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I Differentiating a particular integral (retarded potential)

  1. Jul 30, 2017 #1
    Hi, friends! Under particular conditions on ##\phi:\mathbb{R}^3\times\mathbb{R}\to\mathbb{R}## - I think, as said here, that it is sufficient that ##\phi\in C_c^1(\mathbb{R}^4)##: please correct me if I am wrong - the following equality holds$$\frac{\partial}{\partial r_k}\int_{\mathbb{R}^3} \frac{\phi(\boldsymbol{l},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}= \int_{\mathbb{R}^3} \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|}\right) d\mu_{\boldsymbol{l}}$$where the integral is a Lebesgue integral.
    Let $$V(\boldsymbol{x},t):=\frac{1}{4\pi \varepsilon_0}\int_{\mathbb{R}^3} \frac{\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)}{\|\boldsymbol{x}-\boldsymbol{y}\|} d\mu_{\boldsymbol{y}},$$ which can be interpretated as a Lorenz gauge retarded electric potential in physics if ##\varepsilon_0## is permittivity, and let ##\rho## be under the assumptions stated for ##\phi## above. Then, by differentiating under the integral sign, we get$$\nabla_x V(\boldsymbol{x},t)=\frac{1}{4\pi \varepsilon_0}\int_{\mathbb{R}^3} -\frac{\dot\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)}{c} \frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^2} -\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|) \frac{\boldsymbol{x}-\boldsymbol{y}}{\|\boldsymbol{x}-\boldsymbol{y}\|^3} d\mu_{\boldsymbol{y}}$$where ##\dot\rho## is the partial derivative taken with respect to the second argument.

    I would like to prove to myself that ##V## satisfies the equality $$\nabla_x^2 V(\boldsymbol{x},t)=-\frac{\rho(\boldsymbol{x},t)}{\varepsilon_0}+\frac{1}{c^2}\frac{\partial^2 V(\boldsymbol{x},t)}{\partial t^2}$$which I think to be satisfied by imposing such assumptions on ##\rho##, usual in physics.
    I think that we cannot differentiate another time under the integral with respect to ##x_1##, ##x_2##, ##x_3## because informal derivations that I have found of the above equality (as in D.J. Griffiths' Introduction to Electrodynamics) introduce ##\delta## not to get a zero Lebesgue integral. Of course I am trying to get a mathematical proof, a rigourous one, and I know that, if we write a ##\delta##, there must be a mathematical justification for that: ##\forall\varphi\in C_c^2(\mathbb{R}^3)## ## \int_{\mathbb{R}^3}\frac{\nabla_y^2\varphi(\mathbf{y})}{\| \mathbf{x}-\mathbf{y}\|}d\mu_{\mathbf{y}}=-4\pi \varphi(\mathbf{x}),## while ##\int_{\mathbb{R}^3} \varphi(\mathbf{y}) \nabla_x^2 \left( \frac{1}{\| \mathbf{x}-\mathbf{y}\|} \right) d\mu_{\mathbf{y}}=\int_{\mathbb{R}^3} \varphi(\mathbf{y})\cdot 0 \,d\mu_{\mathbf{y}}####=0##.
    Could anybody help me to take the divergence of ##\nabla_x V## to prove that ##\nabla_x^2 V=-\frac{\rho}{\varepsilon_0}+\frac{1}{c^2}\frac{\partial^2 V}{\partial t^2}##?
    I ##\infty##-ly thank you!
     
    Last edited: Jul 30, 2017
  2. jcsd
  3. Aug 4, 2017 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
  4. Aug 6, 2017 #3
    I think that another lemma (proved here) useful to prove the desired equality may be: if ##\rho\in C_c^k(\mathbb{R}^4)## and ##f:\mathbb{R^3}\mapsto \mathbb{R}## is locally summable (as ##\boldsymbol{z}\mapsto 1/\|\boldsymbol{y}\|## is), then for any , then we can differentiate under the integral sign in the following way:$$D_{(\boldsymbol{x},t)}^\alpha \int_{\mathbb{R}^3} f(\boldsymbol{x}-\boldsymbol{y}) \rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)d\mu_{\boldsymbol{y}}= \int_{\mathbb{R}^3}f(\boldsymbol{x}-\boldsymbol{y})D_{(\boldsymbol{y},t)}^\alpha\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|) d\mu_{\boldsymbol{y}}$$where ##D^\alpha## are the partial derivatives of order ##\le k##.
    Nevertheless I have not been able to reach the desired result yet...
    Could anybody give me help? I heartily thank you all!
     
  5. Aug 6, 2017 #4
    Errata corrige: I think that another lemma (proved here) useful to prove the desired equality may be: if ##\rho\in C_c^k(\mathbb{R}^4)## and ##f:\mathbb{R^3}\mapsto \mathbb{R}## is locally summable (as ##\boldsymbol{z}\mapsto 1/\|\boldsymbol{y}\|## is), then for any , then we can differentiate under the integral sign in the following way:$$D_{(\boldsymbol{x},t)}^\alpha \int_{\mathbb{R}^3} f(\boldsymbol{x}-\boldsymbol{y}) \rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|)d\mu_{\boldsymbol{y}}= \int_{\mathbb{R}^3}f(\boldsymbol{x}-\boldsymbol{y})D_{(\boldsymbol{\xi},t)}^\alpha\rho(\boldsymbol{y},t-c^{-1}\|\boldsymbol{x}-\boldsymbol{y}\|) d\mu_{\boldsymbol{y}}$$where ##D^\alpha## are the partial derivatives of order ##\le k## and ##\boldsymbol{\xi}=(\xi_1,\xi_2,\xi_3)## are the first three variables of ##\rho:(\boldsymbol{\xi},\tau)\mapsto\rho(\boldsymbol{\xi},\tau) ##.
    Nevertheless I have not been able to reach the desired result yet...
    Could anybody give me help? I heartily thank you all![/QUOTE]
     
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