User:John R. Brews/CZ psychology authors

CZ psychology authors

• User:John_Calvin_Moore
• User:Douglas_Griffith
• User:Michael J. Formica
• User:Patrick Malone
• User:Richard Pettitt
• User:Michael S. Everhart

http://www.charlierose.com/view/interview/11891

Before we conclude that clocks we perceive as moving actually run slower than clocks that appear stationary, so they count two events as taking less time if the clocks are moving, there is a wrinkle to examine. On the train, the light is sent and received at the same location, so the same clock records departure time and arrival time. On the track, though, the location of the flashlight changes with time, so two clocks are needed, one at the sending location and one at the receiving location. Can we be sure the two clocks are synchronized?

To synchronize the clocks on the ground, a light beam is sent from one location to the other. Because light travels at speed c, it will arrive at the other location a time L/c later. Thus, we set the distant clock to time L/c when the light arrives. That is the method for synchronizing separated clocks.

From the train however, these clocks are not synchronized. The distance L=vt used by the track observers is not the correct separation, which is a shorter distance vt' . Consequently, the distant clock is set to the wrong time of arrival for the synchronizing light, namely vt/c instead of vt'/c=vt √(1-v²/c²)/c. Instead of L, the track observers should have used L√(1-v²/c²), the so-called Lorentz contraction of L.

This example can be converted to a comparison of clocks. Suppose that a unit of time is that for the light to go from the flashlight to the mirror and back. The flashlight and mirror become a clock. The same clock is used on the train and on the track. Both the clock on the track and that on the train count one unit of time as t = 2h/c. But to those on the track, the traveling clock on the train counts one unit of time as the longer time t' = 2√(h²+(L/2)²)/c. Substituting L=vt and h=ct/2, we find:

$t^{\prime }={\frac {2{\sqrt {(ct)^{2}/4+(vt)^{2}/4}}}{c}}=t{\sqrt {1+v^{2}/c^{2}}}\ .$