§8 · finite light energy and moving mirrors · Special relativity: from clock operations to electrodynamics

§8 · finite light energy and moving mirrors

Density is not the energy of the whole packet

Why does a finite light complex need a volume calculation?

\frac{u'}{u}=q^2,\qquad\frac{V_{\mathrm{packet}}'}{V_{\mathrm{packet}}}=\frac{1}{q},\qquad\frac{E'}{E}=q

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

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

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

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

Return to why a light complex needs its volume. (included in this file)

Letters: in §8 the paper's light has amplitude A in K and $A'$ in k, and $A^2/8\pi$ is its energy per unit volume in Gaussian units; a, b, c are the direction cosines of the wave normal in K, R the radius of a sphere that moves with the light, S and $S'$ the volumes enclosed, E and $E'$ the energy of the light complex, and V the speed of light. This explanation writes u for the energy density, $V_{\mathrm{packet}}$ for the volume, and q for the frequency factor $\gamma(1-(v/c)\cos\varphi)$ of §7.

The density. §7 found that the amplitude changes by the same factor q as the frequency, so the energy density, which goes as the amplitude squared, changes by $q^2$ .
The paper points out that $A'^2/A^2$ would be the ratio of the energies only if the light complex had the same volume in K and k. It does not.
A boundary around the same light: the sphere $(x-Vat)^2+(y-Vbt)^2+(z-Vct)^2=R^2$ moves with the light, so no energy crosses its surface, and it always encloses the same light complex.
The same surface seen in k at the instant τ = 0: substitute $x=\beta ξ$ (since τ = 0 means $t=vx/V^2$ ), $t=\beta vξ/V^2$ , y = η, z = ζ. The paper prints the result, an ellipsoid: $(\beta ξ-a\beta\frac{v}{V}ξ)^2+(η-b\beta\frac{v}{V}ξ)^2+(ζ-c\beta\frac{v}{V}ξ)^2=R^2$ .
Its volume. The left side is a sphere of radius R in new coordinates $u_1=\beta(1-av/V)ξ$ , $u_2=η-b\beta(v/V)ξ$ , $u_3=ζ-c\beta(v/V)ξ$ . That change of coordinates is triangular, so it multiplies volumes by the product of its diagonal, $\beta(1-av/V)\cdot1\cdot1$ .
So the ellipsoid's volume is the sphere's divided by $\beta(1-a\,v/V)$ , which is q: $S'/S=\sqrt{1-(v/V)^2}/(1-(v/V)\cos\varphi)=1/q$ , as the paper prints it.
Multiply density by volume: $q^2\times1/q=q$ . Together, the three formulas below.

\frac{u'}{u}=q^2,\qquad\frac{V_{\mathrm{packet}}'}{V_{\mathrm{packet}}}=\frac{1}{q},\qquad\frac{E'}{E}=q

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

The paper's result, printed as $E'/E=(1-\frac{v}{V}\cos\varphi)/\sqrt{1-(v/V)^2}$ , becomes $\sqrt{(1-v/V)/(1+v/V)}$ for φ = 0. It remarks that the energy and the frequency of a light complex change with the observer's motion by the same law.
Worked at 0.6c, light travelling the same way as the observer: q = 0.5. The density falls to 0.25 of its value, the volume doubles, and the energy is $0.25\times2=0.5$ of what it was.
The tempting mistake: treat the light complex as a body at rest in K and shrink its volume by 1/1.25 = 0.8. That gives $0.25\times0.8=0.2$ , not 0.5. The boundary moves with the light, not with a body at rest in either frame, so the contraction of a material body does not apply to it.
A case that cannot catch the mistake: light crossing the motion at right angles in K has q = 1.25, and 1/q = 0.8 happens to equal the material factor. Test with light along the motion.
The calculation needs no light quanta. The mass-energy paper of September 1905 starts from exactly this transformation of light energy, and does not assume its own conclusion in advance.

Return to why a light complex needs its volume. (included in this file)

Energy density: q squared

Modern teaching equation, not the equation as printed.

\frac{u'}{u} = q^{2}

Energy density scales by q squared.

The energy per cubic metre of the light changes by the square of the frequency ratio. At q = 0.5 it falls to a quarter.

Total energy: q

Modern teaching equation, not the equation as printed.

\frac{E'}{E} = q

Total energy scales by q.

Density times volume: q squared times one over q leaves q. At q = 0.5 the packet's energy halves. The mass-energy paper imports exactly this ratio.

Packet volume: one over q

Modern teaching equation, not the equation as printed.

\frac{V_{\mathrm{packet}}'}{V_{\mathrm{packet}}} = \frac{1}{q}

Packet volume scales by inverse q.

The volume of the bounded piece of light changes by one over q. Its surface travels with the light, so this is not the material contraction by gamma: at q = 0.5 the volume doubles.

Assumptions

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

Limits

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

Reflection starts with interception

Can the incident ray reach a receding mirror?

\frac{\nu_{\mathrm{ref}}}{\nu}=\frac{1-2(v/c)\cos\varphi+(v/c)^2}{1-(v/c)^2}

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

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

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

Return to whether the light can reach a receding mirror. (included in this file)

Letters: the paper's mirror is the plane ξ = 0 of k, moving at v along x; the incident light has amplitude A, direction cosine cos φ and frequency ν in K, and the reflected light $A'''$ , $\varphi'''$ and $\nu'''$ ; P is the light pressure and V the speed of light. $\nu_{\mathrm{ref}}$ below is the paper's $\nu'''$ .

First, does the light reach the mirror? The light's speed along the mirror's normal is c cos φ, and the mirror recedes at v. If $c\cos\varphi$ is not greater than v, the light never catches the mirror and there is no reflection to calculate.
Otherwise go to the mirror's frame k with §7's rules: $A'=A\,q$ , $\cos\varphi'=(\cos\varphi-v/V)/(1-(v/V)\cos\varphi)$ , $\nu'=\nu\,q$ , with $q=(1-(v/V)\cos\varphi)/\sqrt{1-(v/V)^2}$ .
Reflect in k, where the mirror is at rest: $A''=A'$ , $\cos\varphi''=-\cos\varphi'$ , $\nu''=\nu'$ .
Transform back to K, with −v: each quantity gains the factor $(1+(v/V)\cos\varphi'')/\sqrt{1-(v/V)^2}$ . Put in $\cos\varphi''$ and collect: $A'''=A\frac{1-2\frac{v}{V}\cos\varphi+(v/V)^2}{1-(v/V)^2}$ , as the paper prints it, and the frequency changes by the same factor.

\frac{\nu_{\mathrm{ref}}}{\nu}=\frac{1-2(v/c)\cos\varphi+(v/c)^2}{1-(v/c)^2}

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

Worked at normal incidence, v = 0.6c: $(1-1.2+0.36)/(1-0.36)=0.16/0.64$ = 0.25. The reflected light has a quarter of the incident frequency.
Energy arriving at unit area of the mirror per unit time, measured in K: $\frac{A^2}{8\pi}(V\cos\varphi-v)$ . The mirror recedes, so it intercepts light at the rate $V\cos\varphi-v$ , not V. At normal incidence and 0.6c that is 0.4 of a fixed mirror's rate.
Energy leaving unit area per unit time: $\frac{A'''^2}{8\pi}(-V\cos\varphi'''+v)$ . By the energy principle, the difference is the work the light pressure does per unit time, which the paper sets equal to P times v.
Solving gives the pressure, as printed: $P=2\frac{A^2}{8\pi}\frac{(\cos\varphi-v/V)^2}{1-(v/V)^2}$ .
Worked at normal incidence and 0.6c, with u the incident energy density $A^2/8\pi$ : $P=2u\times0.16/0.64$ = 0.5 u.
To first order in v/V the paper recovers $P=2\frac{A^2}{8\pi}\cos^2\varphi$ , in agreement with experience and with other theories.
The paper's method in one line: transform the light to a frame in which the body is at rest, solve the problem of optics there, and transform back. It says every problem of the optics of moving bodies can be reduced this way to problems of the optics of bodies at rest.
The mirror here is ideal: its motion is prescribed, and it has no recoil, absorption or thickness.

Return to whether the light can reach a receding mirror. (included in this file)

Light reflected from a moving mirror

Modern teaching equation, not the equation as printed.

\frac{\nu_{\mathrm{ref}}}{\nu} = \frac{1 - 2\,\frac{v}{c}\,\cos\left(\varphi\right) + \left(\frac{v}{c}\right)^{2}}{1 - \left(\frac{v}{c}\right)^{2}}

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

Transform into the mirror's rest system, reflect there, and transform back. At normal incidence and v = 0.6c, receding, the reflected frequency is a quarter of the incident one.

Assumptions

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

Limits

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

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

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

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

Density is not probability

What does the height of a probability curve mean?

Mark where 100 particles end up on a ruler divided into 2 μm bins. If 30 land in one bin, that bin holds a probability of about 0.3, and the density there is 0.3 divided by 2 μm, or 0.15 per micrometre. Measured in metres the same density is 150,000 per metre, while the bin still holds 0.3: the height depends on the unit of length, and the probability does not.

P(a\le X\le b)=\int_a^b p(x)\,dx

The probability of the interval from a to b is the integral of the density over that interval.

Under a smooth curve, any single exact position has probability zero, since an interval of zero width has no area. At the starting instant of the Brownian problem every particle sits at one point. No curve can draw that: the whole probability, 1, sits on the single point, and a curve appears only once the particles have begun to spread.

Worked example: A flat density across four micrometres

Spread the probability evenly over 4 μm. The density is 1 divided by 4 μm, or 0.25 per micrometre.
The area over the whole 4 μm is 0.25 per μm × 4 μm = 1.
A 1 μm piece has probability 0.25 × 1 = 0.25, and a 2 μm piece has 0.5.
Squeeze the same probability into 2 μm and the height doubles to 0.5 per micrometre, while the total stays 1. The height was never a probability.

Where this lesson stops

The height of the curve is a density, a probability per unit length. Only the area over an interval is a probability.

This lesson builds on

Adding and averaging

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

Return to the chapter

Adding continuously

How do you add up something that varies continuously?

Chance spread along a line is measured per unit length: a density. Where the density is the same all along a stretch, the chance of the stretch is the density times its length. A particle equally likely to be anywhere in a 4 µm stretch has density 0.25 per micrometre, and the chance of finding it in the first 1.5 µm is 0.25 × 1.5 = 0.375.

Where the density changes from place to place, cut the line into narrow strips. Each strip contributes its height times its width, as if the density were flat across it. Add the strips, then cut them narrower and add again. For a smooth curve the totals settle on one value, and that value is the integral:

\int_a^b p(x)\,dx = \lim_{\Delta x \to 0} \sum_i p(x_i)\,\Delta x

The integral of p of x from a to b is the value that the sum of p of x i times delta x settles on as delta x shrinks.

The units multiply. A probability density is a probability per metre, so a strip of it, times a width in metres, is a plain probability. For a density the whole area must be 1, because the particle has to be somewhere. The Brownian paper's §4 writes exactly that for the chances of a jump Δ, ∫φ(Δ) dΔ = 1, and the light-quanta paper's §3 adds entropy over every frequency, S = v∫φ dν.

One more tool recurs in the papers: integration by parts. It is the product rule for derivatives added up over an interval, and it moves a derivative from one factor to the other at the price of an endpoint term:

\int_a^b u\frac{dv}{dx}\,dx = [uv]_a^b - \int_a^b v\frac{du}{dx}\,dx

The integral of u times the derivative of v equals u v at b minus u v at a, minus the integral of v times the derivative of u.

The endpoint term, uv at b minus uv at a, is part of the answer. It disappears only when uv is zero at both ends, as it is for a density that falls to zero far out on both sides.

Worked example: A rising density, added up in narrower strips

A particle's density rises steadily across a 4 µm stretch, from 0 at the left end to 0.5 per micrometre at the right. The whole area is a triangle, ½ × 4 × 0.5 = 1.
Four strips 1 µm wide, each taking the height at its left edge: 0 + 0.125 + 0.25 + 0.375 = 0.75.
Eight strips 0.5 µm wide: the heights add to 1.75, and times 0.5 that is 0.875.
Sixteen strips 0.25 µm wide: 0.9375. The shortfall halves each time the strips do: 0.25, then 0.125, then 0.0625.
The totals settle on 1, the area of the triangle. That settled value is the integral.

A flat density gives the same total for every cut, which is why the uniform stretch above needed no strips. A curved one, such as the Brownian bell curve, does, and the integral is the value the totals settle on.

Where this lesson stops

An integral is a total of height times width over a range, in the units of both. The widths and any endpoint term are part of the calculation, not decoration on the integral sign.

This lesson builds on

Density is not probability

Return to the chapter

The push of light

How hard does a beam of light push on what it hits?

Sunlight on your hand warms it but does not seem to push it. It does push. Light carries momentum as well as energy, and hands it to whatever absorbs it. The push is tiny: full sunlight, about 1,000 watts on each square metre, presses on a square metre of black cloth with about 3.3 millionths of a newton.

Maxwell's electrodynamics gives light this momentum. For light travelling in one direction, the momentum it carries is its energy divided by the speed of light, c:

p = \frac{E}{c}

Momentum equals energy divided by c.

A force is momentum delivered per second. A beam that delivers power P, energy per second, to a surface that absorbs it therefore pushes with force P/c. A mirror that sends the light straight back reverses its momentum and so receives twice as much each second, 2P/c. The flow of energy in a beam, in watts per square metre, is the Poynting vector, after John Henry Poynting (1884).

The push is tiny, and it was measured before 1905: by Pyotr Lebedev in 1901, and by Ernest Nichols and Gordon Hull between 1901 and 1903. Both reported forces that agreed with Maxwell's value within their experimental error.

Einstein's mass-energy paper of September 1905 does not use this momentum. It compares the energy of two emitted light pulses as reckoned from two frames, using the energy transformation of §8 of the relativity paper. Light's momentum enters his argument in 1906, in his paper on the motion of the centre of gravity, where light crossing a closed box carries momentum from one end to the other.

Worked example: The push of a one-watt beam, absorbed and reflected

A surface absorbs a beam of 1 W: one joule of light energy arrives each second.
Each second it receives momentum of 1 J divided by c, which is 1 ÷ 299 792 458, about 3.34 × 10⁻⁹ kilogram metres per second.
Momentum received per second is a force, so the push is about 3.34 × 10⁻⁹ newtons.
A mirror that reflects the same beam straight back receives twice the momentum each second: about 6.67 × 10⁻⁹ newtons.

F_{\text{absorbed}} = \frac{P}{c}, \qquad F_{\text{reflected}} = \frac{2P}{c}

The force on an absorbing surface is the power divided by c; on a mirror that reflects the light straight back, it is twice the power divided by c.

For scale, 3.34 × 10⁻⁹ newtons is about the weight of a third of a microgram. That is why the push was first measured in 1901, nearly thirty years after Maxwell predicted it in his Treatise of 1873.

Where this lesson stops

Light's momentum is the push a beam gives when absorbed or reflected. It follows from Maxwell's electrodynamics, with no quantum hypothesis.

Return to the chapter

Static laboratory examples

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

Build-time scalar readouts
Quantity	Value	Unit	Status
frameSpeed	0.6	c	value
propagationAngleStationary	0	deg	value
propagationAngleMoving	0	deg	value
lightComplexEnergyStationary	1	J	value
lightComplexEnergyMoving	0.5	J	value
lightComplexVolumeStationary	1	m^3	value
lightComplexVolumeMoving	2	m^3	value
lightAmplitudeStationary	1	V/m	value
lightAmplitudeMoving	0.5	V/m	value
dopplerFactor	0.5	1	value
lorentzFactor	1.25	1	value
energyDensityFactor	0.25	1	value
volumeFactor	2	1	value
materialVolumeFactor	0.8	1	value

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

Evaluator: source:sha256:891a6ad3e6dab00a3fb56fde13eb772ee92b9900f77701de0091dbe8ac5733ae

sr-11: 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-11 online

Build-time scalar readouts
Quantity	Value	Unit	Status
frequencyRatio	0.25000000000000006	1	value
cosPhiReflected	-0.9999999999999993	1	value
phiReflectedDeg	179.9999979086902	deg	value
amplitudeRatio	0.25000000000000006	1	value
radiationPressure	0.5000000000000001	Pa	value
radiationForce	0.5000000000000001	N	value
incidentPower	119916983.2	W	value
reflectedPower	29979245.799999997	W	value
workRate	89937737.4	W	value
energyBalanceResidual	0	W	value

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

Settings (canonical units)

beta: 0.6
frame: lab
incidentAngleDeg: 0
incidentEnergyDensity: 1
mirrorArea: 1
unitLayer: si

Evaluator: source:sha256:3e0e31631a37bf13d942bd1b664e4b2425eda11ab0bdcd18ca122c2f9acb17bd

Source references

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