an electric charge in its rest frame is ; if we boost it to a frame K′ moving in the positive x-direction, it is the electromagnetic field of a charge moving at constant v. [J. D. Jackson, Classical Electrodynamics, Sect :] Consider an observer at the point P = (0,b,0) in the frame K: y x′xx′x z z′z′ y′y′

The origins coincide at time t = 0 = t′. Coordinates of P in reference frame K′: (–vt′,b,0) Distance of P from charge: Components of E′: Components of B′: B′ = 0 y x′xx′x z z′z′ y′y′

The nonvanishing field components in K′ are y x′xx′x z z′z′ y′y′ and in K they are

We can also freeze t and vary b to map out the field at a given time t. From the figure we see that r sin ψ = b, r cos ψ = –vt, which allows us to write y x′xx′x z z′z′ y′y′

v Lines of force for v = 0