Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculation of an integral with retarded time

  1. Dec 17, 2011 #1
    Good morning.

    I would like to prove that the integral

    [itex]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}|}[/itex]

    where [itex]\tilde{\tau} = t - |\vec{r}-\vec{y}|[/itex], is equal to

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

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

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

    [itex]\delta^{(3)} ( \vec{x} - \vec{a}) = \frac{\delta^{(3)} ( \vec{\xi} - \vec{\alpha})}{|J|}[/itex]

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

    Thank you in advance for your help.
  2. jcsd
  3. Dec 17, 2011 #2
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook