# B Bell's Spaceship Paradox & Length Contraction

#### Ibix

Science Advisor
I didn't think of Bell's scenario as implying that the ships themselves had rockets on both ends.
I agree - it's just my writing that was unclear. Nugatory apparently managed to puzzle out what I meant, but I've slightly edited my post to be clearer (I hope).

• Nelli and FactChecker

#### FactChecker

Science Advisor
Gold Member
2018 Award
So a single long rocket which accelerates with no internal stress is not experiencing the same acceleration at the front and the back. I'll buy that since the length must change. (Although I will have to think about whose reference frame I am talking about.)

#### Ibix

Science Advisor
So a single long rocket which accelerates with no internal stress is not experiencing the same acceleration at the front and the back. I'll buy that since the length must change. (Although I will have to think about whose reference frame I am talking about.)
Some care is needed with "no internal stress". A rocket is usually under compressive stress because of the rocket motor at the back. I suspect what you mean is under an unchanging strain, in the sense that a crewmember with a ruler will measure the rocket's length to be constant (transients when the engine is initially lit aside).

You also need to be careful about what you mean by acceleration - proper or coordinate? The proper acceleration does not change along the rocket. The coordinate acceleration does, in most frames.

The difference between the long rocket and Bell is in the startup. Bell's rockets start simultaneously in the initial rest frame. The nose of the long rocket starts moving when shock waves from the ignition of the rocket in the tail reach it - later than Bell's lead rocket starts moving. Hence the different behaviour later on, despite all proper acceleration gauges in both setups showing the same thing in steady state.

• FactChecker

#### stevendaryl

Staff Emeritus
Science Advisor
You also need to be careful about what you mean by acceleration - proper or coordinate? The proper acceleration does not change along the rocket.
If we're talking about a normal rocket, propelled from the rear, the proper acceleration will vary from back to front, with the acceleration being greater in the rear.

It's a little bit misleading to attribute this to the fact that it is pushed from the rear. The same thing would be true if the rocket were pulled from the front. Immediately after launching, the rocket is compressed when pushed from the rear and stretched when pulled from the front, but that's a transient effect. After the rocket reaches an equilibrium length, the proper acceleration of the rear will be greater than the proper acceleration of the front in either case.

• Ibix

#### stevendaryl

Staff Emeritus
Science Advisor
If we're talking about a normal rocket, propelled from the rear, the proper acceleration will vary from back to front, with the acceleration being greater in the rear.

It's a little bit misleading to attribute this to the fact that it is pushed from the rear. The same thing would be true if the rocket were pulled from the front. Immediately after launching, the rocket is compressed when pushed from the rear and stretched when pulled from the front, but that's a transient effect. After the rocket reaches an equilibrium length, the proper acceleration of the rear will be greater than the proper acceleration of the front in either case.
Here's a way to see that this must be true. Suppose that the rocket's length soon after launching and after it has reached its equilibrium length is $L$ according to the launch frame. Now, wait until the rocket is moving relativistically at some speed $v$. Its length when measured from the launch frame will be $L/\gamma$, where $\gamma = 1/sqrt{1-\frac{v^2}{c^2})$. If you think about it, the only way for the length of an object to decrease is if the rear gets closer to the front. That's only possible if (according to the launch frame) the rear is traveling slightly faster than the front. That implies that the coordinate acceleration of the rear is always a little greater than the coordinate acceleration of the front.

#### Ibix

Science Advisor
If we're talking about a normal rocket, propelled from the rear, the proper acceleration will vary from back to front, with the acceleration being greater in the rear.
Ack! You're right. In order for the length to remain invariant the worldlines of the nose and tail, which are hyperbolae, must have a common focus. Which means different proper accelerations.
It's a little bit misleading to attribute this to the fact that it is pushed from the rear.
That wasn't my intention. I was pointing out that the rocket isn't unstressed, and that the two ends of the rocket don't start moving at the same time in the initial rest frame, unlike Bell's rockets. I agree that the same is true of rockets "hanging" from engines near their nose, and didn't intend to imply otherwise.

I was also claiming that the difference between Bell's spaceships and a long rocket is just the timing of the movement starts. That was wrong, as you pointed out. They are different, but the proper accelerations are also different, contrary to what I said above.

#### FactChecker

Science Advisor
Gold Member
2018 Award
Here's a way to see that this must be true. Suppose that the rocket's length soon after launching and after it has reached its equilibrium length is $L$ according to the launch frame. Now, wait until the rocket is moving relativistically at some speed $v$. Its length when measured from the launch frame will be $L/\gamma$, where $\gamma = 1/sqrt{1-\frac{v^2}{c^2})$. If you think about it, the only way for the length of an object to decrease is if the rear gets closer to the front. That's only possible if (according to the launch frame) the rear is traveling slightly faster than the front. That implies that the coordinate acceleration of the rear is always a little greater than the coordinate acceleration of the front.
That is how I imagined it. Which is why I don't understand why having a string in the middle changes so that the string is snapped. But this is probably where I have to put my trust more in the diagrams and figure it out from there.

#### PeterDonis

Mentor
That is how I imagined it.
How you imagined which scenario?

If you have a single long rocket that doesn't stretch, that's not the same scenario as the Bell spaceship paradox. And a string attached at both ends (front and rear) of such a long rocket would not stretch, but that's because such a string is not being subjected to the same motion as the string in the Bell spaceship paradox.

To realize a scenario like the Bell spaceship paradox with a single long rocket, you would have to have rocket engines at both ends, and both providing the same proper acceleration. In that scenario (which is different from the one @stevendaryl and @Ibix have been discussing in their last few posts), the spaceship would stretch (more precisely, internal stresses in the ship would continuously increase) and eventually break, just as the thread does in the standard Bell spaceship paradox. In this scenario, the length of the spaceship as seen in the original rest frame would not decrease; it would stay the same (which is why the word "stretch" can be misleading--it's better to not use length-related terminology at all since length is frame-dependent).

• Ibix

#### Ibix

Science Advisor
That is how I imagined it. Which is why I don't understand why having a string in the middle changes so that the string is snapped. But this is probably where I have to put my trust more in the diagrams and figure it out from there.
Draw a spacetime diagram in the initial rest frame. Bell's ships follow identical hyperbolae, one simply offset to the right. The nose and tail of the long rocket follow hyperbolae with a common focus. If you pick the back hyperbola to have its focus being the origin of coordinates, you will find that it is invariant under Lorentz transform (i.e., points on the hyperbola are mapped onto points on the same hyperbola). In the long rocket case the nose's path has the same focus so has the same property. So nothing changes in the rocket frame. But in Bell's case the front rocket's hyperbola has a different focus and it doesn't map onto itself under Lorentz transform (or, more precisely, there's no choice of origin such that both front and rear ships' hyperbolic worldlines map onto themselves). So the front ship moves compared to the back rocket.

• FactChecker

#### stevendaryl

Staff Emeritus
Science Advisor
That is how I imagined it. Which is why I don't understand why having a string in the middle changes so that the string is snapped. But this is probably where I have to put my trust more in the diagrams and figure it out from there.
You're mixing up two different scenarios: Let $F$ be the initial launch frame. Let $L$ be the initial distance between the rockets. Let the two rockets accelerate for a while until the rear rocket is traveling at speed $v$ relative to $F$. Let $F'$ be the frame in which the rear rocket is momentarily at rest. Then we have two different scenarios:
1. The distance between the rockets, as measured in frame $F$, is still $L$. Then as measured in frame $F'$, the distance between the rockets will be larger than $L$, and the string will break.
2. The distance between the rockets, as measured in frame $F'$, is still $L$. Then as measured in frame $F$, the distance between the rockets will be contracted, and the string will not break.
What's important for the string to break or not is the distance between the rockets, as measured in frame $F'$.

In scenario 1, the front rocket and rear rocket experience the same (proper) acceleration. In scenario 2, the rear rocket experiences greater proper acceleration.

• FactChecker

#### jartsa

That implies that the coordinate acceleration of the rear is always a little greater than the coordinate acceleration of the front.
The rear would get ahead of the front after some finite time, if the coordinate acceleration of the rear was always a little greater than the coordinate acceleration of the front.

#### stevendaryl

Staff Emeritus
Science Advisor
The rear would get ahead of the front after some finite time, if the coordinate acceleration of the rear was always a little greater than the coordinate acceleration of the front.
No. The distance between front and back is $L \sqrt{1-\frac{v^2}{c^2}}$. That gets smaller and smaller, but it never goes to zero (and certainly never becomes negative).

#### Ibix

Science Advisor
The rear would get ahead of the front after some finite time, if the coordinate acceleration of the rear was always a little greater than the coordinate acceleration of the front.
Draw a spacetime diagram and add hyperbolae with their foci at the origin. The distance between any two timelike hyperbolae that curve in the same direction remains constant in their instantaneous rest frames. The coordinate distance falls and the coordinate accelerations are different, yet they never cross.

#### jartsa

No. The distance between front and back is $L \sqrt{1-\frac{v^2}{c^2}}$. That gets smaller and smaller, but it never goes to zero (and certainly never becomes negative).

I know the rear of an accelerating rod does not pass the front of the rod. And from that I (correctly) conclude that at some time the coordinate acceleration of the rear must get smaller than the coordinate acceleration of the front - to avoid the rear passing the front.

If Bob moves faster than Alice and Bob also accelerates more than Alice, then Bob will go past Alice at some time.

#### sweet springs

The rear would get ahead of the front after some finite time, if the coordinate acceleration of the rear was always a little greater than the coordinate acceleration of the front.
This is in accordance with Rindler coordinate where lower the position, greater the proper gravity force.

#### FactChecker

Science Advisor
Gold Member
2018 Award
I know the rear of an accelerating rod does not pass the front of the rod. And from that I (correctly) conclude that at some time the coordinate acceleration of the rear must get smaller than the coordinate acceleration of the front - to avoid the rear passing the front.

If Bob moves faster than Alice and Bob also accelerates more than Alice, then Bob will go past Alice at some time.
There are all sorts of space -time distortions going on during acceleration that make one's intuition very treacherous. IMO, the key to keeping things straight is to use the Minkowski diagram.

#### stevendaryl

Staff Emeritus
Science Advisor
If Bob moves faster than Alice and Bob also accelerates more than Alice, at some time Bob will go past Alice.
Ah! You're actually right. The relationship between the coordinate acceleration and the proper acceleration is a little complicated. Assuming that Alice and Bob are traveling so that they are a constant distance from each other, according to their own instantaneous reference frame, and so that their proper acceleration is constant:
1. Alice is always ahead of Bob.
2. Bob's velocity is always greater than Alice's. (according to the initial launch frame)
3. Bob's proper acceleration is greater than Alice's
4. Bob's coordinate acceleration starts off greater than Alice's (according to the initial launch frame)
5. But eventually, Alice's coordinate acceleration gets larger than Bob's (according to the initial launch frame)
It's not immediately obvious how to reconcile #3 with #5. The formula for constant proper acceleration is this (choosing the zero of the x-coordinate to make the formula simpler)

$x = \sqrt{c^2 t^2 + \frac{c^4}{g^2}} \equiv \sqrt{c^2 t^2 + X^2}$

where $g$ is the proper acceleration and $X = \frac{c^2}{g}$ is the initial location. It's immediate from this that Bob, whose initial location is behind Alice's has a larger proper acceleration: $X_a > X_b$ so $g_a < g_b$. If we take some derivatives, we find:
• $x = \sqrt{c^2 t^2 + \frac{c^4}{g^2}}$
• When $t$ is small, $x \approx \frac{c^2}{g} + \frac{1}{2} g t^2$
• When $t$ is large, $x \approx ct$
• $v = \frac{c^2 t}{x}$
• When $t$ is small, $v \approx g t$.
• When $t$ is large, $v \approx c$.
• $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{gx}{c^2}$
• When $t$ is small, $\gamma \approx 1$
• When $t$ is large, $\gamma \approx \frac{gt}{c}$
• $a = \frac{c^6}{x^3 g^2} = \frac{g}{\gamma^3}$
• When $t$ is small, $a \approx g$
• When $t$ is large, $a \approx \frac{c^3}{g^2 t^3}$
Alice has a smaller value of $g$, but also a smaller value of $\gamma$, so the ratio $a = \frac{g}{\gamma^3}$ is smaller than Bob's when $\gamma \approx 1$, but gets larger than Bob's when $\gamma$ gets large.

#### David Lewis

The distance between front and back is L (1−v2/c2)1/2.
Let L be the initial distance between the rockets.
L is the distance between the rockets, if I understand correctly.
L stays the same in frame F.

#### Attachments

• 26.7 KB Views: 198

#### stevendaryl

Staff Emeritus
Science Advisor
L is the distance between the rockets, if I understand correctly.
L stays the same in frame F.
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.

#### votingmachine

I've asked this before and had it answered, but I'm not satisfied yet ... I expect I have to work thru the math:

The tensile strength of the thread is neglected in this thought experiments. If there is tension in the thread, it will not just pull itself apart, but it will pull on the two spaceship masses. My intuition is that connecting the ships into that single traveling system, with the thread, they just move together. If the thread was an unbreakable steel bar, with the two rockets part of a single ship, it would be expected to stay together. The fragility of the thread is always stressed as important in the conditions, but even a small force seems sufficient to move the ships together.

I can theoretically tow an aircraft carrier with a thread, as long as my force is lower than the tensile strength of the thread. The length contraction seems so gradual that I (intuitively) expect it to always be below that tensile strength.

If it was a single ship, with a rocket at the front and a rocket at the rear, would it snap in half? Does it matter that the tensile strength force changes either is accelerating the trailing ship, or braking the leading ship? That sort of changes the definition of the acceleration force experienced ... but not by much. It seems that the thought experiment relies on an extraordinarily precise acceleration definition. How many zeroes are needed in the in 1.0000...000 g? It seems like the ships will just be very gently pulled together.

And if the force transferred by the string must be omitted from the trailing rocket, then it is also obvious that an external viewer would see a rocket in front with more power pulling a slower rocket by a thread, and then the thread breaking is obvious ... it was not a strong enough tow rope. The viewer is puzzled by the string breaking until told that the rockets will be adjusted to prevent the string forces from changing the acceleration. 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.

The thought experiment seems set up to require that the rockets can overly precisely control acceleration. If I am holding a small helium balloon on a thread and descend stairs, I don't feel that "lift". But it is there. If the balloon remains at a fixed height, the thread breaks. But that never happens ... the balloon is towed.

#### jbriggs444

Science Advisor
Homework Helper
The tensile strength of the thread is neglected in this thought experiments.
The tensile strength is assumed low enough to have negligible effect on the motion of the spacecraft. It is a thought experiment. Imaginary string with low tensile strength and imaginary motors that are both powerful and identical are as cheap as a piece of paper and a number two pencil.

If the tensile strength is large enough to be significant compared to the rocket motor thrust then it is a different thought experiment.

• Ibix

#### Justin Hunt

While the picture shows the rocket engines being located in the back, the thought experiment question makes no such designation. Considering this is a thought experiment, shouldn't we consider an engine that applies the thrust uniformly across the entire ship??

Also, is it from the perspective from the initial state that the distance between the ships is constant? Wouldn't this mean that from the perspective from someone on one of the ships, the distance between the ships is increasing? A stationary observer would see the spaceships being length contracted and the space between the ships being length contracted as well in order for the people in the ships to see no difference. Isn't this paradox just an impossibility?

#### PeterDonis

Mentor
If it was a single ship, with a rocket at the front and a rocket at the rear, would it snap in half?
If the proper acceleration at both ends was the same, yes.

All of the considerations you are raising are reasons why, in an actual scenario, the proper acceleration at both ends would not be the same. But that means, as @jbriggs444 has pointed out, that you're now talking about a different thought experiment. The point of Bell's spaceship paradox is not to describe a practical scenario; it's a highly idealized thought experiment intended to illustrate particular theoretical and pedagogical issues that Bell was interested in.

The thought experiment seems set up to require that the rockets can overly precisely control acceleration.
Yes, of course; the specification of the experiment is that both ships (or both ends of a very long ship, if we're talking about that alternative scenario) have the same proper acceleration. That defines what the term "Bell's spaceship paradox thought experiment" means. If you break that condition, again, you're talking about a different thought experiment.

#### stevendaryl

Staff Emeritus
Science Advisor
I've asked this before and had it answered, but I'm not satisfied yet ... I expect I have to work thru the math:

The tensile strength of the thread is neglected in this thought experiments. If there is tension in the thread, it will not just pull itself apart, but it will pull on the two spaceship masses. My intuition is that connecting the ships into that single traveling system, with the thread, they just move together. If the thread was an unbreakable steel bar, with the two rockets part of a single ship, it would be expected to stay together. The fragility of the thread is always stressed as important in the conditions, but even a small force seems sufficient to move the ships together.

I can theoretically tow an aircraft carrier with a thread, as long as my force is lower than the tensile strength of the thread. The length contraction seems so gradual that I (intuitively) expect it to always be below that tensile strength.
Bell's argument is that if the string is flimsy enough that it has no effect on the rocket motion, then it will stretch and break. If you assume, to the contrary, that the string is very strong, then the same reasoning would imply that the string would pull the front rocket backwards and pull the back rocket forwards. So the distance between the rockets would decrease (as measured in the initial launch frame).

• Grinkle

#### PeterDonis

Mentor
shouldn't we consider an engine that applies the thrust uniformly across the entire ship??
If the ship is idealized as being short enough that it can move in a Born rigid manner (i.e., with the instantaneous distances between all of its parts remaining the same as it moves) with negligible variation in proper acceleration over its length, then this is what the engine is doing in the idealized thought experiment. As far as I know, Bell's original formulation meets this requirement (he was basically idealizing each ship as being a point).

If the ship is long enough that moving in a Born rigid manner requires the proper acceleration to vary measurably along its length, then you cannot consider the engine thrust to be applied uniformly, because that would produce the same proper acceleration all along the ship and the ship would not move in a Born rigid manner; it would stretch.

is it from the perspective from the initial state that the distance between the ships is constant?
Yes. More precisely, the distance between the ships is constant as seen in the inertial frame in which both ships are initially at rest (before they turn their engines on).

Wouldn't this mean that from the perspective from someone on one of the ships, the distance between the ships is increasing?
Yes. Have you read the FAQ entry I linked to earlier in this thread? It discusses this.

A stationary observer would see the spaceships being length contracted
Yes.

and the space between the ships being length contracted as well
No.

in order for the people in the ships to see no difference
The people on the ships do see a difference; they see the distance between the ships increasing. See above.

Isn't this paradox just an impossibility?
No. You apparently have not read the FAQ or any of a number of articles on the paradox, all of which discuss and resolve the issues you raise.

### Want to reply to this thread?

"Bell's Spaceship Paradox & Length Contraction"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving