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I need to show that $$\partial_{\mu}\delta(R^2(\tau)) ~=~ -\frac{R_{\mu}}{R \cdot u} \frac{\partial}{\partial \tau}\delta(R^2(\tau)),$$ where $$R_{\mu} ~=~ x_{\mu} - r_{\mu}(t),$$ and $$u_{\mu} ~=~ \frac{d}{d\tau}r_{\mu}.$$ I'm really just not at all sure how to approach this problem. |
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Let $${\bf R} ~:=~ {\bf x} -{\bf r}(t), \qquad {\bf u}~:=~ \frac{d{\bf r}}{dt}, \qquad {\bf \nabla} ~:=~ \frac{\partial}{\partial {\bf x}}. $$ Now use the chain rule, $$ \left(({\bf R} \cdot {\bf u}) {\bf \nabla} +{\bf R}\frac{d}{dt}\right) \delta(R^2) ~=~\left(({\bf R} \cdot {\bf u}) 2{\bf R} +{\bf R}(2{\bf R}\cdot\frac{d{\bf R}}{dt})\right) \delta(R^2) $$ $$~=~2\left(({\bf R} \cdot {\bf u}) {\bf R} - {\bf R}({\bf R}\cdot {\bf u})\right) \delta(R^2)~=~0. $$ The same calculation can be repeated more carefully with the help of test functions. |
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