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Suppose we have two particles with initial separation of $|x_2^\text{init}-x_1^\text{init}|=d$. At time $t=0$ they start accelerating with proper acceleration of $w$. Then position of the $i$th particle in laboratory frame will be:

$$x_i(t)=\frac{c^2}w\left(\sqrt{1+\frac{w^2t^2}{c^2}}-1\right)+x_i^\text{init}.$$

But then we can see that $$x_2(t)-x_1(t)=d=\text{const}.$$

So, if both particles uniformly accelerate starting from some equal moments of time, distance between them doesn't Lorentz-contract. Why is it so? How should they instead accelerate to have the distance Lorentz-contract? Should they continuously synchronize their velocities according to laboratory frame times instead of their proper times?

Ruslan
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  • Can you clarify in which direction the particles are accelerating? If they both accelerate in the direction perpendicular to their initial displacements then you wouldn't expect that distance to length contract. – Jordan Aug 01 '14 at 07:27
  • @Jordan The proper accelerations are equal. – Ruslan Aug 01 '14 at 07:29
  • Yeah, they're equal but in which direction are they accelerating? Sorry, I just realized that you probably didn't see my edit to my comment. I got tongue/finger tied and said something stupid, but that's fixed now. – Jordan Aug 01 '14 at 07:30
  • @Jordan I consider one-dimensional setting, so they are separated along $x$ axis and accelerate along this same axis. – Ruslan Aug 01 '14 at 07:35
  • The length would only change in the reference frame of the particles in question. From the laboratory point of view, the distance would indeed remain the same. – ticster Aug 01 '14 at 11:37
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    It sounds to me like this is basically the Bell spaceship paradox: http://www.physicsforums.com/showthread.php?t=742729 –  Aug 01 '14 at 17:57

3 Answers3

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Suppose we have two particles [...] How should they instead accelerate to have the distance Lorentz-contract?

A solution is arguably described in J. Franklin, "Rigid body motion in special relativity" (arxiv.org/abs/1105.3899).

Using the notation from the question above: Given the trajectory, wrt. the (inertial) "laboratory frame", of "particle $2$" as

$$x_2[ t ] := \frac{c^2}{w} \left( \sqrt{1 + \frac{w^2~t^2}{c^2}} - 1\right) + x_2[ 0 ],$$

i.e. uniform acceleration of "particle $2$" with proper acceleration $ w $,
and provided that

$$0 \lt x_2[ 0 ] - x_1[ 0 ] \lt \frac{c^2}{w},$$

set the trajectory of "particle $1$" to

$$x_1[ t ] := \left( \frac{c^2}{w} + x_1[ 0 ] - x_2[ 0 ] \right) ~ \left( \sqrt{1 + \frac{w^2~t^2~c^2}{(c^2 + (x_1[ 0 ] - x_2[ 0 ]) ~w)^2}} - 1\right) + x_1[ 0 ],$$

i.e. uniform acceleration of "particle $1$" with proper acceleration $ w ~ \frac{c^2}{c^2 + (x_1[ 0 ] - x_2[ 0 ]) ~w}$.

Then it can be shown that a "family of momentarily co-moving pairs", $\{ (A_t, B_t) \}$, can be considered (or even be found) such that

(a) for any parameter value $t \ge 0$ the pair $(A_t, B_t)$ is at rest to each other,

(b) with distances between any such pair equal to each other:
$AB_t = AB_0 := x_2[ 0 ] - x_1[ 0 ]$

(c) the speed of pair $(A_t, B_t)$ wrt. the (inertial) "laboratory frame" is $\frac{w~t}{ \sqrt{1 + \frac{w^2~t^2}{c^2}} },$

(d) particle $1$ met $A_t$ as it "turned around" wrt. pair $(A_t, B_t)$ (i.e. after "having approached $A_t$ from the direction of $B_t$", and before "going to further separate from $A_t$ towards $B_t$"),

(e) particle $2$ met $B_t$ as it "turned around" wrt. pair $(A_t, B_t)$ (i.e. after "having approached $B_t$ from the direction opposite of $A_t$", and before "going to further separate from $B_t$ in the direction opposite of $A_t$"), and

(f) $A_t$'s indication of having met particle $1$ and $B_t$'s indication of having met particle $2$ were simultaneous to each other.

Since the distance between any pair of members of the (inertial) "laboratory frame" whose indications of "having been met and passed by $A_t$" and "having been met and passed by $B_t$", respectively, were simultaneous to each other is

$$ \frac{x_2[ 0 ] - x_1[ 0 ]}{ \sqrt{1 + \frac{w^2~t^2}{c^2}} } = (x_2[ 0 ] - x_1[ 0 ]) ~ \sqrt{1 - \left( \frac{w~t}{ \sqrt{1 + \frac{w^2~t^2}{c^2}} } ~\frac{1}{c}\right)^2}, $$

where the right-hand-side is stated in terms of the speed value which appears in (c),
there applies the familiar "length-contraction" relation at least between the (inertial) "laboratory frame" and each "momentarily co-moving pair, $(A_t, B_t)$".

However, obviously, since with the described "solution" particle $1$ and particle $2$ are not at rest to each other, there is no value of "distance between them" defined to begin with. Indeed, not even the ping durations of either of these two particles with respect to other remain constant; i.e. unlike the two "ends of a train" which were described in this/my answer to the question about "an accelerating and shrinking train" (PSE/a/63110).

Community
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user12262
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"Length contraction", is the measure of a umproper length of an object, in a inertial frame $O'$ which has a relative velocity to a inertial frame $O$, where the coordinates of the object are fixed in this frame $O$

More precisely, one measures a proper time $\Delta \tau'$ in $O'$ (so at $x'=0$), which corresponds to the synchronisation with the beginning and the end of the object, and this proper time is traduced into an unproper length : $L' = v \Delta \tau'$. Lorentz transformation implies $c^2\Delta \tau' = c^2 (\dfrac{L}{v})^2 - L^2 $. The result is then a "length contration" : $L' = \dfrac{L}{\gamma(v)} =\sqrt{1-\dfrac{v^2}{c^2}} L$

So, first, unless you modify your question, you are not in this situation, because you have not $2$ different inertial frames moving with a constant speed relatively to each other. Secondly, as indicated above, what is measuring is not directly a length, but a proper time in the moving (relatively to the object) frame. This proper time is traduced into a unproper length as indicated above.

Trimok
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  • That the particles' comoving frames are not inertial shouldn't make any problem. Just let the particles stop accelerating at $t=t_1$ in laboratory frame, and then we'll have two inertial frames to consider. – Ruslan Aug 01 '14 at 10:40
  • The particle comoving frames, for a precise time $t$, are, by definition, inertial. There is in fact, one comoving inertial frame by time $t$, which moves with the velocity $v(t)$ relatively to the original frame. If the particles have the same velocity, then, at each time, they have the same comoving inertial frame. – Trimok Aug 01 '14 at 10:51
  • OK, but now they are moving with the constant non-zero velocity with respect to lab frame, but still distance between them is not contracted. Why? This is the core of the question. – Ruslan Aug 01 '14 at 10:54
  • The "length contraction" protocol does not apply, because, if it is true, that the particles have zero velocity, at some time $t$, in a comoving inertial reference frame $C_t$, the problem is that this comoving inertial reference frame $C_t$ changes for each $t$. So you are not able to consider $2$ reference frames with constant relative velocity, during some amount of time. – Trimok Aug 01 '14 at 10:59
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To make the example (and the question) clear, the particles should be on different frames, where when the 2 frames are at rest and coincide, the particles have the given separation.

Then when one frame moves relatively (or accelerates) to the other with a given velocity (what velocity exactly?) the Lorentz-contraction appears but only with respect to the frames each particle is in.

Nikos M.
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