§6 · field equations and field components · Special relativity: from clock operations to electrodynamics

§6 · field equations and field components

Transform derivatives before naming fields

How do the field equations change coordinates?

\partial_t=\gamma(\partial_{t'}-v\partial_{x'}),\qquad\partial_x=\gamma(\partial_{x'}-v\partial_{t'}/c^2)

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.

Return to how the field equations look in the moving frame. (included in this file)

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(\partial_{t'}-v\partial_{x'}),\qquad\partial_x=\gamma(\partial_{x'}-v\partial_{t'}/c^2)

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.

Return to how the field equations look in the moving frame. (included in this file)

The time derivative, rewritten

Modern teaching equation, not the equation as printed.

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

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.

The x derivative, rewritten

Modern teaching equation, not the equation as printed.

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

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.

Assumptions

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.

Limits

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.

Electric and magnetic components mix together

What does a new inertial observer call electric?

\begin{aligned}E_x'&=E_x,&B_x'&=B_x\\E_y'&=\gamma(E_y-vB_z),&B_y'&=\gamma(B_y+vE_z/c^2)\\E_z'&=\gamma(E_z+vB_y),&B_z'&=\gamma(B_z-vE_y/c^2)\end{aligned}

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.

Return to what a new observer calls electric. (included in this file)

Return to the field components along the motion and across it. (included in this file)

Return to the field transformation in SI units, which the paper writes in Gaussian units. (included in this file)

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.

\begin{aligned}E_x'&=E_x,&B_x'&=B_x\\E_y'&=\gamma(E_y-vB_z),&B_y'&=\gamma(B_y+vE_z/c^2)\\E_z'&=\gamma(E_z+vB_y),&B_z'&=\gamma(B_z-vE_y/c^2)\end{aligned}

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.

Return to what a new observer calls electric. (included in this file)

The magnetic field along the motion

Modern teaching equation, not the equation as printed.

B'_{x} = B_{x}

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.

Magnetic field along y

Modern teaching equation, not the equation as printed.

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

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.

Magnetic field along z

Modern teaching equation, not the equation as printed.

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

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.

The electric field along the motion

Modern teaching equation, not the equation as printed.

E'_{x} = E_{x}

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.

Electric field along y

Modern teaching equation, not the equation as printed.

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

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.

Electric field along z

Modern teaching equation, not the equation as printed.

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

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.

Assumptions

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.

Limits

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.

Foundations included in this file

Reading a graph

How does a curve show the relationship between two physical quantities?

Walk for a minute and note how far you have gone every ten seconds: 0 m at the start, 14 m after ten seconds, 28 m after twenty, and so on. The first drawing below puts time along the bottom and distance up the side, with a dot for each pair. That is a graph: each dot is one moment, read across for the time and up for the distance.

The dots of a steady walk fall on a straight line, and the steeper the line, the faster the walk. This one climbs 14 metres for every 10 seconds: 1.4 metres per second. A flat line would mean standing still, and a line that bends upward would mean speeding up.

The papers draw other pairs the same way. Along the bottom goes what you choose or wait for, such as a position or a time; up the side goes what you then find, such as how many particles sit there. A bell-shaped curve, high in the middle and falling away on both sides, says that most particles are near the centre and fewer are farther out.

Worked example: Four particles that average to zero

Four particles end at −3, −1, +1 and +3 micrometres from where they started.
The second drawing puts those positions along the bottom, with a bar one particle tall above each.
The bars stand two on each side of zero, at the same distances out. The picture is symmetric, so the average position is 0.
Yet no bar stands at zero: every particle moved. The graph shows at a glance what the average hides.

Where this lesson stops

Before reading a curve, read its two labels and their units, and ask whether its height is a count, a density or a running total.

This lesson builds on

A sign records direction

Return to the chapter

Fractions and ratios

What does dividing one quantity by another tell you?

The fraction 6/2 is 3, because 2 fits into 6 three times; it also says that 6 is three times 2. Every fraction a/b works the same way. The bottom number can never be zero, since no count of zeros adds up to 6.

When the two quantities have different units, their ratio is a new quantity with a unit of its own: 12 micrometres travelled in 4 seconds is 3 micrometres per second. When they have the same unit, the units cancel and the ratio is a plain number: 2 particles out of 8 is 0.25 of the group.

Worked example: Fourteen metres in ten seconds

A steady walk covers 14 metres in 10 seconds.
Divide the distance by the time: 14 divided by 10 is 1.4, and the unit divides with it, so the rate is 1.4 metres per second.
After 60 seconds the same walk has covered 84 metres, and 84 divided by 60 is 1.4 again. The walk is steady, so the ratio does not depend on how long it is watched.
Metres and seconds are different units, so neither cancels and what comes out is a new quantity that keeps a unit of its own.
The drawing below is the other case: 2 particles out of 8 has the same unit above and below, they cancel, and 0.25 is left as a plain number.

Where this lesson stops

A ratio of two quantities in the same unit is a plain number. A ratio of different units, such as micrometres per second, is a new quantity and keeps its unit.

This lesson builds on

Adding and averaging

Return to the chapter

A sign records direction

How can movement add up to zero?

On the ruler below, the starting point is zero. A final mark three units to the right has displacement +3; one three units to the left has displacement −3. Adding the signed displacements gives zero. Adding the distances from the start gives six.

Displacement compares where something ends with where it started, whatever it did in between. A walker who goes three units right and comes back has displacement zero after walking six, which is the lower pair of arrows in the drawing.

Worked example: Two walkers, opposite directions, average zero

Two walkers both finish one unit from their start, one at −1 and the other at +1. The signed average is zero. The average distance is one.

Where this lesson stops

The sign names a direction relative to a chosen axis; it does not mean a negative distance.

This lesson builds on

Adding and averaging

Return to the chapter

Powers of ten and physical units

How do you write a very small length without counting zeros?

The Brownian paper works its example, in §5, for particles 0.001 mm across. In metres that is 0.000001 m: easy to miscount by one zero, which is a factor of ten. Written as 10⁻⁶ m, the −6 says the decimal point has moved six places to the left, and that length has a name of its own, the micrometre (μm).

The unit goes through the arithmetic with the number. A micrometre is 10⁻⁶ m, so a square micrometre is 10⁻¹² m², not 10⁻⁶ m²: squaring a length squares its unit, power of ten included.

Worked example: Squaring six micrometres

1 micrometre (1 μm) is 10⁻⁶ metres, or 0.000001 m.
The paper's displacement after one minute, about 6 μm, is 6 × 10⁻⁶ m.
Squaring it squares the number and the unit: (6 × 10⁻⁶ m)² = 36 × 10⁻¹² m² = 3.6 × 10⁻¹¹ m².

Where this lesson stops

A measurement is a number and a unit together. Squaring it squares both: (6 × 10⁻⁶ m)² is 3.6 × 10⁻¹¹ m².

This lesson builds on

Fractions and ratios

Return to the chapter

Adding and averaging

What does an average keep, and what does it lose?

Four jars hold 3, 1, 1 and 3 marbles. Pour them together and there are 8. Share the 8 equally between the four jars and each holds 2. Two is the average, although no jar held 2 to begin with.

The average keeps the total and the count and nothing else, so 3, 1, 1, 3 and 2, 2, 2, 2 have the same average. Listing every value twice doubles both the total and the count, and the average stays at 2.

Worked example: Three mornings with the same average

One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
Another morning they hold 8, 0, 0 and 0. Add them: 8 again. Divide by four: 2 again, and now three jars are empty.
A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
Write any of those three lists out twice over. The total doubles to 16 and the count doubles to 8 jars, and 16 divided by 8 is still 2.
Four descriptions and one average. It keeps the total and the count, and the differences between the mornings are what it leaves behind.

Where this lesson stops

An average is the total shared equally among the observations, and two very different lists can have the same one.

Return to the chapter

Rates of change and derivatives

How can a body have a speed at a single instant?

A speed needs two readings of position and the time between them: divide the distance covered by the time taken, and you have the average speed over that interval. An instant has no duration, so that recipe cannot be used at an instant as it stands.

So shorten the interval and watch what happens. A ball rolls so that after t seconds it has gone 3t² metres: 3 m after one second, 12 m after two. Start at the one-second mark and average over shorter and shorter stretches:

Over the next second it goes from 3 m to 12 m: 9 m in 1 s, an average of 9 m/s.
Over the next tenth of a second it reaches 3.63 m: 0.63 m in 0.1 s, an average of 6.3 m/s.
Over the next hundredth it reaches 3.0603 m: 0.0603 m in 0.01 s, an average of 6.03 m/s.
Over the next thousandth it reaches 3.006003 m: 0.006003 m in 0.001 s, an average of 6.003 m/s.

The averages close in on 6 m/s. That number is the derivative: the ball's speed at the instant t = 1 s. The interval is never set to zero, which would mean dividing zero distance by zero time. It only shrinks, and the averages settle.

In symbols, with Δt for the short interval, the same recipe reads:

\frac{dx}{dt} = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\Delta t}

d x by d t is the value that the change in position divided by the change in time settles on as delta t shrinks towards zero.

A derivative carries units: those of the quantity that changes, divided by those of the quantity you vary. Position in metres, varied over time in seconds, gives metres per second. On a graph of position against time, the derivative at a point is the steepness of the curve there.

Einstein's Brownian paper needs rates like this for a concentration of particles, which changes along a tube and in time at once. With two things changing, a rate has to say which one it follows, so the paper writes a curly ∂ and holds the other fixed:

\frac{\partial f}{\partial t} = D\,\frac{\partial^2 f}{\partial x^2}

The partial derivative of f with respect to t equals D times the second partial derivative of f with respect to x.

The lesson on partial derivatives, listed at the end of this page, reads that equation one piece at a time.

Worked example: Why the averages settle on 6 m/s

The list of averages can be explained, not only watched. Start at any time t and let the interval be Δt. In that interval the ball moves 3(t + Δt)² − 3t² metres. Since (t + Δt)² is t² + 2tΔt + (Δt)², that distance is:

3(t+\Delta t)^2 - 3t^2 = 6t\,\Delta t + 3(\Delta t)^2

Three times t plus delta t, squared, minus three t squared, equals six t delta t plus three delta t squared.

Divide that distance by the time Δt to get the average speed over the interval:

\frac{6t\,\Delta t + 3(\Delta t)^2}{\Delta t} = 6t + 3\,\Delta t

Six t delta t plus three delta t squared, divided by delta t, equals six t plus three delta t.

The 3Δt term shrinks with the interval and leaves 6t metres per second. At t = 1 s that is 6 m/s. The extra 3, 0.3, 0.03 and 0.003 in the list above are that 3Δt term, for Δt of 1, 0.1, 0.01 and 0.001 seconds.

Where this lesson stops

A derivative is the value an average rate settles on as the interval shrinks, in units of one quantity per unit of the other. The interval shrinks but is never set to zero, so nothing is ever divided by zero.

This lesson builds on

Functions and graphs

Return to the chapter

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.

Return to the chapter

Functions and graphs

How does a graph record the way one quantity depends on another?

A bath fills from a tap. After one minute it holds 10 litres, after two 20, after three 30. For each time there is exactly one amount of water, and that pairing, one output for each input, is a function. Plot the pairs with time along the bottom and litres up the side and they fall on a straight line that climbs 10 litres for every minute. The slope, in litres per minute, is how fast the bath fills.

In the papers the pairs are physical: the mean square displacement of a Brownian particle for each elapsed time, or the concentration for each position along a tube. A function says how one thing responds when another changes.

A graph puts the input along the horizontal axis and the output up the vertical one, so each point is one pair. Both axes carry units, and so does everything read off the graph: the slope of mean square displacement against time is in square micrometres per second, and the area under a probability density is a plain probability.

A formula such as ⟨x²⟩ = 2Dt describes an average over many particles. One particle's squared displacement scatters widely around the line; the average over many lies close to it. The Brownian paper predicts the average, and ends by hoping that a researcher will soon decide the question by observation.

Worked example: Mean square displacement plotted against time

\langle x^2\rangle = 2Dt

The mean square displacement equals two times the diffusion coefficient times the time.

Take D = 0.5 square micrometres per second, so 2D = 1 square micrometre per second.
At t = 1 s the mean square displacement is 1 µm²; at t = 4 s it is 4 µm²; at t = 9 s, 9 µm².
Plotted against time, these points lie on a straight line through the origin whose slope is 2D, 1 µm² per second.
Their square roots, the root-mean-square displacements, are 1, 2 and 3 µm: four times the time gives twice the distance.

Where this lesson stops

A graph is a record of paired quantities, each with its units. A slope read off it has units too, and a slope stated without them explains nothing.

This lesson builds on

Reading a graph

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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(x,t) = \frac{1}{\sqrt{4\pi D t}}\,e^{-x^2/4Dt}

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}{2Dt}\,f(x,t)

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}{2t} + \frac{x^2}{4Dt^2}\right) f(x,t)

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.

Where this lesson stops

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

This lesson builds on

Rates of change and derivatives

Return to the chapter

Quantities and units

Can a number of metres be compared with a number of seconds?

"The particle moved 6" says nothing until the unit is added: 6 micrometres, or 6 seconds? A physical quantity is a number times a unit, and the number alone changes when the unit does. Six micrometres is 0.006 millimetres: the same length, written with another unit.

Quantities of different kinds cannot be added or compared: 3 metres plus 2 seconds has no meaning. So in a correct equation every term, on both sides, carries the same units. Checking that is the quickest test a formula can pass or fail.

Take the Brownian diffusion coefficient, $D = k_BT/(6\pi\eta a)$ . Boltzmann's constant is in joules per kelvin, T in kelvins, the viscosity η in pascal-seconds and the radius a in metres. The units work out to square metres per second, which is what a diffusion coefficient must be. The worked example below goes through it.

Units also tell apart quantities that share a letter. The light paper's $\rho_\nu$ is an energy per unit volume per unit of frequency, joules per cubic metre per hertz, not per unit of wavelength: two different quantities with different units. And its L is the speed of light in §§1 and 2 but an absorbed energy in §9. A unit check starts from what the quantity is, not from its letter.

Worked example: Checking that D comes out in square metres per second

$k_BT$ : joules per kelvin times kelvins is joules, which are newton-metres.
$6\pi\eta a$ : 6π has no units, and pascal-seconds times metres is (newtons per square metre) × seconds × metres, which is newton-seconds per metre.
Divide: newton-metres divided by newton-seconds per metre is square metres per second.
With numbers: 1.380649 × 10⁻²³ × 290.15 ÷ (6π × 1.35 × 10⁻³ × 5 × 10⁻⁷) = 3.15 × 10⁻¹³ m²/s. That is the modern Boltzmann constant (set modern-si-2019) with the paper's water at 17 °C and its particle.

Where this lesson stops

This lesson stops at checking units. Converting the papers' own units into SI units is a separate step, with its own conventions.

This lesson builds on

Powers of ten and physical units
Fractions and ratios

Return to the chapter

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.

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.

This lesson builds on

Powers of ten and physical units
Fractions and ratios
Quantities and units

Return to the chapter

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.

Where this lesson stops

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

This lesson builds on

Quantities and units
Functions and graphs

Return to the chapter

Static laboratory examples

sr-02: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open sr-02 online

Build-time scalar readouts
Quantity	Value	Unit	Status
magneticFieldStationary	1	T	value
electricFieldStationary	0	V/m	value
magneticFieldMoving	1.0000000000000007	T	value
electricFieldMoving	-10.000000000000007	V/m	value
transverseForceLaboratory	-1.602176634e-18	N	value
transverseForceComoving	-1.6021766340000011e-18	N	value
electromotiveForceMagnetFrame	1	V	value
electromotiveForceConductorFrame	1.0000000000000007	V	value
electromotiveForceExcess	5.563250280268097e-16	1	value
fieldInvariantEDotB	0	T V/m	value
fieldInvariantE2MinusC2B2	-89875517873681760	V^2/m^2	value
lorentzFactor	1.0000000000000007	1	value
pathBoostParallelComponent	0	1	value
endpointSimultaneityOffset	0	s	value
inducedCircuitCurrent	No numeric value	A	not-applicable

0 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

descriptionFrame: magnet-rest
dipoleMoment: 1
fieldModel: uniform
magneticField: 1
mode: analytic
motionState: magnet-moves
pathOrientation: transverse
resolution: shown
segmentLength: 0.1
sliceDeclared: false
speed: 10
testCharge: 1.602176634e-19
testPointDistance: 0.05

Evaluator: source:sha256:5c3f6559691b253e6123c3652e5426f10dc086beb947cf540d4a61aa86b6a325

sr-07

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open sr-07 online

sr-08: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open sr-08 online

Build-time scalar readouts
Quantity	Value	Unit	Status
fieldInvariantEDotB	0	T V/m	value
fieldInvariantE2MinusC2B2	1	V^2/m^2	value
lorentzFactor	1.25	1	value
particleTimeJacobian	1.25	1	value
fieldInvariantEDotBMoving	0	T V/m	value
fieldInvariantE2MinusC2B2Moving	1	V^2/m^2	value

14 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

boost: 179875474.79999998
chargeVelocityX: 0
chargeVelocityY: 0
chargeVelocityZ: 0
decomposeComponents: true
descriptionFrame: stationary
detectorMotion: false
detectorSpeed: 0
electricFieldX: 0
electricFieldY: 1
electricFieldZ: 0
magneticFieldX: 0
magneticFieldY: 0
magneticFieldZ: 0
testCharge: 1.602176634e-19
unitLayer: si

Evaluator: source:sha256:f3775f6b7e890749b3fca8c7e0acb8b6616f3158712e26ab56ef2655d95c9ce8

Source 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).