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In special relativity, an object (a box, perhaps) travelling at 0.5c relative to us, if it thinks it's 1lightsecond long in its own reference frame, will look 0.866 lightseconds long to us.

My question can be phrased in a few ways:

If that box stops in front of us - it decelerates from 0.5c in our ref frame to 0 in our ref frame - will it look 1 light second long to us?

Conversely, imagine a box that is 1 light second long in our reference frame and stops - and then it starts accelerating, and accelerates to 0.5c in our reference frame and then stops accelerating, it stays at 0.5c. Is it still 1 light second long in its own reference frame, or has it seen itself change length?

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  • $\begingroup$ Does this answer your question? In the twin paradox does the returning twin also come back permanently length contracted flatter than the twin on Earth? $\endgroup$
    – Dale
    Commented Feb 6 at 15:22
  • $\begingroup$ See Bell's Spaceship Paradox. If you figure that out, the answer to this question will be obvious. $\endgroup$
    – JEB
    Commented Feb 6 at 15:42
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    $\begingroup$ Why was this downvoted? It's not the best written, but physics-wise, it relates to some top SR paradoxes...aforementioned Bell's Spaceship, and maybe Pole-Barn/Train Tunnel. I suppose a 186,282 mile long box is not the best, when poles, ladders and trains are available. In fact, train is the best, because it has brakes on all wheels, controllable from track of train..so it can decelerate simultaneously in either frame. $\endgroup$
    – JEB
    Commented Feb 6 at 15:47
  • $\begingroup$ @JEB people are very eager to interpret a question in the least charitable way possible across all stack exchange ecosystems. I'm quite happy f or someone to say, "actually, you can accelerate it so that it remains the same size in S, or the same size in S', it depends on how you accelerate it". That's a valid answer, and the answer makes total sense in relativity given the question. Bonus points for "... but if you accelerate it so that it stays the same length in S, it will probably undergo stresses that break it apart because it's being pulled further and further apart in S'" $\endgroup$
    – TKoL
    Commented Feb 7 at 10:52
  • $\begingroup$ Double points for, "but the natural assumed way for most thought experiments is to accelerate it such that it stays the same length in S' and not in S - or that it stays the same length in its own reference frame, for all reference frames that it travels through as it accelerates". $\endgroup$
    – TKoL
    Commented Feb 7 at 10:56

2 Answers 2

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That depends on the question: In which frame did the parts of the box accelerate simultaneously?

Case 1: All parts of the box are being accelerated simultaneously in the rest frame of the box. This means its points will not be accelerated simultaneously in our frame (due to Relativity of simultaneity). The box parts more to the left will stop first, then the parts to the right of the box will stop later, leading to an expansion in the direction of motion, so that its length indeed becomes 1 lightsecond in our frame.

Case 2: All parts of the box are being accelerated simultaneously in our frame. What happens afterwards depends on the material constitution of the box, and the strength of the acceleration. a) If the box's material is very weak, it will be compressed by the simultaneous acceleration in our frame, the box will be broken and/or squeezed, so the length stays at 0.866 lightseconds at the time it stops, or b) the box's material is strong enough to stay intact, thus it will expand as its velocity decreases, because the forces that hold its material together tend to restore its proper length of 1 lightsecond.

Of course, there are many other cases and outcomes depending on the acceleration profile you choose.....

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  • $\begingroup$ Thank you, do you have an opinion on, in relativistic thought experiments, what the most natural way to accelerate an object is? Would you accelerate it such that it stays the same length in it's final velocity/frame S'? I hope that question makes sense. $\endgroup$
    – TKoL
    Commented Feb 7 at 10:56
  • $\begingroup$ @TKoL Usually one accelerates an object such that it stays the same length in $S'$. Hence it undergoes length contraction in $S$. Bell's Spaceship Paradox flips the script, and ship-string-ship system remains the same length in $S$, hence it expands in $S'$ and the string breaks. In the train/tunnel paradox, ppl ask: what happens if the train stops right before the exit door. On can image 4 scenarios: 1) Only the engine brakes, 2) (1) + a $c$ signal is sent back for each car to brake, or, all brakes are applied simultaneously in the 3) tunnel, 4) train frame. All are very instructive. $\endgroup$
    – JEB
    Commented Feb 7 at 23:29
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This is not a relativity question; it is an elementary math question!

The reason it's not a relativity question is that you're asking what happens in just one frame (namely the one you call "ours"). Relativity is for predicting how measurements in one frame relate to measurements in another.

So: There are forces acting on both ends of the box causing it to decelerate. The new length could be anything depending on the nature of those forces. They might cause the box to stretch, or they might cause it to squish. To determine what happens, you need to specify the time path of the acceleration at the left and right ends of the box.

Say that the acceleration of the left and right ends are given by $a_L(t)$ and $a_R(t)$ at time $t$, where $t$ ranges from (say) $0$ to $1$. At time $0$, the distance between the left and right ends is (by your assumption) 1 light second. Now use calculus to compute the distance at time 1.

For example, if $a_L$ and $a_R$ are the same function, then the box length (quite obviously) cannot change at all.

Change the functions $a_L$ and $a_R$, and you'll get a different answer.

If you did everything in another frame, you'd have to use different acceleration functions. In particular, if $a_L$ and $a_R$ are the same function (so that the box length doesn't change during deceleration in your frame) then in another frame they will typically NOT be the same function (because in the other frame, one end of the box will start moving before the other). But you can bypass those computations, because relativity predicts the new length of the box in the other frame, which must turn out to be (in your example) .866 times whatever the new length is in our own frame.

But your question is only about our own frame, where the question comes down to: A box of given length is moving. It decelerates. What is its new length? And the answer comes down to: How does the left side decelerate? How does the right side decelerate? Make your assumptions and calculus gives you the answer.

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