Timeline for The "speed of length contraction" in switching between inertial frames
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2022 at 3:38 | comment | added | R. Rankin | 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 $FM$ | |
| Dec 25, 2022 at 20:40 | comment | added | Yukterez | @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. | |
| Dec 25, 2022 at 4:00 | comment | added | R. Rankin | 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) | |
| Dec 25, 2022 at 3:56 | comment | added | R. Rankin | 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? | |
| Dec 25, 2022 at 3:54 | comment | added | Yukterez | 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. | |
| Dec 25, 2022 at 3:41 | comment | added | R. Rankin | 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) | |
| Dec 25, 2022 at 3:39 | comment | added | R. Rankin | Roger, thank you, that was my first answer in my question, until I started overthinking it | |
| Dec 25, 2022 at 3:36 | comment | added | Yukterez | @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. | |
| Dec 25, 2022 at 3:33 | comment | added | R. Rankin | 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 | |
| Dec 25, 2022 at 3:15 | comment | added | Yukterez | 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. | |
| Dec 25, 2022 at 3:00 | comment | added | R. Rankin | 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) | |
| Dec 25, 2022 at 2:29 | history | edited | Yukterez | CC BY-SA 4.0 |
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| Dec 25, 2022 at 2:18 | history | answered | Yukterez | CC BY-SA 4.0 |