Introduction · magnet, conductor, and the postulates · Special relativity: from clock operations to electrodynamics

Introduction · magnet, conductor, and the postulates

One apparatus, two descriptions

Why should changing the observer change the explanation?

In the usual account of 1905, a magnet moving past a conductor at rest creates an electric field, and that field drives the current. A conductor moving past a magnet at rest meets no electric field at all; the charges moving with it feel a force from the magnetic field instead, and that force drives the current. The two accounts predict the same current for the same relative motion. The paper sets out to remove the asymmetry of the explanation, not to show that the two observers measure the same fields: they do not.

Einstein raises two conjectures to the status of postulates. First, the laws of electrodynamics and optics hold in the same form in every frame in which the laws of mechanics hold. Second, light in empty space always travels with a definite speed $V$ , whatever the motion of the body that emits it. The paper takes them as its starting point; the unsuccessful attempts to detect motion relative to the ether are cited as what suggests the first, not as a proof of it.

The magnet-and-conductor laboratory compares the two descriptions for the same relative motion. At low speed the electric force in one frame equals the magnetic force in the other; at higher speed they differ by the factor $\gamma$ derived in §6, which is how a force across the motion changes between frames.

Return to the magnetic force on the moving charges, and which way it pushes. (included in this file)

Return to why changing the observer should change the explanation. (included in this file)

Letters: the paper writes V for the speed of light in empty space; this explanation writes c. It calls the frame in which Newton's equations hold the stationary system, K from §3 on, and a system moving uniformly relative to it k. The factor this explanation writes γ is the paper's β, which first appears in §3.

The apparatus: a magnet and a conductor, a loop of wire, in motion relative to each other, with the current in the conductor as the thing observed.
The first case, as the physics of 1905 explained it: the magnet moves and the conductor rests. Around the moving magnet an electric field arises, carrying a definite energy, and where parts of the conductor are, that field drives a current.
The second case: the magnet rests and the conductor moves. No electric field arises around the magnet. In the conductor there is instead an electromotive force, to which no energy corresponds in itself, and it drives a current.
Compare the observations: for the same relative motion, the current has the same size and the same course in both cases. What is observed depends only on the relative motion of conductor and magnet.
Compare the explanations: one case has an electric field and the other has none. The asymmetry lies in the account, not in what is observed.
Worked low-speed case: the conductor moves at 10 metres per second across a magnetic field of 0.5 tesla. On a charge q carried with it the magnetic force is qvB, and per unit charge that is $vB=10\times0.5=5$ volts per metre. In the conductor's own frame §6 gives an electric field of γ times vB, and at this speed γ − 1 is about $v^2/(2c^2)=5.6\times10^{-16}$ : no measurement could tell the two numbers apart.
At 0.6c they differ by the factor γ = 1.25. §6 derives it: a force across the motion changes between frames by that factor, so the two accounts describe one push.
The paper adds a second kind of evidence: the unsuccessful attempts to detect the earth's motion relative to the medium of light. Together these lead to the conjecture that absolute rest corresponds to no property of the phenomena, in electrodynamics as in mechanics, and that the same laws of electrodynamics and optics hold in every coordinate system in which the equations of mechanics hold, as had already been shown for quantities of the first order.
The first premise, the principle of relativity: the paper raises that conjecture to a premise. It is assumed, not proved from the examples.
The second premise: light in empty space always travels with a definite speed V, independent of the state of motion of the body that emits it. The paper says the two premises are only apparently incompatible.
What the two premises are said to give: a simple and consistent electrodynamics of moving bodies, built on Maxwell's theory for bodies at rest. A light ether becomes superfluous, since no absolutely stationary space is introduced and no velocity is assigned to a point of empty space in which electromagnetic processes take place.
Where the work starts: every such theory rests on the kinematics of rigid bodies, since its statements concern rigid bodies used as coordinate systems, clocks, and electromagnetic processes. So §1 begins with what time means.
The status of the laboratory: a simulation built on the two premises shows their consequences. It cannot prove them.

Return to the magnetic force on the moving charges, and which way it pushes. (included in this file)

Return to why changing the observer should change the explanation. (included in this file)

Assumptions

Both descriptions concern the same apparatus and the same events, seen from two inertial frames.
The principle of relativity and the constancy of the speed of light are assumptions, and stated as such.

Limits

A simulation built on the postulates cannot prove them.
The model of a magnet and a conductor loop does not cover every induction experiment.

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

Dot and cross products

How much of one arrow lies along another, and which way does a magnetic force push?

The dot product multiplies two arrows into a single number: $\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$ , which also equals the two lengths times the cosine of the angle between them. With an arrow of length 1 along a chosen direction, the dot product is how much of the other arrow lies along that direction: its projection. Arrows at right angles have a dot product of zero.

\begin{aligned} \mathbf{a}\cdot\mathbf{b} &= a_x b_x + a_y b_y + a_z b_z \\ &= |\mathbf{a}|\,|\mathbf{b}|\cos θ \end{aligned}

a dot b equals a x b x plus a y b y plus a z b z, which equals the length of a times the length of b times the cosine of the angle between them.

For a boost along x, the part of an electric field along the motion is its dot product with an arrow of length 1 along x. A field of 5 units pointing 3 along x and 4 along y has 3 along the boost; that part is the same in both frames, and only the rest changes.

The cross product $\mathbf{a}\times\mathbf{b}$ is an arrow at right angles to both a and b. Its length is the area of the parallelogram they span, the two lengths times the sine of the angle between them, and its direction follows the right-hand rule: curl the fingers of the right hand from a toward b, and the thumb points along the product. Swapping the order reverses the arrow.

\mathbf{F} = q\,\mathbf{v}\times\mathbf{B}

The force F equals q times v cross B.

The magnetic force on a charge q moving with velocity v through a field B is $q\,\mathbf{v}\times\mathbf{B}$ . It is at right angles to the motion, so it changes the charge's direction and not its speed. For a positive charge moving along +x through a field along +z, $\mathbf{v}\times\mathbf{B}$ points along −y. In the frame moving with the charge the same push is electric: the relativity paper's §6 gives that frame an electric field along −y.

One combination of the two fields, $\mathbf{E}\cdot\mathbf{B}$ , comes out the same in every frame. That is a modern check on the field transformation, useful for catching an error in a calculation. It is not a premise of the 1905 argument, which derives the transformation from the equations of electrodynamics.

Worked example: A projection, a force direction and a quantity that stays put

Projection: $\mathbf{E} = (3, 4, 0)$ and the boost direction $(1, 0, 0)$ give $\mathbf{E}\cdot(1, 0, 0) = 3$ , the part along the boost. The field's length is $\sqrt{3^2 + 4^2} = 5$ .
Direction: with v along +x and B along +z, $(1, 0, 0)\times(0, 0, 1) = (0, -1, 0)$ , so a positive charge is pushed along −y.
Invariance, measuring fields so that c = 1: take $\mathbf{E} = (0, 0.6, 0.8)$ and $\mathbf{B} = (0, 0, 1)$ , so $\mathbf{E}\cdot\mathbf{B} = 0.8$ .
Seen from a frame moving at 0.6 along x, where γ = 1.25 (the factor 1/√(1 − v²/c²), which the relativity paper prints as β), the §6 rules give $\mathbf{E}' = (0, 0, 1)$ and $\mathbf{B}' = (0, 0.6, 0.8)$ . Both fields changed, and $\mathbf{E}'\cdot\mathbf{B}' = 1 \times 0.8 = 0.8$ did not.

Where this lesson stops

The dot product is how much of one arrow lies along another; the cross product is an arrow for the area two arrows span. This lesson stops at three dimensions.

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.

This lesson builds on

Functions and graphs
Rates of change and derivatives

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

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

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

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).