Time Dilation

Derivation

Time Dilation

D-093 Home PU-105 Threads light · symmetry Depends on Lorentz Transformation from the Two Postulates, Invariance of the Spacetime Interval

Statement

Let a single ideal clock be at rest at a fixed spatial point in an inertial frame \(S'\) that moves with constant velocity \(v\) along the common \(x\)-axis of a second inertial frame \(S\) (standard configuration). If the clock advances by a proper-time interval \(\Delta\tau\) between two of its own ticks — two events that occur at the same place in \(S'\), so \(\Delta x' = 0\) — then the coordinate-time interval \(\Delta t\) assigned to those same two events in \(S\) is \(\Delta t = \gamma\,\Delta\tau\), with \(\gamma = \left(1 - v^2/c^2\right)^{-1/2} \ge 1\). Since \(\gamma \ge 1\), the moving clock runs slow: it is observed from \(S\) to advance by only \(1/\gamma\) seconds for each second of \(S\)-time.

Why it matters

Time dilation is the first and most direct physical consequence of the two postulates: it shows that the rate of an ideal clock is not absolute but depends on its state of motion relative to the observer, dissolving the Newtonian notion of one universal time flowing identically for everyone. It fixes the operational meaning of proper time \(\tau\) as the invariant a clock actually accumulates along its worldline.

It is verified continuously, not merely in thought experiments: atmospheric muons reach the ground only because their decay clocks run slow in the Earth frame; atomic clocks flown around the world return with the predicted offset; and satellite navigation must correct for it, together with the gravitational shift, to remain accurate to metres rather than kilometres. Every relativistic dynamical quantity built on \(d\tau\) — four-velocity, four-momentum, proper acceleration — inherits its consistency from this result.

Assumptions

The frames \(S\) and \(S'\) are inertial and in standard configuration, related by a Lorentz transformation for a boost of speed \(v\) along \(x\).Without inertial frames the linear Lorentz map does not apply; an accelerating clock requires \(d\tau\) integrated along its worldline instead, and a non-standard configuration mixes in rotations that clutter the algebra without changing the result.

The two ticks occur at the same spatial point in the clock's rest frame \(S'\), i.e. \(\Delta x' = 0\).If \(\Delta x' \neq 0\) the two events are not both on one clock's worldline, so \(\Delta t'\) is a coordinate time rather than a proper time; the mixing term \(v\,\Delta x'/c^2\) survives and \(\Delta t = \gamma\,\Delta\tau\) is replaced by the full interval relation.

An ideal clock measures proper time along its own worldline independently of its acceleration history (the "clock hypothesis").If the rate depended on acceleration, the reading would carry extra terms beyond \(\int d\tau\) and the twin bookkeeping would fail; experiment (muons at \(\sim 10^{18}\,g\) in storage rings) confirms the rate depends only on instantaneous speed.

Spacetime is flat: gravitational fields are negligible over the region and duration considered.In curved spacetime the metric coefficients vary and a gravitational term adds to the rate difference; the velocity-only factor \(\gamma\) is then just the local, special-relativistic part of the total redshift.

Derivation

Route 1 — from the Lorentz transformation. Use the inverse boost expressing \(S\)-time in terms of \(S'\)-coordinates.

\[ t = \gamma\left(t' + \frac{v\,x'}{c^2}\right), \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \]

Prior result (Lorentz transformation from the postulates): the inverse boost, obtained by swapping primed/unprimed and \(v \to -v\). A

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

The transformation is linear with constant coefficients, so it applies verbatim to the coordinate differences between the two tick-events. A

\[ \Delta x' = 0 \quad\Longrightarrow\quad \Delta t = \gamma\,\Delta t' \]

Impose the assumption that both ticks occur at the same place in \(S'\); the spatial mixing term vanishes identically. A

\[ \Delta t' \equiv \Delta\tau \quad\Longrightarrow\quad \boxed{\,\Delta t = \gamma\,\Delta\tau\,} \]

The interval read on the co-moving clock, at fixed \(x'\), is by definition the proper time between the events. A

Route 2 — from invariance of the spacetime interval. A coordinate-free derivation using only the invariant.

\[ c^2\,d\tau^2 = c^2\,dt^2 - dx^2 - dy^2 - dz^2 \]

Prior result (invariance of the interval): definition of proper time \(d\tau\) along a timelike worldline; the right side has the same value in every inertial frame. B

\[ d\tau^2 = dt^2\left[\,1 - \frac{1}{c^2}\left(\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2\right)\right] \]

Divide through by \(c^2\) and factor out \(dt^2\), introducing the coordinate-velocity components \(dx^i/dt\). B

\[ d\tau = dt\,\sqrt{1 - \frac{u^2}{c^2}} = \frac{dt}{\gamma} \]

Identify the clock's speed in \(S\) as \(u^2 = (dx/dt)^2 + \dots\); for the clock at rest in \(S'\) this is \(u = v\), giving the Lorentz factor. Take the positive (future-directed) root. B

\[ \Delta\tau = \int \frac{dt}{\gamma} = \frac{\Delta t}{\gamma} \quad\Longrightarrow\quad \Delta t = \gamma\,\Delta\tau \]

Integrate over the tick interval; \(v\) (hence \(\gamma\)) is constant for uniform motion, so it comes out of the integral. Manifestly coordinate-independent. B

Route 3 — the light clock. A concrete mechanism making the factor visible geometrically, using only the constancy of \(c\).

\[ \Delta\tau = \frac{2L}{c} \]

In \(S'\) a pulse bounces straight up and back between mirrors a distance \(L\) apart (aligned perpendicular to the motion); one round trip defines a tick. A

\[ \left(\frac{c\,\Delta t}{2}\right)^2 = L^2 + \left(\frac{v\,\Delta t}{2}\right)^2 \]

In \(S\) the mirrors drift sideways by \(v\,\Delta t\) per round trip, so the pulse follows the hypotenuse; its speed is still \(c\) (second postulate) and the transverse gap \(L\) is unchanged (lengths perpendicular to a boost are invariant). Pythagoras on the half-trip. B

\[ c^2\,\Delta t^2 - v^2\,\Delta t^2 = 4L^2 \quad\Longrightarrow\quad \Delta t = \frac{2L}{c\,\sqrt{1 - v^2/c^2}} \]

Collect the \(\Delta t^2\) terms, factor out \(c^2\), and solve for \(\Delta t\) (positive root) — symbols rearranged before any numbers. A

\[ \Delta t = \gamma\,\frac{2L}{c} = \gamma\,\Delta\tau \]

Substitute the definition of \(\gamma\) and the \(S'\) tick \(2L/c = \Delta\tau\); \(L\) cancels, so the factor is universal and not a property of this particular clock. All three routes agree. A

Result

\[ \Delta t = \gamma\,\Delta\tau, \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} \ge 1 \]

Reading. The proper time \(\Delta\tau\) is the interval measured by the single clock present at both events (both ticks happen where it sits). Any frame in which that clock moves assigns a longer coordinate time \(\Delta t = \gamma\,\Delta\tau\) between the same two events, because \(\gamma \ge 1\). Equivalently, watched from \(S\) the moving clock ticks slow, advancing by only \(1/\gamma\) seconds per second of \(S\)-time. Proper time is the smallest of all the coordinate times: the straight (inertial) worldline between two timelike-separated events is the longest in elapsed proper time. The effect is reciprocal — each inertial observer finds the other's clock slow — because "the same two events" differ between the two comparisons.

Units check. \(v^2/c^2\) is \((\mathrm{m\,s^{-1}})^2/(\mathrm{m\,s^{-1}})^2\), dimensionless, so \(\gamma\) is a pure number and the root is real for \(v < c\). Then \([\Delta t] = [\gamma][\Delta\tau] = \mathrm{s}\), consistent. In the light-clock form, \(2L/c\) has units \(\mathrm{m}/(\mathrm{m\,s^{-1}}) = \mathrm{s}\).

Limiting cases

\(v \to 0\): \(\gamma \to 1\) and \(\Delta t \to \Delta\tau\); coordinate and proper time coincide and Newtonian absolute time is recovered.
\(v \ll c\): \(\gamma \approx 1 + \tfrac{1}{2}v^2/c^2\), so the fractional slowing is \(\tfrac{1}{2}v^2/c^2\) — second order in \(v/c\), which is why it is invisible in everyday life.
\(v \to c\): \(\gamma \to \infty\); a finite proper interval maps to unbounded coordinate time. A photon has \(d\tau = 0\): no proper time elapses along a null worldline.
Sign of \(v\): \(\gamma\) depends only on \(v^2\), so the slowing is identical for approach and recession — unlike the first-order Doppler shift (light and symmetry threads meet here).

Breaks when

The two events are not colocated in the clock's frame (\(\Delta x' \neq 0\)). Then \(\Delta t'\) is a coordinate time, not a proper time, and \(\Delta t = \gamma\,\Delta\tau\) conflates dilation with the relativity of simultaneity; the full transformation with the term \(v\,\Delta x'/c^2\) must be restored, and the naive factor can even give the wrong sign.
The clock accelerates. With non-constant \(v\), \(\gamma\) varies along the path and the elapsed time is \(\Delta\tau = \int dt/\gamma(t)\), not a single factor times \(\Delta t\). This asymmetry between the inertial and accelerating worldlines is exactly what resolves the twin paradox.
Strong gravity / curved spacetime. Near a mass the metric is not flat Minkowski; a gravitational potential term \(\Phi/c^2\) adds to the rate difference, so the velocity-only \(\gamma\) captures just part of the total time dilation (and the gravitational part has the opposite sign for a satellite above the Earth).
Superluminal or lightlike relative speed (\(v \ge c\)). The radicand \(1 - v^2/c^2\) is \(\le 0\) and \(\gamma\) is singular or imaginary; no material clock can occupy such a frame, so the formula has no physical content there.

Failure modes

Inverting the factor. Writing \(\Delta\tau = \gamma\,\Delta t\) (moving clock reads more) instead of \(\Delta t = \gamma\,\Delta\tau\). Anchor it: the single-clock proper interval is always the shorter one, so \(\gamma \ge 1\) multiplies it up to coordinate time.
Assigning proper time to the wrong frame. Proper time belongs to the clock present at both events, not to whichever frame is loosely called "stationary". Identify \(\Delta x = 0\) first; that frame's time is \(\Delta\tau\).
Mistaking dilation for a signal-delay illusion. Time dilation is not the Doppler/light-travel effect; it survives after correcting for propagation delay and is symmetric, whereas the first-order Doppler shift is not.
Treating reciprocity as a contradiction. Concluding "each sees the other slow, so it is inconsistent." The two statements compare different event pairs and different simultaneity slices; both are true.
Mixing frames in the muon problem. Using the dilated lifetime together with the rest-frame distance (or vice versa) — this double-counts or omits a factor of \(\gamma\). Length contraction and time dilation must be applied within one consistent frame.
Inconsistent units in \(v/c\). Plugging \(v\) in \(\mathrm{km\,h^{-1}}\) next to \(c\) in \(\mathrm{m\,s^{-1}}\); the ratio must use one unit system or \(\gamma\) is meaningless.

Discussion

The deepest content of \(\Delta t = \gamma\,\Delta\tau\) is geometric, not mechanical: nothing has physically slowed inside the moving clock. Proper time is the "length" of a worldline in Minkowski spacetime, measured with the indefinite metric \(ds^2 = c^2\,dt^2 - dx^2\). Because of the minus sign, a worldline that detours through space accumulates less proper time than the straight inertial worldline joining the same endpoints — the reverse of the Euclidean triangle inequality. Time dilation is simply the statement that the inertial path is the longest in proper time, and \(\gamma\) is the ratio of coordinate length to proper length.

Reciprocity, which troubles most newcomers, is a direct expression of the relativity of simultaneity (the light and symmetry threads meet here). When \(S\) times the moving clock, it compares one event on that clock's worldline against two of its own synchronized clocks at different places; when \(S'\) does the reverse, it uses a different pair of simultaneous readings, because the two frames disagree about which distant events are simultaneous. Each frame is correct within its own slicing of spacetime, and \(\gamma\) is identical both ways precisely because no inertial frame is preferred.

Time dilation is inseparable from its partners. The same \(\gamma\) that stretches time contracts lengths along the motion, \(\Delta x = \Delta x_0/\gamma\), and the two together keep \(c^2\,\Delta\tau^2\) invariant. In the muon example the effect reads either as a dilated lifetime (Earth frame) or a contracted atmosphere (muon frame); both give the identical survival fraction, which is the invariance made physical. The light-clock route further shows the factor is universal: if two colocated clocks agreed at rest but disagreed in motion, an observer could detect absolute motion, violating the first postulate — so every clock must dilate by the same \(\gamma\).

At the level of the metric, proper time is the parameter along a timelike geodesic, and \(d\tau = dt/\gamma\) is the flat-space limit of \(d\tau = \sqrt{-g_{\mu\nu}\,dx^\mu dx^\nu}/c\). In general relativity the components \(g_{\mu\nu}\) carry the gravitational potential, so a static observer at potential \(\Phi\) sees \(d\tau \approx dt\,(1 + \Phi/c^2)\) in addition to the velocity term. GPS satellites need both: the gravitational (blueshift) piece dominates and the velocity (redshift) piece partly cancels it, netting about \(+38\ \mu\mathrm{s}\) per day. This is the clearest statement that special-relativistic time dilation is the local, weak-field, velocity-only corner of a single spacetime-geometric law.

Common misconceptions. Time dilation is not caused by clocks physically malfunctioning under stress, is not the light-travel (Doppler) delay, and does not single out an absolute "moving" clock — it is fully reciprocal for inertial observers. The asymmetry that resolves the twin paradox comes only from one twin's acceleration, i.e. a genuine change of inertial frame, not from time dilation by itself.

Worked examples

Example 1 — Atmospheric muons reaching the ground. A muon has proper mean lifetime \(\tau_0 = 2.20\ \mu\mathrm{s}\) and travels at \(v = 0.998\,c\). How far does it travel, in the Earth frame, in one mean lifetime?

\[ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}, \qquad \Delta t = \gamma\,\tau_0, \qquad d = v\,\Delta t \]

Set up symbolically: the Earth-frame lifetime is the dilated interval, and distance is speed times Earth-frame time. A

\[ \gamma = \frac{1}{\sqrt{1 - (0.998)^2}} = \frac{1}{\sqrt{1 - 0.996004}} = \frac{1}{\sqrt{0.003996}} \approx 15.82 \]

Insert \(v/c = 0.998\) only now; the small radicand gives a large factor. A

\[ \Delta t = \gamma\,\tau_0 = 15.82 \times 2.20\ \mu\mathrm{s} \approx 34.8\ \mu\mathrm{s} \]

Dilate the proper lifetime into the Earth frame. A

\[ d = v\,\Delta t = (0.998)(3.00\times10^{8}\ \mathrm{m\,s^{-1}})(34.8\times10^{-6}\ \mathrm{s}) \]

Use the dilated (Earth-frame) time, not the proper time, to stay within one frame. A

\[ d \approx 1.04\times10^{4}\ \mathrm{m} \approx 10.4\ \mathrm{km} \]

Reading. Without dilation the muon would cover only \(v\tau_0 \approx 0.66\ \mathrm{km}\) before decaying and almost none would reach sea level; dilation extends its range by \(\gamma \approx 16\), explaining the observed ground-level flux. In the muon's own frame the same conclusion follows from the atmosphere being length-contracted to \(\approx 0.66\ \mathrm{km}\).

Example 2 — A fast interstellar probe. A probe flies to a star \(4.37\ \mathrm{ly}\) away (Earth frame) at \(v = 0.90\,c\). How much time elapses on Earth, and on the probe's onboard clock?

\[ \Delta t = \frac{d}{v}, \qquad \Delta\tau = \frac{\Delta t}{\gamma} \]

Earth clocks time the launch and arrival events (distance over speed); the probe is present at both, so its clock reads the proper time. A

\[ \Delta t = \frac{4.37\ \mathrm{ly}}{0.90\,c} = \frac{4.37}{0.90}\ \mathrm{yr} \approx 4.86\ \mathrm{yr} \]

A light-year divided by a speed in units of \(c\) gives years directly. A

\[ \gamma = \frac{1}{\sqrt{1 - (0.90)^2}} = \frac{1}{\sqrt{0.19}} \approx 2.29 \]

Lorentz factor at \(v = 0.90\,c\). A

\[ \Delta\tau = \frac{\Delta t}{\gamma} = \frac{4.86\ \mathrm{yr}}{2.29} \approx 2.12\ \mathrm{yr} \]

The onboard clock reads the proper time, shorter than Earth time by \(1/\gamma\). A

\[ \Delta t \approx 4.86\ \mathrm{yr}\ \text{(Earth)}, \qquad \Delta\tau \approx 2.12\ \mathrm{yr}\ \text{(probe)} \]

Reading. The crew ages about \(2.12\ \mathrm{yr}\) while \(4.86\ \mathrm{yr}\) pass on Earth. Consistency via length contraction: the probe sees the distance contracted to \(4.37/2.29 \approx 1.91\ \mathrm{ly}\), covered at \(0.90\,c\) in \(1.91/0.90 \approx 2.12\ \mathrm{yr}\) — the same proper time.

Problems

(Warm-up.) A spaceship passes Earth at \(v = 0.60\,c\). Its onboard clock records \(10.0\ \mathrm{s}\) between two events on board. How long is that interval in the Earth frame?
Solution
\(\gamma = 1/\sqrt{1 - 0.36} = 1/\sqrt{0.64} = 1/0.80 = 1.25\). The onboard interval is proper time, \(\Delta\tau = 10.0\ \mathrm{s}\), so \(\Delta t = \gamma\,\Delta\tau = 1.25 \times 10.0\ \mathrm{s} = \mathbf{12.5\ s}\).
(Inverse.) An unstable particle has a measured laboratory lifetime of \(6.0\ \mu\mathrm{s}\), while its proper lifetime is \(2.0\ \mu\mathrm{s}\). Find its speed.
Solution
\(\gamma = \Delta t/\Delta\tau = 6.0/2.0 = 3.0\). From \(\gamma = 1/\sqrt{1 - v^2/c^2}\): \(1 - v^2/c^2 = 1/\gamma^2 = 1/9\), so \(v/c = \sqrt{8/9} = 0.943\). Thus \(\mathbf{v \approx 0.943\,c = 2.83\times10^{8}\ m\,s^{-1}}\).
(Small-velocity expansion.) A jet flies at \(250\ \mathrm{m\,s^{-1}}\) for \(8.0\ \mathrm{hours}\) of onboard (proper) time. Using \(\gamma \approx 1 + \tfrac{1}{2}v^2/c^2\), estimate how much less time elapses on the plane's clock than on a ground clock (ignore gravity).
Solution
\(v/c = 250/(3.00\times10^{8}) = 8.33\times10^{-7}\), so \(\tfrac{1}{2}(v/c)^2 = \tfrac{1}{2}(6.94\times10^{-13}) = 3.47\times10^{-13}\), the fractional slowing. Over \(\Delta\tau = 8.0\ \mathrm{h} = 2.88\times10^{4}\ \mathrm{s}\): \(\Delta t - \Delta\tau \approx (3.47\times10^{-13})(2.88\times10^{4}\ \mathrm{s}) \approx \mathbf{1.0\times10^{-8}\ s = 10\ ns}\). The ground clock reads about \(10\ \mathrm{ns}\) more — the velocity part of the Hafele–Keating result.
(Reciprocity / simultaneity.) Frames \(S\) and \(S'\) move at \(v = 0.80\,c\). Observer \(A\) in \(S\) says \(B\)'s clock (in \(S'\)) runs slow by \(\gamma\). Does \(B\) agree that \(B\)'s own clock is slow? Compute \(\gamma\) and explain how both can claim the other's clock is slow.
Solution
\(\gamma = 1/\sqrt{1 - 0.64} = 1/\sqrt{0.36} = 1/0.60 = \mathbf{1.667}\). \(B\) does not agree that \(B\)'s own clock is slow; by symmetry \(B\) claims \(A\)'s clock runs slow by the same \(1.667\). No contradiction: when \(A\) times \(B\)'s clock, \(A\) compares one event on \(B\)'s worldline against two \(A\)-synchronized clocks at different places, and \(B\) judges those two \(A\)-clocks to be out of sync (relativity of simultaneity). Each measurement uses a different pair of events / a different simultaneity slice, so both conclusions hold within their own frame.
(Twin paradox, quantitative.) Twin \(B\) travels to a star \(6.0\ \mathrm{ly}\) away (Earth frame) at \(0.80\,c\) and returns immediately at \(0.80\,c\). Find the elapsed time for Earth twin \(A\) and travelling twin \(B\), and state why the situation is not symmetric.
Solution
Round trip \(d_{\text{total}} = 12.0\ \mathrm{ly}\) at \(0.80\,c\): \(\Delta t_A = 12.0/0.80 = \mathbf{15.0\ yr}\). With \(\gamma = 1/\sqrt{1 - 0.64} = 1.667\), \(B\) accumulates proper time \(\Delta t/\gamma\) on each inertial leg, so \(\Delta\tau_B = 15.0/1.667 = \mathbf{9.0\ yr}\); \(B\) is \(6.0\ \mathrm{yr}\) younger on return. The situation is not symmetric because \(B\) must turn around, switching inertial frames (accelerating) at the star, while \(A\) stays in one inertial frame throughout. Only \(A\)'s worldline is straight (geodesic) between the two meeting events, and the straight timelike worldline has the greatest proper time — so \(A\) ages more.