[MikeFontenot (OP)]

Here is the graph for D = 0.5 and 1.0, and the data from the program.

[MikeFontenot (OP)]

Here is the graph for D = 0.5 and 1.0, and the data from the program.

I wanted full size.  I'll try again.

Scan 2023-7-9 12.59.51.jpg (651.8 kB . 1700x2338 - viewed 665 times)

Scan 2023-7-9 12.59.51.jpg (651.8 kB . 1700x2338 - viewed 665 times)

[MikeFontenot (OP)]

Trying again to get the new diagram.

[MikeFontenot (OP)]

Trying again to get full size

Scan 2023-7-9 13.37.06.jpg (518.85 kB . 1700x2338 - viewed 680 times).

[MikeFontenot (OP)]

Here is the diagram with D = 2 and D = 3 added:


Scan 2023-7-10 10.34.13.jpg (534.35 kB . 1700x2338 - viewed 596 times)

[MikeFontenot (OP)]

And here is the computer output for the D = 3 case:


Scan 2023-7-10 10.35.22.jpg (670.34 kB . 1700x2338 - viewed 605 times)

[Halc]

Note that all the [ships with greater initial separation] 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.

Your own graph contradicts your own denial. Ships at 2 and 3 are both moving backwards. A ship at 4 would do so faster than c. You seem to not find any of this problematic. Faster than light communication, and even faster than light ships, accelerating in the opposite direction of where their accelerometers indicate. Ship at 1 accelerating at 1g but not actually going anywhere for weeks.

So what's up with all this nonsense? Why are you sticking with it?

[MikeFontenot (OP)]

I was surprised that all of the curves for various choices of the initial separation "D"  apparently DON'T have the same qualitative shape ... at least that's what my computer program says, and I haven't been able to find any errors in it, so far.  The more accurate smooth curves are consistent with your straight-line approximations.  That IS a big surprise to me.   In my original analysis (quite a while ago, now), I THOUGHT it would be possible to factor out the original distance parameter "D", and that the remaining basic curvature would apply to all choices of "D".  I need to look back at that analysis, and try to see where it goes wrong.  At least, the latest results still DO say that the thread doesn't break!

[Halc]

at least that's what my computer program says, and I haven't been able to find any errors in it, so far.

I've pointed out about 6 of them so far. If you paste your code, I can be more specific. As I said, excepting the d1 numbers, I reproduced all your numbers (every one of which is wrong).. Code would show how you got those. None of it is consistent with SR.

The more accurate smooth curves are consistent with your straight-line approximations.

My approximations took only 4-5 data points each, not 30, and they worked with only 1.5 digits of precision since you had not yet put out the number table when I drew that. I simply did it by eye.

I need to look back at that analysis, and try to see where it goes wrong.

Doesn't occur to you to just look at the posts in this thread, which says pretty much what goes wrong? Even if you got the d1, v, and gamma numbers correct, if you still do the same trick to get d2 numbers, you'll still get the backwards, faster than light motion. You're completely disregarding relativity of simultaneity in the whole analysis.  Under SR, if a rigid accelerating object is stationary (has everywhere identical velocity) for a moment in an inertial frame, it cannot have identical velocity anywhere in any other frame, so the gamma cannot apply along its entire length like you are attempting. It would be a contradiction if this were not so.

At least, the latest results still DO say that the thread doesn't break!

Your program shows that the thread would have to move faster than light in order to be that length in the original inertial frame. I notice you don't compute its length in any other frame.  Guess what?  It breaks (or bunches up, depending on frame of choice). The SR solution has none of these problems.

[MikeFontenot (OP)]

I'll try to print out my program:


Scan 2023-7-10 14.53.42.jpg (302.93 kB . 1700x2338 - viewed 624 times)

[Halc]

I'll try to print out my program:

One programming nit bug: d2 is assigned twice, the first one being immediately discarded. It doesn't affect the results, but that line can be deleted.

Thanks for the code. I updated one line of mine and it now reproduces yours.
I don't have any input, just a couple of defines at the top for D0 and MF, so a recompile is needed to change these. You can add the scanf to make that easier.
The MF variable is boolean, 1 to do it your way, and 0 to do it the SR way.
I don't specify A. Since it is always 1, the math is a little simpler. It can be added in to get a different curve than this sort of unit acceleration.

And my compiler will not compile yours without stdio.h. Yours is apparently fine with referencing undeclared functions, or maybe your math.h includes stdio. Mine doesn't.

Code: [Select]

#include <stdio.h>
#include <math.h>
#define DD 0.5
#define MF 1        // 1 to print Mike's number, 0 to print mine.
int main()
{
  double ctime;        // Coordinate time of inertial frame
  double ptime;        // Proper time of ship
  double md1;          // Computed distance per Mike
  double srd1;         // Computed distance per SR
  double vm, vsr;      // Speed per Mike and SR
  double gim, giSR;    // Gamma factor inverted, per Mike & SR
  int    mf = MF;
  printf("ctim%s   d1     d2      v       gamma\n", mf ? "" : "  ptime ");
  for (ctime = 0.1; ctime <= 3.05; ctime += 0.1)
  {
     ptime = asinh(ctime);  // proper time
     // Compute speeds as a function of time
     vm = tanh(ctime);
     vsr = tanh(ptime);
     // Compute gamma as a function of those speeds
     gim = sqrt(1 - vm * vm);
     giSR = sqrt(1 - vsr * vsr);
     // Compute distance traveled by lower ship
     md1 = log(cosh(ctime));
     srd1 = cosh(ptime) - 1.;
     // I think we're good. Print results
     if (mf)
         printf("%.1f %.5f %.5f %.6f %7.4f\n",
                ctime, md1, md1 + gim*DD, vm, 1./gim);
     else
         printf("%.1f %.5f %.5f %.5f %.6f %8.4f\n",
                ctime, ptime, srd1, srd1 + DD, vsr, 1./giSR);
  }
}
Note the corrections in computing
time:  ptime vs ctime,
velocity: vm and vsr,
gamma gim and giSR,
md1 and srd1 (distance of rear ship).
In my code d2 is just d1 + D. Any other curve and the 2nd ship simply doesn't have 1G of acceleration.
Your have d2 being a function of what d1 does, which is instant cause/effect at a distance, which is, as I've pointed out, faster than light communication, and every retrocausality in different frames. This (plus the faster than light travel, backwards motion when accelerating forward, etc) all don't seem to strike you as contradictory.

[MikeFontenot (OP)]

In my code d2 is just d1 + D.

The whole point of this exercise is that the length contraction equation (LCE) of special relativity REQUIRES that the initial inertial observers say the distance between the spaceships MUST decrease by the gamma factor as the speed increases.  That's one of the two most important equations in all of special relativity (the other being the time dilation equation (TDE) ).  And this diagram IS from the point of view of the initial inertial observers.  So the appropriate equation is

  d2  =  d1 + (D / gamma).

You aren't working the appropriate problem.

[MikeFontenot (OP)]

There's GOT to be a mistake somewhere in my program, because for the leading rocket (at "t" just after the rockets are fired) to be moving TOWARD the initial inertial observer at the starting point of the trailing rocket, the leading rocket would have to have been pointing back toward the trailing rocket.  And that contradicts the statement of the scenario.  So that's my job today (and maybe for many days): finding that mistake.

[MikeFontenot (OP)]

I've decided that those curves are correct.  I was wrong to say that the backward movement of the leading rocket required that the rocket be turned around ... it happens solely because of the length contraction equation, the leading rocket doesn't turn around or reduce its thrust.  The trailing rocket can't go upward on the chart to reduce the separation, because that would result in a speed greater than light.  It's the leading rocket that must go downward on the chart (for D > 1).

[MikeFontenot (OP)]

But you're already positing going FTL, [...]

Why do you say that?

Update: OK, I see why you said that.  I ran the case where D = 6.  It shows a maximum velocity for that leading rocket to be about 2.3 ly/y between t = 0.8 and t =  0.9.

But I'm not sure that actually violates any rules.  Lets just start with the length contraction equation.  It says an inertial observer (he) at some time t1 will say a yardstick moving at speed v1 wrt himself is only 1/gamma_1 yards long.  Suppose that observer then instantaneously changes his speed with respect to that yardstick to v2 (but then becoming inertial again), so that the yardstick is instantly 1/gamma_2 yards long.  So in an infinitesimal time, the yardstick has instantaneously gotten shorter, which means that one or both ends of that yardstick just moved at an infinite speed (drastically greater than the speed of light).

My question to you (Halc) is, how would YOU apply the length contraction equation in this case?  And, a more general question for you, is what does the SR theory that you mentioned have to say about this scenario?  I.e., what is your alternative solution?

New addition:
My GUESS is that your alternative solution is that the upper curve should look just like the bottom curve, just shifted up by "D".  (i.e., like the original chart that started this whole discussion).  If I'm right about that, how do you square that chart with the length contraction equation, which says the inertial observer MUST conclude that the separation of the two rockets must get smaller by the factor gamma?

Still newer addition:

That original chart was claimed to be the perspective of the initial inertial observers, who were stationary wrt the rockets before the rockets were fired. For that to be true, it means that the people on the rockets would say that the separation of the two rockets was INCREASING with time, which means that the leading rocket was producing more thrust that the trailing rocket.  So that is a different scenario from what we've been talking about.

What that original chart ACTUALLY showed was the perspective of the people on the trailing rocket.  THEY said the separation of the rockets is constant, and that accelerometers on the two rockets show the same acceleration.  And for that chart, the viewpoint of the initial inertial observers is the one we've been discussing (where the upper curve is above the lower curve by the distance D / gamma).

[MikeFontenot (OP)]

Halc, I just updated my previous response to you, so take a look again at that response.

[MikeFontenot (OP)]

And I just updated that post again.

[MikeFontenot (OP)]

And I just updated it again.

[Halc]

But I'm not sure that actually violates any rules.

Yea, I noticed that violations of causality and locality don't seem to bother you. You don't seem to know the difference between an abstract coordinate choice and physical causation. You don't mind an accelerometer on a ship that lies and says the ship is accelerating forward when in fact it is accelerating the other way.

how would YOU apply the length contraction equation in this case?

I wouldn't of course. The equation is only applicable to describe the coordinate separation between two parallel straight (unaccelerating) worldlines. There are none of those in this case.

I.e., what is your alternative solution?

It's not my solution. My program prints the numbers per SR, as did your chart from 20 years ago. Read any website on Bell's spaceships, since that is exactly this scenario. None of it is anything I'm personally speculating.

If I'm right about that, how do you square that chart with the length contraction equation

Since the equation is entirely inapplicable to accelerating worldlines, there's no conflict with it.

which says the inertial observer MUST conclude that the separation of the two rockets must get smaller by the factor gamma?

This is your assertion, completely unbacked, and driven to obvious self contradiction, which doesn't seem to bother you.

For that to be true, it means that the people on the rockets would say that the separation of the two rockets was INCREASING with time

Which is why the 'string' breaks, yes. The rest of your addition to this post is just more trolling.

I don't really think you care what anybody says. You seem entirely closed to actually learning something (that you've apparently forgotten over time). I see little point in going on. Post your nonsense. Put it online and sell it on Amazon, so everybody can have a good laugh. Post on another forum and ignore their corrections. But remember: by definition, you cannot be wrong about this. It's the rest of the world that has gotten it all wrong for over a century.

[MikeFontenot (OP)]

Instead of modifying the incorrect diagram (in which the initial inertial observers say the two spaceships have a constant separation) to get a diagram where their separation decreases by the factor gamma by lowering only the upper curve, I modified the incorrect diagram by keeping the center of mass of the two spaceships unchanged (moving the lower curve upward, and the upper curve downward, by the same amount).  That produces a MUCH nicer result.  I'll try to attach the diagram below.


Scan 2023-7-18 17.19.20.jpg (716.51 kB . 1700x2338 - viewed 438 times)