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§5 · velocity composition

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Detail

Letters

Show modern qualifications

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

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.

u_x' = \frac{u_x - v}{1 - \frac{v\,u_x}{c^{2}}}

u_y' = \frac{u_y}{\gamma\,\left(1 - \frac{v\,u_x}{c^{2}}\right)}

Shown in modern letters; Einstein's are in §5 of the German source face.

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

Put light in, $u_x = c$ , and the formula returns $c$ 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 $c$ .

Open the foundation: Rates of change and derivatives

Show every step here: The denominator changes as well

Letters: in §5 a point moves in k with constant velocity components $w_ξ$ and $w_η$ , so that $ξ=w_ξ\tau$ , $η=w_η\tau$ , ζ = 0; the resulting speed in K is U, w is the speed in k, α the angle between the velocities v and w, and V the speed of light. The formula below goes the other way, from K to the moving frame, and writes $u_x$ and $u_y$ in K and $u'_x$ and $u'_y$ in the moving frame.

A velocity is a small displacement divided by the small time it takes. Both change between frames, so transform both before dividing.
Along the object's path, take the change of $x'=\gamma(x-vt)$ : $dx'=\gamma(dx-v\,dt)$ .
Take the change of $t'=\gamma(t-vx/c^2)$ along the same path: $dt'=\gamma(dt-v\,dx/c^2)$ .
Divide, and then divide numerator and denominator by dt: $u'_x=\frac{dx'}{dt'}=\frac{u_x-v}{1-vu_x/c^2}$ .
Across the motion $dy'=dy$ , but the time still changes, so $u'_y=\frac{dy}{\gamma(dt-v\,dx/c^2)}=\frac{u_y}{\gamma(1-vu_x/c^2)}$ .

u_x' = \frac{u_x - v}{1 - \frac{v\,u_x}{c^{2}}}

u_y' = \frac{u_y}{\gamma\,\left(1 - \frac{v\,u_x}{c^{2}}\right)}

Shown in modern letters; Einstein's are in §5 of the German source face.

The paper's direction, from k to K, uses the inverse transformation and so carries plus signs. For the point above it prints $x=\frac{w_ξ+v}{1+vw_ξ/V^2}t$ and $y=\frac{\sqrt{1-(v/V)^2}}{1+vw_ξ/V^2}w_ηt$ .
So the law of the parallelogram of velocities holds, in this theory, only as a first approximation. The paper's general result for the resulting speed is symmetric in v and w.
With w along the X axis too, the paper's formula becomes $U=\frac{v+w}{1+vw/V^2}$ .
Worked: v = w = 0.6c. Plain addition would give 1.2c. The formula gives $1.2c/(1+0.36)=1.2c/1.36$ , which is 15c/17, about 0.882c.
Two speeds below light always give a speed below light. The paper shows it by writing v = V − κ and w = V − λ with κ and λ positive and less than V: then $U=V\frac{2V-κ-\lambda}{2V-κ-\lambda+κ\lambda/V}$ , which is less than V because the denominator is larger than the numerator. In the worked case κ = λ = 0.4V and $U=V\times1.2/1.36$ .
Light stays at V: put w = V and $U=\frac{V+v}{1+v/V}=V$ , as the second principle requires.
A worked sideways case in the paper's direction: $w_η=0.5c$ , $w_ξ=0$ , v = 0.6c. Then the sideways speed in K is $\sqrt{1-0.36}\times0.5c=0.4c$ , and the point also moves along x at 0.6c.
The paper remarks that the same formula comes from composing two transformations of §3: a third system moving at w relative to k is related to K by the same equations with v replaced by $(v+w)/(1+vw/V^2)$ .
Speeds in different directions combine by the general formula, not by this aligned one.

A tool for this step

Rates of change and derivatives

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

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

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

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

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

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

\frac{\mathrm{d} x}{\mathrm{d} t} = \lim_{\Delta t \to 0} \frac{x\left(t + \Delta t\right) - x\left(t\right)}{\Delta t}

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

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

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

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

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

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

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

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

3\,\left(t + \Delta t\right)^{2} - 3\,t^{2} = 6\,t\,\Delta t + 3\,\left(\Delta t\right)^{2}

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

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

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

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

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

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.

The chord, and what it settles onto

The ball's position against time, with the chord over a whole second and the tangent at 1 second drawn through the same point.

The broken chord climbs 9 metres a second; the tangent climbs 6. The chord over a tenth of a second climbs 6.3, which at this size is already the tangent: no drawing that fits a page can separate them.

So the settling is plotted instead, one point for each interval the lesson averages over.

What it shows, in words

The first drawing is the ball's position against time, a curve rising more and more steeply. A point is marked at 1 second, 3 metres. The broken line through it meets the curve again a second later, at 12 metres, so it climbs 9 metres a second. The solid line through the same point is less steep and climbs 6: that is the tangent, and its steepness is the speed at the instant.

The second drawing plots the four averages against the interval each was taken over: 9, 6.3, 6.03, 6.003 metres per second for intervals of 1, 0.1, 0.01, 0.001 of a second. The first point sits well above the line at 6 and the other three lie on it. The interval is never set to zero; it only shrinks, and the averages settle.

What it does not show: the averages settle here because this curve is smooth at this instant. A curve with a corner would give one answer from the left and another from the right, and no tangent to draw, and nothing in either drawing rules that out. The lesson's worked example is what shows why these particular averages settle, by cancelling the 3Δt that the shrinking interval leaves behind.

Try it: nudge the frequency, watch the stopping voltage

In §8 of the light-quanta paper, Einstein predicts that the potential V needed to stop the electrons light frees from a metal rises in a straight line with the light's frequency ν, with the same slope for every metal. The derivative dV/dν is that slope: how many volts the stopping potential rises for each hertz of frequency.

Choose a frequency step Δν:

What the model computes

Baseline frequency (ν₀):: 6.00 × 10¹⁴ Hz
Frequency nudge (Δν):: +1.00 × 10¹³ Hz
Potential change (ΔV):: +0.04135668 V
Sensitivity ratio (ΔV / Δν):: 4.135667696 × 10⁻¹⁵ V·s (or V/Hz)
Modern h/e (2019 SI):: 4.135667696 × 10⁻¹⁵ V·s

Summary of frequency steps and sensitivity ratio
Nudge size	Δν (Hz)	ΔV (V)	Ratio ΔV / Δν (V·s)
+1.0 × 10¹⁴ Hz	1.0 × 10¹⁴	0.41357	4.135668 × 10⁻¹⁵
+5.0 × 10¹³ Hz	5.0 × 10¹³	0.20678	4.135668 × 10⁻¹⁵
+1.0 × 10¹³ Hz	1.0 × 10¹³	0.041357	4.135668 × 10⁻¹⁵
+2.0 × 10¹² Hz	2.0 × 10¹²	0.0082713	4.135668 × 10⁻¹⁵

What it shows, in words

A derivative carries units: volts divided by hertz gives volt-seconds. This model is Einstein's §8 straight line, with its slope set to the modern value of h/e, about 4.14 × 10⁻¹⁵ V·s. On a straight line every step size gives the same ratio ΔV / Δν, which is why the table repeats one number four times; on a curve the ratio would change with the step and settle only as Δν shrinks. The model was built with one slope for every metal, so it cannot show that real metals share one. Millikan measured that in 1916.

Where this lesson stops

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

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 notation

Velocity along the motion

u_x' = \frac{u_x - v}{1 - \frac{v\,u_x}{c^{2}}}

Shown in modern letters; Einstein's are in §5 of the German source face.

is minus , divided by one minus v times that velocity 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 velocity along x in the moving system is the velocity along x in the stationary system minus the speed of the moving system, divided by one minus that speed times the stationary velocity along x over the speed of light squared.

Speeds do not simply subtract. An object moving at 15/17 of c in the stationary system moves at 0.6c in a system that itself moves at 0.6c; light, moving along x at c in one, moves at c in both.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c; the object's velocity in the stationary system has components along x and y.
Modern teaching notation: gamma is the paper's beta, c its V. The paper composes speeds the other way round, from the moving system to the stationary one; this is the same relation inverted.

One minus v times the velocity along x, over c²: Close to 1 at everyday speeds, which is why speeds seem simply to subtract. Read the prerequisite
Frame speed times object speed: Both along x. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
v times the velocity along x, over c²: The correction Galilean subtraction leaves out. Read the prerequisite
Velocity along x, minus v: The Galilean answer, what subtraction alone would give. Read the prerequisite
Divided by the correction: Dividing by a number below 1 raises the result, and it can never pass c. Read the prerequisite
What it asserts: The velocity along x one system assigns, from the velocity the other assigns. For a velocity of c along x the result is c. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite
Velocity along x, stationary system: The object's velocity along the motion, in the stationary system. Read the prerequisite
Velocity along x, moving system: The same object's velocity along the motion, in the moving system. 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 notation

Velocity across the motion

u_y' = \frac{u_y}{\gamma\,\left(1 - \frac{v\,u_x}{c^{2}}\right)}

Shown in modern letters; Einstein's are in §5 of the German source face.

is divided by times one minus 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

Transverse velocity has the same denominator and an additional gamma.

Even a velocity at right angles to the motion changes, because the two systems' clocks run at different rates and are set differently along x.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c; the object's velocity in the stationary system has components along x and y.
Modern teaching notation: gamma is the paper's beta, c its V. The paper composes speeds the other way round, from the moving system to the stationary one; this is the same relation inverted.

One minus v times the velocity along x, over c²: Close to 1 at everyday speeds, which is why speeds seem simply to subtract. Read the prerequisite
Frame speed times object speed: Both along x. Read the prerequisite
c squared: The square of the speed of light. Read the prerequisite
v times the velocity along x, over c²: The correction Galilean subtraction leaves out. Read the prerequisite
Velocity along y, over that: Positions across the motion are the same in both systems; the times are not. Read the prerequisite
What it asserts: The velocity across the motion one system assigns, from the velocities the other assigns. Read the prerequisite
Gamma times the correction: The extra gamma comes from the moving system's time. Read the prerequisite
Lorentz factor: One over the square root of one minus v squared over c squared. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Speed of the moving system: Along x. Its sign gives the direction. Read the prerequisite
Velocity along x, stationary system: The object's velocity along the motion, in the stationary system. Read the prerequisite
Velocity along y, stationary system: The object's velocity across the motion, in the stationary system. Read the prerequisite
Velocity along y, moving system: The same object's velocity across the motion, in the moving system. 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

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.

Open the foundation: Hyperbolic functions and rapidity

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: 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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§5 · velocity composition

Original journal scan

The denominator changes as well

Velocity along the motion

Velocity across the motion

Modern qualifications

What this relies on

Assumed here

What this does not establish

References

Reading a graph

Fractions and ratios

A letter stands for a quantity

Marks that look alike and do different jobs

A sign records direction

Powers of ten and physical units

Squares and square roots

Adding and averaging

Rates of change and derivatives

Density is not probability

Dot and cross products

Charges, fields and light

Exponential change

Fields and waves

Counting what crosses a boundary

Events and distant clocks

Functions and graphs

Hyperbolic functions and rapidity

Adding continuously

Logarithms: turning products into sums

Matrices and linear maps

The push of light

Partial derivatives and held-fixed quantities

Quantities and units

Approximating a curve near a point

The units of 1905

Vectors and their components

Energy of motion and inertia

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