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The four-current of a particle moving along a worldine $X^\nu(s)$ is defined as $$j^\mu(x^\nu) = ec \int u^\mu(s)\, \delta^4(x^\nu - X^\nu(s)) \, ds$$

So here's my proof that this is conserved: \begin{align} \partial_\mu j^\mu&= ec\partial_\mu \int u^\mu(s) \,\delta^4(x^\nu - X^\nu(s)) \, ds \\ &= ec \int u^\mu(s) \,\partial_\mu \delta^4(x^\nu - X^\nu(s)) \, ds \\ &= ec \int \frac{\partial x^\mu}{\partial s} \frac{\partial s}{\partial x^\mu}\frac{\partial (\delta^4(x^\nu - X^\nu(s)))}{\partial s} \, ds \\ &= ec \int \frac{\partial}{\partial s} \delta^4(x^\nu - X^\nu(s)) \, ds \\ &= ec \bigg|_{-\infty}^\infty \delta^4(x^\nu - X^\nu(s)) \\ &= 0 \end{align} Each step sort of makes sense but working with delta functions makes me feel deeply uneasy and I'm not sure whether this is "valid" by the physicist's standard of rigour or just some hack that appears to give the correct answer.

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Possible duplicate: – Qmechanic Feb 16 '14 at 10:27

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up vote 2 down vote accepted

A "kosher" way to do this employs test functions. Consider a test function $\phi:\mathbb R^4\to \mathbb R$. Notice that \begin{align} \int_{\mathbb R^4} d^4 x\, \partial_\mu j^\mu(x) \phi(x) &= ec\int_{-\infty}^{\infty} ds\,u^\mu(s)\int_{\mathbb R^4} d^4x\,\partial_\mu\delta^4(x - X(s))\phi(x) \\ &= -ec\int_{-\infty}^{\infty} ds\, u^\mu(s)\int_{\mathbb R^4} d^4 x\,\delta^4(x-X(s))\partial_\mu\phi(x) \\ &= -ec \int_{-\infty}^{\infty} ds\, u^\mu(s)\partial_\mu\phi(X(s)) \\ &= -ec\int_{-\infty}^{\infty} ds \frac{d}{ds}\phi(X(s)) \\ &= -ec \left[\lim_{s\to\infty}\phi(X(s))-\lim_{s\to-\infty}\phi(X(s))\right] \end{align} Now, if we assume that \begin{align} \lim_{s\to\pm\infty}|X(s)|\to\infty \end{align} namely that the particle starts and ends at infinity, then since, by definition, test functions vanish at infinity, we find that \begin{align} \int_{\mathbb R^4} d^4 x\, \partial_\mu j^\mu(x) \phi(x) = 0 \end{align} for all test functions $\phi$, and therefore, by definition \begin{align} \partial_\mu j^\mu = 0 \end{align} in the sense of distributions.

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