Invariance of the Spacetime Interval

Derivation

Invariance of the Spacetime Interval

D-091 Home PU-105 Threads light · symmetry Depends on Lorentz Transformation from the Two Postulates

Statement

For any two events in flat spacetime, the quantity \( s^2 = -c^2\,\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 \) takes the same numerical value in every inertial frame related by a Lorentz transformation (boost, rotation, or any composition thereof). This frame-independent \( s^2 \) is the spacetime interval, and its sign partitions event pairs into timelike (\(s^2<0\)), null (\(s^2=0\)), and spacelike (\(s^2>0\)) classes that fix the causal structure of spacetime.

Why it matters

The interval is the metric of special relativity: it is to spacetime what the Pythagorean distance is to Euclidean space. Once we know that observers disagree about elapsed time \( \Delta t \) and about spatial separation \( \Delta \mathbf{r} \), the interval is the object they agree on. Every Lorentz-invariant physical statement — proper time along a worldline, the mass shell \( E^2 = (pc)^2 + (mc^2)^2 \), the phase of a wave — is ultimately built from this one invariant.

Its sign, not merely its magnitude, is invariant. That is what makes causality objective: if event \(A\) can send a signal to event \(B\) in one frame, it can in every frame, because "\(A\) precedes \(B\) within the light cone" is a statement about the invariant interval, not about any single observer's clock.

Assumptions

Spacetime is flat (special-relativistic), so coordinates are global and the transformation between inertial frames is the Lorentz transformation derived from the postulates.In curved spacetime there is no global linear Lorentz map; only the infinitesimal interval \( ds^2 = g_{\mu\nu}\,dx^\mu dx^\nu \) is invariant, and even then under general coordinate changes rather than global Lorentz transformations.

The two frames \(S\) and \(S'\) are inertial and in standard configuration: \(S'\) moves at constant velocity \( v \) along the common \(x\)-axis, axes parallel, origins coincident at \( t=t'=0 \).Drop constant velocity and the map acquires acceleration-dependent terms (Rindler/Fermi coordinates); the global interval formula no longer holds between the frames.

The speed of light \( c \) has the same value in both frames — the second postulate — which is what fixes the specific coefficients of the Lorentz transformation.If \(c\) transformed like a Newtonian velocity, the correct invariant would be the separate Newtonian quantities \( \Delta t \) and \( \Delta \mathbf{r}^2 \), and \( s^2 \) would not be preserved.

We work with coordinate differences \( \Delta x^\mu \) between two events; because the Lorentz transformation is linear and homogeneous, differences transform exactly as coordinates.For an inhomogeneous (affine) map the constant translation cancels in differences, so this is safe; but a genuinely non-linear map would spoil the quadratic-form invariance shown below.

Derivation

\[ \Delta t' = \gamma\left(\Delta t - \frac{v\,\Delta x}{c^2}\right),\quad \Delta x' = \gamma\left(\Delta x - v\,\Delta t\right),\quad \Delta y' = \Delta y,\quad \Delta z' = \Delta z \]

Assumed prior result: the Lorentz boost from the postulates, with \( \gamma = \left(1 - v^2/c^2\right)^{-1/2} \). Linearity lets us apply it to coordinate differences. A

\[ s'^2 = -c^2\,\Delta t'^2 + \Delta x'^2 + \Delta y'^2 + \Delta z'^2 \]

Definition of the interval in the primed frame — the object we must show equals \( s^2 \). A

\[ \Delta y'^2 + \Delta z'^2 = \Delta y^2 + \Delta z^2 \]

Transverse coordinates are untouched by a boost along \(x\); set them aside and focus on the longitudinal pair. A

\[ -c^2\,\Delta t'^2 = -c^2\gamma^2\left(\Delta t - \frac{v\,\Delta x}{c^2}\right)^2 = -\gamma^2\left(c^2\,\Delta t^2 - 2\,v\,\Delta t\,\Delta x + \frac{v^2}{c^2}\,\Delta x^2\right) \]

Substitute \( \Delta t' \) and expand the square, distributing the \( -c^2 \). Symbols only. A

\[ \Delta x'^2 = \gamma^2\left(\Delta x - v\,\Delta t\right)^2 = \gamma^2\left(\Delta x^2 - 2\,v\,\Delta t\,\Delta x + v^2\,\Delta t^2\right) \]

Substitute \( \Delta x' \) and expand. Note the cross term \( -2v\,\Delta t\,\Delta x \) matches the one in Step 4 up to sign. A

\[ -c^2\,\Delta t'^2 + \Delta x'^2 = \gamma^2\Big[\,-c^2\,\Delta t^2 + 2v\,\Delta t\,\Delta x - \tfrac{v^2}{c^2}\Delta x^2 + \Delta x^2 - 2v\,\Delta t\,\Delta x + v^2\,\Delta t^2\,\Big] \]

Add Steps 4 and 5. The cross terms \( +2v\,\Delta t\,\Delta x \) and \( -2v\,\Delta t\,\Delta x \) cancel exactly — this cancellation is the heart of the invariance. B

\[ = \gamma^2\Big[\,-c^2\,\Delta t^2\left(1 - \frac{v^2}{c^2}\right) + \Delta x^2\left(1 - \frac{v^2}{c^2}\right)\,\Big] \]

Group the surviving \( \Delta t^2 \) terms (\(-c^2\Delta t^2 + v^2\Delta t^2\)) and the surviving \( \Delta x^2 \) terms (\(\Delta x^2 - \tfrac{v^2}{c^2}\Delta x^2\)), factoring the common \( \left(1 - v^2/c^2\right) \) from each. B

\[ \gamma^2\left(1 - \frac{v^2}{c^2}\right) = 1 \quad\Longrightarrow\quad -c^2\,\Delta t'^2 + \Delta x'^2 = -c^2\,\Delta t^2 + \Delta x^2 \]

By the definition of \( \gamma \), the prefactor collapses to unity. This single identity is what forces the longitudinal part of the interval to be preserved. B

\[ s'^2 = -c^2\,\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 = s^2 \]

Add the untouched transverse terms (Step 3) back to Step 8. Invariance under a single standard boost is established. A

\[ \Lambda^{\mathsf{T}}\,\eta\,\Lambda = \eta,\qquad \eta = \mathrm{diag}(-1,+1,+1,+1) \]

A boost in any direction is a rotation of the standard boost, and spatial rotations manifestly preserve \( \Delta x^2+\Delta y^2+\Delta z^2 \) and leave \( \Delta t \) alone. Any Lorentz transformation \( \Lambda \) is a composition of boosts and rotations, so it satisfies this metric-preservation condition; hence \( s'^2 = \Delta x'^{\mathsf T}\eta\,\Delta x' = \Delta x^{\mathsf T}\Lambda^{\mathsf T}\eta\,\Lambda\,\Delta x = \Delta x^{\mathsf T}\eta\,\Delta x = s^2 \) for the full Lorentz group. C

Result

\[ \boxed{\,s^2 = -c^2\,\Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 \;=\; -c^2\,\Delta t'^2 + \Delta x'^2 + \Delta y'^2 + \Delta z'^2\,}\]

Reading. The interval between two events is a single number that all inertial observers compute to be the same, even though each measures different \( \Delta t \) and different \( \Delta \mathbf{r} \). Its sign is invariant too: \( s^2<0 \) (timelike) means one event lies inside the other's light cone and the two can be causally connected by a sub-light signal; \( s^2=0 \) (null) means they are joined by a light ray; \( s^2>0 \) (spacelike) means no signal can connect them and their time-order is frame-dependent.

Units check. Every term is a squared length: \( c^2\,\Delta t^2 \) has units \( (\mathrm{m/s})^2\,(\mathrm{s})^2 = \mathrm{m}^2 \), and \( \Delta x^2,\Delta y^2,\Delta z^2 \) are already \( \mathrm{m}^2 \). So \( s^2 \) carries units \( \mathrm{m}^2 \), consistent across the equation, while \( \gamma \) and \( v^2/c^2 \) are dimensionless. (In the \( +{-}{-}{-} \) convention one writes \( s^2 = c^2\,\Delta t^2 - \Delta\mathbf r^2 \); the physics is identical, only the overall sign flips.)

Limiting cases

Low speed, \( v \ll c \): \( \gamma \to 1 \), \( \Delta t' \to \Delta t \), \( \Delta x' \to \Delta x - v\,\Delta t \). Time becomes absolute and the interval's content reduces to the Galilean-invariant pieces \( \Delta t \) and \( \Delta\mathbf r^2 \) separately.
Ultrarelativistic, \( v \to c \): \( \gamma \to \infty \) and individual coordinates blow up, yet the product \( \gamma^2(1-v^2/c^2) \) stays exactly \(1\), so invariance survives the limit.
Null separation, \( s^2 = 0 \): \( |\Delta\mathbf r| = c\,|\Delta t| \) in every frame; the light cone is Lorentz-invariant, which is the second postulate re-expressed geometrically.
Pure time separation (\( \Delta\mathbf r = 0 \)): \( s^2 = -c^2\,\Delta\tau^2 \), so \( \Delta\tau = |\Delta t| \) is the proper time read by a clock present at both events.
Pure space separation (\( \Delta t = 0 \)): \( s^2 = \Delta\ell^2 \), the invariant proper length between simultaneous events.

Breaks when

Gravity / curved spacetime. With a non-trivial metric \( g_{\mu\nu}\neq\eta_{\mu\nu} \), only the infinitesimal interval \( ds^2 = g_{\mu\nu}\,dx^\mu dx^\nu \) is meaningful; the finite formula \( -c^2\,\Delta t^2 + \Delta\mathbf r^2 \) is coordinate-dependent and not invariant between distant events.
Non-inertial (accelerating or rotating) frames. The map connecting an inertial frame to an accelerating one is not a Lorentz transformation, so the constant-coefficient quadratic form is not preserved; one must integrate \( d\tau \) along the actual worldline.
Non-Lorentz linear maps. A shear or dilation of coordinates fails \( \Lambda^{\mathsf T}\eta\,\Lambda=\eta \) and changes \( s^2 \); only the Lorentz group leaves it invariant.
Deformed / Lorentz-violating kinematics. In theories with a modified dispersion relation (some quantum-gravity phenomenology), the transformations are no longer Lorentz and \( s^2 \) is not conserved.

Failure modes

All-plus metric. Writing \( s^2 = c^2\,\Delta t^2 + \Delta\mathbf r^2 \). This is the four-dimensional Euclidean distance and is not Lorentz-invariant; the relative minus sign is the entire content of the result.
Dropping the cross-term cancellation. Expanding \( \Delta t'^2 \) and \( \Delta x'^2 \) but mis-signing or omitting the \( 2v\,\Delta t\,\Delta x \) terms; the cancellation in Step 6 then fails and the interval appears not to be invariant.
Confusing coordinate time with proper time. Treating \( \Delta t \) in one frame as \( \Delta\tau \). Proper time is \( \Delta\tau = \sqrt{-s^2}/c \) and equals coordinate time only in the clock's rest frame.
Wrong power in \( \gamma \). Using \( \gamma = (1-v^2/c^2)^{+1/2} \) instead of the \( -1/2 \) power breaks Step 8 and the whole identity.
Believing the sign can flip. Assuming a boost can turn a timelike pair spacelike. The sign of \( s^2 \) is invariant; only the ordering of spacelike-separated events can be reversed.
Applying the finite formula across a gravitational field, where only the integrated \( \int ds \) is meaningful.

Discussion

The invariance of \( s^2 \) is the geometric restatement of the two postulates. The relativity principle demands that the transformation between inertial frames form a group preserving some quadratic form; the constancy of \( c \) fixes that form to be \( \eta = \mathrm{diag}(-1,+1,+1,+1) \) rather than the Euclidean \( \delta_{ij} \). Minkowski's insight was that this makes spacetime a genuine geometry — with the interval as its "distance" — in which Lorentz transformations are precisely the isometries, the analogue of rotations and translations in Euclidean space.

Because the interval, not time or length separately, is the invariant, all the famous relativistic effects are shadows of one geometric fact. Time dilation is the statement that a straight (inertial) worldline maximizes proper time \( \Delta\tau=\sqrt{-s^2}/c \) between two timelike-separated events; length contraction is the interval read on the spacelike slice of simultaneity. The twin "paradox" is simply that two worldlines between the same events have different \( \int d\tau \), exactly as two paths between two points in the plane have different lengths.

Formally, the Lorentz group is defined as the set of linear maps \( \Lambda \) satisfying \( \Lambda^{\mathsf T}\eta\,\Lambda = \eta \). The Step-8 identity \( \gamma^2(1-\beta^2)=1 \) is precisely the boost's instance of this condition, and the hyperbolic-rotation view — \( x' = x\cosh\varphi - ct\sinh\varphi \), \( ct' = ct\cosh\varphi - x\sinh\varphi \) with \( \tanh\varphi = v/c \) — makes invariance manifest as the Minkowski analogue of \( \cos^2+\sin^2=1 \), here \( \cosh^2\varphi - \sinh^2\varphi = 1 \). Continuity from the identity singles out the proper orthochronous subgroup \( \mathrm{SO}^+(1,3) \), double-covered by \( \mathrm{SL}(2,\mathbb C) \) — the origin of spinors. The invariant \( \eta \) is what raises and lowers indices, so every Lorentz scalar (\( a_\mu b^\mu \), the d'Alembertian \( \partial_\mu\partial^\mu \), the phase \( k_\mu x^\mu \)) inherits its invariance directly from the invariance of the interval; in general relativity this local Minkowski structure survives as the tangent-space metric.

Common misconceptions. A frequent error is to think the interval says "everyone measures the same thing." They do not: observers genuinely disagree about durations and lengths. What they agree on is the specific combination \( -c^2\,\Delta t^2 + \Delta\mathbf r^2 \). A second misconception is that a negative \( s^2 \) is unphysical or that \( \sqrt{s^2} \) must be a real spacetime length; for timelike pairs it is imaginary, which is exactly why one speaks there of proper time \( c\,\Delta\tau = \sqrt{-s^2} \), not proper length.

Worked examples

A muon is created and decays. In the lab the two events are separated by \( \Delta t = 6.0\ \mu\mathrm{s} \) and \( \Delta x = 1.6\ \mathrm{km} \) (with \( \Delta y=\Delta z=0 \)). Find the invariant interval and the muon's proper lifetime.

Set up symbolically first. A

\[ s^2 = -c^2\,\Delta t^2 + \Delta x^2 \]

Interval definition; transverse terms vanish. A

\[ c\,\Delta t = (2.998\times10^8\ \mathrm{m/s})(6.0\times10^{-6}\ \mathrm{s}) = 1.799\times10^3\ \mathrm{m} \]

Convert the time term to a length so both terms share units \( \mathrm m \). A

\[ s^2 = -(1799\ \mathrm{m})^2 + (1600\ \mathrm{m})^2 = -3.236\times10^6 + 2.560\times10^6 = -6.76\times10^5\ \mathrm{m}^2 \]

Insert numbers. \( s^2<0 \): the pair is timelike, as it must be for a single particle present at both events. A

\[ \Delta\tau = \frac{\sqrt{-s^2}}{c} = \frac{\sqrt{6.76\times10^5}\ \mathrm{m}}{2.998\times10^8\ \mathrm{m/s}} = \frac{822\ \mathrm m}{2.998\times10^8\ \mathrm{m/s}} \]

Proper time from the invariant; this is what a comoving clock reads. B

\[ s^2 = -6.76\times10^{5}\ \mathrm{m}^2,\qquad \Delta\tau \approx 2.74\ \mu\mathrm{s} \]

Reading. Though the lab clock ticks \(6.0\ \mu\mathrm s\), the muon ages only \(2.74\ \mu\mathrm s\) — the invariant interval delivers the proper lifetime directly, with no need to find \( v \) or \( \gamma \) first.

Units check. \( s^2 \) in \( \mathrm m^2 \); \( \sqrt{-s^2}/c \) gives \( \mathrm m /(\mathrm{m/s}) = \mathrm s \). Consistent.

Verify invariance numerically. Two events have \( (c\,\Delta t,\ \Delta x)=(5.0,\ 3.0)\ \mathrm m \) in \(S\) (transverse parts zero). Frame \(S'\) moves at \( v = 0.60\,c \) along \(x\). Compute \( s^2 \) in both frames.

Symbolic plan: transform, then re-evaluate the interval. A

\[ s^2 = -(c\,\Delta t)^2 + \Delta x^2 = -(5.0)^2 + (3.0)^2 = -25 + 9 = -16\ \mathrm{m}^2 \]

Interval in \(S\). Timelike. A

\[ \gamma = \frac{1}{\sqrt{1-0.60^2}} = \frac{1}{\sqrt{0.64}} = 1.25,\qquad \beta=0.60 \]

Lorentz factor for the boost. A

\[ c\,\Delta t' = \gamma\left(c\,\Delta t - \beta\,\Delta x\right) = 1.25\,(5.0 - 0.60\cdot 3.0) = 1.25\,(3.2) = 4.0\ \mathrm m \]

Boost of the time coordinate (written with \( \beta \) so both entries are lengths). B

\[ \Delta x' = \gamma\left(\Delta x - \beta\,c\,\Delta t\right) = 1.25\,(3.0 - 0.60\cdot 5.0) = 1.25\,(0.0) = 0.0\ \mathrm m \]

Boost of the space coordinate. Here \(S'\) is the frame in which the two events occur at the same place. B

\[ s'^2 = -(c\,\Delta t')^2 + \Delta x'^2 = -(4.0)^2 + (0.0)^2 = -16\ \mathrm{m}^2 \]

Re-evaluate the interval in \(S'\). A

\[ s^2 = s'^2 = -16\ \mathrm{m}^2\quad\Longrightarrow\quad c\,\Delta\tau = \sqrt{16}=4.0\ \mathrm m \]

Reading. The two frames disagree on both coordinates (\( c\,\Delta t:5.0\to4.0 \), \( \Delta x:3.0\to0.0 \)) yet agree exactly on \( s^2=-16\ \mathrm m^2 \). \(S'\) is the rest frame, where \( \Delta x'=0 \) and \( c\,\Delta t'=\sqrt{-s^2} \) equals the proper separation.

Units check. Every quantity is in metres; \( s^2 \) in \( \mathrm m^2 \). The invariance is exact, not approximate.

Problems

Two events in frame \(S\) have \( (c\,\Delta t,\ \Delta x,\ \Delta y,\ \Delta z) = (10,\ 6,\ 0,\ 0)\ \mathrm m \). Classify the separation and compute \( s^2 \).
Solution
\( s^2 = -(10)^2 + (6)^2 = -100+36 = -64\ \mathrm m^2 \). Since \( s^2<0 \) the separation is timelike. Proper time: \( c\,\Delta\tau=\sqrt{64}=8.0\ \mathrm m \), i.e. \( \Delta\tau = 8.0/(2.998\times10^8) = 2.67\times10^{-8}\ \mathrm s \).
For the events of Problem 1, a frame \(S'\) moves at \( v=0.80\,c \) along \(x\). Compute \( c\,\Delta t' \) and \( \Delta x' \), then verify \( s'^2 = s^2 \).
Solution
\( \gamma = 1/\sqrt{1-0.64} = 1/0.6 = 1.667 \). \( c\,\Delta t' = 1.667(10 - 0.80\cdot6) = 1.667(5.2) = 8.667\ \mathrm m \). \( \Delta x' = 1.667(6 - 0.80\cdot10) = 1.667(-2.0) = -3.333\ \mathrm m \). Then \( s'^2 = -(8.667)^2 + (-3.333)^2 = -75.11 + 11.11 = -64.0\ \mathrm m^2 = s^2 \). Invariance confirmed.
Two flashes occur at \( (c\,\Delta t,\ \Delta x)=(4,\ 4)\ \mathrm m \). Show they are null and give a frame-independent statement about them.
Solution
\( s^2 = -(4)^2 + (4)^2 = 0 \): null. Because \( |\Delta x| = c\,\Delta t \), the two events are connected by a light ray moving at \(c\); by the invariance of \(s^2=0\) this is true in every inertial frame. No massive signal can connect them (that would require timelike separation), and no frame can make them simultaneous or co-located.
A spacelike pair has \( (c\,\Delta t,\ \Delta x)=(3,\ 5)\ \mathrm m \). Find the frame in which the two events are simultaneous, and the proper length between them.
Solution
\( s^2 = -(3)^2+(5)^2 = -9+25 = +16\ \mathrm m^2 \) (spacelike). Simultaneity requires \( c\,\Delta t' = \gamma(c\,\Delta t - \beta\,\Delta x)=0 \Rightarrow \beta = c\,\Delta t/\Delta x = 3/5 = 0.60 \), so \( v=0.60\,c \). Then \( \gamma = 1.25 \) and \( \Delta x' = 1.25(5 - 0.60\cdot3) = 1.25(3.2) = 4.0\ \mathrm m \). The proper length is \( \sqrt{s^2}=\sqrt{16}=4.0\ \mathrm m \), matching \( \Delta x' \) as it must. The time-order here is frame-dependent, consistent with no causal link.
Using the rapidity form \( x' = x\cosh\varphi - ct\sinh\varphi \), \( ct' = ct\cosh\varphi - x\sinh\varphi \), prove directly that \( -c^2 t'^2 + x'^2 = -c^2 t^2 + x^2 \), and derive the infinitesimal interval's role in proper time along an arbitrary worldline \( \mathbf r(t) \) of speed \( u(t) \).
Solution
Expand: \( x'^2 = x^2\cosh^2\varphi - 2x(ct)\sinh\varphi\cosh\varphi + c^2t^2\sinh^2\varphi \) and \( c^2t'^2 = c^2t^2\cosh^2\varphi - 2x(ct)\sinh\varphi\cosh\varphi + x^2\sinh^2\varphi \). Subtract: the cross terms cancel and \( -c^2t'^2 + x'^2 = x^2(\cosh^2\varphi-\sinh^2\varphi) - c^2t^2(\cosh^2\varphi-\sinh^2\varphi) = x^2 - c^2t^2 \) using \( \cosh^2\varphi-\sinh^2\varphi=1 \). For differentials, \( ds^2 = -c^2\,dt^2 + d\mathbf r^2 \), invariant by the same algebra. Along a worldline \( d\mathbf r = \mathbf u\,dt \), so \( ds^2 = -c^2\,dt^2(1 - u^2/c^2) \) and \( d\tau = \sqrt{-ds^2}/c = dt\,\sqrt{1-u^2/c^2} = dt/\gamma(t) \). Hence \( \Delta\tau = \int \sqrt{1-u^2(t)/c^2}\;dt \); for a straight inertial path \( \Delta\tau=\Delta t/\gamma \), and any accelerated path between the same endpoints has smaller \( \Delta\tau \) — the twin-paradox result, showing the inertial worldline maximizes proper time.