I'm working through Einstein's original 1905 paper*, and I'm having trouble with the section on the transformation of Maxwell's equations from rest to moving frame.

The paper proceeds as follows:

Let the Maxwell-Hertz equations for empty space hold good for the stationary system K, so that we have

IMAGE: Untransformed Maxwell equations

where (X, Y, Z) denotes the vector of the electric force, and (L, M, N) that of the magnetic force.

If we apply to these equations the transformation developed in § 3, by referring the electromagnetic processes to the system of co-ordinates there introduced, moving with the velocity v, we obtain the equations

IMAGE: Transformed Maxwell equations

I do not understand just where this second set of equations comes from.

The only derivations of this transformation that I have been able to find involve either potentials, four-vectors, or both. However since four-vectors had not been invented in 1905, and because the statement of the transformation is so blunt, it seems that Einstein is using or appealing to a simpler method for finding the transformation.

So my question is: What is the simplest derivation of the transformation rules for Maxwell's equations in special relativity? Is Einstein appealing to a pre-existing Lorentz method here, or is there a trick to accomplish all of this quickly? Or does the text here simply belie the true work involved in the derivation?

  • I cannot give a proper link to the paper owing to link restrictions, but it's at hxxp://www.fourmilab.ch/etexts/einstein/specrel/www/#SECTION21
  • 2
    $\begingroup$ This for sure won't answer your question, but it is rather the transformation rules of Maxwell's equations what motivated special relativity. $\endgroup$
    – c.p.
    Commented Jan 27, 2013 at 23:43
  • $\begingroup$ Although Jackson's E&M book tends to be off putting for some people due to its complexity/difficulty, it does have a very good and physically intuitive discussion of relativity. If all else fails, Jackson provides lots of good references on the topic as well (especially some historically significant works, which are interesting to read). $\endgroup$ Commented Aug 10, 2020 at 13:05
  • $\begingroup$ You wrote ... since four-vectors had not been invented in 1905..., but by that time the mathematics of general vector spaces (including 4-dimensional vector spaces) was in fact already well known for at least a few decades. $\endgroup$
    – Lee Mosher
    Commented Aug 27, 2023 at 22:38

4 Answers 4


There exists a remarkably simple explanation for how electromagnetic phenomena must be transformed in a moving frame if: (1) the physics of that frame is to remain invariant; and (2) the speed of light c must remain invariant for both the at-rest observer and within the moving frame. These are Einstein's original two SR postulates, of course. (There is actually a third assumption needed about dimensional scaling orthogonal to the direction of motion. Lorentz discussed that issue explicitly, but Einstein seems to assumed as a given.)

The technique is to use light pendulums to define proper time and length in both frames, then show geometrically how this affects electromagnetics phenomena. I've only used it myself to bring out the energy relations, which is part of Einstein's subsequent very short paper that led to $E=mc^2$ (what a delightfully strange technique that paper uses!). All of the Maxwell transformations are necessarily implicit those energy transformations of moving light pendulums. They have to be, else SR would not work.

I have some light pendulum diagrams on hand that I had intended for use in time dilation discussions. I'll append them to this answer shortly, and explain at least briefly how they can be used to re-interpret electromagnetics and Maxwell's equations. I don't have time for a full treatment (flying tomorrow), but the light pendulum perspective is both helpful and deeply linked to Einstein's very first (and also much later) explanations of why special relativity is an unavoidable consequence of holding c invariant in all frames of motion.

So, the promised Graphical Overkill ensues below. I don't even have time tonight to explain the graphics, but I to promise I'll get back to them.

The initial foray into EM implications is, alas, the very last bullet in the very last slide. But it does show how transformations of both wavelength and energy are inherent and unavoidable within the curious constraints of having to preserve the physics and speed of light in both the frame doing the observing, and in the frame being observed.

3D Light Pendulum, Outbound Phase

3D Light Pendulum, Inbound Phase

2D Light Pendulum

1D Light Pendulum

1D Rho Clock

A Human-Scale Rho Clock

This final graphic is at the very heart of how Einstein first came up with the theory of special relativity. He postulated two things: (1) that the speed of light remains invariant in all unaccelerated frames of reference, and (2) that all physics, including not just mechanics (Galilean relativity) but also electromagnetics, must remain unchanged regardless of frame.

Oh my... what those two postulates do in combination!

Light pendulums give a nicely succinct way of exploring all of those implications using a single device. The rho clocks I use here are simply light pendulums with enough design constraints and frame-specific labels attached to allow those implications to be explored.

So, take a look at this final figure:

When Light Pendulums Move

Notice that the sides of the gray rectangles represent light paths drawn out in spacetime. So, if the speed of light is invariant, guess what? An object that is moving cannot have the shortest possible inbound an outbound light paths, because the light has to constantly pace the object to keep up with it. If the object moves very, very fast, the time it takes for the outbound light to catch up with its leading edge can become very long indeed, as viewed by an observer the "rest" frame labeled 0 ($\phi_0$).

How long? Well, look at the figure: It's just the height of the scaled gray box, which is what the rho clock ($\rho_1$) for frame 1 ($\phi_1$) looks like when viewed from the rest frame $\phi_0$. For shorthand, I call that situation $\rho_{1:>0}$, where "1:>0" just means that frame 1 is being observed by frame 0 (the ":>" is two eyes looking left). And yes, the label on that height is the traditional $\gamma$ of special relativity. The idea of $\gamma$ just emerges a lot more naturally (and a lot more geometrically) in rho diagrams.

As for electromagnetics, recall that both frames must see their physics unchanged. But notice how the pulses of blue light on the left side of the gray rectangle get crowded together to make that true. That doesn't happen just for pulse spacing, it happens for all of electromagnetic theory. So, for example, light traveling along the left branch of the frame 1 rho rectangle ("rho" actually stands for rectangle, not relativity) must increase in frequency to keep the internal physics of frame 1 rho clock unchanged (shorthand: $\rho_{1:>1}$ must remain invariant).

That has huge energy implications over in frame 0, however, since the faster the object travels, the more energetic that side of the rho clock rectangle becomes. If you combine that with some very unusual arguments from Einstein in his second special relativity paper (I like to call it his "asymptotic tautology argument"), you wind up eventually with the famous equation $E=mc^2$. (Side comment: That equation is more famous, but $E^2 = (pc)^2 + (mc^2)^2$ is a lot more useful; just ask any particle physicist.)

The nice thing about Maxwell's equations is that they already accommodated and allowed such unusual forms of scaling long before Einstein and special relativity came around. That is one of the reason why you will see almost breathless praise for Maxwell from Einstein and other physicists involved in the early days of special relativity. The new theory helped them appreciate in new ways just how deep Maxwell's insights had been.

  • $\begingroup$ I'm afraid this doesn't help answer my original question. $\endgroup$ Commented Jan 28, 2013 at 10:35
  • $\begingroup$ Obsessive: Guilty as charged, the absolute most I could get with this quick drop (leaving now) was show the underlying argument and framework that Einstein used to derive the transformations, plus some quick feel for how they work. Vijay, thanks! $\endgroup$ Commented Jan 28, 2013 at 14:58
  • 2
    $\begingroup$ @Terry Bollinger This quote from Mark Twain seems pertinent here: “I didn't have time to write a short letter, so I wrote a long one instead.” $\endgroup$
    – D. Halsey
    Commented Aug 9, 2020 at 21:31
  • $\begingroup$ D. Halsey, thanks for reminding me of this entry. I was trying to explain a point about special relativity just last week, and was wishing I had some online explanation of my rho clock explanation. I had totally forgotten posting my notes on it here. Now I have the link again -- thanks! $\endgroup$ Commented Aug 10, 2020 at 2:08

I've also been going through (the English translation of) Einstein's original 1905 paper "On the Electrodynamics of Moving Bodies". In the context of his paper, the derivation is actually quite straightforward. You only need Maxwell's equations, the Lorentz transformation (which Einstein derived in the preceding segment of the paper), and the chain rule of partial derivatives for composite multivariable functions.

In his paper, Einstein considered two frames of reference K and $k$. K is the (relatively) stationary frame, while $k$ is moving relative to K at velocity $v$ in linear uniform motion. Co-ordinates in K are expressed as ($x$, $y$, $z$, $t$), and in $k$ as ($\xi$, $\eta$, $\zeta$, $\tau$). For simplicity, Einstein assumed that at $t = 0, \tau = 0$, and the origins of both K and $k$ overlapped at $t = \tau = 0$. To further simplify things, Einstein assumed that $k$ and K had parallel axes (and assumed that the axes were in the right-hand configuration), and $k$ is moving along the X axis of K.

From these above assumptions, Einstein derived the Lorentz transformation to convert the co-ordinates ($x$, $y$, $z$, $t$) to ($\xi$, $\eta$, $\zeta$, $\tau$):

$$\textit{$\xi$} = \textit{$\beta$} (\textit{x} - \textit{vt})$$ $$\textit{$\eta$} = \textit{y}$$ $$\textit{$\zeta$} = \textit{z}$$ $$\textit{$\tau$} = \textit{$\beta$} (\textit{t} - \textit{vx}/ \textit{c}^2)$$


$$\textit{$\beta$} = 1 / \sqrt{1 - \textit{v}^2/\textit{c}^2}$$

From these equations we can easily derive the reverse transformation from ($\xi$, $\eta$, $\zeta$, $\tau$) to ($x$, $y$, $z$, $t$):

\begin{align*} x &= \beta (\xi + v \tau)\label{a}\tag{1}\\y &= \eta\label{b}\tag{2}\\z &= \zeta\label{c}\tag{3}\\t &= \beta (\tau + v \xi/ c^2)\label{d}\tag{4} \end{align*}

In the electrodyamical section of Einstein's paper, we now consider a (classical) electromagnetic field in empty space:

Let $\vec{\mathrm{E}}$($x$, $y$, $z$, $t$) be the vector function describing the electric field in frame K. Likewise, let $\vec{\mathrm{B}}$($x$, $y$, $z$, $t$) describe the magnetic field in frame K. $\vec{\mathrm{E}}$ = (X, Y, Z), where X, Y, Z are scalar functions describing the vector component in the x, y, z axes respectively. Similarly, $\vec{\mathrm{B}}$ = (L, M, N), where L, M, N are scalar functions of the component in the x, y, z axes respectively.

Maxwell's equations in free space are:

$$\nabla \cdot \vec{\mathrm{E}} = \frac{\textit{$\rho$}}{ \textit{$\epsilon$}_0}$$ $$\nabla \cdot \vec{\mathrm{B}} = 0$$ $$\nabla \times \vec{\mathrm{E}} = - \frac{\partial\vec{\mathrm{B}}}{\partial\textit{t}}$$ $$\nabla \times \vec{\mathrm{B}} = \textit{$\mu$}_0 (\vec{\mathrm{J}} + \textit{$\epsilon$}_0 \frac{\partial\vec{\mathrm{E}}}{\partial\textit{t}})$$

Where $\rho$ is the electric charge density, $\vec{\mathrm{J}}$ is the electric current density, $\mu_0$ is the permeability of free space, and $\epsilon_0$ is the permittivity of free space.

Taking advantage of the fact that in empty space, $\rho = 0$ and $\vec{\mathrm{J}} = 0$, and switching over to Guassian Units (which is what I suspect Einstein did), we arrive at the following equations: \begin{align*} \nabla \cdot \vec{\mathrm{E}} &= 0\label{e}\tag{5}\\\nabla \cdot \vec{\mathrm{B}} &= 0\label{f}\tag{6}\\\nabla \times \vec{\mathrm{E}} &= - \frac{1}{c}\frac{\partial\vec{\mathrm{B}}}{\partial\ t}\label{g}\tag{7}\\ \nabla \times \vec{\mathrm{B}} &= \frac{1}{c} \frac{\partial\vec{\mathrm{E}}}{\partial t}\label{h}\tag{8} \end{align*}

Apply equation (\ref{g}) to $\vec{\mathrm{E}}$($x$, $y$, $z$, $t$) and $\vec{\mathrm{B}}$($x$, $y$, $z$, $t$), and we get:

\begin{align*} \frac{1}{c}\frac{\partial \mathrm{X}}{\partial t} &= \frac{\partial \mathrm{N}}{\partial y} - \frac{\partial \mathrm{M}}{\partial z}\label{xnm}\tag{9}\\\frac{1}{c}\frac{\partial \mathrm{Y}}{\partial t} &= \frac{\partial \mathrm{L}}{\partial z} - \frac{\partial \mathrm{N}}{\partial x}\label{yln}\tag{10}\\\frac{1}{c}\frac{\partial \mathrm{Z}}{\partial t} &= \frac{\partial \mathrm{M}}{\partial z} - \frac{\partial \mathrm{L}}{\partial y}\label{zml}\tag{11} \end{align*}

Similarly, apply equation (\ref{h}) to $\vec{\mathrm{E}}$($x$, $y$, $z$, $t$) and $\vec{\mathrm{B}}$($x$, $y$, $z$, $t$), and we get:

\begin{align*} \frac{1}{c}\frac{\partial \mathrm{L}}{\partial t} &= \frac{\partial \mathrm{Y}}{\partial z} - \frac{\partial \mathrm{Z}}{\partial y}\label{lyz}\tag{12}\\\frac{1}{c}\frac{\partial \mathrm{M}}{\partial t} &= \frac{\partial \mathrm{Z}}{\partial x} - \frac{\partial \mathrm{X}}{\partial z}\label{mzx}\tag{13}\\\frac{1}{c}\frac{\partial \mathrm{N}}{\partial t} &= \frac{\partial \mathrm{X}}{\partial y} - \frac{\partial \mathrm{Y}}{\partial z}\label{nxy}\tag{14} \end{align*}

These relations hold in frame K. (These equations also assume that the x, y, z axes of frame K are in the right-hand configuration)

Now, we consider how the vector functions $\vec{\mathrm{E}}$($x$, $y$, $z$, $t$) and $\vec{\mathrm{B}}$($x$, $y$, $z$, $t$) are expressed in frame $k$ using the co-ordinates ($\xi$, $\eta$, $\zeta$, $\tau$). This can be accomplished by substituting $x$, $y$, $z$, $t$ with $\xi$, $\eta$, $\zeta$, $\tau$ using equations (\ref{a}) - (\ref{d}).

Thus we get: \begin{align*} \vec{\mathrm{E}}(x, y, z, t) &= \vec{\mathrm{E}}(\beta (\xi + v\tau), \eta, \zeta, \beta (\tau + v\xi/ c^2))\label{Etrans}\tag{15}\\\vec{\mathrm{B}}(x, y, z, t) &= \vec{\mathrm{B}}(\beta (\xi + v\tau), \eta, \zeta, \beta (\tau + v\xi/ c^2))\label{Btrans}\tag{16} \end{align*}

Using the chain rule of composite multivariable functions: $$ \frac{\partial}{\partial x}f(u(x, t), v(x, t)) = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x}$$

We apply it to any general continuous function $$\mathrm{F}(x, y, z, t) = \mathrm{F}(\beta (\xi + v\tau), \eta, \zeta, \beta (\tau + v\xi/c^2))$$

Thus $$\frac{\partial \mathrm{F}}{\partial\xi} = \frac{\partial\mathrm{F}}{\partial x} \frac{\partial x}{\partial\xi} + \frac{\partial\mathrm{F}}{\partial y} \frac{\partial y}{\partial\xi} + \frac{\partial\mathrm{F}}{\partial z} \frac{\partial z}{\partial\xi} + \frac{\partial\mathrm{F}}{\partial t} \frac{\partial t}{\partial\xi}$$ According to equations (\ref{a}) - (\ref{d}), $\frac{\partial x}{\partial \xi} = \beta$, $\frac{\partial y}{\partial\xi} = 0$, $\frac{\partial z}{\partial\xi} = 0$, $\frac{\partial t}{\partial\xi} = \beta \frac{v}{c^2}$, we get:

$$\frac{\partial\mathrm{F}}{\partial\xi} = \beta (\frac{\partial\mathrm{F}}{\partial x} + \frac{v}{c^2} \frac{\partial\mathrm{F}}{\partial t})$$

This relation ship holds for any general continuous function, thus: \begin{equation} \frac{\partial}{\partial\xi} = \beta (\frac{\partial}{\partial x} + \frac{v}{c^2} \frac{\partial}{\partial t})\label{i}\tag{17} \end{equation} Similarly: \begin{align*} \frac{\partial}{\partial\eta} &= \frac{\partial}{\partial y}\label{j}\tag{18}\\\frac{\partial}{\partial\zeta} &= \frac{\partial}{\partial z}\label{k}\tag{19}\\\frac{\partial}{\partial\tau} &= \beta (\frac{\partial}{\partial t} + v \frac{\partial}{\partial x})\label{l}\tag{20} \end{align*}

From equation (\ref{l}), we get \begin{equation} \frac{\partial}{\partial t} = \frac{1}{\beta} \frac{\partial}{\partial\tau} - v \frac{\partial}{\partial x}\label{m}\tag{21} \end{equation}

Plug equation (\ref{m}) into equation (\ref{i}), we get \begin{equation} \frac{\partial}{\partial x} = \beta (\frac{\partial}{\partial\xi} - \frac{v}{c^2} \frac{\partial}{\partial\tau})\label{n}\tag{22} \end{equation}

Plug equation (\ref{n}) back into equation (\ref{m}), we get \begin{equation} \frac{\partial}{\partial t} = \beta (\frac{\partial}{\partial\tau} - v \frac{\partial}{\partial\xi})\label{o}\tag{23} \end{equation}

Now we have all the tools we need to derive the transformation of $\vec{\mathrm{E}}$($x$, $y$, $z$, $t$) and $\vec{\mathrm{B}}$($x$, $y$, $z$, $t$) from frame K to frame $k$:

Apply equation (\ref{m}) to equation (\ref{xnm}): $$\frac{1}{c} \frac{1}{\beta} \frac{\partial\mathrm{X}}{\partial\tau} - \frac{v}{c} \frac{\partial\mathrm{X}}{\partial x} = \frac{\partial \mathrm{N}}{\partial\eta} - \frac{\partial \mathrm{M}}{\partial\zeta}$$

From equation (\ref{e}): $$ \frac{\partial\mathrm{X}}{\partial x} + \frac{\partial\mathrm{Y}}{\partial y} + \frac{\partial\mathrm{Z}}{\partial z} = 0$$ Thus $$ -\frac{\partial\mathrm{X}}{\partial x} = \frac{\partial\mathrm{Y}}{\partial\eta} + \frac{\partial\mathrm{Z}}{\partial\zeta}$$

Then it follows that

$$\frac{1}{c} \frac{1}{\beta} \frac{\partial\mathrm{X}}{\partial\tau} + \frac{v}{c} \frac{\partial\mathrm{Y}}{\partial\eta} + \frac{v}{c} \frac{\partial\mathrm{Z}}{\partial\zeta} = \frac{\partial \mathrm{N}}{\partial\eta} - \frac{\partial \mathrm{M}}{\partial\zeta}$$ Rearrange the terms, and we arrive at \begin{equation} \frac{1}{c}\frac{\partial\mathrm{X}}{\partial\tau} = \frac{\partial}{\partial\eta}[\beta(\mathrm{N} - \frac{v}{c}\mathrm{Y})] - \frac{\partial}{\partial\zeta}[\beta(\mathrm{M} + \frac{v}{c}\mathrm{Z})]\tag{24} \end{equation}

Apply equation (\ref{n}) and (\ref{o}) to equation (\ref{yln}): $$\frac{1}{c}\beta(\frac{\partial\mathrm{Y}}{\partial\tau} - v \frac{\partial\mathrm{Y}}{\partial\xi}) = \frac{\partial \mathrm{L}}{\partial\zeta} - \beta (\frac{\partial\mathrm{N}}{\partial\xi} - \frac{v}{c^2} \frac{\partial\mathrm{N}}{\partial\tau})$$ Rearrange the terms, and we arrive at \begin{equation} \frac{1}{c}\frac{\partial}{\partial\tau}[\beta (\mathrm{Y} - \frac{v}{c}\mathrm{N})] = \frac{\partial \mathrm{L}}{\partial\zeta} - \frac{\partial}{\partial\xi}[\beta (\mathrm{N} - \frac{v}{c}\mathrm{Y})]\tag{25} \end{equation}

Likewise, apply equation (\ref{n}) and (\ref{o}) to equation (\ref{zml}), and rearrange the terms, we obtain: \begin{equation} \frac{1}{c}\frac{\partial}{\partial\tau}[\beta (\mathrm{Z} + \frac{v}{c}\mathrm{M})] = \frac{\partial}{\partial\xi}[\beta (\mathrm{M} + \frac{v}{c}\mathrm{Z})] - \frac{\partial \mathrm{L}}{\partial\eta}\tag{26} \end{equation}

We then go through the exact same process, also utilizing the fact that from equation (\ref{f}) we have the relation: $$ \frac{\partial\mathrm{L}}{\partial x} + \frac{\partial\mathrm{M}}{\partial y} + \frac{\partial\mathrm{N}}{\partial z} = 0$$

Thus we derive \begin{align*} \frac{1}{c}\frac{\partial\mathrm{L}}{\partial\tau} = \frac{\partial}{\partial\zeta}[\beta(\mathrm{Y} - \frac{v}{c}\mathrm{N})] - \frac{\partial}{\partial\eta}[\beta(\mathrm{Z} + \frac{v}{c}\mathrm{M})]\tag{27}\\\frac{1}{c}\frac{\partial}{\partial\tau}[\beta (\mathrm{M} + \frac{v}{c}\mathrm{Z})] = \frac{\partial}{\partial\xi}[\beta (\mathrm{Z} + \frac{v}{c}\mathrm{M})] - \frac{\partial \mathrm{X}}{\partial\zeta}\tag{28}\\\frac{1}{c}\frac{\partial}{\partial\tau}[\beta (\mathrm{N} - \frac{v}{c}\mathrm{Y})] = \frac{\partial \mathrm{X}}{\partial\eta} - \frac{\partial}{\partial\xi}[\beta (\mathrm{Y} - \frac{v}{c}\mathrm{N})]\tag{29} \end{align*} where

$$\beta = 1 / \sqrt{1 - v^2/c^2}$$


I chanced to land here while seeking historical papers...

Nice graphics from Terry Billinger but a plain answer is far duller and just a hard slog. No doubt it was the work of Einstein's friend Besso and so no trace of his effort is left in the published paper as you noted. However this is pure speculation on my part ...

OK I'll use more modern notations than Einstein but no higher mathematics...

Let $X$ and $Y$ be the two Cartesian frames in uniform motion with respect to each other. They share $O$ as common origin of space-time and their Cartesian frames coincide there and then. Their atomic clocks are synchronised while still in the state of rest with respect to each other before moving away and then they use in their respective frames the same definition for the unit of length: for each frame, the proper space crossed by light in vacuum in the proper unit time; thus c=1 in both frames. Their relative motion is along the common axis: $Ox_1 = Oy_1$ with velocity $v = \tanh \theta$, $\theta$ is the additive characteristic of the transformation. With these notations the transformation between $X$ and $Y$ is $$y_0 = x_0 \cosh \theta - x_1 \sinh \theta,$$ $$y_1= -x_0 \sinh \theta + x_1 \cosh \theta,$$ $$ y_2 = x_2 \quad \text{and} \quad y_3=x_3.$$ If you still prefer $c \neq 1$ and the mess of square roots, $\beta$, $\gamma$, etc..., the conversion back to these horrors is trivial.

Now the conversion of the partial differential operators between these frames is easily obtained from the given transformation of coordinates. I use $\partial$ to denote them (with the appropriate index) in the $X$ frame and $\partial'$ for those in the $Y$ frame: $$\partial_0 = \partial'_0 \cosh \theta - \partial'_1 \sinh \theta,$$ $$\partial_1= -\partial'_0 \sinh \theta + \partial'_1 \cosh \theta,$$ $$\partial_2 = \partial'_2 \quad \text{and} \quad y_3=\partial'_3.$$

If you blindly substitute these differential operators in Maxwell's equations written in the $X$ frame $$ \partial_0 H + \nabla \times E = 0,$$ $$ \partial_0 E - \nabla \times H = 0,$$ $$ \nabla \cdot H = \nabla \cdot E =0.$$ you end up with a real mess of 8 equations (2 divergences and 6 components for the two curls).

However a mere look at two pairs of equations, each made of a divergence and its matching component of the curl in the direction of motion, leaves no degree of freedom apart from a kind of scaling function that Einstein sets to one for reason of symmetry (reciprocity of the transformation).

Let's start with the matching pair: $$\partial_0 E_1 = \partial_2 H_3 - \partial_3 H_2,$$ $$\partial_1 E_1 + \partial_2 E_2 + \partial_3 E_3 =0.$$

Using the conversion of partial derivatives given above these two equations become: $$ \cosh \theta\,\partial'_0 E_1 - \sinh \theta\,\partial'_1 E_1 = \partial'_2 H_3 - \partial'_3 H_2,$$ $$-\sinh \theta\,\partial'_0 E_1 + \cosh \theta\,\partial'_1 E_1 = -\partial'_2 E_2 - \partial'_3 E_3.$$

Eliminating the term $\partial'_1 E_1$ between these two equations yields: $$\partial'_0 E_1 = \partial'_2(H_3 \cosh \theta - E_2 \sinh \theta) - \partial'_3(H_2 \cosh \theta + E_2 \sinh \theta),$$ which already solves half the problem (modulo some possible scaling functions) if one must end up with $$\partial'_0 E'_1 = \partial'_2 H'_3 - \partial'_3 H'_2.$$

Thus, neglecting these possible scalings: $$E'_1 = E_1,$$ $$H'_2 = H_2 \cosh \theta + E_3 \sinh \theta,$$ $$H'_3 = H_3 \cosh \theta - E_2 \sinh \theta.$$

The other matching pair $$\partial_0 H_1 = -\partial_2 E_3 + \partial_3 E_2,$$ $$\partial_1 H_1 + \partial_2 H_2 + \partial_3 H_3 =0,$$ similarly leads to the other half of the transformation: $$H'_1 = H_1,$$ $$E'_2 = E_2 \cosh \theta - H_3 \sinh \theta,$$ $$E'_3 = E_3 \cosh \theta + H_2 \sinh \theta.$$

But now one might well wonder if the 4 equations left behind (the components of the two curls transverse to the motion) are going to be consistent (the scaling factors are hardly in measure to help). Here a miracle happens and indeed after some tedious algebra the pedestrian proof is completed and left as exercise...

Of course there is no miracle as easily seen if tensor analysis or differential forms or Clifford algebra, etc... is used to represent Minkowskian space-time and also when it is realised that the electromagnetic field is not made of pairs of "vector" fields, the electric and magnetic fields, as thought at the time Einstein wrote his paper, but is a single bi-vector field written for instance as $$F= e_0 \wedge (E_1 e_1 + E_2 e_2 + E_3 e_3) + H_1 e_2 \wedge e_3 + H_2 e_3 \wedge e_1 + H_3 e_1 \wedge e_2$$ in the appropriate Clifford formalism.

My apologies for the tedious length of this answer which is so elementary that I am surprised it was not given here ages ago. Meanwhile you might have figured it out yourself I hope. My apologies also for some possible typos and other mistakes, hopefully none that cannot be easily corrected by the reader if any.

René Grognard

e-mail: [email protected]


For the following, I will use the vectors $𝐞 = (X, Y, Z)$, $𝐦 = (L, M, N)$, respectively for the "electric" and "magnetic" force vectors that Einstein referred to only in component form in his paper

Zur Elektrodynamik bewegter Körper; von A. Einstein.

as well as the vector $𝐫 = (x, y, z)$, the operator $∇ = (∂/∂x, ∂/∂y, ∂/∂z)$ and their respective transforms, $𝛒 = (ξ, η, ζ)$ and $𝞉 = (∂/∂ξ, ∂/∂η, ∂/∂ζ)$; and will denote the invariant speed (i.e. "light speed") referred to in his paper as $V$ by our contemporary $c$. His $β = 1/\sqrt{1 - (v/c)^2}$ is referred to in contemporary literature as $γ$, but I'll just leave it as is.

Your question is that when he turned this $$\frac 1 c \frac{∂𝐞}{∂t} = ∇×𝐦, \hspace 1em \frac 1 c \frac{∂𝐦}{∂t} = -∇×𝐞$$ into this $$\frac 1 c \frac{∂𝐞̄}{∂τ} = 𝞉×𝐦̄, \hspace 1em \frac 1 c \frac{∂𝐦̄}{∂τ} = -𝞉×𝐞̄$$ under the coordinate transform $$𝛒 = \left(β\left(x - v t\right), y, z\right), \hspace 1em τ = β\left(t - \frac v {c^2} x\right),$$ how did he manage to get these expressions: $$𝐞̄ = \left(X, β\left(Y - \frac v c N\right), β\left(Z + \frac v c M\right)\right), 𝐦̄ = \left(L, β\left(M + \frac v c Z\right), β\left(N - \frac v c Y\right)\right)$$ after carrying out the transformation. How do you derive the second equations from the first?

The answer is: you can't exactly, because you also need these equations: $$∇·𝐞 = 0, \hspace 1em ∇·𝐦 = 0,$$ which he left out (though he hinted at it by saying that he was dealing with "free fields"), and (more importantly) you also need to show that they transform $$𝞉·𝐞̄ = 0, \hspace 1em 𝞉·𝐦̄ = 0,$$ for consistency, which he failed to do. It was a gap.

In fact, you'll see no explicit reference to either $∇·𝐦 = 0$ or $𝞉·𝐦̄ = 0$ in the paper at all. He never actually said or proved anywhere in the paper that $∇·𝐦 = 0$ transforms into $𝞉·𝐦̄ = 0$. The analysis is incomplete.

In component form, the transformations on the coordinates and operators are $$\begin{align} 𝛒 &= \left(β\left(x - v t\right), y, z\right), & τ &= β\left(t - \frac v {c^2} x\right), \\ 𝞉 &= \left(β\left(\frac∂{∂x} + \frac v {c^2} \frac∂{∂t}\right), \frac∂{∂y}, \frac∂{∂z}\right), & \frac∂{∂τ} &= β\left(\frac∂{∂t} + v \frac∂{∂x}\right) \end{align}.$$

You can derive the transformed set of equations from the original set by directly substituting in the transformed operators. But to do so, you also need to combine the extra two equations with the $x$ components of the cited equations to get the $ξ$ components of the transformed equations, as well as the transforms of the extra equations. They mix together.

$$\begin{align} \frac 1 c \frac{∂X}{∂τ} &= \frac 1 c β\left(\frac{∂}{∂t} + v \frac{∂}{∂x} \right) X \\ &= β \frac 1 c \frac{∂X}{∂t} + β \frac v c \frac{∂X}{∂x} \\ &= β \left(\frac{∂N}{∂y} - \frac{∂M}{∂z}\right) + β \frac v c \left(-\frac{∂Y}{∂y} - \frac{∂Z}{∂z}\right) \\ &= \frac{∂\left(β\left(N - \frac v c Y\right)\right)}{∂η} - \frac{∂\left(β\left(M + \frac v c Z\right)\right)}{∂ζ}, \\ \frac {∂X}{∂ξ} &= β\left(\frac∂{∂x} + \frac v {c^2} \frac∂{∂t}\right) X \\ &= β \frac{∂X}{∂x} + β \frac v c \frac 1 c \frac{∂X}{∂t} \\ &= β \left(-\frac{∂Y}{∂y} - \frac{∂Z}{∂z}\right) + β \frac v c \left(\frac{∂N}{∂y} - \frac{∂M}{∂z}\right) \\ &= -\left(\frac{∂\left(β\left(Y - \frac v c N\right)\right)}{∂η} + \frac{∂\left(β\left(Z + \frac v c M\right)\right)}{∂ζ}\right). \end{align}$$ The equations involving $∂L/∂τ$ and $∂L/∂ξ$ follow similarly.


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