Read · Special relativity: from clock operations to electrodynamics

§8 · finite light energy and moving mirrors

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Detail

Letters

Show modern qualifications

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

A derivation

Density is not the energy of the whole packet

Why does a finite light complex need a volume calculation?

The 3 printed paragraphs this passage explains

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

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

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

\frac{E'}{E} = q

Shown in modern letters; Einstein's are in §8 of the German source face.

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.

Open the foundation: Adding continuously

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

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}

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

\frac{E'}{E} = q

Shown in modern letters; Einstein's are in §8 of the German source face.

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.

A tool for this step

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.

The same triangle, cut two ways

A density rising from 0 to 0.5 per micrometre across 4 μm. Each strip is as tall as the curve at its own left edge, which is why every strip falls short and why the shortfall shrinks when the strips do.

4 strips add to 0.75; 8 strips add to 0.875. The triangle itself is 1. Cut narrower and the total climbs toward it without ever passing it, and the value it climbs toward is the integral.

What it shows, in words

Both drawings are the same density, rising in a straight line from 0 at the left end to 0.5 per micrometre at the right, across 4 micrometres. The first is cut into 4 strips of 1 micrometre each and the second into 8 strips of 0.5.

Each strip is as tall as the line at its own left edge, so each leaves a wedge uncovered above it. The 4 strips add to 0.75 and the 8 strips to 0.875, against the triangle's area of 1. Narrower strips leave smaller wedges, so the totals climb toward 1 and never past it. That limit is what the integral names.

What it does not show: these strips take the height at their left edge, which is one choice among several. Taking the right edge would overshoot 1 and climb down to it instead, and taking the middle of each strip would land much closer from the start. All three settle on the same 1, and that they agree is the fact the integral rests on. Nothing in either drawing demonstrates it.

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.

Explore the equations in this step

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

Explore the equation · Modern model notation

Energy density: q squared

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

Shown in modern letters; Einstein's are in §8 of the German source face.

over is squared.

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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.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
A bounded piece of plane light whose surface moves with the light. q is the frequency ratio of section 7 for this light's direction.
Energy density, volume and total energy are three different quantities, and each has its own factor.

Moving over stationary: How the density changes between the systems. Read the prerequisite
What it asserts: A statement about density only. The whole packet's energy changes differently, because its volume changes too. Read the prerequisite
q squared: The density goes as the square of the amplitude, and the amplitude changes by q. Read the prerequisite
Energy density, stationary system: The light's energy per cubic metre in the stationary system. Read the prerequisite
Energy density, moving system: The light's energy per cubic metre in the moving system. Read the prerequisite
The frequency ratio: Section 7's q, for this light's direction. At 0.6c, for light along the motion, q = 0.5. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Total energy: q

\frac{E'}{E} = q

Shown in modern letters; Einstein's are in §8 of the German source face.

over is .

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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.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
A bounded piece of plane light whose surface moves with the light. q is the frequency ratio of section 7 for this light's direction.
Energy density, volume and total energy are three different quantities, and each has its own factor.

Moving over stationary: How the whole packet's energy changes. Read the prerequisite
What it asserts: The packet's energy changes by the same factor as its frequency. The paper remarks on this, and the September paper builds on it; it does not need light quanta. Read the prerequisite
Packet energy, stationary system: The whole energy in the stationary system. Read the prerequisite
Packet energy, moving system: The whole energy of the piece of light in the moving system. Read the prerequisite
The frequency ratio: Energy and frequency change by the same factor. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Explore the equation · Modern model notation

Packet volume: one over q

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

Shown in modern letters; Einstein's are in §8 of the German source face.

over is one over .

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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.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
A bounded piece of plane light whose surface moves with the light. q is the frequency ratio of section 7 for this light's direction.
Energy density, volume and total energy are three different quantities, and each has its own factor.

One over q: The volume changes the opposite way to the frequency. Using gamma here, as for a rod, gives the wrong total energy. Read the prerequisite
Moving over stationary: How the volume changes between the systems. Read the prerequisite
What it asserts: The volume of this particular region, whose boundary moves with the light. Read the prerequisite
The frequency ratio: Section 7's q, for this light's direction. At 0.6c, for light along the motion, q = 0.5. Read the prerequisite
Packet volume, stationary system: The volume at one time of the stationary system. Read the prerequisite
Packet volume, moving system: The volume the light occupies at one time of the moving system. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Modern qualifications

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

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

Assumed here

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

What this does not establish

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

Earlier step: One phase fixes frequency and direction

Source context: German source · English · Interlinear gloss · Facsimile

A derivation

Reflection starts with interception

Can the incident ray reach a receding mirror?

The 4 printed paragraphs this passage explains

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

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

Shown in modern letters; Einstein's are in §8 of the German source face.

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.

Open the foundation: The push of light

Show every step here: Reflection starts with interception

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\,\frac{v}{c}\,\cos\left(\varphi\right) + \left(\frac{v}{c}\right)^{2}}{1 - \left(\frac{v}{c}\right)^{2}}

Shown in modern letters; Einstein's are in §8 of the German source face.

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.

A tool for this step

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}

F_{\text{reflected}} = \frac{2\,P}{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.

JavaScript is off, so the controls below cannot change anything. The construction shows its first setting, and its closing paragraph, “What it shows, in words”, describes the rest.

Try it: how hard the beam pushes

Set the power of the beam and say what the surface does with it. The push is the momentum the light delivers each second.

Power of the beam

Or type a power, in watts:

The surface

A beam of 1 W on a surface that absorbs it pushes with 3.3356 × 10⁻⁹ newtons. A surface that sent it straight back instead would feel 6.6713 × 10⁻⁹ N, a factor of 2.

What it shows, in words

The push is the power divided by the speed of light, and a mirror receives twice as much because it reverses the light's momentum instead of merely stopping it. At the setting above, 1 watts on a surface that absorbs the light, the push is 3.3356 × 10⁻⁹ newtons. Doubling the power doubles the push, and switching between the two surfaces changes it by exactly a factor of 2, at any power.

One watt absorbed gives about 3.34 × 10⁻⁹ newtons, which is the number the lesson prints and why the push went unnoticed for so long.

What it does not show: the surface is at rest and stays at rest. A mirror that is moving changes the light's frequency as well as its direction, which is §8 of the relativity paper and a harder problem than this; nothing here bears on it. Nor does it say anything about how the momentum gets there, only how much arrives each second.

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.

Explore the equations in this step

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

Explore the equation · Modern model notation

Light reflected from a moving mirror

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

Shown in modern letters; Einstein's are in §8 of the German source face.

over is one minus twice over times the cosine of , plus v over c squared, all over one minus v over c squared.

Select a term or operation. In the formula, Down enters an operation, Up returns to its parent, and Left/Right move between siblings. Escape clears selection. Tab leaves the formula.

Explain this equation

Each term and operation, in words

No live values

This equation states a relation. No instrument on this page computes its terms.

Read the equation aloud in words

The 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.

Model assumptions and every term’s meaning

The moving system moves along x at speed v relative to the stationary system, with |v| < c. Modern teaching notation: gamma is the paper's beta, c its V.
A plane mirror at rest in the moving system, facing the incoming light; phi is the angle of incidence in the stationary system.
The light reaches the mirror. For some angles and directions of motion it never does, and then there is no reflection to describe.

v over c: The mirror's speed as a fraction of c. Read the prerequisite
v over c: The same fraction. Read the prerequisite
Its square: In the numerator. Read the prerequisite
Its square: In the denominator. Read the prerequisite
Cosine of the incidence angle: 1 for light meeting the mirror head-on along the motion. Read the prerequisite
One minus v squared over c squared: The same combination that sits under gamma's square root. Read the prerequisite
The whole factor: At normal incidence it simplifies to (1 − v/c)/(1 + v/c): 0.25 at 0.6c. Read the prerequisite
The numerator: One minus twice the projected speed plus its square. Read the prerequisite
Reflected over incident: A pure number. Read the prerequisite
What it asserts: The frequency change on reflection. Intercepted power and radiation pressure are different quantities with different factors. Read the prerequisite
Twice v over c times the cosine: The main first-order change: the light is shifted once going into the mirror's system and once coming back. Read the prerequisite
Angle of incidence: Between the incoming light and the direction of motion, in the stationary system. Read the prerequisite
Incident frequency: The frequency arriving at the mirror, in the stationary system. Read the prerequisite
Speed of light: The paper's V. Read the prerequisite
Reflected frequency: The frequency after reflection, in the stationary system. Read the prerequisite
Speed of the moving system: Signed along x: the sign says which way the system moves. Read the prerequisite

Every term’s units were checked when this page was built. That checks the units, not the model. This equation is written for this edition in modern notation, not transcribed from the paper.

Modern qualifications

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

Assumptions and limits: Reflection starts with interception

Assumed here

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

What this does not establish

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

Earlier step: One phase fixes frequency and direction

Source context: German source · English · Interlinear gloss · Facsimile

References

A. Einstein, Über einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt. Annalen der Physik (4), 17, 132–148 (1905).

A. Einstein, On the motion of particles suspended in liquids at rest required by the molecular-kinetic theory of heat. Annalen der Physik (4), 17, 549–560 (1905), §§1–5. Bibliographic pointer; this preview is not a source transcription or translation.

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

A. Einstein, Does the inertia of a body depend upon its energy content? Annalen der Physik (4), 18, 639–641 (1905). External 1923 Perrett–Jeffery translation, electronically transcribed by John Walker; its notation was modernized. A reference for this explanatory preview, not this edition’s translation or its pinned facsimile.

28 foundation readings sit behind this argument.

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