Time independent Maxwell’s equations

$$ \begin{matrix} \vec \nabla \cdot \vec E=\frac{\rho}{\epsilon_0} & \vec \nabla \times \vec E = 0 \\ \vec \nabla \cdot \vec B=0 & \vec \nabla \times \vec B = \mu_0 \vec j \end{matrix} $$

🗒️ Notes:

• $\vec E$ and $\vec B$ have no more influence on each other
• $\vec \nabla \times \vec E=0$ so $\vec E$ is irrotational
• $\vec \nabla\cdot \vec j=0$ so the current flow is incompressible

In integral form the time independent Maxwell’s equations become:

$$ \begin{aligned} \oint_S \vec E \cot \text d \vec S &= \frac{Q}{\epsilon_0} \\ \oint_L \vec B \cdot \text d \vec l &= \mu_0 \sum I_i \end{aligned} \qquad \begin{aligned} \oint_L \vec E \cdot \text d\vec l &=0 \\ \oint_S \vec B \cdot \text d \vec S&=0

\end{aligned} $$

🗒️ Notes:

• Electric field lines end on point charges
• Magnetic field lines never end

Rearranging the time independent Maxwell equations

• Using $\vec \nabla \times \vec \nabla\phi=0$ and $\vec \nabla\cdot (\vec \nabla\times \vec v)=0$ for all $\phi$ and $v$ we can derive

  $$ \vec B =\vec \nabla \times \vec A \qquad \vec E = -\vec \nabla \phi $$

  where $\phi$ is the scalar electromagnetic potential and $A$ is the vector magnetic potential.

  $\vec A$ is not unique because $\vec A'=\vec A + \vec \nabla \phi$ gives the same magnetic field as $\vec \nabla \phi$ gets canceled when the curl of $\vec A'$ is taken

  🪴 Definition: this extra term is called the gauge degree of freedom

  We can remove this by the choice of the Coulomb gauge

  $$ \vec\nabla \cdot \vec A =0 $$

• Substituting these relations into the Maxwell equation we get

  $$ \nabla^2 \phi = -\frac{\rho}{\epsilon_0} \qquad \nabla^2 \vec A=-\mu_0 \vec j $$

  These are both Poisson equations with $4\pi g=\rho/\epsilon_0$ and $4\pi g=\mu \vec j$ respectively

• We can define the potential difference

  $$ \text dV=\phi (\vec r+\text d \vec l )-\phi(\vec r)= \text d\vec l\cdot \vec \nabla \phi=-\text d \vec l \cdot \vec E $$

• And thus the potential going from $A\to B$ is found via

  $$ \Delta V_{{A} \to {B}}=\int^B_A\text dV=-\int^B_A\vec E\cdot \text d\vec l $$

• Substituting expression for $B$ into that for the magnetic flux $\phi_B$ we get

  $$ \phi_B=\int_S \vec B \cdot \text d \vec S=\int_S \vec \nabla \times \vec A \cdot \text d\vec S=\int_L \vec A\cdot \text d \vec l $$

  This is invariant under the transformation $\vec A'= A+\nabla \psi$ since the line integral of $\nabla\psi$ around a closed curve is zero

Electrostatics

Using the relation we found above $\nabla^2 \phi = -{\rho}/{\epsilon_0}$ we can find

$$ \phi(\vec r)=\frac{1}{4\pi\epsilon_0}\int_V\text dV'\frac{\rho(\vec r')}{|\vec r- \vec r'|} $$