Bell's Paradox Question

Nugatory

Ahhhh.... the Minkowski diagram is the same for all observers. The way you introduce another observer into a Minkowski diagram is by drawing that a new set of x and t axes on the diagram to represent the new observer's coordinate system.

But I suspect that what you're really asking for is a Minkowski diagram in which that other observer's x and t axes make a 90-degree angle on the sheet of paper? That's a fair request, as it's lot easier to visualize.

A.T.

See post #41

The string is not shortening in the “stationary” frame. It can only break. It breaks if its atoms cannot span the separation distance anymore.

If contracting atoms are too abstract, then replace the string with a chain. The accelerating chain links will get shorter and shorter, so at some point they cannot span the constant separation distance, and the chain breaks somwhere.

Eli Botkin

Nugatory:
Not so. One diagram can contain any number of observer coordinates. The x,t axes for a particular observer need not be at a right angle to each other. The scale on any axis is determined by the hyperbolae that are asymptotic to the light ray at 45 degrees. Such a single diagram is usefull in showing how things appear to all observers. There is nothing special about the observer assigned to the right angle coordinates.

Nugatory

I certainly agree with everything you say after the words "not so"... but it is precisely because one diagram can contain any number of observer coordinates and there is nothing special about the 90-degree pair that I didn't understand what you meant when you asked for "the Minkowski diagrams ... for observers other than those co-moving with the two ships'.

DaleSpam

You are welcome.

That and the fact that the limit is infinite, therefore it will eventually break regardless of the actual values of L or d. It is possible for a function to be monotonically increasing to a finite limit, in which case the breakage would depend on the details of how big L was compared to the limiting value of d'.

Excellent, therefore the string breaks and Bell's paradox is resolved using only SR.

The laws of physics are frame invariant, therefore if it breaks in one frame then it breaks in all frames.

I leave the details of any other frame as an exercise for the interested reader (you). I would be glad to look over your efforts if you get stuck somewhere.

Eli Botkin

A.T.:
Let’s try this. Have the string attached to the two ships only loosely. Say each end is firmly in a sleeve such that when you tug the string it will slide in the sleeve.

With the ships always maintaining the same separation why will the string slide out of a sleeve? And which sleeve, fore or aft? And why is the space between the string atoms being reduced but not the space between the ships?

schaefera

I think you may be confused on the same point as I was! Try considering the two frames: S which is stationary and S' which move with the ships after they finish their acceleration.

In S, the distance between the ships remains constant because this is the frame in which they accelerate the same, so the statement of the paradox requires this. But the string is length contracted and therefore breaks. It's not that the space between the ships remains constant but the space between the string atoms is shrunk-- it's that the atoms themselves are shrunk, but not the space they are in.

In S', the ships do not accelerate the same way, and the distance between the two is increasing. Since the front ship begins accelerating first, I'd imagine that the string has to slip from the front ship's sheath. This agrees with the fact that (I'm assuming the sheaths to be frictionless) since the ships accelerate forwards the string would slip backwards in the sheaths.

Eli Botkin

DaleSpam:
It must surely also follow that if it doesn't break in one frame then it doesn't break in all frames ;-)

The main point of frame invariance is that if you can show that it breaks in one frame then you should be able to show that it breaks in other frames.

Eli Botkin

schaefera:

Please tell me why you think "...that the atoms themselves are shrunk, but not the space they are in."

DaleSpam

Indeed. If you think that you can show that it doesn't break in any frame, then please post your work and I am sure we can figure out where you made the mistake.

The main point of frame invariance is that you can always choose to work the problem in the easiest frame.

Eli Botkin

DaleSpam:
I agree "...that you can always choose to work the problem in the easiest frame." However, with this type of problem, where there are questions of interpretation, it can't hurt (and may help) to seek that same solution, from other directions, in other frames.

An even "easier" frame than the ship's co-moving frame would have the observer moving at some constant speed, V, in the direction opposed to the ship's motion. Can your solution be found there?

Bear in mind that I've never claimed that the string doesn't break. My claim is that string breakage has not been proven, not even by Bell himself. Bell did opine that it would break because of the internal EM field distortion (which I find strange since , through V, the distortion is also frame dependent). But that doesn't make a proof.

I must admit, though I have great respect and admiration for Bell's contributions to the understanding of QM, I am puzzled by his claim that breakage is due to Fitzgerald contraction though he knows there is no such contraction of the separation between ships.

If the correct solution is that breakage must take place, then I think the culprit is more likely to be the ships' acceleration, a component of the problem that exists in all frames.

Thanks for "listening."

Austin0

Hi First I will say that I personally have the view that the light speed electromagnetic interactions involved in maintaining physical structure do result in a contraction through the tensile forces. I.e. ; the string will break.
Even so I feel impelled to play Devil's advocate here.
Just looking at the scenario from a target frame of say 0.8c it is clear that the leading ship initiates acceleration before the trailing ship. This not only means increasing the separation but also creating a velocity differential that will persist even after the trailing ship is also accelerating.
Therefore relative to that frame, the distance between the ships must continue to increase without bounds as you calculated with your analysis.
This itself seems problematic. How do you reconcile frame agreement of time and position observations of the ships when the distance remains constant in one frame and is indefinitely expanding in the other frame??
Or conversely: Assume the ships started accelerating from signals that were simultaneous
in the 0.8c frame. Now in that frame the distance remains constant, but in the launch frame the trailing ship fires up first and must eventually intercept and overtake the lead ship.
How do we reconcile these contradictory expectations???

Adopting for the moment a purely kinematic interpretation of SR which has been expressed by many knowledgeable people on this forum in the past.
Assuming no contraction of either the distance or the string. At a velocity of 0.8 c what would be the measurement of that distance?
According to an 0.8c inertial frame the clock in the launch frame would be running ahead at the instantaneous location of the lead ship.Therefore the trailing ship and string would move forward an additional distance before reaching a clock with the same proper time. I.e.; Would be measured as being contracted relative to the initial separation.

As I said I don't necessarily accept this interpretation but I can't dismiss it out of hand.

I might point out that if the physical reality of contraction as implied by the Maxwell maths is actually valid this leads strongly to the logical conclusion that all motion is actual and in a sense absolute, even if it is not determinable or measurable in non-relative quantitative terms. Would you agree?? ;-)

DaleSpam

Then you must agree that proving an outcome in any one frame is sufficient to prove the outcome.

I proved it above. If you disagree then please point out any mistake I made and I will correct it. If I made no mistakes, then I have proven it.

A.T.

Beacuse it will contract, so it cannot span the distance between the sleeves.

An ideal mass-less string? From both I guess.

Because the EM-fields that hold the atoms together are contracting, and are pulling the atoms closer together. If the atoms are forced to span a constant length (like in the original scenario) the bounds between the atoms will break somewhere.

That is given in the scenario. They accelerate such that the distance between the string attachments stays constant.

DaleSpam

Why should this be problematic or need reconciliation? The worldlines of the spaceships are curved. This same effect happens with curved lines in Euclidean geometry.

Take a piece of paper and draw two copies of the same curved shape separated by some fixed horizontal distance. Then draw a set of parallel oblique lines and see how the distances are not fixed on the oblique lines even though they are fixed on the horizontal lines.

I don't know what you are asking here, nor why you would assume no contraction.

I don't know what you mean by this either.

Austin0

You missed this one.

The assumption of no contraction was the kinematic interpretation of SR. That all effects were purely the result of relative motion. Coordinate evaluations without necessary physical imp0lications. As I said this is not my assumption but an assumption made by others who adopt this interpretation.
How many times over the years has someone asked , "is contraction real or what causes it" and the expert response has been that it is not necessarily physical but merely an effect of relative motion?
I merely noted that employing this assumption to clock desynchronization could produce a coordinate contraction without an assumption of actual physical contraction.

Likewise, this is a perennial question, "Is motion real " and the majority response has been a resounding NO , it is purely relative with no physical actuality.
But if one accepts the validity of the physical contraction implied by Maxwell then logically it should be said that, yes it is real, with actual physical consequences but is simply undetectable and unquantifiable. So I will ask you , in your view is motion real or purely relative??

DaleSpam

It appears to be the same question as the other one. Similarly with the question below.

My usual response to such questions, and to this one, is to ask the person to define "real" and "actual" and "physical". Those terms are quite ambiguous and without a solid definition of them the question cannot be answered.

If you have a scenario which is described completely in one inertial frame, then you can use the Lorentz transform to obtain the equivalent scenario in any other inertial frame and be assured that the same laws of physics explain the scenario in both frames.