§3 · deriving and inverting the coordinate map · Special relativity: from clock operations to electrodynamics

§3 · deriving and inverting the coordinate map

Light constraints leave a scale to determine

What map can respect both light directions?

The moving origin follows $x = vt$ , so write $x' = a(v)(x - vt)$ . Requiring $x' = ct'$ on $x = ct$ and $x' = -ct'$ on $x = -ct$ fixes the time combination: $t' = a(v)(t - vx/c^2)$ . An overall scale $a(v)$ remains.

x'=a(v)(x-vt),\qquad t'=a(v)\left(t-\frac{vx}{c^2}\right)

The light constraints give a common coefficient multiplying x minus vt and t minus vx over c squared.

Going to the moving frame and back, with speed $v$ and then $-v$ , must return every event unchanged, which gives $a(v)\,a(-v)\,(1 - v^2/c^2) = 1$ . Space has no preferred direction, so $a(v) = a(-v)$ , and since $a(0) = 1$ the positive root is the one: $a(v) = \gamma$ .

\gamma=\frac{1}{\sqrt{1-v^2/c^2}},\qquad x'=\gamma(x-vt),\quad t'=\gamma\left(t-\frac{vx}{c^2}\right)

The Lorentz factor is one over the square root of one minus v squared over c squared, giving the standard space and time transformation.

Return to the map x′ = γ(x − vt), t′ = γ(t − vx/c²), with its factor γ. (included in this file)

Return to finding the map that respects both light directions. (included in this file)

Letters: the paper's stationary system K has coordinates x, y, z and time t; the moving system k has ξ, η, ζ and time τ, where τ is the time read on clocks at rest in k and set by §1's rule: a time coordinate, not a reading carried by one clock. V is the speed of light (c below), and the paper's β is the modern γ. The explanation writes $x'$ and $t'$ for ξ and τ. In §3 the paper also writes $x'=x-vt$ for a different quantity, a Galilean helper coordinate, which is not ξ.

The shorter route this explanation takes first. k moves at speed v along K's x axis, and its origin follows x = vt. A point at rest in k keeps a fixed value of $x-vt$ , so the moving coordinate must be a multiple of it: $x'=a(v)(x-vt)$ , with a(v) not yet known.
The paper states why the map is linear: space and time are the same everywhere, so equal steps in one system correspond to equal steps in the other.
Write the time map as a general linear combination, $t'=Pt+Qx$ , and ask light to keep its speed c in k in both directions along x.
Light going right, x = ct: then $x'=a(c-v)t$ and $t'=(P+Qc)t$ , and $x'=ct'$ requires $a(c-v)=c(P+Qc)$ .
Light going left, x = −ct: then $x'=-a(c+v)t$ and $t'=(P-Qc)t$ , and $x'=-ct'$ requires $a(c+v)=c(P-Qc)$ .
Add the two conditions: $2ac=2cP$ , so P = a. Subtract them: $-2av=2Qc^2$ , so $Q=-av/c^2$ . That gives the time map below, with the same unknown factor.

x'=a(v)(x-vt),\qquad t'=a(v)\left(t-\frac{vx}{c^2}\right)

The light constraints give a common coefficient multiplying x minus vt and t minus vx over c squared.

Go to k and back. Transforming with speed v and then with −v must return every event unchanged. Put the first map into the second: $x''=a(-v)a(v)\left[(x-vt)+v\left(t-\frac{vx}{c^2}\right)\right]=a(v)a(-v)\left(1-\frac{v^2}{c^2}\right)x$ . So $a(v)a(-v)(1-v^2/c^2)=1$ .
Space has no preferred direction, so the factor cannot depend on which way k moves: a(v) = a(−v). Then $a(v)^2=1/(1-v^2/c^2)$ .
At v = 0 nothing changes, so a(0) = 1 and the positive root is the one. That root is γ, which is real only for speeds below c.

\gamma=\frac{1}{\sqrt{1-v^2/c^2}},\qquad x'=\gamma(x-vt),\quad t'=\gamma\left(t-\frac{vx}{c^2}\right)

The Lorentz factor is one over the square root of one minus v squared over c squared, giving the standard space and time transformation.

Worked at v = 0.6c: $1-v^2/c^2=0.64$ , $\sqrt{0.64}=0.8$ , γ = 1/0.8 = 1.25. Check going there and back: 1.25 × 1.25 × 0.64 = 1.
The paper's own route, which this explanation shortens. It writes $x'=x-vt$ , so that a point at rest in k has fixed $x'$ , y, z, and first finds τ as a function of $x'$ , y, z and t.
A light ray leaves the origin of k at τ₀, is reflected at $x'$ at τ₁, and is back at τ₂; §1's rule in k requires $\frac{1}{2}(\tau_0+\tau_2)=\tau_1$ . In K the ray goes out at V − v and back at V + v, and taking $x'$ very small turns the condition into a differential equation for τ: $\frac{\partial\tau}{\partial x'}+\frac{v}{V^2-v^2}\frac{\partial\tau}{\partial t}=0$ .
Light along the other two axes crosses, seen from K, at $\sqrt{V^2-v^2}$ , which gives $\partial\tau/\partial y=0$ and $\partial\tau/\partial z=0$ . Since τ is linear, $\tau=a\left(t-\frac{v}{V^2-v^2}x'\right)$ .
A ray sent along ξ at τ = 0 has ξ = Vτ, and in K it moves away from k's origin at V − v, so $x'/(V-v)=t$ . Substituting gives $ξ=a\frac{V^2}{V^2-v^2}x'$ , and rays along the other axes give η and ζ. Putting back $x'=x-vt$ , the paper reaches $τ=\varphi(v)\beta\left(t-\frac{v}{V^2}x\right)$ , $ξ=\varphi(v)\beta(x-vt)$ , $η=\varphi(v)y$ , $ζ=\varphi(v)z$ , with $\beta=1/\sqrt{1-(v/V)^2}$ and φ an unknown function of v.
The check that the two principles agree: a spherical light wave sent from the common origin at t = τ = 0, $x^2+y^2+z^2=V^2t^2$ in K, becomes $ξ^2+η^2+ζ^2=V^2\tau^2$ in k, a spherical wave of speed V there too.
Fixing φ: a third system moving at −v relative to k must be at rest relative to K, which gives $\varphi(v)\varphi(-v)=1$ . A rod along η, moving across its own length, has length $l/\varphi(v)$ in K, and by symmetry that cannot depend on the direction of motion, so $\varphi(v)=\varphi(-v)$ . Together, φ(v) = 1, and the paper's printed result is $τ=\beta(t-vx/V^2)$ , $ξ=\beta(x-vt)$ , η = y, ζ = z.

Return to the map x′ = γ(x − vt), t′ = γ(t − vx/c²), with its factor γ. (included in this file)

Return to finding the map that respects both light directions. (included in this file)

The scale, fixed

Modern teaching equation, not the equation as printed.

\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}

The Lorentz factor is one over the square root of one minus v squared over c squared.

Going to the moving system and back must return every event unchanged, and the scale must not depend on the direction of motion or jump at v = 0. Those three conditions fix the factor as this.

Position in the moving system, up to a scale

Modern teaching equation, not the equation as printed.

x' = a(v)\,\left(x - v\,t\right)

The moving position is a common coefficient times x minus v t.

The moving system's origin is at x = vt, so the moving position must vanish there: it is x − vt times a factor the light conditions do not yet fix.

Time in the moving system, up to the same scale

Modern teaching equation, not the equation as printed.

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

The moving time is the same coefficient times t minus v x over c squared.

Requiring light to go at c in both directions in both systems fixes the time combination: t − vx/c², times the same factor as the position.

Position in the moving system

Modern teaching equation, not the equation as printed.

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

The moving position is gamma times x minus v t.

The position half of the Lorentz transformation: the distance from the moving origin, scaled by gamma.

Time in the moving system

Modern teaching equation, not the equation as printed.

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

The moving time is gamma times t minus v x over c squared.

The time half of the Lorentz transformation. The vx/c² term makes the moving system's time depend on where an event is, which is why two systems disagree about what is simultaneous.

Assumptions

The two frames’ axes are parallel, and their origins coincide at time zero.
Space and time are the same everywhere, so the map between coordinates is linear.
The two postulates hold, together with going there and back, the symmetry of space, and continuity at zero speed.

Limits

Keeping one light ray at speed c does not fix every coefficient of the map.
The spacetime interval and matrices check the result later; the 1905 argument does not assume them.

The sideways step needs its own condition

Why do the transverse coordinates stay unchanged?

Write $y' = b(v)\,y$ for a transverse axis. On a ray with $x = 0$ , $y = ct$ , the already determined longitudinal map gives $x' = -\gamma v t$ and $t' = \gamma t$ . Requiring the transformed ray’s speed to be $c$ gives $\gamma^2 v^2 + b^2 c^2 = \gamma^2 c^2$ , hence $b^2 = 1$ .

Taking the positive root, since at $v = 0$ nothing changes, gives $y' = y$ , and $z' = z$ the same way. The inverse map swaps the primed and unprimed letters and replaces $v$ with $-v$ .

x=\gamma(x'+vt'),\qquad t=\gamma\left(t'+\frac{vx'}{c^2}\right)

The inverse transformation uses plus v and exchanges the two frames.

A check with numbers: at $v = 0.6c$ , $\gamma = 1.25$ , and in units where $c = 1$ the event $(t, x) = (2, 1)$ becomes $(1.75, -0.25)$ . The inverse map takes $(1.75, -0.25)$ back to $(2, 1)$ . The coordinate-map laboratory runs the same check in either direction.

Return to why the sideways coordinates stay unchanged. (included in this file)

Letters: in the paper the moving system's sideways coordinates are η and ζ, and the explanation writes $y'$ and $z'$ . β is the modern γ and V the speed of light, c here. The explanation's primed letters, $x'$ and $t'$ , are the paper's ξ and τ.

Allow a sideways scale factor rather than assume there is none: $y'=b(v)\,y$ .
Follow a ray that moves straight across the motion in K: x = 0 and y = ct.
The longitudinal map, already fixed, gives for that ray $x'=\gamma(0-vt)=-\gamma vt$ and $t'=\gamma(t-0)=\gamma t$ . In k the ray also drifts backwards along $x'$ .
Its sideways position in k is $y'=b\,ct$ . Its speed in k must be c, so after a time $t'$ it has gone $ct'$ : $(\gamma vt)^2+(bct)^2=(c\gamma t)^2$ .
Divide by $t^2$ : $\gamma^2v^2+b^2c^2=\gamma^2c^2$ , so $b^2c^2=\gamma^2(c^2-v^2)$ . Since $\gamma^2=c^2/(c^2-v^2)$ , the right side is $c^2$ , and $b^2=1$ .
Take b = 1, the root that is 1 at rest: $y'=y$ , and the same argument along z gives $z'=z$ .
The paper reaches the same place by its own steps: seen from K, light along the η axis travels at $\sqrt{V^2-v^2}$ , which gives $η=a\frac{V}{\sqrt{V^2-v^2}}y$ , and after the unknown factor φ(v) is fixed at 1, η = y and ζ = z.
The inverse map. Transforming from k back to K is the same kind of transformation with speed −v, since K moves at −v relative to k. So swap the primed and unprimed letters and replace v by −v.

x=\gamma(x'+vt'),\qquad t=\gamma\left(t'+\frac{vx'}{c^2}\right)

The inverse transformation uses plus v and exchanges the two frames.

Worked at v = 0.6c, γ = 1.25, in units where c = 1. Take the event (t, x) = (2, 1). Forward: $x'=1.25(1-0.6\times2)=−0.25$ and $t'=1.25(2-0.6\times1)=1.75$ .
Back again: $x=1.25(-0.25+0.6\times1.75)=1.25\times0.8=1$ and $t=1.25(1.75+0.6\times(-0.25))=1.25\times1.6=2$ . The event returns to (2, 1).
Check a whole event, time and position together. Inverting one coordinate alone can pass while the pair fails.
The paper's own check of consistency uses a third system moving at −v relative to k: going there and back must be the identity, which in its notation is $\varphi(v)\varphi(-v)=1$ .

Return to why the sideways coordinates stay unchanged. (included in this file)

Back to the stationary position

Modern teaching equation, not the equation as printed.

x = \gamma\,\left(x' + v\,t'\right)

The stationary position is gamma times x prime plus v t prime.

The inverse map: seen from the moving system, the stationary system moves at −v, so the sign of v changes and the primed and unprimed coordinates swap places.

Back to the stationary time

Modern teaching equation, not the equation as printed.

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

The stationary time is gamma times t prime plus v x prime over c squared.

The time half of the inverse map, with v replaced by −v. At v = 0.6c the event (t, x) = (2, 1) in units with c = 1 goes to (1.75, −0.25), and this pair of equations brings it back.

Assumptions

The factor along the motion is already fixed at γ.
Light has the same speed in every direction, not only along x.

Limits

Checking a ray along the motion cannot fix the sideways scale.
Checks by determinant or interval confirm the result; they do not replace the physical premises it rests on.

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

Squares and square roots

If the mean square grows four times, why does the typical distance only double?

Both +3 and −3 have square 9, so squaring removes the sign, and it weights a large distance more than a small one. A square root returns to the original unit: the square root of a square micrometre is a micrometre.

The root mean square, or RMS, uses both steps: square each displacement, average the squares, then take the square root. The result is a typical distance from the start, in the original unit.

\sqrt{4q}=2\sqrt q\quad(q\ge 0)

The square root of four q is twice the square root of q, for nonnegative q.

Worked example: Doubling every displacement, and what the RMS does

The displacements −3, −1, +1, +3 have squares 9, 1, 1, 9. Their average is 5, so the RMS is √5, about 2.236.
Double each displacement: −6, −2, +2, +6. The squares are 36, 4, 4, 36, and their average is 20.
The mean square went from 5 to 20, four times as large. The RMS went from about 2.236 to √20, about 4.472: twice as large.

Where this lesson stops

Four times the mean square is twice the RMS. That is why, in the Brownian paper, waiting four times as long doubles the typical distance.

This lesson builds on

Adding and averaging

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

Events and distant clocks

What does it mean to say that two distant events happen at the same time?

An event is something that happens at one place at one instant: a spark, a collision, a flash leaving a lamp. To give it a time, you read a clock at that place. Reading a distant clock by looking at it would mix in the time the light takes to reach you.

A reference frame is a set of rigid measuring rods and clocks at rest, with a clock at every place, so that each event gets three coordinates of position and one of time from the rods and the clock beside it. Before that works, the clocks must agree, and for clocks far apart agreeing needs a rule.

§1 of the relativity paper gives the rule. Send a flash of light from clock A to clock B and back. Note A's reading when it leaves, B's reading when it arrives, and A's reading when it returns. The clocks are called synchronous if the trip out takes as long, by the clocks, as the trip back:

t_B - t_A = t'_A - t_B

t B minus t A equals t A prime minus t B: the trip out takes as long as the trip back.

The paper calls this a definition. It cannot be checked by measurement, because checking it would need clocks that are already synchronous. §1 adds two assumptions: that the rule is consistent, so that if B agrees with A then A agrees with B, and two clocks that agree with a third agree with each other; and that light's speed measured out and back is a universal constant, which the paper writes V.

Worked example: Setting a distant clock with one light signal

Clock A reads 0 s as a flash leaves it for clock B.
The flash is reflected at B and arrives back at A when A reads 10 s.
By the rule the two trips take equal times, so the reflection happened at 5 s on A's time.
If B's clock read 5 s at the reflection, it is synchronous with A. If it read 7 s, it is 2 s fast, and setting it back 2 s makes the two agree.

t_B = \frac{1}{2}\left(t_A + t'_A\right)

The time at B equals one half of the sum of A's reading when the flash left and A's reading when it came back.

The rule needs no distance. If B is known to be 1.5 × 10⁹ metres away, the same round trip also gives light's speed: 2 × 1.5 × 10⁹ m divided by 10 s is 3 × 10⁸ metres per second.

Where this lesson stops

An event is something that happens at one place and one time. Whether two distant events are simultaneous depends on a rule for setting clocks, which §1 of the relativity paper supplies.

This lesson builds on

Functions and graphs

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

Matrices and linear maps

How can a table of four numbers turn one frame's coordinates into another's?

Some rules turn inputs into outputs by multiplying and adding, and nothing else: $x' = p\,x + q\,t$ and $t' = r\,x + s\,t$ , with the same four numbers p, q, r and s for every input. Such a rule is a linear map. Set the four numbers in a table, one row per output, and the table is its matrix. Each row says how much of each input goes into that output.

\begin{aligned} x' &= p\,x + q\,t \\ t' &= r\,x + s\,t \end{aligned}

x prime equals p x plus q t, and t prime equals r x plus s t.

The change from a frame at rest to one moving at v along x is a linear map: $x' = \gamma(x - vt)$ and $t' = \gamma(t - vx/c^2)$ , with $\gamma = 1/\sqrt{1 - v^2/c^2}$ . The relativity paper prints this factor as β; here it is written γ, the modern letter, and the ratio v/c is written out. At v = 0.6c, γ = 1.25 and γv = 0.75c, so the table of multipliers is:

\left(\begin{aligned} 1.25 &\quad -0.75c \\ -0.75/c &\quad 1.25 \end{aligned}\right)

A two by two table: first row 1.25 and minus 0.75 c; second row minus 0.75 over c and 1.25.

Two numbers read off the table describe what the map does. The determinant, ps − qr, is the factor by which it scales areas in the plane of x and t; for this map it is $1.25^2 - 0.75^2 = 1$ , so areas are kept. The eigenvectors are the directions the map only stretches. For this map they are the two light lines, $x = ct$ and $x = -ct$ , stretched by 0.5 and by 2: a flash sent either way still moves at c in the new frame.

The older Galilean rule, $x' = x - vt$ and $t' = t$ , is also linear, with the table $p = 1$ , $q = -v$ , $r = 0$ , $s = 1$ , and its determinant is also 1. What separates the two is the light lines. The Galilean map sends $x = ct$ to $x' = (c - v)t$ : the flash now moves at c − v, not c.

Writing the transformation as a matrix, with determinants and eigenvectors, is a later aid, tied to the geometry Minkowski gave the theory in 1908. §3 of the paper writes the transformation as equations and derives them from light signals and clocks; the table adds a way of reading them, not a premise.

Worked example: Two simultaneous events, seen from 0.6c

Two events happen at the same time, 10 light-seconds apart along x: $\Delta x = 10$ light-seconds and $\Delta t = 0$ .
The first row gives $\Delta x' = 1.25 \times 10 + (-0.75c) \times 0 = 12.5$ light-seconds.
The second row gives $\Delta t' = (-0.75/c) \times 10$ light-seconds $+\ 1.25 \times 0$ . Ten light-seconds divided by c is 10 seconds, so $\Delta t' = -7.5$ s.
In the moving frame the events are not simultaneous: the one farther along x happened 7.5 seconds earlier.

Where this lesson stops

A matrix is a table of multipliers applied to inputs. This lesson stops at two inputs and two outputs.

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-04

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open sr-04 online

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).