Proper Time — Physica

Equivalent forms

\tau = \int \sqrt{1 - v^2/c^2}\, dt

c^2 d\tau^2 = -ds^2

The age a clock would read along any path — the spacetime equivalent of arc length on a curve.

Symbol	Name	SI unit	Dimension	Range
$tau$	Proper Timeoutput Time elapsed on a clock moving with the object	s	T	0 – 100
$t$	Coordinate Time Time elapsed in the lab frame	s	T	0 – 100
$v$	Speed Speed of the moving clock in lab frame	m/s	LT^-1	0 – 299792457
$c$	Speed of Light Speed of light in vacuum	m/s	LT^-1	299792458 – 299792458

Unit systems

Symbol	SI	CGS	Imperial
$tau$	s	s	s
$t$	s	s	s
$v$	m/s	cm/s	ft/s
$c$	m/s	cm/s	ft/s

Where it holds

Valid for any timelike worldline. For null paths (photons

) \tau = 0

identically. In GR generalizes to dtau^

2 = -g_\mu \nu dx^\mu dx^\nu / c^2

.

Dimensional analysis

\tau \to [T]

[T] \sqrt{1 - [LT^-1]^2 / [LT^-1]^2} = [T]

Both sides time.

\checkmark

Discovery

Hermann Minkowski · 1908

Defined by Minkowski as the invariant arc length along a timelike worldline in spacetime — the universal aging clock of any observer.

Try this

A muon survives 2.2 microseconds in its own frame. How does it reach Earth's surface from 15 km altitude?

Proper time tau is what the muon's own clock reads — independent of frame. While lab observers measure dilated time, the muon experiences only Δtau = sqrt(1 - v^2/c^2) Δt, surviving its trip from upper atmosphere.

Research status: stable

Real-world applications

GPS satellite timing
Muon storage ring decay rate predictions
Cosmological age of the universe (proper time of comoving observer)
Spacecraft mission scheduling at relativistic speeds

Common misconceptions

Proper time is not the 'real' time — it is the time read by a comoving clock
All observers agree on someone's proper time (it is invariant)
The twin paradox is resolved by integrating dtau along each worldline

Experimental verification

Muon lifetime extension in cosmic rays and at storage rings (CERN g-2: muons survive 30x longer at

\gamma = 29)

. Hafele-Keating jets carried atomic clocks; East/West flights measured proper-time differences matching predictions.

Derivation

The invariant interval along a worldline satisfies c^2 dtau^

2 = c^2\,\mathrm{dt}^2 - dx^2 = c^2\,\mathrm{dt}^2 (1 - v^2/c^2)

.

Taking the square root: dtau

= dt \sqrt{1 - v^2/c^2} = dt / \gamma

.

Integrating along the worldline gives total elapsed proper time

\tau = \int dt/\gamma

.

Limiting cases

v = 0

⟶

\tau = t

A stationary clock reads coordinate time.

v \to c

⟶

\tau \to 0

Light rays accumulate zero proper time.

accelerated path⟶ tau is path-dependentTwin paradox: the traveling twin ages less because their worldline has shorter proper-time integral.

What if…

What if a photon could have proper time?

It can't — null paths have dtau $= 0$ . Photons experience no aging between emission and absorption.

What if you accelerated at 1g for a year?

Proper time elapsed $is \sim 1$ year; coordinate time on Earth slightly more $(\sim 1.0006\,\mathrm{yr})$ ; but the integrated effect grows exponentially with sustained acceleration.

What if you fell into a black hole?

Your proper time to reach the singularity is finite (typically microseconds for stellar-mass BHs), even though external observers never see you cross.

1

Cosmic ray muon

Given ·

t:: 0.00005
v:: 297000000

Find · tau

Steps

$\beta = 2.97e_{8} / 3e_{8} = 0.99$
$\gamma = 1/\sqrt{0.0199} \approx 7.09$
$\tau = t/\gamma = 5e-5 / 7.09 \approx 7.05e-6\,\mathrm{s}$

Result ·

\gamma \approx 7.09 \to \tau = 5e-5 / 7.09 \approx 7.05

microseconds — well within the 2.2 microsecond rest lifetime extended in lab frame.

2

Twin paradox traveler

Given ·

t:: 20
v:: 260000000

Find · tau

Steps

$\beta = 2.6e_{8} / 3e_{8} = 0.866$
$\gamma = 1/\sqrt{1 - 0.75} = 2$
$\tau = 20/2 = 10$ years

Result ·

\gamma = 2

→ traveler ages 10 years while Earth twin ages 20.

Lorentz Factor→Time Dilation→Spacetime Interval→Lorentz Transformation→