Proper Time Article Index for
Proper 		Website Links For
Time 		 

Information About

Proper Time
  CATEGORIES ABOUT PROPER TIME
   
minkowski spacetime
   
relativity
   
timekeeping
   
APPAREL
BABY
BEAUTY
BOOKS
CAR TOYS
CELL PHONES
DVD'S
ELECTRONICS
GOURMET FOOD
GROCERIES
HEALTH & PERSONAL
HOME & GARDEN
JEWELRY
MUSIC
MUSIC INSTRUMENTS
OFFICE PRODUCTS
SOFTWARE
SPORTING GOODS
TOOLS & HARDWARE
TOYS
VIDEO GAMES
SHOPPING HOME
MORE SHOPPING...




MATHEMATICAL FORMALISM


The formal definition of proper time involves describing the path that a clock, observer, or test particle is taking through spacetime and the Metric Structure of that spacetime.

Using mapped with a Coordinate System x^\mu and equipped with a corresponding Metric Tensor g_{\mu
u}, the proper time au\ experienced in moving between two Event s along a timelike path ''P'' is given by the Path Integral

au = \int_P \;\! d au

where

d au = \sqrt{dx_\mu \; dx^\mu} = \sqrt{g_{\mu
u} \; dx^\mu \; dx^
u}


Derivation


For any spacetime, there is an incremental invariant interval ''ds'' between events with an incremental coordinate separation ''dxμ'' of

ds^2 = g_{\mu
u} \; dx^\mu \; dx^
u.

This is referred to as the line element of the spacetime. ''s'' may be Spacelike , Lightlike , or Timelike . Spacelike paths cannot be physically traveled (as they require moving faster than light). Lightlike paths can only be followed by light beams, for which there is no passage of proper time. Only timelike paths can be traveled by massive objects, in which case the invariant interval becomes the proper time au\ . So for our purposes au\ \equiv ''s''.

Taking the square root of each side of the line element gives the above definition of d au\ . After that, take the path integral of each side to get au\ as described by the first equation.


USAGE IN SPECIAL RELATIVITY


In Special Relativity spacetime is mapped with a Four-vector coordinate system x^\mu = (t,x,y,z) where
: ''t'' is a Temporal coordinate and
: ''x'',''y'', and ''z'' are Orthogonal Spatial coordinates.
This spacetime and mapping are described with the Minkowski Metric :


g_{\mu
u} = \left (
\begin{matrix}
1 & 0 & 0 & 0 \
0 & - rac{1}{c^2} & 0 & 0 \
0 & 0 & - rac{1}{c^2} & 0 \
0 & 0 & 0 & - rac{1}{c^2}
\end{matrix} ight ) .


(Note: The +--- Metric Signature is used in this article so that d au\ will always be Positive definite for timelike paths.)

In special relativity, the proper time equation becomes

d au = \int_P \sqrt {dt^2 - dx^2/c^2 - dy^2/c^2 - dz^2/c^2}.

To make things even easier, Inertial Motion in special relativity is where the spatial coordinates change at a constant rate with respect to the temporal coordinate. This further simplifies the proper time equation to

\Delta au = \sqrt{\Delta t^2 - \Delta x^2/c^2 - \Delta y^2/c^2 - \Delta z^2/c^2},

where \Delta means "the change in" between two Event s.


Example 1: The twin "paradox"


For a Twin "paradox" scenario, let there be an observer ''A'' who moves between the coordinates (0,0,0,0) and (10 years, 0, 0, 0) inertially. This means that ''A'' stays at x=y=z=0 for 10 years of coordinate time. The proper time for ''A'' is then

\Delta au = \sqrt{(10\; \mathrm{years})^2} = 10\; \mathrm{years}

So we find that 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 coordinate time at 0.866''c'' to (5 years, 4.33 light-years, 0, 0). Once there, ''B'' accelerates, and travels in the other spatial direction for 5 years to (10 years, 0, 0, 0). For each leg of the trip, the proper time is

\Delta au = \sqrt{(5\;\mathrm{years})^2 - (4.33\;\mathrm{years})^2} = \sqrt{6.25\;\mathrm{years}^2} = 2.5 \; \mathrm{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) to (10 years, 0, 0, 0) is 5 years. Thus is it is shown that the proper time equation incorporates the time dilation effect. In fact, for an object in a SR spacetime traveling with a velocity of ''v'' for a time ''T'', the proper time experienced is

\Delta au = \sqrt{T^2 - (v_x T/c)^2 - (v_y T/c)^2 - (v_z T/c)^2 } = T \sqrt{1 - 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 (d au\ ) 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 \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 (T, \;\, r\sin(\omega T),\;\, r\cos(\omega T), \;\, 0), where T is the current coordinate time. When ''r'' and \omega are constant, dx = -r \omega \cos(\omega T) \; dT and dy = r \omega \sin(\omega T) \; dT. The incremental proper time formula then becomes

d au = \sqrt{dT^2 - (r \omega /c)^2 \cos^2(\omega T)\; dT^2 - (r \omega /c)^2 \sin^2(\omega T) \; dT^2} = dT\sqrt{1 - \left ( rac{r\omega}{c} ight )^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 T_1 and T_2, the proper time experienced will be

\int_{T_1}^{T_2} d au = (T_2 - T_1) \sqrt{ 1 - \left ( rac{r\omega}{c} ight )^2}.

As ''v''=''rω'' for a rotating observer, this result is as expected given the time dilation formula above, and shows the general application of the integral form of the proper time formula.


USAGE IN GENERAL RELATIVITY


The difference between SR and General Relativity (GR) is that in GR you can use any metric which is a solution of the Einstein Field Equations , not just the Minkowski metric. Because inertial motion in curved spacetimes lacks the simple expression it has in SR, the path integral form of the proper time equation must always be used.


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 r=\sqrt{x^2 + y^2} and heta = \arctan(x/y) - \omega t. The ''t'' and ''z'' coordinates remain unchanged. In this new coordinate system, the incremental proper time equation is

d au^2 = \sqrt{\left - \left ( rac{r \omega}{c} ight )^2 ight dt^2 - rac{dr^2}{c^2} - rac{r^2\, d heta^2}{c^2} - rac{dz^2}{c^2} - 2 rac{r^2 \omega \, dt \, d heta}{c^2}}

With ''r'', ''θ'', and ''z'' being constant over time, this simplifies to d au = dt \sqrt{ 1 - \left ( rac{r \omega}{c} ight )^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 rotataing observer. Now the proper time equation becomes

d au = \sqrt{\left - \left ( rac{R \omega}{c} ight )^2 ight dt^2 - \left ( rac{R\omega}{c} ight ) ^2 \,dt^2 + 2 \left ( rac{R \omega}{c} ight ) ^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.


Example 4: The Schwarzschild solution - Time on Planet Earth


The Schwarzschild Solution has an incremental proper time equation of

d au = \sqrt{\left( 1 - 2m/r ight ) dt^2 - rac{1}{c^2}\left ( 1 - 2m/r ight )^{-1} dr^2 - rac{r^2}{c^2} d heta^2 - rac{r^2}{c^2} \sin^2 heta \; d\phi^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),
: ''θ'' is the latitudinal coordinate, being the angular separation from the North Pole in Radian s.
: \phi\ is a longitudinal coordinate, analogous to the latitude on the Earth's surface but independent of the Earth's Rotation . This is also given in radians.
: ''m'' is the geometrized mass of a central massive object, being ''m=MG/c2'',
:: ''M'' is the mass of the object,
:: ''G'' is the Gravitational Constant .

To demonstrate the use of the proper time relationship, several sub-examples involving the Earth will be used here. It should be noted that the use of the Schwarzschild solution for the Earth is not entirely correct for the following reasons:
  • Due to its rotation, the Earth is an Oblate Spheroid instead of being a true Sphere . This results in the gravitational field also being oblate instead of spherical.

  • In GR, a rotataing object also drags spacetime along with itself. This is described by the Kerr Solution . However, the amount of frame dragging that occurs for the Earth is so small that it can be ignored.


For the Earth , M=5.9736 imes 10^{24}kg, meaning that m = 4.436 imes 10^{-3}m. When standing on the north pole, we can assume dr = d heta\ = d\phi\ (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 d au = dt \sqrt{1 - 2m/r}. Then using the polar radius of the Earth as the radial coordinate (or ''r'' = 6356780 meters), we find that

d au = \sqrt{\left ( 1 - 1.3957 imes 10^{-9} ight ) \;dt^2} = \left (1 - 6.978 imes 10^{-10} ight ) dt.

At the Equator , the radius of the Earth is ''r'' = 63781400 meters. In addition, the rotation of the Earth needs to be taken into account. This imparts on an observer an angular velocity of ''dφ/dt'' of 2''π'' divided by the Sidereal Period of the Earth's rotation, 86162.4 seconds. So d\phi\ = 7.2922×10^{-5} ''dt''. The proper time equation then produces

d au = \sqrt{\left ( 1 - 1.39093 imes 10^{-9} ight ) dt^2 - 2.40 imes 10^{-12} dt^2} = \left( 1 - 6.9667 imes 10^{-10} ight ).

This should have been the same as the previous result, but as noted above the Earth is not spherical as assumed by the Schwarzschild solution. Even so this demonstrates how the proper time equation is used.


SEE ALSO