Timeline for What is physically different about a moving vs still object in space?
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| Jun 4, 2020 at 16:03 | history | edited | CommunityBot |
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| Jan 1, 2019 at 22:01 | comment | added | Colin MacLaurin | Indeed the appearance in a photo will not generally be the $1/\gamma$ Lorentz-contraction, because of different travel times of the photons across the object. Wikipedia calls this effect "Terrell rotation". | |
| Jan 1, 2019 at 10:21 | history | edited | InertialObserver | CC BY-SA 4.0 |
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| Dec 31, 2018 at 4:00 | comment | added | WillO | Okay---I'm not sure we need to belabor this, but for the record I was referring to your first paragraph, not your final paragraph with the explicit assumption. The OP asked, in essence, whether we'd see a difference between a photo taken while the asteroid is in motion and a photo taken after it stops (once again, all at once in the observer's frame). It seemed to me that (even if this was not your intention) your words could have misled a reader into believing that the two photos would show different lengths, which is a common misconception that it's better not to encourage. | |
| Dec 31, 2018 at 3:48 | comment | added | InertialObserver | Okay that's fine, but that was an assumption I stated. | |
| Dec 31, 2018 at 3:48 | comment | added | WillO | @InertialObserver: Yes, if you know the rest length of the asteroid then I absolutely agree with you. But it's hard for me to imagine a circumstance in which I just happen to know the rest length of a random passing asteroid. | |
| Dec 31, 2018 at 3:46 | comment | added | InertialObserver | If you look at my answer I say that we know the rest length of the asteroid, so I don't need two observations.. I only need one to know that it (or was) moving relative to me.. once I know that the asteroid is .96753 m, that's it. I know it must be in relative motion with me, because I know its rest length is 1m.. | |
| Dec 31, 2018 at 3:42 | comment | added | WillO | @Inertialobserver: And how do those two observations tell me that the asteroid decelerated? | |
| Dec 31, 2018 at 3:42 | comment | added | InertialObserver | @WillO because the asteroid decelerated in order to stop.. | |
| Dec 31, 2018 at 3:40 | comment | added | WillO | @InertialObserver: I take a picture of an asteroid and observe that it is .96753 meters long. (An observer co-moving with the asteroid would say that it is one meter long, but in this case there is no such observer, or at least none that I'm in contact with.) Now the asteroid stops suddenly (all at once, in my frame). I take another picture and it's still .96753 meters long. All I have is two observations: .96753 and .96753. From these two observations, how do I infer that the asteroid was moving? | |
| Dec 31, 2018 at 0:56 | comment | added | InertialObserver | When an object is moving relative to an observer it's length gets contracted by a factor of $\frac{1}{\gamma}$. When its at rest $\gamma =1$. I don't think I understand what youre saying. | |
| Dec 31, 2018 at 0:55 | comment | added | WillO | @InertialObserver: I don't know what you mean by "the length contraction will be visible by taking a picture". If you take a picture while it's moving, and then take another picture after it stops dead all at once, its length will be the same in both pictures. | |
| Dec 31, 2018 at 0:48 | comment | added | InertialObserver | The length contraction will be apparent if I take a picture. If it stops obviously things get more complicated. But the length contraction will be visible by taking a picture if it's moving relativistically. | |
| Dec 31, 2018 at 0:38 | comment | added | WillO | The reference to the Lorentz factor and length contraction threatens to be badly misleading. If an asteroid in motion (relative to me) stops dead all at once (relative to me), its length as measured in my frame does not change, so there's no effect for me to observe. (Its length in its own initial frame changes, because in that frame the "front" and "back" ends of the asteroid stopped moving at different times --- but the question is about what the "stationary" observer can discern.) | |
| Dec 30, 2018 at 23:35 | history | edited | InertialObserver | CC BY-SA 4.0 |
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| Dec 30, 2018 at 23:29 | vote | accept | Jack | ||
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| Dec 30, 2018 at 23:25 | comment | added | InertialObserver | @JackNicholson It’s because you have no right to call yourself stationary and the other thing moving. For all you know if you’re on the “stationary” asteroid, you’re moving towards the “moving asteroid” and it’s at rest. This is what’s meant by the “pop science” everything is relative ordeal. | |
| Dec 30, 2018 at 23:09 | history | edited | InertialObserver | CC BY-SA 4.0 |
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| Dec 30, 2018 at 23:07 | comment | added | niels nielsen | Jack, it's actually worse than you think. if you are out in space and observing those two asteroids, there is no way physically possible for you to determine which of the two is standing still and which is moving. If you hitched a ride on first one then the other, and performed any physics experiment you can think of while riding on each, the results would show no differences at all. Out in space, experiencing uniform, unaccelerated motion and being at rest are indistinguishable. This is the crux of the law of special relativity. | |
| Dec 30, 2018 at 23:06 | history | edited | InertialObserver | CC BY-SA 4.0 |
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| Dec 30, 2018 at 23:00 | history | edited | InertialObserver | CC BY-SA 4.0 |
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| Dec 30, 2018 at 22:59 | comment | added | Jack | If it wasn't an image and you could go over to the asteroids and make measurements, would you still not be able to tell? If I resume time, one object will continue moving while the other will stay still - how does it have that memory? Surely there is something physical surrounding it that keeps its motion that could be measured? | |
| Dec 30, 2018 at 22:57 | history | answered | InertialObserver | CC BY-SA 4.0 |