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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 separated by coordinate differences (Δt, Δx, Δy, Δz) in an inertial frame, the quantity s2 = −c2Δt2 + Δx2 + Δy2 + Δz2 takes the same numerical value in every inertial frame related by a Lorentz transformation. Its sign partitions event pairs into timelike (s2 < 0), lightlike (s2 = 0) and spacelike (s2 > 0) classes, and this partition is therefore frame-independent.

Why it matters

The interval is the metric of Minkowski spacetime: it is to relativity what the Pythagorean distance is to Euclidean geometry. Whereas separate lengths and time intervals are frame-dependent (length contraction, time dilation), their particular combination s2 is absolute, and absolutes are what physical laws must be built from.

Because the sign of s2 is invariant, so is the answer to "can event A influence event B?". The interval is thus the geometric carrier of causality: it fixes which events lie inside a given event's light cone and which lie outside, independently of the observer.

Assumptions
The transformation between frames is the Lorentz transformation derived from the two postulates.If we used the Galilean transformation instead, t would be absolute and only the spatial distance Δx2+Δy2+Δz2 would be invariant; the mixing of space and time that makes s2 conserved would not occur.
The speed of light c is the same finite constant in all inertial frames.This is the number that appears in γ and in the coefficient of Δt2; without a single universal c there is no invariant combination and no light cone.
Spacetime is flat and the coordinates are those of a global inertial frame.If gravity curves spacetime, only the differential interval ds2 along a path is invariant and the finite algebraic form s2 = gμνΔxμΔxν with constant gμν no longer holds globally.
We consider a single boost along one axis (here x); the full result follows by composing boosts and rotations.If we forgot that spatial rotations must also preserve s2, we would have proved invariance only for a measure-zero subset of the Lorentz group and could not claim frame-independence in general.
Derivation
1
x′ = γ(x − vt),   t′ = γ(t − vx/c2),   y′ = y,   z′ = z
Standard boost of speed v along x, taken as a prior result; γ = 1/√(1 − v2/c2). We work with coordinate differences, so the same linear relations hold for (Δt, Δx, …); write t, x for the differences to keep notation light. A
2
s′2 = −c2t′2 + x′2 + y′2 + z′2
Definition of the interval evaluated in the primed frame; this is the object we must show equals s2. A
3
y′2 + z′2 = y2 + z2
The transverse coordinates are unchanged by a boost along x, so those terms carry over untouched and can be set aside. A
4
−c2t′2 + x′2 = −c2γ2(t − vx/c2)2 + γ2(x − vt)2
Substitute the boost expressions from Step 1 into the two longitudinal terms. A
5
= γ2[ −c2(t − vx/c2)2 + (x − vt)2 ]
Factor the common γ2 so the bracket can be expanded once. A
6
= γ2[ −c2t2 + 2vtx − v2x2/c2 + x2 − 2vtx + v2t2 ]
Expand both squares. The cross terms are +2vtx (from −c2·−2tvx/c2) and −2vtx (from the second square); they cancel exactly — the hallmark of the mixing that makes the interval work. B
7
= γ2[ −c2t2(1 − v2/c2) + x2(1 − v2/c2) ]
Group the surviving terms: −c2t2 and +v2t2 = −c2t2(1−v2/c2); likewise x2 and −v2x2/c2 = x2(1−v2/c2). B
8
γ2(1 − v2/c2) = 1
By the definition of γ, its square is exactly the reciprocal of (1−v2/c2), so the prefactor collapses to unity — this single identity is what forces invariance. B
9
−c2t′2 + x′2 = −c2t2 + x2
Apply Step 8 to Step 7; the longitudinal part of the interval is unchanged. A
10
s′2 = −c2t2 + x2 + y2 + z2 = s2
Add back the untouched transverse terms from Step 3. Since v was arbitrary, this holds for every boost; composing with spatial rotations (which preserve x2+y2+z2 and leave t alone) extends it to the whole Lorentz group. C
Result
s′2 = −c2t′2 + x′2 + y′2 + z′2 = −c2t2 + x2 + y2 + z2 = s2

Reading. Although each observer disagrees about elapsed time and about spatial separation, they all compute the same value for the particular combination s2. That value, and in particular its sign, is a property of the pair of events, not of the observer. Timelike pairs (s2<0) can be causally connected by a sub-light signal; lightlike pairs (s2=0) lie on each other's light cone; spacelike pairs (s2>0) cannot influence one another.

Units check. c2t2 has units (m/s)2·s2 = m2, matching x2, y2, z2 which are already m2. So s2 has units of m2 throughout and every term is dimensionally consistent; γ and v2/c2 are dimensionless.

Limiting cases
  • v → 0: γ → 1 and the boost becomes the identity, so trivially s′2 = s2 — no content, but a necessary consistency check.
  • v → c: γ → ∞, yet the product γ2(1−v2/c2) stays exactly 1, so invariance survives the ultrarelativistic limit even as individual coordinates blow up.
  • c → ∞ (Galilean limit): the −c2t2 term dominates, the cross-term mixing that cancels in Step 6 disappears, and one recovers separate invariance of time and of Euclidean distance.
  • Purely transverse separation (x = 0, motion along x): only y2+z2 contributes and it is manifestly untouched by the boost.
Breaks when
  • Curved spacetime (general relativity): with gravity present, spacetime is not globally flat. Only the infinitesimal interval ds2 = gμνdxμdxν is invariant along a worldline; the finite algebraic form −c2Δt2+Δx2+… is coordinate-dependent and generally meaningless between distant events.
  • Non-inertial (accelerating or rotating) frames: the coordinate transformation to such a frame is not a Lorentz transformation, so the constant-coefficient interval is not preserved; fictitious metric components appear and one must use the local proper-time integral instead.
  • Frames not related by an orthochronous, proper Lorentz transformation but by an arbitrary linear map (e.g. a shear or a dilation of coordinates): such maps do not satisfy ΛTηΛ = η and change s2; only the Lorentz group leaves it invariant.
Failure modes
  • Writing the interval with all plus signs, s2 = c2t2+x2+…, i.e. treating time as a fourth Euclidean axis — this quantity is NOT Lorentz invariant.
  • Sign-convention confusion: mixing the (−+++) and (+−−−) conventions within one calculation, flipping the physical meaning of "timelike" and "spacelike".
  • Forgetting the c2 on the time term and adding seconds2 to metres2 — a dimensional error that also breaks invariance.
  • Using Galilean t′ = t together with the relativistic x′, producing a hybrid transformation that preserves neither the interval nor causality.
  • Assuming that because Δt and Δx separately change, their combination must change too — missing that the changes are correlated and cancel.
  • Applying the single-boost result to two frames in a general relative-velocity direction without first rotating axes to align the boost with the relative motion.
Discussion

The invariance of s2 is the true content of special relativity. The postulates and the Lorentz transformation are the scaffolding; the payoff is that spacetime carries a metric of signature (−+++), and physics is the study of quantities that transform simply under the group leaving that metric fixed. This is why four-vectors, whose "length" is an interval-type invariant, are the natural language of the theory: energy–momentum pμ has invariant pμpμ = −m2c2, exactly analogous to s2.

Geometrically, the light cone at each event is the locus s2 = 0, and its invariance means all observers agree on it. The interior (s2<0) is the absolute future and past; the exterior (s2>0) is "elsewhere", the set of events no observer can order in time relative to the apex and none can causally link to it. Time dilation and length contraction are then not independent postulated effects but projections of a single rigid invariant onto different observers' axes — the interval stays fixed while the "shadow" it casts on the t-axis or x-axis varies.

For timelike separations the invariant defines proper time via c2τ2 = −s2, the time read by a clock carried between the two events; for spacelike separations, √s2 is the proper distance, the length measured in the frame where the events are simultaneous. Both are what they are because s2 does not depend on who measures it.

Formally, the Lorentz group is defined as exactly the set of linear transformations Λ satisfying ΛTηΛ = η, where η = diag(−1,1,1,1). Our Step-8 identity γ2(1−β2)=1 is precisely the boost's instance of this defining condition, and the hyperbolic-rotation view — x′ = x cosh φ − ct sinh φ, ct′ = ct cosh φ − x sinh φ with tanh φ = v/c — makes invariance manifest as the Minkowski analogue of cos2+sin2=1, here cosh2φ − sinh2φ = 1. The whole of special relativity is thus "the geometry of the group O(1,3)".

Common misconceptions. The interval is not a distance you could lay a ruler along in spacetime; s2 can be negative, and its square root is imaginary for timelike pairs — which is why one speaks of proper time there, not proper length. And invariance of s2 does not mean "nothing changes between frames": lengths and times genuinely differ; it is only their specific combination that is shared.

Worked examples
1
A light pulse is lightlike, in every frame.
Emission event E1 at (t1=0, x1=0); absorption E2 at (t2=10 ns, x2=3 m). Show s2=0 and that it stays 0 in a frame moving at v=0.6c. A
2
s2 = −c2Δt2 + Δx2
Only the x-direction is involved; symbols first. A
3
cΔt = (3.00×108 m/s)(10×10−9 s) = 3.00 m,   Δx = 3.00 m
Insert numbers; note the pulse covers exactly cΔt = Δx, as light must. A
4
s2 = −(3.00 m)2 + (3.00 m)2 = 0 m2
Lightlike separation. A
5
γ = 1/√(1 − 0.36) = 1/√0.64 = 1.25
Boost factor for v = 0.6c. A
6
cΔt′ = γ(cΔt − (v/c)Δx) = 1.25(3.00 − 0.6×3.00) = 1.25(1.20) = 1.50 m
Δx′ = γ(Δx − (v/c)·cΔt) = 1.25(3.00 − 0.6×3.00) = 1.50 m
Transform both events (E1 stays at the origin). B
s′2 = −(1.50 m)2 + (1.50 m)2 = 0 m2 = s2

Reading. The pulse's separation is lightlike for the moving observer too; both frames place E2 exactly on E1's light cone, as required by the constancy of c.

Units check. Every term is in m2; s2=0 is trivially invariant under scaling but here the individual coordinates changed (3.00 m → 1.50 m) while their interval did not.

1
A ticking clock: a timelike interval encodes proper time.
In frame S a clock at rest at x=0 records two ticks: E1=(0,0), E2=(5.00 s, 0). Find s2, the proper time, and verify invariance in a frame with v=0.8c. A
2
s2 = −c2Δt2 + Δx2,   Δx = 0
Events share a location in S, so only the time term survives. A
3
s2 = −(3.00×108)2(5.00)2 = −(9.00×1016)(25.0) = −2.25×1018 m2
Timelike (s2<0). A
4
τ = √(−s2)/c = √(2.25×1018)/(3.00×108) = (1.50×109)/(3.00×108) = 5.00 s
Proper time equals the elapsed time in the clock's rest frame — as it must. B
5
γ = 1/√(1 − 0.64) = 1/0.6 = 1.667
Boost factor for v = 0.8c. A
6
Δt′ = γ(Δt − vΔx/c2) = 1.667(5.00) = 8.33 s  ⇒   cΔt′ = 2.50×109 m
Δx′ = γ(Δx − vΔt) = −1.667(0.8×3.00×108)(5.00) = −2.00×109 m
The moving observer sees the clock displaced and its interval dilated to 8.33 s. B
s′2 = −(2.50×109)2 + (2.00×109)2 = (−6.25 + 4.00)×1018 = −2.25×1018 m2 = s2

Reading. The moving observer measures a longer time (8.33 s vs 5.00 s — time dilation) and a nonzero displacement, yet the interval, and hence the proper time τ = 5.00 s the clock actually reads, is unchanged.

Units check. Both terms are m2; τ = √(m2)/(m/s) = s, correctly a time.

Problems
  1. Two events have separation Δt = 2.0 s, Δx = 9.0×108 m (Δy=Δz=0). Classify the interval and compute s2.
    Solution cΔt = (3.0×108)(2.0) = 6.0×108 m. s2 = −(6.0×108)2 + (9.0×108)2 = (−3.6×1017) + (8.1×1017) = +4.5×1017 m2. Since s2>0 the interval is spacelike; the events cannot be causally connected and are simultaneous in some frame.
  2. For the spacelike pair in Problem 1, find the proper distance and the speed of the frame in which the two events are simultaneous.
    Solution Proper distance L = √s2 = √(4.5×1017) = 6.7×108 m. Simultaneity requires Δt′=0: from Δt′=γ(Δt − vΔx/c2)=0 we need v = c2Δt/Δx = (9.0×1016)(2.0)/(9.0×108) = 2.0×108 m/s = 0.667c. Check: this is sub-light, consistent with a spacelike interval.
  3. A muon is created at (0,0) and decays at (t=2.2 μs, x=0) in its rest frame. Compute s2 and the proper lifetime, then find the decay time and position seen in a lab where the muon moves at v=0.98c.
    Solution Rest frame: Δx=0, s2=−c2Δt2=−(3.0×108)2(2.2×10−6)2=−(9.0×1016)(4.84×10−12)=−4.36×105 m2. Proper lifetime τ=√(−s2)/c=2.2 μs. Lab: γ=1/√(1−0.9604)=1/√0.0396=5.03. Δt′=γΔt=5.03(2.2 μs)=11.1 μs; Δx′=γvΔt=5.03(0.98×3.0×108)(2.2×10−6)=3.25×103 m. Verify: s′2=−(c·11.1 μs)2+(3.25×103)2=−(3.33×103)2+1.06×107=−1.11×107+1.06×107≈−4.4×105 m2=s2 (rounding). The lab sees the muon live 11.1 μs and travel 3.25 km.
  4. Using the rapidity form x′=x cosh φ − ct sinh φ, ct′=ct cosh φ − x sinh φ, prove directly that −c2t′2+x′2=−c2t2+x2.
    Solution x′2 = x2cosh2φ − 2x(ct)sinh φ cosh φ + c2t2sinh2φ. c2t′2 = c2t2cosh2φ − 2x(ct)sinh φ cosh φ + x2sinh2φ. Subtract: −c2t′2+x′2 = x2(cosh2φ − sinh2φ) − c2t2(cosh2φ − sinh2φ). The cross terms cancel identically. Using cosh2φ − sinh2φ = 1 gives −c2t2+x2, as required. (With tanh φ = v/c this reproduces the standard boost.)
  5. Two events are separated by Δt=1.0×10−8 s and Δx=4.0 m. Is there a frame in which they occur at the same place? If so, find its speed; if not, explain.
    Solution cΔt=(3.0×108)(1.0×10−8)=3.0 m. s2=−(3.0)2+(4.0)2=−9.0+16.0=+7.0 m2>0: spacelike. Same place requires Δx′=γ(Δx−vΔt)=0, i.e. v=Δx/Δt=4.0/(1.0×10−8)=4.0×108 m/s=1.33c. This exceeds c and is not achievable, so NO inertial frame co-locates the events — consistent with the interval being spacelike (only spacelike pairs can be made simultaneous, never co-located; only timelike pairs can be co-located, never simultaneous).