Editing Inertial frame of reference (section)

=== Simple example ===
[[File:Two reference frames.PNG|thumb|320px|'''Figure 1''': Two cars moving at different but constant velocities observed from stationary inertial frame ''S'' attached to the road and moving inertial frame ''S′'' attached to the first car.]]
Consider a situation common in everyday life. Two cars travel along a road, both moving at constant velocities. See Figure 1. At some particular moment, they are separated by 200 meters. The car in front is traveling at 22 meters per second and the car behind is traveling at 30 meters per second. If we want to find out how long it will take the second car to catch up with the first, there are three obvious "frames of reference" that we could choose.<ref>{{Cite book |last=Susskind |first=Leonard |title=Special relativity and classical field theory: the theoretical minimum |author2=Art Friedman |date=2017 |publisher=Hachette UK |isbn=978-0-465-09334-2 |location=New York |at=Figure 2.1 |language=en |oclc=968771417}}</ref>

First, we could observe the two cars from the side of the road. We define our "frame of reference" ''S'' as follows. We stand on the side of the road and start a stop-clock at the exact moment that the second car passes us, which happens to be when they are a distance {{nowrap|1=''d'' = 200 m}} apart. Since neither of the cars is accelerating, we can determine their positions by the following formulas, where <math>x_1(t)</math> is the position in meters of car one after time ''t'' in seconds and <math>x_2(t)</math> is the position of car two after time ''t''.

: <math>x_1(t) = d + v_1 t = 200 + 22t,\quad x_2(t) = v_2 t = 30t.</math>

Notice that these formulas predict at ''t'' = 0 s the first car is 200m down the road and the second car is right beside us, as expected. We want to find the time at which <math>x_1=x_2</math>. Therefore, we set <math>x_1=x_2</math> and solve for <math>t</math>, that is:

: <math>200 + 22t = 30t,</math>
: <math>8t = 200,</math>
: <math>t = 25\ \mathrm{seconds}.</math>

Alternatively, we could choose a frame of reference ''S′'' situated in the first car. In this case, the first car is stationary and the second car is approaching from behind at a speed of {{math|1=''v''<sub>2</sub> − ''v''<sub>1</sub> = 8 m/s}}. To catch up to the first car, it will take a time of {{math|1={{sfrac|''d''|''v''<sub>2</sub> − ''v''<sub>1</sub>}} = {{sfrac|200|8}} s}}, that is, 25 seconds, as before. Note how much easier the problem becomes by choosing a suitable frame of reference. The third possible frame of reference would be attached to the second car. That example resembles the case just discussed, except the second car is stationary and the first car moves backward towards it at {{nowrap|8 m/s}}.

It would have been possible to choose a rotating, accelerating frame of reference, moving in a complicated manner, but this would have served to complicate the problem unnecessarily. One can convert measurements made in one coordinate system to another. For example, suppose that your watch is running five minutes fast compared to the local standard time. If you know that this is the case, when somebody asks you what time it is, you can deduct five minutes from the time displayed on your watch to obtain the correct time. The measurements that an observer makes about a system depend therefore on the observer's frame of reference (you might say that the bus arrived at 5 past three, when in fact it arrived at three).