Relationship between current density and velocity 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:

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?
 A: $\newcommand{\bl}[1]{\boldsymbol{#1}} 
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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}
$\hebl$
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 $\il{\bl\varepsilon}$.
$\hebl$
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.
