Consider the rail car pictured in Figure 26.7, with two observers, one in the middle of the moving rail car at O', and the other on the ground at location O. Lightning strikes both ends of the rail car, leaving marks at A, A', B, and B', timed such that the observer on the ground at O sees the light from both strikes simultaneously.
Does the observer in the rail car see the strikes simultaneously? No. The rail car is moving forward at speed v. The light from the lightning strikes propagates from A' and B' towards O'. Since O' is moving towards B', the light from that end reaches the observer first; the light from the other end arrives a bit later.
The observer on the ground says that the lightning strikes were simultaneous, and the observer in the rail car says that lightning hit at B' before A', and both are correct. In relativity, simultaneity is not absolute.
Note from this example that an observation is made when light (or some other signal) reaches an observer. This is an important subtlety in relativity.
To understand time dilation, let's do what Einstein called a "thought experiment". Imagine two observers, observer 1 inside a vehicle which is moving with respect to observer 2, as shown in Figure 26.8. Observer 1 makes a simple clock by sending a pulse of light from a bulb, reflecting it from a mirror on the top of the vehicle, and measuring the time when it returns to where it started. The time it takes for the light to travel from bulb to mirror to bulb is Dt1 = 2d / c. Think of this as one tick of this clock.
How long does it take for the light to travel from bulb to mirror to bulb for observer 2 who is outside the moving vehicle? According to observer 2, the light leaves the bulb at an angle, travels up to the mirror, reflects at an angle, and returns to the bulb. Let's say that the time it takes is Dt2. We can find Dt2 in terms of d and v, by considering the path of the light as seen by observer 2. The path is a triangle, and if we divide it in half, we have two identical right triangles (Figure 26.8c). The length of one hypotenuse is cDt2/2, since it takes the light equal times to travel to the mirror and back. The lengths of the sides are d, and vDt2/2, the distance the vehicle moves in time Dt2/2. Using Pythagorean's theorem:
The g term is always greater than or equal to 1. g =1 only when the relative speed between two observers is 0. This result means that observer 2 will say that observer 1's clock runs slow. If 1 second passes in observer 1's frame (Dt1 = 1sec.) then observer 2 will say that it actually took a time Dt2 = g Dt_1 = g sec. The exact value of g depends on the speed of observer 1 relative to observer 2:
| description | speed v | g |
|---|---|---|
| satellite in geosynchronous orbit | 3.07×103m/s = 11,000 km/hr | 1.0000000001 |
| half the speed of light | 0.5c | 1.15 |
| nine-tenths the speed of light | 0.9c | 2.29 |
| ninety-nine-hundreths the speed of light | 0.99c | 7.09 |
Note that for a satellite in geosynchronous orbit, the difference one day is shorter than one day on earth by 4.3ms.
The time that it takes for something to occur in a frame at rest with the action (the rest frame) is called the proper time. Observer 1 measures the proper time of the light clock, but observer 2 is in motion relative to the clock and therefore sees the effect of time dilation.
An astronaut at rest on Earth has a heartbeat rate of 70 beats/min. When the astronaut is traveling in a spaceship at 0.90c, what will this rate be as measured by (a) an observer also in the ship and (b) an observer at rest on the Earth?
(a) An observer in the ship is at rest relative to the astronaut, so their relative speed is 0. Thus the rate must be the same. Note that for v=0, g=1.
(b) The astronaut is moving with speed 0.90c relative to the observer on Earth. Thus, from the table above we find that g = 2.29. Since one heartbeat takes (1/70)min = (6/7)sec = 0.857sec in proper time, then to the observer on Earth it takes (2.29)(0.857sec) = 1.96 sec, and therefore her heartrate is 30 beats/min.
The following is a famous paradox of special relativity. This being a physics course, there is a definite answer and a clear way to resolve the paradox, and that resolution may give you some additional insight.
Consider this situation involving two 20 year old twin brothers, Speedo and Goslo. Goslo is a stay-at-home kinda guy while Speedo signs up for a space trip to a planet 30 lightyears from Earth. Speedo takes off, accelerates to nearly the speed of light, travels to the planet, then turns around, and comes back to Earth. On his return he finds that 60 years have elapsed on Earth, yet he has aged just 10 years! While Goslo would claim that Speedo was traveling at a very high relative speed, Speedo can equally well claim that he was at rest while Goslo and the rest of Earth was moving at high speed relative to him. Given this symmetry, why is it that Speedo aged less?
The resolution is to note that there are times during Speedo's voyage when special relativity doesn't apply. For instance, he must accelerate, then decelerate. He is the one who turns around after reaching the distant planet. Goslo, and the rest of Earth continued on their normal routine during the entire time. This difference breaks the symmetry mentioned above, and lets us differentiate Speedo's time in space from the rest of Earth.
Not only does time change, as measured by observers in relative motion, but also distances. The proper length of an object is its length measured in its rest frame. When the length is measured from a frame moving relative to the object, the measurement is less than the proper length, hence the term length contraction.
An object of proper length Lp moving relative to an observer is measured by the observer to have a length: