Calculation of an integral with retarded time

In summary, the conversation discusses how to prove the equality of two integrals involving the Liénard-Wiechert potentials. The speaker suggests using a Dirac function and provides a helpful resource for further understanding.
  • #1
FunWarrior
4
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.
 
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  • #2

FAQ: Calculation of an integral with retarded time

1. How is the integral with retarded time calculated?

The integral with retarded time is calculated by taking the time delay into account when evaluating the integral. This is done by substituting the variable of integration with a new variable that takes into account the time delay.

2. What is the purpose of calculating an integral with retarded time?

The purpose of calculating an integral with retarded time is to take into account the time delay in physical processes. This is particularly important in fields such as electromagnetism and quantum mechanics, where the speed of light is a limiting factor.

3. How does the calculation of an integral with retarded time differ from a regular integral?

The calculation of an integral with retarded time differs from a regular integral by incorporating the time delay into the integration. This means that the variable of integration is substituted with a new variable that takes into account the time delay.

4. What are some applications of calculating integrals with retarded time?

Calculating integrals with retarded time has various applications in physics, particularly in fields such as electromagnetism and quantum mechanics. It is used to accurately model the time-dependent behavior of physical systems, taking into account the time delay in processes.

5. Are there any challenges in calculating integrals with retarded time?

Yes, there can be challenges in calculating integrals with retarded time. This is because the equations and variables involved can become more complex due to the incorporation of the time delay. Additionally, numerical methods may be required for more complex integrals, which can introduce errors and require careful analysis.

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