I have a question about the four current in covariant representation. the four current is defined as

$$ J^{\alpha} = \binom{c\rho}{\vec{j}} $$

and i'm having a point charge, $$\rho(\vec{x},t)=e\delta(\vec{x}-\vec{r}(t)),$$ $$\vec{j}(\vec{x},t)=e\frac{d\vec{r}}{dt}\delta(\vec{x}-\vec{r}(t)).$$ Now, the four-vector $r=(r^0, \vec{r})$ is the space-time-point on the trajectory of my point charge and $x=(x^0, \vec{x})$ is the observation point.

With $r^{\alpha}(\tau)$ as function of proper time and the four-velocity $V^{\alpha}$ the four current can be written like $$J^{\alpha} = e \int{d\tau V^{\alpha}\delta^{(4)}(x-r(\tau))}.$$

My Question: why and how? I just can't verify this expression, particulary i don't understand where the integral is coming from, and i'm a little bit confused about how $r^{\alpha}(\tau)$ looks like. I mean i know $$r^{\alpha} = \binom{ct}{\vec{r}(t)},$$ but how to handle this with proper time and also what's about $V^{\alpha}(\tau)$ and $V^{\alpha}(t)$?


1 Answer 1


Note: $c=1$ in the following.

Every time-like worldline can be parametrized by its proper time. If you are given $\vec r(t)$, then the proper time at $t_0$ is given by $$\tau(t_0) = \int_0^{t}\sqrt{1-\left(\frac{\mathrm{d}\vec r}{\mathrm{d}t}\right)^2}\mathrm{d}t $$ and inverting this expression to get $t(\tau)$ gives you the worldline $r^\mu = (t(\tau),\vec r(t(\tau))$ parametrized by its proper time.

Now, insert $1 =\frac{\mathrm{d}t}{\mathrm{d}t}$ into your expression for $\rho$. Then, your four-current becomes $$ j^\mu = e \frac{\mathrm{d}r^\mu}{\mathrm{d}t}\delta(\vec x-\vec r(t))$$ and using the chain rule $$ j^\mu(\vec x,t) = e \frac{\mathrm{d}r^\mu}{\mathrm{d}\tau}\frac{\mathrm{d}\tau}{\mathrm{d}t}\delta(\vec x - \vec r(t)) = e u^\mu \frac{\mathrm{d}\tau}{\mathrm{d}t}\delta(\vec x - \vec r(t))$$ Now, we use $$ j^\mu(\vec x,t')= = \int j^\mu(\vec x,t)\delta(t'-t)\mathrm{d}t=\int e u^\mu \frac{\mathrm{d}\tau}{\mathrm{d}t}\delta(\vec x - \vec r(t))\delta(t'-t)\mathrm{d}t$$ and again by the chain rule $$ j^\mu(\vec x ,t') = \int e u^\mu \delta(\vec x - \vec r(t))\delta(t'-t)\mathrm{d}\tau$$ Finally, $t= r^0$, $t' = x^0$ and $\delta(x - r) = \delta(\vec x - \vec r)\delta(x^0 - r^0)$ give the formula you asked about.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.