# Comparison of E & M¶

## Useful Tricks¶

$\partial_i(x_k j_i) = j_k + x_k \partial_i j_i$

This is useful because we have

$\nabla \cdot \vec j = \partial_i j_i = \frac{d}{dt} \rho = 0,$

and the LHS can be turned into a surface integral as one wish and disappears.

A similar one is

$\partial_i(x_k x_l j_i) .$

## Statics in Vacuum¶

### Source of Fields¶

1. Source of Electric Field in Electrostatics

Source of electric field is charge

$\nabla \cdot \vec E = 4\pi \rho$
1. Source of Maganetic field in Magnetostatics

Current is the source of maganetic field

$\nabla \times \vec B = \frac{4\pi}{c} \vec j .$

### Potentials¶

1. Electric Potential

Electric potential is given by

$\vec E = -\nabla \phi.$

Imediately, we have the curl of electrical field being 0, i.e.,

$\nabla\times \vec E = 0.$

By implementing Gauss’s law, the equation for potential becomes

$\nabla^2 \phi = - 4\pi \rho .$

The solution, apply the Green’s function to Laplace equation,

$\phi(\vec x) = \int d^3x' \frac{\rho(\vec x')}{\lvert \vec x - \vec x' \rvert}$
1. Magnetic Potential

Magnetic potential is given by

$\vec B = \nabla \times \vec A .$

Applying curl of magnetic field and solving the equation,

$\vec A = \frac{1}{c}\int \frac{\vec j}{\lvert \vec x - \vec x' \rvert} d^3 x' .$

### Gauge of fields¶

By definition, electric potential and maganetic potential are, repectively,

$\vec E = -\nabla \phi ,$
$\vec B = - \nabla \times \vec A .$

Electric field is invariant under a transform

$\phi' = \phi + \phi_0,$

where $$\nabla\phi_0=0$$.

Similarly, the potential for magnetic field is

$\vec A' = \vec A + \nabla \psi,$

in which $$\psi$$ can be any scalar fields.

Gauge

As expected, these definitions of fields do not determine the potential completely. This is gauge freedom.

It might seem strange to talk about such a freedom. As we would ask why we have such freedom for potentials?

In class electrodynamics, potentials are merely mathematical tools. So the notion that potental has gauge freedom comes only from the mathematical definition of potentials.

However, we do expect such a freedom is part of nature as we step into quantum ralm. In quantum world, Aharonov-Bohm effect proves that potentials are actually real existance. In such cases, the gauge freedom do have a very important impact on our theory. Gauge freedom is part of the internal structure of fields and goes deep into group theory, topology and differential geometry.

### Multipole Expansion¶

Requirement

One should be able to derive these multipole expansions of fields without refering to any material.

In the expression for potentials,

$\phi(\vec x) = \int d^3x' \frac{\rho(\vec x')}{\lvert \vec x - \vec x' \rvert},$

and

$\vec A = \frac{1}{c}\int \frac{\vec j}{\lvert \vec x - \vec x' \rvert} d^3x',$

the term

$\frac{\vec j}{\lvert \vec x - \vec x' \rvert}$

can be Taylor expanded when $$\vec x' \ll \vec x$$,

$\begin{split}&\frac{1}{\lvert \vec x - \vec x' \rvert} \\ =& \frac{1}{r} - x'_i \partial_i \frac{1}{r} + \frac{1}{2} x'_i x'_j \partial_i \partial_j \frac{1}{r} + cdots\end{split}$

where $$1/r$$ is $$1/\lvert x \rvert$$ .

Apply this expansion, we can find the dipole and quadrapole of a charge distribution, which are

$\vec p = \int \rho(\vec x') \vec x' d^3 x',$
$Q_{ij} = \frac{1}{2} \int \rho(\vec x') ( 3x'_i x'_j - r'^2 \delta_{ij}) d^3x'.$

The corresponding potentials are

$\vec \phi_d = \frac{\vec p\cdot \vec x}{r^3},$

and

$\vec \phi_q = \frac{x_i Q_{ij} x_j }{r^5}.$

The electric field can be calculated using $$\vec E = -\nabla \phi$$.

For magnetic field, a dipole expansion shows that

$\vec A = \frac{\mu\times \vec x}{r^3},$

where

$\vec \mu = \frac{1}{2c} \int \vec x' \times \vec j(\vec x') d^3 x'.$

### Force and Torque¶

Requirement

1. Write the most general expression for force and torque.
2. Derive the expression with dipole and quadrapole approximations.

Among the tricks, virtual work principle could be a nice one.

Force and torque can be calculated using virtual work principle. However, for dipoles, they can be calculated directly.