16.1 Space, time and the invariant interval

(a) Events and the position four-vector

In Newtonian mechanics space and time are separate: the distance between two points and the time between two instants are each the same for every observer. Special relativity replaces them by a single four-dimensional spacetime, whose points are events, things that happen at a definite place and a definite time. An event is labelled in an inertial frame \(S\) by four real coordinates, which we combine into the position four-vector

\[\boxed{x^\mu=(x^0,x^1,x^2,x^3)=(ct,\,x,\,y,\,z)},\qquad \mu=0,1,2,3.\]

The time coordinate is multiplied by \(c\) so that all four components have the dimensions of length. Greek indices run over \(0,1,2,3\), and Latin indices \(i,j,\dots\) over the spatial values \(1,2,3\) only.

Let a second frame \(S'\) move with speed \(v\) along the common \(x\)-axis of \(S\), the origins coinciding at \(t=t'=0\). The Lorentz transformation derived in Chapter 14 reads, with \(\beta=v/c\) and \(\gamma=1/\sqrt{1-\beta^2}\),

\[ct'=\gamma(ct-\beta x),\qquad x'=\gamma(x-\beta ct),\qquad y'=y,\qquad z'=z,\]

with the inverse obtained by changing the sign of \(\beta\): \(ct=\gamma(ct'+\beta x')\) and \(x=\gamma(x'+\beta ct')\). Unlike the Galilean transformation, it mixes \(t\) and \(x\), and neither the time between two events nor the distance between them is the same in the two frames. A certain combination of the two, however, is the same.

(b) The invariant interval

Consider the quantity \(c^2t^2-x^2-y^2-z^2\) for the event with coordinates \((ct,x,y,z)\). Substituting the inverse transformation,

\[\begin{aligned}c^2t^2-x^2&=\gamma^2\left[(ct'+\beta x')^2-(x'+\beta ct')^2\right]\\&=\gamma^2\left[c^2t'^2(1-\beta^2)-x'^2(1-\beta^2)\right]=c^2t'^2-x'^2,\end{aligned}\]

since the cross terms \(2\beta ct'x'\) cancel and \(\gamma^2(1-\beta^2)=1\). Together with \(y=y'\) and \(z=z'\) this gives

\[c^2t^2-x^2-y^2-z^2=c^2t'^2-x'^2-y'^2-z'^2.\]

Applied to the differences of the coordinates of two events, the same algebra shows that the spacetime interval

\[\boxed{(\Delta s)^2=c^2(\Delta t)^2-(\Delta x)^2-(\Delta y)^2-(\Delta z)^2=c^2(\Delta t)^2-(\Delta r)^2}\]

has the same value in every inertial frame. It plays the role in spacetime that the squared distance \((\Delta x)^2+(\Delta y)^2+(\Delta z)^2\) plays in Euclidean space, which is invariant under rotations. The invariance of the interval is the geometrical content of the Lorentz transformation; indeed, the Lorentz transformations can be defined as the linear transformations that leave \((\Delta s)^2\) unchanged.

The interval can be written compactly with the Minkowski metric

\[\eta_{\mu\nu}=\operatorname{diag}(1,-1,-1,-1),\qquad (\Delta s)^2=\sum_{\mu,\nu}\eta_{\mu\nu}\,\Delta x^\mu\Delta x^\nu=\eta_{\mu\nu}\,\Delta x^\mu\Delta x^\nu,\]

where in the last form we use the summation convention: an index repeated once up and once down is summed from 0 to 3. Many texts use the opposite overall sign, \(\operatorname{diag}(-1,1,1,1)\); the physics is the same, and only the signs of squared four-vectors change.

(c) Lorentz boosts as hyperbolic rotations

The analogy with rotations can be made precise. Define the rapidity \(\zeta\) by

\[\tanh\zeta=\beta,\qquad\text{so that}\qquad \cosh\zeta=\gamma,\qquad \sinh\zeta=\beta\gamma.\]

The Lorentz transformation then takes the form

\[\begin{pmatrix}ct'\\x'\end{pmatrix}=\begin{pmatrix}\cosh\zeta&-\sinh\zeta\\-\sinh\zeta&\cosh\zeta\end{pmatrix}\begin{pmatrix}ct\\x\end{pmatrix},\]

which differs from a rotation \(x'=x\cos\phi+y\sin\phi\), \(y'=-x\sin\phi+y\cos\phi\) only in that hyperbolic functions replace circular ones. Since \(\cosh^2\zeta-\sinh^2\zeta=1\), the combination \(c^2t^2-x^2\) is preserved, just as \(x^2+y^2\) is preserved by a rotation because \(\cos^2\phi+\sin^2\phi=1\). Rapidities are additive: two successive boosts along \(x\) with rapidities \(\zeta_1\) and \(\zeta_2\) give a boost with rapidity \(\zeta_1+\zeta_2\), and the identity \(\tanh(\zeta_1+\zeta_2)=(\tanh\zeta_1+\tanh\zeta_2)/(1+\tanh\zeta_1\tanh\zeta_2)\) is precisely the relativistic addition of velocities of Chapter 14.

Older texts make the analogy literal by using an imaginary fourth coordinate \(x_4=ict\). Then \(s^2=x^2+y^2+z^2+(ict)^2=x^2+y^2+z^2-c^2t^2\) looks like a Euclidean length, and a boost becomes a rotation of the \((x,ict)\) axes through the imaginary angle \(i\zeta\), as sketched in Figure 16.1. This convention avoids the distinction between upper and lower indices but at the cost of complex coordinates, and it is no longer used in modern work. In this book we always use the real coordinates \(x^\mu=(ct,\mathbf r)\) with the metric \(\eta_{\mu\nu}\).

(d) Timelike, spacelike and lightlike intervals

Because \((\Delta s)^2\) is not a sum of squares, it can be positive, negative or zero, and its sign is the same in every frame. This gives an absolute classification of pairs of events. For simplicity take \(\Delta y=\Delta z=0\).

If \((\Delta s)^2>0\), that is \(\lvert\Delta x\rvert<c\lvert\Delta t\rvert\), the interval is timelike: the time separation of the two events exceeds the time light needs to cover the distance between them. A material body can then travel from one event to the other, and there is a frame, moving with velocity \(v=\Delta x/\Delta t\), in which both events occur at the same place. In that frame the time between them is the proper time \(\Delta\tau\), and invariance of the interval gives

\[\boxed{(\Delta s)^2=c^2(\Delta t)^2-(\Delta x)^2=c^2(\Delta\tau)^2}.\]

The order in time of two events separated by a timelike interval is the same in all frames, so one can be the cause of the other.

If \((\Delta s)^2<0\), that is \(\lvert\Delta x\rvert>c\lvert\Delta t\rvert\), the interval is spacelike: the time separation is less than the time light needs to cover the distance. No signal can connect the two events. There is a frame, moving with velocity \(v=c^2\Delta t/\Delta x\) (less than \(c\)), in which they are simultaneous, and in that frame their distance apart is the proper distance \(\sqrt{-(\Delta s)^2}\). The time order of such events depends on the frame, which is harmless precisely because neither can influence the other.

If \((\Delta s)^2=0\), that is \(\lvert\Delta x\rvert=c\lvert\Delta t\rvert\), the interval is lightlike or null: the two events can be connected by a light signal, and only by a light signal.

These three cases are pictured on a spacetime diagram with \(ct\) vertical and \(x\) horizontal (Figure 16.2). The lines \(x=\pm ct\) through an event are the paths of light signals through it; in three space dimensions they form a double cone, the light cone of the event. Events inside the upper half (the future light cone) are timelike separated from the vertex and lie in its absolute future; events inside the lower half (the past light cone) are in its absolute past; events outside the cone are spacelike separated from it and form its “elsewhere”. Since the interval is invariant, the light cone is the same in all frames.

Example 16.1 In a frame \(S\) event \(A\) occurs at the origin at \(t=0\). (a) Event \(B\) occurs on the \(x\)-axis at \(x=900\;\text{m}\) at \(t=2\;\mu\text{s}\). Classify the interval, find the speed of the frame in which \(A\) and \(B\) are simultaneous, and their separation in that frame. (b) Event \(C\) occurs at \(x=900\;\text{m}\) at \(t=5\;\mu\text{s}\). Classify the interval \(AC\) and find the proper time between \(A\) and \(C\).

Solution. (a) With \(c=3\times10^8\;\text{m s}^{-1}\), \(c\Delta t=600\;\text{m}<\Delta x=900\;\text{m}\), so the interval is spacelike, with \((\Delta s)^2=600^2-900^2=-4.5\times10^5\;\text{m}^2\). In a frame moving with velocity \(v\) along \(x\), \(\Delta t'=\gamma(\Delta t-v\Delta x/c^2)\), which vanishes for

\[v=\frac{c^2\Delta t}{\Delta x}=c\,\frac{600}{900}=\frac23c.\]

In that frame \(\Delta t'=0\), so the separation is the proper distance \(\sqrt{900^2-600^2}\;\text{m}=671\;\text{m}\).

2. Now \(c\Delta t=1500\;\text{m}>900\;\text{m}\), so the interval is timelike. In the frame moving with \(v=\Delta x/\Delta t=0.6c\) both events occur at the same place, and the time between them is

\[\Delta\tau=\frac{1}{c}\sqrt{c^2(\Delta t)^2-(\Delta x)^2}=\sqrt{5^2-3^2}\;\mu\text{s}=4\;\mu\text{s},\]

since \(900\;\text{m}\) is the distance light travels in \(3\;\mu\text{s}\). As a check, \(\Delta t/\gamma=5\times0.8=4\;\mu\text{s}\).