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The first part of my question is meant to confirm my understanding of length contraction in the context of the following simple thought experiment:

Imagine three particles O(the observer), A and B in a one dimensional space. Initially, they are all stationary with respect to each other. Let the distance between the particle A and B, as observed by O (and consequently, as observed by A or B) be $L$.

  1. Now let's imagine that the particle O starts to move with the velocity $v$. In the observer O's frame of reference, the distance between A and B should now be measured as $L$$\sqrt{1-\frac{v^2}{c^2}}$. The distance between A and B as observed by A and B should of course still be $L$.
  2. Now let's imagine the observer O is stationary, as in the initial setup, and at a certain time instant $t$ in the observer O's frame of reference, both particles A and B start moving from rest with the velocity $v$ at once. My understanding leads me to conclude that distance between A and B as measured by observer O is still $L$. After all, two uniformly accelerating point objects in observer's frame of reference can not come closer to each other. The distance between A and B, as observed by A or B, Would be $L$$\sqrt{1-\frac{v^2}{c^2}}$, wouldn't it? (Please confirm).

This leads us to the second part of the question. The particles A and B were unconnected in the above example, which is why the distance between A and B as measured by O remained $L$. What if the particles were connected?

Common explanations on the internet propound that an object (which may be approximated as a collection of some point particles bound by some force that maintains the distance between them) that begins to accelerate in a frame of reference will be seen by the observer to contract in its length.

What is the reason that the 'observed distance' between two connected particles may undergo length contraction, but the 'observed distance' between two unconnected particles may remain the same?

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"both particles A and B start moving from rest with the velocity v at once" implies event1: A's acceleration and event2: B's acceleration are simultaneous in O's frame $\Rightarrow$ In a frame moving with velocity $v$, these events were not simultaneous (by the relativity of simultaneity), so in A and B's frame, their separation can change since one stated to accelerate earlier than the other. Since length measured by O is Lorentz contracted, this means in A and B's frame, their separation has increased to $\frac{L}{\sqrt(1-v^2/c^2)}$.

If the particle had been connected by a string, results won't change, the string is made of, say 10 particles, then all those particles will start to accelerate at different times by relativity of simultaneity, so by definition, the string would stretch and produce a required increase in length.

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  • $\begingroup$ But that would mean that an accelerating object will find itself increasing in length. $\endgroup$ – Kraken Feb 9 at 14:19
  • $\begingroup$ Yes, but what an accelerating body sees is out of the purview of Special relativity, so we cannot analyze it easily using Special relativity. $\endgroup$ – Kutsit Feb 9 at 15:16
  • $\begingroup$ So what happens when the accelerating body stops accelerating? $\endgroup$ – Kraken Feb 9 at 16:23
  • $\begingroup$ once it stops accelerating, we can apply SL to analyze things as usual $\endgroup$ – Kutsit Feb 9 at 18:10
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All of the following holds whether or not $A$ and $B$ are connected:

Say the particles are lined up in the order $O$, $A$, $B$ and are stationary in what I'll call the "lab frame".

1) Suppose $O$ starts moving to the left in the lab frame. Then:

1a) In $O$'s new frame: $O$ is stationary. $A$ started moving to the right at (say) 12PM and $B$ started moving to the right at (say) 12:01PM (both at the same velocity).

Between 12PM and 12:01 PM, only $A$ was moving, so the distance between $A$ and $B$ was shrinking, say from $L$ to $L'$. Now that they're both moving, the distance remains $L'$.

If $A$ and $B$ are connected by a rod, the length of that rod has shrunk from $L$ to $L'$.

1b) In the lab frame: $O$ is moving. $A$ and $B$ are still stationary. Obviously the distance between them has not changed. If there's a rod connecting them, its length remains as before.

2) Suppose $A$ and $B$ start moving to the right, simultaneously in the lab frame.

2a) In the lab frame: the distance between $A$ and $B$ has (obviously) not changed. Therefore neither has the length of any rod that might be connecting them.

2b) In the new frame of $A$ and $B$: $A$ and $B$ used to be moving leftward. At (say) 12PM $B$ stopped moving, and at (say) 12:01PM, $A$ stopped moving. Between noon and 12:01, the distance between them grew, say from $L$ to $L''$. Now that they're both moving, the distance remains at $L''$. If there's a rod connecting them, it has stretched.

Note that in both cases 1) and 2), once things are underway, $O$ sees the rod connecting $A$ with $B$ as shorter than $A$ and $B$ do.

As to your question in comments about whether an accelerating object finds itself increasing in length, the answer depends entirely on the nature of the acceleration. If, at each instant in the rod's instantaneous frame, all points of the rod are accelerated identically, then the length of the rod won't change. If all points of the rod are accelerated identically in some other frame, then they won't be accelerated identically in the rod's frame. And of course all of this is very easily analyzed with special relativity, so the comment saying otherwise is incorrect.

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