# Calculation of an integral with retarded time

1. Dec 17, 2011

### FunWarrior

Good morning.

I would like to prove that the integral

$h^{\mu \nu} (\vec{r},t) = \int d \zeta \int d^3 \vec{y} \frac{F^{\mu \nu} (\zeta,\tilde{\tau}) \delta^{(3)} (\vec{r} - \vec{x}(\zeta,\tilde{\tau}))}{|\vec{r}-\vec{y}|}$

where $\tilde{\tau} = t - |\vec{r}-\vec{y}|$, is equal to

$\int d \zeta \frac{F^{\mu \nu} (\zeta,\tau)}{|\vec{r}-\vec{x}(\zeta,\tau)| (1-\hat{n} \cdot \dot{\vec{x}}(\zeta,\tau))}$

where $\displaystyle \hat{n}= \frac{\vec{r}-\vec{x}(\zeta,\tau)}{|\vec{r}-\vec{x}(\zeta,\tau)|}$, $\tau = t - |\vec{r}-\vec{x}(\zeta,\tau)|$ and $\dot{\vec{x}}(\zeta,\tau)$ is the derivative of $\vec{x}$ with respect to his second variable.

I would like to integrate with respect to $y$ using the Dirac function $\delta^{(3)}$ but I don't manadge to find the value of $\vec{y}$ such that $\vec{r} - \vec{x}(\zeta,\tilde{\tau})$ vanishes. I also tried to use

$\delta^{(3)} ( \vec{x} - \vec{a}) = \frac{\delta^{(3)} ( \vec{\xi} - \vec{\alpha})}{|J|}$

where $\vec{\xi} = \vec{\xi} (\vec{x})$, $\vec{\alpha} = \vec{\xi} (\vec{a})$ and $J$ is the Jacobian of the transformation of $\vec{x}$ into $\vec{\xi}$ but without success.