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Special relativity: from clock operations to electrodynamics

Clocks, rods and light first, then electricity and magnetism: follow the relativity paper through its introduction and all ten sections, including the later ones most retellings skip.

Draft explanation, not yet reviewed

Newly authored explanatory preview in modern notation, with editorial and physics review pending. Both the kinematic and electrodynamic halves are treated, but this is not a German transcription, an aligned translation, or a complete critical edition. The source paragraphs, footnotes, acknowledgment, and date-lines are not claimed to be fully represented or reviewed here. The headings and the way the argument is divided into passages are ours: Einstein divided his paper into an introduction and ten numbered sections in two parts, under his own titles.

Show me one example before the notation →

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

First encounter · No algebra required

What should the distant clock read when the flash reflects?

Two stations, A and B, stand apart along a straight track. Each has a clock. A flash leaves A when its clock reads 0, reflects at B, and returns to A when its clock reads 10. What should the clock at B have read at the reflection?

An authored clock-setting example. The readings below are not experimental evidence.

What would you say?

Open any answer to examine its assumptions. Nothing must be answered to continue.

0

This uses the departure reading. It would assign no time to the outward trip. Nothing in the two readings requires that assumption.

5

This is the halfway reading. It follows once we agree to assign equal times to the two directions. Without that agreement, the two readings alone do not force it.

10

This uses the return reading at A. It would assign no time to the return trip. Receiving news of an event is not the event itself.

Nothing tells us yet

Exactly: the two local readings do not yet specify how the distant clock should be set. We first need to state a procedure. This is not a failure to find the answer.

State the agreement before assigning the reading

If we agree that the outward trip takes as long as the return trip, assign the reflection the reading halfway between departure and return. This is a stated clock-setting procedure, not a measurement of the separate one-way travel times.

Choose whether to use the equal-travel-time agreement
Change the two local readings, in seconds
Or adjust the accepted round-trip duration

The slider offers durations from 2 to 40 seconds. The editable readings above also admit other durations.

Readings in seconds. The distant entry is an assignment, not a local observation at A.
Event and whose reading it isReading
Flash leaves A: A’s clock0
Flash reflects at B: proposed assignment to B’s clockNot assigned before an agreement
Flash returns to A: A’s clock10

Leaves at 0, returns at 10. No distant reading has been assigned because no agreement is selected.

ABOutward flashReturning flash
A path sketch, not a scale drawing or an animation measuring travel time. The complete readings are in the table.

Load the accepted round trip in the clock laboratory →

The laboratory uses the stated agreement and an ideal stationary separation that reproduces this round trip. No measured separation or independently measured one-way speed is inferred here.

The whole worked route, without controls

Try a second flash leaving at 20 and returning at 30. The same agreement gives a new reading for its reflection, without changing the rule. Both tables here use the equal-travel-time agreement.

Leaves at 0, returns at 10

Readings in seconds. The distant entry is an assignment, not a local observation at A.
Event and whose reading it isReading
Flash leaves A: A’s clock0
Flash reflects at B: proposed assignment to B’s clock5
Flash returns to A: A’s clock10

Leaves at 20, returns at 30

Readings in seconds. The distant entry is an assignment, not a local observation at A.
Event and whose reading it isReading
Flash leaves A: A’s clock20
Flash reflects at B: proposed assignment to B’s clock25
Flash returns to A: A’s clock30

In the first example, the reflection is assigned 5 while its returning signal reaches A at 10. Those are different events. In the second, the reflection is assigned 25 and the signal returns at 30.

What is getting in the way?

Is the event the same as seeing it?

No. The reflection happens at B. A receives the returning signal later. The time of reception at A is not the reading assigned to the earlier reflection at B.

Why agree on a rule?

The departure and return readings use one clock, at A. They do not by themselves tell us how a distant clock at B is set. The equal-travel-time rule supplies that missing agreement.

Does this measure the two one-way travel times?

No. The round-trip readings are the starting observations. Assigning equal time to each direction is a stated procedure, not a separate measurement of each one-way time.

Show me the second flash again →

Next: another pair of clocks glides past

Suppose a moving pair of stations uses the same agreement with its own flashes. Will it agree with us about which distant events happen at the same moment?

Set a distant clock by an agreed procedure instead of assuming it already agrees with yours.

Section 1 defines what “at the same time” means for distant clocks by this agreement. An event ledger keeps the resulting readings separate from the times when signals arrive.

More guidance: an event is not the moment you see it →

Compare a pair moving at sixty percent of light speed →

Continue to the section 1 clock-setting argument →

The linked reading is a draft explanation. A reviewed sentence-aligned original and translation are still in preparation.

Introduction · magnet, conductor, and the postulates

An assumption

One apparatus, two descriptions

Why should changing the observer change the explanation?

The 4 printed paragraphs this passage explains
  • Introduction, paragraph 1: Maxwell's electrodynamics, as it was usually understood, treats a magnet moving past a resting conductor differently from a conductor moving past a resting magnet, although the current observed depends only on their relative motion. Einstein calls this an asymmetry that the phenomena themselves do not seem to have.
  • Introduction, paragraph 2: Such examples, and the failed attempts to detect the earth's motion through the light medium, lead Einstein to conjecture that absolute rest has no counterpart in electrodynamics either. He raises this principle of relativity to a postulate, adds that light in empty space always travels at one speed V whatever the motion of its source, and says a light ether will prove superfluous.
  • Introduction, paragraph 3: Every electrodynamics rests on the kinematics of rigid bodies, since its statements concern rigid bodies used as coordinate systems, clocks and electromagnetic processes. Neglecting this, Einstein says, is the root of the present difficulties.
  • §6, paragraph 7: So the asymmetry noted in the introduction, between a moving magnet and a moving conductor, disappears, and the question of where the electromotive force is seated, as in unipolar machines, loses its point.

The German text of these paragraphs is not on this site yet, so they have no links.

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 VV, 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 γ\gamma derived in §6, which is how a force across the motion changes between frames.

Show every step here: One apparatus, two descriptions

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Modern qualifications

This page reconstructs the argument in modern words; it is not a translation of the paper’s opening. Whether the Michelson–Morley experiment shaped Einstein’s own route is a historical question with its own evidence; the paper itself cites only the unsuccessful attempts to detect the earth’s motion through the ether.

Assumptions and limits: One apparatus, two descriptions

Assumed here

  • 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.

What this does not establish

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

Source context: Facsimile

§1 · operational simultaneity

A definition

Time at a distant clock is an operation

How can clocks at different places share a time coordinate?

The 10 printed paragraphs this passage explains
  • §1, paragraph 1: Einstein starts from a coordinate system in which Newton's equations of mechanics hold, and calls it the stationary system.
  • §1, paragraph 2: The position of a point at rest in that system can be found with rigid measuring rods by the methods of Euclidean geometry.
  • §1, paragraph 3: Describing a motion means giving positions as functions of time, which has a meaning only once it is said what time is. Every statement about time is a statement about simultaneous events, such as a train arriving when the small hand of a watch points to 7.
  • §1, paragraph 4: Reading time from the hand of one clock works only at the place of that clock. It does not connect events that happen at other places, far from the clock.
  • §1, paragraph 5: An observer at the clock could assign each distant event the time at which its light signal arrives, but that assignment depends on where the observer stands. Einstein looks for a more practical rule.
  • §1, paragraph 6: Clocks at A and at B each give a time only for events near them. Einstein defines a common time by declaring that light takes as long to go from A to B as to come back, so the two clocks are synchronous when the outward and the return times agree.
  • §1, paragraph 7: He assumes this definition can be applied without contradiction to any number of clocks: if the clock at B is synchronous with the clock at A, the one at A is synchronous with the one at B, and two clocks synchronous with a third are synchronous with each other.
  • §1, paragraph 8: This fixes what synchronous clocks at rest in different places are, and with it what simultaneous and time mean: the time of an event is the reading, at that event, of a resting clock at its place that is synchronous with one chosen clock.
  • §1, paragraph 9: In agreement with experience, twice the distance between two clocks divided by the time light takes to go there and back is taken to be a universal constant V, the speed of light in empty space.
  • §1, paragraph 10: Because this time is defined by clocks at rest in the stationary system, Einstein calls it the time of the stationary system.

The German text of these paragraphs is not on this site yet, so they have no links.

tB−tA=tA,return−tBt_B - t_A = t_{A,\mathrm{return}} - t_B

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

The outward assigned travel time equals the return assigned travel time.

tB=tA+tA,return2t_B = \frac{t_A + t_{A,\mathrm{return}}}{2}

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

The assigned reflection time is the midpoint of departure and return times.

If clock A sends a flash at time 0 and the reflection comes back at time 10, the rule says the flash reached B at time 5. Seeing B later, when its light arrives, is a separate event and does not change that assignment.

The rule assumes light takes as long to go as to come back, and Einstein states that as a definition rather than a measurement. It sets clocks within one frame; §2 shows that clocks in another frame, set by the same rule, disagree about which distant events happen at the same time. The clock-synchronization laboratory keeps departure, reflection, return and each clock reading as separate records.

Show every step here: Time at a distant clock is an operation

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Modern qualifications

Philosophers since Reichenbach (1928) have debated how much of the equal-travel-time rule is a convention rather than a fact; Einstein calls it a definition. Either way, simultaneity set this way belongs to one frame, which is the point of §2.

Assumptions and limits: Time at a distant clock is an operation

Assumed here

  • The clocks and the endpoints are at rest in one inertial frame.
  • The rule assumes light takes as long to go as to come back.

What this does not establish

  • A round trip read on one clock cannot measure the two one-way times separately.
  • Setting clocks does not move them or change their rates.

Source context: Facsimile

§2 · lengths and measurement events

A definition

A length is a specified pair of events

Which endpoint measurements define a moving rod’s length?

The 9 printed paragraphs this passage explains
  • §2, paragraph 1: Einstein states the two principles the argument uses: the laws by which physical systems change are the same in any two systems in uniform translation relative to each other, and every light ray moves in the stationary system at the speed V whether its source is at rest or moving.
  • §2, paragraph 2: A rigid rod of length l is set moving along the x-axis at speed v. Its length can be found in two ways: by an observer who moves with it and lays a measuring rod beside it, or by finding where its two ends are at one given time with the synchronized clocks of the stationary system.
  • §2, paragraph 3: By the principle of relativity, the first way, measuring alongside the moving rod, must give its length at rest, l.
  • §2, paragraph 4: The second way gives what Einstein calls the length of the moving rod in the stationary system, and he will find that it differs from l.
  • §2, paragraph 5: The usual kinematics silently assumes the two lengths are equal, that is, that a moving rigid body at a given moment can be replaced, geometrically, by the same body at rest.
  • §2, paragraph 6: Clocks are fixed to both ends of the rod and set to agree with the stationary system's clocks wherever they happen to be.
  • §2, paragraph 7: Observers moving with the rod apply the light-signal test of section 1 to its two clocks. Because the rod moves, light takes longer to go one way along it than to come back, so these observers find the clocks out of step, while observers in the stationary system call them synchronous.
  • §2, footnote 1: A footnote says that time here means both the time of the stationary system and the reading of the moving clock at the place in question.
  • §2, paragraph 8: So simultaneity has no absolute meaning: two events that are simultaneous seen from one system are not simultaneous seen from a system moving relative to it.

The German text of these paragraphs is not on this site yet, so they have no links.

A ruler travelling with the rod gives its rest length. An observer the rod moves past has to note where both ends are at the same moment of that observer’s time; marking one end now and the other a little later adds the distance the rod moved in between.

Differences between two events transform by the same map as the events themselves. So a pair with Δt=0\Delta t = 0 and Δx≠0\Delta x \ne 0 has Δt′=−γv Δx/c2\Delta t' = -\gamma v\,\Delta x/c^2, which is not zero for any v≠0v \ne 0: what one frame calls simultaneous, the other does not.

Δt′=−γ v Δxc2\Delta t' = -\frac{\gamma\,v\,\Delta x}{c^{2}}

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

Events simultaneous in the unprimed frame have a primed time difference of minus gamma v delta x over c squared.

In the simultaneity laboratory you pick the two end events yourself and see which frame calls them simultaneous. A pair chosen in the wrong frame gives a larger separation, and that does not mean the rod got longer.

Show every step here: A length is a specified pair of events

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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 a reviewed transcription.

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Modern qualifications

Einstein’s §2 argues with light signals and clocks and reaches the relativity of simultaneity without the transformation, which he derives only in §3. The formula here borrows §3’s result to put a number on it.

Assumptions and limits: A length is a specified pair of events

Assumed here

  • The rod moves at constant velocity along the direction of motion and keeps its rest length.
  • Each frame reads time on its own synchronized clocks.

What this does not establish

  • The measured length is not what a photograph of the moving rod would show.
  • Choosing a different pair of end events for each frame is the point of the measurement, not a correction to it.

Earlier step: Time at a distant clock is an operation

Source context: Facsimile

§3 · deriving and inverting the coordinate map

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 German text of these paragraphs is not on this site yet, so they have no links.

The moving origin follows x=vtx = vt, so write x′=a(v)(x−vt)x' = a(v)(x - vt). Requiring x′=ct′x' = ct' on x=ctx = ct and x′=−ct′x' = -ct' on x=−ctx = -ct fixes the time combination: t′=a(v)(t−vx/c2)t' = a(v)(t - vx/c^2). An overall scale a(v)a(v) remains.

x′=a(v) (x−v t)x' = a(v)\,\left(x - v\,t\right)
ξ=a (x−v t)\xi = a\,\left(x - v\,t\right)
t′=a(v) (t−v xc2)t' = a(v)\,\left(t - \frac{v\,x}{c^{2}}\right)
τ=a (t−v xV2)\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 vv and then −v-v, must return every event unchanged, which gives a(v) a(−v) (1−v2/c2)=1a(v)\,a(-v)\,(1 - v^2/c^2) = 1. Space has no preferred direction, so a(v)=a(−v)a(v) = a(-v), and since a(0)=1a(0) = 1 the positive root is the one: a(v)=γa(v) = \gamma.

γ=11−v2c2\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}
β=11−v2V2\beta = \frac{1}{\sqrt{1 - \frac{v^{2}}{V^{2}}}}
x′=γ (x−v t)x' = \gamma\,\left(x - v\,t\right)
ξ=β (x−v t)\xi = \beta\,\left(x - v\,t\right)
t′=γ (t−v xc2)t' = \gamma\,\left(t - \frac{v\,x}{c^{2}}\right)
τ=β (t−v xV2)\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.

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

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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 τ\tau; the two-direction argument here is a shorter modern route to the same map. His β\beta is the modern γ\gamma, and his τ\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 operationOne apparatus, two descriptions

Source context: 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.

The German text of these paragraphs is not on this site yet, so they have no links.

Write y′=b(v) yy' = b(v)\,y for a transverse axis. On a ray with x=0x = 0, y=cty = ct, the already determined longitudinal map gives x′=−γvtx' = -\gamma v t and t′=γtt' = \gamma t. Requiring the transformed ray’s speed to be cc gives γ2v2+b2c2=γ2c2\gamma^2 v^2 + b^2 c^2 = \gamma^2 c^2, hence b2=1b^2 = 1.

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

x=γ (x′+v t′)x = \gamma\,\left(x' + v\,t'\right)
x=β (ξ+v τ)x = \beta\,\left(\xi + v\,\tau\right)
t=γ (t′+v x′c2)t = \gamma\,\left(t' + \frac{v\,x'}{c^{2}}\right)
t=β (τ+v ξV2)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=0.6cv = 0.6c, γ=1.25\gamma = 1.25, and in units where c=1c = 1 the event (t,x)=(2,1)(t, x) = (2, 1) becomes (1.75,−0.25)(1.75, -0.25). The inverse map takes (1.75,−0.25)(1.75, -0.25) back to (2,1)(2, 1). The coordinate-map laboratory runs the same check in either direction.

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

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Modern qualifications

A modern reader can check the result by confirming that c2t2−x2−y2−z2c^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: Facsimile

§4 · rods, clocks, and the limit of the model

A derivation

Use the same measurement protocol

What do a moving clock and rod actually report?

The 6 printed paragraphs this passage explains
  • §4, paragraph 1: A rigid sphere of radius R at rest in the moving system is, seen from the stationary system at one moment, an ellipsoid whose axis along the motion is shortened.
  • §4, footnote 1: A footnote specifies that the sphere means a body that has the shape of a sphere when examined at rest.
  • §4, paragraph 2: Dimensions across the motion are unchanged, while the one along it is shortened, the more so the faster the body moves. At v = V moving bodies would shrink to flat figures, speeds above V make the considerations meaningless, and in the theory the speed of light plays the part of an infinitely great speed.
  • §4, paragraph 3: The same results hold the other way round, for bodies at rest in the stationary system viewed from a uniformly moving one.
  • §4, paragraph 4: Einstein asks how fast a clock at rest at the origin of the moving system runs, as judged from the stationary system.
  • §4, paragraph 5: Using the transformation at the clock's place, he finds that it falls behind by a fixed fraction of a second each second, to a first approximation half of (v/V) squared.

The German text of these paragraphs is not on this site yet, so they have no links.

Δt=γ Δτ\Delta t = \gamma\,\Delta\tau
L=L0γL = \frac{L_0}{\gamma}

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

Coordinate time between a moving clock’s ticks is gamma times its proper elapsed time; simultaneous moving-rod length is rest length over gamma.

At 0.6c0.6c the factor is 0.8 for both: the moving clock records 0.8 s while the resting frame’s clocks advance 1 s, and a rod 1 m long at rest measures 0.8 m. The same number comes out of two different measurements.

For the clock, the two ticks happen at one place in the clock’s own frame. For the rod, the two end marks happen at one time in the measuring frame. The moving-clock and simultaneity laboratories let you set up each measurement on its own.

1−1γ≈12 v2c21 - \frac{1}{\gamma} \approx \frac{1}{2}\,\frac{v^{2}}{c^{2}}
1−1β≈12 v2V21 - \frac{1}{\beta} \approx \frac{1}{2}\,\frac{v^{2}}{V^{2}}

At small speed, the clock’s fractional loss per coordinate second approaches one half v squared over c squared.

Show every step here: Use the same measurement protocol

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Modern qualifications

The light clock, a flash bouncing between two mirrors, is a later teaching device; the 1905 paper derives these results from the transformation. Einstein prints the moving clock’s lag only to second order, 12t v2/V2\frac{1}{2}t\,v^2/V^2; at 0.6c0.6c that gives 0.18 of each second where the exact value is 0.2.

Assumptions and limits: Use the same measurement protocol

Assumed here

  • The clock is ideal and the rod rigid, both moving at constant velocity, and each measurement uses the right pair of events.
  • γ is computed from the relative speed of the two frames.

What this does not establish

  • Each of two clocks in relative motion runs slow as judged from the other, because the two judgments use different distant clocks.
  • Transforming one pair of events does not by itself give a length or a clock reading; the pair has to be the right one.

Earlier step: Light constraints leave a scale to determine

Source context: Facsimile

A qualification

A reunion compares whole paths

Can reciprocal rate descriptions settle a reunion?

The 3 printed paragraphs this passage explains
  • §4, paragraph 6: If, of two synchronized clocks at A and B, the one at A is carried to B at speed v, on arrival it lags behind the clock at B by about half of t times (v/V) squared seconds, where t is the time the journey took.
  • §4, paragraph 7: The result still holds when the clock travels from A to B along any path made of straight pieces, even when A and B are the same point.
  • §4, paragraph 8: Assuming the result also holds for a smooth closed curve, a clock sent round such a curve and back to A lags behind the one that stayed. Einstein concludes that a balance-wheel clock at the equator must run very slightly slower than an identical clock at a pole.

The German text of these paragraphs is not on this site yet, so they have no links.

Δτ=∫t0t11−u(t)2c2 dt\Delta\tau = \int_{t_{0}}^{t_{1}} \sqrt{1 - \frac{u\left(t\right)^{2}}{c^{2}}}\,\mathrm{d}t

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

An ideal clock’s elapsed proper time is the path integral of the square root of one minus its speed squared over c squared.

Send one clock out at 0.6c0.6c and back at the same speed, 10 seconds in all by the clocks at rest, and ignore the turnaround. The travelling clock records 8 seconds and the clock that stayed records 10. To get those numbers you have to name each leg and the two meetings.

Each of two clocks in steady relative motion finds the other running slow, and there is no contradiction until they meet again; the reunion is decided by the two paths. Einstein’s §4 remark that a clock at the equator should run slower than one at a pole is a different case: on the real rotating Earth gravity enters, and this flat-spacetime model does not decide it. The moving-clock laboratory keeps the two kinds of comparison apart.

Show every step here: A reunion compares whole paths

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Modern qualifications

The integral is modern notation; Einstein reasons with a clock carried around a closed curve and states the result to second order. His equator remark stays as he printed it rather than being quietly corrected, and the laboratory declines to make an Earth prediction from special relativity alone.

Assumptions and limits: A reunion compares whole paths

Assumed here

  • The clocks are ideal, move slower than light along given paths, and gravity is absent.
  • Both clocks start together and meet again at named events.

What this does not establish

  • Comparisons on the real Earth also need gravity.
  • Acceleration changes the path a clock takes; it adds no extra slowing of its own to the formula.

Earlier step: Use the same measurement protocol

Source context: Facsimile

§5 · velocity composition

A derivation

The denominator changes as well

Why do two ordinary-looking speeds not simply add?

The 3 printed paragraphs this passage explains
  • §5, paragraph 1: A point moves uniformly in the moving system k, with constant velocity components along two of k's axes.
  • §5, paragraph 2: Transforming its motion to the stationary system shows that the parallelogram law for adding velocities holds only to a first approximation. For two velocities along one line the result is (v + w)/(1 + vw/V²), and combining two speeds below V always gives a speed below V.
  • §5, paragraph 3: Combining the speed of light with a slower speed still gives the speed of light. The same addition formula follows from applying two transformations of section 3 in turn, which shows that these transformations form a group.

The German text of these paragraphs is not on this site yet, so they have no links.

ux′=ux−v1−v uxc2u_x' = \frac{u_x - v}{1 - \frac{v\,u_x}{c^{2}}}
uy′=uyγ (1−v uxc2)u_y' = \frac{u_y}{\gamma\,\left(1 - \frac{v\,u_x}{c^{2}}\right)}

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

The transformed velocity along x is the velocity along x minus v, divided by one minus v times the velocity along x over c squared; the transverse velocity has the same denominator and an additional gamma.

An object moving at ww in the moving frame, in the same direction, moves in the original frame at U=(v+w)/(1+vw/c2)U = (v + w)/(1 + vw/c^2). With v=w=0.6cv = w = 0.6c, U=15c/17U = 15c/17, about 0.882c0.882c, not the 1.2c1.2c that plain addition would give.

Put light in, ux=cu_x = c, and the formula returns cc in every frame, as the second postulate requires. The velocity-composition laboratory shows the relativistic and the Galilean sums side by side, and prints the Galilean one even when it exceeds cc.

Show every step here: The denominator changes as well

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Modern qualifications

Modern texts often use rapidities, which simply add for boosts along one line, or multiply matrices. Both are later aids, not premises of the 1905 argument, and neither should hide that two boosts in different directions combine into a boost and a rotation, the Thomas–Wigner rotation.

Assumptions and limits: The denominator changes as well

Assumed here

  • The velocity comes from two neighbouring events on the object’s path.
  • The frames move at constant velocity relative to each other, along x.

What this does not establish

  • The plus-sign formula for combining speeds and the minus-sign change of frame run in opposite directions.
  • Two boosts in different directions combine into a boost and a rotation.

Earlier step: Light constraints leave a scale to determine

What is getting in the way?
  • An unfamiliar word or symbol

    v is the speed of the moving frame, measured in the original one. w is the object's speed measured in the moving frame, and U is the same object's speed measured in the original frame. c is the speed of light. uₓ and u′ₓ are the object's velocities along x in the two frames.

    Open the explanation that addresses this

  • An algebraic move

    Transform the two small changes first, then divide. With dx′ = γ(dx − v dt) and dt′ = γ(dt − v dx/c²), the ratio dx′/dt′ is (uₓ − v)/(1 − v uₓ/c²). The factors γ cancel, and the second term of dt′ is what puts 1 − v uₓ/c² in the denominator.

    Open the explanation that addresses this

  • The physical reason for a step

    A speed is a distance divided by the time taken, and the two frames disagree about both. They disagree about the time because each sets its distant clocks by its own light signals. So a speed cannot pass from one frame to the other by adding v; the time between the same two events changes too.

    Open the explanation that addresses this

  • The connection to the picture

    Picture two neighbouring events on the object's path: here at one moment, and a little further along a little later. Each frame measures the distance and the time between those same two events. Both numbers change from frame to frame, and the speed is their ratio.

    Open the explanation that addresses this

  • The purpose of the calculation

    It finds what the two postulates require of speeds. Put in light, uₓ = c, and the formula returns c in every frame, as the second postulate demands, and two speeds below c combine to a speed below c. At everyday speeds v w/c² is tiny and the formula is ordinary addition.

    Open the explanation that addresses this

  • Too much at once

    One example: v = w = 0.6c. Plain addition gives 1.2c. The formula divides that by 1 + 0.6 × 0.6 = 1.36, which gives 15c/17, about 0.882c. The faster the two speeds, the further the divisor rises above 1.

    Open the explanation that addresses this

Source context: Facsimile

§6 · field equations and field components

A derivation

Transform derivatives before naming fields

How do the field equations change coordinates?

The 6 printed paragraphs this passage explains
  • §6, paragraph 1: Einstein assumes that the Maxwell-Hertz equations for empty space hold in the stationary system K, with X, Y, Z the electric force and L, M, N the magnetic force.
  • §6, paragraph 2: Applying the transformation of section 3, he rewrites these equations in the coordinates and time of the moving system k.
  • §6, paragraph 3: The principle of relativity requires that the same equations hold in k, for the electric and magnetic forces measured there by their effects on charges and magnets.
  • §6, paragraph 4: Since the two sets of equations for k must say the same thing, the forces measured in k equal combinations of those in K, up to one common factor ψ(v) that may depend on the speed.
  • §6, paragraph 5: Going to k and back, and symmetry, fix that factor at 1, which gives the transformation of the electric and magnetic forces. Einstein states its meaning two ways: the old way speaks of an electromotive force on a moving charge, to a first approximation, and the new way says the force on it is the electric force in a system in which it is momentarily at rest.
  • §6, footnote 1: A footnote gives the symmetry argument: with only one magnetic component present, reversing the direction of motion must reverse the sign of the transformed electric force without changing its size.

The German text of these paragraphs is not on this site yet, so they have no links.

∂t=γ (∂t′−v ∂x′)\partial_{t} = \gamma\,\left(\partial_{t'} - v\,\partial_{x'}\right)
∂t=β (∂τ−v ∂ξ)\partial_{t} = \beta\,\left(\partial_{\tau} - v\,\partial_{\xi}\right)
∂x=γ (∂x′−v ∂t′c2)\partial_{x} = \gamma\,\left(\partial_{x'} - \frac{v\,\partial_{t'}}{c^{2}}\right)
∂x=β (∂ξ−v ∂τV2)\partial_{x} = \beta\,\left(\partial_{\xi} - \frac{v\,\partial_{\tau}}{V^{2}}\right)

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

Take the simplest case, a wave travelling along xx with an electric component EyE_y and a magnetic component BzB_z. In SI units and empty space the Maxwell-Hertz equations reduce to the pair ∂tEy=−c2 ∂xBz\partial_t E_y = -c^2\,\partial_x B_z and ∂tBz=−∂xEy\partial_t B_z = -\partial_x E_y. Substitute the two derivative rules above into both equations; the next step needs the pair.

After the substitution, EyE_y and BzB_z appear only in the combinations Ey−vBzE_y - vB_z and Bz−vEy/c2B_z - vE_y/c^2. Multiply each by γ\gamma and call the results Ey′E'_y and Bz′B'_z. The pair then reads ∂t′Ey′=−c2 ∂x′Bz′\partial_{t'} E'_y = -c^2\,\partial_{x'} B'_z and ∂t′Bz′=−∂x′Ey′\partial_{t'} B'_z = -\partial_{x'} E'_y, the same form as before. The field-equations laboratory shows this grouping one term at a time.

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

Show every step here: Transform derivatives before naming fields

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Modern qualifications

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

Assumptions and limits: Transform derivatives before naming fields

Assumed here

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

What this does not establish

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

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

Source context: Facsimile

A derivation

Electric and magnetic components mix together

What does a new inertial observer call electric?

The 3 printed paragraphs this passage explains
  • §6, paragraph 5: Going to k and back, and symmetry, fix that factor at 1, which gives the transformation of the electric and magnetic forces. Einstein states its meaning two ways: the old way speaks of an electromotive force on a moving charge, to a first approximation, and the new way says the force on it is the electric force in a system in which it is momentarily at rest.
  • §6, paragraph 6: The same holds for magnetomotive forces. The electromotive force is only an auxiliary concept, needed because electric and magnetic forces do not exist independently of the state of motion of the coordinate system.
  • §6, paragraph 7: So the asymmetry noted in the introduction, between a moving magnet and a moving conductor, disappears, and the question of where the electromotive force is seated, as in unipolar machines, loses its point.

The German text of these paragraphs is not on this site yet, so they have no links.

Ex′=ExE'_{x} = E_{x}
Bx′=BxB'_{x} = B_{x}
Ey′=γ (Ey−v Bz)E'_{y} = \gamma\,\left(E_{y} - v\,B_{z}\right)
By′=γ (By+v Ezc2)B'_{y} = \gamma\,\left(B_{y} + \frac{v\,E_{z}}{c^{2}}\right)
Ez′=γ (Ez+v By)E'_{z} = \gamma\,\left(E_{z} + v\,B_{y}\right)
Bz′=γ (Bz−v Eyc2)B'_{z} = \gamma\,\left(B_{z} - \frac{v\,E_{y}}{c^{2}}\right)

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

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

Take a region with no electric field and a magnetic field BzB_z pointing along positive zz. An observer moving along positive xx measures Ey′=−γvBzE'_y = -\gamma vB_z, an electric field pointing along negative yy. At v=0.6cv = 0.6c, γ=1.25\gamma = 1.25, so Ey′=−0.75c BzE'_y = -0.75c\,B_z.

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

Show every step here: Electric and magnetic components mix together

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Modern qualifications

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

Assumptions and limits: Electric and magnetic components mix together

Assumed here

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

What this does not establish

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

Earlier step: Transform derivatives before naming fields

Source context: Facsimile

§7 · Doppler shift and aberration

A derivation

One phase fixes frequency and direction

Why must a frequency change accompany a direction change?

The 7 printed paragraphs this passage explains
  • §7, paragraph 1: Far from the origin of the stationary system is a source of electromagnetic waves, which near the origin are plane waves with a given amplitude, frequency and direction of the wave normal.
  • §7, paragraph 2: Einstein asks what these waves are like for an observer at rest in the moving system. With the transformations of sections 6 and 3 they are again plane waves, with a new frequency and a new direction.
  • §7, paragraph 3: From the new frequency: an observer moving at speed v relative to a very distant source, at an angle φ to the line joining them, receives light whose frequency depends on v and φ in a way Einstein gives exactly.
  • §7, paragraph 4: This is Doppler's principle for any speed. When the observer moves straight away from the source the formula becomes a simple square-root ratio, and Einstein remarks that, against the usual view, the frequency becomes infinite for an extreme speed of approach; the page prints that speed as v = −∞.
  • §7, paragraph 5: The new direction gives the law of aberration in its most general form: the angle at which the moving observer receives the light differs from the angle in the source's system, and light arriving at right angles in one system arrives slanted in the other.
  • §7, paragraph 6: Einstein also finds how the amplitude of the waves changes for the moving observer, and how simple the change is when the observer moves along the line to the source.
  • §7, paragraph 7: It follows that a light source approached at the speed of light would have to appear infinitely intense.

The German text of these paragraphs is not on this site yet, so they have no links.

q:=ν′νq := \frac{\nu'}{\nu}
q=γ (1−vc cos⁡(φ))q = \gamma\,\left(1 - \frac{v}{c}\,\cos\left(\varphi\right)\right)

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

The frequency ratio is gamma times one minus v over c times the unprimed direction cosine.

cos⁡(φ′)=cos⁡(φ)−vc1−vc cos⁡(φ)\cos\left(\varphi'\right) = \frac{\cos\left(\varphi\right) - \frac{v}{c}}{1 - \frac{v}{c}\,\cos\left(\varphi\right)}
cos⁡(φ′)=cos⁡(φ)−vV1−vV cos⁡(φ)\cos\left(\varphi'\right) = \frac{\cos\left(\varphi\right) - \frac{v}{V}}{1 - \frac{v}{V}\,\cos\left(\varphi\right)}

The aberration direction cosine is cosine phi minus v over c, divided by one minus v over c times cosine phi.

The phase of a plane wave, ωt−kxx−kyy−kzz\omega t - k_x x - k_y y - k_z z, counts wave crests, and every observer counts the same crests at the same event. Substitute the inverse coordinate map, which writes tt, xx, yy and zz in primed coordinates. The coefficient of t′t' is the new angular frequency and the coefficients of x′x', y′y' and z′z' are the new wave vector, so one substitution gives both formulas above. The Doppler effect of sound is an analogy only: it depends on motion through the air, and nothing here does.

At v=0.6cv = 0.6c and φ=0\varphi = 0, q=0.5q = 0.5. At φ\varphi = 90 degrees, q=1.25q = 1.25 and cos⁡φ′=−0.6\cos\varphi' = -0.6: light that crossed the motion at right angles in the original frame travels at about 127 degrees to the x′x' axis in the new one. Each angle belongs to one frame, and the Doppler-and-aberration laboratory labels which.

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Modern qualifications

An observer moving at the speed of light appears in these formulas only as a limit; no inertial frame moves at cc. Stellar aberration had been observed since Bradley’s report of 1729, and the second-order Doppler shift was first measured by Ives and Stilwell in 1938. Both are measurements of the world, separate from this ideal calculation.

Assumptions and limits: One phase fixes frequency and direction

Assumed here

  • The light is a plane wave in empty space, and both observers describe the same events.
  • The angle φ is measured in the original frame, between the direction of the light and the direction of motion.

What this does not establish

  • Light that crosses the motion at right angles in one frame does not, in general, do so in the other.
  • No observer can move at the speed of light, so the formulas’ limit at v = c describes no frame.

Earlier step: Light constraints leave a scale to determineElectric and magnetic components mix together

Source context: Facsimile

§8 · finite light energy and moving mirrors

A derivation

Density is not the energy of the whole packet

Why does a finite light complex need a volume calculation?

The 3 printed paragraphs this passage explains
  • §8, paragraph 1: The light energy per unit volume changes between systems as the square of the amplitude, but the volume of a given portion of light changes too. Following a sphere that moves with the light, through whose surface no energy passes, Einstein asks how much energy it encloses, seen from the moving system.
  • §8, paragraph 2: Seen from the moving system the sphere is an ellipsoid of a different volume, and combining the two changes shows how the energy of the enclosed light changes.
  • §8, paragraph 3: The energy and the frequency of a portion of light change with the observer's motion by the same law.

The German text of these paragraphs is not on this site yet, so they have no links.

u′u=q2\frac{u'}{u} = q^{2}
Vpacket′Vpacket=1q\frac{V_{\mathrm{packet}}'}{V_{\mathrm{packet}}} = \frac{1}{q}
E′E=q\frac{E'}{E} = q

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

Energy density scales by q squared, packet volume by inverse q, and total energy by q.

Einstein bounds the light complex by a sphere that moves with the light. That boundary is not a body at rest in either frame, so the ordinary contraction of a material body does not apply to it. At 0.6c0.6c, for light travelling the same way as the observer, q=0.5q = 0.5: the density falls to one quarter, the volume doubles, and the total energy halves.

Treating the light complex as a body at rest in the original frame, and shrinking its volume by 1/γ=0.81/\gamma = 0.8, would give a total-energy factor of 0.25×0.8=0.20.25 \times 0.8 = 0.2 instead of 0.5. For light crossing the motion at right angles in the original frame the two volume factors happen to agree, both 0.8, so that case alone cannot catch the mistake. The light-packet laboratory includes the case that tells them apart.

The mass-energy paper of September 1905 starts from exactly this transformation of light energy. It does not assume E=mc2E = mc^2 in advance, and it needs no localized light quanta.

Show every step here: Density is not the energy of the whole packet

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Modern qualifications

Here uu is energy density and VpacketV_{\mathrm{packet}} the volume of the complex. The letter LL is avoided: in this paper it names a magnetic component, and in the mass-energy paper the emitted energy. The calculation follows one bounded complex of light; it gives light no rest frame.

Assumptions and limits: Density is not the energy of the whole packet

Assumed here

  • The same finite light complex is compared, each frame measuring its volume at one instant of its own time.
  • The waves transform as plane waves, with q defined from the light’s direction in the original frame.

What this does not establish

  • A light complex is not a rod, and light has no rest frame.
  • The result needs neither light quanta nor the mass-energy relation it later supports.

Earlier step: One phase fixes frequency and direction

Source context: Facsimile

A derivation

Reflection starts with interception

Can the incident ray reach a receding mirror?

The 4 printed paragraphs this passage explains
  • §8, paragraph 4: A perfectly reflecting mirror at rest in the moving system meets the plane waves of section 7. Einstein asks for the pressure of the light on it and for the direction, frequency and intensity of the light after reflection.
  • §8, paragraph 5: He describes the incoming light as seen from the mirror's own system, reflects it there, and transforms the reflected light back into the stationary system.
  • §8, paragraph 6: The energy arriving at the mirror and the energy leaving it differ by the work done by the light's pressure, which gives the pressure exactly. To a first approximation it agrees with experience and with other theories.
  • §8, paragraph 7: By this method every problem in the optics of moving bodies can be solved: the electric and magnetic forces of the light are transformed into a system at rest relative to the body, which turns the problem into one about bodies at rest.

The German text of these paragraphs is not on this site yet, so they have no links.

νrefν=1−2 vc cos⁡(φ)+(vc)21−(vc)2\frac{\nu_{\mathrm{ref}}}{\nu} = \frac{1 - 2\,\frac{v}{c}\,\cos\left(\varphi\right) + \left(\frac{v}{c}\right)^{2}}{1 - \left(\frac{v}{c}\right)^{2}}

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

The reflected frequency ratio is one minus twice v over c cosine phi plus v over c squared, divided by one minus v over c squared.

At normal incidence and v=0.6cv = 0.6c, three different ratios appear. The reflected light has one quarter of the incident frequency. The mirror intercepts incident energy at 0.4 times the rate of a fixed mirror, because it is moving away from the light. And with incident energy density uu, the light presses on the mirror with pressure 0.5u0.5u.

To count the energy the mirror intercepts, the moving-mirror laboratory uses the speed at which light crosses the moving surface, ccos⁡φ−vc\cos\varphi - v, and not cc alone. Light travelling parallel to the mirror in the original frame, with the mirror receding along positive xx, never reaches it, and the laboratory reports that no reflection happens.

Show every step here: Reflection starts with interception

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Modern qualifications

The mirror here is ideal: its motion is prescribed, and it has no recoil, absorption or thickness. Einstein states the pressure in his Gaussian units; the SI form used here changes the units and leaves the result the same.

Assumptions and limits: Reflection starts with interception

Assumed here

  • The mirror is flat and perfectly reflecting, and it moves slower than light.
  • The light must be able to reach the mirror: c cos φ is greater than v.

What this does not establish

  • Where the light only just fails to catch the mirror, a number from the calculation is not a reflection.
  • A reflected ray can still move towards +x in the laboratory, and a mirror receding faster moves away from it.

Earlier step: One phase fixes frequency and direction

Source context: Facsimile

§9 · charge and current

A derivation

Neutrality and current belong to a frame

How can a neutral current acquire charge density?

The 3 printed paragraphs this passage explains
  • §9, paragraph 1: Einstein starts from the Maxwell-Hertz equations with convection currents, electricity carried along by moving charges. With the charges bound to small rigid bodies, ions or electrons, these are the electromagnetic basis of Lorentz's electrodynamics and optics of moving bodies.
  • §9, paragraph 2: Transformed into the moving system, the equations keep their form, with the velocity of the charges and the density of electricity as measured there. So the electrodynamic basis of Lorentz's theory agrees with the principle of relativity.
  • §9, paragraph 3: A further result: if a charged body moves in any way and its charge, measured in a system moving with it, stays the same, then its charge measured from the stationary system stays constant too.

The German text of these paragraphs is not on this site yet, so they have no links.

ρ′=γ (ρ−v Jxc2)\rho' = \gamma\,\left(\rho - \frac{v\,J_x}{c^{2}}\right)
Jx′=γ (Jx−v ρ)J_x' = \gamma\,\left(J_x - v\,\rho\right)

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

The primed charge density is gamma times rho minus v times the current along x over c squared; the primed current along x is gamma times the current along x minus v times rho.

The current components across the motion, JyJ_y and JzJ_z, are unchanged. For a neutral wire, ρ=0\rho = 0, carrying current JxJ_x, the new density is ρ′=−γvJx/c2\rho' = -\gamma vJ_x/c^2, which is not zero for any v≠0v \ne 0. To see why, transform each kind of charge on its own: the positive ions and the moving electrons have different velocities, so their densities change by different factors.

Charge is conserved in every frame: density and current together obey the continuity equation. The total charge of one body is the same in both frames when each frame adds up the charge at one of its own instants. A fixed box of space is a different comparison, because charge flows through its walls. The charge-and-current laboratory compares the densities and the total charge of a body.

Show every step here: Neutrality and current belong to a frame

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Modern qualifications

Einstein treats convection current, charge carried along by moving bodies, and this edition keeps his account as he gives it instead of turning it into a general formula for a single charged fluid. The four-current, which combines ρ\rho and JJ into one four-vector, came later; it is used here only to check results.

Assumptions and limits: Neutrality and current belong to a frame

Assumed here

  • Charge and current are spread through a real body, and the new observer moves along x.
  • When total charge is compared between frames, it is the charge of the same isolated body.

What this does not establish

  • Charge density is not total charge: a density can change between frames while the body’s charge stays fixed.
  • In a neutral wire the positive and negative charges move at different speeds, so there is no single velocity of the charge.

Earlier step: Light constraints leave a scale to determine

Source context: Facsimile

§10 · electron force, work, and closing scope

A derivation

A force-to-acceleration ratio needs two frame labels

Why are there two transverse coefficients?

The 9 printed paragraphs this passage explains
  • §10, paragraph 1: A small particle carrying an electric charge ε, called an electron, moves in an electromagnetic field, and Einstein makes only the following assumption about its law of motion.
  • §10, paragraph 2: When the electron is at rest at some moment, its motion in the next instant follows the ordinary law: its mass μ times its acceleration equals ε times the electric force, as long as it moves slowly.
  • §10, paragraph 3: Now the electron has a speed v at some moment, and Einstein seeks the law of its motion in the next instant.
  • §10, paragraph 4: Without loss of generality the electron is taken to be at the origin at that moment, moving along the x-axis, so that it is momentarily at rest in a system k moving with the same speed.
  • §10, paragraph 5: In k it therefore obeys the law for a resting electron, and the transformations of sections 3 and 6 connect the coordinates, time and forces of k with those of the stationary system.
  • §10, paragraph 6: Transforming the equations of motion back to the stationary system gives the equations Einstein labels (A).
  • §10, paragraph 7: Following the usual approach, Einstein asks for the electron's longitudinal and transverse mass. Taking the force to be the one measured in a system moving with the electron, for example with a spring balance, and measuring accelerations in the stationary system, he obtains one mass along the motion and another across it.
  • §10, paragraph 8: Other definitions of force and acceleration would give other numbers for these masses, so different theories of the electron's motion must be compared with great care.
  • §10, paragraph 9: These results about mass hold for any ponderable material point too, since adding an arbitrarily small charge makes it an electron in this sense.

The German text of these paragraphs is not on this site yet, so they have no links.

Fxax=m γ3\frac{F_x}{a_x} = m\,\gamma^{3}
Fy′ay=m γ2\frac{F_y'}{a_y} = m\,\gamma^{2}
Fyay=m γ\frac{F_y}{a_y} = m\,\gamma

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

Longitudinal laboratory force over acceleration is m gamma cubed. Rest-frame transverse force over laboratory acceleration is m gamma squared. Laboratory transverse force over laboratory acceleration is m gamma.

For an electron moving along xx, a force across the motion is γ\gamma times larger in the electron’s momentary rest frame than in the laboratory. Einstein divides the rest-frame force by the laboratory acceleration and gets mγ2m\gamma^2. Dividing the laboratory force by the same acceleration gives mγm\gamma.

At 0.6c0.6c, γ=1.25\gamma = 1.25. The longitudinal mass is mγ3=1.953125 mm\gamma^3 = 1.953125\,m, Einstein’s transverse mass is mγ2=1.5625 mm\gamma^2 = 1.5625\,m, and the all-laboratory transverse ratio is mγ=1.25 mm\gamma = 1.25\,m. The electron-dynamics laboratory shows each number with the frames it uses, and leaves Einstein’s result as he printed it.

Show every step here: A force-to-acceleration ratio needs two frame labels

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Modern qualifications

Einstein himself warns that other definitions of force and acceleration would give other values for these masses. The modern habit of defining force as the rate of change of momentum gives mγm\gamma for the transverse ratio. It is a later convention, and this edition does not substitute it for his printed result.

Assumptions and limits: A force-to-acceleration ratio needs two frame labels

Assumed here

  • In its momentary rest frame the electron obeys ordinary mechanics for slow motion.
  • Every force and every acceleration is labelled with the frame it is measured in.

What this does not establish

  • The two masses are ratios of force to acceleration under stated conventions; the electron’s own mass does not change with the observer’s speed.
  • The electron is treated as a point driven slowly, without the effects of its own radiation or field.

Earlier step: Electric and magnetic components mix together

Source context: Facsimile

A derivation

The work integral has an observable endpoint

What energy is required to accelerate an electron?

The 5 printed paragraphs this passage explains
  • §10, paragraph 10: Einstein finds the electron's kinetic energy. An electron starting from rest and accelerated slowly by an electrostatic force, so that it loses no energy as radiation, gains energy equal to the work done on it, which he calculates.
  • §10, paragraph 11: That energy grows without limit as the speed approaches V, so, as in the earlier results, speeds greater than light have no possibility of existing.
  • §10, paragraph 12: By the argument given earlier, the same expression for kinetic energy holds for ponderable masses.
  • §10, paragraph 13: Einstein lists three properties of the electron's motion open to experiment: the ratio of how strongly magnetic and electric forces deflect it, which gives its speed; the relation between the potential difference it passes through and the speed it gains; and the radius of its path in a magnetic field.
  • §10, paragraph 14: These three relations, he says, completely express the laws by which the electron must move according to the theory.

The German text of these paragraphs is not on this site yet, so they have no links.

K=∫0vm γ(u)3 u duK = \int_{0}^{v} m\,\gamma\left(u\right)^{3}\,u\,\mathrm{d}u
K=m c2 (γ−1)K = m\,c^{2}\,\left(\gamma - 1\right)

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

Integrating the longitudinal work from rest gives m c squared times gamma minus one.

The integral follows from one derivative, dγ/du=γ3u/c2d\gamma/du = \gamma^3 u/c^2: the integrand mγ3um\gamma^3 u is mc2mc^2 times it, so the integral is mc2(γ−1)mc^2(\gamma - 1). At low speed this approaches the familiar mv2/2mv^2/2. At 0.6c0.6c the exact value is 0.25 mc20.25\,mc^2 and the low-speed formula gives 0.18 mc20.18\,mc^2, a difference large enough to see.

∣q∣ U=K|q|\,U = K
rB=γ m v∣q∣ Br_B = \frac{\gamma\,m\,v}{|q|\,B}

Shown in modern letters; Einstein's are in the facsimile of the printed paper.

For acceleration from rest, charge magnitude times voltage magnitude supplies kinetic energy; in a transverse magnetic field the circular radius is gamma m v over charge magnitude B.

The third relation compares bending by the two kinds of field. In an electric field across the motion the path curves with radius γmv2/(∣q∣E)\gamma mv^2/(|q|E) at that instant; the field also speeds the electron up as the path turns, so the radius keeps changing. In a magnetic field the speed stays constant and the path is a circle of radius γmv/(∣q∣B)\gamma mv/(|q|B). The electron-dynamics laboratory keeps the two cases apart.

Show every step here: The work integral has an observable endpoint

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

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Modern qualifications

The three relations hold within the model’s assumptions: fields as specified, and no energy lost to radiation. After §10 the paper closes with thanks to M. Besso and the date line, Bern, June 1905. Those belong to the German source text, and this explanation does not repeat them.

Assumptions and limits: The work integral has an observable endpoint

Assumed here

  • The electron has a fixed mass m and moves slower than light, and force is defined as stated.
  • For the voltage relation, the electron starts at rest and all the work of the field becomes its energy of motion.

What this does not establish

  • Losses by radiation and interactions with matter are left out.
  • The kinetic energy formula alone does not give the September paper’s result that a body’s mass changes with its energy.

Earlier step: A force-to-acceleration ratio needs two frame labels

Source context: Facsimile

References and source status

Newly authored explanatory preview in modern notation, with editorial and physics review pending. Both the kinematic and electrodynamic halves are treated, but this is not a German transcription, an aligned translation, or a complete critical edition. The source paragraphs, footnotes, acknowledgment, and date-lines are not claimed to be fully represented or reviewed here. The headings and the way the argument is divided into passages are ours: Einstein divided his paper into an introduction and ten numbered sections in two parts, under his own titles.

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 reviewed translation or pinned facsimile.

27 foundation readings sit behind this argument.

The question we were answering:

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