Relativity and Cosmology

ok start explaining! In the first example, the two spaceships are not subject to other external forces such as gravity. They are connected together, thus forming one unit of which the string is part. they both accelerate at the same rate, consequently they will eventually arrive at the same velocity at the same moment with the same disposition ship A to ship B, remaining the same distance apart. The string accelerates as part of either and both ship(s) being pulled by one and pushed by the other. There is an assumption that the string is not elastic and cannot distort for other reasons.

If the string were in stress, then both ships would also be stressed from the there acceleration centre along line X, consequently both ships would also fall to bits eventually.

Alternatively, Both ships plus string could be 'seen' as stationary while the universe accelerates Past them in which case neither string nor ships would be stressed

Is that enough to start?

Hi Guest, you wrote: "... they both accelerate at the same rate, consequently they will eventually arrive at the same velocity at the same moment with the same disposition ship A to ship B, remaining the same distance apart.

This scenario is true in the original reference frame (or in any Newtonian reference frame). However, the ships and the string don't find themselves in inertial reference frames. What matters to the string is not how the inertial observers of the original frame view it, but rather what happens its two ends.

It takes quite fancy math to rigorously prove the relativistic premise that the front end of the string is indeed accelerated faster then the rear end and that the string is stretched or must snap. In simpler terms, one can say that the front clock gains time on the rear clock and hence accelerates faster - to keep up with it's faster clock. It is not too difficult to prove the 'faster front clock' premise in a fairly rigorous way.

Secondly, I do not understand your: "If the string were in stress, then both ships would also be stressed from the there acceleration center along line X, consequently both ships would also fall to bits eventually."

The accelerating ships are not constrained at both the nose and the tail, like the string, so why do you think they would be affected?

Your closing statement: "Alternatively, Both ships plus string could be 'seen' as stationary while the universe accelerates Past them in which case neither string nor ships would be stressed." is not equivalent to the first scenario.

While the ships are accelerating, neither of them is in an inertial reference frame, so they can't say: "I'm stationary, it's the universe accelerating past me". That 'flexibility' is reserved only for un-accelerated observers in free space.

You are welcome to argue it further and I will supply more analysis, but please consider the remarks carefully.

Jorrie

So what exactly is point of having the second ship? Wouldn't this string meet the same end without the second ship present? I am also assuming that the string would snap where it is hitched to the forward ship.

Using the same theory, do you think it would be possible to make bubbles in space using a child's bubble solution and bubble wand? Since the wand would be moving faster than the solution and the nature of the bubble solution to close its self off...

Hi again Guest:

You asked: "So what exactly is point of having the second ship? Wouldn't this string meet the same end without the second ship present?"

No, if attached only to the lead ship it will be fine and it will more or less keep its length, unless the acceleration is vicious enough to tear the string apart due to its inertia.

The point of the puzzle is that the proper distance between the two co-accelerating ships (as measured by a tape measure on any one of them) will increase. If the string is attached to both, it will have to stretch or snap. Where it will snap is not a given, probably dependant on string uniformity and fastening methods.

About the child's bubbles in space, I don't know. Normal inertia would force the bubble material to lag if you accelerate the wand, but the characteristics might be totally different in space. I'll have to think about accelerating the wand to relativistic speeds...

Regards, Jorrie

Jorrie

I find a slightly different interpretation more helpful. In the original inertial frame, the ships start accelerating at the same time. Once they are moving, however, the clock of the front ship becomes ahead of the clock in the rear ship (their times being measured by Einstein's method). So the problem is that the front ship has been accelerating longer in the local frames of reference, so the ships aren't travelling at the same velocity. Although the clocks remain synchronised to the original frame of reference, they are not synchronised in a common frame. Once we maintain the clocks synchronised in the spaceships' frame of motion, the problem goes away. (It's another way of saying that time does funny things under acceleration or in gravitational fields).

BTW, I personally find the connection with a non-uniform gravitational field to be unhelpful - until you can add an equally easy explanation as to why this constant acceleration system is equivalent to non-uniform gravity.

And, in answer to the question "do we need ships", I say no - except in so far as they are used to provide clocks that that are synchronised (wrt the original reference frame) for the ends of the string.

Regards

Fyz

Hi Fyz,

I agree with most of what you said. In my first reply above, I did talk about the slower clock with faster acceleration and speed: "In simpler terms, one can say that the front clock gains time on the rear clock and hence accelerates faster - to keep up with it's faster clock. It is not too difficult to prove the 'faster front clock' premise in a fairly rigorous way." The issue of the equivalence with non-uniform gravity is quite confusing, I must admit.

However, it is at least qualitatively equivalent to Bell's paradox. Hover two ships vertically one above the other in a non-uniform field, like Earth's. The top one will measure less g's and it's clock will run faster than the lower one. In order to experience identical accelerations, the two ships must be accelerating away from each other vertically, so the string will break.

This not the case in a uniform gravity field (if by uniform field is meant the Newtonian potential gradient is constant in the direction of interest). Here the top clock will still run faster, but the accelerations will be identical while they hover motionless, so the string will not break.

But, I think your interpretation is still the simplest by far.

As far as the need for ships is concerned, I would say one also needs a propulsion system tied to each end of the string!(?)

Regards, Jorrie

Hi Jorrie

Then there's the "hard" version - spaceships spaced and accelerating around a circle. None of the above semi-intuitive explanations helps.

Regards

Fyz

Hello,

Let's consider a rod, which center moves with a constant acceleration 'g' directed along the rod. Is it right to say that the front end of the rod will move with acceleration smaller than 'g', and the back end with acceleration higher than 'g'? If so is it due to Lorentz contraction?

Thanks, Vladimir.

Hi Vladimir.

Yes, the rear end of any rod-like thing being accelerated length-wise always experience higher proper acceleration than the front end. One can view it as Lorentz contraction that is responsible, but it is better to draw the hyperbolas for the front and the rear with a common vertex and then observe that the spacetime curvature is less for the front end.

Jorrie