It's all relative.
The electrons in a cathode ray tube television travel from the electron gun to the screen at a speed that is about 1/3 the speed of light.
How long does it take the electron to travel the 1/3 of a meter distance from the electron gun to the screen?
The electron has a speed of v = 1 x 10^8 m/s and it travels a distance d = 1/3 meter so it takes a time, t:
t = d/v = 1/3 / 1 * 10^8 = 0.3 x 10^8 s = 3 ns, three nanoseconds.
The time of travel of the electron must be measured by two different clocks each at rest in your reference frame. One clock measures the time the electron begins its flight and one measures the ending time. Subtract the two times to get the travel time. These clocks must be synchronized.
Proper time and time dilation
The time measured by a stationary clock is very important in relativity it is called the proper time, t0, this is the time measured by a clock on the electron. To this clock the electron is at rest and the television is moving past at 1/3 the speed of light.
In special relativity there is a phenomenon known as time dilation, if a clock moves relative to you, it appears to run more slowly. To find the time interval, t, measured by clocks at rest when a clock moving at speed v measures time t0, multiply t0 by
g = 1/(1-(v/c)^2)^0.5 (g is the Greek letter gamma)
If c is the speed of light, 3 x 10^8 m/s, and the electron is traveling v = 1 x 10^8 m/s, then for the electron in a television set v/c = 1/3 and (v/c)^2 = 1/9 = 0.11
so g = 1/(1-(v/c)^2)^0.5 = 1/(1-0.11)^2 = 1.25
The relation between the time measured in the television frame t to the proper time is
t = g * t0 or also t0 = t//g
The time measured by a clock moving with the electron is t0, the time measured by clocks in the television frame is t = 3 ns. The time measured by the electron clock is
t0 = t/g = 3/1.25 = 2.4 ns
Viewed in the television frame of reference, the clock on the electron records less time than the clocks in the television frame. The moving electron ages less according to its clock than according to the television frame.
Proper length and length contraction
You can measure the length of a stationary object like the television by measuring the positions of the two ends of the object at any time. They don't move so it doesn't matter when you measure them. The length of a stationary object is called the proper length, L0. On the other hand you have to measure the positions of the ends of a moving object at the same time to find its length, L.
A moving object will have a length that is shorter than its proper length by a factor of g.
L = L0/g
So the television is at rest and its length L0 = 0.3 m.
To the electron, the television is moving past at 0.3c.
The length of the television viewed by the electron is
L = L0/g = 0.3 m/1.25 = 0.24 m
So in the electron frame the television moves at a speed
L/T0 = 0.24 m/2.4 ns = 1 * 10^8 m/s = 0.3c just as we said, it is all relative.
Consider the Stanford linear accelerator, SLAC
It is 3 km long and the electrons in the accelerator are traveling 99.99999995% of the speed of light at the end of the accelerator.
How long would it take an electron traveling at this speed to cover the 3 km?
t = d/v = 3*10^3 m/ 3 * 10^8 m/s = 10^-5 s.
What is the length of the accelerator as viewed by the electron?
To find this we'll need to calculate g, you can do this with a calculator but here's a neat trick.
(1 + e)^n = 1 + ne where e is a number that is small compared to 1.
So 99.99999995% c= 0.9999999995c = (1-5 *10^-10)c
and (v/c)^2 =
(1-5 * 10^-10)^2 = 1-2*5 *10^-10 = 0.999999999
and g = 1/(1-(v/c)^2)^0.5 = 1/(10 * 10^-10)^0.5 = 1/(3 *10^-5) = 0.3 * 10^5 = 3 * 10^4
so the electron would see an accelerator that is 30,000 times shorter. or
L = L0/g = 3 km/3*10^4 = 0.1 m, or about 4 inches long!
The time the electron would measure on its clock is
T0 = T/g = 10^-5 s / 3 * 10^4 = 0.3 * 10^-9 s
In the 10^-5 seconds we measure the electron traveling 3 km its clock only records 3 * 10^-10 s, a time interval shorter by 30000 times.
The speed of light is defined to be exactly: c= 2.99792458 x 10^8 m/s
Scientific Explorations with Paul Doherty
8 November 2004