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Susskind Lecture 11 - Electric and Magnetic Forces

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Susskind Lecture 11 - Electric and Magnetic Forces

Lecture 11 closes classical mechanics by showing why magnetic forces force us to use gauge potentials. This is the classical doorway to quantum mechanics and field theory.

Fields and del operators

A scalar field assigns one number to each point, like temperature T(x, t). A vector field assigns a vector, like wind velocity, electric field, or magnetic field.

Useful derivative operations:

gradient:    ∇V
divergence:  ∇ · A
curl:        ∇ × A

Two vector-calculus facts matter here:

∇ · (∇ × A) = 0
∇ × (∇V) = 0

Magnetic field and vector potential

Magnetic fields have zero divergence:

∇ · B = 0

Therefore they can be written as the curl of a vector potential:

B = ∇ × A

The vector potential is not unique. If s(x) is any scalar function, then:

A → A + ∇s

leaves B unchanged. This is a gauge transformation.

Electric and magnetic forces

For a static electric field:

E = -∇Φ
F_electric = e E
potential energy = e Φ

The magnetic part of the Lorentz force is velocity-dependent:

F_magnetic = (e/c) v × B

It is perpendicular to v and B, so a static magnetic field changes the direction of velocity but does no work on the particle.

Lagrangian for a charged particle

To get the magnetic Lorentz force from an action principle, the Lagrangian must use the vector potential:

L = (1/2)m v^2 + (e/c) A(x) · v - e Φ(x)

A gauge transformation changes the action only by an endpoint term. Since stationary-action variations keep endpoints fixed, the equations of motion are gauge invariant.

Mechanical versus canonical momentum

The conjugate momentum is:

p = ∂L/∂v = m v + (e/c) A

This is canonical momentum. It is not the same as mechanical momentum:

p_mechanical = m v

Mechanical momentum is gauge invariant and directly tied to motion. Canonical momentum is the variable used in Lagrangian/Hamiltonian mechanics, and it changes under gauge transformations.

Hamiltonian

Solve for velocity in terms of canonical momentum:

m v = p - (e/c) A

Then the Hamiltonian is:

H = (1/2m) [p - (e/c) A]^2 + e Φ

Although the magnetic vector potential appears, physical predictions depend only on gauge-invariant fields and mechanical motion.

Uniform magnetic field

For a constant B along z, the charged particle moves in circles in the x-y plane while moving uniformly along z. The full motion is a helix unless the z velocity is zero.

The circular motion is not caused by loss of energy. The magnetic force bends the velocity vector while preserving its magnitude.

Common pitfalls

  • Thinking the vector potential is the magnetic field. B is the curl of A.
  • Treating gauge choice as physical. Different A fields can describe the same B.
  • Confusing canonical momentum with mechanical momentum.
  • Thinking a static magnetic field changes kinetic energy. The magnetic force is perpendicular to velocity.
  • Trying to write a magnetic-force Lagrangian using B directly. The action formulation needs A.

See also

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