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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

R. Rankin
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  • 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. – WillO Dec 24 '22 at 15:25
  • @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 – R. Rankin Dec 24 '22 at 15:57
  • @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) – R. Rankin Dec 25 '22 at 03:05
  • ok. I had thought you were boosting the bus, not yourself. if you were boosting the bus i would stand by my comment. – WillO Dec 25 '22 at 03:07
  • No, according to the clock postulate, acceleration does not cause any additional time dilation on top of the time dilation caused by the velocity. See https://physics.stackexchange.com/q/704658/123208 – PM 2Ring Dec 25 '22 at 03:23
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    @PM 2Ring - he wasn't asking about additional time dilation, but about how fast an object contracts. – Yukterez Dec 25 '22 at 03:34
  • @Yukterez The last paragraph specifically mentions "an accelerating observer should have a time dilation due to acceleration and not just velocity". – PM 2Ring Dec 25 '22 at 03:36
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    Ok that is clearly wrong, luckily he didn't bake that into his equations. – Yukterez Dec 25 '22 at 03:37

1 Answers1

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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.

Yukterez
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  • 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) – R. Rankin Dec 25 '22 at 03:00
  • 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. – Yukterez Dec 25 '22 at 03:15
  • 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 – R. Rankin Dec 25 '22 at 03:33
  • @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. – Yukterez Dec 25 '22 at 03:36
  • Roger, thank you, that was my first answer in my question, until I started overthinking it – R. Rankin Dec 25 '22 at 03:39
  • If you want some motivation, I'm looking for a class of tetrad frames in which the observed expansion of the universe is nullified by this effect (clearly noninertial frames) – R. Rankin Dec 25 '22 at 03:41
  • Then you need to define the peculiar velocity v(t) relative to the local CMB frame such that γ'(t)=a'(t)=H(t)a(t), with a(t) now standing for the scale factor, and t the time in the CMB frame. The problem is that due to the relative simultaneity, the Hubbleparameter will have different values in the front and in the back (it will be higher in the back since it was higher in the past), so that will not be watertight and only work locally. – Yukterez Dec 25 '22 at 03:54
  • Not sure what you mean. Here's the route I went:$$\frac{dL}{dt}=0=HL_{0}+\frac{L_{0}}{2c^{2}}\frac{\left(-2v\right)}{\sqrt{1-\frac{v\left(t\right)^{2}}{c^{2}}}}\frac{dv}{dt}$$

    $$c^{2}H=\frac{v}{\sqrt{1-\frac{v\left(t\right)^{2}}{c^{2}}}}a$$

    $$c^{2}H^{2}\left(1-\frac{v\left(t\right)^{2}}{c^{2}}\right)=\left(va\right)^{2}$$

    $$a=Hc\sqrt{\left(\frac{c^{2}}{v^{2}}-1\right)}$$ I'm thinking that the factor of L with the hubble parameter H should be the contracted length and hence have a gamma factor also?

     – R. Rankin Dec 25 '22 at 03:56
  • I'm not to worried about anything but local frames, Id like to define it similarly to the null tetrads in the Newman-Penrose formalism. I don't see a lot of inherently noninertial vielbein out there (actually almost none) – R. Rankin Dec 25 '22 at 04:00
  • @R. Rankin - another thing to consider is that the length contraction only works in one direction, while the width and depth are unaffected, so for the volume to be conserved you'd need γ³ in order to compensate for the other two directions. If you take Null coordinates where the local reference points are photons on the other hand, the volume is 0 all the way, but that might not be the way you'd want to conserve it. – Yukterez Dec 25 '22 at 20:40
  • Thank you, because the frame inherently needs acceleration, I'm working on the second order frame bundle (the bundle of two-frames over M thought of as the prolongation $J^{1}FM$). Such a frame then corresponds to a non-coordinate frame on $F^{(2)}M$ similarly to how "regular" tetrads correspond to (generally) non-coordinate frames on $FM$ – R. Rankin Dec 26 '22 at 03:38