Bell's Paradox Question

A.T.

Is death from rifle bullets physical? Nah, merely an effect of relative motion.

Eli Botkin

DaleSpam:
I've reviewed your math (your earlier reply #43) and find no error. What you've shown for certain is that for a sequence of instantaneously co-moving observers the ships' separation is some d' > d. I'm familiar with the math, having done this myself. And of course your conclusion favoring string breakage is pre-ordained since you've also stipulated that the string must remain L (= d) at any d'.

But why does every co-moving observer "see" d expand to a value d' > d but not see L undergo a proportionate expansion to an L' > L? I think that would be an essential point to address so others couldn't claim that you have, in effect, proved what you assumed.

By selecting the frames of the co-moving observers (who always deal with a d' > d) you have avoided the problem of finding a solution for frames of observers for whom d' < d (and, as you know, there are many such frames.)

I note that you don't claim string contraction, as many others (including Bell) do. Is that because you've selected a massless string (no atoms to contract)? ;-)

DaleSpam

Good, so now you have seen it proven using SR.

We did assume it. We assumed the string was stiff. If we had instead assumed it was elastic then it could have stretched, but then we would have been working a different problem, one requiring a relativistic version of Hookes law and the elasticity of the string material.

Yes, that is why I chose the MCIF.

Eli Botkin

DaleSpam:
Saying "Yes, that is why I chose the MCIF" sounds like an admission that you knew you couldn't get that SR conclusion in the other frames.

Calling the string "stiff" is just another way of saying "all L' = L, regardless of the frame.." I think that SR should decide that. You should be aware that SR tells us that even "proper lengths" of so-called stiff objects can be altered for a particular inertial observer if the object undergoes a history of acceleration. If this is unfamiliar territory, let me know so I would then describe how and why. But that would have to wait until next weekend since I will be sans computer until then.

DaleSpam

Wow! You sound paranoid.

There are an infinite number of equally valid ways of doing a problem, and we agree that we can choose to do it the easy way. There is also an obvious difficulty with some of those ways. So I pick a way which avoids the obvious difficulty precisely because it avoids the obvious difficulty (which you agree is valid to do).

To me the obvious conclusion is that I am lazy and don't want to do things the hard way, but what comes to your mind is instead that I am trying to hide the fact that it can't be done the other way. Sounds like you think I am some sinister agent of a cover up.

The problem can be worked in an infinite number of frames, and because of the principle of relativity we know that the answer must be the same. I welcome you to do it the hard way if that interests you, but I am lazy and will stick with the easy way. If you get stuck (as often happens when doing things the hard way) then post your work, and I have already offered to help get you unstuck.

Eli Botkin

QDaleSpam:
Since no paranoia has been detected in my family it's not likely that I carry that trait, nor do I think that you are lazy :-)

I think you should also address the issues in my 2nd paragraph, they are important for understanding what SR is saying.

DaleSpam

No, stiff means that the proper length is always L, regardless of the forces on the string. In frames other than the MCIF it is certainly possible that L≠L'.

Eli Botkin

DaleSpam:
Returned from vacation, willing to continue discussion of our differences.

You say “… stiff means that the proper length is always L, regardless of the forces on the string.” I would modify that to: stiff means that the length is an unchanging value (regardless of forces applied) in any selected frame. In a different frame it will still be unchanging, but at a different value. SR’s transformation equations require that.

As for the term “proper length,” care must be taken not to think of it as a synonym for “true length.” “Proper length” is only shorthand for “the length measured in a frame wherein the body is at rest.” In SR that’s just another frame like any other.

So, if a log (or string) has “proper length” L in frame A, then, after acceleration to a velocity V, it will have a “proper length” >L in a frame B that has a velocity V relative to frame A.

Therefore it cannot be correct to assume that the stiff string maintains the same length L (or proper length L) throughout its acceleration interval.

DaleSpam

This is only true if the object and the frame are both inertial, which isn't the case here.

Proper length is frame invariant. If it has proper length L then it has proper length L in frame A, B, C, D, ... Proper length only equals coordinate length in the rest frame.

Eli Botkin

DaleSpam:
You say "This is only true if the object and the frame are both inertial,..."

I presume that you would define a non-inertial object as an object under acceleration. But it is reasonable to consider such an object to be "jumping" from one instantaneously co-moving inertial frame to the next such frame. That's what acceleration is, its a transfer from one inertial frame to another. SR deals with coordinate transformations between inertial frames.

I don't know what you mean by "Proper length is frame invariant. If it has proper length L then it has proper length L in frame A, B, C, D, ... " Please tell me what SR says about proper length when it isn't related to the object's rest frame.

yuiop

EDIT: Oops, I had only read the first page of this thread when I posted this so completely missed 4 pages of the discussion. Hope it is still relevant.

It does and it doesn't ...bear with me and I will try and explain. Imagine we have two spaceships parked on the ground that 1 kilometre apart. They are joined by a 1km string. This string is designed to snap when stretched to twice its rest length. In the ground frame both rockets take off simultaneously and accelerate equally, such that they maintain a spatial separation of 1 km at all times (as measured in the ground frame). When the rockets reach a velocity V, which corresponds to a gamma factor of 2, the length of the string in the ground frame should be 1/2km due to length contraction and is stretched over a space of 1km so it snaps. In the ground frame, there is no length contraction of the space between the rockets. Length contraction requires we have a velocity and we cannot assign a velocity to the space (vacuum) between the rockets.

Now let us have a look from the point of view of the rockets. Initially they see the separation as 1 km. As they accelerate the space between them appears to increase and as they arrive at the critical velocity V they measure the space between themselves as 2km. The string is approximately at rest in there reference frame because it is co-moving with the rockets so it should have a length of 1km but it is stretched over a distance of 2km so it also snaps from their point of view. (note that I am using a loose definition of rest frame for the rocket observers, because from their point of view they are are not exactly at rest with respect to each other, but it is a reasonable approximation for our purposes).

Note that according tot he rocket observers the distance between the rockets is 2km and according to the ground observers the distance is 1 km, so there is a sort of "length contraction" of the space between the rockets because different observers disagree on the length, but at no time does the spatial separation between the rockets contract according to any observer. In fact the distance expands according to the rocket observers and remains constant according to the ground observers.

The string on the other hand can be assigned a velocity relative to the ground frame and so it really does physically contract according to the ground observers.

In summary, according to the ground based observers the string contracts, but the space does not and according to the rocket based observers the space expands (because they consider themselves to be moving apart from each other) and the theoretical length of the string remains constant. The space between the rockets according to the ground based observers is 1/2 the distance measured by the rocket based observers, so the Lorentz transformation of space is still satisfied, even though there is no actual "contraction" of the separation space according to any observer.

You cannot accelerate space (vacuum) to make it contract, but your perception of the space between two markers can vary with your velocity relative to the markers.

If we do a variation of the paradox, whereby the rocket captains are instructed to maintain a constant distance of 1km between their rockets (by their own measurements) as they accelerate, then at the critical velocity V, the ground based observers will measure the space between the rockets to be 1/2km and so any string between the rockets will not snap, because this time the string and the separation distance, length contract at the same rate according to the ground based observers, while the rocket based observers say the string length and separation space remain constant so they also agree that the string does not snap.

A.T.

What do you mean by "proper length when it isn't related to the object's rest frame". Proper length is per definition the length measured in the the object's rest frame. From any other frame, the proper length of a moving object is measured by a co-moving ruler.

DaleSpam

See http://math.ucr.edu/home/baez/physic...eleration.html

The proper length is
[tex]\int_P \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu}}[/tex]
where P is the space like path consisting of the intersection of the objects worldsheet with a hyperplane orthogonal to the tangent vector.

Eli, you seem to be grasping for straws now. As I said, the fact that the string breaks can be proven with SR. With this current line of questioning you are straining at very minor details that are already well established in the SR literature. How to handle acceleration and the definition of proper length are well known. The fact that you don't know about them is not a flaw in the proof I gave.

I am willing to continue the conversation in the context of improving your education, but not in the context of defending the proof above. It is a valid proof that the string breaks. Is that acceptable to you?

Eli Botkin

DaleSpam:
Thanks for your reply. I am always willing to continue my education, as we all should.

It seems to me that in your replies there has been an avoidance of a certain SR question, which is not a minor detail.

Why would an inertial observer, who is traveling in a direction opposed to the ships', predict that the string will break? In his frame the ships approach each other. Your own prediction of breakage is based on frames wherein the ships increase their separation while (you say) the string retains its "proper" length (reply #43). After all, a principal SR teaching is that the physics should be coordinate-free.

I would truly appreciate your view of this issue.

DaleSpam

Do you accept the proof as valid?

Eli Botkin

What I accept is that under all your assumptions your mathematical deduction is correct.

You've correctly shown that for frames in which the ships are momentarily at rest, successive frames show an increasing ship separation. One of your assumptions is that the string length, L, stays constant from such frame to frame. Therefore you can correctly conclude that the string will break.

Now to my reply #82, would you be willing to address that? Thanks.

DaleSpam

Fair enough.

Do you think the assumptions I made are the standard ones relevant to Bells spaceships? Do you think they are correct assumptions?