Deriving continuity equation from 4-current of a charged particle

Question

how can i check that following 4-current for a single charged particle $$j^{\mu}(x)=qc\int d\tau u^{\mu}(\tau)\delta^{4}(x-r(\tau))$$ satisfies continuity equation $$\partial_\mu j^\mu = 0.$$

Answer 1

Formally, you can just take the derivative. Parametrising the worldline by $s$ (not necessarily proper time, could be an affine parameter for light-like lines): $r(s)$, and setting the charge to $1$: $$ j = \int \delta(x-r)\frac{dr}{ds}ds $$ I'll rewrite it as: $$ j = \int \delta(x-r)dr $$ to emphasise that the 4-current density is independent of the parametrisation. You can now take the derivative in the integral and recognise an antiderivative: $$ \begin{align} \partial \cdot j &= \int \partial\delta(x-r)\cdot dr \\ &= [\delta(x-r)] \\ &= 0 \end{align} $$ You get zero because the worldine typically starts from the distant pas to the distant future, so the delta's are not supported on the event where you evaluate the continuity equation. To build intuition, I suggest by looking at 1D then 2D Euclidian space which proves the conservation of current density.

Hope this helps.

Answer 2

Let's multiply by $\partial_\mu \epsilon(x)$ and then integrate over $x$. We find \begin{align} \int d^4 x \partial_\mu \epsilon(x) j^\mu(x) &= q c \int d \tau \frac{d x_0^\mu(\tau)}{d\tau} \int d^4 x \partial_\mu \epsilon(x) \delta^4 ( x - x_0(\tau)) \\ &= q c \int d \tau \frac{d x_0^\mu(\tau)}{d\tau} \partial_\mu \epsilon(x_0(\tau)) \\ &= q c \int d \tau \frac{d \epsilon(x_0(\tau)) }{d\tau} \\ &= q c [ \epsilon(x_0(\tau_f)) - \epsilon(x_0(\tau_i)) ] \end{align} This is true for any function $\epsilon(x)$. We now set $\epsilon(x) = \delta^4(x-y)$ where $y \neq x_0^\mu(\tau_{f,i})$. It follows that $$ \partial_\mu j^\mu(y) = 0 $$

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ASIDE: Strictly speaking the identity is true only for smooth functions $\epsilon(x)$. We can still reproduce our result by taking $\epsilon(x)$ to be a smooth function that peaks at $x=y$ and smoothly goes to zero in a ball of radius $r$ and then take $r \to 0$.

Answer 3

Let's do it naively, without introducing distributions, and using Cartesian coordinates, so that an event can be represented by the 4-vector \begin{equation} \begin{aligned} \mathbf{X} & = X^{\mu} \mathbf{b}_{\mu} = \\ & = X^0 \, \mathbf{b}_0 + X^1 \, \mathbf{X}_1 + X^2 \, \mathbf{X}_2 + X^3 \, \mathbf{X}_3 = \\ & = ct \, \mathbf{b}_0 + x \, \mathbf{b}_1 + y \, \mathbf{b}_2 + z \, \mathbf{b}_3 \ , \end{aligned} \end{equation} and the divergence of a 4-vector reads \begin{equation} \nabla \cdot \mathbf{V} = \left(\mathbf{b}^{\nu} \dfrac{\partial}{\partial X^{\nu}} \right) \cdot \left( V^{\mu} \mathbf{b}_{\mu} \right) = \dfrac{\partial V^{\mu}}{\partial X^{\mu}} \ . \end{equation}

It's easy to identify in the expression of the current density the increment of the 4-vector describing the trajectory of the event $P$ \begin{equation} \mathbf{J} = J^{\mu} \mathbf{b}_{\mu} = \mathbf{b}_{\mu} qc \int \delta^{4}(X^{\nu}-X^{\nu}_P(\tau)) \, \underbrace{U_P^{\mu}(\tau) \, d\tau}_{d X^{\mu}_P} \ , \end{equation} and evaluating the 4 integrals (1 of each components), exploiting the relations of the form \begin{equation} \int \delta(X^0 - X_P^0) \delta(X^1- X_P^1) \delta(X^2 - X_P^2) \delta(X_3 - X_P^3) dX^0_P = \delta(X^1- X_P^1) \delta(X^2 - X_P^2) \delta(X_3 - X_P^3) \ , \end{equation} we get \begin{equation}\begin{aligned} \mathbf{J} & = \mathbf{b}_0 \, \delta(X^1- X_P^1) \delta(X^2 - X_P^2) \delta(X_3 - X_P^3) + \\ & + \mathbf{b}_1 \, \delta(X^0- X_P^0) \delta(X^2 - X_P^2) \delta(X_3 - X_P^3) + \\ & + \mathbf{b}_2 \, \delta(X^0- X_P^0) \delta(X^1 - X_P^1) \delta(X_3 - X_P^3) + \\ & + \mathbf{b}_3 \, \delta(X^0- X_P^0) \delta(X^1 - X_P^1) \delta(X_2 - X_P^2) \ . \end{aligned}\end{equation} From the latter expression, it's easy to realize that the $i^{th}$ component doesn't depend on $i^{th}$ component and thus, all the derivatives appearing in the 4-divergence of such a vector fields are identically zero,

\begin{equation} \nabla \cdot \mathbf{J} = \underbrace{\dfrac{\partial J^0}{\partial X^0}}_{=0} + \underbrace{\dfrac{\partial J^1}{\partial X^1}}_{=0} + \underbrace{\dfrac{\partial J^2}{\partial X^2}}_{=0} + \underbrace{\dfrac{\partial J^3}{\partial X^3}}_{=0} = 0 \ , \end{equation} to get the desired result.