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Is there a physical meaning to the equation $$\delta(x-a)=\dfrac{\delta(\xi-\alpha)}{|J|} \, ?$$ In non-rectangular coordinate systems where the transformation is non-singular, what is the implication of dividing the Dirac delta function by the Jacobian of the transformation to the coordinate system?

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We can handle easily integrals where the vector variable of integration, let $\:\mathbf{u}\:$, is the argument of the $\:\delta-$function, for example

\begin{align} \iiint\delta^{3}\left(\mathbf{u}\right)\rm{L}\left(\mathbf{u}\right) \mathrm d^{3}\mathbf{u}&=\rm{L}\left(\mathbf{0}\right) \tag{01}\\ \iiint\delta^{3}\left(\mathbf{u}\right)\mathbf{M}\left(\mathbf{u}\right) \mathrm d^{3}\mathbf{u}&=\mathbf{M}\left(\mathbf{0}\right) \tag{02} \end{align} where $\:\rm{L}\left(\mathbf{u}\right)\:$ and $\:\mathbf{M}\left(\mathbf{u}\right)\:$ scalar and vector functions respectively of the vector variable $\:\mathbf{u}$.

But to handle integrals of the form \begin{align} &\iiint \delta^{3}\bigl[\mathbf{F}\left(\mathbf{x}\right)\bigr]\rm{H}\left(\mathbf{x}\right) \mathrm d^{3}\mathbf{x} \tag{03}\\ &\iiint\delta^{3}\bigl[\mathbf{F}\left(\mathbf{x}\right)\bigr]\mathbf{G}\left(\mathbf{x}\right) \mathrm d^{3}\mathbf{x} \tag{04} \end{align} where $\:\rm{H}\left(\mathbf{x}\right)\:$ and $\:\mathbf{G}\left(\mathbf{x}\right),\mathbf{F}\left(\mathbf{x}\right)\:$ scalar and vector functions respectively of the vector variable $\:\mathbf{x}$, with $\:\mathbf{F}\left(\mathbf{x}\right)\ne \mathbf{x}\:$, that is the argument of the $\:\delta-$function is not the variable of integration, we must proceed to a change of the vector variable from $\:\mathbf{x}\:$ to $\:\mathbf{u}\:$ \begin{equation} \mathbf{u}=\mathbf{F}\left(\mathbf{x}\right) \tag{05} \end{equation} and check with care if we can convert without complications these integrals to expressions like (01) and (02).

Indeed, if the vector function $\:\mathbf{F}\:$ in (05) is invertible then \begin{equation} \mathbf{x}= \mathbf{F}^{-1}\left(\mathbf{u}\right) \tag{06} \end{equation} It remains one step : to find the relation between the infinitesimal volumes $\:\mathrm d^{3}\mathbf{x}=\mathrm dx_{1} \mathrm dx_{2} \mathrm dx_{3}\:$ and $\:\mathrm d^{3}\mathbf{u}= \mathrm du_{1}\mathrm du_{2}\mathrm du_{3}\:$. We have the following linear transformation between infinitesimals

\begin{align} \mathrm dx_{1} &=\dfrac{\partial x_{1}}{\partial u_{1}}\mathrm du_{1}+\dfrac{\partial x_{1}}{\partial u_{2}}\mathrm du_{2}+\dfrac{\partial x_{1}}{\partial u_{3}}\mathrm du_{3} \tag{07}\\ \mathrm dx_{2} &=\dfrac{\partial x_{2}}{\partial u_{1}}\mathrm du_{1}+\dfrac{\partial x_{2}}{\partial u_{2}}\mathrm du_{2}+\dfrac{\partial x_{2}}{\partial u_{3}}\mathrm du_{3} \tag{08}\\ \mathrm dx_{3} &=\dfrac{\partial x_{3}}{\partial u_{1}}\mathrm du_{1}+\dfrac{\partial x_{3}}{\partial u_{2}}\mathrm du_{2}+\dfrac{\partial x_{3}}{\partial u_{3}}\mathrm du_{3} \tag{09} \end{align} or \begin{equation} \mathrm d\mathbf{x}= \begin{bmatrix} \mathrm dx_{1}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm dx_{2}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm dx_{3}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial x_{1}}{\partial u_{1}}&\dfrac{\partial x_{1}}{\partial u_{2}}&\dfrac{\partial x_{1}}{\partial u_{3}}\\ \dfrac{\partial x_{2}}{\partial u_{1}}&\dfrac{\partial x_{2}}{\partial u_{2}}&\dfrac{\partial x_{2}}{\partial u_{3}}\\ \dfrac{\partial x_{3}}{\partial u_{1}}&\dfrac{\partial x_{3}}{\partial u_{2}}&\dfrac{\partial x_{3}}{\partial u_{3}}\\ \end{bmatrix} \begin{bmatrix} \mathrm du_{1}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm du_{2}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm du_{3}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}} \end{bmatrix} = \mathbf{J}\left(\mathbf{F}^{-1}\right)\mathrm d\mathbf{u} \tag{10} \end{equation} where \begin{equation} \mathbf{J}\left(\mathbf{F}^{-1}\right)\stackrel{\text{def}}{\equiv} \begin{bmatrix} \dfrac{\partial x_{1}}{\partial u_{1}}&\dfrac{\partial x_{1}}{\partial u_{2}}&\dfrac{\partial x_{1}}{\partial u_{3}}\\ \dfrac{\partial x_{2}}{\partial u_{1}}&\dfrac{\partial x_{2}}{\partial u_{2}}&\dfrac{\partial x_{2}}{\partial u_{3}}\\ \dfrac{\partial x_{3}}{\partial u_{1}}&\dfrac{\partial x_{3}}{\partial u_{2}}&\dfrac{\partial x_{3}}{\partial u_{3}}\\ \end{bmatrix} \tag{11} \end{equation} the so-called Jacobian matrix of the vector function $\:\mathbf{F}^{-1}$, a matrix function of $\:\mathbf{u}\:$. We know that for an invertible linear transformation the ratio of the algebraic values of the transformed to the initial volume is equal to the determinant of the respective matrix. So \begin{equation} \mathrm d^{3}\mathbf{x}=\left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert\mathrm d^{3}\mathbf{u} \tag{12} \end{equation} where \begin{equation} \left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert\stackrel{\text{def}}{\equiv}\det\left[\mathbf{J}\left(\mathbf{F}^{-1}\right)\right]= \begin{vmatrix} \dfrac{\partial x_{1}}{\partial u_{1}}&\dfrac{\partial x_{1}}{\partial u_{2}}&\dfrac{\partial x_{1}}{\partial u_{3}}\\ \dfrac{\partial x_{2}}{\partial u_{1}}&\dfrac{\partial x_{2}}{\partial u_{2}}&\dfrac{\partial x_{2}}{\partial u_{3}}\\ \dfrac{\partial x_{3}}{\partial u_{1}}&\dfrac{\partial x_{3}}{\partial u_{2}}&\dfrac{\partial x_{3}}{\partial u_{3}}\\ \end{vmatrix} \tag{13} \end{equation} the so-called Jacobian of the vector function $\:\mathbf{F}^{-1}$, the determinant of the Jacobi matrix $\:\mathbf{J}\left(\mathbf{F}^{-1}\right)\:$. The Jacobian is a scalar function of $\:\mathbf{u}\:$.

Now, if in equations (03) and (04) we make the following substitutions

\begin{equation} \mathbf{F}\left(\mathbf{x}\right) \longrightarrow \mathbf{u}\;, \quad \mathbf{x}\longrightarrow \mathbf{F}^{-1}\left(\mathbf{u}\right)\;, \quad \mathrm d^{3}\mathbf{x}\longrightarrow \left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert\mathrm d^{3}\mathbf{u} \tag{14} \end{equation} according to equations (05),(06) and (12) respectively, then these integrals are converted to the form of (01) and (02), that is

\begin{align} \iiint \delta^{3}\bigl[\mathbf{F}\left(\mathbf{x}\right)\bigr]\mathrm{H}\left(\mathbf{x}\right) \mathrm{d}^{3}\mathbf{x} & =\iiint \delta^{3}\left(\mathbf{u}\right)\mathrm{H}\left[\mathbf{F}^{-1}\left(\mathbf{u}\right)\right] \left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert\mathrm d^{3}\mathbf{u} \nonumber\\ &=\mathrm{H}\left[\mathbf{F}^{-1}\left(\mathbf{0}\right)\right]\left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert_{\mathbf{u}=\mathbf{0}} \tag{15}\\ \nonumber\\ \iiint \delta^{3}\bigl[\mathbf{F}\left(\mathbf{x}\right)\bigr]\mathbf{G}\left(\mathbf{x}\right) \mathrm d^{3}\mathbf{x} & =\iiint \delta^{3}\left(\mathbf{u}\right)\mathbf{G}\bigl[\mathbf{F}^{-1}\left(\mathbf{u}\right)\bigr] \left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert\mathrm d^{3}\mathbf{u} \nonumber\\ &=\mathbf{G}\bigl[\mathbf{F}^{-1}\left(\mathbf{0}\right)\bigr]\left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert_{\mathbf{u}=\mathbf{0}} \tag{16} \end{align}

Note that starting from equation (06) we found equations (07) to (013) concerning the vector function $\:\mathbf{F}^{-1}$. With similar steps we can start from (05) and find the respective equations for the vector function $\:\mathbf{F}$. Indeed

\begin{align} \mathrm du_{1} &=\dfrac{\partial u_{1}}{\partial x_{1}}\mathrm dx_{1}+\dfrac{\partial u_{1}}{\partial x_{2}}\mathrm dx_{2}+\dfrac{\partial u_{1}}{\partial x_{3}}\mathrm dx_{3} \tag{17}\\ \mathrm du_{2} &=\dfrac{\partial u_{2}}{\partial x_{1}}\mathrm dx_{1}+\dfrac{\partial u_{2}}{\partial x_{2}}\mathrm dx_{2}+\dfrac{\partial u_{2}}{\partial x_{3}}\mathrm dx_{3} \tag{18}\\ \mathrm du_{3} &=\dfrac{\partial u_{3}}{\partial x_{1}}\mathrm dx_{1}+\dfrac{\partial u_{3}}{\partial x_{2}}\mathrm dx_{2}+\dfrac{\partial u_{3}}{\partial x_{3}}\mathrm dx_{3} \tag{19} \end{align} or \begin{equation} \mathrm d\mathbf{u}= \begin{bmatrix} \mathrm du_{1}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm du_{2}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm du_{3}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}} \end{bmatrix} = \begin{bmatrix} \dfrac{\partial u_{1}}{\partial x_{1}}& \dfrac{\partial u_{1}}{\partial x_{2}}&\dfrac{\partial u_{1}}{\partial x_{3}}\\ \dfrac{\partial u_{2}}{\partial x_{1}}& \dfrac{\partial u_{2}}{\partial x_{2}}& \dfrac{\partial u_{2}}{\partial x_{3}}\\ \dfrac{\partial u_{3}}{\partial x_{1}}& \dfrac{\partial u_{3}}{\partial x_{2}}& \dfrac{\partial u_{3}}{\partial x_{3}}\\ \end{bmatrix} \begin{bmatrix} \mathrm dx_{1}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm dx_{2}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}}\\ \mathrm dx_{3}\vphantom{\dfrac{\partial x_{1}^{\prime}}{\partial u_{1}}} \end{bmatrix} = \mathbf{J}\left(\mathbf{F}\right)\mathrm d\mathbf{x} \tag{20} \end{equation} where \begin{equation} \mathbf{J}\left(\mathbf{F}\right)\stackrel{\text{def}}{\equiv} \begin{bmatrix} \dfrac{\partial u_{1}}{\partial x_{1}}& \dfrac{\partial u_{1}}{\partial x_{2}}&\dfrac{\partial u_{1}}{\partial x_{3}}\\ \dfrac{\partial u_{2}}{\partial x_{1}}& \dfrac{\partial u_{2}}{\partial x_{2}}& \dfrac{\partial u_{2}}{\partial x_{3}}\\ \dfrac{\partial u_{3}}{\partial x_{1}}& \dfrac{\partial u_{3}}{\partial x_{2}}& \dfrac{\partial u_{3}}{\partial x_{3}}\\ \end{bmatrix} \tag{21} \end{equation} the Jacobian matrix of the vector function $\:\mathbf{F}$, a matrix function of $\:\mathbf{x}\:$. So \begin{equation} \mathrm d^{3}\mathbf{u}=\left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1},x_{2},x_{3}\right)}\right\vert \mathrm d^{3}\mathbf{x} \tag{22} \end{equation} where \begin{equation} \left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1},x_{2},x_{3}\right)}\right\vert\stackrel{\text{def}}{\equiv}\det\left[\mathbf{J}\left(\mathbf{F}\right)\right]= \begin{vmatrix} \dfrac{\partial u_{1}}{\partial x_{1}}& \dfrac{\partial u_{1}}{\partial x_{2}}&\dfrac{\partial u_{1}}{\partial x_{3}}\\ \dfrac{\partial u_{2}}{\partial x_{1}}& \dfrac{\partial u_{2}}{\partial x_{2}}& \dfrac{\partial u_{2}}{\partial x_{3}}\\ \dfrac{\partial u_{3}}{\partial x_{1}}& \dfrac{\partial u_{3}}{\partial x_{2}}& \dfrac{\partial u_{3}}{\partial x_{3}}\\ \end{vmatrix} \tag{23} \end{equation} the Jacobian of the vector function $\:\mathbf{F}$, the determinant of the Jacobi matrix $\:\mathbf{J}\left(\mathbf{F}\right)\:$. The Jacobian is a scalar function of $\:\mathbf{x}\:$.

From equations (11) and (21) we have \begin{align} \mathbf{J}\left(\mathbf{F}^{-1}\right)\cdot \mathbf{J}\left(\mathbf{F}\right) & = \begin{bmatrix} \dfrac{\partial x_{1}}{\partial u_{1}}&\dfrac{\partial x_{1}}{\partial u_{2}}&\dfrac{\partial x_{1}}{\partial u_{3}}\\ \dfrac{\partial x_{2}}{\partial u_{1}}&\dfrac{\partial x_{2}}{\partial u_{2}}&\dfrac{\partial x_{2}}{\partial u_{3}}\\ \dfrac{\partial x_{3}}{\partial u_{1}}&\dfrac{\partial x_{3}}{\partial u_{2}}&\dfrac{\partial x_{3}}{\partial u_{3}}\\ \end{bmatrix} \begin{bmatrix} \dfrac{\partial u_{1}}{\partial x_{1}}& \dfrac{\partial u_{1}}{\partial x_{2}}&\dfrac{\partial u_{1}}{\partial x_{3}}\\ \dfrac{\partial u_{2}}{\partial x_{1}}& \dfrac{\partial u_{2}}{\partial x_{2}}& \dfrac{\partial u_{2}}{\partial x_{3}}\\ \dfrac{\partial u_{3}}{\partial x_{1}}& \dfrac{\partial u_{3}}{\partial x_{2}}& \dfrac{\partial u_{3}}{\partial x_{3}}\\ \end{bmatrix} \nonumber\\ &= \begin{bmatrix} \dfrac{\partial x_{1}}{\partial x_{1}}&0&0\\ 0&\dfrac{\partial x_{2}}{\partial x_{2}}&0\\ 0&0&\dfrac{\partial x_{3}}{\partial x_{3}} \end{bmatrix} = \begin{bmatrix} \:\:1 \:\:&\:\:0\:\:& \:\:0\:\:\vphantom{\dfrac{\partial x_{1}}{\partial x_{1}}}\\ \:\:0 \:\:&\:\:1\:\:& \:\:0\:\:\vphantom{\dfrac{\partial x_{1}}{\partial x_{1}}}\\ \:\:0\:\:&\:\:0\:\:& \:\:1\:\:\vphantom{\dfrac{\partial x_{1}}{\partial x_{1}}}\\ \end{bmatrix} = \mathbf{I} \nonumber \end{align} so \begin{equation} \mathbf{J}\left(\mathbf{F}^{-1}\right)= \left[\mathbf{J}\left(\mathbf{F}\right)\right]^{-1} \tag{24} \end{equation} This means that for the Jacobian determinants we have \begin{equation} \left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert=\left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1},x_{2},x_{3}\right)}\right\vert^{-1} \tag{25} \end{equation} and equation (12) is completed to \begin{equation} \mathrm d^{3}\mathbf{x}=\left\vert\dfrac{\partial\left(x_{1},x_{2},x_{3}\right)}{\partial\left(u_{1},u_{2},u_{3}\right)}\right\vert\mathrm d^{3}\mathbf{u}=\left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1},x_{2},x_{3}\right)}\right\vert^{-1}\mathrm d^{3}\mathbf{u} \tag{26} \end{equation}


$\color{blue}{\textbf{Example A :}}$

Suppose we want to determine the integral \begin{equation} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\delta\left(\alpha x+\beta\right)\mathrm{h}\left(x\right)\mathrm dx\,, \quad \alpha \in \mathbb{R}\setminus\lbrace 0 \rbrace \tag{A-01} \end{equation} Under the variable change \begin{equation} u=\alpha x+\beta \tag{A-02} \end{equation} we have \begin{equation} x=\dfrac{u-\beta}{\alpha} \tag{A-03} \end{equation} and \begin{equation} \mathrm d x=\dfrac{\mathrm d u}{\alpha} \tag{A-04} \end{equation} The Jacobian here is constant, that is independent of $\:u$ \begin{equation} \mathrm{J}\left(u\right)=\dfrac{\partial\left(x\right)}{\partial\left(u\right)}=\dfrac{1}{\alpha} \tag{A-05} \end{equation} So \begin{align} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(\alpha x+\beta\right)\mathrm{h}\left(x\right)\mathrm dx & =\int\limits_{\boldsymbol{-} \mathrm{sign}(\alpha) \, \infty}^{\boldsymbol{+}\mathrm{sign}(\alpha) \, \infty}\!\!\!\!\!\!\!\!\delta\left(u\right)\mathrm{h}\left(\dfrac{u-\beta}{\alpha}\right)\dfrac{\mathrm d u}{\alpha} \nonumber\\ & = \dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}}\int\limits_{\boldsymbol{-} \infty}^{\boldsymbol{+} \infty}\!\!\delta\left(u\right)\mathrm{h}\left(\dfrac{u-\beta}{\alpha}\right)\mathrm d u =\dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}}\mathrm{h}\left(\boldsymbol{-}\dfrac{\,\beta\,}{\alpha}\right) \nonumber\\ & = \dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\mathrm{h}\left(x\right)\mathrm dx \tag{A-06} \end{align} that is \begin{equation} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(\alpha x+\beta\right)\mathrm{h}\left(x\right)\mathrm dx=\dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \int\limits_{\boldsymbol{-}\infty}^{\boldsymbol{+}\infty}\!\delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right)\mathrm{h}\left(x\right)\mathrm dx \tag{A-07} \end{equation} expressed as \begin{equation} \delta\left(\alpha x+\beta\right) \doteq \dfrac{1}{\boldsymbol{\vert}\alpha\boldsymbol{\vert}} \delta\left(x\boldsymbol{+}\frac{\,\beta\,}{\alpha}\right) \tag{A-08} \end{equation} Equality in (A-08) is valid in the sense of equation (A-07).


$\color{blue}{\textbf{Example B :}}$

In this Example we'll find an important physical meaning of the Jacobian in relation to Dirac delta function in classical electrodynamics and more precisely in the electric and magnetic fields produced by a point charge moving in any arbitrary way.

Suppose that we want to determine the electric and magnetic fields produced by a point charge moving in any arbitrary way. We start with the retarded potentials, scalar and vector : \begin{align} \phi\left(\mathbf{x},t\right)&=\dfrac{1}{4\pi\epsilon_{o}}\iiint \dfrac{\rho\left(\mathbf{x}^{\prime},t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)}{\|\mathbf{x}^{\prime}-\mathbf{x}\|}\mathrm d^{3}\mathbf{x}^{\prime}\:, \quad \text{scalar potential} \tag{B-01}\\ \mathbf{A}\left(\mathbf{x},t\right)&=\dfrac{\mu_{o}}{\:\:4\pi\:\:}\iiint \dfrac{\mathbf{j}\left(\mathbf{x}^{\prime},t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)}{\|\mathbf{x}^{\prime}-\mathbf{x}\|}\mathrm d^{3}\mathbf{x}^{\prime}\:, \quad \text{vector potential} \tag{B-02} \end{align} By these two equations we'll find the potentials at field point $\:\mathbf{x}= \left(x_{1},x_{2},x_{3}\right)\:$ and time $\:t\:$, taking into account the contributions of charges and their currents from all points $\:\mathbf{x}^{\prime}= \left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)\:$ at the retarded time \begin{equation} t^{\prime}=t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c} \tag{B-03} \end{equation} since a time period $\:\Delta t =\|\mathbf{x}^{\prime}-\mathbf{x}\|/c\:$ is needed for this contribution to travel with the speed of light $\:c\:$ from $\:\mathbf{x}^{\prime}\:$ to $\:\mathbf{x}\:$.

So, let a point charge $\:q\:$ moving with position vector $\:\boldsymbol{\xi}\left(t\right)$, see $\color{blue}{\textbf{Figure-01}}$. We suppose that \begin{equation} \left\Vert \dfrac{\mathrm d\boldsymbol{\xi}\left(t\right)}{\mathrm d t} \right\Vert <c \tag{B-04} \end{equation} The volume charge density would be expressed via Dirac $\:\delta-$function \begin{equation} \rho\left(\mathbf{x},t\right)=q\cdot\delta^{3}\bigl[\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr] \tag{B-05} \end{equation} as well as the charge current density \begin{equation} \mathbf{j}\left(\mathbf{x},t\right)=q\cdot\delta^{3}\bigl[\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr] \cdot\dfrac{\mathrm d\boldsymbol{\xi}\left(t\right)}{\mathrm dt}=q\cdot\delta^{3}\bigl[\mathbf{x}-\boldsymbol{\xi}\left(t\right)\bigr]\cdot\mathbf{v}\left(t\right) \tag{B-06} \end{equation} where \begin{equation} \mathbf{v}\left(t\right)= \bigl[\upsilon_{1}\left(t\right),\upsilon_{2}\left(t\right),\upsilon_{3}\left(t\right)\bigr]= \biggl[\dfrac{\mathrm d \xi_{1}\left(t\right)}{\mathrm d t},\dfrac{\mathrm d \xi_{2}\left(t\right)}{\mathrm d t},\dfrac{\mathrm d\xi_{3}\left(t\right)}{\mathrm d t}\biggr]= \dfrac{\mathrm d\boldsymbol{\xi}\left(t\right)}{\mathrm dt} \tag{B-07} \end{equation} the velocity of the charge.

The potentials have the following expressions now \begin{align} \phi\left(\mathbf{x},t\right)&=\dfrac{q}{4\pi\epsilon_{o}}\iiint \dfrac{\delta^{3}\Biggl[\mathbf{x}^{\prime}-\boldsymbol{\xi}\left(t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)\Biggr]}{\|\mathbf{x}^{\prime}-\mathbf{x}\|}\mathrm d^{3}\mathbf{x}^{\prime} \tag{B-08} \\ & \nonumber\\ \mathbf{A}\left(\mathbf{x},t\right)&=\dfrac{\mu_{o}q}{\:\:4\pi\:\:}\iiint \dfrac{\delta^{3}\Biggl[\mathbf{x}^{\prime}-\boldsymbol{\xi}\left(t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)\Biggr]\cdot\mathbf{v}\left(t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)}{\|\mathbf{x}^{\prime}-\mathbf{x}\|}\mathrm d^{3}\mathbf{x}^{\prime} \tag{B-09} \end{align} Except constants, the scalar and vector integrals (B-08),(B-09) are identical to (03),(04) in main section respectively if to the latter we make the following substitutions : \begin{align} \mathbf{x} & \quad -\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow \quad \mathbf{x}^{\prime} \tag{B-10a}\\ \mathbf{F}\left(\mathbf{x}\right) & \quad -\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow \quad \mathbf{F}\left(\mathbf{x}^{\prime}\right)=\mathbf{x}^{\prime}-\boldsymbol{\xi}\left(t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right) \tag{B-10b}\\ \mathrm{H}\left(\mathbf{x}\right) & \quad -\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow \quad \mathrm{H}\left(\mathbf{x}^{\prime}\right)=\dfrac{1}{\|\mathbf{x}^{\prime}-\mathbf{x}\|} \tag{B-10c}\\ \mathbf{G}\left(\mathbf{x}\right) & \quad -\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!-\!\!\!\longrightarrow \quad \mathbf{G}\left(\mathbf{x}^{\prime}\right)=\dfrac{\mathbf{v}\left(t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)}{\|\mathbf{x}^{\prime}-\mathbf{x}\|} \tag{B-10d} \end{align} Attention must be given to the fact that in (B-08),(B-09) and (B-10) the vector $\:\mathbf{x}^{\prime}\:$ is the variable of integration while $\:\mathbf{x}\:$ is a constant vector, the position vector of the point where we try to find the field potentials.

According to the steps in main section, see equations (05) and (06), we make in (B-08) and (B-09) the following change of the integration variable \begin{equation} \mathbf{u}=\mathbf{x}^{\prime}-\boldsymbol{\xi}\left(t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)=\mathbf{F}\left(\mathbf{x}^{\prime}\right) \tag{B-11} \end{equation} and so (B-08),(B-09) are expressed respectively as follows \begin{align} \phi\left(\mathbf{x},t\right)&=\dfrac{q}{4\pi\epsilon_{o}}\iiint \dfrac{\delta^{3}\left(\mathbf{u}\right)} {\left\Vert\mathbf{F}^{-1}\left(\mathbf{u}\right)-\mathbf{x}\right\Vert\cdot\left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)}\right\vert}d^{3}\mathbf{u} \tag{B-12}\\ & \nonumber\\ \mathbf{A}\left( \mathbf{x},t \right)&=\dfrac{\mu_{o}q}{\:\:4\pi\:\:}\iiint \dfrac{\delta^{3}\left(\mathbf{u}\right)\cdot\mathbf{v}\left(t-\dfrac{\|\mathbf{F}^{-1}\left(\mathbf{u}\right)-\mathbf{x}\|}{c}\right)} {\left\Vert\mathbf{F}^{-1}\left(\mathbf{u}\right)-\mathbf{x}\right\Vert\cdot\left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)}\right\vert}d^{3}\mathbf{u} \tag{B-13} \end{align} and after these \begin{align} \phi\left(\mathbf{x},t\right)&=\dfrac{q}{4\pi\epsilon_{o}}\dfrac{1} {\left\Vert\mathbf{F}^{-1}\left(\mathbf{0}\right)-\mathbf{x}\right\Vert\cdot \left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)}\right\vert_{\mathbf{u}=\mathbf{0}}} \tag{B-14}\\ & \nonumber\\ \mathbf{A}\left(\mathbf{x},t\right)&=\dfrac{\mu_{o}q}{\:\:4\pi\:\:} \dfrac{\mathbf{v}\left(t-\dfrac{\|\mathbf{F}^{-1}\left(\mathbf{0}\right)-\mathbf{x}\|}{c}\right)} {\left\Vert\mathbf{F}^{-1}\left(\mathbf{0}\right)-\mathbf{x}\right\Vert\cdot \left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)}\right\vert_{\mathbf{u}=\mathbf{0}}} \tag{B-15} \end{align} After a tedious and not so easy elaboration we end up with the following expressions for the potentials \begin{equation} \phi\left(\mathbf{x},t\right)=\dfrac{q}{4\pi\epsilon_{o}}\dfrac{1} {\Vert \mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\Vert\cdot \left[1-\mathbf{v}\left(t-\dfrac{\|\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\|}{c}\right)\boldsymbol{\cdot}\left(\dfrac{\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}}{c\left\Vert\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right\Vert}\right)\right]} \tag{B-16} \end{equation} \begin{equation} \mathbf{A}\left(\mathbf{x},t\right)=\dfrac{\mu_{o}q}{\:\:4\pi\:\:} \dfrac{\mathbf{v}\left(t-\dfrac{\Vert \mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\Vert}{c}\right)} {\Vert \mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\Vert\cdot \left[1-\mathbf{v}\left(t-\dfrac{\|\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\|}{c}\right) \boldsymbol{\cdot} \left(\dfrac{\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}}{c\left\Vert\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right\Vert}\right)\right]} \tag{B-17} \end{equation} and with a next step to \begin{equation} \phi\left(\mathbf{x},t\right)=\dfrac{q}{4\pi\epsilon_{o}}\dfrac{1} {\left[1-\dfrac{\mathbf{v}\left(t^{\boldsymbol{*}}\right)}{c}\boldsymbol{\cdot}\mathbf{n}_{_{\mathbf{R}}}\right]\cdot\Vert \mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\Vert}=\dfrac{1}{4\pi\epsilon_{o}}\dfrac{\left(\dfrac{q}{\kappa}\right)}{R} \tag{B-18} \end{equation} \begin{equation} \mathbf{A}\left(\mathbf{x},t\right)=\dfrac{\mu_{o}q}{\:\:4\pi\:\:} \dfrac{\mathbf{v}\left(t^{\boldsymbol{*}}\right)} { \left[1-\dfrac{\mathbf{v}\left(t^{\boldsymbol{*}}\right)}{c}\boldsymbol{\cdot}\mathbf{n}_{_{\mathbf{R}}}\right]\cdot\Vert \mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\Vert}=\dfrac{\mu_{o}}{\:\:4\pi\:\:}\dfrac{\left(\dfrac{q}{\kappa}\right)}{R}\mathbf{v}\left(t^{\boldsymbol{*}}\right) = \dfrac{\mathbf{v}\left(t^{\boldsymbol{*}}\right)} {c^{2}}\phi\left(\mathbf{x},t\right) \tag{B-19} \end{equation} Note that in equation (B-18) the scalar potential $\:\phi\left(\mathbf{x},t\right)\:$ seems to be the electrostatic one, not caused by the charge $\:q\:$ but by a charge $\:q/\kappa\:$ that is greater than, less than or equal to $\:q\:$ depending upon the relation of $\:\kappa\:$ to $\:1 \:$ : $\:\kappa<1\:$, $\:\kappa >1\:$, $\:\kappa =1\:$ respectively. That is if instantaneously the charge is coming closer, is running away or keep the same distance from the field point $\:\rm{A}\:$ respectively.

Beyond the details of these calculations, the essential and more important conclusions relevant to Jacobian-Dirac delta function of the question herein are :

  1. This strange positive factor $\:\kappa\:$ inserted automatically in our equations is nothing more than the Jacobian of the transformation $\:\mathbf{x}^{\prime}=\mathbf{F}^{-1}\left(\mathbf{u}\right)\:$ \begin{align} \left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)}\right\vert_{\mathbf{u}=\mathbf{0}} &=\left\vert\dfrac{\partial\left(u_{1},u_{2},u_{3}\right)}{\partial\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)}\right\vert_{\mathbf{x}^{\prime}=\mathbf{x}^{\boldsymbol{*}}} =1-\dfrac{\mathbf{v}\left(t^{\boldsymbol{*}}\right)\boldsymbol{\cdot}\mathbf{n}_{_{\mathbf{R}}}}{c} \nonumber\\ &=1-\dfrac{\mathbf{v}\left(t-\dfrac{\|\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\|}{c}\right)\boldsymbol{\cdot} \left(\dfrac{\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}}{\left\Vert\mathbf{x}-\mathbf{x}^{\boldsymbol{*}}\right\Vert}\right)}{c} \nonumber\\ &=1-\dfrac{\mathbf{v}\left(t-\dfrac{R}{c}\right)\boldsymbol{\cdot}\left(\dfrac{\mathbf{R}}{R}\right)}{c} \stackrel{\text{def}}{\equiv}\kappa \tag{B-20} \end{align} what Feynman explains in his Lectures in $\S$21-5. The potentials of a moving charge; the general solution of Lienard and Wiechert (Volume II, Electromagnetism and Matter).

  2. The variable change (B-11) is an invertible function and more precisely the equation \begin{equation} \mathbf{u}=\mathbf{x}^{\prime}-\boldsymbol{\xi}\left(t-\dfrac{\|\mathbf{x}^{\prime}-\mathbf{x}\|}{c}\right)=\mathbf{F}\left(\mathbf{x}^{\prime}\right)=\mathbf{0} \tag{B-21} \end{equation} has one and only one solution $\:\mathbf{x}^{\boldsymbol{*}}=\mathbf{F}^{-1}\left(\mathbf{0}\right)\:$ with respect to $\:\mathbf{x}^{\prime}\:$, the so called retarded position, satisfying the equation \begin{equation} \mathbf{x}^{\boldsymbol{*}}-\boldsymbol{\xi}\left(t-\dfrac{\|\mathbf{x}^{\boldsymbol{*}}-\mathbf{x}\|}{c}\right)=\mathbf{F}\left(\mathbf{x}^{\boldsymbol{*}}\right)=\mathbf{0} \tag{B-22} \end{equation} The retarded position $\:\mathbf{x}^{\boldsymbol{*}}\:$ and retarded time $\:t^{\boldsymbol{*}}\:$ are where and when the point charge was emitting its signal that is arriving at field point $\:\mathbf{x}\:$ at the present time $\:t\:$. The invertibility of the vector function $\:\mathbf{F}\:$ is due to the fact that to every space-time point $\:\left[\boldsymbol{\xi}(t),t\right]\:$ of the charge there corresponds one and only one retarded space-time point $\:\left[\mathbf{x}^{\boldsymbol{*}}=\boldsymbol{\xi}(t^{\boldsymbol{*}}),t^{\boldsymbol{*}}\right]\:$, which in turn is due to the fact that the point charge is moving with subluminal speed (or what is the same : it's impossible two light signals emitted by the charge from two different positions-time moments of its motion to reach simultaneously at a field point $\:\mathbf{x}$).


FIGURES

$\color{blue}{\textbf{Figure-01}}$

$\color{blue}{\textbf{Figure-02a}}$

$\color{blue}{\textbf{Figure-02b}}$

$\color{blue}{\textbf{Figure-03}}$

$\endgroup$
  • $\begingroup$ Two notes: First, there is no need to consider the scalar $L(\boldsymbol u)$ and vector $\boldsymbol M(\boldsymbol u)$ separately, inasmuch as a vector can always be thought of as a tuple of scalars. You may restrict yourself to scalar functions WLOG. Second, a necessary and sufficient condition for $\delta(F(x))$ be well-defined is that $F$ is locally a Submersion. $\endgroup$ – AccidentalFourierTransform Dec 4 '17 at 13:31

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