§8 · photoelectric energy and counts · Light quanta: from entropy to an energy scale

§8 · photoelectric energy and counts

More electrons is not more energy per electron

What changes when frequency rises, and what changes when power rises?

The most energetic electron receives a complete quantum and escapes with the least modeled energy loss. Subtracting the escape work $\Phi$ gives the maximum kinetic energy, provided emission is energetically allowed.

K_{\mathrm{max}}=h\nu-\Phi,\qquad\nu_0=\frac{\Phi}{h}

Maximum kinetic energy equals h frequency minus escape work, with threshold frequency Phi over h.

For a hypothetical work function of 2 electron volts and a frequency of 600 terahertz, the maximum is about 0.4814 electron volts using modern SI constants. Doubling optical power while retaining that frequency and escape work leaves the maximum unchanged.

Power is energy arriving per second. At fixed frequency, dividing it by $h\nu$ gives the incoming quantum rate. Under a fixed yield, twice the power can therefore produce twice the electron count per second. Below threshold, negative $h\nu - \Phi$ is an energy deficit, not a negative kinetic energy of an emitted electron.

Return to the slope, threshold, and the distinction between energy and count. (included in this file)

Letters: the paper takes the exciting light to consist of energy quanta of size $\frac{R}{N}\beta\nu$ , writes P for the work an electron must do to leave the body (Φ here), and writes ε for the electron's charge. This explanation writes the quantum as hν, with h a later name for Rβ/N.

The simplest picture, which the paper assumes occurs: a light quantum gives its whole energy to a single electron. The paper also leaves open that some electrons take up only part of a quantum's energy.
An electron excited inside the body loses part of its kinetic energy before it reaches the surface. The fastest electrons are those excited right at the surface and moving straight out.
Each electron must do a work P, characteristic of the body, to leave it. So the kinetic energy of the fastest electrons is the quantum minus that work: $\frac{R}{N}\beta\nu-P$ .
In the explanatory letters, with hν for the quantum and Φ for P, that is the first half of the formula below. Setting it to zero gives the lowest frequency at which the fastest electron can still escape, the threshold $\nu_0=\Phi/h$ .

K_{\mathrm{max}}=h\nu-\Phi,\qquad\nu_0=\frac{\Phi}{h}

Maximum kinetic energy equals h frequency minus escape work, with threshold frequency Phi over h.

Worked case with modern SI constants, for a hypothetical work function of 2 electron volts at a frequency of 600 terahertz: $h\nu=6.62607\times10^{-34}\times6.00\times10^{14}=3.97564\times10^{-19}$ joule, which divided by $1.602177\times10^{-19}$ joule per electron volt is 2.4814 electron volts. Subtract 2: the maximum kinetic energy is 0.4814 electron volts.
The threshold for that work function is $\nu_0=2$ electron volts divided by h, about 484 terahertz.
Below the threshold, $h\nu-\Phi$ is negative. That is a shortfall of energy: no electron escapes by this route, and there is no negative kinetic energy to report.
The paper's own check, in its letters: charge the body to a positive potential Π that just prevents it from losing electricity. Then Π times the electron's charge ε equals the kinetic energy $\frac{R}{N}\beta\nu-P$ .
Multiply by N: the charge becomes E, the charge of one gram-equivalent of a monovalent ion, and the right side becomes Rβν minus the escape work for that amount of charge. The paper sets that work to zero for an estimate, takes $E=9{,}6\cdot10^{3}$ in electromagnetic units, $\nu=1{,}03\cdot10^{15}$ , the limit of the solar spectrum toward the ultraviolet, and $\beta=4{,}866\cdot10^{-11}$ .
Then Π = Rβν/E: $8.31\times10^{7}\times4.866\times10^{-11}\times1.03\times10^{15}/(9.6\times10^{3})=4.34\times10^{8}$ electromagnetic units, with R the standard value of 1905. One volt is $10^{8}$ of those units, so Π is about 4.34 volts. The page gives 4,3 volts and says the result agrees in order of magnitude with Lenard's.
The paper's further prediction: Π, plotted against the frequency of the exciting light, must be a straight line whose slope does not depend on the substance. In the explanatory letters the slope of $K_{\mathrm{max}}$ against ν is h for every metal, and only the intercept, set by Φ, changes.
Power sets the count, not the energy of each electron. Optical power is energy per second; divided by hν it gives quanta per second. At 600 terahertz, one milliwatt carries $2.5\times10^{15}$ quanta per second, and two milliwatts carry twice as many, each still of 2.4814 electron volts.
So, if each quantum gives up its energy independently of all the others, the speeds of the electrons do not depend on the intensity of the light, while the number leaving the body is proportional to it, other things being equal (p. 147). Twice the power can give twice the electrons per second; it cannot raise the maximum energy of one.
The laboratory's straight line follows from the equation it is programmed with. It shows a consequence of the model; only measurements made independently can test the model.

Return to the slope, threshold, and the distinction between energy and count. (included in this file)

Return to the straight line of electron energy against frequency. (included in this file)

The most an escaping electron can keep

Modern teaching equation, not the equation as printed.

K_{\mathrm{max}} = h\,\nu - \Phi

The largest kinetic energy equals h times the frequency, minus the escape work.

One quantum pays the escape work, and what is left is the most energy of motion an electron can carry away. The frequency sets it; the light's power does not appear.

The frequency below which no electron escapes

Modern teaching equation, not the equation as printed.

\nu_0 = \frac{\Phi}{h}

The threshold frequency equals the escape work divided by h.

Setting the largest kinetic energy to zero gives the frequency whose quantum only just pays the escape work. Below it the model emits nothing, however bright the light.

Assumptions

One electron can receive one quantum of energy hν.
Escape requires positive work Φ, with any additional loss nonnegative.
Frequency, work function and yield stay fixed when the power changes.

Limits

In the laboratory the threshold and the straight line follow from the model’s assumptions; they are not experimental proof.
The work function used in a hypothetical example is not a sourced value for a named metal.
Real collection and emission rates require additional material and apparatus assumptions.

Einstein's own limits on the photoelectric laws

Where does Einstein expect his photoelectric regularities might stop holding?

After working out how the escaping electrons should behave, Einstein adds one sentence of caution: about the presumable limits of validity of these regularities, remarks would have to be made like those about the presumable deviations from Stokes's rule. His word, 'mutmaßlich', means presumable: he expects limits, and he does not locate them.

The regularities are those of the paragraph before. If each energy quantum gives its energy to electrons independently of all the others, the speeds of the escaping electrons do not depend on the intensity of the exciting light, while their number is proportional to it. The energy relation behind them assumes one quantum for one electron:

ΠE=R\beta\nu-P'

Pi times E equals R beta nu minus P prime.

The remarks on Stokes's rule, in §7, named two conditions. First, so many quanta are being transformed at once in each unit of volume that one produced quantum can take its energy from several exciting quanta. Second, the light is not of the kind black radiation has within the range of Wien's law, for instance light from a body so hot that Wien's law no longer holds at the wavelength concerned.

Carried over, as the sentence invites: in very intense light an electron might gather energy from more than one quantum, and then its energy is no longer bounded by one quantum's (R/N)βν less the work P it spends leaving. For light outside Wien's range, the entropy argument of §6 that gave quanta of size Rβν/N does not apply, so it does not secure the size of the energy portions. Einstein spells out neither case for the photoelectric effect; the caution stays a caution.

Return to the energy one quantum can give an electron. (included in this file)

Letters: Π is the potential the illuminated body reaches, E the charge of a gram-equivalent of a univalent ion, $P'$ the work for a gram-equivalent of electrons to leave the body, R the gas constant, N the number of real molecules in a gram-equivalent, β the constant from Planck's formula in §2, and ν the frequency. (R/N)βν is the energy of one quantum; later physics writes it hν.

The regularities being qualified come from the paragraph before: each quantum gives its energy to electrons independently of all the others.
Consequence one: the speeds of the escaping electrons, the quality of the cathode rays, do not depend on the intensity of the light, because each electron still receives the energy of a single quantum.
Consequence two: the number of electrons leaving is proportional to the intensity, other things equal, because doubling the intensity doubles the number of quanta arriving.
Both rest on one quantum for one electron, which gives the stopping condition, for a gram-equivalent of electrons:

ΠE=R\beta\nu-P'

Pi times E equals R beta nu minus P prime.

Worked case, with the paper's own check: $P'=0$ , $R=8.31\times10^{7}$ , $\beta=4.866\times10^{-11}$ , $\nu=1.03\times10^{15}$ and $E=9.6\times10^{3}$ . Then $R\beta\nu=8.31\times10^{7}\times4.866\times10^{-11}\times1.03\times10^{15}\approx4.16\times10^{12}$ , and $Π=R\beta\nu/E\approx4.3\times10^{8}$ , which the paper converts to about 4.3 volts. One quantum, one electron.
The first condition from §7, carried over: if so many quanta are being transformed at once in each unit of volume that one electron could take energy from two quanta, the energy available to it could reach twice (R/N)βν, and the equation above would no longer bound it. The paper names no intensity at which that begins.
The second condition from §7, carried over: the quantum picture came from the entropy of radiation within the range of Wien's law (§6). Light well outside that range, such as light from a very hot source at the wavelength used, is not covered by that argument, so the argument does not fix the size of its energy portions.
What Einstein asserts is only that such remarks would have to be made. He calls the limits presumable and leaves their details open, so neither carried-over case is a prediction of the paper.

Return to the size of the stopping potential in the paper's check. (included in this file)

Assumptions

The regularities in question are those of the paragraph before: the speeds of the escaping electrons do not depend on the intensity of the light, and their number is proportional to it.
In §7 Einstein named two conditions under which Stokes's rule might fail.
§8 says only that similar remarks would apply to the photoelectric laws; it does not spell them out.

Limits

The two conditions are carried over from §7 by the analogy Einstein's sentence invites; §8 does not state them itself, so this passage marks them as a reading of his remark.
The paper gives no intensity and no temperature at which the laws fail, and neither does this passage.

A stopping voltage is a magnitude with a sign convention

Why does retarding an electron not make its charge positive?

eV_s=K_{\mathrm{max}}=h\nu-\Phi

The positive charge magnitude times the stopping-voltage magnitude equals maximum kinetic energy, h frequency minus escape work.

$V_s$ is a nonnegative magnitude. The sign of a laboratory electrode voltage must also specify which electrode is the reference. Using $e$ for a positive magnitude does not change the negative charge of the electron.

If the electron receives only part of the available energy or loses energy inside matter, its outgoing kinetic energy is at most $h\nu - \Phi$ . These losses do not reverse the frequency dependence or turn extra intensity into extra energy for one electron.

A footnote to the stopping condition allows that the light may also have to free each electron from a neutral molecule, at the cost of some work, before the electron can leave. The paper says nothing in the derived relation changes: its $P'$ , which it defines as the potential of a gram-equivalent of negative electricity with respect to the body, is then read as a sum of two terms, the work to free the electron and the work to leave the body. In the letters used here, $\Phi$ stands for that sum, and the stopping potential still rises in a straight line with frequency, with the same slope.

An inverse energy-budget question asks for the smallest accelerating-potential magnitude capable of supplying a quantum $h\nu$ . Within that idealized conversion, $eV_{\mathrm{acc}}$ must be at least $h\nu$ . This is an energetic lower bound, not a promise that every collision produces light.

Return to charge magnitude, signed voltage, and the retarding energy budget. (included in this file)

Return to the stopping voltage, beside the paper's 4.3-volt check in electromagnetic units. (included in this file)

Letters: the paper charges the illuminated body to a positive potential Π and writes ε for the electron's charge. This explanation writes $V_s$ for the size of the stopping potential and e for the positive size of the electron's charge; the electron's charge itself is −e.

Set the arrangement and the sign: the body is at the potential $V_s$ above the conductors around it, which are at zero. An electron leaving the body is pulled back toward it, so the potential retards the electron.
An electron moving from the body out to the surroundings, across a potential difference $V_s$ , loses kinetic energy $eV_s$ . If it had kinetic energy K at the surface, it arrives with $K-eV_s$ , or is turned back if that would be negative.
The body stops losing electricity when even the fastest electron is turned back, that is, when $eV_s$ equals the largest kinetic energy $K_{\mathrm{max}}$ .
Insert the one-quantum budget of the previous passage for $K_{\mathrm{max}}$ .

eV_s=K_{\mathrm{max}}=h\nu-\Phi

The positive charge magnitude times the stopping-voltage magnitude equals maximum kinetic energy, h frequency minus escape work.

Divide by e: $V_s=(h\nu-\Phi)/e$ . Worked with the previous passage's hypothetical numbers: $K_{\mathrm{max}}$ is 0.4814 electron volts, so $V_s$ is 0.4814 volts, since one electron volt is e times one volt.
Keep sizes and signs apart. $V_s$ is a size, zero or positive. The electron's charge is negative, and writing e for its size does not change that. A voltage read from an instrument also needs its reference electrode named.
The paper's gram-equivalent form of the stopping condition is $ΠE=R\beta\nu-P'$ , with $P'$ the potential of a gram-equivalent of negative electricity with respect to the body: the work for that much electricity to leave it. A footnote attached to it takes up one more possibility.
Suppose the light must first free each electron from a neutral molecule, at the cost of a certain work, before the electron can make its way out. The footnote says nothing in the derived relation changes; $P'$ is then to be read as a sum of two terms.
Why nothing changes: one quantum's energy, (R/N)βν per electron ( $h\nu$ in the explanatory letters), now pays two works in turn, freeing the electron from its molecule and taking it out of the body. Both come out of the same quantum, so the fastest electron keeps the quantum's energy less their sum. Write $\Phi$ for the sum, and $eV_s=h\nu-\Phi$ stands as before.
Worked case, with hypothetical numbers: freeing the electron costs 1.0 electron volt and leaving the body 2.0, so $\Phi$ is 3.0 electron volts. A quantum of 3.5 electron volts leaves the fastest electron 3.5 − 1.0 − 2.0 = 0.5 electron volts, and $V_s$ is 0.5 volts. A quantum of 2.5 electron volts releases no electron from the body.
What stays: the frequency enters only through the quantum's energy, so the stopping potential plotted against frequency is still a straight line whose slope does not depend on the substance. The two works together only move the frequency at which the line reaches zero.
Partial transfer and losses: if an electron receives only part of a quantum, or loses energy inside the body, it reaches the surface with less than $h\nu-\Phi$ . The paper then replaces the equation by an inequality: in its letters, Π times E, plus the escape work for a gram-equivalent, is at most Rβν.
In the explanatory letters: $eV_s\le h\nu-\Phi$ . Losses can only lower the stopping potential; they cannot give one electron more than the quantum it received.
The inverse process, cathode luminescence: electrons that have fallen through a potential produce light. Reversing the budget, the energy supplied per electron must be at least the energy of the quantum produced, $eV_{\mathrm{acc}}\ge h\nu$ in the explanatory letters.
Worked scale from the paper: in the substances Lenard examined, the cathode rays had to fall through several hundred, in some cases several thousand, volts to produce visible light, far more than the few volts one quantum of visible light needs. The paper concludes that the kinetic energy of one electron is used to produce many light quanta.
A bound is not a promise: $eV_{\mathrm{acc}}\ge h\nu$ states what is needed in energy, not that every collision produces light.

Return to charge magnitude, signed voltage, and the retarding energy budget. (included in this file)

Return to the stopping voltage, beside the paper's 4.3-volt check in electromagnetic units. (included in this file)

What the stopping potential measures

Modern teaching equation, not the equation as printed.

e\,V_{\mathrm{s}} = K_{\mathrm{max}} = h\,\nu - \Phi

The electron's charge times the stopping potential equals the largest kinetic energy, which equals h times the frequency minus the escape work.

A retarding potential that just stops the fastest electrons measures their energy: charge times potential. The one-quantum budget says what that energy should be.

Assumptions

e denotes a positive charge magnitude; the electron charge is −e.
A retarding potential removes kinetic energy, with the sign convention of the electrodes stated.

Limits

No stopping endpoint for emitted electrons is supplied when emission is not allowed.
A negative energy deficit and a signed collector voltage are different quantities.
The footnote gives no size for the work of freeing an electron from a molecule; the passage's figures for it are hypothetical.

Foundations included in this file

Reading a graph

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

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

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

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

Worked example: Four particles that average to zero

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

Where this lesson stops

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

This lesson builds on

A sign records direction

Return to the chapter

Fractions and ratios

What does dividing one quantity by another tell you?

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

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

Worked example: Fourteen metres in ten seconds

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

Where this lesson stops

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

This lesson builds on

Adding and averaging

Return to the chapter

A sign records direction

How can movement add up to zero?

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

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

Worked example: Two walkers, opposite directions, average zero

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

Where this lesson stops

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

This lesson builds on

Adding and averaging

Return to the chapter

Powers of ten and physical units

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

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

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

Worked example: Squaring six micrometres

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

Where this lesson stops

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

This lesson builds on

Fractions and ratios

Return to the chapter

Adding and averaging

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

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

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

Worked example: Three mornings with the same average

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

Where this lesson stops

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

Return to the chapter

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{dx}{dt} = \lim_{\Delta t \to 0} \frac{x(t+\Delta t) - x(t)}{\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(t+\Delta t)^2 - 3t^2 = 6t\,\Delta t + 3(\Delta t)^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{6t\,\Delta t + 3(\Delta t)^2}{\Delta t} = 6t + 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.

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.

This lesson builds on

Functions and graphs

Return to the chapter

Charges, fields and light

What does an electric or a magnetic field do to a charge?

A comb run through dry hair lifts small scraps of paper before it touches them. The charge on the comb fills the space around it with an electric field, and the field pushes on any charge placed in it, harder for a larger charge. For a charge q in a field E the force is qE, along the field for a positive charge and against it for a negative one.

A magnet does nothing to a charge at rest. A charge moving with velocity v through a magnetic field B feels a push at right angles both to its motion and to the field, which is what turns an electric motor. Together these are the Lorentz force:

\mathbf{F} = q\left(\mathbf{E} + \mathbf{v} \times \mathbf{B}\right)

The force equals q times the sum of the electric field and the cross product of the velocity with the magnetic field.

Carrying a charge q across a potential difference ΔV changes its energy by qΔV. The electron's charge has magnitude e = 1.602 × 10⁻¹⁹ coulombs, so one volt changes an electron's energy by 1.602 × 10⁻¹⁹ joules. That amount is called one electron volt, 1 eV. §8 of the light-quanta paper states the energy of electrons that light frees from a metal in these terms: by the potential needed to stop them.

An electron held near a point by a force proportional to its displacement, like a mass on a spring, vibrates at a natural frequency of its own, and it absorbs and emits light near that frequency. §1 of the light-quanta paper fills a closed box with such bound electrons, which it calls resonators, together with free electrons, gas molecules and radiation, and asks what equilibrium requires. Using Planck's condition of 1900 for resonators and radiation, the answer gives the radiation an infinite total energy. That is the difficulty in the section's title.

Maxwell's equations tie fields to charges and currents in four statements: charges are the sources of electric field; there are no magnetic charges; a changing magnetic field drives a circulating electric field, which is Faraday's induction; and currents and changing electric fields drive circulating magnetic fields. Together they allow waves of field that cross empty space at the speed of light, c.

The relativity paper opens with a case from this theory. Move a magnet past a conductor, and the theory describes an electric field that drives a current. Move the conductor past the magnet instead, and it describes no electric field, only a magnetic force on the conductor's charges. The current is the same either way. The paper counts this difference of description, with no difference in what is observed, among its grounds for taking only relative motion to matter.

Worked example: An electron across one volt

\Delta W = q\,\Delta V

The change in energy equals the charge times the potential difference.

An electron starts at rest and crosses a potential difference of 1 V.
It gains energy e × 1 V = 1.602 × 10⁻¹⁹ joules: one electron volt.
At low speed that energy is ½mv², with the electron's mass m = 9.109 × 10⁻³¹ kg. So v² = 2 × 1.602 × 10⁻¹⁹ ÷ 9.109 × 10⁻³¹ = 3.52 × 10¹¹ m²/s².
The square root gives v ≈ 5.93 × 10⁵ m/s, about 600 kilometres per second.

That is fast, but only 0.2 per cent of the speed of light, so the low-speed formula ½mv² is safe here. The charge and mass used are modern values.

Where this lesson stops

Charge is the property that makes electric forces. Fields carry the force between separated charges, without action at a distance.

Return to the chapter

Functions and graphs

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

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

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

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

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

Worked example: Mean square displacement plotted against time

\langle x^2\rangle = 2Dt

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

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

Where this lesson stops

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

This lesson builds on

Reading a graph

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Orders of magnitude

How big, how fast and how energetic are the things in the 1905 papers, measured against each other?

Two numbers of the same order differ by less than a factor of ten. Numbers several orders apart belong to different worlds, and comparing them usually tells you which effect can be ignored.

Sizes. A water molecule is about 0.3 nanometres across and a Brownian grain about 1 micrometre: the grain is roughly 3,000 times larger. The light-quanta paper prints the mass of a hydrogen atom as 1/N gram, 1.62 × 10⁻²⁴ g.

Speeds against light. The ratio v/c is about 3.3 × 10⁻⁶ for a rifle bullet at 1,000 m/s, 8.3 × 10⁻⁷ for a jet at 250 m/s, and 9.9 × 10⁻⁵ for the Earth in its orbit at 29.8 km/s. The corrections of relativity go as the square of v/c, so for the Earth they are about 10⁻⁸. Only Kaufmann's fast electrons, a large fraction of the speed of light, reached speeds where they are not small.

Energies of light. One quantum hν of red light (650 nm) carries about 1.91 electron volts, green (530 nm) 2.34 eV, and ultraviolet (250 nm) 4.96 eV. The light paper's photoelectric estimate, about 4.3 volts, is of the same order.

Molecular kicks. An estimate of how often water molecules strike a grain 1 μm across gives about 1.6 × 10¹⁹ blows a second, and about 6 × 10¹⁹ for a grain of 1 μm radius. The assumptions are rough, so the honest statement is 10¹⁹ to 10²⁰ blows a second: far too many for any single blow to be seen.

Worked example: Two estimates, done with powers of ten

The Earth: v/c = 29,800 ÷ 299,792,458 ≈ 9.9 × 10⁻⁵. Squared, that is about 9.9 × 10⁻⁹, near 10⁻⁸.
Water's molecules: 1,000 kg per cubic metre divided by 0.018015 kg per gram-molecule, times 6.02 × 10²³, is 3.34 × 10²⁸ molecules per cubic metre.
Molecules near 600 m/s crossing a surface at a quarter of n times their mean speed give about 5 × 10³⁰ strikes per square metre each second.
A sphere 1 μm across has a surface of 3.1 × 10⁻¹² m², so it takes about 1.6 × 10¹⁹ strikes a second. The inputs are rough, so the result is an order of magnitude, 10¹⁹ to 10²⁰.

Where this lesson stops

This lesson places numbers on a scale of powers of ten. Converting the papers' printed units into SI units is a separate step, with its own conventions.

This lesson builds on

Powers of ten and physical units
Fractions and ratios

Return to the chapter

The units of 1905

How do Einstein's numbers, printed in centimetres, grams, ergs and the old electrical units, become SI units?

Einstein's papers measure length in centimetres, mass in grams and time in seconds, the CGS system. Energy is in ergs, and one erg is 10⁻⁷ joules. Viscosity is in grams per centimetre per second, the unit now called the poise, and one poise is 0.1 pascal-seconds. These mechanical factors are exact, because they follow from 1 cm = 0.01 m and 1 g = 0.001 kg.

To convert, multiply by a ratio equal to one. Since 1 poise is 0.1 pascal-seconds, the ratio (0.1 Pa s)/(1 P) equals one, and multiplying by it changes the unit without changing the viscosity. The Brownian paper gives water's viscosity as 1,35 · 10⁻², with no unit name; it is in poise, so the viscosity is 1.35 × 10⁻³ Pa s. The paper calls this viscosity k, which is not Boltzmann's constant.

\begin{aligned} &1.35\times10^{-2}\,\mathrm{P}\times\frac{0.1\,\mathrm{Pa\,s}}{1\,\mathrm{P}} \\ &\quad = 1.35\times10^{-3}\,\mathrm{Pa\,s} \end{aligned}

1.35 times ten to the minus two poise, times 0.1 pascal-seconds per poise, equals 1.35 times ten to the minus three pascal-seconds.

The same ratio works for compound units. The gas constant R = 8.31 × 10⁷ erg per mole per kelvin is 8.31 joules per mole per kelvin. Neither the light paper nor the Brownian paper prints this number; the edition supplies the standard value of 1905 as an editorial input and labels it that way.

Electricity had two CGS systems. The electrostatic system builds its charge unit, the statcoulomb, from the force between charges. The electromagnetic system builds its current unit from the force between currents; its units are the abcoulomb and the abvolt. One abcoulomb is 10 coulombs, one abvolt is 10⁻⁸ volts, one statvolt is 299.792458 volts, and one gauss is 10⁻⁴ tesla. Planck's 1901 elementary charge, 4,69 · 10⁻¹⁰ statcoulombs, is 1.56 × 10⁻¹⁹ coulombs.

These electrical correspondences are conventional, not exact. They assume the magnetic constant $\mu_0$ is exactly 4π × 10⁻⁷ henry per metre. Since the 2019 revision of the SI it is a measured quantity, equal to that value within about one part in a billion. The site marks every electrical conversion conventional.

The light paper's photoelectric check in §8 uses the electromagnetic system. It takes E = 9,6 · 10³ for the charge of one gram-equivalent of a monovalent ion, so a potential Π comes out in abvolts, and the paper multiplies by 10⁻⁸ to get volts. It prints the result as about 4,3 volts. On the result line the factor is set as 10⁷; the edition records this as a misprint, because the sentence four lines earlier gives 10⁻⁸. The check can also be run per electron, with Planck's charge in statcoulombs and the result in statvolts, and it gives 4.31 volts; the edition keeps that as a documented alternative and follows the page.

The paper's R, E and J are amounts per gram-equivalent, a mole of monovalent ions. Dividing by N gives the amount per molecule, R/N, or per electron, ε. With the paper's E and the N it prints in §2, 6,17 · 10²³, the charge on one ion is 9.6 × 10³ ÷ 6.17 × 10²³ = 1.56 × 10⁻²⁰ abcoulombs, which is 1.56 × 10⁻¹⁹ coulombs. The paper does not print this quotient.

The relativity paper writes its fields in Gaussian units, where electric and magnetic fields share one unit. That is why its transformation reads β(Y − (v/V)N), with the dimensionless ratio v/V; here N is a magnetic field component, not Avogadro's number. SI measures the electric field in volts per metre and the magnetic field in tesla, so the same law reads $\gamma(E_y - vB_z)$ , with the speed itself multiplying the magnetic field. One statvolt per centimetre is 2.998 × 10⁴ volts per metre.

\begin{gathered} 1\,\mathrm{abV} = 10^{-8}\,\mathrm{V}, \\ 1\,\mathrm{abC} = 10\,\mathrm{C}, \\ 1\,\mathrm{statV} = 299.792458\,\mathrm{V}, \\ 1\,\mathrm{G} = 10^{-4}\,\mathrm{T} \end{gathered}

One abvolt corresponds to ten to the minus eight volts, one abcoulomb to ten coulombs, one statvolt to 299.792458 volts, and one gauss to ten to the minus four tesla, all by convention.

Worked example: The photoelectric check, converted to volts

The inputs: β = 4,866 · 10⁻¹¹ and ν = 1,03 · 10¹⁵ per second are printed in §8, and so is E = 9,6 · 10³. R = 8.31 × 10⁷ erg per mole per kelvin is the edition's editorial input, because the paper does not print it.
With P′ = 0, Π = Rβν/E = 8.31 × 10⁷ × 4.866 × 10⁻¹¹ × 1.03 × 10¹⁵ ÷ 9.6 × 10³ = 4.34 × 10⁸ abvolts.
Multiply by the ratio 10⁻⁸ volts per abvolt: 4.34 volts. The paper prints about 4,3 volts.
In §9 the conversion runs the other way. An energy per gram-equivalent divided by E is a potential: the printed Rβν = 6,4 · 10¹² erg gives 6.4 × 10¹² ÷ 9.6 × 10³ = 6.67 × 10⁸ abvolts, or 6.67 volts.
Stark's ionization voltage of 10 volts is 10⁹ abvolts, and 10⁹ × 9.6 × 10³ = 9.6 × 10¹² erg per gram-equivalent, the upper bound for J that §9 prints.
E itself, 9.6 × 10³ abcoulombs, is 9.6 × 10⁴ coulombs per gram-equivalent. Beside it, the modern Faraday constant is 96 485 coulombs per mole (set modern-si-2019); Einstein's value is 0.50 percent below it.

Where this lesson stops

A conversion factor is a ratio equal to one. This lesson converts units only: it never turns a historical constant's value into a modern one, and it treats every electrical correspondence as a convention.

This lesson builds on

Powers of ten and physical units
Fractions and ratios
quantities-units

Return to the chapter

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.

K=\frac{1}{2}mv^2

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

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

K_0-K_1=\frac{1}{2}(m_0-m_1)v^2

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

bridge-squaring-square-roots
Fractions and ratios

Return to the chapter

Static laboratory examples

lq-08: 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 lq-08 online

Build-time scalar readouts
Quantity	Value	Unit	Status
incidentPower	0.001	W	value
frequency	600000000000000	Hz	value
workFunction	2.2	eV	value
quantumEfficiency	0.1	1	value
collectorPotential	0	V	value
quantumEnergy	3.9756420899999997e-19	J	value
thresholdFrequency	531957633258682	Hz	value
maxKineticEnergy	4.508534951999996e-20	J	value
stoppingPotentialMagnitude	0.28140061815431494	V	value
quantumRate	2515316966070253.5	s^-1	value
emissionRate	251531696607025.38	s^-1	value
photocurrent	0.00004029982070141531	A	value

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

Settings (canonical units)

collectorPotential: 0
frequency: 600000000000000
incidentPower: 0.001
quantumEfficiency: 0.1
workFunction: 2.2

Evaluator: source:sha256:cb3a52018327607422cd26eae5cd67dbef5a987c590c9dfb64309da1c96bb3f7

Source references

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).
A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.
A. Einstein, Zur Elektrodynamik bewegter Körper. Annalen der Physik (4), 17, 891–921 (1905).
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.
R. A. Millikan, A Direct Photoelectric Determination of Planck's "h". Physical Review (2), 7 (3), 355–388 (1916).