Proper time

From Wikipedia, the free encyclopedia

Proper time is time as measured by the clock for an observer who is traveling through spacetime. The concept of proper time is necessary in Einstein's theories of relativity because of effects such as time dilation, which result in observers travelling between the same positions along different paths/world lines through spacetime experiencing time differently.

1 Mathematical formalism
- 1.1 Derivation
2 Usage in special relativity
- 2.1 Example 1: The twin "paradox"
- 2.2 Example 2: The rotating disk
3 Usage in general relativity
- 3.1 Example 3: The rotating disk (again)
- 3.2 Example 4: The Schwarzschild solution - Time on Planet Earth
4 See also

[edit] 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 tensor calculus, proper time is defined as follows: Given a spacetime which is a pseudo-Riemannian manifold mapped with a coordinate system $x μ$ and equipped with a corresponding metric tensor $g μν$ , the proper time $\tau\$ experienced in moving between two events along a timelike path P is given by the line integral

$\tau = \int_P \;\! d\tau$

where

$d\tau = \sqrt{dx_\mu \; dx^\mu} = \sqrt{g_{\mu\nu} \; dx^\mu \; dx^\nu}$

[edit] Derivation

For any spacetime, there is an incremental invariant interval ds between events with an incremental coordinate separation dx^μ of

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

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 $\tau\$ . So for our purposes $\tau\ \ \stackrel{\mathrm{def}}{=}\ s$ .

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

[edit] Usage in special relativity

In special relativity spacetime is mapped with a four-vector coordinate system $x μ = (t, x, y, z)$ 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:

$g_{\mu\nu} = \left ( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -\frac{1}{c^2} & 0 & 0 \\ 0 & 0 & -\frac{1}{c^2} & 0 \\ 0 & 0 & 0 & -\frac{1}{c^2} \end{matrix} \right ) .$

(Note: The +--- metric signature is used in this article so that $d\tau\$ will always be positive definite for timelike paths.)

In special relativity, the proper time equation becomes

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

This can also be written as

$\tau = \int_P \sqrt {1 - \frac{1}{c^2} \left ( \left (\frac{dx}{dt}\right)^2 + \left (\frac{dy}{dt}\right)^2 + \left ( \frac{dz}{dt}\right)^2 \right) } dt = \int_P \sqrt {1 - \frac{v(t)^2}{c^2}} dt$ ,

where $v (t)$ is the coordinate speed at coordinate time $t$ .

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

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

where $Δ$ means "the change in" between two events.

[edit] 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

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

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 0.866c 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

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

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 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

$\Delta \tau = \sqrt{T^2 - (v_x T/c)^2 - (v_y T/c)^2 - (v_z T/c)^2 } = T \sqrt{1 - v^2/c^2}$ ,

which is the SR time dilation formula.

[edit] 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\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 $ω$ 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 $ω$ are constant, $dx = r \omega \cos(\omega T) \; dT$ and $dy = -r \omega \sin(\omega T) \; dT$ . The incremental proper time formula then becomes

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

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 $T 1$ and $T 2$ , the proper time experienced will be

$\int_{T_1}^{T_2} d\tau = (T_2 - T_1) \sqrt{ 1 - \left ( \frac{r\omega}{c} \right )^2}$ .

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.

[edit] 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.

[edit] 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 $θ = arctan(x / y) - ω t$ . The t and z coordinates remain unchanged. In this new coordinate system, the incremental proper time equation is

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

With r, θ, and z being constant over time, this simplifies to $d\tau = dt \sqrt{ 1 - \left (\frac{r \omega}{c} \right )^2 }$ , 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

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

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.

[edit] Example 4: The Schwarzschild solution - Time on Planet Earth

The Schwarzschild solution has an incremental proper time equation of

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

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 radians.

$\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. 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 rotating 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, $M=5.9742 \times 10^{24}$ kg, meaning that $m = 4.4354 \times 10^{-3}$ m. When standing on the north pole, we can assume $dr = d\theta\ = d\phi\ = 0$ (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 $d\tau = dt \sqrt{1 - 2m/r}$ . Then using the polar radius of the Earth as the radial coordinate (or $r\ = 6,356,752$ meters), we find that

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

At the equator, the radius of the Earth is $r\ = 6,378,137$ meters. In addition, the rotation of the Earth needs to be taken into account. This imparts on an observer an angular velocity of $\ d\phi/dt$ of $\ 2\pi$ divided by the sidereal period of the Earth's rotation, $\ 86162.4$ seconds. So $d\phi = 7.2923 \times 10^{-5} dt$ . The proper time equation then produces

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

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.

Categories: Minkowski spacetime | Relativity | Timekeeping

Proper time

From Wikipedia, the free encyclopedia

Contents

[edit] Mathematical formalism

[edit] Derivation

[edit] Usage in special relativity

[edit] Example 1: The twin "paradox"

[edit] Example 2: The rotating disk

[edit] Usage in general relativity

[edit] Example 3: The rotating disk (again)

[edit] Example 4: The Schwarzschild solution - Time on Planet Earth

[edit] See also

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