# Does the Ampère-Maxwell law fail for the field of a uniformly moving point charge?

I think I found a terrible contradiction in the Ampère-Maxwell law (the 4th Maxwell equation) that really baffles me.

If a point charge q is in motion in vacuo with constant velocity $\vec v$, it causes an electric field $\vec E$ as well as a magnetic field $\vec B$ at a point P. The vector-position $\vec r$ of the point P with respect to the charge, the charge itself and $\vec v$ determine a plane. We can imagine the charge moving along the x-axis of a cartesian plane, its equation of motion given by $x = x_0 + vt$, $v = \dfrac {dx}{dt}$ = const. while the point P is located at a permanent position on the y-axis, y(P) = const.

The electric field caused by the charge q at P is given by

$$\vec E(P) = \dfrac{q}{4 \pi \epsilon_0} \dfrac{1 - v^2/c^2}{[1 - (v^2/c^2) sin^2 \phi]^{3/2}} \dfrac{1}{r^2} \hat r \tag{01}$$

and the magnetic field is given by

$$\vec B(P) = \dfrac{\mu_0 q}{4 \pi} \dfrac{1 - v^2/c^2}{[1 - (v^2/c^2) sin^2 \phi]^{3/2}} \dfrac{\vec v \times \hat r}{r^2} \tag{02}$$

So the electric field, being radial to the charge, is contained in the x0y plane, while the magnetic field is perpendicular to the same.

The Ampère-Maxwell law states that $$\operatorname{curl} \vec B = \epsilon_0 \mu_0 \dfrac{\partial \vec E}{\partial t}$$

And we have

$$\vec B = \epsilon_0 \mu_0 \vec v \times \vec E = \dfrac{\vec v \times \vec E}{c^2}$$

$$\vec v \times \vec E = vE_y \hat z$$

$$B_x = 0$$ $$B_y = 0$$ $$B_z = \dfrac{v}{c^2} E_y$$

Therefore $$\operatorname{curl} \vec B = \dfrac{v}{c^2} \left[ \dfrac{\partial E_y}{\partial y} \hat x - \dfrac{\partial E_y}{\partial E_x} \hat y \right]$$

Calculating the derivative of $\vec E$ with respect to time, we get

$$\dfrac{dE_x}{dt} = \dfrac{qv}{4 \pi \epsilon_0} \left( 1 - \dfrac{v^2}{c^2} \right) \left[ x^2 + y^2 \left( 1 - \dfrac{v^2}{c^2} \right) \right]^\dfrac{-3}{2} \left[ 3x^2 \left[ x^2 + y^2 \left( 1 - \dfrac{v^2}{c^2} \right) \right]^{-1} -1 \right]$$

$$\dfrac{dE_y}{dt} = - \dfrac{qv}{4 \pi \epsilon_0} \left( 1 - \dfrac{v^2}{c^2} \right) 3xy \left[ x^2 + y^2 \left( 1 - \dfrac{v^2}{c^2} \right) \right]^\dfrac{-5}{2}$$

Then, when I calculate $\dfrac{\partial E_y}{\partial y}$ and $\dfrac{\partial E_y}{\partial x}$ to insert in the above formula for $\operatorname{curl} \vec B$, I do not obtain an equality to my surprise, which indicates that the law of Ampère-Maxwell may not be valid in this case.

So the following three propositions cannot be all true:

1. The formulas given above to calculate the electric and magnetic field at point P are correct.
2. The derivatives above were calculated correctly.
3. The law of Ampère-Maxwell is correct as stated.

Which one of them is false?

• 3.) is true because countless experiments over the past 150 years have verified its truth. Either 1.) or 2.) is false. – JM1 Dec 17 '17 at 5:44
• Now that I have performed more than a cursory evaluation of what you've written, I can say with confidence that 1.) is true as well. Try reevaluating your derivatives. – JM1 Dec 17 '17 at 6:24
• The fields of a moving point charge are given by the Liénard-Wiechert potentials, which satisfy the Maxwell equations by construction. Anything different from that is an approximation. – Emilio Pisanty Dec 17 '17 at 7:37
• The Ampere law has a current in it. Your moving charge is a current. – Rob Jeffries Dec 17 '17 at 8:11
• Some weeks ago I asked in PSE about the magnetic field of a charge moving near c. Your equation B(P) seems to be the answer. Now I’m struggling with the solution for $v \rightarrow c$ and have some questions here – HolgerFiedler Dec 17 '17 at 15:53

The complete Ampère-Maxwell is $$\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} = \mu_{0}\,\boldsymbol{\jmath}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{01}$$ where $\:\boldsymbol{\jmath}\:$ the electric current density vector. In our case $$\boldsymbol{\jmath}\left(\mathbf{r},t\right)=q\cdot\delta\left(\mathbf{r}-\mathbf{x}\right)\cdot\dfrac{\mathrm d\mathbf{x}}{\mathrm dt}=q\cdot\delta\left(\mathbf{r}-\mathbf{x}\right)\cdot\boldsymbol{\upsilon} \tag{02}$$ where $\:\mathbf{x}\left(t\right),\boldsymbol{\upsilon}\left(t\right)= \mathrm d\mathbf{x}/\mathrm dt\:$ the position and velocity of the point charge. At all field points except the singular point where the charge is on we have $$\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} = \frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{03}$$

More exactly from above Figure we have $$\mathbf{E}\left(\mathbf{r},t\right)=\dfrac{q}{4\pi \epsilon_{0}\gamma^{2}}\dfrac{1}{\left(1\!-\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}}\dfrac{\mathbf{r}_{\bf o}}{\:\:\Vert\mathbf{r}_{\bf o}\Vert^{3}}\ \tag{04}$$

$$\beta=\dfrac{\upsilon}{\:c\:}\,,\quad \gamma=\left(1-\beta^{2}\right)^{-\frac12} \tag{05}$$ Now $$\mathbf{r}_{\bf o}=\mathbf{r}-\mathbf{x}= \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac12}\\ y \vphantom{\dfrac12}\\ z\vphantom{\dfrac12} \end{bmatrix}\,, \quad \Vert\mathbf{r}_{\bf o}\Vert=\Vert\mathbf{r}-\mathbf{x}\Vert=\left[\left(x\!-\!\upsilon\,t\right)^{2}+y^{2}+z^{2}\right]^{\frac12} \tag{06}$$ and $$\sin^{2}\!\phi=\dfrac{y^{2}+z^{2}}{\Vert\mathbf{r}_{\bf o}\Vert^{2}}=\dfrac{y^{2}+z^{2}}{\left(x\!-\!\upsilon\,t\right)^{2}+y^{2}+z^{2}} \tag{07}$$ So $$\mathbf{E}\left(x,y,z,t\right)=\dfrac{Q}{\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac12}\\ y \vphantom{\dfrac12}\\ z \vphantom{\dfrac12} \end{bmatrix}\,,\quad Q\equiv \dfrac{q}{4\pi \epsilon_{0}\gamma^{2}} \tag{08}$$ For the magnetic field $$\mathbf{B}\left(x,y,z,t\right) =\dfrac{1}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right) \tag{09}$$ so we have \begin{align} \mathbf{B}\left(x,y,z,t\right) &=\dfrac{1}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right) \nonumber\\ &=\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \begin{bmatrix} \:\:\upsilon \:\:\vphantom{\dfrac{\partial}{\partial x}}\\ 0 \vphantom{\dfrac{\partial}{\partial x}} \\ 0 \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \boldsymbol{\times} \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac{\partial}{\partial x}}\\ y \vphantom{\dfrac{\partial}{\partial x}}\\ z \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \nonumber\\ & =\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \vphantom{\dfrac{\partial}{\partial x}}\\ \upsilon & 0 & 0 \vphantom{\dfrac{\partial}{\partial x}} \\ x\!-\!\upsilon\,t & \hphantom{x\!-} y \hphantom{x\!-} & \hphantom{x\!-}z\hphantom{x\!-}\vphantom{\dfrac{\partial}{\partial x}} \end{vmatrix} \tag{10} \end{align} or $$\mathbf{B}\left(x,y,z,t\right) =\dfrac{1}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right)=\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \begin{bmatrix} \hphantom{-} 0 \vphantom{\dfrac{\partial}{\partial x}}\\ -\upsilon\,z \vphantom{\dfrac{\partial}{\partial x}}\\ \hphantom{-}\upsilon\,y \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \tag{11}$$

From (08) with differentiations I prove(1) that $$\dfrac{1}{c^{2}}\dfrac{\partial\mathbf{E}}{\partial t} =\dfrac{1}{4\pi \epsilon_{0}\gamma^{2}}\dfrac{q\upsilon}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} 2\left(x\!-\!\upsilon\,t\right)^{2}-\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right) \vphantom{\dfrac12}\\ 3\left(x\!-\!\upsilon\,t\right)y \vphantom{\dfrac12}\\ 3\left(x\!-\!\upsilon\,t\right)z \vphantom{\dfrac12} \end{bmatrix} \tag{12}$$ and from (11) I prove(2) $$\boldsymbol{\nabla} \boldsymbol{\times}\mathbf{B}=\dfrac{1}{4\pi \epsilon_{0}\gamma^{2}}\dfrac{q\upsilon}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} 2\left(x\!-\!\upsilon\,t\right)^{2}\!-\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\\ 3\left(x\!-\!\upsilon\,t\right)y \vphantom{\dfrac{\partial}{\partial x}} \\ 3\left(x\!-\!\upsilon\,t\right)z \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \tag{13}$$ that is equation (03) $$\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} = \frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{03}$$

EDIT

The electromagnetic field vectors $\:\mathbf{E},\mathbf{B}\:$ given by equations (04) and (09) respectively satisfy the four Maxwell equations without charge and charge current densities \begin{align} \rho\left(\mathbf{r},t\right) & = 0 \tag{14a}\\ \boldsymbol{\jmath}\left(\mathbf{r},t\right) & = \boldsymbol{0} \tag{14b} \end{align} at all field points except at the singular point $$\mathbf{r}_{\bf o}=\mathbf{r}-\mathbf{x}=\boldsymbol{0} \tag{15}$$ that is at the point where the charge is on.

In order to include this singularity in our equations we must use the following equations $$\dfrac{\mathbf{r}_{\bf o}}{\:\:\Vert\mathbf{r}_{\bf o}\Vert^{3}}=\dfrac{\mathbf{r}-\mathbf{x}}{\:\:\Vert\mathbf{r}-\mathbf{x}\Vert^{3}}=-\boldsymbol{\nabla}\left(\!\dfrac{1}{\Vert\mathbf{r}-\mathbf{x}\Vert}\right) \tag{16}$$

$$\boldsymbol{\nabla}\boldsymbol{\cdot}\boldsymbol{\nabla}\left(\!\dfrac{1}{\Vert\mathbf{r}-\mathbf{x}\Vert}\right)=\nabla^{2}\left(\!\dfrac{1}{\Vert\mathbf{r}-\mathbf{x}\Vert}\right)=-4\pi\delta\left(\mathbf{r}-\mathbf{x}\right) \tag{17}$$ where the components of the 3-vector operator $\:\boldsymbol{\nabla}\:$ are the partial derivatives with respect to the components $\:(x,y,z)\:$ of the position vector $\:\mathbf{r}\:$ of the field point.

Then the electromagnetic field vector $\:\mathbf{E}\:$ of equation (04) is written $$\mathbf{E}\left(\mathbf{r},t\right)=\boldsymbol{-}\dfrac{q}{4\pi \epsilon_{0}\gamma^{2}}\dfrac{1}{\left(1\!-\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}}\cdot\boldsymbol{\nabla}\left(\!\dfrac{1}{\Vert\mathbf{r}-\mathbf{x}\Vert}\right) \tag{18}$$ I believe that the vector $\:\mathbf{E}\:$ of equation (04) satisfies the Maxwell equation $$\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{E}\left(\mathbf{r},t\right)=0 \tag{19}$$ while the vector $\:\mathbf{E}\:$ of equation (18) satisfies the Maxwell equation $$\boldsymbol{\nabla}\boldsymbol{\cdot}\mathbf{E}\left(\mathbf{r},t\right)=\dfrac{\rho\left(\mathbf{r},t\right)}{\epsilon_{0}}=\dfrac{q\,\delta\left(\mathbf{r}-\mathbf{x}\right)}{\epsilon_{0}} \tag{20}$$ Also the vector $\:\mathbf{B}\:$ of equation (09) is expressed with respect to $\:\mathbf{E}\:$ of equation (18) $$\mathbf{B}\left(\mathbf{r},t\right) =\dfrac{1}{c^{2}}\left[\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\left(\mathbf{r},t\right)\right]=\boldsymbol{-}\dfrac{ \mu_{0}\,q}{4\pi\gamma^{2}}\dfrac{1}{\left(1\!-\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}}\cdot\left[\boldsymbol{\upsilon}\boldsymbol{\times}\boldsymbol{\nabla}\left(\!\dfrac{1}{\Vert\mathbf{r}-\mathbf{x}\Vert}\right)\right] \tag{21}$$ and since $\:\boldsymbol{\upsilon}\:$ is a constant vector $$\mathbf{B}\left(\mathbf{r},t\right)=\dfrac{ \mu_{0}\,q}{4\pi\gamma^{2}}\dfrac{1}{\left(1\!-\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}}\cdot\left[\boldsymbol{\nabla}\boldsymbol{\times}\left(\!\dfrac{\boldsymbol{\upsilon}}{\Vert\mathbf{r}-\mathbf{x}\Vert}\right)\right] \tag{22}$$ The vectors $\:\mathbf{E},\mathbf{B}\:$ of equations (04),(09) satisfy the Maxwell equation (03) $$\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} = \frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{03}$$ while the vectors $\:\mathbf{E},\mathbf{B}\:$ of equations (18),(22) satisfy (I believe) the Maxwell equation (01) with the electric current density of equation (02) $$\boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} = \mu_{0}\,\boldsymbol{\jmath}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t}=\mu_{0}\,q\,\delta\left(\mathbf{r}-\mathbf{x}\right)\boldsymbol{\upsilon}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{23}$$

(1) Proof of equation (12)

From (08) \begin{align} \dfrac{1}{c^{2}}\dfrac{\partial\mathbf{E}}{\partial t} & = \dfrac{\partial}{\partial t}\Biggl(\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}}\Biggr) \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac12}\\ y \vphantom{\dfrac12}\\ z \vphantom{\dfrac12} \end{bmatrix} +\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}}\dfrac{\partial}{\partial t} \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac12}\\ y \vphantom{\dfrac12}\\ z \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & =\dfrac{3Q\upsilon\left(x\!-\!\upsilon\,t\right)}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac12}\\ y \vphantom{\dfrac12}\\ z \vphantom{\dfrac12} \end{bmatrix} +\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \begin{bmatrix} -\upsilon \vphantom{\dfrac12}\\ \hphantom{-} 0 \vphantom{\dfrac12}\\ \hphantom{-} 0 \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & =\dfrac{3Q\upsilon\left(x\!-\!\upsilon\,t\right)}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac12}\\ y \vphantom{\dfrac12}\\ z \vphantom{\dfrac12} \end{bmatrix} +\dfrac{Q\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} -\upsilon \vphantom{\dfrac12}\\ \hphantom{-} 0 \vphantom{\dfrac12}\\ \hphantom{-} 0 \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ & =\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} 3\upsilon\left(x\!-\!\upsilon\,t\right)^{2}-\upsilon\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right] \vphantom{\dfrac12}\\ 3\upsilon\left(x\!-\!\upsilon\,t\right)y \vphantom{\dfrac12}\\ 3\upsilon\left(x\!-\!\upsilon\,t\right)z \vphantom{\dfrac12} \end{bmatrix} \nonumber\\ &=\dfrac{Q\upsilon}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} 2\left(x\!-\!\upsilon\,t\right)^{2}\!-\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\\ 3\left(x\!-\!\upsilon\,t\right)y \vphantom{\dfrac{\partial}{\partial x}} \\ 3\left(x\!-\!\upsilon\,t\right)z \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \tag{12} \end{align}

(2) Proof of equation (13)

From (11) $$\mathbf{B}\left(x,y,z,t\right)= f\left(x,y,z\right)\mathbf{M}\left(x,y,z\right) \tag{13.1}$$ where $$f\left(x,y,z\right) \equiv\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \,,\qquad \mathbf{M}\left(x,y,z\right) \equiv \begin{bmatrix} \hphantom{-} 0 \\ -\upsilon\,z \\ \hphantom{-}\upsilon\,y \end{bmatrix} \tag{13.2}$$ If $\: f\left(x,y,z\right)\:$ and $\:\mathbf{M}\left(x,y,z\right)\:$ are scalar and vector functions respectively of the coordinates in $\:\mathbb{R}^{3}\:$ then $$\boldsymbol{\nabla}\boldsymbol{\times}\left(f\, \mathbf{M}\right)=\left( \boldsymbol{\nabla} f\right)\boldsymbol{\times}\mathbf{M}+ f \left(\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{M}\right) \tag{13.3}$$ For a short proof of identity (13.3) see my answer here : Passing from curl to vector product, equation (04).

Now $$\boldsymbol{\nabla}\boldsymbol{\times}\mathbf{M} = \boldsymbol{\nabla}\boldsymbol{\times} \begin{bmatrix} \: \hphantom{-} 0 \:\vphantom{\dfrac{\partial\hphantom{x}}{\partial x}}\\ \:-\upsilon\,z \: \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}}\\ \:\hphantom{-}\upsilon\,y \: \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}} \end{bmatrix} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}}\\ \dfrac{\partial\hphantom{x}}{\partial x} & \dfrac{\partial\hphantom{y}}{\partial y} & \dfrac{\partial\hphantom{z}}{\partial z}\\ 0 &\!\! -\upsilon\,z & \upsilon\,y \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}} \end{vmatrix} = \begin{bmatrix} \: 2\upsilon \:\vphantom{\dfrac{\partial\hphantom{x}}{\partial x}}\\ \:0 \: \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}}\\ \:0 \: \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}} \end{bmatrix} =2\,\boldsymbol{\upsilon} \tag{13.4}$$ and $$\boldsymbol{\nabla} f=\dfrac{\partial f}{\partial x}\mathbf{i}+\dfrac{\partial f}{\partial y}\mathbf{j}+\dfrac{\partial f}{\partial z}\mathbf{k} \tag{13.5}$$ where \begin{align} \dfrac{\partial f}{\partial x} & =\dfrac{-3Q\left(x\!-\!\upsilon\,t\right)}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \tag{13.5a}\\ \dfrac{\partial f}{\partial y} & =\dfrac{-3Q\left(1\!-\!\beta^{2}\right)y}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \tag{13.5b}\\ \dfrac{\partial f}{\partial z} & =\dfrac{-3Q\left(1\!-\!\beta^{2}\right)z}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \tag{13.5c} \end{align} so $$\boldsymbol{\nabla} f= \dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} -3\left(x\!-\!\upsilon\,t\right) \:\vphantom{\dfrac{\partial\hphantom{x}}{\partial x}}\\ \:-3\left(1\!-\!\beta^{2}\right)y \: \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}}\\ \:-3\left(1\!-\!\beta^{2}\right)z \: \vphantom{\dfrac{\partial\hphantom{x}}{\partial x}} \end{bmatrix} \tag{13.6}$$ From (13.2), (13.6) $$\left(\boldsymbol{\nabla} f\right)\boldsymbol{\times}\mathbf{M}=\dfrac{-3Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \vphantom{\dfrac{\partial}{\partial x}}\\ \left(x\!-\!\upsilon\,t\right) & \left(1\!-\!\beta^{2}\right)y & \left(1\!-\!\beta^{2}\right)z \vphantom{\dfrac{\partial}{\partial x}} \\ 0 & \hphantom{x\!-} -\upsilon\,z \hphantom{x\!-} & \hphantom{x\!-}\upsilon\,y\hphantom{x\!-}\vphantom{\dfrac{\partial}{\partial x}} \end{vmatrix} \tag{13.7}$$ that is
$$\left(\boldsymbol{\nabla} f\right)\boldsymbol{\times}\mathbf{M}=\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} -3\upsilon\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\\ 3\upsilon\left(x\!-\!\upsilon\,t\right)y \vphantom{\dfrac{\partial}{\partial x}} \\ 3\upsilon\left(x\!-\!\upsilon\,t\right)z \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \tag{13.8}$$
From (13.2), (13.4) $$f \left(\boldsymbol{\nabla} \boldsymbol{\times}\mathbf{M}\right)=\dfrac{Q}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} 2\upsilon\left(x\!-\!\upsilon\,t\right)^{2}\!+2\upsilon\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\\ 0 \vphantom{\dfrac{\partial}{\partial x}} \\ 0 \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \tag{13.9}$$
and from (13.1),(13.8),(13.9) we have finally $$\boldsymbol{\nabla} \boldsymbol{\times}\mathbf{B}=\left(\boldsymbol{\nabla} f\right)\boldsymbol{\times}\mathbf{M}+ f\left(\boldsymbol{\nabla} \boldsymbol{\times}\mathbf{M}\right)=\dfrac{Q\upsilon}{c^{2}\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\right]^{\frac52}} \begin{bmatrix} 2\left(x\!-\!\upsilon\,t\right)^{2}\!-\left(1\!-\!\beta^{2}\right)\left(y^{2}+z^{2}\right)\\ 3\left(x\!-\!\upsilon\,t\right)y \vphantom{\dfrac{\partial}{\partial x}} \\ 3\left(x\!-\!\upsilon\,t\right)z \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \tag{13}$$

• Where did eq 4 and eq 9 come from? – Emil Dec 23 '17 at 9:01
• @Emil : In a system $\:S'(x',y',z',t')\:$ position the charge $\:q\:$ on the origin $\:O'\:$ at $\:t'=0\:$ and write down the electromagnetic field vectors (electrostatic $\:\mathbf{E'}\:$ and $\:\mathbf{B'}\equiv\boldsymbol{0}\:$). Then take a system $\:S(x,y,z,t)\:$ moving with velocity $\:\boldsymbol{\upsilon}\:$ along the common $\:x-,x'\:$-axis and $\:y'-,z'\:$-axis parallel to $\:y-,z\:$-axis respectively. Apply the Lorentz Transformation to $\:\mathbf{E'},\mathbf{B'}$. You will find $\:\mathbf{E},\mathbf{B}\:$ you are asking for. – Frobenius Dec 23 '17 at 10:16
• @Emil : May be this excerpt from "Relativity - Special, General, and Cosmological" [Rindler] could be useful. – Frobenius Dec 23 '17 at 10:48
• Sorry, I just don't understand what δ(r−x) means (in eq. 01). – Luiz de Assis Netto Dec 27 '17 at 0:10
• "But surprisingly enough elaborating on the equations I found that $$∇×B=1/c^2∂E/∂t$$ (03)" - which is not surprising because the point P is in empty space, where there are no charges or currents. – Luiz de Assis Netto Dec 27 '17 at 1:34

This is more a contribution to the discussion as an answer.

In math.stackexchange I asked about the extreme value for the equation (2). For the charges velocity near c and a small distance to the observer there is an answer that the magnetic field tends to infinity. This seems to be strange because a charge (including its field) has a limited energy content only and according the theory the rise of the magnetic field is accompanied by the weakening of the electric field.

But there is another point with the equations (1) and (2). We can rewrite them

$$\vec E(P) = \dfrac{q}{4 \pi \epsilon_0} C \dfrac{1}{r^2} \hat r \tag{01}$$

and

$$\vec B(P) = \dfrac{\mu_0 q}{4 \pi} C \dfrac{\vec v \times \hat r}{r^2} \tag{02}$$

Now neglecting C (which perhaps is not allowed, but this is for discussion) B tends to infinity for v near c. E stays unchanged. But by theory E has to decrease with increasing v?

Anymore, I’ve never heard that the magnetic field of charges in particle accelerators increases to infinity. On the other side the field has to increase due to theory.

Why I made this contribution to a discussion? Because my point of view is that a charge has not only an permanent electric field and a permanent magnetic dipole moment. The magnetic dipole moments of charges are randomly distributed and their influence is unnoticed. Under motion the magnetic dipole moments get aligned and a macroscopic magnetic field occurs. The strength of the magnetic field depends from the alignment of the charges only and not from the absolute value of their velocity.