Relativistic Maxwell Equations
\( \def\ds{\displaystyle} \)Introduction
In this section, we will derive the formula for the transformation of the Maxwell equations for a general boost in an arbitrary direction.
In the literature, the derivation of the relativistic Maxwell equations is generally derived using differential geometry. Einstein used standard multivariate calculus in his original article, but in the case where the boost is along with one of the axes. When the velocity has an arbitrary direction, the relativistic Maxwell equations’ derivation using multivariate calculus is much more challenging, and it is not generally found in the literature and on the web.
The derivation of the relativistic Maxwell equations in this section will make heavy use of multivariate calculus: definitions and useful formulas are essential to follow the passages and are presented in the first section.
In the second section we will discuss in detail how the relevant differential operators of divergence and curl transform under the Lorentz transformation.
Finally, in the last section, we will get into the derivation of the relativistic Maxwell equations for a space free of charges and currents.
Definitions and Useful Formulas from Multivariate Calculus
We will consider a stationary system of reference where the coordinates of an event are represented by the four-vector \((x_1,x_2,x_3,t)\), and a moving frame where the same event has coordinates \((x’_1,x’_2,x’_3,t’)\). The fourth coordinate in both four-vectors has units of time, not space, as the mathematics of the Maxwell equations’ derivation becomes a little less notation-heavy this way.
A vector field \(A\) is a three-dimensional valued function acting on a four-vector,
\begin{align*} {\vb A} &= {\vb A}(x_1,x_2,x_3,t) \\ {\vb A} &: R^4 \rightarrow R^3 \\ \end{align*}A few differential operators of \(\vb A\) are extensively used in the derivation below and deserve to be explicitly defined to minimize confusion when the notation will get a little heavy,
- the Jacobian,
- the Jacobian only for the spatial dimensions,
- the Jacobian for the spatial dimension applied to a vector, typically a velocity \(\vb v\),
- the divergence,
\(\ds \div\vb A = \pdv{A_1}{x_1}+\pdv{A_2}{x_2}+\pdv{A_3}{x_3}\)
- and finally the curl,
\(\ds \curl \vb A = \left(\pdv{A_3}{x_2}\;-\pdv{A_2}{x_3} \right) \vb e_1 – \left(\pdv{A_3}{x_1}-\pdv{A_1}{x_3} \right) \vb e_2 + \left(\pdv{A_2}{x_1}-\pdv{A_1}{x_2} \right) \vb e_3\)
where \(\vb e_1, \vb e_2\) and \(\vb e_3\) are unitary vectors along the three spatial axes. The divergence and the curl are assumed to be operators only for the spatial coordinates.
A few identities1,2 from calculus are key to derive the relativistic Maxwell equations,
- the divergence and the curl of the cross product of a constant vector \(\vb v\) and a field \(\vb A\)
- the curl of the triple product of a constant vector \(\vb v\) with a field \(\vb A\), which has two different expressions, the latter derived with the use of equation (\ref{id2})
Transformation of Differential Operators
Before working on the Maxwell equations, we need to understand how the differential operators transform between \(F\) and \(F’\). We consider a field \(A\) in \(F\),
\begin{align*} {\vb A} &= {\vb A}(x_1,x_2,x_3,t) \\ {\vb A} &: R^4 \rightarrow R^3 \\ \end{align*}which we compose with the inverse of the Lorentz transformation from \(F’\) to \(F\),
\begin{align} \vb \Lambda({-\vb v})& = \frac{\beta}{v^2} \ket{\vb v,0}\bra{\vb{v},0} +\ds{\beta}\ket{\vb{v},0}\bra{\vb{0},1}+ (\vb{I’}-\frac{1}{v^2}\ket{\vb{v},0}\bra{\vb{v},0}) +\frac{ \beta} {c^2} \ket{\vb{0},1}\bra{\vb{v},0} +\beta \ket{\vb{0},1}\bra{\vb{0},1} \\ \vb \Lambda(-{\vb v}) &: R^4 \rightarrow R^4 \\ \label{lambda} \end{align}obtaining a field \(\vb A’\) defined on the space \(F’\),
\begin{align} {\vb A’} & = {\vb A}\circ \vb \Lambda(-{\vb v})\nonumber\\ {\vb A’} & = {\vb A’}({\vb x’},t’) = {\vb A}(\; \Lambda(-{\vb v})({\vb x’},t’)) \label{tr}\\ {\vb A’} &: R^4 \rightarrow R^3 \nonumber \\ \end{align}Since the Maxwell equations are in \(F\) coordinates, we need to understand how the partial derivatives of the field \(A’\) transform under the Lorentz transformation, which we are able to compute from the Jacobian chain rule,
\begin{equation} \grad_{x’,t’}{\vb A’} = \grad_{x,t}{\vb A}\cdot \grad_{x’,t’}{\Lambda} \end{equation}where \(\grad_{x’,t’}{\vb A’}\), \(\grad_{x,t}{\vb A}\) and \(\grad_{x’,t’}{\Lambda’}\) are the Jacobians of \(\vb A’\), \(\vb A\) and \(\Lambda\) respectively, computed in their own coordinate systems. The Lorentz transformation is linear, hence the Jacobian is \(\Lambda\), and the chain rule becomes explicitly in coordinate form
\begin{equation} \mqty( \ds\pdv{A’_1}{x’_1} &\ds\pdv{A’_1}{x’_2} &\ds\pdv{A’_1}{x’_3} &\ds\pdv{A’_1}{t’}\\ \ds\pdv{A’_2}{x’_1} &\ds\pdv{A’_2}{x’_2} &\ds\pdv{A’_2}{x’_3} &\ds\pdv{A’_2}{t’}\\ \ds\pdv{A’_3}{x’_1} &\ds\pdv{A’_3}{x’_2} &\ds\pdv{A’_3}{x’_3}&\ds\pdv{A’_3}{t’} \\ ) = \mqty( \ds\pdv{A_1}{x_1} &\ds\pdv{A_1}{x_2} &\ds\pdv{A_1}{x_3} &\ds\pdv{A_1}{t}\\ \ds\pdv{A_2}{x_1} &\ds\pdv{A_2}{x_2} &\ds\pdv{A_2}{x_3} &\ds\pdv{A_2}{t}\\ \ds\pdv{A_3}{x_1} &\ds\pdv{A_3}{x_2} &\ds\pdv{A_3}{x_3}&\ds\pdv{A_3}{t} \\ ) \cdot \Lambda'(-{\vb v}) \end{equation}To get to the partial derivatives of \(\vb A\) we need to invert the equation above,
\begin{equation} \mqty( \ds\pdv{A’_1}{x’_1} &\ds\pdv{A’_1}{x’_2} &\ds\pdv{A’_1}{x’_3} &\ds\pdv{A’_1}{t’}\\ \ds\pdv{A’_2}{x’_1} &\ds\pdv{A’_2}{x’_2} &\ds\pdv{A’_2}{x’_3} &\ds\pdv{A’_2}{t’}\\ \ds\pdv{A’_3}{x’_1} &\ds\pdv{A’_3}{x’_2} &\ds\pdv{A’_3}{x’_3}&\ds\pdv{A’_3}{t’} \\ )\cdot \Lambda'({\vb v}) = \mqty( \ds\pdv{A_1}{x_1} &\ds\pdv{A_1}{x_2} &\ds\pdv{A_1}{x_3} &\ds\pdv{A_1}{t}\\ \ds\pdv{A_2}{x_1} &\ds\pdv{A_2}{x_2} &\ds\pdv{A_2}{x_3} &\ds\pdv{A_2}{t}\\ \ds\pdv{A_3}{x_1} &\ds\pdv{A_3}{x_2} &\ds\pdv{A_3}{x_3}&\ds\pdv{A_3}{t} \\ ) \end{equation}The bra-ket formalism helps us here once again to compute the product of the Jacobian matrices. Any row \(j\) of \(\grad_{x’,t’}{\vb A’} \cdot \grad_{x’,t’}{\Lambda}(\vb v)\) is the vector
\begin{equation} \bra{\ds\pdv{A’_j}{x’_1},\ds\pdv{A’_j}{x’_2}, \ds\pdv{A’_j}{x’_3} ,\ds\pdv{A’_j}{t’}} \frac{\beta}{v^2} \ket{\vb v,0}\bra{\vb{v},0} -\ds{\beta}\ket{\vb{v},0}\bra{\vb{0},1}+ (\vb{I’}-\frac{1}{v^2}\ket{\vb{v},0}\bra{\vb{v},0}) -\frac{ \beta} {c^2} \ket{\vb{0},1}\bra{\vb{v},0} +\beta \ket{\vb{0},1}\bra{\vb{0},1} \end{equation}which is,
\begin{equation} \frac{\beta}{v^2}\bra{\grad_{x’}{\vb A’_j} }\ket{\vb{v}}\bra{\vb{v},0} -\ds{\beta}\bra{\grad_{x’}{\vb A’_j} }\ket{\vb{v}}\bra{\vb{0},1}+ \bra{\grad_{x’}{\vb A’_j},0}-\frac{1}{v^2}\bra{\grad_{x’}{\vb A’_j}}\ket{\vb{v}}\bra{\vb{v},0} -\frac{ \beta} {c^2} \ds\pdv{A’_j}{t’} \bra{\vb{v},0} +\beta \ds\pdv{A’_j}{t’}\bra{\vb{0},1} \end{equation}The spatial gradient of any of the component of the field \(\vb A\) computed in \((\vb x,t)= \Lambda'(-\vb v)(\vb x’,t’)\), reads
\begin{equation} \grad_x{\vb A}_j = \grad_{x’}{\vb A’_j} + \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_j} }\ket{\vb{v}}\bra{\vb{v}} -\frac{ \beta} {c^2} \ds\pdv{A’_j}{t’} \bra{\vb{v}} \label{e1} \end{equation}and the time derivative is
\begin{equation} \ds\pdv{A_j}{t} = -\ds{\beta}\bra{\grad_{x’}{\vb A’_j} }\ket{\vb{v}} +\ds\beta\pdv{A’_j}{t’} \label{e2} \end{equation}where
\begin{align} \grad_{x’} \vb A’_j & = \bra{ \ds\pdv{A’_j}{x’_1}, \ds\pdv{A’_j}{x’_2}, \ds\pdv{A’_j}{x’_3} } \\ \grad_x \vb A_j & = \bra{ \ds\pdv{A_j}{x_1}, \ds\pdv{A_j}{x_2}, \ds\pdv{A_j}{x_3} } \end{align}Divergence in \(F\) and \(F’\)
We use now equation(\ref{e1}) to compute the partial derivatives that are part of the divergence operator in \(F\),
\begin{align*} \ds\pdv{A_1}{x_1} &= \pdv{A’_1}{x’_1}+ \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_1} }\ket{\vb{v}}{{v}_1} -\frac{ \beta} {c^2} \ds\pdv{A’_1}{t’} v_1 \\ \ds\pdv{A_2}{x_2} &= \pdv{A’_2}{x’_2}+ \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_2} }\ket{\vb{v}}{{v}_2} -\frac{ \beta} {c^2} \ds\pdv{A’_2}{t’} v_2 \\ \ds\pdv{A_3}{x_3} &= \pdv{A’_3}{x’_3}+ \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_3} }\ket{\vb{v}}{{v}_3} -\frac{ \beta} {c^2} \ds\pdv{A’_3}{t’} v_3 \\ \end{align*}To derive an expression for the field’s divergence, we expand the scalar products in the equation above and group the terms on the partial derivative along each coordinate. We will work out the details for the first equation, as the others are identically derived by just changing the indexes.
\begin{align*} \ds\pdv{A_1}{x_1} &= \pdv{A’_1}{x’_1}+ \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_1} }\ket{\vb{v}}{{v}_1} -\frac{ \beta} {c^2} \ds\pdv{A’_1}{t’} v_1 \\ & = \pdv{A’_1}{x’_1}+ \frac{\beta-1}{v^2} \left(\ds\pdv{A’_1}{x’_1}v_1+\ds\pdv{A’_1}{x’_2}v_2+ \ds\pdv{A’_1}{x’_3} v_3\right)v_1-\frac{ \beta} {c^2} \ds\pdv{A’_1}{t’} v_1 \\ &= \qty(1 + \frac{\beta-1}{v^2}v_1^2 ) \pdv{A’_1}{x’_1} + \frac{\beta-1}{v^2} \left(\ds\pdv{(A’_1\cdot v_1)}{x’_2}v_2+ \ds\pdv{(A’_1\cdot v_1)}{x’_3} v_3\right) -\frac{ \beta} {c^2} \ds\pdv{(A’_1\cdot v_1)}{t’} \\ \end{align*}The divergence of the field in \(F\) is,
\begin{align*} \ds\pdv{A_1}{x_1}+ \ds\pdv{A_2}{x_2}+\ds\pdv{A_2}{x_2} &= \qty(1 + \frac{\beta-1}{v^2}v_1^2 ) \pdv{A’_1}{x’_1} + \frac{\beta-1}{v^2} \left(\ds\pdv{(A’_1\cdot v_1)}{x’_2}v_2+ \ds\pdv{(A’_1\cdot v_1)}{x’_3} v_3\right) -\frac{ \beta} {c^2} \ds\pdv{(A’_1\cdot v_1)}{t’} +\\ &\qty(1 + \frac{\beta-1}{v^2}v_2^2 ) \pdv{A’_2}{x’_2} + \frac{\beta-1}{v^2} \left(\ds\pdv{(A’_2\cdot v_2)}{x’_1}v_1+ \ds\pdv{(A’_2\cdot v_2)}{x’_3} v_3\right) -\frac{ \beta} {c^2} \ds\pdv{(A’_2\cdot v_2)}{t’} +\\ &\qty(1 + \frac{\beta-1}{v^2}v_3^2 ) \pdv{A’_3}{x’_3} + \frac{\beta-1}{v^2} \left(\ds\pdv{(A’_3\cdot v_3)}{x’_1}v_1+ \ds\pdv{(A’_3\cdot v_3)}{x’_2} v_2\right) -\frac{ \beta} {c^2} \ds\pdv{(A’_3\cdot v_3)}{t’} +\\ \end{align*}by grouping the terms by the components of the field \(A\),
\begin{align*} \ds\pdv{A_1}{x_1}+ \ds\pdv{A_2}{x_2}+\ds\pdv{A_2}{x_2} & = \pdv{A’_1}{x’_1}+\pdv{A’_2}{x’_2}+\pdv{A’_3}{x’_3} \\ & + \frac{\beta-1}{v^2}\qty( \pdv{A’_1}{x’_1}\cdot v_1 + \pdv{A’_1}{x’_2}\cdot v_2 + \pdv{A’_1}{x’_3}\cdot v_3) v_1 \\ &+ \frac{\beta-1}{v^2}\qty( \pdv{A’_2}{x’_1}\cdot v_1 + \pdv{A’_2}{x’_2}\cdot v_2 + \pdv{A’_2}{x’_3}\cdot v_3) v_2 \\ &+ \frac{\beta-1}{v^2}\qty( \pdv{A’_3}{x’_1}\cdot v_1 + \pdv{A’_3}{x’_2}\cdot v_2 + \pdv{A’_3}{x’_3}\cdot v_3) v_2 \\ & -\frac{ \beta} {c^2} \ds\pdv{( A’_1\cdot v_1 + A’_2\cdot v_2+A’_3\cdot v_3)}{t’} \end{align*}we get the compact equation for the transformation of the divergence of \(\vb A\) to \(\vb A’\),
\begin{equation} \div{\vb A} = \div \vb A’ + \frac{\beta-1}{v^2}\bra{\grad_{x’}\vb A’\cdot \vb v}\ket{\vb v} -\frac{ \beta} {c^2} \ds\pdv{\bra{\vb A’}\ket{\vb v}}{t’} \end{equation}If we consider a divergence-free field \(A\) in \(F\), the equation above gives an important condition on the divergence of \(\vb A’\),
\begin{equation} \div\vb A’ = – \frac{\beta-1}{v^2}\bra{\grad_{x’}\vb A’\cdot \vb v}\ket{\vb v} +\frac{ \beta} {c^2} \ds\pdv{\bra{\vb A’}\ket{\vb v}}{t’} \label{div2} \end{equation}Curl in \(F\) and \(F’\)
The other operator needed for the Maxwell equations is the curl,
\begin{align*} (\curl{\vb A})_1 &= \ds\pdv{A_3}{x_2} -\ds\pdv{A_2}{x_3} = \pdv{A’_3}{x’_2}+ \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_3} }\ket{\vb{v}}{{v}_2} – \ds\frac{\beta}{c^2}\pdv{A’_3}{t’} v_2 – \pdv{A’_2}{x’_3} – \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_2} }\ket{\vb{v}}{{v}_3} +\frac{\beta}{c^2} \ds\pdv{A’_2}{t’} v_3 \\ (\curl{\vb A})_2 &= -\ds\pdv{A_3}{x_1} +\ds\pdv{A_1}{x_3} =- \pdv{A’_3}{x’_1}- \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_3} }\ket{\vb{v}}{{v}_1} – \ds\frac{\beta}{c^2}\pdv{A’_3}{t’} v_1 – \pdv{A’_1}{x’_3} – \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_1} }\ket{\vb{v}}{{v}_3} +\frac{\beta}{c^2} \ds\pdv{A’_1}{t’} v_3 \\ (\curl{\vb A})_3 &= \ds\pdv{A_2}{x_1} -\ds\pdv{A_1}{x_2} = \pdv{A’_2}{x’_1}+ \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_2} }\ket{\vb{v}}{{v}_1} – \ds\frac{\beta}{c^2}\pdv{A’_2}{t’} v_1 – \pdv{A’_1}{x’_2} – \frac{\beta-1}{v^2}\bra{\grad_{x’}{\vb A’_1} }\ket{\vb{v}}{{v}_2} +\frac{\beta}{c^2} \ds\pdv{A’_1}{t’} v_2 \\ \end{align*}which is in vector form easily recognizable as,
\begin{equation} \curl{\vb A} = \curl{\vb A’}+ \frac{\beta-1}{v^2} {\vb v}\cross \qty(\grad_{x’}\vb A’\cdot\vb v)-\frac{\beta}{c^2}\pdv{\vb v \cross\vb A’}{t’} \label{rot} \end{equation}Relativistic Maxwell Equations
We consider an electric field \(\vb E\) and a magnetic field \(\vb B\) in a region of the system of reference \(F\) free of charges and currents. The Maxwell equations in this case are
\begin{align} \ds\pdv{\vb B}{t}&=-\curl{\vb E} \label{Max_B} \\ \ds\pdv{\vb E}{t}&=c^2 \curl{\vb B} \label{Max_E} \\ \div \vb B &= 0 \\ \div \vb E&=0 \end{align}The fields \(\vb b’\) and \(\vb e’\), obtained by applying the composition of the Lorentz transformation as in equation (\ref{tr}),
\begin{align*} {\vb b’} & = {\vb b’}({\vb x’},t’) = {\vb B}(\; \Lambda(-{\vb v})({\vb x’},t’)) \\ {\vb e’} & = {\vb e’}({\vb x’},t’) = {\vb E}(\; \Lambda(-{\vb v})({\vb x’},t’)) \end{align*}fulfill the equations obtained by applying directly the Lorentz transformation to equations (\ref{Max_B}) and (\ref{Max_E}),
\begin{align} -\beta \qty(\grad_{x’}\vb b’\cdot\vb v) + \beta \pdv{\vb b’}{t’} &= – \curl{\vb e’}- \frac{\beta-1}{v^2} {\vb v}\cross \qty(\grad_{x’}\vb e’\cdot\vb v)+\frac{\beta}{c^2}\pdv{\vb v \cross\vb e’}{t’} \label{MX1_B} \\ -\frac{\beta}{c^2} \qty(\grad_{x’}\vb e’\cdot\vb v) + \frac{\beta}{c^2} \pdv{\vb e’}{t’} &= \curl{\vb b’}+ \frac{\beta-1}{v^2} {\vb v}\cross \qty(\grad_{x’}\vb b’\cdot\vb v)-\frac{\beta}{c^2}\pdv{\vb v \cross\vb b’}{t’} \label{MX1_E} \\ \end{align}As these equations are not in the form of the Maxwell equations, the fields \(\vb b’\) and \(\vb e’\) are not the transformed electric and magnetic fields we are looking for. There is quite some work to be done to get to the right fields.
Step 1: divergence-free fields
The scalar product of equations (\ref{MX1_B}) and (\ref{MX1_E}) with the vector \(\vb v\),
\begin{align} -\beta \bra{\grad_{x’}\vb b’\cdot\vb v}\ket{\vb v} + \beta \pdv{\bra{\vb b’}\ket{\vb v}}{t’} &= – \bra{\curl{\vb e’}}\ket{\vb v} \\ -\frac{\beta}{c^2} \bra{\grad_{x’}\vb e’\cdot\vb v}\ket{\vb v} + \frac{\beta }{c^2}\pdv{\bra{\vb e’}\ket{\vb v}}{t’} &= \bra{\curl{\vb b’}}\ket{\vb v} \end{align}together with the divergence-free condition (\ref{div2}), gives a very useful expression for the divergence of \(\vb b’\) and \(\vb e’\),
\begin{align} \div \vb b’ &= \frac{\beta-1}{\beta v^2}\qty(\pdv{\bra{\vb b’}\ket{\vb v}}{t’} – \bra{\curl{\vb e’}}\ket{\vb v}) \label{divB} \\ \div\vb e’ &= \frac{\beta-1}{\beta v^2}\qty(\pdv{\bra{\vb e’}\ket{\vb v}}{t’} + c^2 \bra{\curl{\vb b’}}\ket{\vb v}) \label{divE} \end{align}The terms \(\grad_{x’} \vb b’\cdot \vb v\) and \(\grad_{x’} \vb e’\cdot \vb v\) in left side of equations (\ref{MX1_B}) and (\ref{MX1_E}) can be substituted from equation (\ref{id1}), and using the divergence of \(\vb b’\) and \(\vb e’\) just calculated, we obtain,
\begin{align} \pdv{}{t’}\qty( \beta\; \vb b’ – \frac{\beta-1}{ v^2} \bra{\vb b’}\ket{\vb v} \vb v -\frac{\beta}{c^2} \vb v \cross\vb e’ ) = -\beta\curl (\vb v \cross \vb b’) -\beta \curl{\vb e’} – \frac{\beta-1}{ v^2} \bra{\curl{\vb e’}}\ket{\vb v} \vb v- \frac{\beta-1}{v^2} {\vb v}\cross \qty(\grad_{x’}\vb e’\cdot\vb v) \label{MX2_B} \\ \pdv{}{t’}\qty( \beta\; \vb e’ – \frac{\beta-1}{ v^2} \bra{\vb e’}\ket{\vb v} \vb v +\beta \vb v \cross\vb b’ ) = -\beta\curl (\vb v \cross \vb e’) +\curl{(c^2\vb b’)} +\frac{\beta-1}{ v^2} \bra{\curl{(c^2\vb b’)}}\ket{\vb v} \vb v+ \frac{\beta-1}{v^2} {\vb v}\cross \qty(\grad_{x’}(c^2\vb b’)\cdot\vb v) \label{MX2_E} \\ \end{align}We are close to the solution: we need now to use the curl of the triple product.
Step 2: the curl of the triple product
Equations (\ref{MX2_B}) and (\ref{MX2_E}) are getting us closer to the solution. We have on the left side the partial derivative with respect to the time \(t’\) of two fields. On the right side, we have a few curl operators, which it is what we need, but we still have terms that are not in curl form
\begin{equation} – \frac{\beta-1}{ v^2} \bra{\curl{\vb e’}}\ket{\vb v} \vb v- \frac{\beta-1}{v^2} {\vb v}\cross \qty(\grad_{x’}\vb e’\cdot\vb v) \end{equation}The key to the solution here is to use the curl of the triple product,
\begin{equation} \curl(\vb v\cross (\vb v\cross\vb e’)) =\curl \qty( \bra{\vb v}\ket{\vb e’}\vb v – v^2\vb e’) = – \vb v \cross \qty(\grad_x\vb e’\cdot \vb v) -\bra{\vb v}\ket{\curl \vb e’}\vb v \end{equation}which substituted in (\ref{MX2_B}) and (\ref{MX2_E}) yields finally equations in the same operator form as equations (\ref{Max_B}) and (\ref{Max_E}), but now in the coordinates of \(F’\);
\begin{align} \pdv{}{t’}\left( \beta\; \vb b’ – (\beta-1) \bra{\vb b’}\ket{\frac{\vb v}{v}} \frac{\vb v}{v} -\frac{\beta}{c^2} \vb v \cross\vb e’\right) &= – \curl \left( \beta\; \vb e’ \; – \; (\beta-1) \bra{\vb e’}\ket{\frac{\vb v}{v}}\frac{\vb v}{v} + \beta\;\vb v \vb\cross \vb b’ \right) \label{MXB_v1} \\ \pdv{}{t’}\left( \beta\; \vb e’ – (\beta-1) \bra{\vb e’}\ket{\frac{\vb v}{v}} \frac{\vb v}{v} +\beta \vb v \cross\vb b’ \right) & = – c^2 \curl \left(\beta\; \vb b’ \; – \; (\beta-1) \bra{\vb b’}\ket{\frac{\vb v}{v}}\frac{\vb v}{v} -\frac{ \beta}{c^2}\;\vb v \vb\cross \vb e’ \right) \label{MXB_v2} \end{align}Hence for the principle of relativity, the electric and magnetic field in \(F’\) must be
\begin{align} \vb B’&=\phi(\vb v)\left(\beta\; \vb b’ – (\beta-1) \bra{\vb b’}\ket{\frac{\vb v}{v}} \frac{\vb v}{v} -\frac{\beta}{c^2} \vb v \cross\vb e’\right) & \\ \vb E’ & = \phi(\vb v)\left( \beta\; \vb e’ – (\beta-1) \bra{\vb e’}\ket{\frac{\vb v}{v}} \frac{\vb v}{v} +\beta \vb v \cross\vb b’ \right) \end{align}where \(\phi(\vb v)\) is a scalar function depending only on the velocity.
The condition on the possible values of \(\phi(\vb v)\) can be found by requiring that the inversion of the electric and magnetic fields in \(F’\) is the identity,
\begin{align} \vb B &=\phi(-\vb v)\left(\beta\; \vb B’ – (\beta-1) \bra{\vb B’}\ket{\frac{\vb v}{v}} \frac{\vb v}{v} +\frac{\beta}{c^2} \vb v \cross\vb E’ \right) & \\ \vb E & = \phi(-\vb v)\left( \beta\; \vb E’ – (\beta-1) \bra{\vb E’}\ket{\frac{\vb v}{v}} \frac{\vb v}{v}-\beta \vb v \cross\vb B’ \right) \end{align}which is equivalent to
\begin{equation} \phi(-\vb v)\cdot\phi(\vb v)=1 \end{equation}The transformed fields approach the original ones for \(\vb v\) going to zero,
\begin{equation} \lim_{\vb v \to \vb 0} \vb B’ = \vb B \end{equation} \begin{equation} \lim_{\vb v \to \vb 0} \phi(\vb v) = 1 \end{equation}hence, for reason of continuity, \(\phi(\vb v)=1\). The final form of the relatistic electric and magnetic fields is,
\begin{align} \vb B’&= \left(\beta\; \vb B- (\beta-1) \bra{\vb B}\ket{\frac{\vb v}{v}} \frac{\vb v}{v} -\frac{\beta}{c^2} \vb v \cross\vb E\right) & \\ \vb E’ & = \left( \beta\; \vb E – (\beta-1) \bra{\vb E}\ket{\frac{\vb v}{v}} \frac{\vb v}{v} +\beta \vb v \cross\vb B \right) \end{align}