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Suppose I want to measure the length of an object in front of me, along the axis separating us (for instance looking headlong at a bus, we could talk about someone standing inside the middle of the bus with a light signal and mirror, but you get the idea).

Let's say in the rest frame of the bus I measure a length $\mathcal{L}_{0}$. Now, suppose I boost into an accelerated frame with relative velocity v, the length is now observed to be:

$$\mathcal{L}=\mathcal{L}_{0}\sqrt{1-\left(v/c\right)^{2}}$$ I could imagine repeating the boosting several times, while only going into an inertial frame momentarily to measure the length. It would appear as though the “bus” is shortening with a measurable speed:

$$\frac{d\mathcal{L}}{dt}=\frac{d}{dt}\mathcal{L}_{0}\sqrt{1-\left(v/c\right)^{2}}$$

$$=\frac{\mathcal{L}_{0}}{c^{2}}\frac{\left(-v\right)}{\sqrt{1-\left(v/c\right)^{2}}}\frac{dv}{dt}$$

I understand special relativisticaly how this occurs, but now we have acceleration and noninertial reference frames. This is more the purview of General relativity. How would I go about calculating this observed “speed of length contraction” properly for some constant acceleration and relative velocity? I'm guessing there's more to the story? I'm very familiar with GR, but for some reason this is stumping me.

Also, an accelerating observer should have a time dilation due to acceleration and not just velocity, I'm not sure where that fits in

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  • $\begingroup$ This has nothing to do with general relativity. Calculate the trajectories of the two ends of the bus separately; then the length of the bus at a given moment is the difference between the positions of the ends. Those trajectories in turn depend on the details of how you boost. (Do you apply a force at the front, or at the back, or all along the length, or something else?) The answer will depend on your auxiliary assumptions about that. $\endgroup$
    – WillO
    Commented Dec 24, 2022 at 15:25
  • $\begingroup$ @WillO I'm an external ideal pointlike accelerating observer measuring the bus's length. The bus itself is in an inertial frame. The bus is just a prop, I could be talking about the distance between two stars instead, just empty space $\endgroup$
    – R. Rankin
    Commented Dec 24, 2022 at 15:57
  • $\begingroup$ @WillO is there not also a time dilation ocurring just from acceleration itself irrespective of velocity (for example time runs slower on the surface of earth) $\endgroup$
    – R. Rankin
    Commented Dec 25, 2022 at 3:05
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    $\begingroup$ @PM 2Ring - he wasn't asking about additional time dilation, but about how fast an object contracts. $\endgroup$
    – Yukterez
    Commented Dec 25, 2022 at 3:34
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    $\begingroup$ Ok that is clearly wrong, luckily he didn't bake that into his equations. $\endgroup$
    – Yukterez
    Commented Dec 25, 2022 at 3:37

1 Answer 1

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If $\rm \ L(t)=L_0/\gamma(t)=L_0 \sqrt{1-v(t)^2/c^2}$

then $\rm \ L'(t)=dL/dt=-L_0 v'(t) \ v(t) \ \gamma(t)/c^2$

see here and here.

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  • $\begingroup$ ok, that's the equation I first came up with. Is that for the accelerating observer watching other lengths parallel to their velocity, or for an inertial observer watching the accelerating person length contract in the direction of motion (they should differ at least by a factor of gamma due to the time shift) $\endgroup$
    – R. Rankin
    Commented Dec 25, 2022 at 3:00
  • $\begingroup$ The v(t) is the speed of the observed object in the frame of the observer, who was moving in your example, so x and t are the moving observer's coordinates in which he rests. If the observed object experiences proper acceleration a=F/m and the observer is at rest, that translates to a=v'(t)γ(t)³. In that case you need to time the acceleration in such a way that no material stress occurs due to Bell's paradoxon, which you avoid by accelerating the back of the object sooner than the front. $\endgroup$
    – Yukterez
    Commented Dec 25, 2022 at 3:15
  • $\begingroup$ I'm just trying to find the proper speed of "universal" length contraction as observed by an accelerating observer. I say universal because everything, including the empty space between objects should be observed to contract in the direction of acceleration $\endgroup$
    – R. Rankin
    Commented Dec 25, 2022 at 3:33
  • $\begingroup$ @R. Rankin - in that case use the formula with the gamma, since then you use your own comoving rulers and clocks to measure the universe. $\endgroup$
    – Yukterez
    Commented Dec 25, 2022 at 3:36
  • $\begingroup$ Roger, thank you, that was my first answer in my question, until I started overthinking it $\endgroup$
    – R. Rankin
    Commented Dec 25, 2022 at 3:39

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