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I Differentiation under the integral in retarded potentials

  1. Jun 23, 2017 #1
    Hello, friends! I know, thanks to @Hawkeye18 who proved this identity to me, that, if ##\phi:V\to\mathbb{R}## is a bounded measurable function defined on the bounded measurable domain ##V\subset\mathbb{R}^3##, then, for any ##k\in\{1,2,3\}##,
    $$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{r})}{\|\boldsymbol{r}-\boldsymbol{r}\|} d\mu_{\boldsymbol{r}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{r})}{\|\boldsymbol{r}-\boldsymbol{r}\|}\right) d\mu_{\boldsymbol{r}}.$$
    An important application of this result is the proof that the magnetostatic vector potential ##\frac{\mu_0}{4\pi}\int_\mathbb{R}^3 \frac{\boldsymbol{J}(\boldsymbol{r})}{\|\boldsymbol{r}-\boldsymbol{r}\|} d\mu_{\boldsymbol{r}}## satisfies Ampère's law.
    Since I am, fuitlessly until now, trying to prove to myself that the Lorenz gauge magnetodynamic retarded vector potential satisfies Maxwell's equations, I am wondering whether, given a constant ##c##, it is also true that $$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{r},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{r}\|)}{\|\boldsymbol{r}-\boldsymbol{r}\|} d\mu_{\boldsymbol{r}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{r},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{r}\|)}{\|\boldsymbol{r}-\boldsymbol{r}\|}\right) d\mu_{\boldsymbol{r}}$$for, say ##\phi\in C^2(\mathbb{R}^4)## and ##\phi(-,t)## compactly supported as a function of the first vector variable.
    If it is true, how could we prove it?
    I ##\infty##-ly thank any answerer!
     
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  3. Jun 24, 2017 #2

    stevendaryl

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    I assume that you mean [itex]\| \mathbf{r} - \mathbf{r'} \|[/itex], and that you mean [itex]\frac{\partial}{\partial r'_k}[/itex]?

    I have not worked out the extent to which [itex]\phi[/itex] being compactly supported affects whether you can pull the partial derivative inside the integral. However, your modified problem is obtained from the original by replacing [itex]\phi(\mathbf{r})[/itex] by [itex]\tilde{\phi}(\mathbf{r}) \equiv \phi(\mathbf{r}, t - \frac{1}{c} \| \mathbf{r} - \mathbf{r'} \|)[/itex]. But knowing that [itex]\phi(\mathbf{r}, t)[/itex] is compactly supported in the first argument doesn't necessarily imply that [itex]\tilde{\phi}[/itex] is compactly supported.

    Here's a analogous problem:

    Let [itex]f(x,y)[/itex] be a function that is zero unless [itex]|x| < 2|y|[/itex]. So that's compactly supported in the first argument. But if we define [itex]\tilde{f}(x,y) = f(x, y-x)[/itex], then [itex]\tilde{f}[/itex] is not compactly supported in the first argument.
     
  4. Jun 24, 2017 #3
    Thank you very much! Yes, @stevendaryl, I meant:

    Hello, friends! I know, thanks to @Hawkeye18 who proved this identity to me, that, if ##\phi:V\to\mathbb{R}## is a bounded measurable function defined on the bounded measurable domain ##V\subset\mathbb{R}^3##, then, for any ##k\in\{1,2,3\}##,
    $$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|}\right) d\mu_{\boldsymbol{l}}.$$
    An important application of this result is the proof that the magnetostatic vector potential ##\frac{\mu_0}{4\pi}\int_\mathbb{R}^3 \frac{\boldsymbol{J}(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}## satisfies Ampère's law.
    Since I am, fuitlessly until now, trying to prove to myself that the Lorenz gauge magnetodynamic retarded vector potential satisfies Maxwell's equations, I am wondering whether, given a constant ##c##, it is also true that $$\frac{\partial}{\partial r_k}\int_V \frac{\phi(\boldsymbol{l},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}}= \int_V \frac{\partial}{\partial r_k}\left(\frac{\phi(\boldsymbol{l},t-c^{-1}\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|}\right) d\mu_{\boldsymbol{l}}$$for, say ##\phi\in C^2(\mathbb{R}^4)## and ##\phi(-,t)## compactly supported as a function of the first vector variable.
    If it is true, how could we prove it?
    I ##\infty##-ly thank any answerer!
     
  5. Jul 1, 2017 #4
    Dear @stevendaryl , I have tried to adapt the proof by Hawkeye18 to this problem, and, as far as I understand, it can used in an almost identical way, with ##\frac{\partial }{\partial r_k}\left(\frac{\phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} \right)## instead of ##\frac{\partial }{\partial r_k}\left(\frac{\phi(\boldsymbol{l})}{\|\boldsymbol{r}-\boldsymbol{l}\|} \right)##, provided for example that ##\phi\in C^2(\mathbb{R}^4)## and that the two functions $$\boldsymbol{l}\mapsto \phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)$$$$\boldsymbol{l}\mapsto \dot\phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)$$ are compactly supported. This last condition is met if ##\phi\in C_2^2(\mathbb{R}^4)##.
    By following Hawkeye18's reasoning we also see that the derivatives of ##V:=\frac{1}{4\pi\varepsilon}\int_{\mathbb{R}^3} \frac{\phi(\boldsymbol{l},t-c^1\|\boldsymbol{r}-\boldsymbol{l}\|)}{\|\boldsymbol{r}-\boldsymbol{l}\|} d\mu_{\boldsymbol{l}} ## are continuous and also that ##V\in C^1(\mathbb{R}^3)##.
    Am I right? I ##\infty##-ly thank you!
     
  6. Jul 7, 2017 #5
    @stevendaryl and anyone else reading: Are my conclusions in the preceding post correct?
     
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