原時

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原時是在相對論中與事件在同處的時鐘所測量的唯一時間，他不僅取決於事件，時鐘也在事件的行動之中。對同一個事件，一個加速中的時鐘所測得的原時會比在非加速（慣性）中時鐘的原時為短。雙生子佯謬就是其中的一個例子。

相對的，協調時能由一個與事件有一段距離的觀測者來應用。在狹義相對論中，協調時總是由在慣性系統內有關聯的觀測者計算，而原時則由同在加速中的觀測者測量。

在四維時空中，原時類似在三維空間（歐幾里德空間）的弧長。

在習慣上，原時通常使用大寫希臘字母 $τ$ 來標示，以與協調時 $t$ 或 $T$ .有所區別。

[编辑] 數學的形式

原時的定義形式中，包含repesents的時鐘、觀測者或測試的粒子在時空中的路徑描述，和那個時空的度量結構。

[编辑] 在狹義相對論

在狹義相對論，原時的定義如下：

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

此處， $v (t)$ 是在協調時 $t$ 的座標速度， $x$ 、 $y$ 和 $z$ 是空間中的正交座標。

如果 $t$ 、 $x$ 、 $y$ 和 $z$ 都用一個參量 $λ$ 的參數，公式可以簡化為：

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

以微分的型式可以寫成路徑的積分：

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

此處， $P$ 是時鐘在時空中的路徑。

為讓事件簡化，在特殊相對論中的慣性運動可以轉化成對瞬時座標成常數比的空間座標。這進一步簡化了原時方程式：

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

此處， $Δ$ 的意思是在兩個事件的變化。

特殊相對論的方程式是後續的一般狀況中的特例。

[编辑] 在廣義相對論

一些非中文的文字因为尚未翻譯而被隐藏，歡迎參與翻譯。

Using tensor calculus, proper time is more rigorously 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

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

[编辑] 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.

[编辑] Derivation for 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

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

as above.

[编辑] 在狹義相對論中的例子

[编辑] 例一：雙生子的"佯謬"

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 $Δ T$ , the proper time experienced is

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

which is the SR time dilation formula.

[编辑] 例二：旋轉盤

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.

[编辑] 廣義相對論的例子

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.

[编辑] 例三：旋轉盤 (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.

[编辑] 例四：史瓦西解 – 地球上的時間

原時方程式有一個新增的史瓦西解：

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

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.

[编辑] 相關條目

狹義相對論
廣義相對論
時間膨脹
羅倫茲轉換
Four-vector
閔考夫斯基空間

[编辑] 資料來源

譯自英文維基百科

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原時