Two-spaceship paradox This solution incorrect

[yuiop]

Yes, they mean the all points accelerated with constant and equal proper acceleration by law

x(t,s)=s+\frac{c^{2}}{w}\sqrt{1+(wt/c)^{2}}-
-\frac{c^{2}}{w}

0{\leq}s{\leq} L

Ok, they are essentially using the same formula used by Baez in "The Relativistic Rocket Equations" See http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

I quote from Baez:

"First of all we need to be clear what we mean by continuous acceleration at 1g. The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference traveling at the same instantaneous speed as the rocket (see relativity FAQ on accelerating clocks). This acceleration will be denoted by a. The proper time as measured by the crew of the rocket (i.e. how much they age) will be denoted by T, and the time as measured in the non-accelerating frame of reference in which they started (e.g. Earth) will be denoted by t. We assume that the stars are essentially at rest in this frame. The distance covered as measured in this frame of reference will be denoted by d and the final speed v. The time dilation or length contraction factor at any instant is the gamma factor γ."

His formula is d = (c2/a) (sqrt[1 + (at/c)2] - 1).

Say we have two rockets initially at rest in frame S. We take one rocket r1 to be at the origin of frame S (x=0) and the other rocket r2 is at x = L. Both accelerate with constant and equal proper acceleration (a) in the x direction. At time t as measured in the frame S they will be a distance d2-d1 apart

= [L + (c2/a) (sqrt[1 + (at/c)2] - 1)] - [((c2/a) (sqrt[1 + (at/c)2] - 1)]

= L.

An observer in an inertial frame that is momentarily comoving with accelerating rockets, measures them to be a distance L/sqrt(1-(v/c)^2), which is of course greater than their initial separation.

 

[J.F.]

An observer in an inertial frame that is momentarily comoving with accelerating rockets, measures them to be a distance L/sqrt(1-(v/c)^2), which is of course greater than their initial separation.

But inertial frame S' that is momentarily comoving with both accelerating rockets is impossible. Inertial frame S' that is momentarily comoving only with one from two rockets is possible.
An observer in an inertial frame S' that is momentarily comoving only with one from two rockets measures them to be a distance L(t') by law:
L(t^{'}) = (2/\sqrt{1-t^{'}^{2}})-1
w=c=1

 

[yuiop]

But inertial frame S' that is momentarily comoving with both accelerating rockets is impossible. Inertial frame S' that is momentarily comoving only with one from two rockets is possible.
An observer in an inertial frame S' that is momentarily comoving only with one from two rockets measures them to be a distance L(t') by law:
L(t^{'}) = (2/\sqrt{1-t^{'}^{2}})-1
w=c=1

I disagree. Two rockets accelerating with constant equal proper acceleration starting simultaneously from rest in Frame S will be accelerating with equal acceleration from the point of view of someone in frame S and they will maintain constant separation and equal velocity at all times according to the observer in frame S. The separation distance measured by either rocket pilot will appear to be increasing if they make the measurement by radar signals because of light travel times and time dilation. If they have equal velocity at any infinitesimal instant in frame S then it is easy to show that there is an inertial reference frame that is instantaneously co-moving with both rockets simultaneously.

Taking the view that the rockets do not maintain the same constant separation in frame S requires assuming the motion of the front rocket somehow affects the motion of the rear rocket. How do you propose that happens if the two rockets are not physically connected?

 

[J.F.]

I disagree. Two rockets accelerating with constant equal proper acceleration starting simultaneously from rest in Frame S will be accelerating with equal acceleration from the point of view of someone in frame S and they will maintain constant separation and equal velocity at all times according to the observer in frame S. The separation distance measured by either rocket pilot will appear to be increasing if they make the measurement by radar signals because of light travel times and time dilation. If they have equal velocity at any infinitesimal instant in frame S then it is easy to show that there is an inertial reference frame that is instantaneously co-moving with both rockets simultaneously.

Taking the view that the rockets do not maintain the same constant separation in frame S requires assuming the motion of the front rocket somehow affects the motion of the rear rocket. How do you propose that happens if the two rockets are not physically connected?

I disagree. You claims in contradiction with Lorentz trasforms.

 

[Doc Al]

I disagree. You claims in contradiction with Lorentz trasforms.

Where's the contradiction? Are you saying that the Lorentz transforms prevent two rockets (moving at the same speed and thus in the same frame) from accelerating equally (constant proper acceleration) and thus maintaining equal separation as measured in their own frame? I don't think so!

FYI: There's no problem whatsoever in defining a co-moving inertial frame at any moment.

 

[J.F.]

Where's the contradiction? Are you saying that the Lorentz transforms prevent two rockets (moving at the same speed and thus in the same frame) from accelerating equally (constant proper acceleration) and thus maintaining equal separation as measured in their own frame? I don't think so!

FYI: There's no problem whatsoever in defining a co-moving inertial frame at any moment.

Which frame? Inertial frame a co-moving only with one roket, or with both rockets simultaneously?

 

[yuiop]

I disagree. You claims in contradiction with Lorentz trasforms.

Hi J.F.

First, an apology. Upon further reflection you appear to be right to claim that two rockets undergoing constant and equal proper acceleration would not have a common instantaneous co-moving inertial frame at any time. At any given instant in the momentary reference frame of one rocket, the other rocket is moving away from it and so they do not consider themselves to be stationary with respect to each other so there is no common instantaneous inertial reference frame for both rockets.

To avoid talking at cross purposes I think we should try and set some clear definitions for the types of accelerations methods applied.

Type 1 acceleration:

The rockets are not connected and are instructed to launch simultaneously in the same direction and maintain constant proper acceleration (as measured by accelerometers onboard each rocket. This is the acceleration method used in Bell’s original two rocket paradox except that a flimsy string (that snaps) connects the two rockets. It is better to replace the string with a tape measure that has drum that can feed out extra tape to maintain constant tension in the tape.

As the rockets accelerate in this scheme an observer on either rocket notices that the rockets appear to be getting further apart as measured by bouncing light signals off each other and as confirmed by the tape measure bailing out extra tape to span the gap while maintaining constant tension on the tape.

The velocity and distance of each rocket in the launch frame is given the relative rocket equations (See Baez http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html ) where a is the same for each rocket. At all times the rockets maintain constant separation as measured by an observer in the launch frame.

On a Minkowski diagram the paths of the rockets are parallel As seen from the launch frame and a line of simultaneity drawn from the origin to one rocket does not pass through the other rocket.


Type 2 acceleration:

The rockets are connected by a strong (but not perfectly rigid) rod. The two rockets are instructed to maintain constant proper separation at all times. To a certain extent this will happen naturally under gentle acceleration with only one of the rockets providing thrust and the rod undergoing natural length contraction. This acceleration scheme is sometimes know as Born-rigid transportation.

As the rockets accelerate according to this scheme the separation distance remains constant (by definition) so they are at rest with respect to each other. However the two rockets do not experience the same proper acceleration. This is the equivalence principle. To first order approximation they appear to be stationary in a gravitational field, the rear rocket “feeling” a greater gravitational force than the front rocket.

On a Minkowski diagram the paths converge and a line of simultaneity drawn from the origin to one rocket passes through the other rocket.

-------------------------------------------------------------------------------------

In summary, if both rockets experience constant and equal proper acceleration then they do not maintain constant and equal proper separation and if they measure constant and equal separation they do not measure equal proper acceleration.
In summary, if both rockets experience constant and equal proper acceleration then they do not maintain constant and equal proper separation and if they measure constant and equal separation they do not measure equal proper acceleration.

 

[Doc Al]

Oops... I forgot to correct my error here (which kev has already explained) and add my own apology.

Where's the contradiction? Are you saying that the Lorentz transforms prevent two rockets (moving at the same speed and thus in the same frame) from accelerating equally (constant proper acceleration) and thus maintaining equal separation as measured in their own frame?

The rockets can certainly arrange to maintain constant proper separation (the situation I had in mind), but as kev points out that does not mean that the rockets have equal proper accelerations. (My bad.)

FYI: There's no problem whatsoever in defining a co-moving inertial frame at any moment.

For the case I had in mind--constant proper separation--this is true.

Which frame? Inertial frame a co-moving only with one roket, or with both rockets simultaneously?

You are correct that if the two rockets have the same proper acceleration, then there is no single inertial frame that they share. My apologies!

 

[jcsd]

Two rockets starting with the same velocity with the same proper accelartion do share the same mometraily comoving inertial frame.

At any given instant they have the same velocity in the original inertial frame S. (Ignoring translation) which comoving inertial frame they are in is purely a function of their velocity in a given inertial frame and as their velocity is equal in an inertial frame they are in the same momentrailry comoving inertial frame at all times.

Lorentz contraction is the difference between the 'same' lengths measured in different inertial frames. The proper length between them changes because they are changing inertial frames.

 

[Doc Al]

At any given instant they have the same velocity in the original inertial frame S.

At any given instant according to frame S they have the same speed, but not according to each other. From the rear rocket's viewpoint, the front rocket is getting further away. (The definition of "same instant" is different for the moving rockets than for frame S.)

Lorentz contraction is the difference between the 'same' lengths measured in different inertial frames. The proper length between them changes because they are changing inertial frames.

Not sure I understand this statement. You agree that the proper separation really does change, right?

 

[jcsd]

At any given instant according to frame S they have the same speed, but not according to each other. From the rear rocket's viewpoint, the front rocket is getting further away. (The definition of "same instant" is different for the moving rockets than for frame S.)

Yes, but they are not in an inertial frame.

Take an arbitary inertial frame S. Any frame traveling at constant velocity in S is also an inertial frame. Each velocity in S defines a unique inertial frame. This is true in Newtonian physics (it's a postulate/definition) and it's true in special relativity.

Our two spaceships are at any given instant moving with the same velcoity in an inertial frame (we cna see this just by considering the orignal frame), so at any given instant they have the same comoving inertial frame.

Not sure I understand this statement. You agree that the proper separation really does change, right?

What I'm trying to illustarte is that length contraction is something that happens when you compare lengths in different inertial frames. The proper length at any given instant is the length between the ships in the comoving inertial frame, as they change comoving inertial frame, the proper length changes.

 

[Doc Al]

Yes, but they are not in an inertial frame.

Exactly my point.

Take an arbitary inertial frame S. Any frame traveling at constant velocity in S is also an inertial frame. Each velocity in S defines a unique inertial frame. This is true in Newtonian physics (it's a postulate/definition) and it's true in special relativity.

This is true. But the rockets are not moving with constant velocity, they are accelerating.

Our two spaceships are at any given instant moving with the same velcoity in an inertial frame (we cna see this just by considering the orignal frame), so at any given instant they have the same comoving inertial frame.

That would be true if they were moving at constant velocity, but they are not. To see this, imagine that the acceleration happens in bursts. Initially, have the rockets moving at some speed v. At the same moment in the original inertial frame S, we give each rocket a small burst of speed \Delta v. But according to the frame of the moving rockets, the rocket in front received its burst of speed before the rocket in back. As long as the rockets continue with the same acceleration, they will continue to separate. In order to maintain constant proper separation, the rear rocket must have greater acceleration than the lead rocket.

What I'm trying to illustarte is that length contraction is something that happens when you compare lengths in different inertial frames. The proper length at any given instant is the length between the ships in the comoving inertial frame, as they change comoving inertial frame, the proper length changes.

Again, I'm not sure I understand what you are saying here. I agree that length contraction is a change in perspective due to switching frames, but there's more going on here that mere length contraction. Since this is the setup for the Bell spaceship "paradox" (with the flimsy rope connecting the rockets), would you say that the rope breaks or not? (If it breaks, that's not just a disagreement between reference frames.)

 

[J.F.]

AAPPS Bulletin Vol. 15 No. 5, October 2005
http://www.aapps.org/archive/bulletin/vol15/15-5/15_5_p17p21abs.html

Jong-Ping Hsu
Nobuhiro Suzuki

We demonstrate a resolution to the “two-spaceship paradox” by explicit calculation using coordinate transformations with at least one frame undergoing constant linear acceleration. A metric such as ds2 = (1 + Kx)2dw2—dx2—dy2—dz2 can lead to a coordinate transformation between an inertial frame and a frame moving with a constant linear acceleration. This coordinate transformation reduces to the Lorentz transformation in the limit of zero acceleration.

This solution incorrect!

I think, that a Møller metric such as
ds^{2}=(1+kx)^{2}dw^{2}-dx^{2}-dy^{2}-dz^{2}
is not completely adequate for the “two-spaceship paradox”.
To the given problem more corresponds uniformly accelerated frame:

ds^{2}=c^{2}dt^{2}-\left{[dx-\frac {wtdt} {\sqrt{1+(wt/c)^{2}}} \right]^{2}-
dy^{2}-dz^{2}

This metrics turns out as result of a shift
x\to x-(c^2/w){\sqrt{1+(wt/c)^{2}
from the Minkowski frame:
ds^2=c^{2}dt^{2}- dx^{2}-dy^2-dz^2

 

[yuiop]

I think, that a metric such as
ds^{2}=(1+kx)^{2}dw^{2}-dx^{2}-dy^{2}-dz^{2}
is not completely adequate for the “two-spaceship paradox”.
To the given problem more corresponds uniformly accelerated frame:

ds^2=c^{2}dt^{2}- \left[dx-\frac {wtdt} {\sqrt{1+(wt/c)^{2}} \right]^2-dy^2-dz^2

This metrics turns out as result of a shift
x\to x-(c^2/w){\sqrt{1+(wt/c)^{2}
from the Minkowski frame:
ds^2=c^{2}dt^{2}- dx^{2}-dy^2-dz^2

Your point is ...?

Do you agree that a string connecting two rockets undergoing the Type 1 acceleration method described in post #37 will snap due to length contraction?

 

[J.F.]

Your point is ...?

Do you agree that a string connecting two rockets undergoing the Type 1 acceleration method described in post #37 will snap due to length contraction?

Hi kev.
Let's consider a case when the ships are accelerated according to the laws:

x_{1}(t)= \int_{0}^{t} v( \tau )d\tau

x_{2}(t)= L +\int_{0}^{t} v( \tau )d\tau

v(t)\to v=const

Finally, both of the ship, does not get on the Minkowski frame:

ds^2=c^{2}dt^{2}- dx^{2}-dy^2-dz^2

As consequence Lorentz length contraction is broken

 

[yuiop]

Hi kev.
Let's consider a case when the ships are accelerated according to the laws:

x_{1}(t)= \int_{0}^{t} v( \tau )d\tau

x_{2}(t)= L +\int_{0}^{t} v( \tau )d\tau

v(t)\to v=const

Finally, both of the ship, does not get on the Minkowski frame:

ds^2=c^{2}dt^{2}- dx^{2}-dy^2-dz^2

As consequence Lorentz length contraction is broken

Hi J.F.

Sorry I did not see your post earlier. I take it that you have concluded that if two rockets with initial separation L in the launch frame, take off simultaneously with equal proper acceleration, then their mutual proper separation will be increasing while their separation measured by an observer that remains in the launch frame remains constant. That, I think is the whole point of the "two spaceship paradox" and is the reason why Bell concluded (correctly in my opinion) that a string connecting the two rockets will snap.

 

[J.F.]

Hi J.F.

Sorry I did not see your post earlier. I take it that you have concluded that if two rockets with initial separation L in the launch frame, take off simultaneously with equal proper acceleration, then their mutual proper separation will be increasing while their separation measured by an observer that remains in the launch frame remains constant. That, I think is the whole point of the "two spaceship paradox" and is the reason why Bell concluded (correctly in my opinion) that a string connecting the two rockets will snap.

Agreed. But I wish to discuss only a method offered Jong-Ping Hsu.
Hsu has started own method from the transformations:(1)-(3) see URL:
http://www.geocities.com/jaykovf/a127.JPG

The transformations:(1)-(3) preserve the metric (4) see URL:

http://www.geocities.com/jaykovf/a124.JPG

But the metric (4) does not concern to a case when a two rockets with initial separation L in the launch frame, take off simultaneously with equal proper acceleration w.From what it suddenly Hsu postulates, what rockets after the beginning of movement have dropped in on FR (4)?