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How do I derive that the 4-current of a point charge is $$j^{\mu}(x)=ec\int_{-\infty}^{+\infty}\dot{z}^{\mu}(s)\delta(x-z(s))ds$$ where $\dot{z}^{\mu}(s)$ is the 4-velocity of the charge and $s$ is the proper time?

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What definition of 4-current are you using? The equation you wrote could be just the definition. If it is not, we need to know the starting point for the derivation. –  Chris White Mar 18 '13 at 7:38
    
It couldn't be a definition. And even if it were the definition of 4-current, I'd like to know how did they come with this. @ChrisWhite –  Anuar Mar 18 '13 at 17:50
    
@Anuar Didn't my comment help in any way? The expression you wrote can be seen as a "generalisation" of the 4-current in terms of the 4-velocity $j^{\mu}=\rho u^{\mu}$ –  nijankowski Mar 18 '13 at 19:14
    
@NijankowskiV. Yes it helped, but I still don't understand how did you get the integral expression. –  Anuar Mar 19 '13 at 3:05
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@Anuar Maybe this will help, take the 4-current $j^{\mu}(x)$ and you can rewrite it as $j^{\mu}(x)=\int dy j^{\mu}(y)\delta^{4}(x-y)$. This is just the property of the delta function $f(T)=\int dt f(t)\delta(T-t)$. And with the arguments in my answer you can make the corresponding identifications. –  nijankowski Mar 19 '13 at 3:36

1 Answer 1

$j^{\mu}(x)$ is defined as the source 4-current. Take an arbitrary coordinate system and suppose that the path of the charge is ${\bf r}={\bf r}(t)$. If you take the point charge to be $q$ its corresponding charge density is

$\rho=q\delta({\bf x}-{\bf r}(t))$

and the current density is

$\rho{\bf v}=q{\bf v}\delta({\bf x}-{\bf r}(t))$

To have relativistic invariance, you have to parametrize the trajectory in terms of the proper time of the charge (i.e. $\tau$ instead of $t$). Thus, ${\bf r}(\tau)$ specifies an invariant world line that does not depend on the coordinate system. Now, for $j^{\mu}(ct,{\bf x})$ you want to pick out the time $t$ that corresponds to any point ${\bf x}(\tau)$ on the world line. This can be achieved with the use of a delta function $\delta(t-ct(\tau))$. Here $t(\tau)$ returns the time corresponding to ${\bf r}(\tau)$ and is the 0-component of the position 4-vector. Thus, you are lead to

$j^{\mu}(x)=qc\int d\tau u^{\mu}(\tau)\delta^{4}(x-r(\tau))$

To emphasize, the role played by the $\delta$ function in this expression is to simply force the particle to be found at the correct location at each proper time.

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