From Light Signals to Spacetime
Consider an idealized experiment in a region where gravitational curvature can be neglected over its size and duration. Whether this approximation is adequate must be checked for an actual experiment; the restriction concerns physical conditions, not whether the experiment takes place on Earth.
Take two long railcars moving along parallel tracks. Each railcar carries clocks fixed at several positions. We call each railcar together with its clocks a laboratory. Laboratory \(S\) calls its positions \(x\) and its clock readings \(t\). Laboratory \(S'\) uses \(x'\) and \(t'\).
Each observer records events locally, where they are, and may choose that location as the origin of their own coordinates. We later compare these records to determine how the observers’ descriptions relate. In this experiment, the clocks along each railcar share one coordinate system, with its origin at the railcar’s center; they occupy different positions within that system.
Place ourselves at the center of \(S\), where \(x=0\). We directly observe events here; observers or instruments elsewhere make their own local records. Equip each clock with a light detector and memory, forming a recording station. Each station stays at a fixed position \(x\) on \(S\) or \(x'\) on \(S'\).
When a station on \(S'\) passes a station on \(S\), a light pulse at their encounter triggers each station to record its own clock reading and position label. The timestamp records the pulse’s local arrival at the detector. We idealize their detectors as colocated at a single event \(P\) shared by both stations; the clock housings do not need to touch. The station on \(S\) stores \((t_P,x_P)\), and the station on \(S'\) stores \((t'_P,x'_P)\). These pairs are the event’s coordinates in the two laboratories.
We compare only events within the region covered by both recording networks. To record events beyond that region using the same procedure, we must extend the networks, calibrate the added stations, and synchronize their clocks.
The synchronized network is one way to assign coordinates to events. Alternatively, a single station at the origin could send a light pulse, have it reflected by a distant object, and record its return using the same clock. Assuming the vacuum light-speed rule introduced in Section 1, the object’s radar distance at the reflection event is
\[\text{distance}=\frac{c\left(t_{\mathrm{return}}-t_{\mathrm{send}}\right)}{2},\]where \(t_{\mathrm{send}}\) and \(t_{\mathrm{return}}\) are the emission and return readings of that one clock. No distant clock needs to be synchronized. We use distributed stations here because they let us compare records made locally at the same event. See this radar construction for the alternative procedure.
Both the synchronized-network procedure and the radar alternative rely on a physical assumption: every inertial laboratory obtains the same constant round-trip vacuum light speed \(c\), in every direction throughout the modeled region. We keep this model within that region. Extending it to the whole universe would assume that flat spacetime applies everywhere; the general relativity post will explain why gravity can require a different description. The results we derive in this post depend on this light-speed assumption.
This model uses pointlike detectors with their response delays corrected. Real instruments require calibration, and any remaining timing error contributes measurement uncertainty.
We collect the records afterward. A stored timestamp describes the local event, not when its message reaches us at the origin. Transmission delays change when we obtain the records, not the timestamps already stored in them. The records can be sent to us; the clocks themselves need not be brought together.
Follow the center station of \(S'\) as it passes successive stations on \(S\). Each station on \(S\) stores its local \(t\) and \(x\) at the encounter. From these collected records, reconstruct \(x(t)\), the position of the center of \(S'\) at time \(t\) according to \(S\). The measurements fit
\[x(t)=x(0)+vt.\]This is uniform relative motion. In \(S\) coordinates, \(S\) has velocity zero and \(S'\) has the measured velocity \(v\). The value of \(v\) is an input to the experiment, not a universal constant.
We ask:
What rule relates the pairs \((t,x)\) and \((t',x')\) recorded locally at the same event?
1. How does one laboratory synchronize its clocks?
The clocks on one railcar need a rule for assigning common time labels. Its center clock emits one light pulse when it reads \(t_{\mathrm{emit}}\). A clock at position \(x\) reflects the pulse immediately, and the pulse returns to the center clock when it reads \(t_{\mathrm{return}}\). Define
\[\begin{aligned} t_{\mathrm{reflect}} &:=\frac{t_{\mathrm{emit}}+t_{\mathrm{return}}}{2},\\ c &:=\frac{2\lvert x\rvert}{t_{\mathrm{return}}-t_{\mathrm{emit}}}. \end{aligned}\]The first line synchronizes the remote clock: the remote clock is adjusted to read \(t_{\mathrm{reflect}}\) when it reflects the pulse. In the second, \(c\) is the measured round-trip speed of light.
We assume that each inertial laboratory obtains the same constant round-trip speed \(c\) for light in vacuum, using its own clocks and rulers, regardless of the apparatus’s orientation or the source’s motion. This is a physical input to the model, not something the following derivation proves. The synchronization convention then assigns that same speed to one-way propagation; it cannot force different laboratories to obtain the same round-trip result.
Establishing how accurately nature obeys this assumption requires actual experiments, with instrument uncertainties and environmental effects accounted for. Here we derive its consequences. The railcar thought experiment does not establish its validity everywhere in the universe, and the gravity sequel will retain the corresponding local principle rather than prove it without assumptions.
2. Which laboratories provide the simple comparison?
Release several small test carts inside one laboratory after shielding them as well as possible from springs, air resistance, electric fields, and other detectable interactions. Record each cart’s passages at the laboratory’s stations and reconstruct its position function \(x(t)\) from those local records. The observations may fit
\[x(t)=x(0)+ut,\]where the constant
\[u:=\frac{dx}{dt}\]is the test cart’s measured velocity relative to the laboratory. A position function \(x(t)\) of this form is called a uniform trajectory.
A synchronized clock-and-position system in which sufficiently isolated carts obey this constant-velocity rule is called an inertial reference system. “Inertial” is not another word for stationary. Two such systems may move uniformly relative to one another.
This definition is operational but approximate. A laboratory is treated as inertial only over the region and duration in which free test carts show no detectable departure from the rule.
For the experiment in the opening, require both clock-and-position systems to be inertial. Choose the encounter of their center stations as the shared zero event:
\[t=t'=0, \qquad x=x'=0.\]The origin choice sets \(x(0)=0\). The measured rule from the opening therefore becomes
\[x(t)=vt.\]No new motion assumption has been added here; the origin choice only removes the constant \(x(0)\).
3. Why does this first rule fail for light?
The origin of \(S'\) has position function \(x(t)=vt\) according to \(S\). Now consider any event to which \(S\) assigns coordinates \((t,x)\). Subtracting the origin’s position \(vt\) from the event’s position \(x\) gives the first proposed spatial rule. This does not determine the time assigned by \(S'\). As a separate first hypothesis, suppose both laboratories assign the same time to every event. Consequently, if two events are simultaneous in \(S\), they are also simultaneous in \(S'\): \(\Delta t=0\) implies \(\Delta t'=0\).
\[x'=x-vt, \qquad t'=t.\]For an event on the moving origin, its position coordinate is \(x=x(t)=vt\). Substitution gives
\[x'=vt-vt=0.\]At the center-station encounter, a local detector triggers a lamp fixed to \(S\) to emit a light pulse toward increasing \(x\), with trigger delay neglected in our ideal model. Both laboratories assign the emission \((t,x)=(0,0)\) and \((t',x')=(0,0)\). Laboratory \(S\) assigns each later event on the pulse’s path coordinates satisfying
\[x=ct.\]This rule predicts
\[x'=ct-vt=(c-v)t'.\]Thus it predicts the light speed \(c-v\) in \(S'\). A pulse traveling in the opposite direction similarly acquires speed magnitude \(c+v\).
That dependence conflicts with the equal round-trip speeds assumed in Section 1. Combining that physical input with our synchronization convention requires each laboratory to represent its light rays by
\[x=\pm ct\]or, in its own coordinates,
\[x'=\pm ct'.\]This first rule cannot preserve both light-propagation rules. We therefore ask:
What coordinate rule maps every uniform trajectory to a uniform trajectory (because both \(S\) and \(S'\) were defined as inertial laboratories) and maps both light rays \(x=ct\) and \(x=-ct\) to light rays of the same speed in \(S'\)?
4. Construct the coordinate rule
Within our idealized region, we assume that no position along the tracks or clock reading is physically distinguished: repeating the experiment elsewhere or later under the same conditions gives the same coordinate rule. We therefore seek a linear rule, meaning that
\[\begin{aligned} x'&=ax+bt,\\ t'&=dx+et, \end{aligned}\]where \(a\), \(b\), \(d\), and \(e\) are constant during the experiment. They may depend on the measured relative velocity \(v\), but not on the event’s coordinates \(x\) and \(t\).
Every event at the center of \(S'\) has \(x=vt\) and \(x'=0\). Substituting these values into \(x'=ax+bt\) gives
\[0=a(vt)+bt=(av+b)t.\]This must hold for every \(t\), so \(b=-av\). The coordinate rule is now
\[\begin{aligned} x'&=a(x-vt),\\ t'&=dx+et. \end{aligned}\]For the positive light ray \(x=ct\), the condition \(x'=ct'\) gives
\[a(c-v)=ce+c^2d.\]For the negative light ray \(x=-ct\), the condition \(x'=-ct'\) gives
\[a(c+v)=ce-c^2d.\]Adding these two equations yields
\[e=a.\]Subtracting the second equation from the first yields
\[d=-\frac{av}{c^2}.\]The rule must therefore have the form
\[\begin{aligned} x'&=a(x-vt),\\ t'&=a\left(t-\frac{vx}{c^2}\right). \end{aligned}\]It remains to determine the coefficient \(a\). Experiments performed entirely inside either inertial laboratory are assumed to obey the same physical rules. This assumption is called the principle of relativity. The labels \(S\) and \(S'\) and the choice of positive direction are arbitrary. Exchanging the labels leaves \(a\) unchanged and replaces \(v\) with \(-v\). Apply these replacements to the spatial rule:
\[x=a\bigl(x'-(-v)t'\bigr)=a(x'+vt').\]Substitute the expressions for \(x'\) and \(t'\):
\[\begin{aligned} x &=a\left[ a(x-vt) +va\left(t-\frac{vx}{c^2}\right) \right]\\ &=a^2\left(1-\frac{v^2}{c^2}\right)x\\ &\Longrightarrow a^2\left(1-\frac{v^2}{c^2}\right)=1. \end{aligned}\]Choose the positive root so that the rule approaches the identity when \(v=0\). Thus \(a=\gamma(v)\), where
\[\gamma:(-c,c)\longrightarrow\mathbb R_{>0}, \qquad \gamma(v):=\frac{1}{\sqrt{1-v^2/c^2}}.\]This real factor is defined for \(\lvert v\rvert<c\). The resulting rule therefore treats \(c\) as a limiting relative speed for the inertial laboratories considered here.
For the chosen pair of laboratories, \(v\) is fixed, so \(\gamma(v)\) is constant across events. Write \(\gamma:=\gamma(v)\).
The required coordinate update is therefore
\[\boxed{ \begin{aligned} x'&=\gamma(x-vt),\\ t'&=\gamma\left(t-\frac{vx}{c^2}\right). \end{aligned}}\]This is the Lorentz transformation in one spatial direction.
To state its input and output explicitly, define the dimensionless relative speed
\[\beta:=\frac{v}{c}.\]Then the coordinate-pair transformation is the linear map
\[\Lambda_v^{(1+1)}:\mathbb R^2\longrightarrow\mathbb R^2,\]We use the ordered coordinates \((ct,x)\) so that both coordinates have units of length and every matrix entry is dimensionless. In these coordinates,
\[\begin{pmatrix} ct'\\ x' \end{pmatrix} = \underbrace{ \begin{pmatrix} \gamma&-\gamma\beta\\ -\gamma\beta&\gamma \end{pmatrix}}_{[\Lambda_v^{(1+1)}]} \begin{pmatrix} ct\\ x \end{pmatrix}.\]The matrix is a coordinate representation of the map; the two event-coordinate pairs are its input and output.
5. What does the rule predict for clocks?
Choose two locally recorded events in which the center station of \(S'\) passes two different stations on \(S\). Its clock is present at both events, so their position difference in \(S'\) is
\[\Delta x'=0.\]The inverse time transformation is
\[\Delta t =\gamma\left(\Delta t'+\frac{v\Delta x'}{c^2}\right).\]Substituting \(\Delta x'=0\) gives
\[\Delta t=\gamma\Delta t'.\]The one clock from \(S'\) supplies both readings used in \(\Delta t'\). Two different clocks from \(S\) supply the readings used in the larger interval \(\Delta t\). This prediction is called time dilation.
The reading accumulated by a clock along its own path is called its proper time, denoted \(\tau\). For this clock,
\[\Delta\tau=\Delta t',\]so
\[\Delta t=\gamma\Delta\tau.\]No clock is claimed to malfunction. The two quantities are produced by different measurement procedures: one clock accompanies both events, while a synchronized pair of clocks in the other laboratory records them at different positions.
6. What does the rule predict for simultaneous measurements?
Consider a rod at rest in \(S'\). Let its endpoint separation measured there be
\[L_0:=\Delta x'\]for two endpoint events with \(\Delta t'=0\). This is the rod’s rest length.
To measure the moving rod’s length in \(S\), record both endpoint positions at the same \(S\) time:
\[\Delta t=0.\]The spatial transformation then gives
\[L_0=\Delta x'=\gamma\Delta x.\]Therefore the length assigned by \(S\) is
\[L:=\Delta x=\frac{L_0}{\gamma}.\]This is length contraction along the direction of relative motion. It is a prediction about two simultaneous endpoint observations, not a claim that every observer selects the same pair of endpoint events.
Indeed, if two separated events are simultaneous in \(S\), then \(\Delta t=0\) and
\[\Delta t' =\gamma\left(\Delta t-\frac{v\Delta x}{c^2}\right) =-\gamma\frac{v\Delta x}{c^2}.\]Unless \(\Delta x=0\), laboratory \(S'\) does not call those events simultaneous. Relativity of simultaneity is not an additional postulate. It is required by the same coordinate rule that preserved the light observations.
7. How does the coordinate rule extend to three spatial dimensions?
To describe positions away from the \(x\)-axis, we now include \(y\) and \(z\) in the laboratories’ records.
Start with the \(x\)–\(y\) plane. Set \(t=t'=0\) when the two center stations meet at the same position; both laboratories label that position \(x=y=0\) and \(x'=y'=0\). Their axes remain aligned, and their relative motion is along \(x\). A general linear transformation of \((ct,x,y)\) has nine coefficients. The earlier longitudinal calculation fixes four of them, leaving
\[\begin{aligned} ct'&=\gamma ct-\gamma\beta x+Ay,\\ x'&=-\gamma\beta ct+\gamma x+By,\\ y'&=Cct+Dx+qy. \end{aligned}\]Here \(\beta=v/c\) as before; \(A,B,C,D,q\) are dimensionless unknown coefficients, constant for a fixed relative velocity \(v\).
First use records made at the centers of \(S\) and \(S'\) after they have passed each other, so \(t\ne0\).
For an event at \(S\)’s center, \(S\) records \(x=0\) and \(y=0\). The corresponding transverse coordinate in \(S'\) is \(y'=0\), since \(S\)’s center lies on the \(x'\)-axis. Substituting these coordinates into the third equation gives
\[0=Cct+D\cdot0+q\cdot0=Cct.\]Since \(ct\ne0\), this gives \(C=0\).
For an event at \(S'\)’s center, \(S\) records \(x=vt=\beta ct\) and \(y=0\). Laboratory \(S'\) assigns \(y'=0\) because the event occurs at its own center. Substituting these coordinates and using \(C=0\) gives
\[0=0\cdot ct+D\beta ct+q\cdot0=D\beta ct.\]For \(v\ne0\), we have \(\beta ct\ne0\), so \(D=0\).
Thus \(y'=qy\). For \(v=0\), the aligned laboratories use the identity transformation.
Next assume that neither transverse direction is physically preferred. Reflect the setup across its longitudinal axis, replacing \(y\) and \(y'\) by \(-y\) and \(-y'\). The time and longitudinal-coordinate rules must stay unchanged, but the terms \(Ay\) and \(By\) reverse sign. Therefore \(A=B=0\).
Only the transverse scale remains. Write it as \(q(v)\) to show its possible dependence on relative velocity. As in Section 4, exchanging the laboratories must give the inverse rule:
\[y'=q(v)y, \qquad y=q(-v)y', \qquad q(-v)q(v)=1.\]Reversing the longitudinal direction replaces \(v\) by \(-v\) without changing the transverse coordinates. Assuming no preferred longitudinal direction therefore gives \(q(-v)=q(v)\). Combining these conditions yields
\[q(v)^2=1.\]The positive \(y\) directions agree, so \(q(v)=1\) and \(y'=y\). The same argument applies to \(z\); reflection symmetry also rules out mixing between \(y\) and \(z\). Therefore
\[y'=y, \qquad z'=z.\]8. How does the clock result define a spacetime interval?
Section 5 already predicted the elapsed time on \(S'\)’s center clock. We now express that result using the three spatial coordinates introduced in Section 7. The labels \(S\) and \(S'\) can stand for any two inertial laboratories satisfying our setup.
How do \(S\) and \(S'\) express the elapsed time on the same clock using their own records?
Take the two encounters on \(S'\)’s center clock from Section 5. Its elapsed time is \(\Delta\tau\). Laboratory \(S\) records the time difference \(\Delta t>0\) and the center’s displacement \(\Delta x=v\Delta t\). Because the relative motion is along \(x\), \(\Delta y=\Delta z=0\). Consequently,
\[\Delta x^2+\Delta y^2+\Delta z^2 =(v\Delta t)^2+0+0 =v^2\Delta t^2.\]Substitute this into the clock result from Section 5:
\[\begin{aligned} \Delta\tau &=\frac{\Delta t}{\gamma(v)} =\Delta t\sqrt{1-\frac{v^2}{c^2}},\\ c^2\Delta\tau^2 &=c^2\Delta t^2-v^2\Delta t^2\\ &=c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2. \end{aligned}\]We can replace \(S\) in this calculation with any other inertial laboratory that records the same two encounters. That laboratory uses its own time and position differences, and the clock’s velocity relative to it, in the same time-dilation rule. The result is again \(c^2\Delta\tau^2\): all these calculations predict the elapsed proper time on the same clock in \(S'\).
In particular, using the records of \(S\) and \(S'\) gives
\[c^2\Delta t^2-\Delta x^2-\Delta y^2-\Delta z^2 =c^2\Delta t'^2-\Delta x'^2-\Delta y'^2-\Delta z'^2.\]The two laboratories assign different time and position differences, but this combination gives the same value \(c^2\Delta\tau^2\).
Using the convention with a negative time term, the spacetime interval is \(\Delta s^2:=-c^2\Delta t^2+\Delta x^2+\Delta y^2+\Delta z^2=-c^2\Delta\tau^2\).
What do the differentials act on?
We have written the clock rule using finite time and position differences. To understand its differential notation, follow \(S'\)’s center clock using the time and position coordinates recorded by \(S\).
Represent the modeled region as a smooth space of events \(M\). Laboratory \(S\) assigns coordinates through the map
\[\chi:M\longrightarrow\mathbb R^4, \qquad \chi(p)=\bigl(c\,t(p),x(p),y(p),z(p)\bigr).\]Here \(p\) is an event, and \(t,x,y,z\) are the coordinate functions assigning its timestamp and position. Multiplying \(t(p)\) by \(c\) expresses all four components in units of length.
Write the center clock’s path as \(\zeta:I\to M\), where \(I\subseteq\mathbb R\) is the interval of \(S\)’s laboratory timestamps covered by the journey. The path parameter \(\lambda\in I\) is \(S\)’s recorded time, in seconds, at the clock’s position. Reserve \(t:M\to\mathbb R\) for the function assigning \(S\)’s timestamp to each spacetime event. Along the clock’s path,
\[t\bigl(\zeta(\lambda)\bigr)=\lambda,\qquad \chi\bigl(\zeta(\lambda)\bigr)=(c\lambda,v\lambda,0,0).\]At \(p=\zeta(\lambda)\), the path has a tangent vector
\[V_p:=\frac{d\zeta}{d\lambda}\in T_pM.\]The space \(T_pM\), called the tangent space, consists of tangents to smooth paths through \(p\). This is the construction used for configuration space in Energy Beyond Particle Mechanics, applied here to paths of events. In \(S\)’s coordinates, our clock’s tangent has components
\[\frac{d}{d\lambda}\chi\bigl(\zeta(\lambda)\bigr) =\frac{d}{d\lambda}(c\lambda,v\lambda,0,0) =(c,v,0,0).\]These components are coordinate rates per unit of \(S\)’s time. The capital \(V_p\) denotes the tangent; \(v\) remains the relative velocity of the laboratories.
For a smooth scalar function \(f:M\to\mathbb R\), its differential at \(p\) is a linear map
\[df_p:T_pM\longrightarrow\mathbb R, \qquad df_p(V_p) =\frac{d}{d\lambda}f\bigl(\zeta(\lambda)\bigr).\]To evaluate it on any tangent at \(p\), differentiate \(f\) along a path with that tangent. For our clock, choosing the position function \(f=x\) gives
\[dx_p(V_p) =\frac{d}{d\lambda}x\bigl(\zeta(\lambda)\bigr) =\frac{d}{d\lambda}(v\lambda) =v.\]The time-coordinate differential acts on the spacetime tangent by differentiating \(t\circ\zeta\); the transverse coordinates stay zero:
\[dt_p(V_p)=\frac{d}{d\lambda}t\bigl(\zeta(\lambda)\bigr)=1, \qquad dy_p(V_p)=0, \qquad dz_p(V_p)=0.\]A linear map from tangent vectors to real numbers is called a covector. The space of these maps is the dual space \(T_p^*M\), so \(dx_p\in T_p^*M\). The differential \(dx_p\) is the operation; \(v\) is its value on this particular clock’s tangent.
How do these differentials express the metric?
We now use the coordinate rates recorded by \(S\) to express how fast \(S'\)’s center clock ticks relative to \(S\)’s clocks. We write each squared rate as a product with two tangent inputs at the same event \(p\). For a second tangent \(W_p\in T_pM\), the product \(dx_p(V_p)\,dx_p(W_p)\) gives \(dx_p(V_p)^2\) when we use the clock’s tangent for both inputs, setting \(W_p=V_p\). This represents the clock rule through a map
\[\eta_p:T_pM\times T_pM\longrightarrow\mathbb R.\]The map will be linear in each input separately, which is the meaning of bilinear. The tensor product denotes these products of coordinate rates:
\[(dx_p\otimes dx_p)(V_p,W_p) :=dx_p(V_p)\,dx_p(W_p).\]The interval’s coefficients then give the Minkowski metric
\[\eta_p =-c^2\,dt_p\otimes dt_p +dx_p\otimes dx_p +dy_p\otimes dy_p +dz_p\otimes dz_p.\]This expression defines \(\eta_p\) at each event using the coordinate differentials there. We write \(\eta\) for the metric whose value at event \(p\) is \(\eta_p\). Exchanging its two inputs leaves the result unchanged, so the map is symmetric.
Use our clock’s tangent for both inputs, substituting the coordinate rates above:
\[\begin{aligned} \eta_p(V_p,V_p) &=-c^2\bigl(dt_p(V_p)\bigr)^2 +\bigl(dx_p(V_p)\bigr)^2 +\bigl(dy_p(V_p)\bigr)^2 +\bigl(dz_p(V_p)\bigr)^2\\ &=-c^2\cdot1^2+v^2+0+0\\ &=-c^2+v^2. \end{aligned}\]Define \(\tau:I\to\mathbb R\) so that \(\tau(\lambda)\) is the reading, in seconds, of \(S'\)’s center clock at the event \(\zeta(\lambda)\). To calculate its rate, choose a nonzero parameter increment \(\Delta\lambda\) with both \(\lambda\) and \(\lambda+\Delta\lambda\) in \(I\). Using \(t(\zeta(\lambda))=\lambda\) and Section 5’s clock rule gives
\[\begin{aligned} &\Delta t =t\bigl(\zeta(\lambda+\Delta\lambda)\bigr)-t\bigl(\zeta(\lambda)\bigr) =\Delta\lambda,\\ &\tau(\lambda+\Delta\lambda)-\tau(\lambda) =\frac{\Delta t}{\gamma(v)} =\frac{\Delta\lambda}{\gamma(v)}. \end{aligned}\]Divide the clock-reading difference by \(\Delta\lambda\) and take the limit. Then substitute the definition of \(\gamma(v)\) and the metric value \(\eta_p(V_p,V_p)=-c^2+v^2\):
\[\begin{aligned} \frac{d\tau}{d\lambda} &=\lim_{\Delta\lambda\to0} \frac{\tau(\lambda+\Delta\lambda)-\tau(\lambda)}{\Delta\lambda}\\ &=\lim_{\Delta\lambda\to0} \frac{\Delta\lambda/\gamma(v)}{\Delta\lambda} =\frac{1}{\gamma(v)}\\ &=\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{c}\sqrt{-\eta_p(V_p,V_p)}. \end{aligned}\]This derivative measures the moving clock’s seconds per second recorded by \(S\). We now want to express the interval along this same path using either \(S\)’s time or the moving clock’s own time.
How can we express the interval using either clock’s time?
Changing the time parameter changes the path’s tangent and hence the scalar \(\eta_p(V_p,V_p)\). We will express the full bilinear rule on the parameter interval \(I\), so that we can write it in either time coordinate. Since \(\eta_p\) accepts spacetime tangents, we first need to send tangents of \(I\) to spacetime along the clock’s path. Fix a parameter value \(\lambda_0\) and let \(p=\zeta(\lambda_0)\). The interval’s tangent space is one-dimensional:
\[T_{\lambda_0}I =\left\{a\left.\frac{\partial}{\partial\lambda}\right|_{\lambda_0} \;\middle|\;a\in\mathbb R\right\}.\]Here \(\partial/\partial\lambda\) is the tangent that differentiates functions of the parameter. Its dual, the cotangent space, consists of linear maps on these tangents:
\[T_{\lambda_0}^*I :=\{\alpha:T_{\lambda_0}I\to\mathbb R\mid\alpha\text{ is linear}\}.\]The coordinate function on \(I\) is \(\operatorname{id}_I:I\to\mathbb R\), with \(\operatorname{id}_I(\lambda)=\lambda\). Apply our differential definition to this function to obtain the coordinate covector on the interval:
\[\begin{aligned} &(d\lambda)_{\lambda_0} :=d(\operatorname{id}_I)_{\lambda_0}\in T_{\lambda_0}^*I,\\ &(d\lambda)_{\lambda_0}\left( a\left.\frac{\partial}{\partial\lambda}\right|_{\lambda_0} \right)=a. \end{aligned}\]Thus \(dt_p\) acts on spacetime tangents in \(T_pM\), while \((d\lambda)_{\lambda_0}\) acts on parameter tangents in \(T_{\lambda_0}I\).
The derivative of the path transfers a parameter tangent to spacetime:
\[\begin{aligned} &d\zeta_{\lambda_0}:T_{\lambda_0}I\longrightarrow T_pM,\\ &d\zeta_{\lambda_0}\left( a\left.\frac{\partial}{\partial\lambda}\right|_{\lambda_0} \right)=aV_p. \end{aligned}\]Use this map on both inputs before applying \(\eta_p\). For any \(A,B\in T_{\lambda_0}I\), define
\[(\zeta^*\eta)_{\lambda_0}(A,B) :=\eta_p\bigl(d\zeta_{\lambda_0}(A),d\zeta_{\lambda_0}(B)\bigr).\]This bilinear form on \(I\) is the pullback metric. To calculate its coefficient, write \(A=a\,\partial/\partial\lambda\) and \(B=b\,\partial/\partial\lambda\) at \(\lambda_0\). Linearity of \(d\zeta_{\lambda_0}\) and bilinearity of \(\eta_p\) give
\[\begin{aligned} &(\zeta^*\eta)_{\lambda_0}(A,B) =\eta_p(aV_p,bV_p) =ab\,\eta_p(V_p,V_p),\\ &\bigl((d\lambda)_{\lambda_0}\otimes(d\lambda)_{\lambda_0}\bigr)(A,B) =(d\lambda)_{\lambda_0}(A)\,(d\lambda)_{\lambda_0}(B) =ab. \end{aligned}\]These evaluations determine the bilinear form on every pair of inputs. We define the line element along the path, denoted \(ds^2\), by
\[\begin{aligned} (ds^2)_{\lambda_0} &:=(\zeta^*\eta)_{\lambda_0}\\ &=\eta_p(V_p,V_p)\, (d\lambda)_{\lambda_0}\otimes(d\lambda)_{\lambda_0} \in T_{\lambda_0}^*I\otimes T_{\lambda_0}^*I. \end{aligned}\]The scalar \(\eta_p(V_p,V_p)\) is the coefficient of this tensor, with units \(\mathrm{m^2/s^2}\) for our time parameter.
The moving clock’s reading \(\tau\) is already a function on \(I\). Its differential is
\[d\tau_{\lambda_0} =\left.\frac{d\tau}{d\lambda}\right|_{\lambda_0}(d\lambda)_{\lambda_0}.\]Square \(d\tau/d\lambda=1/\gamma(v)\) at \(\lambda_0\) and use \(\eta_p(V_p,V_p)=-c^2+v^2\):
\[\begin{aligned} \left(\left.\frac{d\tau}{d\lambda}\right|_{\lambda_0}\right)^2 &=\frac{1}{\gamma(v)^2} =1-\frac{v^2}{c^2} =-\frac{\eta_p(V_p,V_p)}{c^2},\\ \eta_p(V_p,V_p) &=-c^2\left(\left.\frac{d\tau}{d\lambda}\right|_{\lambda_0}\right)^2. \end{aligned}\]Substitute this coefficient into \((ds^2)_{\lambda_0}\), then use the expression for \(d\tau_{\lambda_0}\) above:
\[\begin{aligned} (ds^2)_{\lambda_0} &=\eta_p(V_p,V_p)\,(d\lambda)_{\lambda_0}\otimes(d\lambda)_{\lambda_0}\\ &=-c^2\left(\left.\frac{d\tau}{d\lambda}\right|_{\lambda_0}\right)^2 (d\lambda)_{\lambda_0}\otimes(d\lambda)_{\lambda_0}\\ &=-c^2\,d\tau_{\lambda_0}\otimes d\tau_{\lambda_0}. \end{aligned}\]Write \(d\tau^2:=d\tau\otimes d\tau\). Then \(ds^2=-c^2d\tau^2\).
What if the clock’s velocity changes?
Now allow the clock to move independently of its railcar while both recording networks remain inertial.
How much time does the clock accumulate along its recorded path when its velocity changes?
Keep \(\lambda\) equal to \(S\)’s time along the new path. Write \(u_x(\lambda),u_y(\lambda),u_z(\lambda)\) for the clock’s velocity components in \(S\). At each \(p=\zeta(\lambda)\), with \(V_p=d\zeta/d\lambda\), the differentials give
\[dt_p(V_p)=1, \qquad dx_p(V_p)=u_x(\lambda), \qquad dy_p(V_p)=u_y(\lambda), \qquad dz_p(V_p)=u_z(\lambda).\]The speed is \(u(\lambda)=\sqrt{u_x(\lambda)^2+u_y(\lambda)^2+u_z(\lambda)^2}\). Substituting these rates into the metric gives
\[\eta_p(V_p,V_p) =-c^2+u_x(\lambda)^2+u_y(\lambda)^2+u_z(\lambda)^2 =-c^2+u(\lambda)^2.\]We assume an ideal clock: during acceleration, its instantaneous ticking rate follows the constant-velocity rule at its current velocity, with no additional acceleration-dependent change. This extends the clock model beyond the uniform motion considered in Section 5. For a smooth path with \(u(\lambda)<c\), its rate is therefore
\[\frac{d\tau}{d\lambda} =\frac{1}{c}\sqrt{-\eta_p(V_p,V_p)} =\sqrt{1-\frac{u(\lambda)^2}{c^2}}.\]Integrate between parameter values \(\lambda_1\) and \(\lambda_2\), which are the starting and ending laboratory times, to predict the change on the moving clock:
\[\tau(\lambda_2)-\tau(\lambda_1) =\int_{\lambda_1}^{\lambda_2} \sqrt{1-\frac{u(\lambda)^2}{c^2}}\,d\lambda.\]The integral adds the clock increments along the full path. Applying the finite constant-velocity formula once to the journey’s endpoints generally gives a different result: different paths between those endpoints can accumulate different elapsed times.
9. How do coordinate transformations act on differentials?
Both laboratories assign coordinates to the same events in \(M\). To relate their coordinate differentials, use \(S\)’s chart \(\chi\) and define \(S'\)’s chart in the same coordinate order:
\[\chi':M\longrightarrow\mathbb R^4, \qquad \chi'(p)=\bigl(c\,t'(p),x'(p),y'(p),z'(p)\bigr).\]Here \(t',x',y',z':M\to\mathbb R\) assign \(S'\)’s timestamp in seconds and positions in metres. As coordinate charts, \(\chi\) and \(\chi'\) have smooth inverses on their coordinate images. The coordinate transition map converts \(S\)’s coordinates of an event into \(S'\)’s coordinates of that same event:
\[F:=\chi'\circ\chi^{-1}: \chi(M)\longrightarrow\chi'(M).\]Write \(X=(X^0,X^1,X^2,X^3)\in\chi(M)\) for a coordinate tuple, with all four entries measured in metres. The Lorentz transformation from Sections 4 and 7 gives
\[F(X)= \begin{pmatrix} \gamma(v)\bigl(X^0-(v/c)X^1\bigr)\\ \gamma(v)\bigl(X^1-(v/c)X^0\bigr)\\ X^2\\ X^3 \end{pmatrix}.\]Let \(F^\mu:\chi(M)\to\mathbb R\) denote component \(\mu\) of \(F\), with \(\mu=0,1,2,3\). Differentiate these component functions with respect to the input coordinates \(X^\nu\), keeping \(v\) constant. At \(X=\chi(p)\), the resulting Jacobian matrix \(J_p\in\mathbb R^{4\times4}\) is
\[J_p:=\left. \left(\frac{\partial F^\mu}{\partial X^\nu}\right)_{\mu,\nu=0}^{3} \right|_{X=\chi(p)} = \begin{pmatrix} \gamma(v) & -\gamma(v)v/c & 0 & 0\\ -\gamma(v)v/c & \gamma(v) & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}.\]The matrix \(J_p\) represents the derivative \(DF_{\chi(p)}:\mathbb R^4\to\mathbb R^4\). The chart derivatives \(d\chi_p,d\chi'_p:T_pM\to\mathbb R^4\) assign coordinate components to a spacetime tangent. Since \(\chi'=F\circ\chi\), the chain rule gives
\[d\chi'_p=DF_{\chi(p)}\circ d\chi_p.\]Writing these linear maps in components gives the matrix multiplication
\[\begin{pmatrix}c\,dt'_p\\dx'_p\\dy'_p\\dz'_p\end{pmatrix} =J_p \begin{pmatrix}c\,dt_p\\dx_p\\dy_p\\dz_p\end{pmatrix}.\]The first row, followed by division by \(c\), gives the time-coordinate differential:
\[\begin{aligned} c\,dt'_p&=\gamma(v)c\,dt_p-\gamma(v)\frac{v}{c}dx_p,\\ dt'_p&=\gamma(v)\left(dt_p-\frac{v}{c^2}dx_p\right). \end{aligned}\]10. What velocity does \(S'\) measure for a test cart?
Let a small test cart move along the common \(x\) direction. Write \(u\) and \(u'\) for its signed instantaneous velocities in \(S\) and \(S'\), measured in metres per second along their positive \(x\) axes. The laboratories retain their constant relative velocity \(v\).
What velocity \(u'\) does \(S'\) calculate for the same cart?
Write the cart’s smooth path as \(\rho:I\to M\), where \(I\) is its interval of laboratory timestamps in \(S\). Choose \(\lambda\) to equal that timestamp. At \(p=\rho(\lambda)\), define the cart’s tangent \(V_p\):
\[\begin{aligned} &t\bigl(\rho(\lambda)\bigr)=\lambda,\\ &V_p:=\frac{d\rho}{d\lambda}\in T_pM. \end{aligned}\]The coordinate differentials give the timestamp and position rates along this path:
\[\begin{aligned} &dt_p(V_p)=\frac{d}{d\lambda}t\bigl(\rho(\lambda)\bigr)=1,\\ &dx_p(V_p)=\frac{d}{d\lambda}x\bigl(\rho(\lambda)\bigr)=u. \end{aligned}\]Evaluate the first two rows of Section 9’s Jacobian relation on \(V_p\), dividing the time row by \(c\). Substituting these rates gives
\[\begin{aligned} dt'_p(V_p) &=\gamma(v)\left(dt_p(V_p)-\frac{v}{c^2}dx_p(V_p)\right) =\gamma(v)\left(1-\frac{vu}{c^2}\right),\\ dx'_p(V_p) &=\gamma(v)\bigl(dx_p(V_p)-v\,dt_p(V_p)\bigr) =\gamma(v)(u-v). \end{aligned}\]The functions \(x'\circ\rho,t'\circ\rho:I\to\mathbb R\) give \(S'\)’s position and timestamp records along the cart’s path. Its measured velocity \(u'\) is the limiting ratio of their increments. Where \(dt'_p(V_p)\ne0\), divide both increments by \(\Delta\lambda\) and take the limit:
\[\begin{aligned} u' &=\lim_{\Delta\lambda\to0} \frac{x'(\rho(\lambda+\Delta\lambda))-x'(\rho(\lambda))} {t'(\rho(\lambda+\Delta\lambda))-t'(\rho(\lambda))}\\ &=\frac{\dfrac{d}{d\lambda}x'\bigl(\rho(\lambda)\bigr)} {\dfrac{d}{d\lambda}t'\bigl(\rho(\lambda)\bigr)} =\frac{dx'_p(V_p)}{dt'_p(V_p)}. \end{aligned}\]Substitute the evaluated rates and cancel \(\gamma(v)\). The nonzero-denominator condition is \(1-uv/c^2\ne0\):
\[u'=\frac{\gamma(v)(u-v)}{\gamma(v)(1-uv/c^2)} =\frac{u-v}{1-uv/c^2}.\]This predicts \(u'\) at each event on the cart’s path from its instantaneous velocity \(u\) and the laboratory velocity \(v\) measured by \(S\). For a cart at rest in \(S\), substituting \(u=0\) gives \(u'=-v\). When both \(\lvert u\rvert\) and \(\lvert v\rvert\) are much smaller than \(c\), the denominator is nearly one and \(u'\approx u-v\).
For a light pulse travelling along positive \(x\), substitute \(u=c\):
\[u'=\frac{c-v}{1-v/c}=c,\]as required by the light-propagation assumption.
11. How do momentum and energy fit this geometry?
We have answered the coordinate-comparison question without a force law. In Rediscovering Mechanics and Energy from Questions, integrating force over time gave a momentum difference; integrating force along a displacement gave an energy difference.
Can the spacetime geometry suggest a momentum that lets us repeat those endpoint calculations?
What geometric object could represent momentum?
Keep the cart’s path \(\rho:I\to M\) and laboratory-time parameter \(\lambda\) from Section 10. Model the cart as a particle with fixed rest mass \(m>0\), measured in kilograms by the low-speed inertial calibration of the first post and assumed identical in both laboratories. Allow its signed velocity \(u:I\to(-c,c)\) in \(S\) to vary. For this cart, let \(\tau:I\to\mathbb R\) record an ideal clock carried along its path, in seconds.
To describe motion per unit of the cart’s own elapsed time, divide its laboratory-time tangent by the clock rate from Section 8. At \(p=\rho(\lambda)\), write \(u=u(\lambda)\) and define the four-velocity \(U_p\) and a mass-weighted candidate \(P_p\):
\[\begin{aligned} \frac{d\tau}{d\lambda}&=\sqrt{1-\frac{u^2}{c^2}} =\frac{1}{\gamma(u)},\\ U_p&:=\frac{V_p}{d\tau/d\lambda} =\gamma(u)V_p\in T_pM,\\ P_p&:=mU_p,\\ d\chi_p(P_p)&=m\gamma(u) \begin{pmatrix}c\\u\\0\\0\end{pmatrix}. \end{aligned}\]All four components of \(P_p\) have momentum units, \(\mathrm{kg\,m/s}\). Its spatial component suggests the function \(p_x:(-c,c)\to\mathbb R\),
\[p_x(u):=m\gamma(u)u, \qquad p_x(u)\approx mu\quad\text{when }\lvert u\rvert\ll c.\]Because \(P_p\) is a tangent vector, Section 9’s chain rule gives its components in the other chart:
\[d\chi'_p(P_p)=J_p\,d\chi_p(P_p).\]This is the transformation law for a contravariant vector, or tensor of type \((1,0)\). The vector \(P_p\) is the same object in \(T_pM\); the two charts assign different component columns to it. The chain rule gives this Jacobian law for any smooth change of coordinates, including the Lorentz transformation used here.
The metric \(\eta_p\) is a tensor of type \((0,2)\): it takes two tangent vectors and returns a scalar. Evaluating the same metric on the same vectors gives the same number in either chart. For example,
\[\begin{aligned} \eta_p(P_p,P_p) &=m^2\gamma(u)^2\eta_p(V_p,V_p)\\ &=m^2\gamma(u)^2(-c^2+u^2) =-m^2c^2. \end{aligned}\]Which endpoint quantities follow from a force law?
Geometry alone does not establish that \(P_p\) describes momentum exchanged in interactions. We adopt conservation of total four-momentum as an additional physical law: summing the particles’ \(P_p\) components in a single inertial chart gives the same totals before and after an isolated interaction.
Let \(F_x:I\to\mathbb R\) be the signed force in \(S\), in newtons, supplied by an independently calibrated interaction law. We also adopt the relativistic force equation. Integrating it from laboratory time \(\lambda_1\) to \(\lambda_2\), with \(u_i:=u(\lambda_i)\), gives
\[\begin{aligned} \frac{d}{d\lambda}p_x(u(\lambda))&=F_x(\lambda),\\ p_x(u_2)-p_x(u_1) &=\int_{\lambda_1}^{\lambda_2}F_x(\lambda)\,d\lambda. \end{aligned}\]This is the first post’s impulse calculation with the new momentum function. To calculate the work \(W\) in joules, use \(d(x\circ\rho)/d\lambda=u(\lambda)\), substitute the force equation, and apply the chain rule:
\[\begin{aligned} W&:=\int_{\lambda_1}^{\lambda_2} F_x(\lambda)\frac{d(x\circ\rho)}{d\lambda}\,d\lambda\\ &=\int_{\lambda_1}^{\lambda_2} \frac{d}{d\lambda}p_x(u(\lambda))\,u(\lambda)\,d\lambda\\ &=\int_{\lambda_1}^{\lambda_2} \frac{dp_x}{du}\bigl(u(\lambda)\bigr) \frac{du}{d\lambda}\,u(\lambda)\,d\lambda. \end{aligned}\]To find an endpoint expression, differentiate \(\gamma(u)=(1-u^2/c^2)^{-1/2}\):
\[\begin{aligned} \frac{d\gamma}{du}&=\frac{u}{c^2}\gamma(u)^3,\\ \frac{dp_x}{du} &=m\gamma(u)+mu\frac{d\gamma}{du}\\ &=m\gamma(u)^3\left(1-\frac{u^2}{c^2}\right)+\frac{mu^2}{c^2}\gamma(u)^3 =m\gamma(u)^3,\\ \frac{d}{du}\bigl(mc^2\gamma(u)\bigr) &=mu\gamma(u)^3 =u\frac{dp_x}{du}. \end{aligned}\]Substituting into the work integral gives
\[\begin{aligned} W&=\int_{\lambda_1}^{\lambda_2} mu(\lambda)\gamma(u(\lambda))^3\frac{du}{d\lambda}\,d\lambda\\ &=\int_{u_1}^{u_2}mu\gamma(u)^3\,du\\ &=\left[mc^2\gamma(u)\right]_{u_1}^{u_2}. \end{aligned}\]Define the cart’s energy in \(S\) by \(E:(-c,c)\to\mathbb R\), measured in joules:
\[E(u):=mc^2\gamma(u), \qquad W=E(u_2)-E(u_1).\]The time component of \(P_p\) is therefore \(E/c\). Write \(E'\) and \(p'_x\) for the cart’s energy and momentum in \(S'\). The first two rows of the Lorentz Jacobian give
\[\begin{pmatrix}E'/c\\p'_x\end{pmatrix} = \begin{pmatrix}\gamma(v)&-\gamma(v)v/c\\-\gamma(v)v/c&\gamma(v)\end{pmatrix} \begin{pmatrix}E/c\\p_x\end{pmatrix}, \qquad E'=\gamma(v)(E-vp_x).\]Energy and momentum components depend on the laboratory. Conservation of total four-momentum includes conservation of the sum of these energies in each inertial laboratory.
The work integral at fixed mass determines energy differences; identifying \(E/c\) with the time component of the conserved four-vector fixes the reference used here. It assigns rest energy \(E(0)=mc^2\). In reactions that change the particles’ rest masses, these rest terms enter the same conserved energy sum. The kinetic energy \(K:(-c,c)\to\mathbb R\), measured in joules relative to rest, is
\[K(u):=E(u)-E(0)=mc^2\bigl(\gamma(u)-1\bigr) \approx\frac12mu^2\quad\text{when }\lvert u\rvert\ll c.\]These relations have experimental support: Bertozzi’s electron measurements tested the speed–kinetic-energy relation, and Rainville and colleagues compared atomic mass differences with emitted gamma-ray energies to test the rest-energy relation.
12. Answer to the main question
Under the stated assumptions, the Lorentz transformation relates the two inertial laboratories’ coordinates for the same event:
\[\begin{aligned} x'&=\gamma(v)(x-vt),\\ t'&=\gamma(v)\left(t-\frac{vx}{c^2}\right),\\ y'&=y,\\ z'&=z. \end{aligned}\]Time dilation, length contraction, and relativity of simultaneity follow from this rule. For a uniformly moving clock, the spacetime interval recovers the same elapsed time from either laboratory’s records. The metric lets us apply the clock rule along a path with changing velocity.
With the additional momentum and force laws, we recovered momentum and energy as endpoint quantities for impulse and work, extending the first post’s construction to relativistic speeds.
This is the flat-spacetime description of special relativity, used here within the modeled region. To account for gravity beyond a region where curvature can be neglected, we face a further question:
What rule predicts clocks, light, and freely falling objects when gravitational effects vary across a region?
From Free Fall to Curved Spacetime begins with that question.