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I was having a little thought experiment about Lorentz contraction, and I couldn't really figure out what would actually happen. Note that I'm not looking for a answer 'this effect is barely noticeable'; consider I have a cat that steps on my keyboard.

Say, we are a stationary observer, and we observe a rod continuously accelerating along its axis. As far as the rod is concerned, it is accelerating uniformly.

Being a stationary observer, my guess is that we don't see it accelerating uniformly. It is exhibiting Lorentz contraction, so the rod seems to be getting shorter as it accelerates. In order for the tip and the end of the rod to end up closer to each other after some time accelerating, the end of the rod must appear to be moving faster than the tip of the rod.

Now, stuff gets weird. The end appears to be moving faster for us, stationary observers, so it must exhibit more Lorentz contraction than the tip, repeat ad nausea. The opposite holds for the tip of the rod. So, am I right in presuming that the rod will end up as such (moving left-to-right)

->) [=======-======-======-======-======] < Stationary rod (equal spacing of '-')
->) [=-==-===-====-=====]                 < Accelerating rod (unequal spacing of '-')

And then there is also the relativity of simultaneity... is that another factor affecting the apparent shape of the rod, or is that another way of arriving at the same result?

Alternatively, do I have all of this wrong, and am I applying all the wrong principles to the poor rod?

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  • $\begingroup$ If the rod is not to stretch and eventually break, the rear of the rod must have greater proper acceleration than the front of the rod. See, for example, Rindler observers $\endgroup$
    – Alfred Centauri
    Commented Apr 12, 2015 at 13:24
  • $\begingroup$ @AlfredCentauri Thanks, so at least I got that bit right. If I read the Rindler chart correctly on wikipedia, does that mean the answer to my question is 'yes, your rod will end up like that'? $\endgroup$
    – Sanchises
    Commented Apr 12, 2015 at 13:33
  • $\begingroup$ See also Bell's spaceship paradox. Searching this site for the phrase will find some related questions. $\endgroup$
    – John Rennie
    Commented Apr 12, 2015 at 17:29
  • $\begingroup$ If the rod is not to stretch not to be compressed, but if its two ends ($A$ and $B$) maintain constant ping durations between each other (which are of course not equal but $\tau A_{BA}$ and $\tau B_{AB}$, respectively) and if $A$ maintains hyperbolic motion with constant acceleration $a$ then $B$ also maintains hyperbolic motion with constant acceleration $b := a~\text{Exp}[ \frac{a}{2~c}~\tau A_{BA} ]$, where vice versa $a := b~/~\text{Exp}[ \frac{b}{2~c}~\tau B_{AB} ]$. Cmp. physics.stackexchange.com/questions/38377/… $\endgroup$
    – user12262
    Commented Apr 13, 2015 at 22:56

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This is correct. Each point on the rod will follow a hyperbola, so the path of the rod through time will look like this. When b (time) is zero, the rod is at rest and evenly spaced, but at any other value the rod is contracted more in the direction it's accelerating away from.

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