To keeps thing simpler, we ignore any mechanical compression or stretch that a 'rigid' rod may suffer when pushed or pulled from one end.* On a Minkowski spacetime diagram, the end where the rod is being pushed with a constant force will follow a hyperbolic worldline, like hyperbola A in this diagram.**

The distance of the vertex of the hyperbola from the origin will be c

^{2}/a, where a is the acceleration measured with an accelerometer at the point where the force is applied. The origin is where the two blue lines cross. If the rod has a rest length L, the front end of the rod will follow the hyperbola B, with vertex a distance a+L from the origin. Hence, the acceleration that an accelerometer fixed to the front of the rod will register is c

^{2}/(a+L). This is algebraically smaller than the acceleration of the rear end.

Now, we must not be confused by the fact that the rod is temporarily compressed when the force is applied (as described in **); the reason for the smaller proper acceleration is that relative to the Minkowski coordinates, the rod Lorentz contracts as it picks up speeds. This means that during the acceleration the front end moves marginally slower than the rear end, as observed in the coordinate system; hence the front end suffers less time dilation than the rear end and clocks that were synchronized before the acceleration will gradually lose sync. Relative to coordinate clocks, both clocks will record less elapsed time, but the (faster) rear clock (A) will lose more time than the (slower) front clock (B).

As observed by persons riding on the rod, the accelerating rod does not Lorentz contract, but keeps its proper length L (ignoring mechanical compression as in *). In the original coordinate system the rod will however gradually Lorentz contract. If the acceleration is stopped after some coordinate time t, the whole rod will move uniformly at speed v = a t and the rod will have a constant Lorentz contracted length: L' = L, where is the Lorentz factor.***

By Einstein's equivalence principle, the above scenario is qualitatively the same as if the rod has been sitting vertically on its rear end on Earth's surface. The rod will experience marginally less gravity at the top than at the bottom and apart from a minor constant compression, the length of the rod will be L, as measured by us on Earth. But to an inertial observer free-falling at a speed v past the rod, the rod will appear to be length contracted to the same extent as above.

Since the distance a for an acceleration of 1g is about one light year, the difference in acceleration for a practical length rod is obviously pretty small, but if it is kept up for a long time, the difference in time can be significant. This has been practically demonstrated by clocks on Earth and in space.

I hope this clears up any confusion that might have arisen out of the aborted thread. If not, please shout.

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Regards

Jorrie

* If we keep the acceleration mild and the rod reasonably short, it makes little difference to the situation. Forces propagate through the rod at the speed of sound in the rod, so when abruptly accelerated by pushing it, the rod will compress/expand, i.e its length will oscillate for a short time and quickly settle into a very slightly compressed length state. The reverse will happen when the acceleration abruptly stops.

** This slide is from prof John Mallinckrodt's talk: Simple, Interesting, and Unappreciated Facts about Relativistic Acceleration. If the link does not work, Google the title and you should get it. The "Identical Asymptotic Light Cones" are explained in the slides, but essentially means that both ends of the rod will eventually approach the speed of light and the rod will appear to have zero length in the original inertial frame.

*** The Lorentz factor