Mass and energy: two accounts, one subtraction: The unsectioned argument · explanatory reading

Reading a graph

How does a curve show the relationship between two physical quantities?

Walk for a minute and note how far you have gone every ten seconds: 0 m at the start, 14 m after ten seconds, 28 m after twenty, and so on. The first drawing below puts time along the bottom and distance up the side, with a dot for each pair. That is a graph: each dot is one moment, read across for the time and up for the distance.

The dots of a steady walk fall on a straight line, and the steeper the line, the faster the walk. This one climbs 14 metres for every 10 seconds: 1.4 metres per second. A flat line would mean standing still, and a line that bends upward would mean speeding up.

The papers draw other pairs the same way. Along the bottom goes what you choose or wait for, such as a position or a time; up the side goes what you then find, such as how many particles sit there. A bell-shaped curve, high in the middle and falling away on both sides, says that most particles are near the centre and fewer are farther out.

Worked example: Four particles that average to zero

1. Four particles end at −3, −1, +1 and +3 micrometres from where they started.
2. The second drawing puts those positions along the bottom, with a bar one particle tall above each.
3. The bars stand two on each side of zero, at the same distances out. The picture is symmetric, so the average position is 0.
4. Yet no bar stands at zero: every particle moved. The graph shows at a glance what the average hides.

Where this lesson stops

Before reading a curve, read its two labels and their units, and ask whether its height is a count, a density or a running total.

This lesson builds on

• A sign records direction

An equals sign states a balance

Is an equals sign an instruction to calculate, or a statement?

In school, 7 + 5 = asks for an answer. In physics the equals sign more often states a fact: two ways of counting the same thing agree.

Here is an account in made-up units. A body starts with 100 units of energy, sends 30 away as light and keeps 70. Energy before equals energy after plus energy sent out: 100 = 70 + 30. The line is a balance to check, not a sum to do, and it is just as true written 100 − 70 = 30.

The mass-energy paper writes the same kind of balance: E₀ − E₁ = L. The body's energy before, minus its energy after, equals the energy L that the light carries away. It is a statement about the two accounts. Nothing in it says to work out E₀ first.

Because an equation states a balance, it can be read from either side and rearranged without changing what it says. E₁ = E₀ − L tells the same story.

Worked example: Rearranging a balance without changing it

1. Start from 100 = 70 + 30, in made-up units.
2. Take 70 from both sides: 100 − 70 = 30. Still true, and still the same account.
3. Put the paper's letters in place of the numbers: E₀ = E₁ + L, or E₀ − E₁ = L.
4. Whichever form is written, it states one fact: what the body lost is what the light carried away.

Where this lesson stops

An equals sign states that two quantities are the same. Which quantities, measured how, is what the paper around it has to say.

This lesson builds on

• bridge-letter-for-quantity

Fractions and ratios

What does dividing one quantity by another tell you?

The fraction 6/2 is 3, because 2 fits into 6 three times; it also says that 6 is three times 2. Every fraction a/b works the same way. The bottom number can never be zero, since no count of zeros adds up to 6.

When the two quantities have different units, their ratio is a new quantity with a unit of its own: 12 micrometres travelled in 4 seconds is 3 micrometres per second. When they have the same unit, the units cancel and the ratio is a plain number: 2 particles out of 8 is 0.25 of the group.

Worked example: Fourteen metres in ten seconds

1. A steady walk covers 14 metres in 10 seconds.
2. Divide the distance by the time: 14 divided by 10 is 1.4, and the unit divides with it, so the rate is 1.4 metres per second.
3. After 60 seconds the same walk has covered 84 metres, and 84 divided by 60 is 1.4 again. The walk is steady, so the ratio does not depend on how long it is watched.
4. Metres and seconds are different units, so neither cancels and what comes out is a new quantity that keeps a unit of its own.
5. The drawing below is the other case: 2 particles out of 8 has the same unit above and below, they cancel, and 0.25 is left as a plain number.

Where this lesson stops

A ratio of two quantities in the same unit is a plain number. A ratio of different units, such as micrometres per second, is a new quantity and keeps its unit.

This lesson builds on

• Adding and averaging

A sign records direction

How can movement add up to zero?

On the ruler below, the starting point is zero. A final mark three units to the right has displacement +3; one three units to the left has displacement −3. Adding the signed displacements gives zero. Adding the distances from the start gives six.

Displacement compares where something ends with where it started, whatever it did in between. A walker who goes three units right and comes back has displacement zero after walking six, which is the lower pair of arrows in the drawing.

Worked example: Two walkers, opposite directions, average zero

Two walkers both finish one unit from their start, one at −1 and the other at +1. The signed average is zero. The average distance is one.

Where this lesson stops

The sign names a direction relative to a chosen axis; it does not mean a negative distance.

This lesson builds on

• Adding and averaging

Powers of ten and physical units

How do you write a very small length without counting zeros?

The Brownian paper works its example, in §5, for particles 0.001 mm across. In metres that is 0.000001 m: easy to miscount by one zero, which is a factor of ten. Written as 10⁻⁶ m, the −6 says the decimal point has moved six places to the left, and that length has a name of its own, the micrometre (μm).

The unit goes through the arithmetic with the number. A micrometre is 10⁻⁶ m, so a square micrometre is 10⁻¹² m², not 10⁻⁶ m²: squaring a length squares its unit, power of ten included.

Worked example: Squaring six micrometres

1. 1 micrometre (1 μm) is 10⁻⁶ metres, or 0.000001 m.
2. The paper's displacement after one minute, about 6 μm, is 6 × 10⁻⁶ m.
3. Squaring it squares the number and the unit: (6 × 10⁻⁶ m)² = 36 × 10⁻¹² m² = 3.6 × 10⁻¹¹ m².

Where this lesson stops

A measurement is a number and a unit together. Squaring it squares both: (6 × 10⁻⁶ m)² is 3.6 × 10⁻¹¹ m².

This lesson builds on

• Fractions and ratios

Squares and square roots

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

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

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

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

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

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

Where this lesson stops

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

This lesson builds on

• Adding and averaging

Adding and averaging

What does an average keep, and what does it lose?

Four jars hold 3, 1, 1 and 3 marbles. Pour them together and there are 8. Share the 8 equally between the four jars and each holds 2. Two is the average, although no jar held 2 to begin with.

The average keeps the total and the count and nothing else, so 3, 1, 1, 3 and 2, 2, 2, 2 have the same average. Listing every value twice doubles both the total and the count, and the average stays at 2.

Worked example: Three mornings with the same average

1. One morning the four jars hold 3, 1, 1 and 3 marbles. Add them: 8. Divide by the four jars: an average of 2.
2. Another morning they hold 8, 0, 0 and 0. Add them: 8 again. Divide by four: 2 again, and now three jars are empty.
3. A third morning every jar holds 2, which is the average itself. Add them: 8. Divide by four: 2.
4. Write any of those three lists out twice over. The total doubles to 16 and the count doubles to 8 jars, and 16 divided by 8 is still 2.
5. Four descriptions and one average. It keeps the total and the count, and the differences between the mornings are what it leaves behind.

Where this lesson stops

An average is the total shared equally among the observations, and two very different lists can have the same one.

Conservation and symmetry

When an experiment is turned, moved or watched from a moving frame, what should stay the same?

Turn an apparatus to face another way, move it to another bench, or run it tomorrow instead of today, and it gives the same result. Each of those changes is a symmetry of the experiment: a change of setting that leaves the measured outcome unchanged. The relativity paper's first postulate adds one more: describing the experiment from a frame in uniform motion leaves the laws in the same form.

What stays the same is often a relation, not a number. The second postulate is a relation kept in both frames: light moves at the same speed c. §3 of the paper checks it on a flash from the origin. A sphere of light, $x^2 + y^2 + z^2 = c^2t^2$, is also a sphere in the moving frame, with the moving frame's own coordinates.

x squared plus y squared plus z squared equals c squared t squared, and the same with every coordinate primed.

Conservation works the same way. An energy is not the same number in every frame, but in each frame the energy before equals the energy after. The mass-energy paper uses exactly this. A body at rest sends out two equal pulses of light, L in all, in opposite directions. Seen from a frame moving along the pulses, the pulses carry different energies, and their total is γL rather than L, where γ = 1/√(1 − v²/c²) is the factor the relativity paper prints as β. Each frame keeps its own balance, and comparing the two balances is the whole argument.

Modern texts add one more quantity that every frame agrees on: $c^2\Delta t^2 - \Delta x^2$ for any two events, the spacetime interval. It is a useful check on a calculation, and its home is Minkowski's geometry of 1908. It is not a premise of the 1905 route, which starts from the two postulates.

Worked example: Two pulses from 0.6c, and one interval checked

1. In the body's frame, two pulses of energy $L/2$ leave in opposite directions along x: $L/2 + L/2 = L$ leaves the body.
2. From a frame moving at 0.6c along x, where γ = 1.25, a pulse sent along the motion carries $\frac{L}{2}\gamma\left(1 - \frac{v}{c}\right) = 0.5L \times 1.25 \times 0.4 = 0.25L$, and the pulse sent back carries $0.5L \times 1.25 \times 1.6 = L$.
3. Their total is $1.25L = \gamma L$, not L. Each frame's account still balances, because each is compared only with itself, before and after.
4. The interval as a check: two events at the same time 10 light-seconds apart have $c^2\Delta t^2 - \Delta x^2 = 0 - 100 = -100$ square light-seconds. From 0.6c they are 7.5 seconds and 12.5 light-seconds apart, and $7.5^2 - 12.5^2 = 56.25 - 156.25 = -100$ again.

Where this lesson stops

A symmetry is a change that leaves the measured outcome unchanged. This lesson stops at naming what stays the same; showing why it must is the papers' work.

This lesson builds on

• Quantities and units
• Functions and graphs
• Events and distant clocks
• matrices-linear-maps

Events and distant clocks

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

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

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

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

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

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

Worked example: Setting a distant clock with one light signal

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

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

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

Where this lesson stops

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

This lesson builds on

• Functions and graphs

Functions and graphs

How does a graph record the way one quantity depends on another?

A bath fills from a tap. After one minute it holds 10 litres, after two 20, after three 30. For each time there is exactly one amount of water, and that pairing, one output for each input, is a function. Plot the pairs with time along the bottom and litres up the side and they fall on a straight line that climbs 10 litres for every minute. The slope, in litres per minute, is how fast the bath fills.

In the papers the pairs are physical: the mean square displacement of a Brownian particle for each elapsed time, or the concentration for each position along a tube. A function says how one thing responds when another changes.

A graph puts the input along the horizontal axis and the output up the vertical one, so each point is one pair. Both axes carry units, and so does everything read off the graph: the slope of mean square displacement against time is in square micrometres per second, and the area under a probability density is a plain probability.

A formula such as ⟨x²⟩ = 2Dt describes an average over many particles. One particle's squared displacement scatters widely around the line; the average over many lies close to it. The Brownian paper predicts the average, and ends by hoping that a researcher will soon decide the question by observation.

Worked example: Mean square displacement plotted against time

The mean square displacement equals two times the diffusion coefficient times the time.

1. Take D = 0.5 square micrometres per second, so 2D = 1 square micrometre per second.
2. At t = 1 s the mean square displacement is 1 µm²; at t = 4 s it is 4 µm²; at t = 9 s, 9 µm².
3. Plotted against time, these points lie on a straight line through the origin whose slope is 2D, 1 µm² per second.
4. Their square roots, the root-mean-square displacements, are 1, 2 and 3 µm: four times the time gives twice the distance.

Where this lesson stops

A graph is a record of paired quantities, each with its units. A slope read off it has units too, and a slope stated without them explains nothing.

This lesson builds on

• Reading a graph

Quantities and units

Can a number of metres be compared with a number of seconds?

"The particle moved 6" says nothing until the unit is added: 6 micrometres, or 6 seconds? A physical quantity is a number times a unit, and the number alone changes when the unit does. Six micrometres is 0.006 millimetres: the same length, written with another unit.

Quantities of different kinds cannot be added or compared: 3 metres plus 2 seconds has no meaning. So in a correct equation every term, on both sides, carries the same units. Checking that is the quickest test a formula can pass or fail.

Take the Brownian diffusion coefficient, $D = k_BT/(6\pi\eta a)$. Boltzmann's constant is in joules per kelvin, T in kelvins, the viscosity η in pascal-seconds and the radius a in metres. The units work out to square metres per second, which is what a diffusion coefficient must be. The worked example below goes through it.

Units also tell apart quantities that share a letter. The light paper's $\rho_\nu$ is an energy per unit volume per unit of frequency, joules per cubic metre per hertz, not per unit of wavelength: two different quantities with different units. And its L is the speed of light in §§1 and 2 but an absorbed energy in §9. A unit check starts from what the quantity is, not from its letter.

Worked example: Checking that D comes out in square metres per second

1. $k_BT$: joules per kelvin times kelvins is joules, which are newton-metres.
2. $6\pi\eta a$: 6π has no units, and pascal-seconds times metres is (newtons per square metre) × seconds × metres, which is newton-seconds per metre.
3. Divide: newton-metres divided by newton-seconds per metre is square metres per second.
4. With numbers: 1.380649 × 10⁻²³ × 290.15 ÷ (6π × 1.35 × 10⁻³ × 5 × 10⁻⁷) = 3.15 × 10⁻¹³ m²/s. That is the modern Boltzmann constant (set modern-si-2019) with the paper's water at 17 °C and its particle.

Where this lesson stops

This lesson stops at checking units. Converting the papers' own units into SI units is a separate step, with its own conventions.

This lesson builds on

• Powers of ten and physical units
• Fractions and ratios

Approximating a curve near a point

Why can a curve be replaced, close to one point, by a few simple terms?

What is 2.1 squared, without a calculator? Start from what you know: 2 squared is 4. Picture a square 2 wide growing to 2.1 wide. It gains two strips, each 2 long and 0.1 thick, which add 0.4, and a small corner, 0.1 by 0.1, which adds 0.01. So 2.1 squared is 4.41. The first correction, 0.4, did almost all the work; the second, 0.01, is forty times smaller.

The same recipe works for any smooth curve, close to a point where you know it. Start with its value there. Add its slope times the step Δ: the slope, or first derivative, is the change in output per unit of input. Then add half its curvature times Δ²: the curvature, or second derivative, says how fast the slope itself changes. Further terms carry higher powers of Δ, so they are small only when Δ is:

The value a step delta away equals the value, plus delta times the slope, plus half of delta squared times the curvature, plus further terms.

The Brownian paper does exactly this in §4. It expands the number of particles at x + Δ in powers of the jump Δ and keeps the terms up to Δ², because only very small jumps contribute and each further surviving term is much smaller than the one before. The Δ term averages away, since jumps to the left and to the right are equally likely, and the Δ² term becomes the diffusion coefficient. A real particle does not make independent jumps over arbitrarily short times, so the expansion holds for times long enough for successive jumps to be independent and short enough for each jump to be small.

Worked example: What the dropped term costs

For the square function, the local expansion is exact through the quadratic term.

Keep only the value and the slope term, 4 + 2 × 2 × Δ, and see what dropping Δ² costs as the step grows:

1. Δ = 0.1: 4.4 instead of 4.41. The error, 0.01, is about 0.2 per cent.
2. Δ = 0.5: 6 instead of 6.25. The error, 0.25, is 4 per cent.
3. Δ = 2: 12 instead of 16. The error, 4, is a quarter of the answer.
4. Halving the step quarters the error, because the dropped term is Δ².

Where this lesson stops

An expansion is a claim about size. Say which terms you keep, and why the ones you drop are smaller: here, because they carry higher powers of a small step.

This lesson builds on

• Squares and square roots

Energy of motion and inertia

If a body moves at the same speed but carries less energy of motion, what has changed?

A force that pushes a body through a distance does work on it, and the work becomes the body's energy of motion, its kinetic energy. That is separate from energy stored inside the body, as heat or in its chemistry.

For one body at everyday speeds, doubling the speed multiplies the energy of motion by four. At a fixed speed, doubling the mass doubles it. The mass-energy paper's title asks about a body's Trägheit, its inertia: its resistance to a change in its motion, which is measured by pushing it, not by weighing it.

In Newtonian mechanics, kinetic energy is one half times inertial mass times speed squared.

This formula holds at speeds small compared with the speed of light, and it is not exact close to that speed. So the mass-energy argument works with the exact expression and reads the mass from its low-speed limit, where it takes this form.

Worked example: Two bodies at the same speed, different masses

1. Two bodies move at 2 metres per second.
2. The 3 kg body has ½ × 3 × 2² = 6 joules of energy of motion; the 2 kg body has ½ × 2 × 2² = 4 joules.
3. The speed is the same, and the energies differ by 2 joules because the masses differ by 1 kg.

At the same low speed, the kinetic-energy drop equals one half times the mass decrease times the squared speed.

Here the two masses are given, so the example proves nothing about mass and energy. The 1905 argument runs the other way: it computes the energy difference from the light the body gives off, finds that it has this form, and reads the mass difference from the coefficient of ½v².

Where this lesson stops

At the same low speed, less energy of motion means less mass. The comparison gives only a difference in mass, never the total energy stored inside the body.

This lesson builds on

• Squares and square roots
• Fractions and ratios

me-01

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open me-01 online

me-02: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open me-02 online

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Settings (canonical units)

beta

0.6

emittedEnergy

1

energyUnit

normalized

notation

printed

showNaive

false

speedAxis

linear

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me-03

This laboratory is not included in this file. The chapter's authored foundation examples remain available above. Open me-03 online

sr-10: a built-in worked result

Static host-calculated example, not a running experiment or an observation. Canonical units and owner-supplied statuses are retained. Open sr-10 online

0 array outputs are not reproduced. This table does not claim to reproduce the interactive plot.

Settings (canonical units)

beta

0.6

initialAmplitude

1

initialEnergyJ

1

initialVolumeM3

1

propagationAngleDeg

0

showCountermodel

true

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