# Extreme values for a vector equation

For a question on physics.stackexchange about Does the Ampère-Maxwell law fail for the field of a uniformly moving point charge? with

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

for this sketch

I want to know the extreme value for $v \rightarrow c$ and have doubts about my thoughts:

A possible case is $\phi =0°$. For this angle the equation gives

$$\vec B(P) = \dfrac{\mu_0 q}{4 \pi} \dfrac{1 - v^2/c^2}{1} \dfrac{\vec v \times \hat r}{r^2}$$

But what is about the term $\dfrac{\vec v \times \hat r}{r^2}$? It is a vector product and for 0° it is zero and $\vec B(P)$ equals zero?

Taking $\phi =90°$ I’m lost because I don’t know what is the solution for $v \rightarrow c$ of

$$\dfrac{1 - v^2/c^2}{[1 - (v^2/c^2)]^{3/2}}$$

• What is $\hat r$? – Alex Ravsky Dec 20 '17 at 1:14
• @AlexRavsky This seems to be an unit vector en.wikipedia.org/wiki/Unit_vector . So for an angle of 0° v and r are pointing in the same direction and the expression ${\vec v \times \hat r}$ is zero? – HolgerFiedler Dec 20 '17 at 18:26

From the picture we have $0\le y=r\sin(\pi-\phi)$, so $\sin\phi=-\frac yr$. Put $A=\frac{\mu_0qc^2}{4\pi}$, $\vec u=\frac {\vec v}{c}$. I assume that $\mu_0\ne 0$, $r\ne 0$, and $\hat r=\frac {\vec r}{r}$. Then

$$\vec B(P)=A\cdot\frac{1-u^2}{\left(1-u^2\frac {y^2}{r^2}\right)^{3/2}}\cdot \frac{\vec u\times\vec r}{r^3}.$$

Assume first that the charge needs to go a long way to be accelerated to $c$, so $r$ tends to infinity. Then $\vec B(P)$ tend to zero because $A$ is constant, $u^2\frac {y^2}{r^2}$ tends to zero, $1-u^2$ is bounded, and $|\vec u\times\vec r |=O(r)=o(r^3)$.

As an opposite case from now I assume that $\vec r$ is constant.

Assume first that $x\ne 0$. Then $y<r$, so the denominator $1-u^2\frac {y^2}{r^2}\ne 0$ and $B(\vec P)$ is a continuous function of $u$. Therefore $\lim_{u\to 1} \vec B(P)=\vec B(P)|_{u=1}=0$, because when $u=1$ we have $1-u^2=0$.

Assume now that $x=0$. Then $y=r$. Since $\vec u=(u,0,0)$ and $\vec r=(0,r,0)$, we have

$$\frac{\vec u\times\vec r}{r^3}=\frac 1{r^3} \cdot \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ u & 0 & 0 \\ 0 & r & 0 \end{vmatrix}=\frac{u\mathbf{k}}{r^2}.$$

For $v=c$ the value of $\vec B(P)$ is undefined, because the denominator $\left(1-u^2\frac {y^2}{r^2}\right)^{3/2}=0$. For $v<c$ we have

$$\vec B(P)=A\cdot\frac{1-u^2}{\left(1-u^2\right)^{3/2}}\cdot \frac{\vec u\times\vec r}{r^3}= \frac{Au\mathbf{k}}{\left(1-u^2\right)^{1/2}r^2}.$$

$$\lim_{u\to 1, u\ne 1} \vec B(P)= \lim_{u\to 1, u\ne 1} \frac{Au\mathbf{k}}{\left(1-u^2\right)^{1/2}r^2}=\infty.$$

Now about the growth of $\vec B(P)$ when $v$ tends to $c$, that is $u$ tends to $1$. We have

$$|\vec B(P)|=\left|A\cdot\frac{1-u^2}{\left(1-u^2\frac {y^2}{r^2}\right)^{3/2}}\cdot \frac{\vec u\times\vec r}{r^3}\right|=\frac{C(1-u^2)u}{\left(1-a u^2\right)^{3/2}},$$ where $C=\left|A\cdot \frac{(1,0,0)\times\vec r}{r^3}\right|$ and $a=\frac {y^2}{r^2}\le 1$.

If $x\ne 0$ then $0\le a<1$, and the derivative $\left(\frac{(1-u^2)u}{\left(1-a u^2\right)^{3/2}}\right)'$ is $\frac{1-(3-2a)u^2}{(1-au^2)^{5/2}}$, so $|\vec B(P)|$ monotonically grows when $u$ grows from $0$ to $\frac{1}{\sqrt{3-2a}}$ and then monotonically decreases to $0$ when $u$ grows to $1$.

If $x=0$ then $a=1$ and $|\vec B(P)|=\frac{Cu}{\left(1-u^2\right)^{1/2}}$. This value monotonically grows to the infinity when $u$ monotonically tends to $1$.

• Alex, thanks for your answer. Why r tends to infinity? – HolgerFiedler Dec 20 '17 at 18:27
• Ah, your answer is for r near infinity. That’s not the question. I need an answer for r near the observer, it does not matter for an angle of zero or 90° or any other angle. – HolgerFiedler Dec 22 '17 at 15:20
• @HolgerFiedler OK, then how are related $\vec r$ and $\vec v$? May I assume that $\vec r$ is constant, as in my new answer? – Alex Ravsky Dec 22 '17 at 16:32
• Alex, you wrote „because the denominator $1-u^2\frac {y^2}{r^2}=0$“, but in the nominator is the same expression. So zero divided by zero? – HolgerFiedler Dec 22 '17 at 17:16
• Alex, see my contribution to a discussion here physics.stackexchange.com/questions/374862/… – HolgerFiedler Dec 23 '17 at 8:21

In terms of my answer here :Does the Ampère-Maxwell law fail for the field of a uniformly moving point charge?, the electric field intensity vector $\:\mathbf{E}\:$ and the magnetic-flux density vector $\:\mathbf{B}\:$ at the field point $\:\mathrm P(\mathbf{r})=\mathrm P(x,y,z) \:$ are : $$\mathbf{E}\left(\mathbf{r},t\right)=\dfrac{q}{4\pi \epsilon_{0}\gamma^{2}}\dfrac{1}{\left(1\!-\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}}\dfrac{\mathbf{r}_{\bf o}}{\:\:\Vert\mathbf{r}_{\bf o}\Vert^{3}} \tag{01}$$ and $$\mathbf{B}\left(\mathbf{r},t\right)=\dfrac{1}{c^{2}}\left(\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{E}\right)=\dfrac{\mu_{0}q}{4\pi \gamma^{2}}\dfrac{1}{\left(1\!-\!\beta^{2}\sin^{2}\!\phi\right)^{\frac32}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{r}_{\bf o}}{\:\:\Vert\mathbf{r}_{\bf o}\Vert^{3}} \tag{02}$$ where $$\beta=\dfrac{\upsilon}{\:c\:}\,,\quad \gamma=\left(1-\beta^{2}\right)^{-\frac12} \tag{03}$$ Note that $\:\mathbf{r}_{\bf o}\:$ is the position vector of the field point $\:\mathrm P(x,y,z)\:$ with respect to the present position $\:\mathbf{x}(t)\:$ of the charge $\:q\:$ on the $\:x-$axis $$\mathbf{r}_{\bf o}=\mathbf{r}-\mathbf{x} \tag{04}$$ In terms of the symbols in the question : $$r\equiv\Vert\mathbf{r}_{\bf o}\Vert\,,\quad \hat r \equiv \dfrac{\mathbf{r}_{\bf o}}{\Vert\mathbf{r}_{\bf o}\Vert} \tag{05}$$ that is $\:\hat r \:$ is the unit vector along $\:\mathbf{r}_{\bf o}\:$. The angle $\:\phi\:$ is that of $\:\mathbf{r}_{\bf o}\:$ with respect to the constant velocity vector of the charge $\:\boldsymbol{\upsilon}\:$ along the $\:x-$axis.

Equation (02) in full analysis is $$\mathbf{B}\left(\mathbf{r},t\right)=\dfrac{\mu_{0}q}{4\pi}\dfrac{1-\dfrac{\upsilon^{2}}{c^{2}}}{\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\sin^{2}\!\phi\right)^{\frac32}}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{r}_{\bf o}}{\:\:\Vert\mathbf{r}_{\bf o}\Vert^{3}} \tag{06}$$

We have the following limits for $\:\phi\;\longrightarrow\;0\:$ \begin{align} \lim_{\phi\rightarrow 0}\dfrac{\mathbf{r}_{\bf o}}{\Vert\mathbf{r}_{\bf o}\Vert} & =\mathbf{i}=\text{unit vector along the }x\!-\!\text{axis, so} \tag{07a}\\ \lim_{\phi\rightarrow 0}\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{r}_{\bf o}}{\Vert\mathbf{r}_{\bf o}\Vert} & =\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{i}=\boldsymbol{0} \tag{07b}\\ \lim_{\phi\rightarrow 0}\Vert\mathbf{r}_{\bf o}\Vert^{2} & = \boldsymbol{+}\infty\Longrightarrow \lim_{\phi\rightarrow 0}\left(\dfrac{1}{\Vert\mathbf{r}_{\bf o}\Vert^{2}}\right)=0 \tag{07c} \end{align} so $$\lim_{\phi\rightarrow 0}\mathbf{B}\left(\mathbf{r},t\right)=\dfrac{\mu_{0}q}{4\pi}\dfrac{1-\dfrac{\upsilon^{2}}{c^{2}}}{\left[1\!-\!\dfrac{\upsilon^{2}}{c^{2}} \underbrace{\lim_{\phi\rightarrow 0}(\sin^{2}\!\phi)}_{0} \right]^{\frac32}}\underbrace{\lim_{\phi\rightarrow 0}\left(\dfrac{\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{r}_{\bf o}}{\:\:\Vert\mathbf{r}_{\bf o}\Vert}\right)}_{\boldsymbol{0}}\underbrace{ \lim_{\phi\rightarrow 0}\left(\dfrac{1}{\Vert\mathbf{r}_{\bf o}\Vert^{2}}\right)}_{0}=\boldsymbol{0} \tag{08}$$ as expected since for $\:\phi\;\longrightarrow\;0\:$ the charge tends to be at $\:\boldsymbol{-}\infty\:$ of the $\:x-$axis, see Figure above.

For $\:\phi =90^{\rm o}\:$ the vector $\:\mathbf{r}_{\bf o}\:$ is the projection of the position vector $\:\mathbf{r}=(x,y,z)\:$ of the field point $\:\mathrm P\:$ on the $\:yz-$plane $$(\mathbf{r}_{\bf o})_{\phi =90^{\rm o}}=y\mathbf{j}+z\mathbf{k}\,, \quad \Vert\mathbf{r}_{\bf o}\Vert_{\phi =90^{\rm o}}=\sqrt{y^{2}+z^{2}} \tag{09}$$ and $$\boldsymbol{\upsilon}\boldsymbol{\times}\mathbf{r}_{\bf o}= \begin{vmatrix} \:\:\mathbf{i} & \mathbf{j} & \mathbf{k}\:\: \vphantom{\tfrac12}\\ \:\:\upsilon & 0 & 0 \:\:\vphantom{\tfrac12}\\ \:\: 0 & y & z \:\: \vphantom{\tfrac12} \end{vmatrix} =\upsilon(-z\mathbf{j}+y\mathbf{k}) \tag{10}$$ so $$\left[\mathbf{B}\left(\mathbf{r},t\right)\right]_{\phi =90^{\rm o}}=\dfrac{\mu_{0}q}{4\pi}\dfrac{\upsilon}{\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)^{\frac12}}\dfrac{(-z\mathbf{j}+y\mathbf{k})}{\left(y^{2}+z^{2}\right)^{\frac32}} \tag{11}$$ and $$\lim_{\upsilon\rightarrow c}\left[\mathbf{B}\left(\mathbf{r},t\right)\right]_{\phi =90^{\rm o}}=\dfrac{\mu_{0}q}{4\pi}\dfrac{(-z\mathbf{j}+y\mathbf{k})}{\left(y^{2}+z^{2}\right)^{\frac32}} \underbrace{\lim_{\upsilon\rightarrow c}\left[\dfrac{\upsilon}{\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)^{\frac12}}\right]}_{\boldsymbol{+}\infty}=\boldsymbol{\infty} \tag{12}$$

EDIT

In my answer here :Does the Ampère-Maxwell law fail for the field of a uniformly moving point charge?, there are given the following alternative expressions for the electric field intensity vector $\:\mathbf{E}\:$ and the magnetic-flux density vector $\:\mathbf{B}\:$ at the field point $\:\mathrm P(\mathbf{r})=\mathrm P(x,y,z) \:$ : $$\mathbf{E}\left(x,y,z,t\right)=\dfrac{q}{4\pi \epsilon_{0}}\dfrac{\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)}{\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \begin{bmatrix} x\!-\!\upsilon\,t \vphantom{\dfrac12}\\ y \vphantom{\dfrac12}\\ z \vphantom{\dfrac12} \end{bmatrix} \tag{13}$$

$$\mathbf{B}\left(x,y,z,t\right) =\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)}{\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)\left(y^{2}+z^{2}\right)\right]^{\frac32}} \begin{bmatrix} \hphantom{-} 0 \vphantom{\dfrac{\partial}{\partial x}}\\ -\upsilon\,z \vphantom{\dfrac{\partial}{\partial x}}\\ \hphantom{-}\upsilon\,y \vphantom{\dfrac{\partial}{\partial x}} \end{bmatrix} \tag{14}$$

For the magnitudes we have $$\Vert\mathbf{E}\Vert=\dfrac{q}{4\pi \epsilon_{0}}\dfrac{\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!R^{2}\right]^{\frac12}}{\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)R^{2}\right]^{\frac32}} \tag{15}$$

$$\Vert\mathbf{B}\Vert =\dfrac{\mu_{0}q}{4\pi }\dfrac{\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)\,\upsilon\,R}{\left[\left(x\!-\!\upsilon\,t\right)^{2}\!+\!\left(1\!-\!\dfrac{\upsilon^{2}}{c^{2}}\right)R^{2}\right]^{\frac32}} \tag{16}$$ where $\:R=\sqrt{y^{2}+z^{2}}\:$ the distance of the observer $\:\mathrm P\:$ from the $\:x-$axis (see Figure).

Now, in (16) take $\:\upsilon/c\:$ as close to $\:1\:$ you want and R as close to $\:0\:$ you want. In any case $\:\Vert\mathbf{B}\Vert\:$ goes closer and closer to $\:0$.

• Frobenius, so for a charge near the spread of light the magnetic field tends to be infinite? – HolgerFiedler Dec 22 '17 at 5:21
• @HolgerFiedler : I think so. It's mathematics. And this result is valid when the charge is on the present position $\:\phi=90^{\rm o}\:$ (constant $\mathbf{r}_{\bf o}$) as you ask and not for the charge moving to the $\:+\infty\:$ of the $\:x-$axis accelerated to $\:c\:$ as in Alex Ravsky answer. Please make clear if you ask for the former or the latter. – Frobenius Dec 22 '17 at 8:30
• Infinite strength of the magnetic field seems not to be what is observed in particle accelerators. But I awaited this result from the equation. So what is wrong? – HolgerFiedler Dec 22 '17 at 9:29
• @HolgerFiedler : I think that fixing in your question $\:\phi=90^{\rm o}\:$ is wrong. What is the meaning and the necessity of such a condition ? If we remove this fixing then IMO the Alex Ravsky answer is right. In terms of my answer as the charge is moving with constant velocity to the $\:+\infty\:$ of the $\:x-$axis then $\:\phi\longrightarrow 180^{\rm o}\:$,$\:\mathbf{r}_{\bf o}/\Vert \mathbf{r}_{\bf o} \Vert \longrightarrow\: (-\mathbf{i})\:$ and $\:\Vert \mathbf{r}_{\bf o} \Vert \longrightarrow\: \infty \:$, so $\:\mathbf{B} \longrightarrow\: \boldsymbol{0}\:$. – Frobenius Dec 22 '17 at 10:31
• The angle is not important, important is the speed of the charge near c and a small distance to the observer. Any solution for B near c with small r is welcome. – HolgerFiedler Dec 22 '17 at 11:22