Following is only the problem definition of this FAQ
Next are my comments of this FAQ.
For the explanation of this FAQ see the original FAQ.
[Relativity FAQ] [Copyright]
Original by Michael Weiss 1995
Bell's Spaceship Paradox
John Bell described this Special Relativity paradox in the essay, "How
to teach special relativity", in his collection "Speakable and
Unspeakable in Quantum Mechanics." He did not originate the puzzle,
but we'll call it Bell's Spaceship Paradox.
To begin, a statement of the paradox and if you notice some
ambiguities in my formulation, that's the point! (That's
always the point in SR paradoxes.) Bell asks us to consider
two rocket ships, each accelerating at the same constant rate, one
chasing the other. The ships start out at rest in some coordinate
system (the "labframe"). Since they have the same acceleration,
their speeds should be equal at all times (relative to the labframe)
and so they should stay a constant distance apart (in the labframe).
But after a time they will acquire a large velocity, and so the
distance between them should suffer Lorentz contraction. Which is it?
The last part of the previous section is not very clear. It does not clearly describe all what is observed in the "lab frame" or rest frame.
Suppose the two spaceships start with no distance between (Nose to Back) then what will happen? Will the distance stay equal and stay zero? Is there any where Lorentz contraction involved? Are there the space ships Lorentz contracted?
Clearly what is the paradox (error) in the rest frame.
For the explanation, see the original FAQ by Michael Weiss
Bell's Spaceship Paradox  Explanation 1
In order to understand the paradox it is important to consider the paradox from two different points of view:
of an observer at rest in the rest frame
of a moving observer at in a moving frame ie the frame of the space ship
First what is the situation?

Starting point of the paradox are two space ships at rest in the rest frame, one behind each other.
spaceship 2 spaceship 1
<><>
> Direction of movement
B2 F2B1 F1
At a certain moment, both space ships start to with the same acceleration. Because they both have the same acceleration they will also both at each instant have the same speed v in the rest frame.
Because their speeds increase the length of each spaceship will contract and the distance between the two space ships will increase. All measured in the rest frame.
spaceship 2 spaceship 1
<> <>
> Direction of movement
B2 F2 B1 F1

What happens in the moving frame ?

In the moving frame you use a moving clock and a moving clock runs slower. If you do that and you measure the length of each space ship you will see that the length of each spaceship has not changed.
On the other hand if you measure the distance between the two space ships than that distance
has increased.

Bell's Spaceship Paradox  Explanation 2
The main idea behind the above exercise is to keep the distance between the back of spaceship 1 (the point B1) and the back of spaceship 2 (the point B2) in the rest frame constant. This is performed with a clock in the rest frame. If you there after use a moving clock (which runs slower) in order to measure that same distance, ofcourse than that distance becomes longer.
A different example is the following: Instead that the points B2 and B1 have the same speed do the experiment such that the front of spaceship 2 (Point F2) and the back of spaceship 1 (the point B1) have the same speed. Physically what you do is you push spaceship 1 and you pull spaceship 2.
The following sketch shows this:
spaceship 2 spaceship 1
<><>
> Direction of movement
B2 F2B1 F1
First what implies to push
Suppose your spaceship or rod is 3 light seconds large.
What happens if you push against that rod for a period of 3 seconds. That means for a period of 3 seconds acceleration is greater than 0 (and constant).
For a description See: Changing Length part 4 Description of an experiment.
What this experiment (on paper) shows is that after 6 seconds the whole rod has a constant speed v to the right, but and that is important the whole rod is contracted. This seems to be in agreement with SR.
Now let us study when you want to pull a rod
What happens if you pull a rod for a period of 3 seconds. That means for a period of 3 seconds acceleration is greater than 0 (and constant), but now in the opposite direction as before.
Again for a description See: Changing Length part 4 Description of an experiment.
What the second part of this experiment (on paper) shows is that after 6 seconds the whole rod has a constant speed v to the left, but and that is important the whole rod is expanded. This is not in agreement with SR. In fact the rod should first become equal to its original length and than even shorter i.e. the rod should contract.
Reflection
A rod that is pulled is physical contracted. A rod that is pushed is physical expanded.
If you push/pull a rod from the middle (which has the speed v), than the total length of the rod stays constant in the rest frame.
You can argue that the contraction (keeping v constant) of a pulled rod is of temporary nature and that the rod will expand partly or completely to its full length.
However is this is the case so will a pushed rod.
The result will be that the total length of the rod stays physical constant in the rest frame, which is not in agreement with SR.
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Created: 18 March 2002
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