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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
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