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Given that the position vector $\textbf{r}$ to be a vector under rotation, we mean that it transforms under rotation as $\textbf{r}^\prime=\mathbb{R}\textbf{r}$. Now, taking two time-derivatives of it, one can easily see that the acceleration $\textbf{a}=\ddot{\textbf{r}}$ transforms as $\textbf{a}^\prime=\mathbb{R}\textbf{a}$ i.e., also behaves as a vector under rotation.

Now a four-vector is something which transforms under Lorentz transformation as $x^\mu$ does. Given the transformation of $x^\mu$: $$x'^\mu=\Lambda^{\mu}{}_{\nu} x^\nu\tag{1}$$ how can one show that the four-current density $j^\mu$ also transforms like (1) preferably from the definition $j^\mu=(c\rho,\textbf{j})$?

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  • $\begingroup$ Would you be happy showing it in a specific example? For instance, the free particle has $j^\mu = Q\int dx^\mu \delta^D ( x^\nu - u^\nu \tau )$. $\endgroup$ – Prahar Mar 19 '18 at 15:23
  • $\begingroup$ I would prefer a more general derivation perhaps. @Prahar $\endgroup$ – SRS Mar 19 '18 at 15:26
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    $\begingroup$ Can you explain where the difficulty here lies? If you take the definition of charge density and 3-current as charge per volume and charge per time and use that charge is invariant under Lorentz transformations while volume and time are not, you should arrive at the wanted expression rather straightforwardly. $\endgroup$ – ACuriousMind Mar 19 '18 at 19:01
  • $\begingroup$ @ACuriousMind I don't understand. $c\rho=cQ/(\Delta x\Delta y\Delta z)$ and $j_x=Q/(\Delta t\Delta y\Delta z)$ and so on. How do you proceed next? $\endgroup$ – SRS Mar 20 '18 at 9:24
  • $\begingroup$ Isn't it as simple as $j^\mu \propto v^\mu$? $\endgroup$ – FGSUZ Jun 15 '18 at 22:32
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$\color{blue}{\textbf{ANSWER A}}\:$ (based on charge invariance, paragraph extracted from Landau)

The answer is given in ACuriousMind's comment as pointed out also by WetSavannaAnimal aka Rod Vance. Simply I give the details copying from "The Classical Theory of Fields", L.D.Landau and E.M.Lifshitz, Fourth Revised English Edition :

$\boldsymbol{\S}\: \textbf{28. The four-dimensional current vector}$

Instead of treating charges as points, for mathematical convenience we frequently consider them to be distributed continuously in space. Then we can introduce the "charge density" $\:\varrho\:$ such that $\:\varrho dV\:$ is the charge contained in the volume $\: dV$. The density $\:\varrho\:$ is in general a function of the coordinates and the time. The integral of $\:\varrho\:$ over a certain volume is the charge contained in that volume.......

.......The charge on a particle is, from its very definition, an invariant quantity, that is, it does not depend on the choice of reference system. On the other hand, the density $\:\varrho\:$ is not generally an invariant--only the product $\:\varrho dV\:$ is invariant.

Multiplying the equality $\:de=\varrho dV\:$ on both sides with $\:dx^{i}\:$: \begin{equation} de\,dx^{i}=\varrho dVdx^{i}=\varrho dVdt\dfrac{dx^{i}}{dt} \nonumber \end{equation} On the left stands a four-vector (since $\:de\:$ is a scalar and $\:dx^{i}\:$ is a four-vector). This means that the right side must be a four-vector. But $\: dVdt\:$ is a scalar(1), and so $\:\varrho dx^{i}/dt\:$ is a four-vector.This vector (we denote it by $\:j^{i}$) is called the current four-vector: \begin{equation} j^{i}=\varrho \dfrac{dx^{i}}{dt}. \tag{28.2} \end{equation}

The space components of this vector form the current density vector, \begin{equation} \mathbf{j}=\varrho \mathbf{v}, \tag{28.3} \end{equation}
where $\:\mathbf{v}\:$ is the velocity of the charge at the given point. The time component of the four vector (28.2) is $\:c\varrho$. Thus \begin{equation} j^{i}=\left(c\varrho ,\mathbf{j}\right) \tag{28.4} \end{equation}


(1) Note by Frobenius : We have \begin{equation} dVd(ct)=dx^{1}dx^{2}dx^{3}dx^{4} \tag{01} \end{equation} Now, for the relation between the infinitesimal 4-volumes in Minkowski space \begin{equation} dx'^{1}dx'^{2}dx'^{3}dx'^{4} =\begin{vmatrix} \dfrac{\partial x'_{1}}{\partial x_{1}}& \dfrac{\partial x'_{1}}{\partial x_{2}}&\dfrac{\partial x'_{1}}{\partial x_{3}}&\dfrac{\partial x'_{1}}{\partial x_{4}}\\ \dfrac{\partial x'_{2}}{\partial x_{1}}& \dfrac{\partial x'_{2}}{\partial x_{2}}&\dfrac{\partial x'_{2}}{\partial x_{3}}&\dfrac{\partial x'_{2}}{\partial x_{4}}\\ \dfrac{\partial x'_{3}}{\partial x_{1}}& \dfrac{\partial x'_{3}}{\partial x_{2}}&\dfrac{\partial x'_{3}}{\partial x_{3}}&\dfrac{\partial x'_{3}}{\partial x_{4}}\\ \dfrac{\partial x'_{4}}{\partial x_{1}}& \dfrac{\partial x'_{4}}{\partial x_{2}}&\dfrac{\partial x'_{4}}{\partial x_{3}}&\dfrac{\partial x'_{4}}{\partial x_{4}} \end{vmatrix} dx^{1}dx^{2}dx^{3}dx^{4}=\left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert dx^{1}dx^{2}dx^{3}dx^{4} \tag{02} \end{equation} where $\:\left\vert\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)/\partial\left(x^{1},x^{2},x^{3},x^{4}\right)\right\vert\:$ the Jacobian, that is determinant of the Jacobi matrix. But the Jacobi matrix is the Lorentz matrix $\:\Lambda\:$ with $\:\det(\Lambda)=+1$, that is \begin{equation} \left\vert\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\right\vert=\det(\Lambda)=+1 \tag{03} \end{equation} so \begin{equation} dx'^{1}dx'^{2}dx'^{3}dx'^{4} =dx^{1}dx^{2}dx^{3}dx^{4}=\text{scalar invariant} \tag{04} \end{equation}

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  • $\begingroup$ Good answer, but it induces another question: why is electric charge Lorentz invariant? which then has to be answered without using the fact that $j$ is a four-vector. Remark on notation: $\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}$ really looks like $\frac{\partial x'^\mu}{\partial x^\nu}$ , that is, like the transformation matrix. Is this a standard notation for Jacobi's determinant? I think $\bigg|\dfrac{\partial\left(x'^{1},x'^{2},x'^{3},x'^{4}\right)}{\partial\left(x^{1},x^{2},x^{3},x^{4}\right)}\bigg|$ would be better. $\endgroup$ – Ján Lalinský Jun 15 '18 at 19:06
  • $\begingroup$ @Ján Lalinský (1) I think that from the beginning the invariance of the electric charge of a particle was an hypothesis proved valid from the experiment. (2) You are right for the notation, I must correct it in a few answers of mine (3) I 'll prepare a second answer on the spirit of yours, which by the way I up-voted. $\endgroup$ – Frobenius Jun 15 '18 at 19:22
  • $\begingroup$ Regarding 1), I think that is true, but somewhat laconic and unsatisfactory pedagogically. I think a better explanation is this: should the electric charge of a body change when it changes speed, the law of conservation of charge in a fixed volume where $\mathbf j=0$ on its boundary could not be valid. So, charge does not depend on its velocity; this I think is a good motivation for why we should assign the same value of charge to a body irrespective of the frame we observe it from. $\endgroup$ – Ján Lalinský Jun 15 '18 at 19:34
  • $\begingroup$ @Ján Lalinský I think I must agree with you. The final conclusion is that the charge of a particle is a scalar Lorentz invariant and till now there is no experimental evidence to the contrary. $\endgroup$ – Frobenius Jun 15 '18 at 21:23
  • $\begingroup$ As stated before, I don't think this is perfectly satisfactory because it requires showing the electric charge is Lorentz invariant. That also requires defining what charge is in a moving frame (which is not as simple as "just use Coulomb's law" because Coulomb's law does not apply to moving charges!). $\endgroup$ – knzhou Jun 17 '18 at 17:41
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You can take charge conservation as your starting point. This can be written as: $$ \frac{\partial\rho}{\partial t} = \partial_{i} j^i = \nabla\cdot \vec{J} $$

Since this is an experimental fact, it is a good starting point. The above equation can now be re-written in a "more" co-variant formulation as: $$ \partial_\mu j^\mu = 0 $$

From this equation you can clearly deduce that $j^\mu$ must transform like $x^\mu$.

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  • $\begingroup$ Is it possible to use the definition of $j^\mu$ and work out from that? ACuriousMind says that it should be straightforward but I can't see how. $\endgroup$ – SRS Mar 22 '18 at 12:53
  • $\begingroup$ @SRS You need to assume that charge is Lorentz invariant to make ACuriousMind's method work. I didn't think of topologically_astounded's suggestion but I think it's the more elegant one: its experimental grounding lends weight too. $\endgroup$ – WetSavannaAnimal Mar 22 '18 at 13:15
  • $\begingroup$ Well. Even then I don't see how that works out. See my comment below ACuriousMind's. @WetSavannaAnimalakaRodVance $\endgroup$ – SRS Mar 22 '18 at 13:17
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    $\begingroup$ If validity of local conservation equation $\partial_\mu j^\mu = 0$ in every frame was enough, then any quantity that obeys that equation in every frame would be four-vector. But that is not so; take 4-tuple [energy density, energy current density]. It obeys the equation, but it is not a four-vector. $\endgroup$ – Ján Lalinský Jun 13 '18 at 23:25
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    $\begingroup$ In general \begin{equation} \require{cancel} (\text{4-divergence of } j^{\mu}=0) \quad\cancel{=\!=\!=\!=\!=\!=\!\Longrightarrow} \quad (j^{\mu} \text{ is four-vector}) \end{equation} Your answer is wrong. $\endgroup$ – Frobenius Jun 15 '18 at 9:48
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Charge density $\rho$ and current density $\mathbf j$ obey Maxwell's equations in all inertial frames. This means that in every inertial frame, the current density 4-tuple obeys the same relation; in the original frame, we have $$ (c\rho,\mathbf j) = (c\epsilon_0\nabla\cdot \mathbf E,\nabla\times\mathbf B/\mu_0 - \epsilon_0\partial_t \mathbf E). $$ and in the primed frame moving with respect to the first frame, we have $$ (c\rho',\mathbf j') = (c\epsilon_0\nabla'\cdot \mathbf E',\nabla'\times\mathbf B'/\mu_0 - \epsilon_0\partial_t' \mathbf E'). $$

We can express fields $\mathbf E',\mathbf B'$ operations $\partial_t',\nabla'$ on the right-hand side withf $\mathbf E,\mathbf B$ and operations $\partial_t,\nabla$, using the transformation formulae for fields $\mathbf E,\mathbf B$ in relativistic theory$^*$. When that is done, it can be inferred that the 4-tuple transforms as a four-vector. This method of proof is tedious but quite convincing.

$^*$Those follow from general relativistic transformation of 3-force in relativistic mechanics; see Frobenius' answer, formula 11, here:

https://physics.stackexchange.com/a/411129/31895

or the paper https://arxiv.org/abs/physics/0507099 . When applied to the Lorentz formula, which defines electric and magnetic field in every inertial frame: $$ \mathbf F =q\mathbf E + q\mathbf v\times\mathbf B. $$ we may derive transformation formulae for the fields.

Easier (but less convincing) way to prove $j$ is a four-vector: Maxwell's equations imply $$ j^\mu = \partial_\nu F^{\nu\mu}. $$ Because $F^{\nu\mu}$ is a four-tensor$^{**}$, the expression $\partial_\nu F^{\nu\mu}$ defines a four-tensor.

$^{**}$ This follows from the definition of $F$ -- antisymmetric tensor whose components are formed from components of electric and magnetic field -- and the transformation formulae for those fields mentioned above. Alternatively, if we accept that there is universal equation of motion of a test particle in EM field for every frame and every four-velocity $$ qF^{\nu\mu}u_\mu = m\,du^\nu/d\tau $$ it seems that $F$ must be a four-tensor. All other-than-$F$ quantities transform as four-tensors ($q,m,\tau$ are invariant, $u$ is a 4-vector by definition), so $F^{\nu\mu}u_\mu$ is a four-tensor. Then, it is plausible that $F$ in this expression is a four-tensor as well (this is the problematic part - how to make sure that F must be tensor here?).

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  • $\begingroup$ The right part of your last equation is a four-vector. So is the left part, since the equation should be covariant (valid equation in any inertial frame). How $F^{\nu \mu}$ cannot be a four-tensor, since $u_{\mu}$ defines a four-vector ? The contraction can't cancel the arbitrary non-tensor part of $F^{\nu \mu}$, if this guy wasn't a tensor. $\endgroup$ – Cham Jun 15 '18 at 13:50
  • $\begingroup$ @Someone that was the direction I suggested, but the problem is it is not directly showing $F$ must be a tensor, only that $F^{\mu\nu}u_\mu$ is a tensor. How do you prove from this $F$ is a tensor? $u$ is not arbitrary four-vector, but must obey $u^\nu u_\nu = -c^2$. $\endgroup$ – Ján Lalinský Jun 15 '18 at 18:20
  • $\begingroup$ I think it is obvious. If $u_{\mu}$ is a tensor and $F^{\nu \mu} \, u_{\mu}$ is also a tensor, then $F^{\nu \mu}$ must also be a tensor. The combination $F^{\nu \mu} \, u_{\mu}$ is just a matrix multiplication. After a coordinates transformation, write this (put in the tensor indices) : \begin{equation} \tilde{F} \, \tilde{u} = \frac{\partial \tilde{x}}{\partial x} F \, u = \tilde{F} \, \frac{\partial x}{\partial \tilde{x}} \, u.\end{equation} The matrix $\frac{\partial \tilde{x}}{\partial x}$ is invertible. Since this is true for any timelike 4-velocity, you should get $\tilde{F} = ..F$. $\endgroup$ – Cham Jun 15 '18 at 18:35
  • $\begingroup$ I can see how this proves $Fu = \Lambda ^{-1}\tilde{F}\Lambda u$ ($\Lambda$ is the Lorentz transformation matrix). But how do you get rid of $u$ to conclude $F = \Lambda ^{-1}\tilde{F}\Lambda$? $\endgroup$ – Ján Lalinský Jun 15 '18 at 18:55
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I think the starting point of this is to see how $j^\mu$ is defined. In the absence of charges the EM action is given by

$$ S= \int d^4 x F_{\mu \nu} F^{\mu \nu} $$

where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$ which comes from gauge invariance. The equation of motion is

$$ \partial_\mu F^{\mu \nu} =0 $$

and introducing charges means that by Lorentz covariance the only possibility is

$$ \partial_\mu F^{\mu \nu} = j^\nu $$

Then writing everything explicitly in terms of electrico-magnetic fields, charges and currents would give the desired relation. I think one ambiguity would be in $A_\mu = ( \pm \Phi,\vec A)$ and a choice would have to be made and as the Lagrangian has $A_\mu j^\mu$. Here one would have to invoke some physical idea like Prahar mentioned above.

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  • $\begingroup$ Can we not proceed from the definition of $j^\mu$ as $j^\mu=(c\rho,\textbf{J})$? @BorunChowdhury $\endgroup$ – SRS Mar 19 '18 at 17:03
  • $\begingroup$ I don't think you can proceed from this definition as you have to justify why this combination of charge and current is a four vector. This comes from the scalar $A_\mu j^\mu$. The fact that $A_\mu$ is a four-vector comes from gauge invariance meaning under gauge transformations $\partial_\mu \to \partial_\mu + A_\mu$. $\endgroup$ – Borun Chowdhury Mar 19 '18 at 18:04
  • $\begingroup$ Minor points. I think you wrote the action $S$ (apart from the usual factor of $-1/4$), instead of the Lagrangian $L$. And also probably instead of Lorentz invariance of the equation of motion, you meant Lorentz covariance. @BorunChowdhury $\endgroup$ – SRS Jun 15 '18 at 17:05
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$\color{blue}{\textbf{ANSWER B}}\:$ (based on the co-variance of Mawxell equations under Lorentz tranformations )

enter image description here

Let the quantities \begin{equation} \mathbf{E}=\left(E_x,E_y,E_z\right), \quad \mathbf{B}=\left(B_x,B_y,B_z\right), \quad \mathbf{j}=\left(j_x,j_y,j_z\right), \quad \rho \nonumber \end{equation} satisfying the Maxwell equations in empty space in an inertial system $\:\mathrm S$ : \begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \tag{01a}\\ \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} & = \mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{01b}\\ \boldsymbol{\nabla} \boldsymbol{\cdot} \mathbf{E} & = \frac{\rho}{\epsilon_{0}} \tag{01c}\\ \boldsymbol{\nabla} \boldsymbol{\cdot} \mathbf{B} & = 0 \tag{01d} \end{align} If we apply the 1+1-dimensional Lorentz transformation : \begin{align} x' & = \gamma\left(x\boldsymbol{-}\upsilon t\right) \tag{02a}\\ y' & = y \vphantom{\left(t\boldsymbol{-}\frac{\upsilon x}{c^{2}} \right)} \tag{02b}\\ z' & = z \vphantom{\left(t\boldsymbol{-}\frac{\upsilon x}{c^{2}} \right)} \tag{02c}\\ t' & = \gamma\left(t\boldsymbol{-}\dfrac{\upsilon x}{c^{2}} \right) \tag{02d} \end{align} for the configuration of the systems $\:\mathrm S\:$ and $\:\mathrm S'\:$ as in Figure-01, then the following defined primed quantities \begin{align} E'_{x} & = E_{x} \tag{16a}\\ E'_{y} & = \gamma \left(E_{y} \boldsymbol{-}\upsilon B_{z}\right) \tag{16b}\\ E'_{z} & = \gamma \left(E_{z} \boldsymbol{+}\upsilon B_{y}\right) \tag{16c}\\ B'_{x} & = B_{x} \tag{17a}\\ B'_{y} & = \gamma\Bigl(B_{y}+\dfrac{\upsilon}{c^{2}} E_{z}\Bigr) \tag{17b}\\ B'_{z} & = \gamma\Bigl(B_{z}-\dfrac{\upsilon}{c^{2}} E_{y}\Bigr) \tag{17c}\\ j'_{x} & = \gamma\left(j_{x}\boldsymbol{-}\upsilon \rho \right) \tag{24a}\\ j'_{y} & = j_{y} \tag{24b}\\ j'_{z} & = j_{z} \tag{24c}\\ \rho' & = \gamma\Bigl(\rho \boldsymbol{-}\dfrac{\upsilon j_{x}}{c^{2}}\Bigr) \tag{18} \end{align} satisfy the primed Maxwell equations in system $\:\mathrm S'\:$ \begin{align} \boldsymbol{\nabla'} \boldsymbol{\times} \mathbf{E'} & = -\frac{\partial \mathbf{B'}}{\partial t'} \tag{22}\\ \boldsymbol{\nabla'} \boldsymbol{\times} \mathbf{B'} & = \mu_{0}\mathbf{j'}+\frac{1}{c^{2}}\frac{\partial \mathbf{E'}}{\partial t'} \tag{25}\\ \boldsymbol{\nabla'} \boldsymbol{\cdot} \mathbf{E'} & = \frac{\rho'}{\epsilon_{0}} \tag{10}\\ \boldsymbol{\nabla'} \boldsymbol{\cdot} \mathbf{B'} & = 0 \tag{13} \end{align} Comparing the set of equations (24),(18) with (02) we conclude that the charge current density vector $\:\mathbf{J}=\left(c\rho,\mathbf{j}\right)\:$ is transformed as the space-time position vector $\:\mathbf{X}=\left(ct,\mathbf{x}\right)$.

So $\:\mathbf{J}\:$ is a 4-vector.


So, under the assumption of Maxwell equations covariance we can prove that the charge 4-current density is a Lorentz 4-vector and based on this we prove the charge invariance, see a related answer of mine here : Why charge is Lorentz invariant but relativistic mass is not?


It's available in $\LaTeX$ the 3+1-dimensional version of this answer.


Proof :

The Maxwell differential equations of electromagnetic field in empty space are \begin{align} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \tag{01a}\\ \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} & = \mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \tag{01b}\\ \boldsymbol{\nabla} \boldsymbol{\cdot} \mathbf{E} & = \frac{\rho}{\epsilon_{0}} \tag{01c}\\ \boldsymbol{\nabla} \boldsymbol{\cdot} \mathbf{B} & = 0 \tag{01d} \end{align} where $\: \mathbf{E} =$ electric field intensity vector, $\:\mathbf{B}=$ magnetic-flux density vector, $\:\rho=$ electric charge density, $\:\mathbf{j} =$ electric current density vector. All quantities are functions of the three space coordinates $\:\left( x,y,z\right)\:$ and time $\:t$.

We will apply on them the following Lorentz transformation and we must define the new variables $\:\mathbf{E'},\mathbf{B'},\mathbf{j'},\rho'\:$ so that the form of equations (01) to remain unchanged (co-variant) in the new reference frame. From the definition of the new current 4-vector we'll prove that it is a Lorentz 4-vector. So, let the usual configuration of two systems $\:\mathrm S,\mathrm S'\:$ the latter moving relatively to the former with velocity $\:\upsilon \in (-c,c)\:$ along the common axis $\:x$, see Figure-01.
The Lorentz transformation equations are \begin{align} x' & = \gamma\left(x\boldsymbol{-}\upsilon t\right) \tag{02a}\\ y' & = y \vphantom{\left(t\boldsymbol{-}\frac{\upsilon x}{c^{2}} \right)} \tag{02b}\\ z' & = z \vphantom{\left(t\boldsymbol{-}\frac{\upsilon x}{c^{2}} \right)} \tag{02c}\\ t' & = \gamma\left(t\boldsymbol{-}\dfrac{\upsilon x}{c^{2}} \right) \tag{02d} \end{align} Now, we must express the partial derivatives with respect to the space-time variables $\:(x,y,z,t)\:$ in terms of the partial derivatives with respect to the space-time variables $\:(x',y',z',t')$. From (02) we have \begin{align} \dfrac{\partial \hphantom{x}}{\partial x} & = \dfrac{\partial \hphantom{x'}}{\partial x'}\dfrac{\partial x'}{\partial x\hphantom{'}}\boldsymbol{+}\dfrac{\partial \hphantom{t'}}{\partial t'}\dfrac{\partial t'}{\partial x\hphantom{'}}=\gamma\dfrac{\partial \hphantom{x'}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial \hphantom{t'}}{\partial t'} \tag{03a}\\ \dfrac{\partial \hphantom{y}}{\partial y} & = \dfrac{\partial \hphantom{y'}}{\partial y'} \tag{03b}\\ \dfrac{\partial \hphantom{z}}{\partial z} & = \dfrac{\partial \hphantom{z'}}{\partial z'} \tag{03c}\\ \dfrac{\partial \hphantom{t}}{\partial t} & = \dfrac{\partial \hphantom{x'}}{\partial x'}\dfrac{\partial x'}{\partial t\hphantom{'}}\boldsymbol{+}\dfrac{\partial \hphantom{t'}}{\partial t'}\dfrac{\partial t'}{\partial t\hphantom{'}}=\boldsymbol{-}\gamma\upsilon\dfrac{\partial \hphantom{x'}}{\partial x'}\boldsymbol{+}\gamma\dfrac{\partial \hphantom{t'}}{\partial t'} \tag{03d} \end{align} Starting with Maxwell equation (01a) we have \begin{equation} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} \Longrightarrow \begin{cases} \dfrac{\partial E_{z}}{\partial y}\boldsymbol{-}\dfrac{\partial E_{y}}{\partial z}=\boldsymbol{-}\dfrac{\partial B_{x}}{\partial t}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \\ \dfrac{\partial E_{x}}{\partial z}\boldsymbol{-}\dfrac{\partial E_{z}}{\partial x}=\boldsymbol{-}\dfrac{\partial B_{y}}{\partial t}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \\ \dfrac{\partial E_{y}}{\partial x}\boldsymbol{-}\dfrac{\partial E_{x}}{\partial y}=\boldsymbol{-}\dfrac{\partial B_{x}}{\partial t}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \end{cases} \tag{04} \end{equation} and using the partial derivative relations (03) \begin{align} \dfrac{\partial E_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial E_{y}}{\partial z'} &=\gamma \upsilon\dfrac{\partial B_{x}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\partial B_{x}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} \tag{05a}\\ \dfrac{\partial E_{x}}{\partial z'}\boldsymbol{-}\gamma\dfrac{\partial E_{z}}{\partial x'}\boldsymbol{+}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial E_{z}}{\partial t'}&=\gamma \upsilon\dfrac{\partial B_{y}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\partial B_{y}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} \tag{05b}\\ \gamma\dfrac{\partial E_{y}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial E_{y}}{\partial t'}\boldsymbol{-}\dfrac{\partial E_{x}}{\partial y'}&=\gamma \upsilon\dfrac{\partial B_{z}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\partial B_{z}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} \tag{05c} \end{align} With Maxwell equation (01b) \begin{equation} \boldsymbol{\nabla} \boldsymbol{\times} \mathbf{B} = \mu_{0}\mathbf{j}+\frac{1}{c^{2}}\frac{\partial \mathbf{E}}{\partial t} \Longrightarrow \begin{cases} \dfrac{\partial B_{z}}{\partial y}\boldsymbol{-}\dfrac{\partial B_{y}}{\partial z}=\mu_{0}j_{x} \boldsymbol{+}\dfrac{1}{c^{2}}\dfrac{\partial E_{x}}{\partial t}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \\ \dfrac{\partial B_{x}}{\partial z}\boldsymbol{-}\dfrac{\partial B_{z}}{\partial x}=\mu_{0}j_{y} \boldsymbol{+}\dfrac{1}{c^{2}}\dfrac{\partial E_{y}}{\partial t}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \\ \dfrac{\partial B_{y}}{\partial x}\boldsymbol{-}\dfrac{\partial B_{x}}{\partial y}=\mu_{0}j_{z} \boldsymbol{+}\dfrac{1}{c^{2}}\dfrac{\partial E_{z}}{\partial t}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \end{cases} \tag{06} \end{equation} and so \begin{align} \dfrac{\partial B_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial B_{y}}{\partial z'} &=\mu_{0}j_{x}\boldsymbol{-}\dfrac{\gamma \upsilon}{c^{2}}\dfrac{\partial E_{x}}{\partial x'}\boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial E_{x}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} \tag{07a}\\ \dfrac{\partial B_{x}}{\partial z'}\boldsymbol{-}\gamma\dfrac{\partial B_{z}}{\partial x'}\boldsymbol{+}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial B_{z}}{\partial t'} & = \mu_{0}j_{y}\boldsymbol{-}\dfrac{\gamma \upsilon}{c^{2}}\dfrac{\partial E_{y}}{\partial x'}\boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial E_{y}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} \tag{07b}\\ \gamma\dfrac{\partial B_{y}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial B_{y}}{\partial t'}\boldsymbol{-}\dfrac{\partial B_{x}}{\partial y'}&=\mu_{0}j_{z}\boldsymbol{-}\dfrac{\gamma \upsilon}{c^{2}}\dfrac{\partial E_{z}}{\partial x'}\boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial E_{z}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} \tag{07c} \end{align} Continuing with (01c) \begin{align} \boldsymbol{\nabla} \boldsymbol{\cdot} \mathbf{E} = \frac{\rho}{\epsilon_{0}} \Longrightarrow \dfrac{\partial E_{x}}{\partial x}\boldsymbol{+}\dfrac{\partial E_{y}}{\partial y}\boldsymbol{+}\dfrac{\partial E_{z}}{\partial z} & =\frac{\rho}{\epsilon_{0}} \Longrightarrow \nonumber\\ \gamma\dfrac{\partial E_{x}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial E_{x}}{\partial t'}\boldsymbol{+}\dfrac{\partial E_{y}}{\partial y}\boldsymbol{+}\dfrac{\partial E_{z}}{\partial z} & = \frac{\rho}{\epsilon_{0}} \nonumber \end{align} so \begin{equation} \dfrac{\partial \gamma E_{x}}{\partial x'}\boldsymbol{+}\dfrac{\partial E_{y}}{\partial y'}\boldsymbol{+}\dfrac{\partial E_{z}}{\partial z'} = \frac{\rho}{\epsilon_{0}}\boldsymbol{+}\dfrac{\gamma\upsilon}{c^{2}}\dfrac{\partial E_{x}}{\partial t'} \tag{08} \end{equation} and finally with (01d) \begin{align} \boldsymbol{\nabla} \boldsymbol{\cdot} \mathbf{B} =0 \Longrightarrow \dfrac{\partial B_{x}}{\partial x}\boldsymbol{+}\dfrac{\partial B_{y}}{\partial y}\boldsymbol{+}\dfrac{\partial B_{z}}{\partial z}&=0\Longrightarrow \nonumber\\ \gamma\dfrac{\partial B_{x}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial B_{x}}{\partial t'}\boldsymbol{+}\dfrac{\partial B_{y}}{\partial y}\boldsymbol{+}\dfrac{\partial B_{z}}{\partial z} & =0 \Longrightarrow \nonumber \end{align} that is \begin{equation} \dfrac{\partial \gamma B_{x}}{\partial x'}\boldsymbol{+}\dfrac{\partial B_{y}}{\partial y'}\boldsymbol{+}\dfrac{\partial B_{z}}{\partial z'} = \dfrac{\gamma\upsilon}{c^{2}}\dfrac{\partial B_{x}}{\partial t'} \tag{09} \end{equation} Now, using the eight (8) scalar equations (05),(07), (08) and (09) we must try to define the 10 scalar primed quantities - the components of $\:\mathbf{E'},\mathbf{B'},\mathbf{j'}\:$ and the scalar $\:\rho'\:$ - in terms of the unprimed ones in such a way that to yield the primed Maxwell equations. Let begin with equation (08). This is candidate for the Maxwell equation \begin{equation} \boldsymbol{\nabla'} \boldsymbol{\cdot} \mathbf{E'} = \frac{\rho'}{\epsilon_{0}} \tag{10} \end{equation} The problem is that equation (10) has partial derivatives with respect to $\:(x',y',z')\:$ but not with respect to $\:t'\:$ as (08) does. But we can see that this partial derivative with respect to $\:t'\:$ in the rhs of (08) could be expressed in terms of partial derivatives with respect to $\:(x',y',z')\:$ from equation (07a). More exactly from (07a) \begin{equation} \dfrac{\gamma\upsilon}{c^{2}}\dfrac{\partial E_{x}}{\partial t'} =\dfrac{\partial (\upsilon B_{z})}{\partial y'}\boldsymbol{-}\dfrac{\partial (\upsilon B_{y})}{\partial z'} \boldsymbol{-}\mu_{0}\upsilon j_{x}\boldsymbol{+}\dfrac{\gamma \upsilon^{2}}{c^{2}}\dfrac{\partial E_{x}}{\partial x'} \tag{11} \end{equation} Inserting this expression in (08) we have \begin{equation} \dfrac{\partial \gamma E_{x}}{\partial x'}\boldsymbol{+}\dfrac{\partial E_{y}}{\partial y'}\boldsymbol{+}\dfrac{\partial E_{z}}{\partial z'} = \frac{\rho}{\epsilon_{0}}\boldsymbol{+}\dfrac{\partial (\upsilon B_{z})}{\partial y'}\boldsymbol{-}\dfrac{\partial (\upsilon B_{y})}{\partial z'} \boldsymbol{-}\mu_{0}\upsilon j_{x}\boldsymbol{+}\dfrac{\gamma \upsilon^{2}}{c^{2}}\dfrac{\partial E_{x}}{\partial x'} \nonumber \end{equation} so \begin{equation} \dfrac{\partial E_{x}}{\partial x'}\boldsymbol{+}\dfrac{\partial \left[\gamma (E_{y} \boldsymbol{-}\upsilon B_{z})\right]}{\partial y'}\boldsymbol{+}\dfrac{\partial \left[\gamma (E_{z} \boldsymbol{-}\upsilon B_{y})\right]}{\partial z'} = \frac{\gamma\Bigl(\rho \boldsymbol{-}\dfrac{\upsilon j_{x}}{c^{2}}\Bigr)}{\epsilon_{0}} \tag{12} \end{equation} Let continue with (09). This is candidate for the Maxwell equation \begin{equation} \boldsymbol{\nabla'} \boldsymbol{\cdot} \mathbf{B'} =0 \tag{13} \end{equation} From (05a) \begin{equation} \dfrac{\gamma\upsilon}{c^{2}}\dfrac{\partial B_{x}}{\partial t'} = \dfrac{\gamma \upsilon^{2}}{c^{2}}\dfrac{\partial B_{x}}{\partial x'}\boldsymbol{-}\dfrac{\upsilon}{c^{2}}\dfrac{\partial E_{z}}{\partial y'}\boldsymbol{+}\dfrac{\upsilon}{c^{2}}\dfrac{\partial E_{y}}{\partial z'} \tag{14} \end{equation} Inserting this expression in (09) we have \begin{equation} \dfrac{\partial \gamma B_{x}}{\partial x'}\boldsymbol{+}\dfrac{\partial B_{y}}{\partial y'}\boldsymbol{+}\dfrac{\partial B_{z}}{\partial z'} = \dfrac{\gamma \upsilon^{2}}{c^{2}}\dfrac{\partial B_{x}}{\partial x'}\boldsymbol{-}\dfrac{\upsilon}{c^{2}}\dfrac{\partial E_{z}}{\partial y'}\boldsymbol{+}\dfrac{\upsilon}{c^{2}}\dfrac{\partial E_{y}}{\partial z'} \nonumber \end{equation} so \begin{equation} \dfrac{\partial B_{x}}{\partial x'}\boldsymbol{+}\dfrac{\partial \left[\gamma\Bigl(B_{y}+\dfrac{\upsilon}{c^{2}} E_{z}\Bigr)\right]}{\partial y'}\boldsymbol{+}\dfrac{\partial \left[\gamma\Bigl(B_{z}-\dfrac{\upsilon}{c^{2}} E_{y}\Bigr)\right]}{\partial z'} = 0 \tag{15} \end{equation} From equations (12) and (15) it seems that till now it would be a good choice to define seven (7) scalar primed quantities - the components of $\:\mathbf{E'},\mathbf{B'}\:$ and the scalar $\:\rho'\:$ - in terms of the unprimed ones as follows \begin{align} E'_{x} & = E_{x} \tag{16a}\\ E'_{y} & = \gamma \left(E_{y} \boldsymbol{-}\upsilon B_{z}\right) \tag{16b}\\ E'_{z} & = \gamma \left(E_{z} \boldsymbol{+}\upsilon B_{y}\right) \tag{16c} \end{align} \begin{align} B'_{x} & = B_{x} \tag{17a}\\ B'_{y} & = \gamma\Bigl(B_{y}+\dfrac{\upsilon}{c^{2}} E_{z}\Bigr) \tag{17b}\\ B'_{z} & = \gamma\Bigl(B_{z}-\dfrac{\upsilon}{c^{2}} E_{y}\Bigr) \tag{17c} \end{align} and \begin{equation} \rho' = \gamma\Bigl(\rho \boldsymbol{-}\dfrac{\upsilon j_{x}}{c^{2}}\Bigr) \tag{18} \end{equation} It remains to define the rest three (3) scalar primed quantities - the components of $\:\mathbf{j'}$ - and to check if all these defined primed quantities are consistent to transform equations (05) and (07) to the primed versions of Maxwell equations (01a) and (01b) respectively. If we think the six (6) scalar equations (16),(17) as a linear system with 6 "unknowns" the unprimed quantities $\:E_{x},E_{y},E_{z},B_{x},B_{y},B_{z}\:$ then, solving with respect to them, we have \begin{align} E_{x} & = E'_{x} \tag{19a}\\ E_{y} & = \gamma \left(E'_{y} \boldsymbol{+}\upsilon B'_{z}\right) \tag{19b}\\ E_{z} & = \gamma \left(E'_{z} \boldsymbol{-}\upsilon B'_{y}\right) \tag{19c} \end{align} \begin{align} B_{x} & = B'_{x} \tag{20a}\\ B_{y} & = \gamma\Bigl(B'_{y}\boldsymbol{-}\dfrac{\upsilon}{c^{2}} E'_{z}\Bigr) \tag{20b}\\ B_{z} & = \gamma\Bigl(B'_{z} \boldsymbol{+}\dfrac{\upsilon}{c^{2}} E'_{y}\Bigr) \tag{20c} \end{align} Replacing them in (05a) we have \begin{align} & \dfrac{\partial \overbrace{\left[\gamma \left(E'_{z} \boldsymbol{-}\upsilon B'_{y}\right)\right]}^{E_{z}}}{\partial y'}\boldsymbol{-}\dfrac{\partial \overbrace{\left[\gamma \left(E'_{y} \boldsymbol{+}\upsilon B'_{z}\right)\right]}^{E_{y}}}{\partial z'} =\gamma \upsilon\dfrac{\partial \overbrace{B'_{x}}^{B_{x}}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\partial \overbrace{B'_{x}}^{B_{x}}}{\partial t'} \quad \stackrel{(15) ,(17)}{=\!=\!=\!\Longrightarrow} \nonumber\\ &\dfrac{\partial E'_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial E'_{y}}{\partial z'} = \upsilon\underbrace{\left(\dfrac{\partial B'_{x}}{\partial x'}\boldsymbol{+}\dfrac{\partial B'_{y}}{\partial y'}\boldsymbol{+}\dfrac{\partial B'_{z}}{\partial z'}\right)}_{0} \boldsymbol{-}\dfrac{\partial B'_{x}}{\partial t'} \nonumber \end{align} so \begin{equation} \dfrac{\partial E'_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial E'_{y}}{\partial z'} = \boldsymbol{-}\dfrac{\partial B'_{x}}{\partial t'} \tag{21a} \end{equation} Replacing them in (05b) we have \begin{align} & \dfrac{\partial \overbrace{E'_{x}}^{E_{x}}}{\partial z'}\boldsymbol{-}\gamma\dfrac{\partial \overbrace{\left[\gamma \left(E'_{z} \boldsymbol{-}\upsilon B'_{y}\right)\right]}^{E_{z}}}{\partial x'}\boldsymbol{+}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma \left(E'_{z} \boldsymbol{-}\upsilon B'_{y}\right)\right]}^{E_{z}}}{\partial t'} = \nonumber\\ &\gamma \upsilon\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{y}\boldsymbol{-}\dfrac{\upsilon}{c^{2}} E'_{z}\Bigr)\right]}^{B_{y}}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{y}\boldsymbol{-}\dfrac{\upsilon}{c^{2}} E'_{z}\Bigr)\right]}^{B_{y}}}{\partial t'} \quad =\!=\!=\!\Longrightarrow \nonumber\\ &\dfrac{\partial E'_{x}}{\partial z'}\boldsymbol{-}\gamma^{2}\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{\partial E'_{z}}{\partial x'}=\boldsymbol{-}\gamma^{2}\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{\partial B'_{y}}{\partial t'}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \nonumber \end{align} so \begin{equation} \dfrac{\partial E'_{x}}{\partial z'}\boldsymbol{-}\dfrac{\partial E'_{z}}{\partial x'} = \boldsymbol{-}\dfrac{\partial B'_{y}}{\partial t'} \tag{21b} \end{equation} and finally replacing them in (05c) \begin{align} & \gamma\dfrac{\partial \overbrace{\left[\gamma \left(E'_{y} \boldsymbol{+}\upsilon B'_{z}\right)\right]}^{E_{y}}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma \left(E'_{y} \boldsymbol{+}\upsilon B'_{z}\right)\right]}^{E_{y}}}{\partial t'}\boldsymbol{-}\dfrac{\partial \overbrace{E'_{x}}^{E_{x}}}{\partial y'} = \nonumber\\ &\gamma \upsilon\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{z} \boldsymbol{+}\dfrac{\upsilon}{c^{2}} E'_{y}\Bigr)\right]}^{B_{z}}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{z} \boldsymbol{+}\dfrac{\upsilon}{c^{2}} E'_{y}\Bigr)\right]}^{B_{z}}}{\partial t'}\quad =\!=\!=\!\Longrightarrow \nonumber\\ &\gamma^{2}\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{\partial E'_{y}}{\partial x'}\boldsymbol{-}\dfrac{\partial E'_{x}}{\partial y'}=\boldsymbol{-} \gamma^{2}\left(1-\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{\partial B'_{z}}{\partial t'} \vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \nonumber \end{align} so \begin{equation} \dfrac{\partial E'_{y}}{\partial x'}\boldsymbol{-}\dfrac{\partial E'_{x}}{\partial y'} = \boldsymbol{-}\dfrac{\partial B'_{z}}{\partial t'} \tag{21c} \end{equation} Equations (21a),(21b) and (21c) is a proof that the primed vectors $\:\mathbf{E'},\mathbf{B'}\:$ defined by (16), (17) satisfy the primed version of Maxwell equation (01a) \begin{equation} \boldsymbol{\nabla'} \boldsymbol{\times} \mathbf{E'} = -\frac{\partial \mathbf{B'}}{\partial t'} \tag{22} \end{equation} We continue now with equation (01b). Replacing in (07a) the unprimed quantities $\:E_{x},E_{y},E_{z},B_{x},B_{y},B_{z}\:$ by their expressions (19),(20) we have \begin{align} &\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{z} \boldsymbol{+}\dfrac{\upsilon}{c^{2}} E'_{y}\Bigr)\right]}^{B_{z}}}{\partial y'}\boldsymbol{-}\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{y}\boldsymbol{-}\dfrac{\upsilon}{c^{2}} E'_{z}\Bigr)\right]}^{B_{y}}}{\partial z'} = \nonumber\\ &\mu_{0}j_{x}\boldsymbol{-}\dfrac{\gamma \upsilon}{c^{2}}\dfrac{\partial \overbrace{E'_{x}}^{E_{x}}}{\partial x'}\boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial \overbrace{E'_{x}}^{E_{x}}}{\partial t'}\quad =\!=\!=\!\Longrightarrow \nonumber\\ & \gamma\left(\dfrac{\partial B'_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial B'_{y}}{\partial z'}\right)=\mu_{0}j_{x}\boldsymbol{-}\dfrac{\gamma \upsilon}{c^{2}} \underbrace{\left(\dfrac{\partial E'_{x}}{\partial x'}\boldsymbol{+} \dfrac{\partial E'_{y}}{\partial y'}\boldsymbol{+} \dfrac{\partial E'_{z}}{\partial z'}\right)}_{(18) :\: \tfrac{\rho'}{\epsilon_{0}} \stackrel{(18)}{=\!=} \tfrac{\gamma}{\epsilon_{0}}\bigl(\rho \boldsymbol{-}\tfrac{\upsilon j_{x}}{c^{2}}\bigr)} \boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial E'_{x}}{\partial t'}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}}\quad =\!=\!=\!\Longrightarrow \nonumber\\ &\gamma\left(\dfrac{\partial B'_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial B'_{y}}{\partial z'}\right)=\mu_{0}j_{x}\boldsymbol{-}\dfrac{\gamma^{2} \upsilon}{\epsilon_{0}c^{2}} \Bigl(\rho \boldsymbol{-}\dfrac{\upsilon j_{x}}{c^{2}}\Bigr) \boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial E'_{x}}{\partial t'}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}}\quad \stackrel{\epsilon_{0}c^{2}=\mu_{0}^{-1}}{=\!=\!=\!\Longrightarrow} \nonumber\\ &\gamma\left(\dfrac{\partial B'_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial B'_{y}}{\partial z'}\right)=\mu_{0}\left(1\boldsymbol{+}\dfrac{\gamma^{2} \upsilon^{2}}{c^{2}}\right)j_{x}\boldsymbol{-}\mu_{0}\gamma^{2} \upsilon \rho \boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial E'_{x}}{\partial t'}\vphantom{\frac{\dfrac{a}{b}}{\frac{c}{d}}} \nonumber \end{align} so \begin{equation} \dfrac{\partial B'_{z}}{\partial y'}\boldsymbol{-}\dfrac{\partial B'_{y}}{\partial z'}=\mu_{0}\bigl[\gamma\left(j_{x}\boldsymbol{-}\upsilon \rho \right) \bigr]\boldsymbol{+}\dfrac{1}{c^{2}}\dfrac{\partial E'_{x}}{\partial t'} \tag{23a} \end{equation} Replacing in (07b) \begin{align} &\dfrac{\partial \overbrace{B'_{x}}^{B_{x}}}{\partial z'}\boldsymbol{-}\gamma\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{z} \boldsymbol{+}\dfrac{\upsilon}{c^{2}} E'_{y}\Bigr)\right]}^{B_{z}}}{\partial x'}\boldsymbol{+}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{z} \boldsymbol{+}\dfrac{\upsilon}{c^{2}} E'_{y}\Bigr)\right]}^{B_{z}}}{\partial t'} = \vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} \nonumber\\ & \mu_{0}j_{y}\boldsymbol{-}\dfrac{\gamma \upsilon}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma \left(E'_{y} \boldsymbol{+}\upsilon B'_{z}\right)\right]}^{E_{y}}}{\partial x'}\boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma \left(E'_{y} \boldsymbol{+}\upsilon B'_{z}\right)\right]}^{E_{y}}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} =\!=\!=\!\Longrightarrow \nonumber\\ &\dfrac{\partial B'_{x}}{\partial z'}\boldsymbol{-}\gamma^{2}\left(1\boldsymbol{-}\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{\partial B'_{z}}{\partial x'}= \mu_{0}j_{y}\boldsymbol{+}\gamma^{2}\left(1\boldsymbol{-}\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{1}{c^{2}}\dfrac{\partial E'_{y}}{\partial t'} \nonumber \end{align} so \begin{equation} \dfrac{\partial B'_{x}}{\partial z'}\boldsymbol{-}\dfrac{\partial B_{z'}}{\partial x'} =\mu_{0}j_{y} \boldsymbol{+}\dfrac{1}{c^{2}}\dfrac{\partial E'_{y}}{\partial t'} \tag{23b} \end{equation} Replacing in (07c) \begin{align} & \gamma\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{y}\boldsymbol{-}\dfrac{\upsilon}{c^{2}} E'_{z}\Bigr)\right]}^{B_{y}}}{\partial x'}\boldsymbol{-}\gamma\dfrac{\upsilon}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma\Bigl(B'_{y}\boldsymbol{-}\dfrac{\upsilon}{c^{2}} E'_{z}\Bigr)\right]}^{B_{y}}}{\partial t'}\boldsymbol{-}\dfrac{\partial \overbrace{B'_{x}}^{B_{x}}}{\partial y'} = \nonumber\\ & \mu_{0}j_{z}\boldsymbol{-}\dfrac{\gamma \upsilon}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma \left(E'_{z} \boldsymbol{-}\upsilon B'_{y}\right)\right]}^{E_{z}}}{\partial x'}\boldsymbol{+}\dfrac{\gamma}{c^{2}}\dfrac{\partial \overbrace{\left[\gamma \left(E'_{z} \boldsymbol{-}\upsilon B'_{y}\right)\right]}^{E_{z}}}{\partial t'}\vphantom{\frac{\frac{a}{b}}{\frac{c}{d}}} =\!=\!=\!\Longrightarrow \nonumber\\ &\gamma^{2}\left(1\boldsymbol{-}\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{\partial B'_{y}}{\partial x'}\boldsymbol{-}\dfrac{\partial B'_{x}}{\partial y'}= \mu_{0}j_{z}\boldsymbol{+}\gamma^{2}\left(1\boldsymbol{-}\dfrac{\upsilon^{2}}{c^{2}}\right)\dfrac{1}{c^{2}}\dfrac{\partial E'_{z}}{\partial t'} \nonumber \end{align} so \begin{equation} \dfrac{\partial B'_{y}}{\partial x'}\boldsymbol{-}\dfrac{\partial B'_{x}}{\partial y'} = \mu_{0}j_{z} \boldsymbol{+}\dfrac{1}{c^{2}}\dfrac{\partial E'_{z}}{\partial t'} \tag{23c} \end{equation} If beyond the definitions (16),(17) and (18) we define also \begin{align} j'_{x} & = \gamma\left(j_{x}\boldsymbol{-}\upsilon \rho \right) \tag{24a}\\ j'_{y} & = j_{y} \tag{24b}\\ j'_{z} & = j_{z} \tag{24c} \end{align} then equations (23a),(23b) and (23c) is a proof that the primed vectors $\:\mathbf{E'},\mathbf{B'},\mathbf{j'}\:$ defined by (16), (17) and (24) satisfy the primed version of Maxwell equation (01b) \begin{equation} \boldsymbol{\nabla'} \boldsymbol{\times} \mathbf{B'} = \mu_{0}\mathbf{j'}+\frac{1}{c^{2}}\frac{\partial \mathbf{E'}}{\partial t'} \tag{25} \end{equation}

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  • $\begingroup$ When you go from 11 to 12 and 16 to 17, you are defining new quantities $\mathbf{E}',\mathbf{B}',\rho',\mathbf j'$ in such a way that they obey the Maxwell equation in the primed frame, but it is not clear that those are actual electric and magnetic fields in the primed frame. The answer should say why that is the correct transformation; I think that Maxwell's equations by themselves aren't enough to prove that. $\endgroup$ – Ján Lalinský Jun 17 '18 at 22:30
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    $\begingroup$ @Ján Lalinský : (1) Many thanks for your valuable comments below my answers. For me, these are good chances (motivations) to reexamine the relevant subjects and check what may be I had learned by the wrong way in the past. (2) My answer is nothing more than what Einstein did in his famous 1905 paper On the Electrodynamics of moving bodies. I skip the steps which demand to have a factor $\:k(\upsilon)\:$ in front of the right hand sides of (12),(13),(17) and (20). This factor by symmetry arguments is $k(\upsilon)=1$ [$\:\psi(\upsilon)=1\:$ in the paper]. $\endgroup$ – Frobenius Jun 18 '18 at 7:16
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    $\begingroup$ @Ján Lalinský : As concerning the '...use relativistic transformation formulae for the fields $\mathbf{E}$,$\mathbf{B}$, which follow from the theory of relativity and the Lorentz force formula' I think I have done the inverse one here : Are magnetic fields just modified relativistic electric fields? but to find the transformation law of the Lorentz force, see equation (11) there in. $\endgroup$ – Frobenius Jun 18 '18 at 8:08
  • $\begingroup$ One must use relativistic transformation formulae for 3-force (which follow from the special theory of relativity) and the Lorentz force formula, which defines fields $\mathbf E,\mathbf B$ in all frames. $\endgroup$ – Ján Lalinský Jun 18 '18 at 9:05
  • $\begingroup$ I meant to say that one must use transf. formulae for 3-force, not for fields $\mathbf E,\mathbf B$ (which is the sought result), but I think you got the idea. I think Einstein's derivation seems OK but it is strange that it is possible to derive how quantities $\mathbf E,\mathbf B$ transform without using their definition (Lorentz force formula) in both frames. $\endgroup$ – Ján Lalinský Jun 18 '18 at 9:36
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Rather than approaching from the fields ($F^{\mu\nu}$, $A^\mu$, etc.), a more direct approach, starting from matter, can be suggested.

In fact, the charge density $\rho (t, x^i )$ and the current density $J^i (t, x^i )$ for a point charge $q$ a charge moving with velocity $V^i (t) = \frac{d}{dt} w^i (t) $ is

$$ \rho (t, x^i) = q \delta^{(3)}(x^i - w^i(t)) $$ $$ J^i (t,x^i) = q V^i (t) \delta^{(3)}(x^i - w^i(t)) $$

and we can combine these and write as

$$ J^\mu (t, x^i) = q \left( 1, V^i (t) \right) \delta^{(3)}(x^i - w^i(t)), $$

where $\mu = 0, \ 1, \ 2, \ 3$ and $ i = 1, \ 2,\ 3 $.

Now, please observe that, if we reparametrize the particle's space-time position by the proper time ($t = t(\tau) := w^0 (\tau)$ and $w^i = w^i(\tau)$),

$$ J^\mu (x^\mu) = q \int d \tau \ u^\mu (\tau) \delta^{(4)}(x^\mu - w^\mu(\tau)) \cdots (\ast)$$

$$ \left( \delta^{(4)}(x^\mu - w^\mu(\tau)) = \delta(t - w^0(\tau) ) \delta^{(3)}(x^i - w^i(\tau)) \right),$$

where $\tau$ and $u^\mu = \frac{d}{d\tau} w^\mu = \frac{dt}{d\tau} ( 1, V^i )$ are the proper time and 4-velocity of the point charge, respectively.

(This equation is introduced not only in relativity texts but also in books regarding electromagnetism (Jackson Ch.12, for example).)

Please notice that from this expression, we can obviously see that $J^\mu$ transforms like $u^\mu$ which is a contravariant quantity ($u^\mu = dx^\mu/d\tau$ and $dx^\mu$ is by definition contravariant and $d\tau$ is Lorentz invariant). This can be the answer of your question. Physically (or geometrically), equation $(\ast)$ provides a picture of "the distribution of charge and current for a charged particle as a superposition of charges that momentarily flash into existence and then flash out of existence." (Misner, Thorne, Wheeler: 120-121) 4-current is just a flow of "electromagnetic existence," so it is plausible that $J^\mu$ follows the transformation properties of $u^\mu$.

For continuous distributions, we just drop the integral and the delta function in equation $(\ast)$ and "continuous-ize" it:

$$ J^\mu = \varrho u^\mu ,$$

where $\varrho$ is the Lorentz invariant charge density ("continuous-ized $q$")-the charge density seen as in the (momentarily co-moving) rest frame.

So evidently, $J^\mu$ is just a multiple of $u^\mu$, which is a contravariant quantity. Thus, $J^\mu$ is contravariant, i.e. "transforms like $dx^\mu$ under Lorentz transformation."

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  • $\begingroup$ > "and "continuous-ize" it" ... the formula $j^\mu = \rho_0 u^\mu$ is valid only for charged fluid. Realistic distributions like current in a wire are often due to negative and positive particles that have different velocity. Your derivation can be altered to take this into account, just "continuous-ize" each group of particles moving as a whole separately and add them up. $\endgroup$ – Ján Lalinský Jun 17 '18 at 22:20

protected by Qmechanic Jun 18 '18 at 19:40

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