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I am confused when reading this section about electromagnetic field in '<Introduction to Many-body quantum theory in condensed matter physics> Henrik Bruus& Karsten Flensberg' Page 21:

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Why here in the last line velocity equal to the integral of the current density in the whole space? $$\vec{v} = \int d^3\vec{r} \vec{J}~?$$ From this relation, it seems like $$\vec{J}(r) = \vec{v}(r') \delta^3(r-r')~?$$ But I'm not sure, why it's defined like this? Could anyone explain it? Why it's not just $-qv\cdot \delta A = -qJ\cdot\delta A $ if elsewhere are zero?

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  • $\begingroup$ are we discussing a point particle here? The above makes sense if we are. $\endgroup$ Commented Nov 17, 2022 at 18:43
  • $\begingroup$ Yes. It's a particle. $\endgroup$ Commented Nov 17, 2022 at 18:52
  • $\begingroup$ It does not say $d^3 r$ under the integral in your reference but only $dr$ $\endgroup$
    – Thomas
    Commented Nov 18, 2022 at 8:23

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$\newcommand{\bl}[1]{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\mb}[1]{\mathbf {#1}} \newcommand{\mc}[1]{\mathcal {#1}} \newcommand{\mr}[1]{\mathrm {#1}} \newcommand{\mf}[1]{\mathfrak{#1}} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\il}[1]{\:#1\:} \newcommand{\plr}[1]{\left(#1\right)} \newcommand{\blr}[1]{\left[#1\right]} \newcommand{\vlr}[1]{\left\vert#1\right\vert} \newcommand{\Vlr}[1]{\left\Vert#1\right\Vert} \newcommand{\lara}[1]{\left\langle#1\right\rangle} \newcommand{\lav}[1]{\left\langle#1\right|} \newcommand{\vra}[1]{\left|#1\right\rangle} \newcommand{\lavra}[2]{\left\langle#1|#2\right\rangle} \newcommand{\lavvra}[3]{\left\langle#1\right|#2\left|#3\right\rangle} \newcommand{\vp}{\vphantom{\dfrac{a}{b}}} \newcommand{\Vp}[1]{\vphantom{#1}} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\x}{\bl\times} \newcommand{\ox}{\bl\otimes} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\qqlraqq}{\qquad\bl{-\!\!\!-\!\!\!-\!\!\!\longrightarrow}\qquad} \newcommand{\qqLraqq}{\qquad\boldsymbol{\e\!\e\!\e\!\e\!\Longrightarrow}\qquad} \newcommand{\tl}[1]{\tag{#1}\label{#1}} \newcommand{\zref}[1]{\text{equation }\eqref{#1}} \newcommand{\hebl}{\bl{=\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!==\!=\!=\!=}} \newcommand{\hmbl}{$\bl{-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-}$}$

The electric charge density scalar $\il{\rho\plr{\mb x,t}}$ and the electric current density vector $\il{\mb j\plr{\mb x,t}}$ appear in the following two of the Maxwell equations in vacuum \begin{align} \bl{\nabla\cdot}\mb E\plr{\mb x,t} & \e \dfrac{\rho\plr{\mb x,t}}{\epsilon_0} \tl{01}\\ \bl{\nabla\times}\mb B\plr{\mb x,t} & \e \mu_0\,\mb j\plr{\mb x,t} \p \dfrac{1}{c^2} \dfrac{\partial \mb E\plr{\mb x,t}}{\partial t} \tl{02} \end{align} where $\il{\mb x\e \plr{x_1,x_2,x_3}}$ the field point.

Note the $\il{\rho,\mb j}$ are volume densities (there are surface densities and line densities also) connected by the relation

\begin{equation} \mb j\plr{\mb x,t}\e \mb v\plr{\mb x,t}\rho\plr{\mb x,t} \tl{03} \end{equation} where $\il{\mb v\plr{\mb x,t}}$ the velocity of electric charges contained in an infinitesimal volume at field point $\il{\mb x\e \plr{x_1,x_2,x_3}}$.

Now, consider that there exists only a single point particle with electric charge $\il{q}$ and position vector \begin{equation} \mb x'\plr{t}\e \blr{\,x'_1\plr{t},x'_2\plr{t},x'_3\plr{t}\Vp{A^2}} \tl{04} \end{equation}

A continuous real function $\il{\rho\plr{\mb x,t}}$ could represent the single point charge $\il{q}$ at $\il{\mb x'\e \plr{x'_1,x'_2,x'_3}}$ if it satisfies the following conditions (we omit the $\il{t\m}$dependence) \begin{equation} \begin{split} \blr{\dfrac{\rho\plr{\mb x}}{q}} &\e 0 \qquad\texttt{for any}\qquad \mb x\bl\ne\mb x' \Vp{\dfrac{a}{\tfrac{a}{b}}}\\ \iiint\limits_{\mc B\plr{\mb x',\bl\varepsilon}}\blr{\dfrac{\rho\plr{\mb x}}{q}}\mr d^3\mb x &\e 1 \qquad\texttt{for any}\qquad \bl\varepsilon\gr 0 \Vp{\dfrac{\dfrac{a}{b}}{b}}\\ \end{split} \tl{05} \end{equation} where $\il{\mc B\plr{\mb x',\bl\varepsilon}}$ a ball with center at $\il{\mb x'}$ and radius $\il{\bl\varepsilon}$.

But these conditions define exactly the 3-dimensional $\il{\delta\m\!}$function, see equation \eqref{A-02} in APPENDIX, so \begin{equation} \blr{\dfrac{\rho\plr{\mb x}}{q}}\e \delta\plr{\mb x\m\mb x'}\quad \bl\implies \quad\rho\plr{\mb x}\e q\,\delta\plr{\mb x\m\mb x'} \tl{06} \end{equation} Restoring the $\il{t\m}$dependence we have \begin{equation} \rho\plr{\mb x,t}\e q\,\delta\blr{\mb x\m\mb x'\plr{t}}\e q\,\delta\blr{x_1\m x'_1\plr{t}}\delta\blr{x_2\m x'_2\plr{t}}\delta\blr{x_3\m x'_3\plr{t}} \tl{07} \end{equation} The velocity field $\il{\mb v\plr{\mb x,t}}$ is everywhere $\il{\bl 0}$ except on the position of the point electric charge $\il{q}$ \begin{equation} \mb v\plr{\mb x,t} \e \left. \begin{cases} \hp{aaaa}\bl 0 &\quad\texttt{if}\quad \mb x\bl\ne\mb x'\vp\\ \dfrac{\mr d\mb x'}{\mr dt}\e \mb v\plr{t}&\quad\texttt{if}\quad \mb x\e\mb x' \end{cases} \right\} \tl{08} \end{equation} So for the electric current density vector $\il{\mb j\plr{\mb x,t}}$ $\zref{03}$ yields \begin{equation} \begin{split} \mb j\plr{\mb x,t}&\e q\,\mb v\plr{t}\delta\blr{\mb x\m\mb x'\plr{t}}\\ &\e q\,\plr{\dfrac{\mr dx_1'}{\mr dt},\dfrac{\mr dx_2'}{\mr dt},\dfrac{\mr dx_3'}{\mr dt}}\delta\blr{x_1\m x'_1\plr{t}}\delta\blr{x_2\m x'_2\plr{t}}\delta\blr{x_3\m x'_3\plr{t}}\\ \end{split} \tl{09} \end{equation}

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APPENDIX : On Dirac $\il{\delta\m\!}$functions

$\square$ The 1-dimensional $\il{\delta\m\!}$function satisfies the following conditions \begin{equation} \begin{split} \delta\plr{x} &\e 0 \qquad\texttt{for any}\qquad x\bl\ne 0\Vp{\dfrac{a}{\tfrac{a}{b}}}\\ \int\limits_{\m\bl\varepsilon}^{\p\bl\varepsilon}\delta\plr{x}\mr dx &\e 1 \qquad\texttt{for any}\qquad \bl\varepsilon\gr 0 \Vp{\dfrac{\dfrac{a}{b}}{b}}\\ \end{split} \tl{A-01} \end{equation}

$\square$ The 3-dimensional $\il{\delta\m\!}$function satisfies the following conditions \begin{equation} \begin{split} \delta\plr{\mb x} &\e 0 \qquad\texttt{for any}\qquad \mb x\bl\ne \bl 0\Vp{\dfrac{a}{\tfrac{a}{b}}}\\ \iiint\limits_{_{\mc B\plr{\mb 0,\bl\varepsilon}}}\delta\plr{\mb x}\mr d^3 \mb x &\e 1 \qquad\texttt{for any}\qquad \bl\varepsilon\gr 0 \Vp{\dfrac{\dfrac{a}{b}}{b}}\\ \end{split} \tl{A-02} \end{equation} where $\il{\mc B\plr{\bl 0,\bl\varepsilon}}$ a ball with center at $\il{\bl 0}$ and radius $\il{\bl\varepsilon}$.

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In my answer here Magnetic field due to a single moving charge the Biot and Savart Law is obtained by use of a Dirac $\:\delta\m$ expression of the electric current $\:I\:$ similar to equation \eqref{09}, see equations (BS-01) to (BS-05) there.

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