Lecture 0: Fields, Forces & Thinking Like a Physicist
We begin a journey into theoretical physics — and it starts with electrodynamics.
Lecture 0 lays the foundation: where physical laws actually come from, how
theoretical physics turns them into predictions, and how the electric and
magnetic fields emerge from a handful of simple experiments.
Starting from one principle — that general physical laws come from experiment,
not logic (after theoretical physicist B. G. Levich) — we build the electro-
and magnetostatic picture step by step: Coulomb’s law, the electric field and
the principle of superposition, why the static electric field is “vortex-free”
(∇×E = 0) and can be written as a potential (E = −∇φ), the magnetic field and
the Lorentz force F = q(E + v×B), Ørsted’s law (∇×B = μ₀J), and charge
conservation via the continuity equation (∂ρ/∂t + ∇·J = 0). Along the way we
lean on Stokes’ and Gauss’ theorems — and one recurring habit: think in
pictures first, then write the math.
By the end you’ll hold the line integrals of E and B, and a clear map of what’s
still missing (their divergences) on the road to the full Maxwell–Lorentz
equations.
Lecture 1: From Gauss’s Law to Maxwell’s Missing Term
Lecture 1 completes the static picture — then breaks it open.
In Lecture 0 we found the circulations (curls) of the electric and magnetic
fields. Now we hunt down the piece we were missing — their divergences — and
see that together, divergence and curl determine a field completely.
We derive Gauss’s law and its differential form ∇·E = ρ/ε₀, turn it into the
Poisson equation ∇²φ = −ρ/ε₀, and show why there are no magnetic monopoles:
∇·B = 0, a solenoidal field that lets us write B = ∇×A. With the scalar
potential φ and the vector potential A in hand — auxiliary fields that aren’t
directly observable — electrostatics and magnetostatics look neatly separate…
…until Faraday. A changing magnetic field induces a circulating electric field
(∇×E = −∂B/∂t), and the two fields start to mix (E = −∇φ − ∂A/∂t). Then Maxwell
makes his leap: by symmetry, a changing electric field should curl the magnetic
field too. With no experiment available at the time, we pin the missing term
down from pure theory — using charge conservation — and arrive at the
displacement current: ∇×B = μ₀J + μ₀ε₀ ∂E/∂t. The electric and magnetic fields
are now fully symmetric, and the road to Maxwell’s equations is open.
Lecture 2: Potentials, Gauge Freedom & the Lorentz Condition
Lecture 2 closes the “picture” chapter and steps into the mathematics.
We open with a recap of everything built so far — the four field equations and
the mental picture behind each one: charges as the sources and sinks of E
(∇·E = ρ/ε₀), the absence of magnetic monopoles (∇·B = 0), a changing magnetic
flux curling the electric field (∇×E = −∂B/∂t), and — after Maxwell’s symmetry
argument and the displacement current — a changing electric flux curling the
magnetic field (∇×B = μ₀J + μ₀ε₀ ∂E/∂t). Divergence and curl together determine
a field completely, and in vacuum we now have the full set.
Then we change worlds. Instead of solving for E and B directly, we introduce the
auxiliary fields — the scalar potential φ and the vector potential A. Because
∇·B = 0 we can write B = ∇×A, and Faraday’s law then lets us write
E = −∇φ − ∂A/∂t. Substituting these back into Maxwell’s equations gives two
coupled equations for φ and A that look, at first, hard to untangle.
The way out is a beautiful idea: the potentials are not unique. Because only E
and B are physical, we’re free to transform A → A + ∇ψ and φ → φ − ∂ψ/∂t without
changing a single measurable field. That freedom is a gauge freedom, and ψ is a
gauge field we can pick to make life easier. Choosing the Lorenz condition
(∇·A + (1/c²) ∂φ/∂t = 0) decouples and symmetrises everything: both potentials
obey the same wave (d’Alembertian) operator, □A = −μ₀J and □φ = −ρ/ε₀. We finish
by proving the choice is legitimate — the Lorenz condition is just a particular
gauge, valid because the gauge field ψ it requires always solves the
d’Alembertian equation. In the static limit this collapses to the Poisson
equation ∇²φ = −ρ/ε₀, which is where the next lecture begins.
Lecture 3: Solving Poisson’s Equation with Fourier Transforms
Lecture 3 moves from the general machinery of the potentials to solving a concrete
problem — and that problem is electrostatics.
We open where Lecture 2 ended: by exploiting gauge freedom we wrote the scalar and
vector potentials as d’Alembertian (wave) equations, □φ = −ρ/ε₀ and □A = −μ₀J, linked
by the Lorenz condition ∇·A + (1/c²) ∂φ/∂t = 0. If we know the charge density and the
current density, solving these gives us the potentials — and with them the whole
electrodynamics of the system.
So we specialise. For stationary charges there is no current and nothing varies in
time, and the general equations collapse: the wave equation for φ becomes the Poisson
equation ∇²φ = −ρ/ε₀, the vector potential obeys ∇²A = 0, and the Lorenz condition
reduces to ∇·A = 0. Electrostatics, then, is fully determined the moment we can solve
the Poisson equation.
The heart of the lecture is that solution. We want φ(r) in terms of the charge density
ρ(r), and we get it with Fourier transforms. After setting up the forward and inverse
transforms between position space and k space, we use the beautiful property that
taking the Laplacian of a function corresponds to multiplying its transform by −k².
That converts the differential equation into a simple algebraic one: φ(k) = ρ(k)/(ε₀ k²).
Bringing this back to position space leaves a single ∫d³k to evaluate — the k integral.
The k integral is done carefully and from scratch. We choose spherical polar
coordinates, knock out the trivial φ-integral, and turn the θ-integral into a sinc
factor, sin(k|r−r′|)/(k|r−r′|). What’s left is ∫ sin u/u, which we evaluate by contour
integration: Cauchy’s integral theorem, a half-circle contour c₁ − c₂, the limits
R → ∞ and r → 0, and a small arc that contributes −iπ. De Moivre’s theorem splits the
result and yields the celebrated ∫₀^∞ sin x/x dx = π/2.
Every factor cancels, and out drops the solution: φ(r) = (1/4πε₀) ∫ ρ(r′)/|r−r′| d³r′ —
simultaneously the general solution of the Poisson equation and, for a point charge,
Coulomb’s law itself. We close by noting we will never need to re-derive it: give any
charge distribution, and it becomes just an integral to evaluate. Next up, point
charges and discrete distributions — and then we step beyond stationary charges to
moving currents.
Lecture 4: From Potential to Field — Gradient Exercises
Lecture 4 shifts from theory to practice: we do exercises.
So far we’ve built up the theoretical framework — Maxwell’s equations, the
potentials φ and A, gauge freedom, and the derivation that in static situations
the scalar potential satisfies the Poisson equation ∇²φ = −ρ/ε₀. Now we turn to
the actual calculations that turn these equations into physics.
We begin with a recap of where we are: the Poisson equation gives the potential
in terms of the charge density, φ(r) = (1/4πε₀) ∫ ρ(r′)/|r−r′| d³r′. The
geometry involves r (test point) and r′ (source point), and the electric field
is obtained by E = −∇φ, meaning we need to compute gradients of expressions
involving |r−r′|.
The lecture contains three key gradient exercises:
- ∇·r — the divergence of the position vector, giving 3 in three dimensions.
- ∇|r−r′| — the gradient of the magnitude, yielding (r−r′)/|r−r′|.
- ∇(1/|r−r′|) — the gradient of the inverse magnitude, yielding
−(r−r′)/|r−r′|³. This is the crucial result: once we have the potential for
a point charge, φ = q/(4πε₀|r−r′|), taking the negative gradient gives us
Coulomb’s law, E = q/(4πε₀) (r−r′)/|r−r′|³.
All calculations are done explicitly in Cartesian coordinates, using the
del operator ∇ = (∂/∂x, ∂/∂y, ∂/∂z) and the Kronecker delta to handle the
index manipulation cleanly. We also prove two vector calculus identities along
the way: ∇×∇ψ = 0 (curl of a gradient is zero) and ∇·(∇×A) = 0 (divergence of
a curl is zero), justifying the potential ansätze used throughout.
Lecture 5: Feynman’s Calculus Trick for Vector Operations
Lecture 5 continues the exercise-based approach, but now we tackle an inverse problem:
given an electric field, find the charge distribution that produces it.
We start with Griffiths’ classic problem: compute the divergence of E = r̂/r².
Direct calculation in Cartesian coordinates gives zero, which seems wrong — we know
a point charge should produce a delta function. The resolution comes from recognizing
that E = −∇φ with φ = 1/r, so ∇·E = −∇²(1/r). The task becomes evaluating the
Laplacian of 1/r.
We use the Fourier transform method developed earlier to solve Poisson’s equation.
After deriving the Fourier transform of 1/r (getting φ(k) = 1/(2π²k²)), we note that
∇² in position space corresponds to −k² in k-space. This cancels the denominator,
leaving an integral that is precisely the three-dimensional delta function definition:
δ³(r) = (1/2π³) ∫ e^(−ik·r) d³k. The result: ∇²(1/r) = −4πδ(r).
With ∇·E = −∇²φ = 4πδ(r), we recover the expected charge distribution: a point
charge at the origin. This is a more direct route than Griffiths’ approach, and it
shows how Fourier methods handle singularities elegantly.
The lecture concludes with Feynman’s calculus trick — a prescription for treating
the del operator as a vector when evaluating expressions like ∇·(A×B), ∇×(φA),
and ∇(A·B). The key rules: split ∇ = ∇ₐ + ∇ᵦ, apply standard vector identities
(back cap rule), then replace ∇ₐ and ∇ᵦ with the proper gradient operator. We work
through several examples, including the hydrodynamic identity ∇(v·v) = 2[v×(∇×v) + v·∇v]
that appears in Bernoulli’s equation.
