Electromagnetism Review Sheet

Michael Li May 20, 2016

Electromagnetism and Relativity Review of Special Relativity; tensors and index notation. Lorentz force law. Electromagnetic tensor. Lorentz transformations of electric and magnetic elds. Currents and the conservation of charge. Maxwell equations in relativistic and non-relativistic forms. [5]

Electrostatics Gauss’s law. Application to spherically symmetric and cylindrically symmetric charge distributions. Point, line and surface charges. Electrostatic potentials; general charge distributions, dipoles. Electrostatic energy. Conductors. [3]

Magnetostatics Magnetic elds due to steady currents. Ampre’s law. Simple examples. Vector potentials and the Biot-Savart law for general current distributions. Magnetic dipoles. Lorentz force on current distributions and force between current-carrying wires. Ohm’s law. [3]

Electrodynamics Faraday’s law of induction for xed and moving circuits. Electromagnetic energy and Poynting vector. 4-vector potential, gauge transformations. Plane electromagnetic waves in vacuum, polarization. [5]

1 Contents

Contents 2

1 Introduction 3

2 Electrostatics 3 2.1 Gauss’ Law ...... 3 2.2 Electrostatic Potential ...... 4 2.3 Field Lines and Equipotentials ...... 5 2.4 Electrostatic Energy ...... 6 2.5 Conductors, or the Image Charge Method ...... 6

3 Magnetostatics 7 3.1 Ampere’s Law ...... 8 3.2 Vector Potential ...... 8 3.3 Magnetic Dipoles ...... 9 3.4 Magnetic Force Examples ...... 10

4 Electrodynamics 11 4.1 Induction ...... 11 4.2 Magnetostatic Energy ...... 13 4.3 Resistance ...... 13 4.4 Displacement Currents ...... 14 4.5 EM Waves. Let there be light!TM ...... 15 4.6 Poynting Vector ...... 16

5 Electromagnetism and Relativity 17 5.1 Vectors and Covectors ...... 17 5.1.1 Lorentz Transformations ...... 17 5.2 Vectors and Covectors ...... 18 5.3 Conserved Currents ...... 18 5.4 Gauge potentials and EM Fields ...... 19 5.4.1 Lorenz Invariants ...... 21 5.5 Maxwell Equations ...... 21 5.6 e Lorentz Force Law ...... 21

2 1 Introduction

Denition (Charge and Current). We dene the charge density, ρ(x, t) as the charge per unit volume, and the current density J = ρv, where v is the velocity. en we dene the charge Q and current I as followed: Z Z I = J · dS Q = ρ(x, t) dV S V Now we would introduce some fundmanetal laws of nature: Law (Continuity equation). ∂ρ dQ Z + ∇ · J = 0 = − J · dS. ∂t dt S Law (Lorentz force law). F = q(E + v × B) while the second aspect is governed by Maxwell’s equations. Law (Maxwell’s Equations). ρ ∇ · E = ∇ · B = 0 ε0 ∂B ∂E ∇ × E + = 0 ∇ × B − µ ε = µ J ∂t 0 0 ∂t 0 −12 −3 −1 2 2 −6 where 0 = 8.85×10 m kg s C is the electric constant and µ0 = 4π∗10 m· kg · C−2 is the magnetic constant.

2 Electrostatics

is is basically the maxwell equations taking B = 0 and J = 0. e rst one we would tackle is ∇ · = ρ : E 0

2.1 Gauss’ Law If we integrate the maxwell equation over a surface V , we obtain Gauss’ law using the divergence theorem: Law (Gauss’ law). Z Q E · dS = , S ε0 where Q is the total charge inside V . Now we dene the term on the le:

Denition (Flux through surface). e ux of E through the surface S is dened to be Z E · dS. S

3 e only important thing in calculating the electric eld for these surfaces is to use the right symmetry and pick the right surface. We would list the results of some common examples: Sphere Using spherical symmetry and picking a radius r from a ρ uniform charge R- radius sphere, we have:

( Q 2 ˆr r > R 4πε0r E = Qr 3 ˆr r ≤ R 4πε0R

Innite line Using circular symmetry, where r is now the r in cylindrical coordinates, we take a cylinder around the line and reach that: η E(r) = ˆr. 2πε0r Innite Plane Now we have vertical symmetry, and we can take a box that is symmetric about the plane to get E(z): σ E(z) = 2ε0 Where it point upwards above the plane and downwards below.

2.2 Electrostatic Potential From ∇ × E = 0, we know that E = −∇φ for some electrostatic potential φ. Substi- tuting into ∇ · E = ρ , we have: 0 ρ ∇2φ = . ε0 We now list some examples: Point Charge is gives ρ = Qδ3(r), where δ3 is the 3D delta function. Now we can solve this to give φ = Q as from 1A Vector Calculus we know the solution is some 4π0r 1 multiple of r . Dierentiating it gives Coulomb’s law. Dipole We have: 1 Q Q φ = − . 4πε0 r |r + d| From superposition of point charges. We taylor expand |R + d to be: 1 1 1 1 1 = − d · ∇ + (d · ∇)2 + ··· |r + d| r r 2 r 1 d · r 1 d · d 3(d · r)2 = − − − + ··· . r r3 2 r3 r5 en this gives: Q 1 1 d · r Q d · r φ = − + 3 + ··· ∼ 3 . 4πε0 r r r 4πε0 r If we dene the electric dipole moment to be p = Qd, then we have: 1 3(p · ˆr)ˆr − p E = −∇φ = 3 . 4πε0 r

4 General To nd φ for a general charge distribution ρ, we use the Green’s function for the Laplacian. e Green’s function is dened to be the solution to 1 1 ∇2G(r, r0) = δ3(r − r0), or G(r, r0) = − . 4π |r − r0| We assume all charge is contained in some compact region V . en 1 Z φ(r) = − ρ(r0)G(r, r0) d3r0 ε0 V Z 0 1 ρ(r ) 3 0 = 0 d r 4πε0 V |r − r | en Z 1 0 1 3 0 E(r) = −∇φ(r) = − ρ(r )∇ 0 d r 4πε0 V |r − r | Z 0 1 0 (r − r ) 3 0 = ρ(r ) 0 3 d r 4πε0 V |r − r | Now, to make it simple, we can ask what φ and E look like very far from V , ie. |r| |r0| using taylor expansion: 1 1 r · r0 = + + ··· . |r − r0| r r3 en we get Z 0 1 0 1 r · r 3 0 φ(r) = ρ(r ) + 3 + ··· d r 4πε0 V r r 1 Q p · ˆr = + 2 + ··· , 4πε0 r r where Z Z r Q = ρ(r0) dV 0 p = r0ρ(r0) dV 0 ˆr = . V V krk So if we have a huge lump of charge, we can consider it to be a point charge Q, plus some dipole correction terms when we are far away from the charge.

2.3 Field Lines and Equipotentials Let’s start from some denitions: Denition (Field line). A eld line is a continuous line tangent to the electric eld E. e density of lines is proportional to |E|. Denition (Equipotentials). Equipotentials are surfaces of constant φ. Because E = −∇φ, they are always perpendicular to eld lines. if we indicate eld lines using straight lines and equipotentials using dashed ones, we have:

+ -

5 2.4 Electrostatic Energy If we want to calculate the amount of energy stored (potential energy), then it is the sum of work done to bring the charges from innity. Since we can bring each charge one by one, and between two point charges i and j, the work done is:

qiqj 1 4π0 |ri − rj e total potential energy is:

1 1 X qiqj 1 1 X qiqj U = = . 4πε0 2 |ri − rj| 4πε0 2 |ri − rj| i6=j i6=j

Since the potential at each point is due to all other points:

1 X qj φ(ri) = . 4πε0 |ri − rj| j6=i

We can write N 1 X U = q φ(r ). 2 i i i=1 us, for continuous charge distributions, we have: 1 Z ε Z U = ρ(r)φ(r) d3r = 0 (∇ · E)φ d3r 2 2 ε Z = 0 [∇ · (Eφ) − E · ∇φ] d3r 2 ε Z = 0 E · E d3r. 2

R ∇(Eφ) vanishes as we assume φ → 0 as we go to innity.

2.5 Conductors, or the Image Charge Method Denition (Conductor). A conductor is a region of space which contains lots of charges that are free to move. In electrostatic situations, we must have E = 0 inside a conductor. Why? Since if there is an electric eld, charges move around to balance it. And aer charges has balanced the eld, there is no eld! erefore all charges in a conductor must be on the surface. To consider what happens when a charge comes near to a surface, we consider an example: Example. Suppose we have a conductor that lls all space x < 0. We ground it such that φ = 0 throughout the conductor. en we place a charge q at x = d > 0.

φ = 0 +

6 We look for a potential that has a source at x = d and satises φ = 0 for x < 0. Since the solution to the Poisson equation is unique, we can use the method of images, which is to pretend that we don’t have a conductor. Instead, we have a charge −q and x = −d. en by symmetry we will get φ = 0 when x = 0. e potential of this pair is " # 1 q q φ = p − p . 4πε0 (x − d)2 + y2 + z2 (x + d)2 + y2 + z2 To get the solution we want, we “steal” part of this potential and declare our potential to be   1 √ q − √ q if x > 0 φ = 4πε0 (x−d)2+y2+z2 (x+d)2+y2+z2 0 if x ≤ 0 Using this solution, we can immediately see that it satises Poisson’s equations both outside and inside the conductor. Now we calculate the surface charge induced. We can calculate the electric eld near the surface, and use the relation σ = Eoutside · nˆ. To nd σ, we only need the component of E in the x direction: ∂φ q x − d x + d E = − = − x 3/2 3/2 ∂x 4πε0 |r − d| |r + d|

for x > 0. en induced surface charge density is given by Ex at x = 0: q d σ = E ε = − . x 0 2π (d2 + y2 + z2)3/2 e total surface charge is then given by Z σ dy dz = −q.

Which is expected. Note. Here are some handy tips for handling other situations: Sphere Image Charge For a grounded sphere with radius R and a charge at (d, 0, 0), we should place R2 the image charge at x = d . Conductor in Field For a spherical conductor in a constant electric eld, we orient the electric eld in the z-axis to give φ0 = −E0r cos θ and add a dipole term to the potential, giving: R3 φ = −E r − cos θ 0 r2 Note. For people learning uids, if this looks like the solution of a sphere pass- ing through a uid, because this is! is is solving the Laplace equation with φ = 0 on the spherical boundary.

3 Magnetostatics

Now similarly, we look at E = 0. en we are le with: ∇ × B = µJ ∇ · B = 0

7 3.1 Ampere’s Law We use Stokes’ Law to integrate the rst equation to get:

Law (Ampere’s law). I B · dr = µ0I, C where I is the current through the surface.

Similarly, we have the classic examples: Long Straight Wire We integrate over a disc to nd: µ I B(r) = 0 ϕˆ. 2πr

Innite surface Now if we have constant x direction current on an innite plane with surface current density k, then we must have y direction magnetic eld. So we choose a loop through the surface to nd: µ k B(z) = 0 for z > 0. 2 Similarly, the direction reverses for above/below the plane.

Solenoid Imagine wrapping N wires per unit length with each wire having current I. en we integrate over a loop that passes the cylinder, as the magnetic eld must be in the z direction inside the cylinder:

b = µ0INz.

inside the cylinder. Outside the cylinder, B is constant, so it is 0.

3.2 Vector Potential Denition (Vector potential). If B = ∇ × A, then A is a vector potential. is is NOT unique! If A is a vector potential, then A0 = A + ∇χ is another one. But we do have some canonical choice: Denition ((Coulomb) gauge). Each choice of A is called a gauge. An A such that ∇ · A = 0 is called a Coulomb gauge.

Proposition. We can always pick χ such that ∇ · A0 = 0. Proof. Suppose that B = ∇ × A with ∇ · A = ψ(x). en for any A0 = A + ∇χ, we have ∇ · A0 = ∇A + ∇2χ = ψ + ∇2χ. So we need a χ such that ∇2χ = −ψ. is is the Poisson equation which we know that there is always a solution by, say, the Green’s function. Hence we can nd a χ that works.

8 Now the other maxwell equation says:

2 2 ∇ × B = −∇ A + ∇(∇ · A) = −∇ A = µ0J. (∗)

Which we can solve this using Green functions to get

µ Z J(r0) A(r) = 0 dV 0, 4π |r − r0|

Note this is integrating over r0, not r. Fortunately, this does satisfy the Coulomb gauge:

µ Z 1 ∇ · A(r) = 0 J(r0) · ∇ dV 0 4π |r − r0| µ Z 1 = − 0 J(r0) · ∇0 dV 0 4π |r − r0|

Here we employed a clever trick — dierentiating 1/|r − r0| with respect to r is the negative of dierentiating it with respect to r0. Now that we are dierentiating against r0, we can integrate by parts to obtain

µ Z J(r0) ∇0 · J(r0) = − 0 ∇0 · − dV 0. 4π |r − r0| |r − r0|

e rst term vanish due to being a total derivative, and the second term vanish as ∇ · J = 0 from the continuity equation, as the current is steady. Now we have the famous Biot-Savart law: Law (Biot-Savart law). e magnetic eld is

µ Z r − r0 B(r) = ∇ × A = 0 J(r0) × dV 0. 4π |r − r0|3

If the current is localized on a curve, this becomes I 0 µ0I 0 r − r B = dr × 0 3 , 4π C |r − r | since J(r0) is non-zero only on the curve.

3.3 Magnetic Dipoles According to Maxwell’s equations, magnetic monopoles don’t exist. However, it turns out that a localized current looks like a dipole from far far away. Example (Current loop). Take a current loop of wire C, radius R and current I.

9 Based on the elds generated by a straight wire, we can guess that B looks like this, but we want to calculate it. By the Biot-Savart law, we know that

Z 0 I 0 µ0 J(r ) µ0I dr A(r) = 0 dV = 0 . 4π |r − r | 4π C |r − r | Far from the loop, |r − r0| is small and we can use the Taylor expansion.

I 0 µ0I 1 r · r 0 A(r) = + 3 + ··· dr . 4π C r r

1 Note that r is a constant of the integral, and we can take it out. e rst r term vanishes because it is a constant, and when we integrate along a closed loop, we get 0. So we only consider the second term. Following from green’s theorem, we have: I Z g · r0 dr0 = g × dS = S × g. C S Using this, we have µ m × r A(r) ≈ 0 , 4π r3 where Denition (Magnetic dipole moment). e magnetic dipole moment is

m = IS.

en µ 3(m · ˆr)ˆr − m B = ∇ × A = 0 . 4π r3 is is the same form as E for an electric dipole Similar to electrostatics, we can do this to a general distribution to get:

Denition (Magnetic dipole moment). 1 Z m = r0 × J(r0) dV 0. 2

3.4 Magnetic Force Examples Now we list two classic examples to help understand the magnetic araction/repulsion: Example (Two parallel wires).

z y x

10 We know that the eld produced by each current is µ I B = 0 ϕ.ˆ 1 2πr e particles on the second wire will feel a force µ I F = qv × B = qv × 0 1 yˆ. 1 2πd

But J2 = nqv and I2 = J2A, where n is the density of particles and A is the cross- sectional area of the wire. So the number of particles per unit length is nA, and the force per unit length is µ I I I I F = naF = 0 1 2 zˆ × yˆ = −µ 1 2 xˆ. 2πd 0 2πd

So if I1I2 > 0, ie. the currents are in the same direction, the force is aractive. Otherwise the force is repulsive. Example (General force). In the general case, following the same procedure, we have, using Biot-Savart’s law: I I µ0 r2 − r1 F = I1I2 dr2 × dr1 × 3 . 4π C1 C2 |r2 − r1|

For well-separated currents, approximated by m1 and m2, we claim that the force can be wrien as µ 3(m · ˆr)(m · ˆr) − (m · m ) F = 0 ∇ 1 2 1 2 , 4π r3 No, we are not going to prove this. It is a complicated mess of indices.

4 Electrodynamics

Now we look at elds that do something slightly more interesting. ey change with time!

4.1 Induction If we integrate the Maxwell equation below ∂B ∇ × E + = 0. ∂t rough a surface, use Stokes’ theorem, then we have: Z d Z E · dr = − B · dS. C dt S Denition (Electromotive force (emf)). e electromotive force (emf) is Z E = E · dr. C

11 Note. is is NOT a force! is is really the voltage of the system, or the work done on a unit charge moving along the curve. Hey, don’t look at me. I didn’t invent the denition. R Denition (Magnetic ux and Faraday’s Law). e magnetic ux is Φ = S B · dS., so the equation above becomes Faraday’s Law of induction: dΦ E = − . dt Note. Minus is natural. at means the magnetic eld generated by the current induced opposes the original magnetic eld, which kinds of prevent run o. Or the world exploding. Also this formula still works in the most general case, when the curve is not constant: Consider a moving loop C(t) bounding a surface S(t). As the curve moves, the curve sweeps out a cylinder Sc.

S(t + δt)

Sc S(t)

e change in ux is Z Z Φ(t + δt) − Φ(t) = B(t + δt) · dS − B(t) · dS S(t+δt) S(t) Z ∂B Z = B(t) + · dS − B(t) · dS + O(δt2) S(t+δt) ∂t S(t) " # Z ∂B Z Z = δt · dS + − B(t) · dS + O(δt2) S(t) ∂t S(t+δt) S(t)

We know that S(t + δt), S(t) and Sc together form a closed surface. Since ∇ · B = 0, the divergence of B is 0, so by the Divergence theorem the integral of B over a closed surface is 0. So we obtain "Z Z # Z − B(t) · dS + B(t) · dS = 0. S(t+δt) S(t) Sc

Hence we have Z ∂B Z Φ(t + δt) − Φ(t) = δt · dS − B(t) · dS S(t) ∂t Sc

We can simplify the integral over Sc by writing the surface element as

dS = (dr × v) δt.

en B · dS = δt(v × B) · dr. So dΦ δΦ Z ∂B Z = lim = · dS − (v × B) · dr. dt δ→0 δt S(t) ∂t C(t)

12 ∂B From Maxwell’s equation, we know that ∂t = −∇ × E. So we have dΦ Z = − (E + v × B) dr. dt C Now dening the emf properly we have: ∂Φ E = − ∂t

4.2 Magnetostatic Energy is work done is identied with the energy stored in the system. Recall that the vector potential A is given by B = ∇ × A. So 1 Z 1 Z 1 I U = I B · dS = I (∇ × A) · dS = I A · dr 2 S 2 S 2 C 1 Z = J · A dV 2 3 R

Using Maxwell’s equation ∇ × B = µ0J, we obtain

1 Z = (∇ × B) · A dV 2µ0 1 Z = [∇ · (B × A) + B · (∇ × A)] dV 2µ0

Assuming that B × A vanishes suciently fast at innity, the integral vanishes. So we are le with 1 Z = B · B dV. 2µ0 us, in general, the energy stored in E and B is

Z ε 1 U = 0 E · E + B · B dV. 2 2µ0 Note this kind of is hand-waving, because we don’t know if weird stu is happening when E and B are together. Turns out there isn’t, but still, be careful.

4.3 Resistance What is the most important law in high school electrical physics? Law (Ohm’s law). E = IR or J = σE. Where R is dened as the resistance. Denition (Resistivity and conductivity). For the wire of length L and cross-sectional AR 1 area A, we dene the resistivity to be ρ = L , and the conductivity σ = ρ . Let’s do a classic high school example:

13 Example.

z y d x

Suppose the bar moves to the le with speed v. Suppose that the sliding bar has resistance R, and everything else has no resistance. You can see we assume this so that the resistance does not change with time. ere are two dynamical variables, the position of the bar x(t), and the current I(t). If a current I = qv ows, using the Lorentz force formula, the force on the bar is

F = IB`xˆ.

So mx¨ = IB`. We can compute the current using dΦ E = − = −B`x.˙ dt So Ohm’s law gives IR = −B`x.˙ Hence B2`2 mx¨ = − x.˙ R Integrating once gives 2 2 x˙(t) = −ve−B ` t/mR.

With resistance, we need to do work to keep a constant current. In time δt, the work needed is δW = EIδt = I2Rδt using Ohm’s law. So Denition (Joule heating). Joule heating is the energy lost in a circuit due to friction. It is given by dW = I2R. dt

4.4 Displacement Currents Let’s have a quick note about the last Maxwell equation:

∂E ∇ × B = µ J + ε 0 0 ∂t

∂E Now what is µ0ε0 ∂t ? is was really discovered mathematically to make the equations consistent with charge conservation and is called a displacement current.

14 4.5 EM Waves. Let there be light!TM Now we let ρ = 0 and J = 0 to nd solutions to Maxwell’s equations in vacuum. We ∂B dierentiate the fourth equation wrt t and use ∇ × E = − ∂t :

2 ∂ E ∂ ∂B 2 2 µ0ε0 = (∇ × B) = ∇ × = ∇( ∇ · E ) + ∇ E = ∇ E. ∂t2 ∂t ∂t | {z } =ρ/ε0=0

Similarly, we have: 1 ∂2B − ∇2B = 0, c2 ∂t2 is is clearly the wave equation, with the speed of the wave being c = √ 1 , the µ00 speed of light! Now the most important solutions are the monochromatic waves with:

E = (0,E0 sin(kx − ωt), 0).

Where we assume the electric eld is oscillating in the y direction, and the wave is traveling in the x direction. Equivalently, we could’ve had a wave with electric eld oscillating in z direction.

Denition (Amplitude, wave number and frequency).

(i) E0 is the amplitude

2π (ii) k is the wave number. λ = k is the wavelength. (iii) ω is the (angular) frequency. Since the wave travels at speed c, we have:

ω2 = c2k2

∂B Now we use ∇ × E = − ∂t to solve for B, giving: E B = 0 sin(kx − ωt). c Now we make out solves easier, in the most general form we would write:

E = E0 exp(i(k · x − ωt)), B = B0 exp(i(k · x − ωt)), where k is the wave vector, or the direction in which the wave is travelling. E, B, and k are orthogonal to each other. Note. THIS IS VERY IMPORTANT! e complex notation is for notation only. e only part that actually exists is the real part, and before we do anything, we need to take the real part of it.

Denition (Polarization). A solution with real E0, B0, k is said to be linearly polarized, which means waves oscillate up and down in a xed plane. If E0 = α + βi and B0 are complex, then it is said to be elliptically polarized. In the special case where |α| = |β| and α · β = 0, this is circular polarization.

15 is says that the waves oscillate up and down in a xed plane. Now if we have a wave Einc = E0yˆ exp(i(kx + ωt)), coming at right angle into a conductor, then we know since E = 0 inside the conductor, this solution clearly doesn’t work. Using the image method again, we have, with Eref = −E0yˆ exp(i(−kx− ωt)).: E = Einc + Eref . Where the reected wave is caused by the surface current. Why?, rst let’s derive the incident and reected magnetic eld: E E B = 0 zˆ exp(i(kx − ωt)) B = 0 zˆ exp(i(−kx − ωt)) inc c ref c en the surface magnetic eld is:

2E0 −iωt B · zˆ| − = e , x=0 c is gives rise to a surface current, as inside the conductor we have B = 0. e surface current is, by the same method as in Ampere’s law:

2E0 −iωt B · zˆ| − = e , x=0 c e surface current is oscillating, so charges are accelerating, and they give out light in the process, which are the reected waves. And that’s why metals are shiny!

4.6 Poynting Vector Now we would derive Poynting’s theorem as followed: First we dierentiate the energy equation for an electric eld E and B: dU Z 1 Z = − J · E dV − (E × B) · dS. dt V µ0 S From vector identities, we have E · (∇ × B) − B · (∇ × E) = ∇ · (E × B): dU Z 1 Z = − J · E dV − (E × B) · dS. dt V µ0 S So: eorem (Poynting theorem). dU Z 1 Z + J · E dV = − (E × B) · dS . dt V µ0 S | {z } | {z } Total change of energy in V (elds + particles) Energy that escapes through the surface S

Denition (Poynting vector). e Poynting vector is 1 S = E × B. µ0 Note. e poynting vector can and should be seen as the energy per unit surface area geiven into the system at any given time. It characterizes the energy transfer.

16 5 Electromagnetism and Relativity

Now the original course ended here. But our great David Tong (rightfully?) decided it was a bit short and added this section. Now, rst things rst:

5.1 Vectors and Covectors Wait….where was this in Special Relativity? Don’t panic, we will go through this again. First o we had the Minkowski metric η, vectors Xµ, Y µ and a dot product: X · Y = XT ηY.

v µ Right? Now we dene Xµ = ηµvX and X = X. So the dot product now becomes µ XµX . Now pure mathematicians will start mumbling about that one is contravariant and one is covariant, but we will skip this part. e only thing we care is that from this point on: - You can only sum i the same indices appear once above and once below.

µv - You can use ηµv to raise or lower indices. To raise them, use η , which is the inverse matrix, but actually the same matrix.

5.1.1 Lorentz Transformations ese are the transformations or matrices that preserve the Minkowski metric. What µ does it mean? In Einstein notation, this means the matrix/transformation Λν satisfy: ρ σ Λ µηρσΛ ν = ηµν . Ok, wait a sec… didn’t we just use 2 lower or upper indices to represent a matrix? Why are we using one upper and one lower now? Hold on that thought, we will come to it very soon. From special relativity, we know there are two classes of Lorentz transformations: (i) Rotations: 1 0 0 0

µ 0  Λ ν =   0 R  0 where RT R = 1, ie. is an orthogonal matrix. (ii) Boosts, eg. a boost in the x direction is  γ −γv/c 0 0

µ −γv/c γ 0 0 Λ ν =    0 0 1 0 0 0 0 1

Now thus, the lower indices one transform as:

0 0ν ν σ ν σρ ρ Xµ 7→ Xµ = ηµν X = ηµν Λ σX = ηµν Λ ση Xρ = Λµ Xρ ρ µ where Λµ , as you have probably guessed, is the inverse of Λ ρ. But also remember, it ρ is also the transpose of Λ µ, as it is an orthogonal matrix (in the Minkowski metric).

17 Denition (Vectors and co-vectors). Vectors have indices up and transform according µ µ ν to X 7→ Λ ν X . ν Co-vectors have indices down and transform according to Xµ 7→ Λµ Xν . We can explore dierent objects and see if they are vectors or co-vectors. Denition (4-derivative). e relativistic generalization of ∇ is the 4-derivative, dened to be ∂ 1 ∂ ∂ = = , ∇ . µ ∂Xµ c ∂t

5.2 Vectors and Covectors is is the ”ocial” explanation of how the indices actually work:

Denition (Vectors and co-vectors). Vectors have indices up and transform according µ µ ν to X 7→ Λ ν X . ν Co-vectors have indices down and transform according to Xµ 7→ Λµ Xν . Now vectors are just the normal vectors we know. ONce we changed the basis, we need to change it in the inverse way of the basis change to get our new vector, because our vector needs to stay the same. But some things change in the same way as a basis, such as: Denition (4-derivative). e relativistic generalization of ∇ is the 4-derivative, dened to be ∂ 1 ∂ ∂ = = , ∇ . µ ∂Xµ c ∂t is is what a co-vector really is, because it co-changes with the basis. Here is the denition of a tensor if anyone forgot: Denition (Tensor). A tensor of type (m, n) is a quantity

µ1···µn T ν1···νn which transforms as

0µ1···µn µ1 µm σ1 σn ρ1,··· ,ρm T ν1···νn = Λ ρ1 ··· Λ ρm Λν1 ··· Λνn × T σ1,··· ,σn .

5.3 Conserved Currents We want the continuity equation: ∂ρ + ∇ · J = 0. ∂t To work in 3+1D, or 4-vectors. Turns out it does, with:

ρc J µ = J

µ And the charge equation becomes, simply, ∂µJ = 0. Nice, huh?

18 5.4 Gauge potentials and EM Fields Now we dene our potentials as before: ∂A E = −∇φ − ∂t B = ∇ × A.

∂A Ok, ok…I know something has changed. e extra − ∂t term is to compensate the fact that when we dened the potential for E we had no magnetic eld, but in reality we do have.Now since we have 4 components of a thing,we make it into a 4-vector! φ/c Aµ = A Trust me this works. Now we dene the anti-symmetric electromagnetic tensor

Fµν = ∂µAν − ∂ν Aµ.

Since this is antisymmetric, the diagonals are all 0, and Aµν = −Aνµ. So this thing has (4×4−4)/2 = 6 independent components. And this encapsulates all information about the EM eld. Writing it out, we have:

 0 Ex/c Ey/c Ez/c −Ex/c 0 −Bz By  Fµν =   −Ey/c Bz 0 −Bx  −Ez/c −By Bx 0 Note this is also invariant under gauge transformations. Under a boost by v in the x-direction, we have: 0 Ex = Ex 0 Ey = γ(Ey − vBz) 0 Ez = γ(Ez + vBy) 0 Bx = Bx v B0 = γ B + E y y c2 z v B0 = γ B − E z z c2 y By multiplying the Lorenz matrix with this one. We look at an example to show the relationship between E and B elds: Example (Boosted line charge). An innite line along the x direction with uniform charge per unit length, η, has electric eld 0 η E = y . 2πε (y2 + z2)   0 z e magnetic eld is B = 0. Plugging this into the equation above, an observer in frame S0 boosted with v = (v, 0, 0), ie. parallel to the wire, sees 0  0  ηγ ηγ E = y = y0 . 2πε (y2 + z2)   2πε (y02 + z02)   0 z 0 z0

19 Also,  0   0  ηγv ηγv B = z = z0 . 2πε σ2(y2 + z2)   2πε σ2(y02 + z02)   0 −y 0 −y0 In frame S0, the charge density is Lorentz contracted to γη. e magnetic eld can be wrien as µ I0 B = 0 ϕˆ0, 2πpy02 + z02  0  1 where ϕˆ0 = −z0 is the basis vector of cylindrical coordinates, and p 02 02   y + z y0 I0 = −γηv is the current. is is what we calculated from Ampere’s law previously. But we didn’t use Am- pere’s law here. We used Gauss’ law, and then applied a Lorentz boost. We see that magnetic elds are relativistic eects of electric elds. ey are what we get when we apply a Lorentz boost to an electric eld. So relativity is not only about very fast objectsTM. It is there when you stick a magnet onto your fridge! Example (Boosted point charge). A point charge Q, at rest in frame S has x Q Q E = ˆr = y , 4πε r2 4πε2(x2 + y2 + z2)3/2 0 0 z and B = 0. In frame S0, boosted with v = (v, 0, 0), we have  x  0 Q E =  γy  . 4πε (x2 + y2 + z2)3/2 0 γz. We need to express this in terms of x0, y0, z0, instead of x, y, z. en we get x0 + vt0 Qγ E0 = y0 . 4πε (γ2(x0 + vt0)2 + y02 + z02)   0 z0 Suppose that the particle sits at (−vt0, 0, 0) in S0. Let’s look at the electric at t0 = 0. x0 en the radial vector is r0 = y0 . In the denominator, we write z0

γ2(x02 + y02 + z02) = (γ2 − 1)x02 + r02 v2γ2 = x02 + r02 c2 ! v2γ2 2 = cos2 θ + 1 r02 c v2 = γ2 1 − sin2 θ r02 c2

20 where θ is the angle between the x0 axis and r0. So 1 Q E0 = rˆ0. 3/2 2 2 v2 2 4πε0r γ 1 − c2 sin θ 3/2 2 v2 2 e factor 1/γ 1 − c2 sin θ squashes the electric eld in the direction of motion. So what we did previously, which would not give the factor is wrong! Well, kind of. Notice the factor is very close to 1 if speeds are small. ere is also a magnetic eld

 0  µ0Qγ B =  z0  . 4π(γ2(x0 + vt0)2 + y02 + z02)3/2 −y0

5.4.1 Lorenz Invariants Now we want something that everyone agrees on. e rst thing is:

1 E2 F F µν = − + B2, 2 µν c2 en we dene the dual electromagnetic tensor, which is:

 0 −Bx −By −Bz  ˜µν 1 µνρσ Bx 0 Ez/c −Ey/c F = ε Fρσ. =   . 2 By −Ez/c 0 Ex/c  Bz Ey/c −Ex/c 0

Where εµνρσ is the natural extension of the Levi-Civita symbol into 4 dimensions. e half comes from the fact that ipping the indices gives 2 times the contribution. is is also a tensor and we have 1 F˜µν F = E · B/c. 4 µν

5.5 Maxwell Equations Now. Show time. e Maxwell equations relativistically are:

µν ν ∂µF = µ0J µν ∂µF˜ = 0.

Wait…thats it? Yup. Well, rst one should notice each equation is 4 equations, but this is it. We can check that we can actually recover the original equations, but we will not. Avoid algebra when you can.

5.6 e Lorentz Force Law e nal aspect of electromagnetism is the Lorentz force law for a particle with charge q moving with velocity u: dp = q(E + v × B). dt

21 To write this in relativistic form, we use the proper time τ (time experienced by particle), which obeys dt 1 = γ(u) = . dτ p1 − u2/c2 dX c E/c We dene the 4-velocity U = = γ , and 4-momentum P = , where dτ u p E is the energy. Note that E is the energy while E is the electric eld. e Lorentz force law can be wrien as dP µ = qF µν U . dτ ν We show that this does give our original Lorentz force law: When µ = 1, 2, 3, we obtain dp = qγ(E + v × B). dτ

dt By the chain rule, since dτ = γ, we have dp = q(E + v × B). dt So the good, old Lorentz force law returns. Note that here p = mγv, the relativistic momentum, not the ordinary momentum. But how about the µ = 0 component? We get

dP 0 1 dE q = = γE · v. dτ c dτ c is says that dE = qE · v, dt which is our good old formula for the work done by an electric eld. Example (Motion in a constant eld). Suppose that E = (E, 0, 0) and u = (v, 0, 0). en d(γu) m = qE. dt So mγu = qEt. So dx qEt u = = . dt pm2 + q2E2t2/c2 Note that u → c as t → ∞. en we can solve to nd r ! mc2 q2E2t2 x = 1 + − 1 q mc2

1 2 For small t, x ≈ 2 qEt , as expected.

22 Example (Motion in constant magnetic eld). Suppose B = (0, 0,B). en we start with dP 0 = 0 ⇒ E = mγc2 = constant. dτ So |u| is constant. en ∂(γu) m = qu × B. ∂t So du mγ = qu × B dt since |u|, and hence γ, is constant. is is the same equation we saw in Dynamics and Relativity for a particle in a magnetic eld, except for the extra γ term. erefore the particle goes in circles with frequency qB ω = . mγ