[MikeFontenot (OP)]

Here is the corrected diagram:

[MikeFontenot (OP)]

Here is the corrected diagram:


Scan 2023-6-27 12.20.28.jpg (643.89 kB . 1700x2338 - viewed 969 times)

[MikeFontenot (OP)]

In the scenario that I am interested in, and which I have analyzed, the
two rockets (immediately after they are ignited) always have the same
constant acceleration, as reported by accelerometers attached to the two
rockets.

In that scenario, the separation of the rockets, according to the accelerating traveler in the trailing rocket, is constant, and the string doesn't break.

According to the initial inertial observers who are stationary wrt the rockets immediately before they are fired, the separation of the two rockets decreases as the acceleration proceeds, as required by the famous length contraction equation (LCE) of special relativity. So they also conclude that the string doesn't break.

The scenario, as given in Bell's Paradox, may be a completely different scenario from the above scenario. As far as I know, there is no mention of rocket accelerometer readings in that Wiki article on Bell's Paradox. If, in Bell's Paradox, the initial inertial observers correctly conclude that the rocket separation doesn't decrease, then the rockets CAN'T be accelerating at the same rate (according to accelerometers attached to the rockets) ... the leading rocket must have a greater acceleration than the trailing rocket. If so, I have no interest in that different scenario.

[paul cotter]

There is no need for accelerometers as the conditions of the experiment state identical acceleration in both rockets.

[pzkpfw]

MikeFontenot thinks with his accelerometers he's found an absolutist loophole to get around relativity.

But two accelerometers on the _same spaceship_ (one at the front, one at the back) won't even agree with each other.

[MikeFontenot (OP)]

It is possible to show that the string doesn't break in Bell's Paradox (assuming that the accelerations of the two spaceships are confirmed with accelerometers to be equal), without referring to ANY inertial observers.

In 1907, Einstein derived the gravitational time dilation equation (GTD), which says that if two clocks are separated  by a fixed distance "L", in a constant gravitational field (with the separation along the direction of the field), such that the field strength doesn't vary with distance from the source of the field, then the clock farther from the source of the field will run faster than the clock closer to source of the field, by the factor f(L*g).

From the equivalence principle, that means that two spaceships (with no gravitational fields anywhere), each accelerating with a constant acceleration "A" (as confirmed by accelerometers), and separated by the distance "L" in the direction of the acceleration at some instant, will always maintain that separation (for as long as that acceleration equals "A").  And clocks in the leading spaceship will run faster than clocks in the trailing spaceship, by the factor f(L*A).

Since the distance between the two spaceships is constant, a thread stretched between them never breaks.

[paul cotter]

Nonsense. Two clocks in identical gravitational fields, regardless of position, will stay synchronised indefinitely. What has the clock rate have to do with the string breaking/not breaking.

[Halc]

I am merging the topics since they're both about the same denial of Bell's string breaking.

Your 'corrected' graph is interesting. I copied it and added some lines for ships at 1, 1.5, and 2. I did my best to keep the separation of successive ships identical. I don't have numbers, so I did it by eye. I didn't take as many data points so my curve isn't as nice as the original ones.  I get this:

FiveShips.jpg (539.02 kB . 1613x1457 - viewed 1110 times)

Note that all the new ships actually, which accelerating forward at first, move backwards initially.
This makes for an interesting way to communicate faster than light.
The superbowl is played at x=0. If the Bills win, the ship there takes off (at exactly kickoff time + 4 hours) at fixed proper acceleration of 1 ly/yr². If they lose, they stay put.
There is another ship at x=1.5 ly. It takes off unconditionally at kickoff time + 4 hours and accelerate at a fixed proper acceleration of 1 ly/yr².  If the ship moves backwards (or for that matter, doesn't reach the 20 meter post in 2 seconds), the Bills have won. If not, they've lost. Presto, faster than light communication.

Concerning your plot:
The ship starting at 0 does seem to be accelerating at the correct rate. Assuming rigid motion (no string breakage), the other ship seems to do ok for at least a year, but looking at year 2, ship0 is moving at 0.9c (hard to measure slope to that precision), location 1.32 (should be 1.25), and gamma of 2.3.  0.5/2.3 is ~0.22 but your contraction is more than that. They seem about 1.4 apart at t=2  Your lower ship is accelerating harder than 1 ly/yr².

All sort of other silliness results from your assertions. You don't care of course. Your reaction to any critique since the beginning is just to repeat the opening argument with the same mistakes, as you have done yet again in what was your new topic.

In the scenario that I am interested in, and which I have analyzed, the two rockets (immediately after they are ignited) always have the same constant acceleration, as reported by accelerometers attached to the two rockets.

Then the accelerometers are broken, because if they look out the window and watch the graph lines go by (one every 0.1 ly), one ship notices a lot more of them going by than the other. Your assertion contradicts your graph.

In 1907, Einstein derived the gravitational time dilation equation (GTD), which says that if two clocks are separated  by a fixed distance "L", in a constant gravitational field (with the separation along the direction of the field)

The paper says no such thing. You're making things up. You've become a crank, which is too bad.
Anyway, citation needed. Where does Einstein associate gravity with the equation mentioned?  What coordinate system is Einstein using? Where does he make any suggestion of the possibility of a uniform gravitational field? Sure, they exist in Newtonian mechanics, but we're not talking about that anymore.

[MikeFontenot (OP)]

Note that all the new ships actually, which accelerating forward at first, move backwards initially.

No they don't.  You are apparently not following my description of how the curves are to be determined.  Try again.

[MikeFontenot (OP)]

Note that all the new ships actually, which accelerating forward at first, move backwards initially.

That doesn't happen.  You have apparently misinterpreted my description of how to produce the new chart.  I'll explain it again.

(In this scenario, two spaceships are initially separated at time t = 0 by distance "D", with their rockets off.  People on the spaceships will say that the separation between the spaceships is constant, before and after the rockets are fired.  After the rockets are fired, the constant acceleration is confirmed by accelerometers on the two spaceships.)

To get a chart that shows the conclusions of INERTIAL OBSERVERS who are stationary wrt the spaceships immediately before the rockets are ignited:

Start out with the two identical curves separated by some fixed distance "D", one starting at the origin, and the other starting a distance "D" up the vertical axis.

The bottom curve is obtained this way:

The rapidity "theta" is

  theta(t) = A * t,

where "A" is the constant acceleration that begins at t = 0.

The velocity is

  v(t)  =  tanh(theta[t]);

The distance traveled is

  d(t)  =  ln(cosh[v{t}]) .

That should allow you to plot the lower curve (which starts at the origin).  And you can plot the upper (incorrect) curve by just moving that lower curve vertically up by the amount "D".

That chart is wrong, because the famous length contraction equation (LCE) of special relativity says that an inertial observer (stationary with the two spaceships immediately before their rockets are fired) will conclude that the spaceships get closer together as their speed increases, by the factor

  gamma  =  1 / sqrt(1  -  v * v) .

Gamma is equal to 1.0 when v = 0, and increases monotonically as "t" increases.
 
To make the two curves get closer together as "t" increases, we can't raise the lower curve, because the lower curve already asymptotically approaches the speed of light as "t" goes to infinity.  So we need to lower that upper curve.  We do that by, at each time "t", dividing the original separation between the two curves "D" by gamma(v[t]).  Gamma starts out equal to 1.0 at t = 0, but then increases (slowly at first, but increasingly fast) as "t" and "v" increase.  For example, when v = 0.866 ly/y, gamma = 2.0 .

If you do the above correctly, you will get the revised diagram that I gave.  Neither of the two curves ever decrease (or stay constant) as "t" increases.

[Origin]

That chart is wrong, because the famous length contraction equation (LCE) of special relativity says that an inertial observer (stationary with the two spaceships immediately before their rockets are fired) will conclude that the spaceships get closer together as their speed increases

Why do you think this?  Clearly each ship will be length contracted, but why would space be length contracted, certainly the space is not moving relative to the stationary observer right?

[MikeFontenot (OP)]

It's understandable that reproducing my diagrams may be difficult for you ... you may not have access to a computer that allows you to compute the tanh, cosh, and log functions.  I can scan a page that gives v, gamma, and d1 and d2 for t = 0 to t = 3 in increments of 0.1, and I will try to post it.

But in the meantime, let me try again to qualitatively explain what's going on.  Regardless of what initial separation you choose (0.5 like I did, or 1.0, or 2.0 ...), the process is the same:  for a given value of "t", you determine "v" on the bottom curve, and then you compute the corresponding gamma for that velocity.  (gamma = 1/sqrt{1-v*v}).  You then take the distance between the two curves at t = 0 (call it D), and divide that by gamma ... call that "d".  Then add "d" to the value of the lower curve at that time "t".  The result is the value of the upper curve at that time "t".  The important point to understand is that that process is the same regardless of how large you chose "D".  The curves you get for the various choices of "D" are all qualitatively the same ... they are just scaled up.

Just FYI:  Here's how the bottom curve is determined:

For each value of "t", determine "theta" (the rapidity) at that time from theta = A * t.  (I had chosen A = 1 ly/y/y.)  Then compute v = tanh(theta).  Then compute d = integral of v.  It's not necessary to do the numerical integration: the closed form solution is d = ln(cosh[theta(t)]).

[MikeFontenot (OP)]

Here's the data to compute my two curves (the bottom one and the modified one starting at D = 0.5 ly.)


Scan 2023-7-5 14.20.13.jpg (652.24 kB . 1700x2338 - viewed 740 times)

[MikeFontenot (OP)]

I don't know why the top part was cut off.  I'll try it again.

[MikeFontenot (OP)]

Scan 2023-7-5 15.47.12.jpg (719.29 kB . 1700x2338 - viewed 732 times)

[MikeFontenot (OP)]

To get a chart that shows the conclusions of INERTIAL OBSERVERS who are stationary wrt the spaceships immediately before the rockets are ignited:

I have no beef with the chart in post 6 that shows this. It seems totally accurate.

In the above, you seem to be OK with the original diagram (the one showing a constant separation), which I now know is incorrect.  The axis labeling uses terms from a long time ago, and don't need to be discussed now.  Basically, the horizontal axis gives the time (or age) of the initial inertial observers who are stationary wrt the two rockets before they are fired, and I would today label that "t".  And similarly for the distance "d" being the distance from the starting locations to the current locations, according to those initial inertial observers (although when I drew that sketch, I used "x" for the vertical axis, not "d").  But my intent was to use the same axes on the revised diagram as I used on the original diagram.  They both show the viewpoint of the initial inertial observers, and I don't discuss the viewpoints of the people in the rockets at all.

Then, you discuss (I think) my hand-drawn revised diagram, and there, you criticize my labeling, etc.  As I look at that hand-drawn diagram now, I agree that I didn't take the time to define the x and t axes as being the viewpoints of the initial inertial observers ... probably because I thought it was obvious that I wanted the same perspective as in the original diagram.

You then said that I use the same symbol t for both Earth (IRF) time and ship time.  That's not true: I don't discuss the viewpoint of the people in the rockets at all in that entire discussion.

[MikeFontenot (OP)]

I HAVE said that, if you take that original diagram, and remove the straight sloped lines, that you DO get the viewpoint of the people in the rear rocket.

[MikeFontenot (OP)]

If you would like me to run my program for a larger choice of the original separation "D", i.e., something larger than D = 0.5, I can do that, and scan the output into a jpeg, and upload it to this forum.  Just tell me what original separation you want.

[MikeFontenot (OP)]

I found your post extremely hard to follow.  But I THINK I see a (really bizarre) mistake you're making here:

Your chart says at t=1, d1 is 0.4338, v=.7616, gamma=1.543
If you take D at 2 (the top line of the edited picture I posted), it starts at x=2 at time 0.
2/gamma is 1.2962 which we add to d1 0.4338 to get 1.73 which is exactly where I drew the data point.

In the above, you use the value of gamma at t = 1 to divide the distance between the curves at t  = 0 to get the new curve at t = 1.  That's completely incoherent!

I think any further discourse would waste both our times.

[MikeFontenot (OP)]

I found your post extremely hard to follow.  But I THINK I see a (really bizarre) mistake you're making here:

Your chart says at t=1, d1 is 0.4338, v=.7616, gamma=1.543
If you take D at 2 (the top line of the edited picture I posted), it starts at x=2 at time 0.
2/gamma is 1.2962 which we add to d1 0.4338 to get 1.73 which is exactly where I drew the data point.

In the above, you use the value of gamma at t = 1 to divide the distance between the curves at t  = 0 to get the new curve at t = 1.  That's completely incoherent!

I think any further discourse would waste both our times.

I said:

"In the above, you use the value of gamma at t = 1 to divide the distance between the curves at t  = 0 to get the new curve at t = 1.  That's completely incoherent!"

A better way for me to say that is:

"In the above, you use the value of gamma at t = 1 to divide the distance "D" between the curves at t  = 0 to get the portion of the new curve for 0 < t < 1."  THAT violates the principle of causality:  effects can't precede causes."  In other words, the value of gamma at t = 1 has NO effect on anything that happens before t = 1.  The value of gamma at any instant can only affect things on or after that instant.

So, what you need to do is, for any t = T where you want a data point for the upper curve, determine gamma(T) for the lower curve, and then divide the original separation "D" by THAT value gamma(T), and add that result to the value of the lower curve at time T, and plot that point.

The result is that no matter what the chosen value of "D" is, the resulting curve will be monotonically increasing everywhere, as time "t" increases.  The curve will not anywhere DECREASE as time increases.  No matter what the choice of "D" is, the shape of the resulting curve will be qualitatively similar to the curve I gave for D = 0.5.  Specifically, it just gets scaled up by the ratio of the two choices of "D".

For example, once you've got the D = 0.5 curve (which we DO have), if you want the D = 1.0 curve, just divide 1.0 by 0.5 to get 2.0, and then,
for each value of "t" you want, multiply the vertical distance from the 0.5 curve down to the bottom curve by 2.0, and THAT gives the vertical distance from the D=1.0 curve down to the bottom curve.

And, to get the D = 2.0 curve, just divide 2.0 by 0.5 to get 4.0, and then,
for each value of "t" you want, multiply the vertical distance from the 0.5 curve down to the bottom curve by 4.0, and THAT gives the vertical distance from the D=2.0 curve down to the bottom curve.

Don't don't lose sight of the original plan.  You start with the original (incorrect) diagram, where the two curves were identical except for their starting point on the vertical axis ... the upper curve was just the lower curve, shifted up by the original distance "D" between the spaceships. That corresponds to the contention that, ACCORDING TO THE INITIAL INERTIAL OBSERVERS WHO ARE STATIONARY WRT THE SPACESHIPS BEFORE THE ROCKETS ARE IGNITED, the two spaceships maintain a constant distance apart during the acceleration. That contention is wrong. 

So, you use the fact that those initial inertial observers MUST instead conclude that that original diagram is incorrect, because the two spaceships MUST get closer together, by the factor gamma, as their speed increases ... that is required by the Length Contraction Equation (LCE) of special relativity.  So those initial inertial observers then conclude that the two spaceships must get closer together, by the gamma factor, as their speed increases ... i.e., the distance "d" between the spaceships must be  d = D/gamma, where gamma is a function of the velocity, and the velocity is a function of the time "t".

But HOW do the two curves change, to get the correct separation "d"?  The lower curve can't be moved upward, toward the upper curve (because it already gets arbitrarily close to the speed of light as the duration of the acceleration approaches infinity).  It is the upper curve which must be modified ... it must not curve upward as much as the lower curve curves upward.  Its curvature must change so that, for each time "t", the vertical distance between the two curves equals the original separation "D" divided by the value that gamma has for the speed at that moment "t".