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6.9* Einstein’s Special Theory of Relativity

As a consequence of physical experiments performed in the latter half of the nineteenth century (most notably the Michelson-Morley experiment of 1887), physicists concluded that the results obtained in measuring the speed of light c are independent of the velocity of the instrument used to measure it. For example, suppose that while on Earth, an experimenter measures the speed of light emitted from the sun and finds it to be 186,000 miles per second. Now suppose that the experimenter places the measuring equipment in a spaceship that leaves Earth traveling at 100,000 miles per second in a direction away from the sun. A repetition of the same experiment from the spaceship yields the same result: Light is traveling at 186,000 miles per second relative to the spaceship, rather than 86,000 miles per second as one might expect!

This revelation led to a new way of relating coordinate systems used to locate events in space-time. The result was Albert Einstein’s special theory of relativity. In this section, we develop via a linear algebra viewpoint the essence of Einstein’s theory.

The basic problem is to compare two different inertial (nonaccelerating) coordinate systems S and $S^{'}$ $S^{'}$ that are in motion relative to each other under the assumption that the speed of light is the same when measured in either system. We assume that $S^{'}$ $S^{'}$ moves at a constant velocity in relation to S as measured from S.

To simplify matters, we assume that the two coordinate systems have parallel axes and share the same x-axis, and that the motion of $S^{'}$ $S^{'}$ relative to S is along this common axis. (See Figure 6.8.)

Two diagrams to compare two different inertial coordinate systems S and S prime to see if the speed of light is the same when measured in either system. — Figure 6.8

Figure 6.8 Full Alternative Text

We also suppose that there are two clocks, C and $C^{'}$ $C^{'}$ , placed in space so that C is stationary relative to S and $C^{'}$ $C^{'}$ is stationary relative to $S^{'}$ $S^{'}$ . These clocks give readings that are real numbers in units of seconds. They are calibrated so that at the instant the origins of S and $S^{'}$ $S^{'}$ coincide, both give the reading of 0.

Given any event p (something whose position and time of occurrence can be described), we may assign a set of space-time coordinates to it. For example, if p is an event that occurs at position (x, y, z) relative to S and at time t as read on clock C, we can assign to p the set of coordinates

(\begin{array}{c} x \\ y \\ z \\ t \end{array}) .

$(\begin{array}{c} x \\ y \\ z \\ t \end{array}) .$

This ordered 4-tuple is called the space—time coordinates of p relative to S and C. Likewise, p has a set of space-time coordinates

(\begin{array}{c} x^{'} \\ y^{'} \\ z^{'} \\ t^{'} \end{array})

$(\begin{array}{c} x^{'} \\ y^{'} \\ z^{'} \\ t^{'} \end{array})$

relative to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ .

Because motion is along a common x-axis, which lies in a common xy-plane, the third component of the space-time coordinates of p is always zero. Thus we consider only the first, second, and fourth coordinates of p, and write

(\begin{array}{r} x \\ y \\ t \end{array}) and (\begin{array}{r} x^{'} \\ y^{'} \\ t^{'} \end{array})

$(\begin{array}{r} x \\ y \\ t \end{array}) and (\begin{array}{r} x^{'} \\ y^{'} \\ t^{'} \end{array})$

to denote the space-time coordinates of an event p relative to S and $S^{'}$ $S^{'}$ , respectively.

As we have mentioned, our unit of time is the second. Our measure of an object’s velocity v is the ratio of its velocity (expressed in miles per second) to the speed of light expressed in the same units (which is approximately 186,000 miles per second). For example, if $S^{'}$ $S^{'}$ is moving at 18,600 miles per second relative to S and the speed of light, c, is 186,000 miles per second, the velocity of $S^{'}$ $S^{'}$ relative to S, v, would have a value of $v = 0.1$ $v = 0.1$ . For this reason, the speed of light c has the value 1.

For a fixed velocity v, let $T_{v} : R^{3} \to R^{3}$ $T_{v} : R^{3} \to R^{3}$ be the mapping defined by

T_{v} (\begin{array}{c} x \\ y \\ t \end{array}) = (\begin{array}{c} x^{'} \\ y^{'} \\ t^{'} \end{array}),

$T_{v} (\begin{array}{c} x \\ y \\ t \end{array}) = (\begin{array}{c} x^{'} \\ y^{'} \\ t^{'} \end{array}),$

where

(\begin{array}{c} x \\ y \\ t \end{array}) and (\begin{array}{c} x^{'} \\ y^{'} \\ t^{'} \end{array})

$(\begin{array}{c} x \\ y \\ t \end{array}) and (\begin{array}{c} x^{'} \\ y^{'} \\ t^{'} \end{array})$

are the space-time coordinates of the same event with respect to S and C and with respect to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ , respectively.

In what follows, we make four assumptions:

The origin of $S^{'}$ $S^{'}$ moves in the positive direction of the common x-axis relative to S at a constant velocity of $v > 0$ $v > 0$ .
The origin of S moves in the negative direction of the common x-axis relative to $S^{'}$ $S^{'}$ at the constant velocity of $- v < 0$ $- v < 0$ .
$T_{v}$ $T_{v}$ is a linear isomorphism.
The speed of any light beam, when measured in either S or $S^{'}$ $S^{'}$ , using the clocks C in S and $C^{'}$ $C^{'}$ in $S^{'}$ $S^{'}$ , is always $c = 1$ $c = 1$ .

Since motion is strictly along the x-axis and we assume that the y-axis is unaffected, we have that for any x, y, and t, there exist $x^{'}$ $x^{'}$ and $t^{'}$ $t^{'}$ such that

T_{v} (\begin{array}{r} x \\ y \\ t \end{array}) = (\begin{array}{r} x^{'} \\ y \\ t^{'} \end{array}) and T_{v} (\begin{array}{r} 0 \\ y \\ 0 \end{array}) = (\begin{array}{r} 0 \\ y \\ 0 \end{array}) .

$T_{v} (\begin{array}{r} x \\ y \\ t \end{array}) = (\begin{array}{r} x^{'} \\ y \\ t^{'} \end{array}) and T_{v} (\begin{array}{r} 0 \\ y \\ 0 \end{array}) = (\begin{array}{r} 0 \\ y \\ 0 \end{array}) .$

Our goal in this section is to calculate the matrix representation of $T_{v}$ $T_{v}$ with respect to the standard basis for $R^{3}$ $R^{3}$ .

Theorem 6.39.

Consider ${e_{1}, e_{2}, e_{3}}$ ${e_{1}, e_{2}, e_{3}}$ , the standard ordered basis for $R^{3}$ $R^{3}$ . Then

(a) $T_{v} (e_{2}) = e_{2} .$ $T_{v} (e_{2}) = e_{2} .$
(b) $T_{v}$ $T_{v}$ maps $span {e_{1}, e_{3}}$ $span {e_{1}, e_{3}}$ into itself.
(c) $T_{v}^{*}$ $T_{v}^{*}$ maps $span {e_{1}, e_{3}}$ $span {e_{1}, e_{3}}$ into itself.

Proof.

Parts (a) and (b) follow immediately from the equations above. For $i = 1$ $i = 1$ and $i = 3$ $i = 3$ ,

〈 T_{v}^{*} (e_{i}), e_{2} 〉 = 〈 e_{i}, T_{v} (e_{2}) 〉 = 〈 e_{i}, e_{2} 〉 = 0,

$〈 T_{v}^{*} (e_{i}), e_{2} 〉 = 〈 e_{i}, T_{v} (e_{2}) 〉 = 〈 e_{i}, e_{2} 〉 = 0,$

and hence (c) follows.

Suppose that, at the instant the origins of S and $S^{'}$ $S^{'}$ coincide, a light flash is emitted from their common origin. The event of the light flash when measured either relative to S and C or relative to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ has space-time coordinates

(\begin{array}{c} 0 \\ 0 \\ 0 \end{array}) .

$(\begin{array}{c} 0 \\ 0 \\ 0 \end{array}) .$

Let P be the set of all events in the xy-plane whose space-time coordinates

(\begin{array}{c} x \\ y \\ t \end{array})

$(\begin{array}{c} x \\ y \\ t \end{array})$

relative to S and C are such that the flash is observable in the common xy-plane at the point (x, y) (as measured relative to S) at the time t (as measured on C). Let us characterize P in terms of x, y, and t. Since the speed of light is 1, at any time $t \geq 0$ $t \geq 0$ the light flash is observable from any point in the plane whose distance to the origin of S (as measured on S) is $t \cdot 1 = t$ $t \cdot 1 = t$ . These are precisely the points in the xy-plane with $x^{2} + y^{2} = t^{2}$ $x^{2} + y^{2} = t^{2}$ , or $x^{2} + y^{2} - t^{2} = 0 .$ $x^{2} + y^{2} - t^{2} = 0 .$ Hence an event lies in P if and only if its space-time coordinates

(\begin{array}{c} x \\ y \\ t \end{array}) (t \geq 0)

$(\begin{array}{c} x \\ y \\ t \end{array}) (t \geq 0)$

relative to S and C satisfy the equation $x^{2} + y^{2} - t^{2} = 0$ $x^{2} + y^{2} - t^{2} = 0$ . Since the speed of light when measured in either coordinate system is the same, we can characterize P in terms of the space-time coordinates relative to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ similarly: An event lies in P if and only if, relative to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ , its space-time coordinates

(\begin{array}{c} x^{'} \\ y \\ t^{'} \end{array}) (t^{'} \geq 0)

$(\begin{array}{c} x^{'} \\ y \\ t^{'} \end{array}) (t^{'} \geq 0)$

satisfy the equation ${(x^{'})}^{2} + y^{2} - {(t^{'})}^{2} = 0$ ${(x^{'})}^{2} + y^{2} - {(t^{'})}^{2} = 0$ .

Let

A = (\begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}) .

$A = (\begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - 1 \end{array}) .$

Theorem 6.40.

If $〈 L_{A} (w), w 〉 = 0$ $〈 L_{A} (w), w 〉 = 0$ for some $w \in R^{3}$ $w \in R^{3}$ , then

〈 T_{v}^{*} L_{A} T_{v} (w), w 〉 = 0.

$〈 T_{v}^{*} L_{A} T_{v} (w), w 〉 = 0.$

Proof.

Let

w = (\begin{array}{c} x \\ y \\ t \end{array}) \in R^{3},

$w = (\begin{array}{c} x \\ y \\ t \end{array}) \in R^{3},$

and suppose that $〈 L_{A} (w), w 〉 = 0$ $〈 L_{A} (w), w 〉 = 0$ .

CASE 1. $t \geq 0$ $t \geq 0$ . Since $〈 L_{A} (w), w 〉 = x^{2} + y^{2} - t^{2}$ $〈 L_{A} (w), w 〉 = x^{2} + y^{2} - t^{2}$ , the vector w gives the coordinates of an event in P relative to S and C. Because

(\begin{array}{l} x^{'} \\ y \\ t^{'} \end{array})

$(\begin{array}{l} x^{'} \\ y \\ t^{'} \end{array})$

are the space-time coordinates of the same event relative to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ , the discussion preceding Theorem 6.40 yields

{(x^{'})}^{2} + y^{2} - {(t^{'})}^{2} = 0.

${(x^{'})}^{2} + y^{2} - {(t^{'})}^{2} = 0.$

Thus $〈 T_{v}^{*} L_{A} T_{v} (w), w 〉 = 〈 L_{A} T_{v} (w), T_{v} (w) 〉 = {(x^{'})}^{2} + y^{2} - {(t^{'})}^{2} = 0$ $〈 T_{v}^{*} L_{A} T_{v} (w), w 〉 = 〈 L_{A} T_{v} (w), T_{v} (w) 〉 = {(x^{'})}^{2} + y^{2} - {(t^{'})}^{2} = 0$ , and the conclusion follows.

CASE 2. $t < 0$ $t < 0$ . The proof follows by applying case 1 to $- w$ $- w$ .

We now proceed to deduce information about $T_{v}$ $T_{v}$ . Let

w_{1} = (\begin{array}{r} 1 \\ 0 \\ 1 \end{array}) and w_{2} = (\begin{array}{r} 1 \\ 0 \\ - 1 \end{array}) .

$w_{1} = (\begin{array}{r} 1 \\ 0 \\ 1 \end{array}) and w_{2} = (\begin{array}{r} 1 \\ 0 \\ - 1 \end{array}) .$

Clearly ${w_{1}, w_{2}}$ ${w_{1}, w_{2}}$ is an orthogonal basis for the span of ${e_{1}, e_{3}}$ ${e_{1}, e_{3}}$ . The next result tells us even more.

Theorem 6.41.

There exists a nonzero scalar a such that $T_{v}^{*} L_{A} T_{v} (w_{1}) = a w_{2}$ $T_{v}^{*} L_{A} T_{v} (w_{1}) = a w_{2}$ and $T_{v}^{*} L_{A} T_{v} (w_{2}) = a w_{1}$ $T_{v}^{*} L_{A} T_{v} (w_{2}) = a w_{1}$ .

Proof.

Because $〈 L_{A} (w_{1}), w_{1} 〉 = 0, 〈 T_{v}^{*} L_{A} T_{v} (w_{1}), w_{1} 〉 = 0$ $〈 L_{A} (w_{1}), w_{1} 〉 = 0, 〈 T_{v}^{*} L_{A} T_{v} (w_{1}), w_{1} 〉 = 0$ by Theorem 6.40. Thus $T_{v}^{*} L_{A} T_{v} (w_{1})$ $T_{v}^{*} L_{A} T_{v} (w_{1})$ is orthogonal to $w_{1}$ $w_{1}$ . Since ${w_{1}, w_{2}}$ ${w_{1}, w_{2}}$ is an orthogonal basis for span ${e_{1}, e_{3}}$ ${e_{1}, e_{3}}$ and each of $T_{v}^{*}, L_{A}$ $T_{v}^{*}, L_{A}$ , and $T_{v}$ $T_{v}$ maps this span into itself, it follows that $T_{v}^{*} L_{A} T_{v} (w_{1})$ $T_{v}^{*} L_{A} T_{v} (w_{1})$ must be a multiple of $w_{2}$ $w_{2}$ , that is, $T_{v}^{*} L_{A} T_{v} (w_{1}) = a w_{2}$ $T_{v}^{*} L_{A} T_{v} (w_{1}) = a w_{2}$ for some scalar a. Since $T_{v}$ $T_{v}$ and A are invertible, so is $T_{v}^{*} L_{A} T_{v}$ $T_{v}^{*} L_{A} T_{v}$ . Thus $a \neq 0$ $a \neq 0$ .

Similarly, there exists a nonzero scalar b such that $T_{v}^{*} L_{A} T_{v} (w_{2}) = b w_{1}$ $T_{v}^{*} L_{A} T_{v} (w_{2}) = b w_{1}$ .

Finally, we show that $a = b$ $a = b$ . Since $T_{v}^{*} L_{A} T_{v} (w_{1}) = a w_{2}$ $T_{v}^{*} L_{A} T_{v} (w_{1}) = a w_{2}$ , we have

2 a = 〈 T_{v}^{*} L_{A} T_{v} (w_{1}), w_{2} 〉 = 〈 w_{1}, T_{v}^{*} L_{A} T_{v} (w_{2}) 〉 = 〈 w_{1}, b w_{1} 〉 = 2 b .

$2 a = 〈 T_{v}^{*} L_{A} T_{v} (w_{1}), w_{2} 〉 = 〈 w_{1}, T_{v}^{*} L_{A} T_{v} (w_{2}) 〉 = 〈 w_{1}, b w_{1} 〉 = 2 b .$

So $a = b$ $a = b$ .

Actually, $a = b = 1$ $a = b = 1$ , as we see in the following result.

For the rest of this section let $B_{v} = {[T_{v}]}_{β}$ $B_{v} = {[T_{v}]}_{β}$ , where $β$ $β$ is the standard ordered basis for $R^{3}$ $R^{3}$ .

Theorem 6.42.

Given $B_{v} = {[T_{v}]}_{β}$ $B_{v} = {[T_{v}]}_{β}$ , as defined above,

(a) $B_{v}^{*} A B_{v} = A .$ $B_{v}^{*} A B_{v} = A .$
(b) $T_{v}^{*} L_{A} T_{v} = L_{A} .$ $T_{v}^{*} L_{A} T_{v} = L_{A} .$

Proof.

Since

e_{1} = \frac{1}{2} (w_{1} + w_{2}) and e_{3} = \frac{1}{2} (w_{1} - w_{2}),

$e_{1} = \frac{1}{2} (w_{1} + w_{2}) and e_{3} = \frac{1}{2} (w_{1} - w_{2}),$

it follows from Theorem 6.41 that

T_{v}^{*} L_{A} T_{v} (e_{1}) = a e_{1} and T_{v}^{*} L_{A} T_{v} (e_{3}) = - a e_{3} .

$T_{v}^{*} L_{A} T_{v} (e_{1}) = a e_{1} and T_{v}^{*} L_{A} T_{v} (e_{3}) = - a e_{3} .$

Furthermore, $T_{v}^{*} L_{A} T_{v} (e_{2}) = e_{2}$ $T_{v}^{*} L_{A} T_{v} (e_{2}) = e_{2}$ , and hence

B_{v}^{*} A B_{v} = (\begin{array}{r} a & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - a \end{array}) .

$B_{v}^{*} A B_{v} = (\begin{array}{r} a & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & - a \end{array}) .$

Let

w = (\begin{array}{l} 0 \\ 1 \\ 1 \end{array}) .

$w = (\begin{array}{l} 0 \\ 1 \\ 1 \end{array}) .$

Then $〈 L_{A} (w), w 〉 = 0$ $〈 L_{A} (w), w 〉 = 0$ , and hence by Theorem 6.40

0 = 〈 T_{v}^{*} L_{A} T_{v} (w), w 〉 = 〈 B_{v}^{*} A B_{v} w, w 〉 = 1 - a .

$0 = 〈 T_{v}^{*} L_{A} T_{v} (w), w 〉 = 〈 B_{v}^{*} A B_{v} w, w 〉 = 1 - a .$

Thus $a = 1$ $a = 1$ . As a consequence, $B_{v}^{*} A B_{v} = A$ $B_{v}^{*} A B_{v} = A$ . This proves (a). Part (b) now follows.

Now consider the situation 1 second after the origins of S and $S^{'}$ $S^{'}$ have coincided as measured by the clock C. Since the origin of $S^{'}$ $S^{'}$ is moving along the x-axis at a velocity v as measured in S, its space-time coordinates relative to S and C are

(\begin{array}{c} v \\ 0 \\ 1 \end{array}) .

$(\begin{array}{c} v \\ 0 \\ 1 \end{array}) .$

Similarly, the space-time coordinates for the origin of $S^{'}$ $S^{'}$ relative to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ must be

(\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array})

$(\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array})$

for some $t^{'} > 0$ $t^{'} > 0$ . Thus we have

T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}) = (\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}) for some t^{'} > 0.

$T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}) = (\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}) for some t^{'} > 0.$ (18)

By Theorem 6.42

〈 T_{v}^{*} L_{A} T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 = 〈 L_{A} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 = v^{2} - 1.

$〈 T_{v}^{*} L_{A} T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 = 〈 L_{A} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 = v^{2} - 1.$ (19)

But also

\begin{array}{rcl} 〈 T_{v}^{*} L_{A} T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 & = & 〈 L_{A} T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 \\ = & 〈 L_{A} (\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}), (\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}) 〉 = - {(t^{'})}^{2} . \end{array}

$\begin{array}{rcl} 〈 T_{v}^{*} L_{A} T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 & = & 〈 L_{A} T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}), T_{v} (\begin{array}{l} v \\ 0 \\ 1 \end{array}) 〉 \\ = & 〈 L_{A} (\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}), (\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}) 〉 = - {(t^{'})}^{2} . \end{array}$ (20)

Combining (19) and (20), we conclude that $v^{2} - 1 = - {(t^{'})}^{2}$ $v^{2} - 1 = - {(t^{'})}^{2}$ , or

t^{'} = \sqrt{1 - v^{2}} .

$t^{'} = \sqrt{1 - v^{2}} .$ (21)

Thus, from (18) and (21), we obtain

T_{v} (\begin{array}{c} v \\ 0 \\ 1 \end{array}) = (\begin{array}{c} 0 \\ 0 \\ \sqrt{1 - v^{2}} \end{array}) .

$T_{v} (\begin{array}{c} v \\ 0 \\ 1 \end{array}) = (\begin{array}{c} 0 \\ 0 \\ \sqrt{1 - v^{2}} \end{array}) .$ (22)

Next, recall that the origin of S moves in the negative direction of the $x^{'}$ $x^{'}$ -axis of $S^{'}$ $S^{'}$ at the constant velocity $- v < 0$ $- v < 0$ as measured from $S^{'}$ $S^{'}$ . Consequently, 1 second after the origins of S and $S^{'}$ $S^{'}$ have coincided as measured on clock C, there exists a time $t^{″} > 0$ $t^{″} > 0$ as measured on clock $C^{'}$ $C^{'}$ such that

T_{v} (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) = (\begin{array}{c} - v t^{″} \\ 0 \\ t^{″} \end{array}) .

$T_{v} (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) = (\begin{array}{c} - v t^{″} \\ 0 \\ t^{″} \end{array}) .$ (23)

From (23), it follows in a manner similar to the derivation of (22) that

t^{″} = \frac{1}{\sqrt{1 - v^{2}}} .

$t^{″} = \frac{1}{\sqrt{1 - v^{2}}} .$ (24)

Hence, from (23) and (24),

T_{v} (\begin{array}{r} 0 \\ 0 \\ 1 \end{array}) = (\begin{array}{r} \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 \\ \frac{1}{\sqrt{1 - v^{2}}} \end{array}) .

$T_{v} (\begin{array}{r} 0 \\ 0 \\ 1 \end{array}) = (\begin{array}{r} \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 \\ \frac{1}{\sqrt{1 - v^{2}}} \end{array}) .$ (25)

The following result is now easily proved using (22), (25), and Theorem 6.39.

Theorem 6.43.

Let $β$ $β$ be the standard ordered basis for $R^{3}$ $R^{3}$ . Then

{[T_{v}]}_{β} = B_{v} = (\begin{array}{c} \frac{1}{\sqrt{1 - v^{2}}} & 0 & \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 & 1 & 0 \\ \frac{- v}{\sqrt{1 - v^{2}}} & 0 & \frac{1}{\sqrt{1 - v^{2}}} \end{array}) .

${[T_{v}]}_{β} = B_{v} = (\begin{array}{c} \frac{1}{\sqrt{1 - v^{2}}} & 0 & \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 & 1 & 0 \\ \frac{- v}{\sqrt{1 - v^{2}}} & 0 & \frac{1}{\sqrt{1 - v^{2}}} \end{array}) .$

Time Contraction

A most curious and paradoxical conclusion follows if we accept Einstein’s theory. Suppose that an astronaut leaves our solar system in a space vehicle traveling at a fixed velocity v as measured relative to our solar system. It follows from Einstein’s theory that, at the end of time t as measured on Earth, the time that passes on the space vehicle is only $t \sqrt{1 - v^{2}}$ $t \sqrt{1 - v^{2}}$ . To establish this result, consider the coordinate systems S and $S^{'}$ $S^{'}$ and clocks C and $C^{'}$ $C^{'}$ that we have been studying. Suppose that the origin of $S^{'}$ $S^{'}$ coincides with the space vehicle and the origin of S coincides with a point in the solar system (stationary relative to the sun) so that the origins of S and $S^{'}$ $S^{'}$ coincide and clocks C and $C^{'}$ $C^{'}$ read zero at the moment the astronaut embarks on the trip.

As viewed from S, the space-time coordinates of the vehicle at any time $t > 0$ $t > 0$ as measured by C are

(\begin{array}{c} v t \\ 0 \\ t \end{array}),

$(\begin{array}{c} v t \\ 0 \\ t \end{array}),$

whereas, as viewed from $S^{'}$ $S^{'}$ , the space-time coordinates of the vehicle at any time $t^{'} > 0$ $t^{'} > 0$ as measured by $C^{'}$ $C^{'}$ are

(\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}) .

$(\begin{array}{l} 0 \\ 0 \\ t^{'} \end{array}) .$

But if two sets of space-time coordinates

(\begin{array}{c} v t \\ 0 \\ t \end{array}) and (\begin{array}{c} 0 \\ 0 \\ t^{'} \end{array})

$(\begin{array}{c} v t \\ 0 \\ t \end{array}) and (\begin{array}{c} 0 \\ 0 \\ t^{'} \end{array})$

are to describe the same event, it must follow that

T_{v} (\begin{array}{c} v t \\ 0 \\ t \end{array}) = (\begin{array}{c} 0 \\ 0 \\ t^{'} \end{array}) .

$T_{v} (\begin{array}{c} v t \\ 0 \\ t \end{array}) = (\begin{array}{c} 0 \\ 0 \\ t^{'} \end{array}) .$

Thus

(\begin{array}{c} \frac{1}{\sqrt{1 - v^{2}}} & 0 & \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 & 1 & 0 \\ \frac{- v}{\sqrt{1 - v^{2}}} & 0 & \frac{1}{\sqrt{1 - v^{2}}} \end{array}) (\begin{array}{c} v t \\ 0 \\ t \end{array}) = (\begin{array}{c} 0 \\ 0 \\ t^{'} \end{array}) .

$(\begin{array}{c} \frac{1}{\sqrt{1 - v^{2}}} & 0 & \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 & 1 & 0 \\ \frac{- v}{\sqrt{1 - v^{2}}} & 0 & \frac{1}{\sqrt{1 - v^{2}}} \end{array}) (\begin{array}{c} v t \\ 0 \\ t \end{array}) = (\begin{array}{c} 0 \\ 0 \\ t^{'} \end{array}) .$

From the preceding equation, we obtain $\frac{- v^{2} t}{\sqrt{1 - v^{2}}} + \frac{t}{\sqrt{1 - v^{2}}} = t^{'}$ $\frac{- v^{2} t}{\sqrt{1 - v^{2}}} + \frac{t}{\sqrt{1 - v^{2}}} = t^{'}$ , or

t^{'} = t \sqrt{1 - v^{2}} .

$t^{'} = t \sqrt{1 - v^{2}} .$ (26)

This is the desired result.

A dramatic consequence of time contraction is that distances are contracted along the line of motion (see Exercise 7).

Let us make one additional point. Suppose that we choose units of distance and time commonly used in the study of motion, such as the mile, the kilometer, and the second. Recall that the velocity v we have been using is actually the ratio of the velocity using these units with the speed of light c, using the same units. For this reason, we can replace v in any of the equations given in this section with the ratio v/c ,where v and c are given using the same units of measurement. Thus, for example, given a set of units of distance and time, (26) becomes

t^{'} = t \sqrt{1 - \frac{v^{2}}{c^{2}}} .

$t^{'} = t \sqrt{1 - \frac{v^{2}}{c^{2}}} .$

Exercises

Complete the proof of Theorem 6.40 for the case $t < 0$ $t < 0$ .
For

$w_{1} = (\begin{array}{r} 1 \\ 0 \\ 1 \end{array}) and w_{2} = (\begin{array}{r} 1 \\ 0 \\ - 1 \end{array}),$ $w_{1} = (\begin{array}{r} 1 \\ 0 \\ 1 \end{array}) and w_{2} = (\begin{array}{r} 1 \\ 0 \\ - 1 \end{array}),$

show that
1. (a) ${w_{1}, w_{2}}$ ${w_{1}, w_{2}}$ is an orthogonal basis for $span ({e_{1}, e_{2}});$ $span ({e_{1}, e_{2}});$
2. (b) $span ({e_{1}, e_{3}})$ $span ({e_{1}, e_{3}})$ is $T_{v}^{*} L_{A} T_{v}$ $T_{v}^{*} L_{A} T_{v}$ -invariant.
Derive (24), and prove that

$T_{v} (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) = (\begin{array}{c} \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 \\ \frac{1}{\sqrt{1 - v^{2}}} \end{array}) .$ $T_{v} (\begin{array}{c} 0 \\ 0 \\ 1 \end{array}) = (\begin{array}{c} \frac{- v}{\sqrt{1 - v^{2}}} \\ 0 \\ \frac{1}{\sqrt{1 - v^{2}}} \end{array}) .$ (25)

Hint: Use a technique similar to the derivation of (22).
Consider three coordinate systems $S, S^{'},$ $S, S^{'},$ and $S^{″}$ $S^{″}$ with the corresponding axes ( $x, x^{'}, x^{″}; y, y^{'}, y^{″};$ $x, x^{'}, x^{″}; y, y^{'}, y^{″};$ and $z, z^{'}, z^{″}$ $z, z^{'}, z^{″}$ ) parallel and such that the $x -, x^{'} -,$ $x -, x^{'} -,$ and $x^{″}$ $x^{″}$ -axes coincide. Suppose that $S^{'}$ $S^{'}$ is moving past S at a velocity $v_{1} > 0$ $v_{1} > 0$ (as measured on S), $S^{″}$ $S^{″}$ is moving past $S^{'}$ $S^{'}$ at a velocity $v_{2} > 0$ $v_{2} > 0$ (as measured on $S^{'}$ $S^{'}$ ), and $S^{″}$ $S^{″}$ is moving past S at a velocity $v_{3} > 0$ $v_{3} > 0$ (as measured on S), and that there are three clocks $C, C^{'},$ $C, C^{'},$ and $C^{″}$ $C^{″}$ such that C is stationary relative to $S, C^{'}$ $S, C^{'}$ is stationary relative to $S^{'}$ $S^{'}$ , and $C^{″}$ $C^{″}$ is stationary relative to $S^{″}$ $S^{″}$ . Suppose that when measured on any of the three clocks, all the origins of $S, S^{'}$ $S, S^{'}$ , and $S^{″}$ $S^{″}$ coincide at time 0. Assuming that $T_{v_{3}} = T_{v_{2}} T_{v_{1}}$ $T_{v_{3}} = T_{v_{2}} T_{v_{1}}$ (i.e., $B_{v_{3}} = B_{v_{2}} B_{v_{1}}$ $B_{v_{3}} = B_{v_{2}} B_{v_{1}}$ ), prove that

$v_{3} = \frac{v_{1} + v_{2}}{1 + v_{1} v_{2}} .$ $v_{3} = \frac{v_{1} + v_{2}}{1 + v_{1} v_{2}} .$

Note that substituting $v_{2} = 1$ $v_{2} = 1$ in this equation yields $v_{3} = 1$ $v_{3} = 1$ . This tells us that the speed of light as measured in S or $S^{'}$ $S^{'}$ is the same. Why would we be surprised if this were not the case?
Compute ${(B_{v})}^{- 1}$ ${(B_{v})}^{- 1}$ . Show ${(B_{v})}^{- 1} = B_{(- v)}$ ${(B_{v})}^{- 1} = B_{(- v)}$ . Conclude that if $S^{'}$ $S^{'}$ moves at a negative velocity v relative to S, then ${[T_{v}]}_{β} = B_{v}$ ${[T_{v}]}_{β} = B_{v}$ , where $B_{v}$ $B_{v}$ is of the form given in Theorem 6.43. Visit goo.gl/9gWNYu for a solution.
Suppose that an astronaut left Earth in the year 2000 and traveled to a star 99 light years away from Earth at 99% of the speed of light and that upon reaching the star immediately turned around and returned to Earth at the same speed. Assuming Einstein’s special theory of relativity, show that if the astronaut was 20 years old at the time of departure, then he or she would return to Earth at age 48.2 in the year 2200. Explain the use of Exercise 4 in solving this problem.
Recall the moving space vehicle considered in the study of time contraction. Suppose that the vehicle is moving toward a fixed star located on the x-axis of S at a distance b units from the origin of S. If the space vehicle moves toward the star at velocity v, Earthlings (who remain “almost” stationary relative to S) compute the time it takes for the vehicle to reach the star as $t = b / v$ $t = b / v$ . Due to the phenomenon of time contraction, the astronaut perceives a time span of $t^{'} = t \sqrt{1 - v^{2}} = (b / v) \sqrt{1 - v^{2}}$ $t^{'} = t \sqrt{1 - v^{2}} = (b / v) \sqrt{1 - v^{2}}$ . A paradox appears in that the astronaut perceives a time span inconsistent with a distance of b and a velocity of v. The paradox is resolved by observing that the distance from the solar system to the star as measured by the astronaut is less than b.

Assuming that the coordinate systems S and $S^{'}$ $S^{'}$ and clocks C and $C^{'}$ $C^{'}$ are as in the discussion of time contraction, prove the following results.
1. (a) At time t (as measured on C), the space-time coordinates of the star relative to S and C are
  
  $(\begin{array}{c} b \\ 0 \\ t \end{array}) .$ $(\begin{array}{c} b \\ 0 \\ t \end{array}) .$
2. (b) At time t (as measured on C), the space-time coordinates of the star relative to $S^{'}$ $S^{'}$ and $C^{'}$ $C^{'}$ are
  
  $(\begin{array}{c} \frac{b - v t}{\sqrt{1 - v^{2}}} \\ 0 \\ \frac{t - b v}{\sqrt{1 - v^{2}}} \end{array}) .$ $(\begin{array}{c} \frac{b - v t}{\sqrt{1 - v^{2}}} \\ 0 \\ \frac{t - b v}{\sqrt{1 - v^{2}}} \end{array}) .$
3. (c) For
  
  $x^{'} = \frac{b - t v}{\sqrt{1 - v^{2}}} and t^{'} = \frac{t - b v}{\sqrt{1 - v^{2}}},$ $x^{'} = \frac{b - t v}{\sqrt{1 - v^{2}}} and t^{'} = \frac{t - b v}{\sqrt{1 - v^{2}}},$
  
  we have $x^{'} = b \sqrt{1 - v^{2}} - t^{'} v$ $x^{'} = b \sqrt{1 - v^{2}} - t^{'} v$ .
  
  This result may be interpreted to mean that at time $t^{'}$ $t^{'}$ as measured by the astronaut, the distance from the astronaut to the star, as measured by the astronaut, (see Figure 6.9) is
  
  Figure 6.9
  
  Figure 6.9 Full Alternative Text
  
  $b \sqrt{1 - v^{2}} - t^{'} v .$ $b \sqrt{1 - v^{2}} - t^{'} v .$
4. (d) Conclude from the preceding equation that
  
  (1) the speed of the space vehicle relative to the star, as measured by the astronaut, is $v$ $v$ ;
  
  (2) the distance from Earth to the star, as measured by the astronaut, is $b \sqrt{1 - v^{2}}$ $b \sqrt{1 - v^{2}}$ .
  
  Thus distances along the line of motion of the space vehicle appear to be contracted by a factor of $\sqrt{1 - v^{2}}$ $\sqrt{1 - v^{2}}$ .

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Table of Contents for 6.9* Einstein&#8217;s Special Theory of Relativity Einstein’s Special Theory of Relativity

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Table of Contents for
6.9* Einstein’s Special Theory of Relativity Einstein’s Special Theory of Relativity