Read · Special relativity: from clock operations to electrodynamics

§3 · deriving and inverting the coordinate map

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Letters

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

A derivation

Light constraints leave a scale to determine

What map can respect both light directions?

The 18 printed paragraphs this passage explains

§3, paragraph 1: Two coordinate systems are set up in the stationary space, with their x-axes along one line, each with a rigid measuring rod and clocks, all of them exactly alike.
§3, paragraph 2: One of them, k, is given a constant speed v along the x-axis of the other, K, and keeps its axes parallel to those of K.
§3, paragraph 3: Each system measures space with its own rod and sets its own clocks by the light signals of section 1, giving coordinates x, y, z and time t in K, and ξ, η, ζ and time τ in k.
§3, paragraph 4: To the values x, y, z, t of every event belong its values ξ, η, ζ, τ in k, and the task is to find the equations that connect them.
§3, paragraph 5: Because of the uniformity attributed to space and time, those equations must be linear.
§3, paragraph 6: Einstein writes x' for x minus vt, which stays fixed for a point at rest in k, and first seeks the time τ of k as what k's clocks show once they are synchronized by the rule of section 1.
§3, paragraph 7: A light signal sent along the x-axis in k and reflected back must, by the rule of section 1, be reflected at the halfway time. Writing this with light's speed in K, and letting the distance shrink, gives an equation that τ must satisfy.
§3, paragraph 8: Any other point could have served as the starting point of the signal, so that equation holds throughout k.
§3, paragraph 9: Light sent along the other two axes shows that τ does not depend on y or z, so τ is an unknown factor times a simple combination of t and x'.
§3, paragraph 10: Requiring that light also travels at V when measured in k, Einstein follows a ray along the ξ-axis and finds ξ as the same unknown factor times a multiple of x'.
§3, paragraph 11: Rays along the other two axes give η and ζ in the same way. Putting back x minus vt for x' yields the transformation up to a still unknown function φ of v, with the factor Einstein calls β, which grows as v approaches V.
§3, paragraph 12: It remains to prove that light also travels at V measured in k, since the two principles have not yet been shown to be compatible.
§3, paragraph 13: A spherical light wave is sent out from the shared origin at the moment both systems' clocks read zero; in K its front is a sphere growing at speed V.
§3, paragraph 14: Transformed into the coordinates of k, the same wavefront is again a sphere growing at speed V.
§3, paragraph 15: So the wave is spherical with speed V in the moving system too, and the two principles are compatible.
§3, paragraph 16: The transformation still contains an unknown function φ of v, which Einstein now sets out to determine.
§3, paragraph 17: He brings in a third system, moving with speed minus v relative to k, and applies the transformation twice.
§3, paragraph 18: Going to k and back must change nothing, so φ(v) times φ(minus v) is 1. A rod lying across the motion shows that φ fixes its length seen from K, which cannot depend on the direction of motion, so φ(v) is 1; this leaves the final transformation equations.

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)\,\left(x - v\,t\right)

\xi = a\,\left(x - v\,t\right)

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

\tau = a\,\left(t - \frac{v\,x}{V^{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 - \frac{v^{2}}{c^{2}}}}

\beta = \frac{1}{\sqrt{1 - \frac{v^{2}}{V^{2}}}}

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

\xi = \beta\,\left(x - v\,t\right)

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

\tau = \beta\,\left(t - \frac{v\,x}{V^{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.

Open the foundation: Matrices and linear maps

Open the foundation: Events and distant clocks

Show every step here: Light constraints leave a scale to determine

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)\,\left(x - v\,t\right)

\xi = a\,\left(x - v\,t\right)

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

\tau = a\,\left(t - \frac{v\,x}{V^{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 - \frac{v^{2}}{c^{2}}}}

\beta = \frac{1}{\sqrt{1 - \frac{v^{2}}{V^{2}}}}

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

\xi = \beta\,\left(x - v\,t\right)

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

\tau = \beta\,\left(t - \frac{v\,x}{V^{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.

A tool for this step

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.

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

Try it: a change of frame as a table of multipliers

Choose the speed of the new frame as a fraction of c, and which rule to use. The table turns four events, measured in light-seconds and seconds, into the new frame's numbers.

Speed of the new frame, v/c

Type a speed and press Enter, or drag the slider from −0.9 to 0.9. A minus sign means the frame moves the other way.

x′ = 1.25 x − 0.75 t
t′ = −0.75 x + 1.25 t

γ = 1.25. Determinant 1. A flash from the origin still moves at c in the new frame, and the two light lines are stretched by 0.5 and 2.

Four events, before and after

The origin: (x, t) = (0, 0) becomes (x′, t′) = (0, 0).
Same moment, 10 light-seconds away: (x, t) = (10, 0) becomes (x′, t′) = (12.5, −7.5).
Same place, 10 seconds later: (x, t) = (0, 10) becomes (x′, t′) = (−7.5, 12.5).
Where a flash from the origin is after 10 s: (x, t) = (10, 10) becomes (x′, t′) = (5, 5).

The first two events happen at the same moment in the old frame. In the new frame the one 10 light-seconds away happens 7.5 seconds earlier than the one at the origin.

What it shows, in words

Each row of the table makes one new coordinate from the old ones by multiplying and adding. For the relativistic rule at 0.6c the factor γ is 1.25, the determinant is 1, and a flash of light still moves at c in the new frame; two events that were simultaneous are 7.5 seconds apart. The Galilean rule also has determinant 1 and keeps simultaneous events simultaneous, but it changes the speed of the flash. Writing the change of frame as a table is a later aid: the 1905 paper writes it as two equations.

Where this lesson stops

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

A tool for this step

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.

One flash, out and back

A flash leaves clock A, turns round at clock B and comes back. By the rule the two trips take equal times, so the turn-round happened half way between leaving and returning: at 5 on A's clock.

B reads 5 at the turn-round, which is what A's clock says the moment was. The two are synchronous.

Now the same flash, against a clock that is running ahead.

B reads 7 where the rule says 5, so it is 2 seconds fast. Set it back 2 and the two agree. Nothing about the flash changed between the drawings; only what B's hands were showing.

What it shows, in words

Two clocks stand apart, drawn as vertical lines with time running down the page. A flash leaves A when A reads 0, reaches B, is reflected, and returns to A when A reads 10. The rule says the trip out took as long as the trip back, so the turn-round was half way between: 5 by A's clock. In the first drawing B reads 5 at that moment and the clocks are synchronous. In the second B reads 7, which is 2 seconds more, so B is 2 seconds fast and setting it back 2 makes them agree.

The same round trip also measures light, if the distance is known separately: with B at 1.5 × 10⁹ metres, a there-and-back of 10 seconds gives 3 × 10⁸ metres per second.

What it does not show, and this is the lesson's own point rather than a limitation of the drawing: the rule is a definition, and no picture can show it being tested. Checking that the trip out really took as long as the trip back would need two clocks already synchronised, which is the thing the rule exists to supply. Nor is there a second observer anywhere here. Only one frame is drawn, so nothing in it bears on how someone moving past would time the same three events; that question needs the paper's §2, and the geometric aid usually reached for is Minkowski's of 1908, which is later than the argument and not its presentation.

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.

Explore the equations in this step

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

Explore the equation · Modern model notationEinstein's letters

The scale, fixed

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

\beta = \frac{1}{\sqrt{1 - \frac{v^{2}}{V^{2}}}}

is one over the square root of one minus squared over squared.

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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.

Model assumptions and every term’s meaning

The moving system k moves along the x axis of the stationary system K at speed v, with |v| < c, and the two origins coincide at time zero.
Modern teaching notation. The paper writes the moving system's coordinates as xi and tau, not x′ and t′, its gamma as beta, and the speed of light as V.
The unknown factor is the same for v and −v (no direction in space is special) and equals 1 at v = 0.

One over it: Dividing 1 by a number below 1 gives a number above 1. At v = 0.6c the root is 0.8 and the factor 1.25. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
Its square root: The positive root, so that the factor is 1 at rest. Read the prerequisite
One minus that: Positive for every speed below c, zero at c, so no system moves at c. Read the prerequisite
How close to light speed: v squared over c squared is 0 at rest and approaches 1 as v approaches c. Read the prerequisite
What it defines: The value of the common factor. From here on the transformation has no unknowns. Read the prerequisite
v squared: Squaring removes the direction: moving either way gives the same factor. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Speed of light: The paper's V. It enters because both systems must find the same speed for light. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. Read the prerequisite

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

Explore the equation · Modern model notationEinstein's letters

Position in the moving system, up to a scale

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

\xi = a\,\left(x - v\,t\right)

is times minus times .

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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.

Model assumptions and every term’s meaning

The moving system k moves along the x axis of the stationary system K at speed v, with |v| < c, and the two origins coincide at time zero.
Modern teaching notation. The paper writes the moving system's coordinates as xi and tau, not x′ and t′, its gamma as beta, and the speed of light as V.

What it asserts: The moving position is proportional to the distance from the moving origin. The light conditions have not yet said by how much. Read the prerequisite
Times the unknown factor: The same factor multiplies the time equation below; a single scale remains to be found. Read the prerequisite
Distance from the moving origin: x minus vt is how far the event is from the moving system's origin, measured in the stationary system. Read the prerequisite
Where the moving origin is: At time t the moving system's origin has travelled vt along x. Read the prerequisite
Position in the stationary system: Where the event is along x, in the stationary system's coordinates. Read the prerequisite
Position in the moving system: Where the same event is along x, in the moving system's coordinates. The paper calls it xi. Read the prerequisite
The factor not yet fixed: The light conditions fix everything but this common scale. The paper carries it as phi(v) times beta and then shows phi(v) is 1. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. Read the prerequisite
Time in the stationary system: When the event happens, by the stationary system's synchronized clocks. Read the prerequisite

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

Explore the equation · Modern model notationEinstein's letters

Time in the moving system, up to the same scale

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

\tau = a\,\left(t - \frac{v\,x}{V^{2}}\right)

is times minus times over squared.

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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.

Model assumptions and every term’s meaning

The moving system k moves along the x axis of the stationary system K at speed v, with |v| < c, and the two origins coincide at time zero.
Modern teaching notation. The paper writes the moving system's coordinates as xi and tau, not x′ and t′, its gamma as beta, and the speed of light as V.

A time that depends on place: vx over c squared is a time: distant clocks in the moving system are set differently. This term is where simultaneity stops being shared. Read the prerequisite
c squared: Divided by the square of the speed of light, the correction is small for everyday v and x. Read the prerequisite
What it asserts: Once light's speed is the same both ways in both systems, only one number, the common factor, remains unknown. Read the prerequisite
Times the same unknown factor: The factor that multiplies the position equation. Read the prerequisite
t corrected for position: Subtracting that place-dependent time from t. Read the prerequisite
Speed times position: v times x: events farther along x get a larger correction. Read the prerequisite
Speed of light: The paper's V. It enters because both systems must find the same speed for light. Read the prerequisite
Position in the stationary system: Where the event is along x, in the stationary system's coordinates. Read the prerequisite
The factor not yet fixed: The light conditions fix everything but this common scale. The paper carries it as phi(v) times beta and then shows phi(v) is 1. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. Read the prerequisite
Time in the stationary system: When the event happens, by the stationary system's synchronized clocks. Read the prerequisite
Time in the moving system: When the same event happens, by the moving system's synchronized clocks. The paper calls it tau; it is a coordinate, not the time one clock records. Read the prerequisite

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

Explore the equation · Modern model notationEinstein's letters

Position in the moving system

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

\xi = \beta\,\left(x - v\,t\right)

is times minus times .

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

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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.

Model assumptions and every term’s meaning

The moving system k moves along the x axis of the stationary system K at speed v, with |v| < c, and the two origins coincide at time zero.
Modern teaching notation. The paper writes the moving system's coordinates as xi and tau, not x′ and t′, its gamma as beta, and the speed of light as V.

What it asserts: How the moving system locates an event, given where and when the stationary system locates it. Read the prerequisite
Scaled by gamma: At everyday speeds gamma is almost exactly 1 and this is the familiar x − vt. Read the prerequisite
Distance from the moving origin: x minus vt, measured in the stationary system. Read the prerequisite
Where the moving origin is: At time t the moving system's origin has travelled vt along x. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Position in the stationary system: Where the event is along x, in the stationary system's coordinates. Read the prerequisite
Position in the moving system: Where the same event is along x, in the moving system's coordinates. The paper calls it xi. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. Read the prerequisite
Time in the stationary system: When the event happens, by the stationary system's synchronized clocks. Read the prerequisite

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

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Time in the moving system

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

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

is times minus times over squared.

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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.

Model assumptions and every term’s meaning

The moving system k moves along the x axis of the stationary system K at speed v, with |v| < c, and the two origins coincide at time zero.
Modern teaching notation. The paper writes the moving system's coordinates as xi and tau, not x′ and t′, its gamma as beta, and the speed of light as V.

A time that depends on place: vx over c squared is a time, and it differs from place to place: simultaneity is not shared between the systems. Read the prerequisite
c squared: Divided by the square of the speed of light, the correction is small for everyday v and x. Read the prerequisite
What it asserts: How the moving system's clocks time an event, given where and when the stationary system locates it. Read the prerequisite
Scaled by gamma: The same factor as in the position equation. Read the prerequisite
t corrected for position: Subtracting that place-dependent time from t. Read the prerequisite
Speed times position: v times x: events farther along x get a larger correction. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Speed of light: The paper's V. It enters because both systems must find the same speed for light. Read the prerequisite
Position in the stationary system: Where the event is along x, in the stationary system's coordinates. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. Read the prerequisite
Time in the stationary system: When the event happens, by the stationary system's synchronized clocks. Read the prerequisite
Time in the moving system: When the same event happens, by the moving system's synchronized clocks. The paper calls it tau; it is a coordinate, not the time one clock records. Read the prerequisite

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

Modern qualifications

Einstein’s own §3 derivation follows a light signal out and back between clocks and solves a differential equation for the moving system’s time $\tau$ ; the two-direction argument here is a shorter modern route to the same map. His $\beta$ is the modern $\gamma$ , and his $\tau$ is a coordinate time, not the proper time of modern texts.

Assumptions and limits: Light constraints leave a scale to determine

Assumed here

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.

What this does not establish

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.

Earlier step: Time at a distant clock is an operation One apparatus, two descriptions

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

A derivation

The sideways step needs its own condition

Why do the transverse coordinates stay unchanged?

The 3 printed paragraphs this passage explains

§3, paragraph 11: Rays along the other two axes give η and ζ in the same way. Putting back x minus vt for x' yields the transformation up to a still unknown function φ of v, with the factor Einstein calls β, which grows as v approaches V.
§3, paragraph 17: He brings in a third system, moving with speed minus v relative to k, and applies the transformation twice.
§3, paragraph 18: Going to k and back must change nothing, so φ(v) times φ(minus v) is 1. A rod lying across the motion shows that φ fixes its length seen from K, which cannot depend on the direction of motion, so φ(v) is 1; this leaves the final transformation equations.

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

x = \beta\,\left(\xi + v\,\tau\right)

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

t = \beta\,\left(\tau + \frac{v\,\xi}{V^{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.

Open the foundation: Squares and square roots

Show every step here: The sideways step needs its own condition

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

x = \beta\,\left(\xi + v\,\tau\right)

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

t = \beta\,\left(\tau + \frac{v\,\xi}{V^{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 $\varphi(v)\varphi(-v)=1$ .

A tool for this step

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.

Four times the area, twice the side

Each cell is the mean square of the four displacements, 5. The big square is the mean square after every displacement is doubled, 20. Four cells fit in it, and its side is two cell sides.

What it shows, in words

The displacements -3, -1, 1, 3 have squares 9, 1, 1, 9, and their average is 5. That is one cell. Doubling every displacement gives a mean square of 20, which is the whole square, and four cells fill it exactly.

The cell’s side is 2.236 and the whole square’s side is 4.472. Four times the area, twice the side: the side did not grow by four, because a side is a length and the 20 is an area. That is the whole difference between a mean square and a root mean square, and it is why waiting four times as long in the Brownian paper doubles the typical distance rather than quadrupling it.

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.

Explore the equations in this step

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

Explore the equation · Modern model notationEinstein's letters

Back to the stationary position

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

x = \beta\,\left(\xi + v\,\tau\right)

is times plus times .

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Each term and operation, in words

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Read the equation aloud in words

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.

Model assumptions and every term’s meaning

The moving system k moves along the x axis of the stationary system K at speed v, with |v| < c, and the two origins coincide at time zero.
Modern teaching notation. The paper writes the moving system's coordinates as xi and tau, not x′ and t′, its gamma as beta, and the speed of light as V.
The inverse is the same map with v replaced by −v and the two systems' coordinates exchanged.

What it asserts: Applying the forward map and then this one returns every event where it started. Read the prerequisite
Scaled by the same gamma: Gamma depends on v squared, so reversing v leaves it unchanged. Read the prerequisite
x′ plus vt′: The moving coordinates combined with the sign of v reversed. Read the prerequisite
Where the stationary origin is, seen from k: In the moving system the stationary origin moves at −v, which is why the sum has a plus. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Position in the stationary system: Where the event is along x, in the stationary system's coordinates. Read the prerequisite
Position in the moving system: Where the same event is along x, in the moving system's coordinates. The paper calls it xi. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. Read the prerequisite
Time in the moving system: When the same event happens, by the moving system's synchronized clocks. The paper calls it tau; it is a coordinate, not the time one clock records. Read the prerequisite

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

Explore the equation · Modern model notationEinstein's letters

Back to the stationary time

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

t = \beta\,\left(\tau + \frac{v\,\xi}{V^{2}}\right)

is times plus times over squared.

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

Explain this equation

Each term and operation, in words

No live values

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

Read the equation aloud in words

The 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.

Model assumptions and every term’s meaning

The moving system k moves along the x axis of the stationary system K at speed v, with |v| < c, and the two origins coincide at time zero.
Modern teaching notation. The paper writes the moving system's coordinates as xi and tau, not x′ and t′, its gamma as beta, and the speed of light as V.
The inverse is the same map with v replaced by −v and the two systems' coordinates exchanged.

The place-dependent time, reversed: Added rather than subtracted, because the stationary system moves at −v as seen from k. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
What it asserts: How the stationary system times an event, given the moving system's coordinates for it. Read the prerequisite
Scaled by the same gamma: Unchanged by reversing v. Read the prerequisite
t′ corrected for position: The moving time plus the place-dependent term. Read the prerequisite
Speed times moving position: v times x′. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared: 1 at rest, larger at any speed, without limit as v approaches c. The paper writes it as beta. Read the prerequisite
Speed of light: The paper's V. It enters because both systems must find the same speed for light. Read the prerequisite
Position in the moving system: Where the same event is along x, in the moving system's coordinates. The paper calls it xi. Read the prerequisite
Speed of the moving system: The speed of k relative to K. Its sign gives the direction. Read the prerequisite
Time in the stationary system: When the event happens, by the stationary system's synchronized clocks. Read the prerequisite
Time in the moving system: When the same event happens, by the moving system's synchronized clocks. The paper calls it tau; it is a coordinate, not the time one clock records. Read the prerequisite

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

Modern qualifications

A modern reader can check the result by confirming that $c^2t^2 - x^2 - y^2 - z^2$ is unchanged, or that the map’s determinant is 1. Those are consequences of the map; the 1905 argument does not start from them.

Assumptions and limits: The sideways step needs its own condition

Assumed here

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

What this does not establish

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.

Earlier step: Light constraints leave a scale to determine

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

References

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

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

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

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

28 foundation readings sit behind this argument.

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