A spacecraft moving at constant velocity .7 * c relative to a nonaccelerating observer keeps time by reflecting a light pulse back and forth between two mirrors oriented at a right angle to the direction of motion. The mirrors are 70 meters apart (big spaceship) and are mounted within a sealed vertical vacuum tube (vertical being perpendicular to the direction of motion) so that the light pulse will travel at its vaccuum speed.

If we assume that the laws of physics are the same in the reference frame of the spaceship as in the frame of the observer, it follows that the speed of light in a vacuum will be identical in both reference frames.

Under this assumption:

• How long does it take a pulse to travel from the lower mirror to the higher, as measured in the spaceship?
• How long does it take the pulse to travel from the lower mirror to the higher as measured by the observer outside the spaceship?
• If the pulse rate of an inhabitant of the spaceship is 64 beats/minute, as measured by that individual, then what pulse rate would be measured by the outside observer, assuming some accurate means of making the observation?

The light will be observed in the reference frame of the spaceship to travel 70 meters at 3*10^8 m/s, which will take .2333 * 10^-6 second.

In the frame of the observer outside the ship the pulse will still travel 70 meters perpendicular to the line of motion, but will also travel some distance parallel to the line of motion.

• The distance travel by the pulse will therefore be greater than the 70 meter distance observed on the spaceship.
• Since the speed of light will be the same in both reference frames, the conclusion must be that the observer outside the ship will measure a longer time interval than an observer on the ship.

If `dt' stands for the time measured by the external observer, the ship and therefore the mirrors will move distance

• distance moved by ship in time `dt':  v `dt' = ( .7 * c ) `dt' = .7 ( 3 * 10^8 m/s ) `dt' = 2.1 * 10^8 m/s * `dt'.

The distance traveled by the pulse is found by the Pythagorean Theorem to be

• distance measured by outside observer = `sqrt( ( 70 m )^2 + ( 2.1 * 10^8 m/s * `dt' )^2).

The time required to travel this distance at the speed of light is

• time for pulse to travel distance = distance / velocity

= `sqrt( ( 70 m )^2 + ( 2.1 * 10^8 m/s * `dt' )^2) / ( 3 * 10^8 m/s).

This time is of course the time `dt' measured by the observer as the light travels from mirror to mirror. Thus we have

• `dt' = `sqrt( ( 70 m )^2 + ( 2.1 * 10^8 m/s * `dt' )^2) / ( 3 * 10^8 m/s).

This could be solved for the only variable `dt', which appears on both sides of the equation.  However, at this point anyone with good judgment will go back and solve the problem using symbols, because carrying through all those numbers and units is going to be a real pain, and we are likely to lose sight of the meanings of the numbers.

In the generalized solution we will solve the problem using v for the speed of the spaceship, c for the speed of light and `dy for the separation of the mirrors. The result will be that any time interval `dt' measured by the outside observer will be 1 / `sqrt( 1 - v^2 / c^2 ) times the time interval `dt observed on the spacecraft. 

Just to bring home the lesson about solving in symbolic form we plug on through the numerical solution:

We can solve this equation for `dt':

square both sides:

• `dt'^2 = ( ( 70 m )^2 + ( 2.1 * 10^8 m/s * `dt' )^2) / ( 3 * 10^8 m/s)^2,

multiply both sides by the denominator and factor out the `dt'^2 in the second term of the right-hand side:

• ( 3 * 10^8 m/s)^2 * `dt'^2 = ( 70 m )^2 + ( 2.1 * 10^8 m/s )^2 * `dt'^2,

collect the `dt'^2 terms on the left-hand side:

• ( 3 * 10^8 m/s)^2 * `dt'^2 - ( 2.1 * 10^8 m/s )^2 * `dt'^2 = ( 70 m )^2

factor out `dt'^2 on left:

• [ ( 3 * 10^8 m/s)^2 - ( 2.1 * 10^8 m/s )^2 ] * `dt'^2 = ( 70 m )^2

(factor out 3 * 10^8 m/s)^2 on the left:

• ( 3 * 10^8 m/s)^2 [ 1 - ( 2.1 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ] * dt'^2 = ( 70 m )^2

divide through by the coefficient of `dt'^2:

• `dt`^2 = ( 70 m ) ^ 2 / [ ( 3 * 10^8 m/s)^2 { 1 - ( 2.1 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 } ],

rearrange the right-hand side, using the knowledge that 70 m / ( 3 * 10^8 m/s) = .2333 * 10^-6 sec, the time measured on the ship:

• `dt'^2 = ( .2333 * 10^-6 sec ) ^ 2 / [ 1 - ( 2.1 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ]

evaluate [ 1 - ( 2.1 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ] :

• [ 1 - ( 2.1 * 10^8 m/s )^2 / ( 3 * 10^8 m/s)^2 ] = .51,

rewrite the equation:

• `dt'^2 = ( .2333 * 10^-6 sec)^2 / .51,

take the square root of both sides and evaluate:

• `dt' = ( .2333 * 10^-6 sec) / .7141 = .3267 sec.

Had we used the result of our symbolic solution we would have obtained the same result directly:

• `dt' = `dt / `sqrt( 1 - v^2 / c^2 ) = .2333 sec / `sqrt ( 1 - ( .7 * c / c ) ^ 2 )

= .2333 sec / `sqrt( 1 - `b^2 ) = .2333 sec / .7141 = .3267 sec.

The time `dt' measured by the 'outside' observer will be 1 / `sqrt( 1 - v^2 / c^2) = 1 / `sqrt ( 1 - ( .7 * c / c ) ^ 2 ) = 1.4 times as great as that observed on the spacecraft.  Note that this is the ratio `dt' / `dt = .3267 sec / ( .2333 sec).

If the velocity of the spacecraft is symbolized by v, the speed of light by c and the separation of the mirrors by `dy, we see that the time required in the spaceship frame is

• `dt = `dy / c.

If the time measured by the outside observer is `dt', then the distance `dx' moved by the spacecraft as the light travels from mirror to mirror is

• `dx' = v * `dt',

and the distance `ds' moved by the light pulse is the hypotenuse of a right triangle with legs `dy and `dx':

• `ds' = `sqrt( `dy^2 + `dx' ^ 2 ) = `sqrt ( `dy^2 + ( v * `dt' ) ^ 2 ).

The time required for light to move this distance, moving as it must at velocity c, is

• `dt' = `ds' / c = `sqrt ( `dy^2 + ( v * `dt' ) ^ 2 ) / c .

We thus have the equation

• `dt' = `sqrt ( `dy^2 + ( v * `dt' ) ^ 2 ) / c ,

which we solve for `dt.

We square both sides:

• `dt' ^ 2= ( `dy^2 + ( v * `dt' ) ^ 2 ) / c^2

then divide through the right-hand side by c^2:

• `dt' ^ 2= `dy^2 / c^2 + ( v * `dt' ) ^ 2 / c^2 = (`dy / c)^2 + ( v / c * `dt' ) ^ 2

We observe that `dy / c = `dt, the time interval measured in the rocket frame. Using this fact and subtracting the second term on the right from both sides we have

• `dt`^2 - ( ( v / c ) `dt' ) ^ 2 = `dt^2.

Factor out `dt'^2 on the left-hand side:

• ( 1 - ( v / c ) ^ 2 ) * `dt'^2 = `dt^2

then dividing by the coefficient of `dt'^2 and taking square roots we have

• `dt' = `dt / `sqrt( 1 - ( v / c ) ^ 2 ).

This could be written as

• `dt' = `dt * [ 1 / `sqrt( 1 - ( v / c ) ^ 2 ) ],

and we see that the time measured by the outside observer will be equal to the product of the time measured by the observer on the rocket and the factor 1 / `sqrt( 1 - ( v / c ) ^ 2 ).

This factor 1 / `sqrt( 1 - ( v / c ) ^ 2 ) is called the 'time dilation factor'.