In special relativity

With the timelike convention for the metric signature, the Minkowski metric is defined by

$${\displaystyle \eta _{\mu \nu }={\begin{pmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{pmatrix}},}$$

and the coordinates by

$${\displaystyle (x^{0},x^{1},x^{2},x^{3})=(ct,x,y,z)}$$

for arbitrary Lorentz frames.

In any such frame an infinitesimal interval, here assumed timelike, between two events is expressed as

$${\displaystyle ds^{2}=c^{2}dt^{2}-dx^{2}-dy^{2}-dz^{2}=\eta _{\mu \nu }dx^{\mu }dx^{\nu },}$$

(1)

and separates points on a trajectory of a particle (think clock). The same interval can be expressed in coordinates such that at each moment, the particle is at rest. Such a frame is called an instantaneous rest frame, denoted here by the coordinates ${\displaystyle (c\tau ,x_{\tau },y_{\tau },z_{\tau })}$ for each instant. Due to the invariance of the interval (instantaneous rest frames taken at different times are related by Lorentz transformations) one may write

$${\displaystyle ds^{2}=c^{2}d\tau ^{2}-dx_{\tau }^{2}-dy_{\tau }^{2}-dz_{\tau }^{2}=c^{2}d\tau ^{2},}$$

since in the instantaneous rest frame, the particle or the frame itself is at rest, i.e., ${\displaystyle dx_{\tau }=dy_{\tau }=dz_{\tau }=0}$. Since the interval is assumed timelike (ie. ${\displaystyle ds^{2}>0}$), taking the square root of the above yields^[10]

$${\displaystyle ds=cd\tau ,}$$

or

$${\displaystyle d\tau ={\frac {ds}{c}}.}$$

Given this differential expression for τ, the proper time interval is defined as

Here P is the worldline from some initial event to some final event with the ordering of the events fixed by the requirement that the final event occurs later according to the clock than the initial event.

Using (1) and again the invariance of the interval, one may write^[11]

where v(t) is the coordinate speed at coordinate time t, and x(t), y(t), and z(t) are space coordinates. The first expression is manifestly Lorentz invariant. They are all Lorentz invariant, since proper time and proper time intervals are coordinate-independent by definition.

If t, x, y, z, are parameterised by a parameter λ, this can be written as

$${\displaystyle \Delta \tau =\int {\sqrt {\left({\frac {dt}{d\lambda }}\right)^{2}-{\frac {1}{c^{2}}}\left[\left({\frac {dx}{d\lambda }}\right)^{2}+\left({\frac {dy}{d\lambda }}\right)^{2}+\left({\frac {dz}{d\lambda }}\right)^{2}\right]}}\,d\lambda .}$$

If the motion of the particle is constant, the expression simplifies to

$${\displaystyle \Delta \tau ={\sqrt {\left(\Delta t\right)^{2}-{\frac {\left(\Delta x\right)^{2}}{c^{2}}}-{\frac {\left(\Delta y\right)^{2}}{c^{2}}}-{\frac {\left(\Delta z\right)^{2}}{c^{2}}}}},}$$

where Δ means the change in coordinates between the initial and final events. The definition in special relativity generalizes straightforwardly to general relativity as follows below.

In general relativity

Proper time is defined in general relativity as follows: Given a pseudo-Riemannian manifold with a local coordinates x^μ and equipped with a metric tensor g_μν, the proper time interval Δτ between two events along a timelike path P is given by the line integral^[12]

${\displaystyle \Delta \tau =\int _{P}\,d\tau =\int _{P}{\frac {1}{c}}{\sqrt {g_{\mu \nu }\;dx^{\mu }\;dx^{\nu }}}.}$

(4)

This expression is, as it should be, invariant under coordinate changes. It reduces (in appropriate coordinates) to the expression of special relativity in flat spacetime.

In the same way that coordinates can be chosen such that x¹, x², x³ = const in special relativity, this can be done in general relativity too. Then, in these coordinates,^[13]

$${\displaystyle \Delta \tau =\int _{P}d\tau =\int _{P}{\frac {1}{c}}{\sqrt {g_{00}}}dx^{0}.}$$

This expression generalizes definition (2) and can be taken as the definition. Then using invariance of the interval, equation (4) follows from it in the same way (3) follows from (2), except that here arbitrary coordinate changes are allowed.

Example 1: The twin "paradox"

For a twin paradox scenario, let there be an observer A who moves between the A-coordinates (0,0,0,0) and (10 years, 0, 0, 0) inertially. This means that A stays at ${\displaystyle x=y=z=0}$ for 10 years of A-coordinate time. The proper time interval for A between the two events is then

$${\displaystyle \Delta \tau _{A}={\sqrt {(10{\text{ years}})^{2}}}=10{\text{ years}}.}$$

So being "at rest" in a special relativity coordinate system means that proper time and coordinate time are the same.

Let there now be another observer B who travels in the x direction from (0,0,0,0) for 5 years of A-coordinate time at 0.866c to (5 years, 4.33 light-years, 0, 0). Once there, B accelerates, and travels in the other spatial direction for another 5 years of A-coordinate time to (10 years, 0, 0, 0). For each leg of the trip, the proper time interval can be calculated using A-coordinates, and is given by

$${\displaystyle \Delta \tau _{leg}={\sqrt {({\text{5 years}})^{2}-({\text{4.33 years}})^{2}}}={\sqrt {6.25\;\mathrm {years} ^{2}}}={\text{2.5 years}}.}$$

So the total proper time for observer B to go from (0,0,0,0) to (5 years, 4.33 light-years, 0, 0) and then to (10 years, 0, 0, 0) is

$${\displaystyle \Delta \tau _{B}=2\Delta \tau _{leg}={\text{5 years}}.}$$

Thus it is shown that the proper time equation incorporates the time dilation effect. In fact, for an object in a SR (special relativity) spacetime traveling with velocity ${\displaystyle v}$ for a time ${\displaystyle \Delta T}$, the proper time interval experienced is

$${\displaystyle \Delta \tau ={\sqrt {\Delta T^{2}-\left({\frac {v_{x}\Delta T}{c}}\right)^{2}-\left({\frac {v_{y}\Delta T}{c}}\right)^{2}-\left({\frac {v_{z}\Delta T}{c}}\right)^{2}}}=\Delta T{\sqrt {1-{\frac {v^{2}}{c^{2}}}}},}$$

which is the SR time dilation formula.

Example 2: The rotating disk

An observer rotating around another inertial observer is in an accelerated frame of reference. For such an observer, the incremental (${\displaystyle d\tau }$) form of the proper time equation is needed, along with a parameterized description of the path being taken, as shown below.

Let there be an observer C on a disk rotating in the xy plane at a coordinate angular rate of ${\displaystyle \omega }$ and who is at a distance of r from the center of the disk with the center of the disk at x = y = z = 0. The path of observer C is given by ${\displaystyle (T,\,r\cos(\omega T),\,r\sin(\omega T),\,0)}$, where ${\displaystyle T}$ is the current coordinate time. When r and ${\displaystyle \omega }$ are constant, ${\displaystyle dx=-r\omega \sin(\omega T)\,dT}$ and ${\displaystyle dy=r\omega \cos(\omega T)\,dT}$. The incremental proper time formula then becomes

$${\displaystyle d\tau ={\sqrt {dT^{2}-\left({\frac {r\omega }{c}}\right)^{2}\sin ^{2}(\omega T)\;dT^{2}-\left({\frac {r\omega }{c}}\right)^{2}\cos ^{2}(\omega T)\;dT^{2}}}=dT{\sqrt {1-\left({\frac {r\omega }{c}}\right)^{2}}}.}$$

So for an observer rotating at a constant distance of r from a given point in spacetime at a constant angular rate of ω between coordinate times ${\displaystyle T_{1}}$ and ${\displaystyle T_{2}}$, the proper time experienced will be

$${\displaystyle \int _{T_{1}}^{T_{2}}d\tau =(T_{2}-T_{1}){\sqrt {1-\left({\frac {r\omega }{c}}\right)^{2}}}=\Delta T{\sqrt {1-v^{2}/c^{2}}},}$$

as v = rω for a rotating observer. This result is the same as for the linear motion example, and shows the general application of the integral form of the proper time formula.

Example 3: The rotating disk (again)

An appropriate coordinate conversion done against the Minkowski metric creates coordinates where an object on a rotating disk stays in the same spatial coordinate position. The new coordinates are

$${\displaystyle r={\sqrt {x^{2}+y^{2}}}}$$

and

$${\displaystyle \theta =\arctan \left({\frac {y}{x}}\right)-\omega t.}$$

The t and z coordinates remain unchanged. In this new coordinate system, the incremental proper time equation is

$${\displaystyle d\tau ={\sqrt {\left[1-\left({\frac {r\omega }{c}}\right)^{2}\right]dt^{2}-{\frac {dr^{2}}{c^{2}}}-{\frac {r^{2}\,d\theta ^{2}}{c^{2}}}-{\frac {dz^{2}}{c^{2}}}-2{\frac {r^{2}\omega \,dt\,d\theta }{c^{2}}}}}.}$$

With r, θ, and z being constant over time, this simplifies to

$${\displaystyle d\tau =dt{\sqrt {1-\left({\frac {r\omega }{c}}\right)^{2}}},}$$

which is the same as in Example 2.

Now let there be an object off of the rotating disk and at inertial rest with respect to the center of the disk and at a distance of R from it. This object has a coordinate motion described by dθ = −ω dt, which describes the inertially at-rest object of counter-rotating in the view of the rotating observer. Now the proper time equation becomes

$${\displaystyle d\tau ={\sqrt {\left[1-\left({\frac {R\omega }{c}}\right)^{2}\right]dt^{2}-\left({\frac {R\omega }{c}}\right)^{2}\,dt^{2}+2\left({\frac {R\omega }{c}}\right)^{2}\,dt^{2}}}=dt.}$$

So for the inertial at-rest observer, coordinate time and proper time are once again found to pass at the same rate, as expected and required for the internal self-consistency of relativity theory.^[14]

Example 4: The Schwarzschild solution – time on the Earth

The Schwarzschild solution has an incremental proper time equation of

$${\displaystyle d\tau ={\sqrt {\left(1-{\frac {2m}{r}}\right)dt^{2}-{\frac {1}{c^{2}}}\left(1-{\frac {2m}{r}}\right)^{-1}dr^{2}-{\frac {r^{2}}{c^{2}}}d\phi ^{2}-{\frac {r^{2}}{c^{2}}}\sin ^{2}(\phi )\,d\theta ^{2}}},}$$

where

• t is time as calibrated with a clock distant from and at inertial rest with respect to the Earth,
• r is a radial coordinate (which is effectively the distance from the Earth's center),
• ɸ is a co-latitudinal coordinate, the angular separation from the north pole in radians.
• θ is a longitudinal coordinate, analogous to the longitude on the Earth's surface but independent of the Earth's rotation. This is also given in radians.
• m is the geometrized mass of the Earth, m = GM/c²,
  • M is the mass of the Earth,
  • G is the gravitational constant.

To demonstrate the use of the proper time relationship, several sub-examples involving the Earth will be used here.

For the Earth, M = 5.9742×10²⁴ kg, meaning that m = 4.4354×10⁻³ m. When standing on the north pole, we can assume ${\displaystyle dr=d\theta =d\phi =0}$ (meaning that we are neither moving up or down or along the surface of the Earth). In this case, the Schwarzschild solution proper time equation becomes ${\textstyle d\tau =dt\,{\sqrt {1-2m/r}}}$. Then using the polar radius of the Earth as the radial coordinate (or ${\displaystyle r={\text{6,356,752 metres}}}$), we find that

$${\displaystyle d\tau ={\sqrt {\left(1-1.3908\times 10^{-9}\right)\;dt^{2}}}=\left(1-6.9540\times 10^{-10}\right)\,dt.}$$

At the equator, the radius of the Earth is r = 6378137 m. In addition, the rotation of the Earth needs to be taken into account. This imparts on an observer an angular velocity of ${\displaystyle d\theta /dt}$ of 2π divided by the sidereal period of the Earth's rotation, 86162.4 seconds. So ${\displaystyle d\theta =7.2923\times 10^{-5}\,dt}$. The proper time equation then produces

$${\displaystyle d\tau ={\sqrt {\left(1-1.3908\times 10^{-9}\right)dt^{2}-2.4069\times 10^{-12}\,dt^{2}}}=\left(1-6.9660\times 10^{-10}\right)\,dt.}$$

From a non-relativistic point of view this should have been the same as the previous result. This example demonstrates how the proper time equation is used, even though the Earth rotates and hence is not spherically symmetric as assumed by the Schwarzschild solution. To describe the effects of rotation more accurately the Kerr metric may be used.