# B Bell's Spaceship Paradox & Length Contraction

#### stevendaryl

Staff Emeritus
Science Advisor
Put a thread between two cars and accelerate the same and you expect the string to remain unbroken. Take your foot off the (trailing car) gas a small amount and you expect the string to pull the trailing car a small amount by tensile strength. Take your foot of a lot, and you expect the string to break, as an inadequate tow rope.
But the difference with the Bell Rocket thought experiment is that even if the two cars/rockets are accelerating identically, the string will break. The only way for the string not to break is if the car/rocket in the rear is accelerating a tiny bit more than the car/rocket in the front.

#### votingmachine

I see the clarifications which correct my understanding. The thought experiment is defined with a perfect acceleration control. Any other consideration is a separate thought experiment.

It is just that 1xG is just so slow. Say that G is 10 m/s2. After 3 million seconds the velocity is 30 million m/s. The speed of light roughly 300 million m/s. So after about 5 weeks of acceleration, the velocity is approaching one tenth light speed.

At 1/10th light speed, the "length contraction" is 0.5% (200 meter thread now at 199 meter length). If across the 5 weeks the trailing spaceship was moved 1 meter closer, the thread is under no additional tension.

The two ships have traveled 5x10exp12 meters. So if all the "extra" acceleration was on the trailing ship. it would have traveled 5,000,000,000,001 meters while the lead ship traveled 5,000,000,000,000 meters, over that 5 weeks.

I know it gets worse, as the next 5 weeks get a little closer to light speed. And gamma is not a linear function. I just find the specification of perfect acceleration measurement and control somewhat bothersome. I can see that if you specified a small enough spaceship, then you might even be able to apply the Heisenburg uncertainty principle to show that the thread would not break. Say the ships were atom sized ... one of the posts mentioned we should think of them as precisely located points ...

#### Ibix

Science Advisor
So what? Just keep accelerating and eventually length contraction will be significant above the noise in your acceleration profile (which you can keep under control with an accelerometer and appropriate averaging, even if not "naturally" by precise engineering).

You can't duck the implications by deciding it's too costly to implement a test and giving up.

#### PeterDonis

Mentor
It is just that 1xG is just so slow.
First, so what? The math doesn't care.

Second, who says the acceleration has to be 1 G? It's a highly idealized thought experiment. You can run it with whatever acceleration you like.

I just find the specification of perfect acceleration measurement and control somewhat bothersome.
You are welcome to start a separate thread if you want to discuss a different thought experiment. This thread's topic is the one Bell proposed.

#### PeterDonis

Mentor
I can see that if you specified a small enough spaceship, then you might even be able to apply the Heisenburg uncertainty principle to show that the thread would not break.
Bell's thought experiment is formulated in classical SR. If you want to drag in quantum mechanics, that's yet another different thought experiment. Which should be discussed in a separate thread if you really want to discuss it. Please stop hijacking this thread with posts about different scenarios than the one we are discussing here.

#### PeterDonis

Mentor
Moderator's note: Some off topic posts and responses to them have been deleted.

#### PeterDonis

Mentor
In the crew's reference frame, both ships will not have the same acceleration. They will feel the same acceleration at different times.
This is not correct. The crews of both ships feel the same proper acceleration at all times. Since the proper acceleration does not vary with time, the fact that, once the ships start accelerating, the instantaneous rest frames of each ship have different simultaneity surfaces, is irrelevant.

#### sweet springs

In the crew's reference frame, both ships will not have the same acceleration. They will feel the same acceleration at different times.
All the time each crew feels independent constant proper gravity force that is caused by proper acceleration of ship according to where he/she is. The more rear ship or lower deck he/she is, the more proper gravity or acceleration works on him/her. I said it in condition that distances between the ships in a row remain constant for observer in IFR and the ships are made Born rigid.

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#### liuxinhua

The spaceship is rest on F′,F′maybe a Rindler_coordinates.
I apologize for talking about two different scenarios without making clear which one I'm talking about. Yes, in Bell's scenario, the distance between the two rockets remains constant in frame F. I'm talking about a different scenario in which the distance remains constant, as measured in frame F′, the frame in which the rockets are momentarily at rest.
I have done a detailed calculation of the speed and acceleration of the two spacecraft.
But I don't know how to paste a picture in the forum.

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#### stevendaryl

Staff Emeritus
Science Advisor
The spaceship is rest on F′,F′maybe a Rindler_coordinates.

I have done a detailed calculation of the speed and acceleration of the two spacecraft.
But I don't know how to paste a picture in the forum.
There should be a button to the lower left called "UPLOAD".

#### liuxinhua

Each spaceship has its own original time τ.
For spaceship A at its every original time, such at time τi, there exists an inertial reference frame Ki spaceship A and spaceship B is rest in Ki. When the original time of spaceship A is τi , corresponding time in Ki is ti.
For spaceship A at time τj, there exists an inertial reference frame Kj spaceship A and spaceship B is rest in K. When the original time of spaceship A isτj , corresponding time in Kj is tj.
Measured in Ki at time ti and measured in Kj at time tj, the acceleration of spaceship A is a constant.
The distance of spaceship A and spaceship B is l0 measured in Kj at time tj.

Thank for stevendaryl.

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#### Mentz114

Gold Member
You and two identical spaceships are all at rest with respect to each other. You note that the two engines start up at the same time, and the thrust curve and acceleration profile of both spaceships are identical. As the ships pick up speed, would you measure the ships to be shorter than their rest length?
These diagrams show that from any instantaneously comoving frame (ICF) on any ship the trailing and leading ships are receeding and the strings must break .
The first diagram is at time zero with all clocks synchronised at $\tau=0$.
The next diagram shows the frame that is comoving with the leftmost clock at its time ≈ 8. This is the top white line. This frame is also comoving with each ship at $\tau$≈8. For each white line the ships to the left of the blue square have negative velocity and those to the right have positive velocity

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#### David Lewis

…shouldn't we consider an engine that applies the thrust uniformly across the entire ship??
I don't think so in this case. The engine (as I've drawn it) applies force just to the rear of the ship. The ship will compress slightly due to classical mechanical deformation but I am ignoring this and for simplicity just focusing on relativistic effects.

The only way for the string not to break is if the car/rocket in the rear is accelerating a tiny bit more than the car/rocket in the front.
You're probably right. Consider, however, that the acceleration of both ships is the same. For the observer (who was initially at rest with respect to the ships) both ships started moving at the same time. For the men in the ship, the front ship got a head start.

#### sweet springs

You're probably right. Consider, however, that the acceleration of both ships is the same. For the observer (who was initially at rest with respect to the ships) both ships started moving at the same time. For the men in the ship, the front ship got a head start.
Same, you say, should be investigated more in detail.
The two engines are the same product and in the same condition. The two pilots are well trained to keep the same starting manual.
The two pilots and a commander on the Earth share the same IFR when the pilots fire engines of the rockets on the Earth or staying still in space with the Earth. The rockets start at the same time for all the three. No head start.

For the commander all the things of the rockets keep same during the flight, constant distance, the same speeds and the same rocket lengths at his Earth time, etc. However, though the same start,
In the front rocket FR, the rear rocket's engine is in sooner phase in accerelatin flight manual than his own. "The rear pilot is reading chapter I of manual though I am reading chapter II"
In the rear rocket FR, the front rocket's engine is in later phase in acceleration flight manual than his own. "The front pilot is reading chapter II of manual though I am reading chapter I"
The two pilots share the same judgement in synchronisity, "the more front, the more future". They observe that the distance between the two strats increasing. So the thead is torn. The distance would be shortened in latter phase for reducing power for inertial flight in space. After all the starting procedures are completed by the with engines cut, the distance of rockets is the same initial value for the three.

In order that the thread is not torn apart the front pilot should reduce power, the rear pilot should increase power or the both is required. The different manual should be given to the pilots for the mission of "no cut of thread during the flight".

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#### stevendaryl

Staff Emeritus
Science Advisor
You're probably right. Consider, however, that the acceleration of both ships is the same. For the observer (who was initially at rest with respect to the ships) both ships started moving at the same time. For the men in the ship, the front ship got a head start.
Not just a head start. From the point of view of the passengers on the ships, the distance between the ships keeps growing. So unless the string can stretch infinitely far, it will break.

Let $L$ be the distance between the two ships as measured in the "launch" frame (where they are initially at rest). Now, wait a while until the ships are traveling at speed $v$. Now, transform to a frame in which the rear ship is momentarily at rest. In this frame:
• The rear ship is (momentarily) at rest.
• The distance between the ships is greater than $\gamma L$
• The front ship is getting farther away from the rear ship.
So whatever is the maximum the string can stretch, eventually $\gamma L$ will get bigger than that, and the string will break.

#### stevendaryl

Staff Emeritus
Science Advisor
Another way to look at the string breaking is in terms of the Rindler horizon. Consider a rocket ship that is traveling according to the trajectory:

$x_{rocket} = (x_0 - \frac{c^2}{g}) + \sqrt{\frac{c^4}{g^2} + c^2 t^2}$

(that's constant proper acceleration $g$ starting at rest at $x=x_0$ at time $t=0$)

Then there is a second trajectory behind the first:

$x_{horizon} = x_0 + ct$

For all $t$, $x_{horizon}(t) < x_{rocket}(t)$.

If you have an observer whose trajectory is such that $x_{observer}(t) < x_{horizon}(t)$, then there is no way for that observer to send a signal to the rocket. That's sort of obvious, because $x_{horizon}$ moves at speed $c$, so something that gets behind the horizon will never be able to catch up with it again.

In the two rocket case, the rear rocket will fall below the Rindler horizon of the front rocket.

#### David Lewis

In the initial launch frame, would it be accurate to say the tension in the thread increases as the ships pick up speed because the molecules of which the thread is composed flatten in the direction of motion?

#### PeterDonis

Mentor
In the initial launch frame, would it be accurate to say the tension in the thread increases as the ships pick up speed because the molecules of which the thread is composed flatten in the direction of motion?
No. The shape of the molecules plays no role, since, as pointed out in post #55, we are talking about SR, not quantum mechanics. In SR, the thread is assumed to be a continuous line extending along the direction of motion--a real thread has a thickness, but the thread's extent in the directions perpendicular to the direction of motion is irrelevant to the SR analysis.

#### A.T.

Science Advisor
In the initial launch frame, would it be accurate to say the tension in the thread increases as the ships pick up speed because the molecules of which the thread is composed flatten in the direction of motion?
The fields that hold the string atoms together are contracted in the initial launch frame, to they produce more attractive force, than in an identical string at rest in the that frame, despite the fact that both strings remain at equal length in the frame.

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#### vanhees71

Science Advisor
Gold Member
You can find some calculations concerning the space-ship paradox in my SRT article here (p. 33):

#### David Lewis

No. The shape of the molecules plays no role, since, as pointed out in post #55, we are talking about SR, not quantum mechanics.
Then I don't understand how an object can get shorter.

#### pervect

Staff Emeritus
Science Advisor
Then I don't understand how an object can get shorter. View attachment 243696
If molecules had a shape, they'd get shorter. In the world of quantum mechanics, though, it's not clear if molecules actually have shapes or not, at least not to me. I would say that molecules have wavefunctions which occupy a non-physical "configuration space", with 3 dimensions for every particle in the molecule (presumably these particles are atoms, but you could break the atoms down into more particles). That's not really a "shape" as far as I am concerned.

Most treatments of introductory QM treat single particle systems, where the wavfunction does occupy normal space. It's when one considers multi-particle systems that one gets into the issue of the wavefunctions not occupying physical space.

But it's much simpler to keep the arcane aspects of QM out of the discussion, which is what the original point was.

#### PeterDonis

Mentor
Then I don't understand how an object can get shorter.
Because that's how the geometry of spacetime works. The SR treatment of the Bell spaceship paradox does not make any hypothesis about the internal structure of the object. It just explores the consequences of the stated conditions, given the geometry of Minkowski spacetime. This geometry puts constraints on any model of an object's internal structure; but it doesn't tell you anything specific about that internal structure.

#### PeterDonis

Mentor
I don't understand how an object can get shorter.
As far as an "object" describable by classical physics is concerned, Lorentz showed in the 1890s (IIRC) that any object made of electric charges bound together by electromagnetic fields would exhibit length contraction in a frame in which it was moving.

#### A.T.

Science Advisor

Then I don't understand how an object can get shorter.
The string in Bell's scenario doesn't get shorter, so the contracted binding EM fields have to span the same distances. Hence the tension. To avoid the complications of QM don't go down to the atomic level, but instead consider the contracting links of a chain that is forced to keep a constant length.

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