View Single Post
Mar29-09, 10:38 PM
P: 645

I assume that you are refering to the section titled "Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks" since that section ends with:

Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.
Shortly before, in that same section:

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2 tv2/c2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.
Now it should be remembered that this is Einstein. If I wrote something similar on this forum, I would probably be appropriatedly chastised for an inaccuracy - perhaps not for being incorrect, but for not clearly stating something.

The time t is the time occupied in the journey from A to B, in the frame of the clock which remains stationary. According to the stationary clock, the clock which was moved travelled a distance of x = v.t.

I'd say that the time t' on the moving clock at the end of that journey would be:

[tex]\gamma ( t - x.v / c^{2} ) = \gamma ( t - t.v^{2} / c^{2} ) = \gamma t (1 - v^{2} / c^{2} ) = t / \gamma [/tex]

The difference is therefore [tex]\Delta = t - t' = t ( 1 - 1 / \gamma) \approx 1/2t . v^{2} / c^{2} [/tex].

When that moving clock stops, there are a number of ticks from the "stationary" clock still travelling to catch up, x/(c-v) = vt/(c-v) worth. Those ticks in transit will, when added to the ticks already received, show that the more time has elapsed has elapsed on the stationary clock, even though that time elapsed at a slower rate.

So Einstein's answer, while possibly not immediately intuitive, is not erroneous.