Static point charge

Newtonian mechanics

Basic Newtonian mechanics

In Newtonian mechanics, the point charge is simply a free particle, ie subjected to no forces

\begin{equation} m \ddot{\vec{x}}(t) = 0 \end{equation}

and the corresponding electric field is generated by the Coulomb law

\begin{equation} \vec{E} (\vec{y}) = \frac{1}{4\pi \varepsilon_0} q \frac{(\vec{x} - \vec{y})}{\| \vec{x} - \vec{y} \|^2} \end{equation}

There is a slight ambiguity - as a charge itself, is the charge $q$ itself affected by the electrostatic field it generates? As the numerator and denominator would both be zero, it is generally disregarded : a point particle does not act upon itself in a static electric field.

Lagrangian mechanics

The Lagrangian of an electrostatic system can simply be written as the Lagrangian of a free particle and the appropriate electric potential,

\begin{equation} S[\vec{x}(t), \phi(\vec{x}(t), t)] = \int_{t_a}^{t_b} \left[ \frac{1}{2} m \dot{\vec{x}}(t) \cdot \dot{\vec{x}}(t) - Q \phi(\vec{x}(t), t) \right] dt \end{equation}

To separate things a bit, let's express our point particle as a density,

\begin{equation} L(\vec{x}(t)) = \int_{\mathbb{R}^3} d^3x\ \delta(\vec{x} - \vec{x}(t)) L(\vec{x}) \end{equation}

Giving us the action

\begin{equation} S[\vec{x}(t), \phi(x, t)] = \int_{t_a}^{t_b} \left[ \frac{1}{2} m \dot{\vec{x}}(t) \cdot \dot{\vec{x}}(t) - \int_{\mathbb{R}^3} d^3x\ Q \delta(\vec{x}(t) - x) \phi(\vec{x}, t) \right] dt \end{equation}

But this is only the Lagrangian for a background electric potential, as there is no term involving derivatives of the field, not allowing any dynamics. What we want is the Lagrangian which will give us in essence the Poisson equation.

\begin{equation} \vec{\nabla} \cdot \vec{E}(\vec{x}, t) = \Delta \phi(\vec{x}, t) = \frac{\rho(\vec{x}, t)}{\varepsilon} \end{equation}

If we consider our previous formula for the Coulomb field, then it can be used to describe any static distribution of charge, simply via adding together point masses, in the continuum limit, using a charge density $\rho$ :

\begin{eqnarray} \vec{E}(\vec{x}, t) &=& \int_{\mathbb{R}^3} \frac{}{} \end{eqnarray}

In our case, $\rho$ being our

Electrodynamics

We have quite a lot of possible formulations for the Maxwell equations. Let's review a few possible ones.

First, let's consider the various differential equations :

\begin{eqnarray} \vec{\nabla} \cdot \vec{E} &=& \frac{\rho}{\varepsilon_0}\\ \vec{\nabla} \cdot \vec{B} &=& 0\\ \vec{\nabla} \times \vec{E} &=& - \frac{\partial \vec{B}}{\partial t}\\ \vec{\nabla} \times \vec{B} &=& \mu_0 \left(\vec{j} + \varepsilon_0 \frac{\partial \vec{E}}{\partial t} \right) \end{eqnarray}

For a static point charge, our charge distribution will be a Dirac delta distribution, with a zero charge current.

\begin{eqnarray} \rho(t, \vec{x}) &=& q \delta^{(n)}(\vec{x} - \vec{x}_0)\\ &=& \end{eqnarray} \begin{equation} \vec{\nabla} \cdot \vec{E} = \frac{q}{\varepsilon} \delta^{(n)}(\vec{x} - \vec{x}_0) \end{equation}

Let's see the integral form of the Maxwell equations now

\begin{eqnarray} \oint_{\partial \Omega} \vec{E} \cdot d\vec{S} &=& \frac{1}{\varepsilon_0} \int_\Omega \rho dV\\ \oint_{\partial \Omega} \vec{B} \cdot d \vec{S} &=& 0\\ \oint_{\partial \Sigma} \vec{E} \cdot d\vec{l} &=& -\frac{d}{dt} \int_\Sigma \vec{B} \cdot d\vec{S}\\ \oint_{\partial \Sigma} \vec{B} \cdot d\vec{l} &=& \mu_0 (\int_\Sigma \vec{J} \cdot d\vec{S} + \varepsilon_0 \frac{d}{dt} \int_\Sigma \vec{E} \cdot d\vec{S}) \end{eqnarray} \begin{equation} \oint_{\partial \Omega} \vec{E} \cdot d\vec{S} = \frac{1}{\varepsilon_0} \int_\Omega q \delta^{(n)}(\vec{x}) dV \end{equation}

For any closed volume around our charge, the right-hand side is always $q$. Let's consider a sphere of arbitrary radius $R$ around, and let's work in $3$ dimensions (this generalizes easily to more dimensions by picking different formulas for the area of a sphere) :

\begin{eqnarray} \oint_{\partial \Omega} E_r R^2 d\theta \sin\theta d\varphi &=& \frac{q}{\varepsilon_0} \\ &=& 4\pi E_r R^2 \end{eqnarray} \begin{equation} E_r = \frac{1}{4\pi R^2} \frac{q}{\varepsilon_0} \end{equation}

Special relativity

\begin{equation} {F^{\mu\nu}}_{,\mu} = \mu_0 J^\nu \end{equation} \begin{equation} J^\nu = (c q \delta(\vec{x}), \vec{0}) \end{equation}

Gauges :

Lorenz

\begin{equation} \partial_\mu A^\mu = 0 \end{equation} \begin{equation} \Box A^\mu = J^\mu \end{equation}

Green's function :

\begin{equation} \Box G(x,y) = \delta(x - y) \end{equation} \begin{equation} G(x,y) = \frac{1}{\| x^\mu - y^\mu \|} \delta (|x^0 - y^0| \mp \frac{\|\vec{x} - \vec{y}\|}{c}) \end{equation}

The bundle method

To consider the case of a point particle without using distributions (this will make the construction of the bundle more complicated), we'll consider Minkowski space from which we removed the origin,

\begin{equation} M = \left\{ (t, \vec{x}) | t \in \mathbb{R}, \vec{x} \in \mathbb{R}^3 \setminus \{ \vec{0} \} \right\} \end{equation}

Electromagnetism in this case is then a gauge theory with a $\operatorname{U}(1)$ principal bundle.

Semiclassical electromagnetism

In the case of semiclassical electromagnetism, we consider the EM field to be classical, but its source is a quantum point particle.

Relativistic quantum mechanics

We can try to perform the quantization of a relativistic point particle coupled to electromagnetism, but things will not go well. Let's see why.

We can remember that, at its core, the action of a point particle is just the length of a string going through spacetime, in other words, the worldvolume of a $0$-brane, at least if we use the Nambu-Goto action :

\begin{equation} S_{\mathrm{NG}}[X, g] = T \int_\Sigma d\mu[X_* g] \end{equation}

$T$ is some proportionality constant corresponding to the tension of the brane (in our case, the mass of our point particle), $\Sigma$ is the path it goes through spacetime, and $g$ the spacetime metric (in our case, we can simply consider it to be non-dynamical and equal to $\eta$).