What a clock actually measures

3.1 Events, clocks, and reference frames#

A particular flash is an event: something happening at one place and one time. To describe it, we need an address and a clock reading. A sequence of events along an object’s motion is its worldline—its history, not a photograph at one instant.

Begin with an ideal laboratory drifting without acceleration or rotation, far from significant gravity. Place mutually stationary rulers and clocks throughout it. Such a network defines an inertial frame. We want every observer using this network to assign the same time to a given distant event, so we must say how its clocks are synchronized.

Send a light pulse from clock A to clock B and immediately reflect it back. If A sends it at $t_1$ and receives it at $t_2$, set B’s reading at the reflection to $(t_1+t_2)/2$. This is Einstein synchronization: the outward and return light travel times are assigned equal values. It accounts for the travel delay; seeing a distant clock now is not the same as assigning a time to the event happening there.

The physical starting points are that the laws of physics are the same in all inertial frames and that light in vacuum has the same speed $c$ in each. These are assumptions supported by experiment, not consequences of a coordinate trick. A laboratory moving relative to the first one builds its own synchronized network using the same procedure.

For an event, collect the labels in one list:

The index $\mu$ runs over $0,1,2,3$, as introduced in Chapter 2. The symbol $c$ is the vacuum speed of light. Multiplying a time by $c$ gives a length: $ct$ is how far light travels during that time. Thus all four entries in this list have length units. We have changed how we label time, not turned a clock into a ruler.

Einstein’s original account starts with this operational treatment of clocks. Einstein’s 1905 paper, in English translation.

3.2 Deriving the Lorentz transformation#

Let frame $S'$ move at speed $v$ in the positive $x$ direction relative to $S$. Their origins meet at $t=t'=0$. Each frame uses its own synchronized clocks and the same units of length and time. We want formulas that turn one frame’s labels for an event into the other’s.

We assume that the laws do not favor a particular position or starting time. These assumptions are called spatial and time homogeneity. With uniform relative motion and the stated synchronization, the relation between the two coordinate lists is linear. We can therefore find it by determining a few coefficients. This change of inertial frame is called a Lorentz boost.

The moving origin obeys $x=vt$ and must have $x'=0$, so

for some factor $A$ depending on $v$. Write the most general linear time transformation as $t'=B t+C x$.

Now send light to the right, so $x=ct$ and $x'=ct'$. Substitution gives

Send light to the left, so $x=-ct$ and $x'=-ct'$. This gives

Add the equations: $2Ac=2cB$, hence $B=A$. Subtract them: $-2Av=2c^2C$, hence $C=-Av/c^2$. Therefore

We have not guessed that time must mix with space. The two light directions forced it.

The moving frame sees the first frame receding at velocity $-v$. Neither frame is privileged, and reversing the spatial direction must not change the scale factor. These requirements are called reciprocity and spatial isotropy. The inverse relation therefore has the form $x=A(x'+vt')$. Substitute the expressions above:

For the inverse to return every value of $x$, we need

Choose the positive root continuously connected to $A=1$ at $v=0$:

For this standard boost, $y'=y$ and $z'=z$. The time transformation has a measurable consequence. Two events with $\Delta t=0$ but different $x$ generally have

For example, take $v=0.6c$, so $\gamma=1.25$. Two flashes separated by one light-second—the distance light travels in one second—along $x$ are simultaneous in $S$. The moving frame assigns them a time difference of $-0.75\,\mathrm s$: the flash at the larger $x$ coordinate happened earlier in its synchronized-clock system.

This is relativity of simultaneity. It concerns the times assigned by synchronized clocks after accounting for signal travel, not merely the order in which someone sees the flashes.

The everyday limit is sensible. If $v/c\ll1$, then $\gamma\approx1$ and $vx/c^2$ becomes negligible for ordinary distances and timing precision. We recover the Galilean approximation $x'\approx x-vt$, $t'\approx t$.

3.3 The spacetime interval#

Choose two events, A and B. Write $\Delta t=t_B-t_A$ for their time difference in one inertial frame, and similarly $\Delta x$, $\Delta y$, and $\Delta z$ for their position differences. Different moving frames generally assign different values to all of these differences. Is there a combination they agree on?

There is a useful clue from ordinary geometry. Rotate a map and a displacement’s horizontal and vertical components change, but the sum of their squares stays equal to the squared length. Try a related combination for time and position: square the spatial differences and subtract the squared time difference expressed as a length.

Define the spacetime interval between the two events in flat spacetime by

The symbol $\Delta s^2$ names this signed quantity. Despite the square in its notation, it can be negative; it is not the square of an ordinary positive distance. Every term has square-length units. The minus sign is a physical distinction between time and space, not a units conversion.

Now test the proposed combination using the Lorentz transformation from §3.2. For the time and $x$ terms,

The expansion produces $+2v\Delta t\Delta x$ from the time square and $-2v\Delta t\Delta x$ from the space square; they cancel. The transverse differences do not change in this boost. All inertial frames therefore assign the same interval, even when they disagree about the separate time and position differences. This agreement is what makes the interval useful.

For a concrete example, one frame assigns $\Delta t=5\,\mathrm s$ and $\Delta x=c(3\,\mathrm s)$, with no sideways separation. The spatial gap is three light-seconds: the distance light travels in three seconds. The interval is $(-25+9)=-16$ square light-seconds. In a frame moving at $v=0.6c$, the Lorentz formulas give $\Delta x'=0$ and $\Delta t'=4\,\mathrm s$. Its answer is also $-16$ square light-seconds. The next section explains why the four seconds have a direct clock interpretation.

The sign tells us which connections are possible:

For a signal emitted at A, also require B to be in A’s future. The possible light signals form a light cone: after elapsed time $\Delta t$, light has reached a sphere of radius $c\Delta t$. Stack these spheres in a diagram that includes time, and they form a cone. Slower objects travel inside it. The surface is null; the interior is timelike. The spatially separated region outside is spacelike. Observers can disagree about the time order of spacelike events, but not about the order of two events joined by a future-directed signal.

For small displacements we use differentials rather than finite changes:

The last expression uses Chapter 2’s summation rule and the coordinates $x^0=ct$. The diagonal matrix packages the coefficients of the measuring rule; its other entries are zero. It is called the Minkowski metric. A metric is a rule for obtaining an interval from small coordinate displacements. Chapter 4 develops that rule on more general spaces. There the local formula cannot in general be turned into a finite separation by simply replacing every $d$ by $\Delta$.

Further calculation: rapidity and successive boosts

To compose boosts conveniently, we can build two new functions from exponentials. For a dimensionless number $\chi$, define $\cosh\chi=(e^\chi+e^{-\chi})/2$ and $\sinh\chi=(e^\chi-e^{-\chi})/2$, then $\tanh\chi=\sinh\chi/\cosh\chi$. Squaring and subtracting gives $\cosh^2\chi-\sinh^2\chi=1$. This resembles the circular identity $\cos^2\theta+\sin^2\theta=1$, with the sign needed for an interval.

The parameter $\chi$ is called rapidity. Choosing it so that $\tanh\chi=v/c$ gives

$$\tanh\chi=\frac vc,\qquad \cosh\chi=\gamma,\qquad \sinh\chi=\gamma\frac vc.$$

Then

$$\begin{pmatrix}ct'\\x'\end{pmatrix} = \begin{pmatrix}\cosh\chi&-\sinh\chi\\-\sinh\chi&\cosh\chi\end{pmatrix} \begin{pmatrix}ct\\x\end{pmatrix}.$$

Multiplying two of these matrices and using the exponential definitions replaces $\chi$ by $\chi_1+\chi_2$. Thus rapidities add for boosts along the same line. Writing the result in terms of the two speeds gives

$$v_{\rm combined}=\frac{v_1+v_2}{1+v_1v_2/c^2}.$$

Two successive boosts of $0.8c$ give $1.6c/1.64\approx0.976c$, not $1.6c$. The combined speed remains below the speed of light.

SPACETIME LAB / 03

The shape of a possible future

A light cone becomes a surface when we restore a second space direction. Change the speed and watch the observer’s worldline tilt.

ExploreDiagram

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Observer speed

At restNear light speed

$$ds^2=-c^2dt^2+dx^2+dy^2,\quad v<c$$

Read the scene. Solid strokes are directly visible; dashed strokes lie behind the cone surface along your sightline. Looking through an open rim can reveal its entire inside. These drawing cues update as you orbit. The brighter upper half is the future cone. Its finite rim is only the edge of the drawing, not a physical boundary. Two space dimensions are shown; the third is suppressed. The cone is a causal boundary, not a physical surface.

WORKED EXAMPLE

Build a boost from exponentials

Can we combine two changes of moving observer by adding a single parameter?

Builds onExponentials and initial data Lorentz transformations

See the idea

Circular functions parameterize $x^2+y^2=1$. A boost preserves a difference of squares, so its natural curve is a hyperbola. We will construct the needed functions from exponentials and check their identities directly.

Work it out

1. 01
   Define the functions

   For a dimensionless real number $\eta$, define $\cosh\eta=(e^\eta+e^{-\eta})/2$ and $\sinh\eta=(e^\eta-e^{-\eta})/2$. Squaring and subtracting cancels the growing and decaying terms.

   $$\cosh^2\eta-\sinh^2\eta=1.$$

   Why this step works The identity is an algebraic consequence of the definitions.

2. 02
   Differentiate and form the velocity ratio

   Differentiating the exponentials gives $(\sinh\eta)'=\cosh\eta$ and $(\cosh\eta)'=\sinh\eta$. Define $\tanh\eta=\sinh\eta/\cosh\eta$. For real finite $\eta$ its magnitude is below one. Write $\beta=v/c$ for velocity in units of light speed; the parameter $\eta$ with $\beta=\tanh\eta$ is called rapidity.

   $$\beta=\tanh\eta,\qquad \gamma=\cosh\eta,\qquad \gamma\beta=\sinh\eta.$$

   Why this step works The hyperbola identity yields γ²(1−β²)=1 with the positive future-directed branch.

3. 03
   Transform an event

   For a primed frame moving at $v=\beta c$ in the positive $x$ direction, use $ct'=ct\cosh\eta-x\sinh\eta$ and $x'=x\cosh\eta-ct\sinh\eta$. Expand $-(ct')^2+(x')^2$: the cross terms cancel and the original interval remains.

   $$\begin{pmatrix}ct'\\x'\end{pmatrix}=\begin{pmatrix}\cosh\eta&-\sinh\eta\\-\sinh\eta&\cosh\eta\end{pmatrix}\begin{pmatrix}ct\\x\end{pmatrix}.$$

   Why this step works The transformation is chosen to preserve the spacetime interval and the stated frame direction.

4. 04
   Invert when a measurement gives the velocity

   Solving $\beta=(e^{2\eta}-1)/(e^{2\eta}+1)$ gives $\eta=\tfrac12\ln[(1+\beta)/(1-\beta)]$. This is $\operatorname{artanh}\beta$. Similarly, solving $y=\sinh\eta$ yields $\operatorname{arsinh}y=\ln(y+\sqrt{1+y^2})$. Its derivative is $1/\sqrt{1+y^2}$, which will evaluate a light-delay integral.

   $$\frac{d}{dy}\operatorname{arsinh}y=\frac1{\sqrt{1+y^2}}.$$

   Why this step works An inverse function answers which parameter produced a measured value.

Go deeper

Multiply two collinear boost matrices and use the exponential definitions to obtain the addition formulas: their rapidities add. Velocities therefore combine as $(\beta_1+\beta_2)/(1+\beta_1\beta_2)$, not by ordinary addition. This simple additive parameter applies to collinear boosts. Noncollinear boosts also produce a spatial rotation; treating all boosts as commuting is a different and incorrect generalization.

Further readingDavid Tong · General relativity

Test the idea

FIRST, PREDICT

What does adding two collinear rapidities accomplish?

It implies every pair of boosts commutes.It composes the boosts while keeping subluminal speeds subluminal.It adds the two ordinary speeds.

Compare the reasoning

It implies every pair of boosts commutes.

Different spatial directions introduce rotations; the collinear assumption matters.

It composes the boosts while keeping subluminal speeds subluminal.

The hyperbolic tangent addition formula gives relativistic velocity addition.

It adds the two ordinary speeds.

Rapidity and speed are different parameters.

A hint

Write velocity as c times tanh of the rapidity.

NOW CHANGE THE EXAMPLE

If $e^\eta=2$, calculate $v/c=\tanh\eta$.

A hint

Use $e^{-\eta}=1/2$ in the definitions.

Work through the solution

$\tanh\eta=(2-1/2)/(2+1/2)=3/5$.

Rapidity adds for collinear boosts because it is the parameter of a hyperbola-preserving transformation.

3.4 Adding up a clock’s elapsed time#

Take a clock on a journey. During a sufficiently short part of the journey, use an inertial frame in which the clock is momentarily at rest. Its spatial displacement is zero in that frame, so the interval is $ds^2=-c^2d\tau^2$, where $d\tau$ is the time recorded by the clock.

Other inertial frames agree on this interval. In a frame where the clock is moving, substitute $dx=v_xdt$, $dy=v_ydt$, and $dz=v_zdt$ into the interval formula:

Writing $v^2=v_x^2+v_y^2+v_z^2$ for the ordinary speed squared, the clock’s elapsed time is therefore

This is called proper time. An ideal clock is assumed to measure it even when the clock accelerates, provided its mechanism is not disturbed. Acceleration changes the journey; we do not add a separate acceleration term to this clock rule. This assumption is often called the clock hypothesis.

For constant speed the square-root factor stays constant, giving

At $v=0.6c$, the factor is 0.8. A journey taking five years according to the frame’s synchronized clocks takes four years on the traveling clock. This is the same calculation as the five-second, three-light-second example in §3.3, with a different unit of time.

If the speed varies, add the contributions from each small part of the journey:

The rule takes a whole path as input; such a rule is called a functional. Two clocks that start together and meet again can compare the accumulated results directly at their reunion.

There is a useful consequence in flat spacetime. Choose the inertial frame in which the departure and reunion occur at the same position. A clock that stays there records $t_B-t_A$. Every other future-directed timelike path between those same events has a square-root factor no greater than one at each step. It therefore records no more time than the stationary clock.

Thus the inertial path between these events gives the greatest elapsed time. The result follows from the clock formula; it differs from the shortest-distance rule for straight lines in ordinary spatial geometry.

3.5 Two clocks meet again#

Two clocks start together. One stays at rest in an inertial frame. The other travels outward at $0.6c$ for five years of that frame’s time and returns at the same speed for another five years. We idealize the turnaround as brief. This is the experiment often described using twins, with their ages playing the role of the clock readings. At reunion,

The traveling clock must change velocity to return. The stay-at-home clock does not, so we cannot exchange their roles while keeping the same experiment. Still, the age difference is accumulated during the journeys: each five-year leg contributes four years to the traveling clock.

A finite turnaround contributes its own elapsed time, found using the variable-speed integral. Making the turn brief makes that contribution small; it does not remove the difference accumulated on the long legs. Neither clock experiences a locally slow mechanism. Each records the proper time along its own path.

This calculation uses special relativity throughout. An accelerated observer can move in flat spacetime; introducing acceleration does not by itself require a gravitational field.

3.6 Four-velocity and four-momentum#

To describe motion using one time measured by the moving particle, differentiate its four position coordinates with respect to proper time. The result is its four-velocity. Since $dt/d\tau=\gamma$, the chain rule gives

Use the Minkowski metric from §3.3 to pair the four-velocity with itself. The result is its squared spacetime norm:

The minus sign comes from the time component. This squared norm stays $-c^2$ even when the ordinary speed changes.

The relativistic extension of momentum for a particle with constant rest mass $m$ is its four-momentum, $p^\mu=mu^\mu$. Its time component is energy divided by $c$, while its three spatial components are ordinary momentum:

We can shorten metric pairings by defining a lowered-index component:

This operation is called lowering an index. In these coordinates it reverses the sign of the time component and leaves the spatial components unchanged. The same rule applies to $u_\mu$ or any other vector. Now its squared spacetime norm can be written

or, after expanding,

The familiar $E=mc^2$ is the special case of zero spatial momentum in the particle’s rest frame. At small speed, the binomial expansion $\gamma=1+\frac12v^2/c^2+O(v^4/c^4)$ gives

Newtonian kinetic energy appears as the first correction to rest energy.

Light exchanges energy and momentum in packets called photons. A photon has zero rest mass, so the energy–momentum relation gives $E=c|\mathbf p|$. Its four-momentum has zero squared spacetime norm: it is null.

A photon has no rest frame. Along its lightlike path, $d\tau=0$, so we cannot define a four-velocity by dividing displacement by proper time. Its energy and momentum remain well-defined; the massive-particle formula $p^\mu=mu^\mu$ is not how we construct photon momentum.

The mass of two light pulses. To find a system’s invariant mass, first add its four-momenta and then use the energy–momentum relation for that total. Consider two photons, each of energy $E_\gamma$, traveling in opposite directions. Total momentum is zero and total energy is $2E_\gamma$, so the system has invariant mass $M=2E_\gamma/c^2$. The combined system has a rest frame even though neither photon does. This example shows why simply adding the individual rest masses would give the wrong result.

3.7 Energy is a measurement made by an observer#

Let an observer have four-velocity $U^\mu$, with $U_\mu U^\mu=-c^2$. The energy that observer measures for a particle with four-momentum $p^\mu$ is

We can establish this formula by checking it in the observer’s rest frame and then using invariance of the pairing under a change of coordinates. In that frame, $U^\mu=(c,0,0,0)$ and $p_0=-E/c$, so $-p_\mu U^\mu=E$. Both four-vectors describe a specified particle and a specified observer, so changing the coordinates leaves their pairing unchanged. Choosing a different observer changes $U$ and can change the measured energy. These are different operations, as in Chapter 2’s distinction between an object and its description.

For a massive particle, define the relative Lorentz factor

Then $E_{(U)}=\gamma_{\rm rel}mc^2$. Decompose the momentum into the observer’s temporal and spatial parts:

The symbol $\perp$ marks the part orthogonal to $U$. Subtracting the first term from $p$ and pairing with $U$ gives $-E_{(U)}+E_{(U)}=0$, which verifies the condition. In the observer’s rest frame, it says $p_\perp^0=0$: the remaining components describe spatial momentum. This three-dimensional set of directions is the observer’s instantaneous rest space.

For a photon moving in the $+x$ direction, take $p^\mu=(E/c,E/c,0,0)$ and an observer chasing it with $U^\mu=\gamma(c,v,0,0)$. The energy measured by that observer is

At $v=0.6c$, the square root is $\sqrt{0.4/1.6}=1/2$: this observer measures half the original photon energy.

To translate that energy change into a color or frequency change, we use a physical input about light: a photon has energy $E=h\nu$, where $\nu$ is frequency (oscillations per second) and $h$ is Planck’s constant, with units of joule seconds. This is a quantum relation, not a result of the Lorentz algebra above. Since $h$ is the same, half the energy means half the frequency. A decrease in light’s frequency is called a redshift. The observer still measures the light’s speed as $c$. Einstein Online: waves, motion, and frequency.

Chapter 11 will extend this measuring procedure from one particle to energy and momentum distributed through a region.

WORKED EXAMPLE

Chase a photon without slowing it down

How can two observers measure the same light speed and different photon energies?

Builds onRapidity

See the idea

A photon’s energy is not a number that floats beside it independently of an observer. It is the contraction of its four-momentum with a particular observer’s four-velocity. When observers move differently, the contractions differ even though both measure the invariant light speed.

Work it out

1. 01
   Specify the photon and observer

   Use $x^0=ct$ and metric $\eta=\operatorname{diag}(-1,1,1,1)$. A photon moving in the positive x direction has $p^\mu=(E/c,E/c,0,0)$. An observer moving at $v=\beta c$ has $u^\mu=\gamma(c,\beta c,0,0)$, with $\gamma=(1-\beta^2)^{-1/2}$.

   $$p\cdot p=0,\qquad u\cdot u=-c^2.$$

   Why this step works The null and timelike normalizations fix which physical objects these vectors represent.

2. 02
   Contract to obtain energy

   Lower the time index with the minus sign and calculate $E_{\rm obs}=-p\cdot u$.

   $$\frac{E_{\rm obs}}E=\gamma(1-\beta)=\sqrt{\frac{1-\beta}{1+\beta}}.$$

   Why this step works Chasing the photon gives a redshift; moving toward it gives a blueshift.

3. 03
   Keep the photon null

   A Lorentz transformation gives $p'^0=\gamma(E/c-\beta E/c)$ and $p'^x=\gamma(E/c-\beta E/c)$. They remain equal. The changed energy and momentum still satisfy the null relation.

   $$p'^x/p'^0=1.$$

   Why this step works The Doppler change alters energy and momentum together, preserving light speed.

Go deeper

For a photon whose direction makes angle θ with the observer’s velocity in the original frame, the factor becomes $\gamma(1-\beta\cos\theta)$. Angle is also observer-dependent: aberration gives $\cos\theta'=(\cos\theta-\beta)/(1-\beta\cos\theta)$ by transforming the temporal and parallel momentum components. These local formulas remain valid in an orthonormal frame in curved spacetime; transporting the photon between events requires the geometry as well.

Further readingDavid Tong · General relativity

Test the idea

FIRST, PREDICT

An observer chasing a photon sees it redshifted. What happens to its measured speed?

It remains c in the observer’s local inertial frame.It becomes c minus the observer’s speed.It falls to zero if the observer accelerates long enough.

Compare the reasoning

It remains c in the observer’s local inertial frame.

The transformed photon momentum remains null.

It becomes c minus the observer’s speed.

Galilean velocity subtraction does not preserve the spacetime interval.

It falls to zero if the observer accelerates long enough.

No massive observer reaches a photon rest frame by a finite physical boost.

A hint

Transform both temporal and spatial momentum components.

NOW CHANGE THE EXAMPLE

An observer chases a photon with $\beta=3/5$. Find the received energy divided by its energy in the original frame.

A hint

Use $\sqrt{(1-\beta)/(1+\beta)}$.

Work through the solution

$\sqrt{(2/5)/(8/5)}=1/2$.

A photon’s locally measured energy changes with the observer; its local speed remains c.

3.8 Proper acceleration#

In inertial Minkowski coordinates, define four-acceleration

Differentiate $u_\mu u^\mu=-c^2$. Because the Minkowski metric is constant,

Four-acceleration is orthogonal to four-velocity. In the instantaneous rest frame, $u^\mu=(c,0,0,0)$, so $a^0=0$ and $a^\mu$ is purely spatial. Its magnitude

is the proper acceleration, measured by an ideal accelerometer. In the instantaneous rest frame, $a^0=0$, so the quantity under the square root is the sum of the three spatial component squares. It is nonnegative.

The coordinate derivative used here works in inertial Cartesian coordinates. If the measuring axes vary from place to place, differentiating components alone also counts the change in the axes. Chapter 6 develops the correction. It will let us calculate an accelerometer reading in general coordinates, including the falling laboratories of Chapter 1.