Read · Special relativity: from clock operations to electrodynamics

§6 · field equations and field components

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Letters

Show modern qualifications

Einstein's letters change 12 of this paper's 40 formulas. 28 stay in today's letters, with a note saying so.

A derivation

Transform derivatives before naming fields

How do the field equations change coordinates?

The 6 printed paragraphs this passage explains

§6, paragraph 1: Einstein assumes that the Maxwell-Hertz equations for empty space hold in the stationary system K, with X, Y, Z the electric force and L, M, N the magnetic force.
§6, paragraph 2: Applying the transformation of section 3, he rewrites these equations in the coordinates and time of the moving system k.
§6, paragraph 3: The principle of relativity requires that the same equations hold in k, for the electric and magnetic forces measured there by their effects on charges and magnets.
§6, paragraph 4: Since the two sets of equations for k must say the same thing, the forces measured in k equal combinations of those in K, up to one common factor ψ(v) that may depend on the speed.
§6, paragraph 5: Going to k and back, and symmetry, fix that factor at 1, which gives the transformation of the electric and magnetic forces. To say what it means, Einstein takes a unit charge at rest in one system or the other, and says the first three equations can be put into words in two ways.
§6, footnote 1: A footnote gives the symmetry argument: with only one magnetic component present, reversing the direction of motion must reverse the sign of the transformed electric force without changing its size.

\partial_{t} = \gamma\,\left(\partial_{t'} - v\,\partial_{x'}\right)

\partial_{t} = \beta\,\left(\partial_{\tau} - v\,\partial_{\xi}\right)

\partial_{x} = \gamma\,\left(\partial_{x'} - \frac{v\,\partial_{t'}}{c^{2}}\right)

\partial_{x} = \beta\,\left(\partial_{\xi} - \frac{v\,\partial_{\tau}}{V^{2}}\right)

Partial by partial t equals gamma times, partial by partial t prime minus v partial by partial x prime. Partial by partial x equals gamma times, partial by partial x prime minus v over c squared partial by partial t prime.

Take the simplest case, a wave travelling along $x$ with an electric component $E_y$ and a magnetic component $B_z$ . In SI units and empty space the Maxwell-Hertz equations reduce to the pair $\partial_t E_y = -c^2\,\partial_x B_z$ and $\partial_t B_z = -\partial_x E_y$ . Substitute the two derivative rules above into both equations; the next step needs the pair.

After the substitution, $E_y$ and $B_z$ appear only in the combinations $E_y - vB_z$ and $B_z - vE_y/c^2$ . Multiply each by $\gamma$ and call the results $E'_y$ and $B'_z$ . The pair then reads $\partial_{t'} E'_y = -c^2\,\partial_{x'} B'_z$ and $\partial_{t'} B'_z = -\partial_{x'} E'_y$ , the same form as before. The field-equations laboratory shows this grouping one term at a time.

The form alone would also allow every new field to carry one common factor that depends on $v$ , and the paper keeps such a factor at first. Transforming there and back must return the original fields, so the factor’s value at $v$ times its value at $-v$ must be 1. By symmetry the factor cannot depend on the direction of motion, so it is the same at $v$ and $-v$ , and the positive root makes it 1.

Open the foundation: Partial derivatives and held-fixed quantities

Show every step here: Transform derivatives before naming fields

Letters: the paper writes the electric force as (X, Y, Z) and the magnetic force as (L, M, N), in Gaussian units, with V for the speed of light and β for the modern γ; in k they are $X'$ to $N'$ . This explanation works in SI units, where the electric field is E, the magnetic field B, and a factor $c^2$ appears that Gaussian units hide. The primed coordinates $x'$ and $t'$ are the paper's ξ and τ.

The paper begins with the Maxwell–Hertz equations for empty space, valid in K. One of them, in its letters, is $\frac{1}{V}\frac{\partial X}{\partial t}=\frac{\partial N}{\partial y}-\frac{\partial M}{\partial z}$ , and there are six in all.
To write them in k's coordinates, every derivative must be rewritten, since $t'$ and $x'$ each depend on both t and x: $t'=\gamma(t-vx/c^2)$ and $x'=\gamma(x-vt)$ .
The chain rule: a change in t changes $t'$ by γ and $x'$ by −γv, so $\partial_t=\gamma\partial_{t'}-\gamma v\partial_{x'}$ . A change in x changes $t'$ by $-\gamma v/c^2$ and $x'$ by γ, so $\partial_x=\gamma\partial_{x'}-\gamma(v/c^2)\partial_{t'}$ . Those are the two rules below.

\partial_{t} = \gamma\,\left(\partial_{t'} - v\,\partial_{x'}\right)

\partial_{t} = \beta\,\left(\partial_{\tau} - v\,\partial_{\xi}\right)

\partial_{x} = \gamma\,\left(\partial_{x'} - \frac{v\,\partial_{t'}}{c^{2}}\right)

\partial_{x} = \beta\,\left(\partial_{\xi} - \frac{v\,\partial_{\tau}}{V^{2}}\right)

Take the simplest case, a wave along x with an electric component $E_y$ and a magnetic component $B_z$ . In SI units and empty space the equations reduce to the pair $\partial_tE_y=-c^2\partial_xB_z$ and $\partial_tB_z=-\partial_xE_y$ .
Substitute into the first: $\gamma(\partial_{t'}-v\partial_{x'})E_y=-c^2\gamma(\partial_{x'}-v\partial_{t'}/c^2)B_z$ . Cancel γ and move the $\partial_{t'}$ terms to the left and the $\partial_{x'}$ terms to the right: $\partial_{t'}(E_y-vB_z)=-c^2\partial_{x'}(B_z-vE_y/c^2)$ .
Substitute into the second: $\gamma(\partial_{t'}-v\partial_{x'})B_z=-\gamma(\partial_{x'}-v\partial_{t'}/c^2)E_y$ . Collect the same way: $\partial_{t'}(B_z-vE_y/c^2)=-\partial_{x'}(E_y-vB_z)$ .
The fields now appear only in two combinations. Multiply both by γ and name them: $E'_y=\gamma(E_y-vB_z)$ and $B'_z=\gamma(B_z-vE_y/c^2)$ . The pair becomes $\partial_{t'}E'_y=-c^2\partial_{x'}B'_z$ and $\partial_{t'}B'_z=-\partial_{x'}E'_y$ , the same form as in K.
The paper does the same with all six equations and finds the combinations $\beta(N-(v/V)Y)$ and $\beta(M+(v/V)Z)$ in place of N and M. The principle of relativity requires that the Maxwell–Hertz equations hold in k too, for the forces the moving system defines by their pushes on electric and magnetic charges.
So the paper sets each new force equal to the matching combination times one common factor ψ(v), which may depend on v but not on the coordinates. The form of the equations cannot fix that factor.
Fixing it: transforming to k and back must give the original fields, so the factor's values at v and at −v multiply to 1. By symmetry, reversing v reverses the sign of a component such as $Y'$ without changing its size, so the factor is the same at v and −v; the positive root makes it 1 (p. 909).
Worked: a wave running along +x has $E_y=cB_z$ in SI. In a frame moving the same way at 0.6c, $E'_y=\gamma(E_y-vB_z)=1.25\,E_y(1-0.6)=0.5\,E_y$ : the field halves, as the frequency will in §7.
Keep two claims apart. That the equations keep their form after the substitution is algebra. That the new combinations are the fields the moving observer measures is the principle of relativity, applied to forces on charges at rest in k.
Changing from the paper's Gaussian units to SI is a change of units, done term by term, not a renaming of letters.

A tool for this step

Partial derivatives and held-fixed quantities

When a quantity depends on two things, what does its rate of change mean?

The concentration of particles in a tube depends on where you look, x, and when you look, t. Ask how fast the concentration changes and there are two different answers: how it changes as you move along the tube at one instant, and how it changes at one place as time passes.

A partial derivative picks one of them. ∂f/∂x is the rate of change along the tube with the time t held fixed, like comparing points in one photograph. ∂f/∂t is the rate of change in time at a fixed position x, like watching one spot under the microscope. The curly ∂ says that something else is being held fixed, and a subscript names it:

\left(\frac{\partial f}{\partial x}\right)_t \qquad \left(\frac{\partial f}{\partial t}\right)_x

The partial derivative of f with respect to x at fixed t, and the partial derivative of f with respect to t at fixed x.

Thermodynamics depends on the same care. How much a gas's volume changes as the pressure changes depends on whether its temperature T is held fixed, with heat flowing in and out, or its entropy S is held fixed, with the gas insulated. The two derivatives have different values, and ∂V/∂p written alone does not say which is meant:

\left(\frac{\partial V}{\partial p}\right)_T \quad \text{versus} \quad \left(\frac{\partial V}{\partial p}\right)_S

The partial derivative of V with respect to p at fixed temperature T, versus the same derivative at fixed entropy S.

In §3 of the light-quanta paper, radiation fills a fixed volume and φ is its entropy per unit volume and per unit interval of frequency, a function of the energy density ρ and of ν. Varying ρ with the frequency ν held fixed, Einstein finds that the derivative is 1/T:

\left(\frac{\partial \varphi}{\partial \rho}\right)_{\nu} = \frac{1}{T}

The partial derivative of phi with respect to rho, at fixed frequency nu, equals one over T.

In §6 of the relativity paper, Maxwell's equations are rewritten in the coordinates ξ, η, ζ, τ of a moving system. A derivative with respect to x taken with y, z and t held fixed then becomes a combination of a derivative with respect to ξ and one with respect to τ, because holding the resting time t fixed does not hold the moving time τ fixed.

Worked example: The same diffusion profile, differentiated two ways

Take the spreading profile of the Brownian paper's §4, written for a single particle:

f\left(x,\,t\right) = \frac{1}{\sqrt{4\,\pi\,D\,t}}\,\exp\left(-\frac{x^{2}}{4\,D\,t}\right)

f of x and t equals one over the square root of four pi D t, times e to the minus x squared over four D t.

Hold t fixed and differentiate with respect to x. Only the exponent depends on x, and the derivative of −x²/4Dt with respect to x is −x/2Dt:

\left(\frac{\partial f}{\partial x}\right)_{t} = -\frac{x}{2\,D\,t}\,f\left(x,\,t\right)

The partial derivative of f with respect to x at fixed t equals minus x over two D t, times f.

That is the slope of the profile in one snapshot: downhill to the right of the centre, uphill to the left. Now hold x fixed and differentiate with respect to t. Both the factor in front and the exponent depend on t:

\left(\frac{\partial f}{\partial t}\right)_{x} = \left(-\frac{1}{2\,t} + \frac{x^{2}}{4\,D\,t^{2}}\right)\,f\left(x,\,t\right)

The partial derivative of f with respect to t at fixed x equals minus one over two t, plus x squared over four D t squared, all times f.

That is what a probe fixed at one place records. Near the centre, where x² is less than 2Dt, the bracket is negative: the concentration falls as the cloud spreads away. Farther out it is positive: the concentration rises as the cloud arrives. Differentiate the first result once more with respect to x and multiply by D, and you get the second. That is the §4 equation ∂f/∂t = D ∂²f/∂x², a rate at one place equal to a curvature at one time.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: hold time fixed, or hold position fixed

In §4 of the Brownian paper, f(x, t) is the number of particles per unit volume at a place x along a tube, at a time t. “How fast does the concentration change?” then has two answers, and which one you mean depends on what you keep still: the moment, or the place.

Keep one of them still:

One moment: t fixed, ∂f/∂x

Freeze the moment and compare neighbouring places, as in a single photograph of the tube. ∂f/∂x says how steeply the concentration changes along it: particles per µm³ for each µm, which is particles per µm⁴.

Particles drift down that slope. The flux, the number crossing each square micrometre in a second, is J = −D ∂f/∂x: Fick’s first law.

One place: x fixed, ∂f/∂t

Keep your eye on one spot under the microscope and let time run. ∂f/∂t says how fast the count there rises or falls: particles per µm³ for each second.

Particles are neither made nor destroyed, so the count at a spot changes only by what flows in and out: ∂f/∂t = −∂J/∂x = D ∂²f/∂x². That is the diffusion equation of §4.

The same question in thermodynamics

Squeeze a gas and its pressure rises, but by how much depends on what else you keep fixed:

Three derivatives of pressure, and what each holds fixed
Held fixed	Derivative	What happens
Temperature T (isothermal)	(∂p / ∂V)_T	Heat flows in or out to keep the temperature constant.
Entropy S: no heat exchanged (adiabatic)	(∂p / ∂V)_S	The gas warms as it is squeezed, so its pressure climbs faster than at fixed temperature.
Volume V (isochoric)	(∂p / ∂T)_V	In a rigid vessel the pressure rises as the gas is heated.

What it shows, in words

∂f/∂x compares places at one moment; ∂f/∂t follows one place through time. They answer different questions and carry different units. In thermodynamics a subscript names what is held fixed, so (∂p/∂V)_T and (∂p/∂V)_S are different numbers for the same gas.

Where this lesson stops

A partial derivative is not defined until you have said which quantities are held fixed.

Explore the equations in this step

Read each operation, check its units and assumptions, or open the mathematical step behind it. These are modern teaching equations, not the equations as printed.

Explore the equation · Modern model notationEinstein's letters

The time derivative, rewritten

\partial_{t} = \gamma\,\left(\partial_{t'} - v\,\partial_{x'}\right)

\partial_{t} = \beta\,\left(\partial_{\tau} - v\,\partial_{\xi}\right)

The derivative with respect to is times the derivative with respect to minus times the derivative with respect to .

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

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Read the equation aloud in words

The time derivative is gamma times the primed time derivative minus v times the primed x derivative.

The rate of change with stationary time, written through the moving system's derivatives. Per second on both sides.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Partial derivatives of a field, taken in each system's own coordinates; the chain rule through the Lorentz transformation relates them.

The two together: The chain rule: a change in one stationary coordinate moves both moving coordinates. Read the prerequisite
The matching rate in the moving system: The same kind of derivative, in the moving system's coordinates. Read the prerequisite
Rate of change in the stationary system: How a field changes with this stationary-system coordinate, the others held fixed. Read the prerequisite
What it asserts: An identity between operators: it holds for any field it is applied to. Substituted into both of Maxwell's equations for a plane wave, it regroups them into the same equations for the primed fields. Read the prerequisite
Times gamma: The same factor as the transformation itself. Read the prerequisite
The other rate in the moving system: Because each moving coordinate depends on both x and t, the other derivative enters too. Read the prerequisite
Times v: How strongly the other coordinate enters. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Time, moving system: t′: the moving system's time. Read the prerequisite
Time, stationary system: t: the derivative is taken as t varies. Read the prerequisite
Position, moving system: x′: the moving system's position. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notationEinstein's letters

The x derivative, rewritten

\partial_{x} = \gamma\,\left(\partial_{x'} - \frac{v\,\partial_{t'}}{c^{2}}\right)

\partial_{x} = \beta\,\left(\partial_{\xi} - \frac{v\,\partial_{\tau}}{V^{2}}\right)

The derivative with respect to is times the derivative with respect to minus times the derivative with respect to over squared.

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The x derivative is gamma times the primed x derivative minus v over c squared times the primed time derivative.

The rate of change along x, written through the moving system's derivatives. Per metre on both sides.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Partial derivatives of a field, taken in each system's own coordinates; the chain rule through the Lorentz transformation relates them.

The two together: The chain rule: a change in one stationary coordinate moves both moving coordinates. Read the prerequisite
The matching rate in the moving system: The same kind of derivative, in the moving system's coordinates. Read the prerequisite
Rate of change in the stationary system: How a field changes with this stationary-system coordinate, the others held fixed. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
Over c squared: Keeps the units: per metre, like the left side. Read the prerequisite
What it asserts: An identity between operators: it holds for any field it is applied to. Substituted into both of Maxwell's equations for a plane wave, it regroups them into the same equations for the primed fields. Read the prerequisite
Times gamma: The same factor as the transformation itself. Read the prerequisite
The other rate in the moving system: Because each moving coordinate depends on both x and t, the other derivative enters too. Read the prerequisite
Times v: How strongly the other coordinate enters. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Position, moving system: x′: the moving system's position. Read the prerequisite
Position, stationary system: x: the derivative is taken as x varies. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Time, moving system: t′: the moving system's time. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Modern qualifications

Einstein writes the electric force as $X, Y, Z$ and the magnetic force as $L, M, N$ , in Gaussian units, with $V$ for the speed of light; his $\beta$ is the modern $\gamma$ , not $v/c$ . The field tensor that packs all six components together came later, with Minkowski in 1908. Where this edition uses it, it checks results; it is never a step in the 1905 argument.

Assumptions and limits: Transform derivatives before naming fields

Assumed here

The coordinates transform by the Lorentz map, and the fields obey Maxwell’s equations in empty space.
The example is in SI units, so factors of c appear explicitly.

What this does not establish

Einstein’s Gaussian equations become SI ones by a change of units, not by renaming symbols.
A common factor on all the new fields is fixed by a separate argument, going there and back and the symmetry of space; rearranging the equations does not fix it.

Earlier step: Light constraints leave a scale to determine The sideways step needs its own condition

Source context: German source · English · Interlinear gloss · Facsimile

A derivation

Electric and magnetic components mix together

What does a new inertial observer call electric?

The 5 printed paragraphs this passage explains

§6, paragraph 5: Going to k and back, and symmetry, fix that factor at 1, which gives the transformation of the electric and magnetic forces. To say what it means, Einstein takes a unit charge at rest in one system or the other, and says the first three equations can be put into words in two ways.
§6, paragraph 6: The old way of putting it: a unit charge moving in an electromagnetic field feels, besides the electric force, an electromotive force, which to a first approximation in v/V is the vector product of its velocity and the magnetic force divided by the speed of light V.
§6, paragraph 7: The new way of putting it: the force on a moving unit charge is the electric force at its position, found by transforming the field to a coordinate system in which the charge is at rest.
§6, paragraph 8: The same holds for magnetomotive forces. The electromotive force is only an auxiliary concept, needed because electric and magnetic forces do not exist independently of the state of motion of the coordinate system.
§6, paragraph 9: So the asymmetry noted in the introduction, between a moving magnet and a moving conductor, disappears, and the question of where the electromotive force is seated, as in unipolar machines, loses its point.

E'_{x} = E_{x}

B'_{x} = B_{x}

E'_{y} = \gamma\,\left(E_{y} - v\,B_{z}\right)

B'_{y} = \gamma\,\left(B_{y} + \frac{v\,E_{z}}{c^{2}}\right)

E'_{z} = \gamma\,\left(E_{z} + v\,B_{y}\right)

B'_{z} = \gamma\,\left(B_{z} - \frac{v\,E_{y}}{c^{2}}\right)

Shown in modern letters; Einstein's are in §6 of the German source face.

E x prime equals E x, and B x prime equals B x. E y prime equals gamma times, E y minus v B z. B y prime equals gamma times, B y plus v E z over c squared. E z prime equals gamma times, E z plus v B y. B z prime equals gamma times, B z minus v E y over c squared.

Take a region with no electric field and a magnetic field $B_z$ pointing along positive $z$ . An observer moving along positive $x$ measures $E'_y = -\gamma vB_z$ , an electric field pointing along negative $y$ . At $v = 0.6c$ , $\gamma = 1.25$ , so $E'_y = -0.75c\,B_z$ .

This is the magnet and conductor from the paper’s introduction, now with equations. A positive charge $q$ carried along at speed $v$ is at rest in the moving frame, so there the whole force on it is electric, of size $\gamma qvB_z$ . In the original frame the same charge feels a magnetic force of size $qvB_z$ . The sizes differ by $\gamma$ , which is how a force across the motion transforms, so both frames describe one push. The field-transformation laboratory places a test charge in each frame.

Open the foundation: Charges, fields and light

Show every step here: Electric and magnetic components mix together

A tool for this step

Vectors and their components

How can one arrow have different numbers in different coordinate systems?

A velocity, a force and an electric field each have a size and a direction. Drawn as an arrow, such a quantity is a vector. To write it down with numbers, choose axes and ask how far the arrow reaches along each one. Those numbers are its components, written $(v_x, v_y)$ in two dimensions or $(v_x, v_y, v_z)$ in three.

The components belong to the axes as much as to the arrow. A particle moving at 0.6c along x has components $(0.6c, 0)$ . Describe the same motion with axes turned so that the new x axis makes an angle θ with the old one, and the components become $(0.6c\cosθ,\ -0.6c\sinθ)$ . The numbers changed; the particle did not. Its speed, the arrow's length, is the same in both descriptions.

|\mathbf{v}| = \sqrt{v_x^{2} + v_y^{2}}

The length of v is the square root of v x squared plus v y squared.

Physics often splits a vector into two parts with respect to a direction that matters: the component along it and the part across it. For a boost along x, an electric field of 5 units pointing 3 along x and 4 along y has a parallel component of 3 and a perpendicular part of 4. The split matters in §6 of the relativity paper, where the component along the motion, which Einstein writes X, is the same in both frames, while the components across it, Y and Z, change.

Two different operations can both look like new numbers for a vector. Turning the axes, or moving the observer, changes the description and leaves the world as it was. Pushing the particle changes the world itself. The site's instruments keep the two apart: an observer change never starts a new experiment.

Worked example: One velocity in two sets of axes

A particle moves at 0.6c along x. In the first axes its components are $(0.6c, 0)$ .
Turn the axes so that $\cosθ = 0.8$ and $\sinθ = 0.6$ , an angle of about 36.9°.
The new components are $0.6c \times 0.8 = 0.48c$ along the new x axis and $-0.6c \times 0.6 = -0.36c$ along the new y axis.
The length is unchanged: $\sqrt{0.48^2 + 0.36^2} = \sqrt{0.2304 + 0.1296} = \sqrt{0.36} = 0.6$ , so the speed is 0.6c in both descriptions.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: one arrow, turned axes

The arrow is fixed: 5 units long, 3 along the original x axis and 4 along y. Turn the axes and read the arrow's components along the new ones.

Angle of the turned axes, in degrees

Type an angle and press Enter, or drag the slider from −90° to 90°. Try 53.13°, where the new x axis lies along the arrow.

In axes turned by 36.87°, the arrow is 4.8 along x′ and 1.4 along y′.

Its length is √(4.8² + 1.4²) = 5, the same 5 as in the original axes, where it is 3 along x and 4 along y.

What it shows, in words

Turning the axes changes the numbers that describe the arrow and leaves the arrow alone. At 36.87° the components are 4.8 and 1.4, at 53.13° they are 5 and 0, and at every angle the length is 5. The figure draws the original axes dashed, the turned axes solid, and dotted lines from the arrow's tip to its component along each turned axis.

Where this lesson stops

A vector is an arrow with a length and a direction. This lesson stops at describing it along different axes.

A tool for this step

The units of 1905

How do Einstein's numbers, printed in centimetres, grams, ergs and the old electrical units, become SI units?

Einstein's papers measure length in centimetres, mass in grams and time in seconds, the CGS system. Energy is in ergs, and one erg is 10⁻⁷ joules. Viscosity is in grams per centimetre per second, the unit now called the poise, and one poise is 0.1 pascal-seconds. These mechanical factors are exact, because they follow from 1 cm = 0.01 m and 1 g = 0.001 kg.

To convert, multiply by a ratio equal to one. Since 1 poise is 0.1 pascal-seconds, the ratio (0.1 Pa s)/(1 P) equals one, and multiplying by it changes the unit without changing the viscosity. The Brownian paper gives water's viscosity as 1,35 · 10⁻², with no unit name; it is in poise, so the viscosity is 1.35 × 10⁻³ Pa s. The paper calls this viscosity k, which is not Boltzmann's constant.

\begin{aligned} &1.35\times10^{-2}\,\mathrm{P}\times\frac{0.1\,\mathrm{Pa\,s}}{1\,\mathrm{P}} \\ &\quad = 1.35\times10^{-3}\,\mathrm{Pa\,s} \end{aligned}

1.35 times ten to the minus two poise, times 0.1 pascal-seconds per poise, equals 1.35 times ten to the minus three pascal-seconds.

The same ratio works for compound units. The gas constant R = 8.31 × 10⁷ erg per mole per kelvin is 8.31 joules per mole per kelvin. Neither the light paper nor the Brownian paper prints this number; the edition supplies the standard value of 1905 as an editorial input and labels it that way.

Electricity had two CGS systems. The electrostatic system builds its charge unit, the statcoulomb, from the force between charges. The electromagnetic system builds its current unit from the force between currents; its units are the abcoulomb and the abvolt. One abcoulomb is 10 coulombs, one abvolt is 10⁻⁸ volts, one statvolt is 299.792458 volts, and one gauss is 10⁻⁴ tesla. Planck's 1901 elementary charge, 4,69 · 10⁻¹⁰ statcoulombs, is 1.56 × 10⁻¹⁹ coulombs.

These electrical correspondences are conventional, not exact. They assume the magnetic constant $\mu_0$ is exactly 4π × 10⁻⁷ henry per metre. Since the 2019 revision of the SI it is a measured quantity, equal to that value within about one part in a billion. The site marks every electrical conversion conventional.

The light paper's photoelectric check in §8 uses the electromagnetic system. It takes E = 9,6 · 10³ for the charge of one gram-equivalent of a monovalent ion, so a potential Π comes out in abvolts, and the paper multiplies by 10⁻⁸ to get volts. It prints the result as about 4,3 volts. On the result line the factor is set as 10⁷; the edition records this as a misprint, because the sentence four lines earlier gives 10⁻⁸. The check can also be run per electron, with Planck's charge in statcoulombs and the result in statvolts, and it gives 4.31 volts; the edition keeps that as a documented alternative and follows the page.

The paper's R, E and J are amounts per gram-equivalent, a mole of monovalent ions. Dividing by N gives the amount per molecule, R/N, or per electron, ε. With the paper's E and the N it prints in §2, 6,17 · 10²³, the charge on one ion is 9.6 × 10³ ÷ 6.17 × 10²³ = 1.56 × 10⁻²⁰ abcoulombs, which is 1.56 × 10⁻¹⁹ coulombs. The paper does not print this quotient.

The relativity paper writes its fields in Gaussian units, where electric and magnetic fields share one unit. That is why its transformation reads β(Y − (v/V)N), with the dimensionless ratio v/V; here N is a magnetic field component, not Avogadro's number. SI measures the electric field in volts per metre and the magnetic field in tesla, so the same law reads $\gamma(E_y - vB_z)$ , with the speed itself multiplying the magnetic field. One statvolt per centimetre is 2.998 × 10⁴ volts per metre.

\begin{gathered} 1\,\mathrm{abV} = 10^{-8}\,\mathrm{V}, \\ 1\,\mathrm{abC} = 10\,\mathrm{C}, \\ 1\,\mathrm{statV} = 299.792458\,\mathrm{V}, \\ 1\,\mathrm{G} = 10^{-4}\,\mathrm{T} \end{gathered}

One abvolt corresponds to ten to the minus eight volts, one abcoulomb to ten coulombs, one statvolt to 299.792458 volts, and one gauss to ten to the minus four tesla, all by convention.

Worked example: The photoelectric check, converted to volts

The inputs: β = 4,866 · 10⁻¹¹ and ν = 1,03 · 10¹⁵ per second are printed in §8, and so is E = 9,6 · 10³. R = 8.31 × 10⁷ erg per mole per kelvin is the edition's editorial input, because the paper does not print it.
With P′ = 0, Π = Rβν/E = 8.31 × 10⁷ × 4.866 × 10⁻¹¹ × 1.03 × 10¹⁵ ÷ 9.6 × 10³ = 4.34 × 10⁸ abvolts.
Multiply by the ratio 10⁻⁸ volts per abvolt: 4.34 volts. The paper prints about 4,3 volts.
In §9 the conversion runs the other way. An energy per gram-equivalent divided by E is a potential: the printed Rβν = 6,4 · 10¹² erg gives 6.4 × 10¹² ÷ 9.6 × 10³ = 6.67 × 10⁸ abvolts, or 6.67 volts.
Stark's ionization voltage of 10 volts is 10⁹ abvolts, and 10⁹ × 9.6 × 10³ = 9.6 × 10¹² erg per gram-equivalent, the upper bound for J that §9 prints.
E itself, 9.6 × 10³ abcoulombs, is 9.6 × 10⁴ coulombs per gram-equivalent. Beside it, the modern Faraday constant is 96 485 coulombs per mole (set modern-si-2019); Einstein's value is 0.50 percent below it.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: converting a printed number

Type a number and choose the unit it was printed in. The calculator multiplies by a ratio equal to one, taken from the site's table of units, and says whether that ratio is exact or a convention.

Number to convert

Write 1.35e-2 or 1,35 · 10^-2. A comma is read as a decimal comma, as the papers print it.

Its unit, and the SI unit to convert to

0.0135 P = 0.00135 Pa s

Ratio used: 1 P = 0.1 Pa s.

Exact: it follows from 1 cm = 0.01 m and 1 g = 0.001 kg.

The papers' own numbers, converted

Water's viscosity, which the paper calls k (Brownian motion, p. 559): 0.0135 P = 0.00135 Pa s. Exact.
The gas constant R (not printed in either paper; supplied by the edition): 8.31 × 10⁷ erg per mole per kelvin = 8.31 J per mole per kelvin. Exact.
E, the charge of one gram-equivalent of a monovalent ion (Light quanta §8, p. 146): 9.6 × 10³ abC per gram-equivalent = 9.6 × 10⁴ C per gram-equivalent. Conventional.
Π = Rβν/E with P′ = 0; the paper prints only the result, about 4,3 volts (Light quanta §8, computed in the worked example): 4.34 × 10⁸ abV = 4.34 V. Conventional.
Rβν for the longest effective wavelength in air (Light quanta §9, p. 148): 6.4 × 10¹² erg per gram-equivalent = 6.4 × 10⁵ J per gram-equivalent. Exact.
The elementary charge (Planck 1901, p. 565): 4.69 × 10⁻¹⁰ statC = 1.56 × 10⁻¹⁹ C. Conventional.

What it shows, in words

A number printed in a CGS unit becomes an SI number when it is multiplied by a ratio equal to one. The mechanical ratios are exact. The electrical ratios are conventional, because they take the magnetic constant as exactly 4π × 10⁻⁷ H/m. The calculator converts units only: it never replaces a historical constant with a modern one.

Where this lesson stops

A conversion factor is a ratio equal to one. This lesson converts units only: it never turns a historical constant's value into a modern one, and it treats every electrical correspondence as a convention.

Letters: the paper's result, printed on p. 909 in Gaussian units with V the speed of light and β the modern γ, is $X'=X$ , $L'=L$ , $Y'=\beta(Y-\frac{v}{V}N)$ , $M'=\beta(M+\frac{v}{V}Z)$ , $Z'=\beta(Z+\frac{v}{V}M)$ , $N'=\beta(N-\frac{v}{V}Y)$ . X, Y, Z are the electric force and L, M, N the magnetic force; primes mark the moving system k. This explanation writes them $E_x$ to $B_z$ in SI units.

Change the units term by term. In Gaussian units electric and magnetic fields share one unit; in SI, $B$ is measured so that $cB$ has the units of E. So $(v/V)N$ becomes $vB_z$ , and $(v/V)Y$ inside a magnetic component becomes $vE_y/c^2$ .
The components along the motion do not change: $E'_x=E_x$ and $B'_x=B_x$ .
Each electric component across the motion picks up a magnetic term: $E'_y=\gamma(E_y-vB_z)$ and $E'_z=\gamma(E_z+vB_y)$ .
Each magnetic component across the motion picks up an electric term divided by $c^2$ : $B'_y=\gamma(B_y+vE_z/c^2)$ and $B'_z=\gamma(B_z-vE_y/c^2)$ . Together, the six lines below.

E'_{x} = E_{x}

B'_{x} = B_{x}

E'_{y} = \gamma\,\left(E_{y} - v\,B_{z}\right)

B'_{y} = \gamma\,\left(B_{y} + \frac{v\,E_{z}}{c^{2}}\right)

E'_{z} = \gamma\,\left(E_{z} + v\,B_{y}\right)

B'_{z} = \gamma\,\left(B_{z} - \frac{v\,E_{y}}{c^{2}}\right)

Shown in modern letters; Einstein's are in §6 of the German source face.

Worked, in units where c = 1: no electric field and a magnetic field $B_z=1$ , seen from a frame moving at 0.6 along x. Then $E'_y=-1.25\times0.6\times1$ = −0.75, an electric field along −y, and $B'_z=1.25$ .
Back again, with −v: $E''_y=1.25(-0.75+0.6\times1.25)=0$ and $B''_z=1.25(1.25-0.6\times0.75)$ = 1. The original fields return.
The paper's reading of the first three equations, for a unit point charge. At rest in K, the force on it is (X, Y, Z) by definition. At rest in k, at least at that moment, the force measured in k is $(X',Y',Z')$ , since the principle of relativity makes the charge a unit charge there too.
The older way of saying it: a unit charge moving in an electromagnetic field feels, besides the electric force, an electromotive force equal to the cross product of its velocity and the magnetic force divided by the speed of light. The paper states that this holds only if terms with the second and higher powers of v/V are neglected.
The new way: the force on a moving unit charge equals the electric force at its place after the field has been transformed to a system at rest relative to the charge. The electromotive force becomes an auxiliary idea, needed only because electric and magnetic forces have no existence independent of the state of motion of the coordinate system.
Worked force: a positive charge q carried along at 0.6c through $B_z$ is at rest in k, so there its whole force is electric, of size 1.25 × q × 0.6 × $B_z$ in these units. In K the same charge feels a magnetic force of size 0.6 × q × $B_z$ . The two differ by the factor 1.25, which is how a force across the motion changes between frames, so both describe one push.
The paper concludes that the asymmetry of the introduction, magnet moving or conductor moving, disappears, and that asking where the electromotive force is seated, as in unipolar machines, becomes pointless.
Before comparing forces, transform the charge's velocity too: a charge moving with the observer is at rest in the new frame.

A tool for this step

Charges, fields and light

What does an electric or a magnetic field do to a charge?

A comb run through dry hair lifts small scraps of paper before it touches them. The charge on the comb fills the space around it with an electric field, and the field pushes on any charge placed in it, harder for a larger charge. For a charge q in a field E the force is qE, along the field for a positive charge and against it for a negative one.

A magnet does nothing to a charge at rest. A charge moving with velocity v through a magnetic field B feels a push at right angles both to its motion and to the field, which is what turns an electric motor. Together these are the Lorentz force:

\mathbf{F} = q\left(\mathbf{E} + \mathbf{v} \times \mathbf{B}\right)

The force equals q times the sum of the electric field and the cross product of the velocity with the magnetic field.

Carrying a charge q across a potential difference ΔV changes its energy by qΔV. The electron's charge has magnitude e = 1.602 × 10⁻¹⁹ coulombs, so one volt changes an electron's energy by 1.602 × 10⁻¹⁹ joules. That amount is called one electron volt, 1 eV. §8 of the light-quanta paper states the energy of electrons that light frees from a metal in these terms: by the potential needed to stop them.

An electron held near a point by a force proportional to its displacement, like a mass on a spring, vibrates at a natural frequency of its own, and it absorbs and emits light near that frequency. §1 of the light-quanta paper fills a closed box with such bound electrons, which it calls resonators, together with free electrons, gas molecules and radiation, and asks what equilibrium requires. Using Planck's condition of 1900 for resonators and radiation, the answer gives the radiation an infinite total energy. That is the difficulty in the section's title.

Maxwell's equations tie fields to charges and currents in four statements: charges are the sources of electric field; there are no magnetic charges; a changing magnetic field drives a circulating electric field, which is Faraday's induction; and currents and changing electric fields drive circulating magnetic fields. Together they allow waves of field that cross empty space at the speed of light, c.

The relativity paper opens with a case from this theory. Move a magnet past a conductor, and the theory describes an electric field that drives a current. Move the conductor past the magnet instead, and it describes no electric field, only a magnetic force on the conductor's charges. The current is the same either way. The paper counts this difference of description, with no difference in what is observed, among its grounds for taking only relative motion to matter.

Worked example: An electron across one volt

\Delta W = q\,\Delta V

The change in energy equals the charge times the potential difference.

An electron starts at rest and crosses a potential difference of 1 V.
It gains energy e × 1 V = 1.602 × 10⁻¹⁹ joules: one electron volt.
At low speed that energy is ½mv², with the electron's mass m = 9.109 × 10⁻³¹ kg. So v² = 2 × 1.602 × 10⁻¹⁹ ÷ 9.109 × 10⁻³¹ = 3.52 × 10¹¹ m²/s².
The square root gives v ≈ 5.93 × 10⁵ m/s, about 600 kilometres per second.

That is fast, but only 0.2 per cent of the speed of light, so the low-speed formula ½mv² is safe here. The charge and mass used are modern values.

Where this lesson stops

Charge is the property that makes electric forces. Fields carry the force between separated charges, without action at a distance.

Explore the equations in this step

Read each operation, check its units and assumptions, or open the mathematical step behind it. These are modern teaching equations, not the equations as printed.

Explore the equation · Modern model notation

The magnetic field along the motion

B'_{x} = B_{x}

Shown in modern letters; Einstein's are in §6 of the German source face.

equals .

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Each term and operation, in words

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The magnetic field component along the motion is the same in both systems.

The component of the magnetic field along the direction of motion is unchanged by the change of system.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Fields in empty space, in SI units; E is the electric field and B the magnetic field, with components along x, y and z.

What it asserts: Only the components across the motion mix. Read the prerequisite
Magnetic field, moving system: The magnetic field in the moving system, one component; the subscript names which. Read the prerequisite
Magnetic field, stationary system: The magnetic field in the stationary system, one component; the subscript names which. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Magnetic field along y

B'_{y} = \gamma\,\left(B_{y} + \frac{v\,E_{z}}{c^{2}}\right)

Shown in modern letters; Einstein's are in §6 of the German source face.

is times plus times over c squared.

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The magnetic field along y in the moving system is gamma times the sum: the stationary magnetic field along y, plus v times the stationary electric field along z divided by c squared.

The magnetic field mixes with the electric field the other way, divided by c squared in SI units.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Fields in empty space, in SI units; E is the electric field and B the magnetic field, with components along x, y and z.

The two together: In the moving system, what was a pure magnetic field is partly electric, and the other way round. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
Over c squared: In SI units the electric field enters the magnetic one divided by c squared. Read the prerequisite
What it asserts: Electric and magnetic fields are not separately the same in both systems; their components across the motion mix. Read the prerequisite
Times gamma: The same factor as for coordinates. Read the prerequisite
Speed times the other field: The other field, carried past at speed v, contributes to this one. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Magnetic field, moving system: The magnetic field in the moving system, one component; the subscript names which. Read the prerequisite
Electric field, stationary system: The electric field in the stationary system, one component; the subscript names which. Read the prerequisite
Magnetic field, stationary system: The magnetic field in the stationary system, one component; the subscript names which. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Magnetic field along z

B'_{z} = \gamma\,\left(B_{z} - \frac{v\,E_{y}}{c^{2}}\right)

Shown in modern letters; Einstein's are in §6 of the German source face.

is times minus times over c squared.

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The magnetic field along z in the moving system is gamma times the difference: the stationary magnetic field along z, minus v times the stationary electric field along y divided by c squared.

The z component, with the opposite sign.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Fields in empty space, in SI units; E is the electric field and B the magnetic field, with components along x, y and z.

The two together: In the moving system, what was a pure magnetic field is partly electric, and the other way round. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
Over c squared: In SI units the electric field enters the magnetic one divided by c squared. Read the prerequisite
What it asserts: Electric and magnetic fields are not separately the same in both systems; their components across the motion mix. Read the prerequisite
Times gamma: The same factor as for coordinates. Read the prerequisite
Speed times the other field: The other field, carried past at speed v, contributes to this one. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Magnetic field, moving system: The magnetic field in the moving system, one component; the subscript names which. Read the prerequisite
Electric field, stationary system: The electric field in the stationary system, one component; the subscript names which. Read the prerequisite
Magnetic field, stationary system: The magnetic field in the stationary system, one component; the subscript names which. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

The electric field along the motion

E'_{x} = E_{x}

Shown in modern letters; Einstein's are in §6 of the German source face.

equals .

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The electric field component along the motion is the same in both systems.

The component of the electric field along the direction of motion is unchanged by the change of system.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Fields in empty space, in SI units; E is the electric field and B the magnetic field, with components along x, y and z.

What it asserts: Only the components across the motion mix. Read the prerequisite
Electric field, moving system: The electric field in the moving system, one component; the subscript names which. Read the prerequisite
Electric field, stationary system: The electric field in the stationary system, one component; the subscript names which. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Electric field along y

E'_{y} = \gamma\,\left(E_{y} - v\,B_{z}\right)

Shown in modern letters; Einstein's are in §6 of the German source face.

is times minus times .

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The electric field along y in the moving system is gamma times the difference: the stationary electric field along y, minus v times the stationary magnetic field along z.

With no electric field and a magnetic field along positive z, an observer moving along positive x finds an electric field along negative y: the force on a moving charge becomes, in its own system, an electric force.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Fields in empty space, in SI units; E is the electric field and B the magnetic field, with components along x, y and z.

The two together: In the moving system, what was a pure magnetic field is partly electric, and the other way round. Read the prerequisite
What it asserts: Electric and magnetic fields are not separately the same in both systems; their components across the motion mix. Read the prerequisite
Times gamma: The same factor as for coordinates. Read the prerequisite
Speed times the other field: The other field, carried past at speed v, contributes to this one. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Electric field, moving system: The electric field in the moving system, one component; the subscript names which. Read the prerequisite
Magnetic field, stationary system: The magnetic field in the stationary system, one component; the subscript names which. Read the prerequisite
Electric field, stationary system: The electric field in the stationary system, one component; the subscript names which. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Electric field along z

E'_{z} = \gamma\,\left(E_{z} + v\,B_{y}\right)

Shown in modern letters; Einstein's are in §6 of the German source face.

is times plus times .

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The electric field along z in the moving system is gamma times the sum: the stationary electric field along z, plus v times the stationary magnetic field along y.

The same mixing for the z component, with the opposite sign.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
Fields in empty space, in SI units; E is the electric field and B the magnetic field, with components along x, y and z.

The two together: In the moving system, what was a pure magnetic field is partly electric, and the other way round. Read the prerequisite
What it asserts: Electric and magnetic fields are not separately the same in both systems; their components across the motion mix. Read the prerequisite
Times gamma: The same factor as for coordinates. Read the prerequisite
Speed times the other field: The other field, carried past at speed v, contributes to this one. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. At 0.6c it is 1.25. Read the prerequisite
Electric field, moving system: The electric field in the moving system, one component; the subscript names which. Read the prerequisite
Magnetic field, stationary system: The magnetic field in the stationary system, one component; the subscript names which. Read the prerequisite
Electric field, stationary system: The electric field in the stationary system, one component; the subscript names which. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Modern qualifications

Einstein’s transformation of the fields is exact. The shorter statement he then makes about the electromotive force on a moving charge holds only to first order in $v/V$ ; at finite speed the factor $\gamma$ stays. His $L$ in this section is a magnetic component, not the light speed $L$ of the light-quanta paper or the emitted energy $L$ of the mass-energy paper.

Assumptions and limits: Electric and magnetic components mix together

Assumed here

Fields are in SI units in empty space, and the new observer moves at speed v along positive x.
Forces are compared at the same event, with the charge’s velocity transformed too.

What this does not establish

An accelerating charge has a different momentary rest frame at each instant, not one frame for its whole path.
Fields, forces and electromotive force each transform by their own rule.

Earlier step: Transform derivatives before naming fields

Source context: German source · English · Interlinear gloss · Facsimile

References

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).

A. Einstein, Does the inertia of a body depend upon its energy content? Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s translation or its pinned facsimile.

28 foundation readings sit behind this argument.

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