2024-04-4 The Potential Formulation

Remaining Course Topics

• Potentials and fields
  • Leads into the Lagrangian and Hamiltonian of EM fields, Quantum electrodynamics
• Relativistic Electrodynamics

Scalar and Vector potentials

• We are seeking the general solution to Maxwell’s Laws
• In a vacuum they are:

$$\vec{\nabla }\cdot \vec{E}=0$$ $$\vec{\nabla }\cdot \vec{B}=0$$ $$\vec{\nabla }\times \vec{E}=-\vec{B}$$ $$\vec{\nabla }\times \vec{B}={\mu }_0{ϵ}_0\vec{\dot{E}}$$

• Given Linear dielectrics /magnetic materials, the first three laws remain the same. The last one becomes

$$\vec{\nabla }\times \vec{B}=\mu ϵ\vec{\dot{E}}$$

• With an electrically conducting material, the amperes law becomes

$$\vec{\nabla }\times \vec{B}=\mu \sigma \vec{E}+\mu ϵ\vec{\dot{E}}$$

Maxwell’s Equations with electric charge and current

• Faraday’s Law always holds
• Maxwell’s laws become:

$$\vec{\nabla }\cdot \vec{E}=\frac{\rho }{{ϵ}_0}$$ $$\vec{\nabla }\cdot \vec{B}=0$$ $$\vec{\nabla }\times \vec{E}=-\vec{B}$$ $$\vec{\nabla }\times \vec{B}=\mu \vec{j}+\mu ϵ\vec{\dot{E}}$$

• Given a charge distribution $\rho (\mathbf{r},t)$ and a current density $\mathbf{J}(\mathbf{r},t)$, what are the electric and magnetic fields at any given point and time?
• Start representing fields as potentials

$$\mathbf{B}=\nabla \times \mathbf{A}$$

and

$$\mathbf{E}=-\nabla V-\frac{\partial \mathbf{A}}{\partial t}$$

Using Gauss’ Law on this:

$$\nabla \cdot \left(-\nabla V-\frac{\partial A}{\partial t}\right)=\frac{1}{{ϵ}_0}\rho$$ $${\nabla }^{2}V+\frac{\partial }{\partial t})(\nabla \cdot \mathbf{A})=-\frac{1}{{ϵ}_0}\rho$$

Looking at the magnetic field:

$$\vec{\nabla }\times \vec{B}=\vec{\nabla }\times \vec{\nabla }\times \vec{A}={\mu }_0\vec{j}+{\mu }_0{ϵ}_0\frac{\partial \vec{E}}{\partial t}={\mu }_0\vec{j}+{\mu }_0{ϵ}_0\frac{d}{dt}\left(-\vec{\nabla }V-\frac{\partial \vec{A}}{\partial t}\right)$$ $$-{\vec{\nabla }}^2\vec{A}+\vec{\nabla }(\vec{\nabla }\cdot \vec{A})={\mu }_0\vec{j}-{\mu }_0{ϵ}_0\vec{\nabla }\frac{\partial V}{\partial t}-{\mu }_0ϵ0-\frac{{\partial }^2\vec{A}}{\partial t^2}$$ $$-{\vec{\nabla }}^2\vec{A}+{\mu }_0{ϵ}_0\frac{{\partial }^2\vec{A}}{\partial t^2}={\mu }_0\vec{j}-\vec{\nabla }(\vec{\nabla }\cdot \vec{A})-{\mu }_0{ϵ}_0\vec{\nabla }\frac{\partial V}{\partial t}$$

The RHS can be written as

$$\vec{\nabla }\left[-\vec{\nabla }\cdot \vec{A}-\frac{1}{c^2}\frac{\partial V}{\partial t}\right]$$

And the LHS + the first term of the right hand side is the inhomogenous wave equation (solvable) We decide the RHS is equal to zero. Because fiskis.

We are working in the Lorentz Gauge in opposition to the Coulomb Gauge used in electrostatics.

$$\vec{E}=-\vec{\nabla }V-\frac{\partial \vec{A}}{\partial t}$$ $$\vec{B}=\vec{\nabla }\times \vec{A}$$ $$\vec{A},{\vec{A}}^{\prime }+\vec{\nabla }\lambda$$ $${\vec{B}}^{\prime }=\vec{\nabla }\times {\vec{A}}^{\prime }=\vec{\nabla }\times \vec{A}+\text{cancelto}0\vec{\nabla }\times (\vec{\nabla }\lambda )=\vec{B}$$ $${\vec{E}}^{\prime }=-\vec{\nabla }{V}^{\prime }-\frac{\partial {\vec{A}}^{\prime \prime }}{\partial t}=-\vec{\nabla }\left(V-\frac{\partial \vec{A}}{\partial t}\right)-\frac{d}{dt}(\vec{A}+\vec{\nabla }\lambda )$$ $${\vec{E}}^{\prime }=-\vec{\nabla }V-\frac{\partial \vec{A}}{\partial t}+\text{cancelto}0\vec{\nabla }\frac{\partial \lambda }{\partial t}-\frac{d}{dt}\vec{\nabla }\lambda -\vec{E}$$

Lorentz Force

The Lorentz force is given by

$$\vec{F}=\frac{d\vec{p}}{dt}=q(\vec{E}+\vec{v}+\vec{B})$$

\vec{E} = -\vec{\nabla} - \frac{ \partial \vec{A} }{ \partial t } , \ \ \vec{B} = \vec{\nabla}\times \vec{A}

\vec{F} = \frac{ d \vec{p} }{ d t } =q\left( -\vec{\nabla}V-\frac{ \partial \vec{A} }{ \partial t } + \vec{v}\times(\vec{V}\times \vec{A}) \right)

$$Rewritetriplecrossproductwithusefulmathematicalformula$$

\vec{F} = \frac{ d \vec{p} }{ d t } = q\left( -\vec{\nabla}V - \frac{ \partial \vec{A} }{ \partial t } + \vec{\nabla} (\vec{v}\cdot \vec{A})-(\vec{v}\vec{\nabla}) \cdot \vec{A} \right)

\vec{F} = \frac{ d \vec{p} }{ d t } = -q \frac{ d \vec{A} }{ d t } -q\vec{\nabla}(V + \vec{v} \cdot \vec{A})

\implies \frac{d}{dt} ( \vec{p} + q\vec{A}) = - q\vec{\nabla}(V-\vec{v} \cdot \vec{A})

the $(\vec{p}+q\vec{A})$ is also known as canonical momentum. Used in Lagrangian and Hamiltonian electrodynamics. This is a velocity dependent potential energy. In class problems 10.3 and 10.7