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

  • 2
    $\begingroup$ 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. $\endgroup$ – user10851 Mar 18 '13 at 7:38
  • $\begingroup$ 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 $\endgroup$ – Ana S. H. Mar 18 '13 at 17:50
  • $\begingroup$ @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 $j^{\mu}=\rho u^{\mu}$ $\endgroup$ – nijankowski Mar 18 '13 at 19:14
  • $\begingroup$ @NijankowskiV. Yes it helped, but I still don't understand how did you get the integral expression. $\endgroup$ – Ana S. H. Mar 19 '13 at 3:05
  • 1
    $\begingroup$ @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. $\endgroup$ – nijankowski Mar 19 '13 at 3:36

$j^{\mu}(x)$ 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.

| cite | improve this answer | |

Another derivation is that the action for the field interacting with a current is $\propto\int d^dx A_\mu j^\mu$ while the action for interaction with a particle is $\propto\int A_\mu dx^\mu$ along the worldline. Taking care of the precise constants in front of these expressions, you can see that the formula you write is tailored precisely so that the two actions coincide.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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