Physics 3 for Electrical Engineering

1. Ben Gurion University of the Negev www.bgu.ac.il/atomchip,www.bgu.ac.il/nanocenter Physics 3 for Electrical Engineering Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin Week 3. Special relativity –transformation of the E and B fields• the field tensor Fmn• 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

2. 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?

3. 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

4. B 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 FL = qv×B . I → FL v

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

6. B 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 FC in this frame. I → FC = 0 v

7. B A paradox: Calculation: in the rest frame of the wire, the field is B = I / 2πrc2ε0 = ρ+ A v / 2πrc2ε0 , where r is the distance from the charged particle to the axis of the wire. Thus the Lorentz force is FL = qvB = qρ+ A v2/ 2πrc2ε0 . I → F FL v

8. 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'

9. 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'

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

11. 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.

12. 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 v0 in the positive x-direction. We have Ex = 0 = Ez while Ey = σ/ε0 and Bx = 0 = By while Bz = σv0/c2ε0 (between the parallel sheets). K

13. v K′ K Now in an inertial frame K′, moving with speed v in the positive x-direction relative to K, the speed of the sheets v0′ is

14. v K′ K In K′ the magnitude of the surface charge density is σ′ = σγ0′/γ0 = σγ(1 – v0v/c2 ), where

15. Some algebra:

16. v K′ K In K′ the surface charge density is σ′= σγ(1 – v0v/c2 ) and the surface current density is σ′v0′ which is

17. 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:

18. 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.

19. Summary: E║′ = E║ E┴′ = γ (E┴ + v × B┴ ) B║′ = B║ B┴′ = γ (B┴ – v × E┴ /c2 )

20. The field tensor Fmn (Here we take c = 1.) We can write the electromagnetic field as a tensorF: If L is any Lorentz transformation in matrix form, then F’ = LFLT.

21. For example, if L is then F’ is

22. 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. 11.10:] Consider an observer at the point P = (0,b,0) in the frame K: y y′ x′ x z′ z

23. 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 y′ x′ x z′ z

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

25. 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 y′ x′ x z′ z

26. v Lines of force for v = 0