# B Bell spaceship paradox - qualitatively

1. Aug 23, 2015

Suppose the string was replaced by some structure which linked the ships together to make a longer ship. If the string breaks then by the same reasoning the joining structure should break also. The first breakage would not necessarily occur in the joining structure but in the weakest point between the front and rear of the ship. In fact we can ignore the joining structure and consider one of the original ships. If the front and rear move with the same acceleration then the ship should break and if any fragments continue accelerating they should break and so on.
Does this mean that any object accelerating to very high speeds will tend to break? What am I overlooking?
Thanks

Last edited by a moderator: Aug 24, 2015
2. Aug 23, 2015

### Staff: Mentor

Nothing. Google for "Born rigid motion", but it would probably be best to start another thread for followup discussion.

[Mentor's note: the stricken text no longer applies because this post has moved to a new thread - this one]

Last edited: Aug 24, 2015
3. Aug 24, 2015

### votingmachine

What I missed/overlooked in this thought experiment is that the rockets are able to deliver constant ACCELERATION not constant FORCE. The increasing force needed to make the acceleration constant makes a difference. (Thanks to PeterDonis for clarifications in a message).

I don't know if this will help, but consider if you have a string that can just exactly hold 10 g before breaking. And it weighs 1 g per meter. If you connected the rockets with 10 meters and they accelerated at just over 1xG, then the string would immediately break. At just under 1xG, it would still be unbroken. But at the moment the force goes over 1xG, the string breaks.

There is more than that, but I made the mistake of reading constant force (a more ordinary rocket has a constant thrust), when it is phrased as constant acceleration.

4. Aug 24, 2015

### loislane

Why do you conclude that for acquiring constant acceleration you need an increasing force instead of a constant force?

5. Aug 24, 2015

### votingmachine

This is one that is hard for me to grasp. But the ordinary relationship is:
F=ma
Consider that the rest mass is the mass in any every day calculation. But mass has more than the rest mass
https://en.wikipedia.org/wiki/Mass_in_general_relativity

EDIT: this is the better/simpler link to look at ... a calculation:
http://www.phy.olemiss.edu/HEP/QuarkNet/mass.html
My poor grasp of it says that the mass increases with velocity. At a level that is insignificant at ordinary velocities. But since it increases, and the acceleration is constant, the force has to increase. The increase in mass of the string is something that also seems to matter ... if the force is larger AND the mass is increased, the ultimate tensile load seems easier to breach.

I see that people are looking at this problem from a wide variety of perspectives, and looking at it as strictly a tensile strength problem is not the one that is generally chosen, and is (I gather) a limited view. But it leads me to see why the string fails with a particular set of initial assumptions, and then I can see why it generalizes. And why the next level analysis is conclusive.

Hopefully I am not the "blind leading the blind" in this answer. There are certainly better answers.

6. Aug 24, 2015

### Staff: Mentor

This would be a good time to ask everyone in the thread to be careful about the distinction between coordinate and proper acceleration - even if you think it's clear which one you mean from the context, it may not be so clear to someone else.

Bell's paradox is stated in terms of both ships experiencing the same non-constant coordinate acceleration, using coordinates in which the observer is at rest. This can be produced using constant thrust and is consistent with constant proper acceleration.

7. Aug 24, 2015

### Staff: Mentor

You're overlooking that in the standard scenario, with two ships linked by a string, each ship is providing thrust and has the same constant proper acceleration. In the alternate scenario with one long ship, you're implicitly thinking of it as the ship having one engine at the rear, providing thrust, but you're still assuming that the front end of the ship would have the same proper acceleration as the rear end. It won't.

Try this comparison:

(1) We have one long ship with a single engine at the rear, producing sufficient thrust that an accelerometer at the rear of the ship registers constant proper acceleration $a$.

(2) We have one long ship with an engine at the front and an engine at the rear, each producing sufficient thrust that an accelerometer co-located with the engine registers the same constant proper acceleration $a$.

In case #1, the ship will be fine; it won't stretch and it won't break. But the front end of the ship will have a proper acceleration smaller than $a$.

In case #2, the ship will stretch and eventually break.

8. Aug 24, 2015

### bcrowell

Staff Emeritus
This thread has had a lot of discussion of dynamics, but the Bell spaceship paradox is purely kinematical. You can strengthen your materials as much as you like, and apply forces wherever you like and according to any system in order to try to relieve the strain, but that doesn't help, because rigid motion is *kinematically* incompatible with the given information.

No. For example, you can rigidly accelerate a rod arbitrarily in any direction perpendicular to its length. You can also rigidly accelerate a rod along its own length, but once you choose the acceleration of one point on the rod, the rest of the rod is constrained to move in a certain way in order to maintain rigidity, and its motion is not the type of motion stated in the Bell spaceship paradox. The exact constraint on what you can do is somewhat complicated, and is given by the Herglotz-Noether theorem. All of these motions would be motions that you would create by applying just the right combination of forces at all the different points on the object; it can't be accomplished simply by making a force at one point and then relying on the internal dynamics of the object itself to make the rest of it come along appropriately.

Last edited: Aug 24, 2015
9. Aug 24, 2015

### votingmachine

Sure, but the thing is there is only one failure mode for any given set up. I can see that the system has an absurd upper bound where there is nothing possible. If the rockets accelerate at 0.111c per second, after one second the rockets are 0.111c. After 2 seconds 0.222 c. After 9 seconds 0.999c. And no matter what the set up, there is a failure of constant acceleration in the next second. The actual failure mode will depend on the properties of the connection. The information I just gave is incompatible with the universe itself. Yet it is within the bounds of the proposed rockets. I think the ordinary rope would snap from the tensile strength being exceeded by the inertial mass at that phenomenal acceleration.

At the point where the connection can no longer withstand the forces, it must break. It might break from a relatavistic mass increase, or it might break from a length contraction. It would depend on the string strength, the string rest mass per unit length, the length of the string, and the acceleration. If the string is poised to break from the rest mass inertial resistance to acceleration, then the likely mode would be from the increasing resistance to acceleration (relatavistic rest mass increase). If the string had strength to handle a load that is many multiples of its total weight, then it seems the length contraction would be the mode by which force exceeds the tensile limit.

I think from what I read in this thread that I must go back and look at the original paper again. There may be more than one element of the set up that I am still mistaken about. But at the moment, I think that the set up allows multiple modes of failure.

10. Aug 24, 2015

### Staff: Mentor

You are confusing coordinate acceleration and proper acceleration here.

11. Aug 25, 2015

### Vmedvil

I don't mean to butt in but wouldn't at those velocities Special Relativity happen and mass increase as you speed up.

Sub

Sub

Making, which is a system of three equations for velocity and acceleration in each dimension, (x,y,z)

∇E(x,y,z) = (1/(1- (u2+2as)/C2)1/2)mc2

Or

∇E(x,y,z) = (1/(1- (u2+a2t2)/C2)1/2)mc2

Then take the sum of the forces for all the sites of acceleration plus is right and minus is left, or whatever your axis is.

∇ΣEf(x,y,z)=∇E1(x,y,z) (-/+) ∇E2(x,y,z) (-/+) ∇E3(x,y,z) ........ (-/+) ∇Ef⇒∞(x,y,z)

change force into energy to be summed for each source Energy input to the system.

Only difference is we are using energy instead of force then calculating in (X,Y,Z) to get the grand total from all sources.
Where,
v = Final Velocity,
u = Initial velocity,
a = acceleration,
s = distance traveled.
m = rest mass
C= Speed of Light in vacuum
t = time taken in 0 Velocity observer clock

Then it would break at the point that is center of energy stress which is the point where net energy is zero.

(∇EΔ(x,y,z) / ∇Ef(x,y,z) ) * ∇L0(x,y,z) = ∇Lb(x,y,z)

∇EΔ(x,y,z) = difference between Final energy and zero

∇Ef(x,y,z) = Final Energy

∇L0(x,y,z) = Total Length of object in that dimension

∇Lb(x,y,z) = Breaking Point this length will be the distance from the middle of the object in that direction for each dimension.

Solve it yourselves that would take hours unless you use a computer program.

Last edited: Aug 25, 2015
12. Aug 25, 2015

### votingmachine

I am going to have to dig up the original paper again. I now see it explicitly as proper in some places but in others it reads as coordinate. I think even in those it is just a combination of ambiguous wording and poor reading on my part.

I think at this point that is the second conceptual error I've had on the set up ... I'm not really comfortable with that error rate ...

I googled and did not find the immediate hit to the original that I know I've seen. Can you post a link to the best spot? I don't mind looking at the many bits of coverage by other people, but I also would like to go back to the beginning ... so any links are appreciated.

13. Aug 25, 2015

### Staff: Mentor

Relativistic mass would increase, but the concept of relativistic mass isn't used much now since it's just a confusing way of saying "energy". The energy, with respect to the initial rest frame, certainly increases, but you don't really need to deal with that to answer the original question, which was simply whether or not the ship breaks, not exactly where, when, and how it breaks.

These equations are not correct; at least, not if $a$ is the proper acceleration. You have failed to take into account relativistic velocity addition. Try taking a look at the relativistic rocket equation:

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]

You appear to be assuming that the net force is zero. That's not the case; if it were, the ship would not be accelerating.

I'm not sure exactly what you're trying to say here, but if you're trying to build a mechanical model of the ship using just relativistic kinematics, it's not going to work. You also need to model the material properties of the ship.

#### Attached Files:

• ###### proxy.php?image=http%3A%2F%2Fi.imgur.com%2F9WRLoj8.png
File size:
1.2 KB
Views:
20
Last edited by a moderator: May 7, 2017
14. Aug 25, 2015

### Vmedvil

v = at / sqrt[1 + (at/c)2] is the new sub equation then and your right I did assume a Zero movement and your right I would also need a internal diagram and need to do the materials. I wasn't given those though. It will break at the weakest point near that location predicted though.

Last edited: Aug 25, 2015
15. Aug 25, 2015

### Staff: Mentor

You don't know where the "weakest point" is unless you have a model of the material properties and internal stresses of the string. You can't build that just from relativistic kinematics.

Also, "weakest point" is a misnomer; the simplest material model of the string would have its properties uniform, meaning there is no "weakest point"; all the points have the same tensile strength.

16. Aug 25, 2015

### Vmedvil

After correcting for the Acceleration, the sum of the energies is point that snaps.

17. Aug 25, 2015

### Staff: Mentor

Your supposed argument for this claim is based on a number of incorrect premises, which I pointed out in a previous post. You only responded to one of them (about the incorrect formula for velocity addition). Fixing that alone doesn't fix your argument.

You also admitted that you haven't modeled the internal stresses in the material, yet you're claiming you can figure out exactly where the string will break without knowing that. That's not correct. Saying that you weren't given a model of the internal stresses doesn't allow you to claim you've magically solved the problem without one.

Finally, you have not addressed at all my observation that you are incorrectly assuming zero net force on the object.

18. Aug 25, 2015

### Vmedvil

Don't make me actually calculate this and it snaps when the Energy-mass of the volume goes above the Max Tensile strength in that Energy/Volume.

19. Aug 25, 2015

### Staff: Mentor

Yes, that's true. But it's not what you said before. If it was what you were trying to say, you weren't making it clear.

I'm not trying to make you calculate anything. I've just pointed out some flaws in your argument.

20. Aug 25, 2015

### Vmedvil

Before, I assumed it was made of many different materials now that it is the same material it doesn't matter, Basically, it breaks at the point of where the original equation with the Relativistic acceleration correction point, then then it breaks when that energy volume goes above the Max tensile strength of that volume.