Bell’s Spaceship Paradox. Does the string break?

I have, possibly in the TNS context, but fixing the equations is off topic. If you got that part right, you'd still think the string doesn't break.

I took my formulas from reviewed sites, still not as good as a textbook, but I have no textbooks, and the numbers from my formulas match those of the calculators available online.

I used the nice natural unit of 1 ly/y², which is about 3% under 1g, making it also a very relatable acceleration.

Still unanswered, 3rd time:

1 How does an engineer know to use the equation for a trailing rocket or not? How does NASA get anything into space if the mathematics requires knowledge of unknowable things?

2) Suppose there is a large number of ships taking off all at once, one every 10 light years. You know the series of ships is there, but you don't know where you are in the list. There might be an unbounded number of ships in the line. Now how far away from the launchpad does our ship get after 2 months?

3) The first derivative of velocity is (coordinate) acceleration. True or false?

 

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Halc, I've looked back to see how I computed the lowest curve in the chart that shows what the initial inertial observers say for rockets starting at zero, 0.5, 1.0, 2.0, and 3.0. That's the chart that shows (for the higher-most curves), a negative slope for "t" around 1.0. I found the C program I used to produce that data, and re-ran it. Everything checks out. For example, the bottom curve (which is the one of interest right now) starts at zero, then is about 0.434 at t = 1, and then about 1.325 at t = 2, and then about 2.309 at t = 3. The two equations I used (with A = 1) were

v(t) = tanh(t)

and

X(t) = ln [ cosh(t) ].

X(t) is just the integral of v(t). That's the way I computed it for many years, doing a numerical integration. I only recently learned about the ln[cosh] closed-form solution for the integral of tanh(t).

I learned that first equation twenty or thirty years ago, from Taylor and Wheeler's book "Spacetime Physics", page 98 of the 1966 edition.

I'll try to attach that chart again.

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What do YOU get for the lowest curve?

 

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And nowhere does your argument about the string not breaking depend on getting those equations right. Hence my not harping on it until you started to.

So I think you're doing something wrong in your calculations.[/QUOTE]Mine agree with the web, including matching the values given by any decent calculator for such things. Your don't, and you've not given any reference for them.

Your v equation is close, and again, works for low v only.

Corrected version is trivially different. v = tanh(A * τ), where τ is proper time. You've been using t (coordinate time) for this calculation. So your v calculation for high values of t returns values that actually would require arbitrarily larger acceleration to achieve in the time indicated. For instance, after 5 years, it would take over 10 g of actual acceleration to achieve the velocity (0.9999c) your formula returns for just 1g for 5 years.

I think you misread it then, mistaking t for τ

The equation with the log in it is alien. Never seen it anywhere.

Mine assumes A=1, (as does yours), so any reference to A and c is dropped. I'd have to put the A back if I was to compute for different accelerations than 1.

newXcoord(t) = X (starting coordinate) + cosh(τ) - 1
where τ = asinh(t) for acceleration of 1.

Note that this works for any ship. The physics doesn't vary from ship to ship.




Still no answer to my questions. This is the 4th asking. I must be hitting pain for you to concentrate so hard on this sidetrack.
Are you perhaps not so confident in your assertions as you profess?

 

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More general form:
Distance at coordinate time t: X(t) = (c²/A) (cosh(asinh(At/c) - 1)
proper time τ: cτ(t) = (c²/A) asinh(At/c)

This can be seen here: https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

Another fun task: Do a Lorentz transform of the picture you have above. I think there's zero chance of you doing that, so maybe I will since it shows the string needing to be very different lengths than you give it at different times.

That comment prompts me to remind you that any equation like LCE is descriptive, not proscriptive. You seem to have gotten that quite backwards.

 

My (Mike's) comments in red:

 

I couldn't find your reference in my Taylor and Wheeler text, but a quick look at my other references supports the position of halc.

v = at/sqrt(1+ (at/c)^2)
and
v = c * tanh(aT/c)

where
c = velocity of light (meters/sec)
v = relative velocity (meters/sec)
a = proper acceleration (meters/sec/sec)
t = elapsed time in inertial frame (sec)
T = elapsed time in accelerating object's frame (proper time) sec

 

On about the second page of my Taylor and Wheeler's Spacetime Physics, down near the bottom it gives two copyright dates: 1963 and 1966. What date or dates does your copy give? If it just says 1963 on your copy, you've got the old edition that probably doesn't discuss the equation of interest to us. The equation of interest is addressed in the Contents section of my 1966 version under "Exercises of Chapter 1", starting on page 60. The equation of interest is in exercise 51 (Clock paradox III, starting on page 64). The actual equation is on page 66. Under item (c), it says:

At any instant the velocity of the spaceship in the laboratory frame is related to its velocity parameter (which is what I call the "rapidity") by the equation dx/dt = tanh(theta), (where theta = A*t), so that the distance dx covered in laboratory time dt is

dx = tanh(theta) dt.

So that says that the velocity v(t) in the lab frame (which is dx/dt) is equal to tanh (A*t), which is what I have been maintaining. And the distance covered (in the lab frame) is just the integral of v(t).

 

I can't find your references again. My copy of Spacetime Physics is copyrighted 1992. I expect it has been extensively rewritten since 1966. In any event, it's not important enough for me to spend any more time searching.

 

You are working (as you have from the beginning) from the delusion that Mike cannot be wrong. Anyone who cannot be wrong also cannot learn anything.
I don't have my copy of T&W with me at the moment, and I find it far more likely that a known troll has reading comprehension skills than the alternative, that T&W actually says what you claim it does.

There's a simple test you know.
Plot the curve for the one lone ship, A=1 ly/y², for 3 years in S (like your chart shows).
Consider the final data point (t=3, x=~2.3) and do a Lorentz transformation of all the data points to the frame of that final one .
In that frame, the ship starts out moving in the negative direction, and should be accelerating at a constant 1 ly/y², coming to a halt 3 years later in S', as measured in S'.
In other words, the two plots should be entirely symmetrical, one being the mirror image of the other.

Does your calculation yield such a mirror image plot?
Do you know how to do a Lorentz transformation? Is the validity of such a transformation something else of which you are in denial?
Is this too convoluted of a task?
I ask because anything following your plot must be accelerating at an ever increasing rate, and thus the Lorentz transform will show it accelerating at far more than 1g just before it comes to a halt. That in turn will manifest as a non-mirror image.

Can you do that?


Or can we get back on track and you answer my questions? 5th asking now.

 

dup post

 

It's in the wiki link I provided.

Agree, so the question is if it is the correct equation being integrated. I took my equation from the wiki site, and it produces values that match that of continuous acceleration calculators online.

Furthermore, I did the Lorentz transform to the final velocity after 3 coordinate years, which should produce a mirror image plot in S' of a ship slowing down from the same distance over 3 coordinate years. I wrote the below code to do it. Mike has written similar code (this very example) which produced identical S numbers, but never S' numbers.

Code:

// Compute lorentz transform of a ship accelerating at 1 ly/y/y
// Frame S is computed, then transformed to S'
#include <stdio.h>
#include <math.h>
#define ITTERS 30      // itterations
#define STEP 0.1
int main(int ac)
{
  int    mf = ac > 1;  // Do it Mike's way
  int    itter = ITTERS;
  double duration = STEP * ITTERS;
  double ctime = 0.;   // Coordinate time of inertial frame
  double ptime;        // Proper time of ship
  double pos;          // Computed position
  double vel;          // Velocity in S
  double lttime;       // Time in S'
  double ltpos;        // Computed position
  double ltvel;        // Velocity of S' relative to S
  double ltgamma;      // Gamma of difference in frame velocities
  double ltpoffset;    // Used to put position origin at last event
  double lttoffset;    // Used to put time origin at last event
  double gamma;        // Gamma factor inverted, per Mike & SR

  // Compute origin of S' frame
  if (mf) {   // Mike's way
     ltvel = tanh(duration);    // speed at coordinate time
     pos = log(cosh(duration)); // distance traveled by ship
  } else {   // SR way
     ptime = asinh(duration);   // proper time
     ltvel = tanh(ptime);       // speed as a function of proper time
     pos = cosh(ptime) - 1.;    // distance traveled by ship
  }
  ltgamma = 1 / sqrt(1 - ltvel * ltvel);
  ltpoffset = ltgamma * (pos - ltvel*duration);
  lttoffset = ltgamma * (duration - ltvel*pos) - duration;

  puts("S time S pos   v       gamma   S'time  S'pos");
  while (itter-- >= 0) {
     if (mf) {   // Mike's way
        vel = tanh(ctime);         // speed at coordinate time
        pos = log(cosh(ctime));    // distance traveled by ship
     } else {   // SR way
        ptime = asinh(ctime);      // proper time
        vel = tanh(ptime);         // speed as a function of proper time
        pos = cosh(ptime) - 1.;    // distance traveled by ship
     }
     gamma = 1 / sqrt(1 - vel * vel);
     // Do Lorentz transform of the data point to the coordinate in S'
     ltpos = ltgamma * (pos - ltvel*ctime) - ltpoffset;
     lttime = ltgamma * (ctime - ltvel*pos) - lttoffset;
     printf(" %.1f %.5f %.6f %7.4f  %7.4f %.5f\n",
            ctime, pos, vel, gamma, lttime, ltpos);
     ctime += STEP;
  }
}

Running it with no arguments produces this:

Code:

> a.exe
S time S pos   v       gamma   S'time  S'pos
0.0 0.00000 0.000000  1.0000   0.0000 2.16228
0.1 0.00499 0.099504  1.0050   0.3013 1.87805
0.2 0.01980 0.196116  1.0198   0.5730 1.62490
0.3 0.04403 0.287348  1.0440   0.8166 1.40151
0.4 0.07703 0.371391  1.0770   1.0338 1.20588
0.5 0.11803 0.447214  1.1180   1.2270 1.03553
0.6 0.16619 0.514496  1.1662   1.3988 0.88782
0.7 0.22066 0.573462  1.2207   1.5516 0.76005
0.8 0.28062 0.624695  1.2806   1.6879 0.64969
0.9 0.34536 0.668965  1.3454   1.8100 0.55441
1.0 0.41421 0.707107  1.4142   1.9196 0.47214
1.1 0.48661 0.739940  1.4866   2.0187 0.40106
1.2 0.56205 0.768221  1.5620   2.1086 0.33964
1.3 0.64012 0.792624  1.6401   2.1906 0.28652
1.4 0.72047 0.813733  1.7205   2.2658 0.24059
1.5 0.80278 0.832050  1.8028   2.3351 0.20088
1.6 0.88680 0.847998  1.8868   2.3993 0.16657
1.7 0.97231 0.861934  1.9723   2.4589 0.13699
1.8 1.05913 0.874157  2.0591   2.5147 0.11153
1.9 1.14709 0.884918  2.1471   2.5671 0.08970
2.0 1.23607 0.894427  2.2361   2.6164 0.07107
2.1 1.32594 0.902861  2.3259   2.6630 0.05527
2.2 1.41661 0.910366  2.4166   2.7072 0.04199
2.3 1.50799 0.917070  2.5080   2.7493 0.03095
2.4 1.60000 0.923077  2.6000   2.7895 0.02192
2.5 1.69258 0.928477  2.6926   2.8279 0.01469
2.6 1.78568 0.933346  2.7857   2.8649 0.00909
2.7 1.87924 0.937749  2.8792   2.9004 0.00494
2.8 1.97321 0.941742  2.9732   2.9347 0.00213
2.9 2.06757 0.945373  3.0676   2.9679 0.00052
3.0 2.16228 0.948683  3.1623   3.0000 0.00000

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In the plot, black is the frame S numbers, and red plots the two S' columns on the right.
That the S' numbers in the first line match the distance and time in S in the bottom line is validation of the formula used.
Running it with arguments uses Mike's equations and produces this:

Code:

> a.exe mike
S time S pos   v       gamma   S'time  S'pos
0.0 0.00000 0.000000  1.0000  -4.0684 6.80409
0.1 0.00499 0.099668  1.0050  -3.1117 5.85255
0.2 0.01987 0.197375  1.0201  -2.2539 5.00054
0.3 0.04434 0.291313  1.0453  -1.4923 4.24513
0.4 0.07795 0.379949  1.0811  -0.8223 3.58175
0.5 0.12011 0.462117  1.1276  -0.2379 3.00442
0.6 0.17014 0.537050  1.1855   0.2678 2.50623
0.7 0.22727 0.604368  1.2552   0.7022 2.07965
0.8 0.29075 0.664037  1.3374   1.0730 1.71699
0.9 0.35983 0.716298  1.4331   1.3877 1.41065
1.0 0.43378 0.761594  1.5431   1.6537 1.15337
1.1 0.51194 0.800499  1.6685   1.8775 0.93842
1.2 0.59369 0.833655  1.8107   2.0653 0.75970
1.3 0.67850 0.861723  1.9709   2.2224 0.61173
1.4 0.76589 0.885352  2.1509   2.3538 0.48974
1.5 0.85544 0.905148  2.3524   2.4634 0.38956
1.6 0.94681 0.921669  2.5775   2.5549 0.30761
1.7 1.03968 0.935409  2.8283   2.6312 0.24086
1.8 1.13381 0.946806  3.1075   2.6950 0.18673
1.9 1.22898 0.956237  3.4177   2.7484 0.14305
2.0 1.32500 0.964028  3.7622   2.7932 0.10802
2.1 1.42174 0.970452  4.1443   2.8309 0.08012
2.2 1.51906 0.975743  4.5679   2.8627 0.05810
2.3 1.61685 0.980096  5.0372   2.8898 0.04092
2.4 1.71505 0.983675  5.5569   2.9128 0.02772
2.5 1.81357 0.986614  6.1323   2.9326 0.01779
2.6 1.91235 0.989027  6.7690   2.9498 0.01055
2.7 2.01136 0.991007  7.4735   2.9647 0.00551
2.8 2.11054 0.992632  8.2527   2.9779 0.00228
2.9 2.20988 0.993963  9.1146   2.9895 0.00053
3.0 2.30933 0.995055 10.0677   3.0000 0.00000

The symmetry is lost this time. Clearly the acceleration vastly is increasing over time.

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Refer to the chart in my post #282. You are (incorrectly) saying that the bottom curve in that chart is according to the people in the trailing rocket, NOT the initial inertial observers. If you make that incorrect assumption, then to get the curve according to the initial inertial observers, you need to divide each point on the curve for the people on the trailing rocket by gamma. If you do that (I just did), you will find that the resulting curve, according to the initial inertial observers, will initially curve upwards, but will then curve downwards, and will then asymptotically approach the horizontal axis as their time gets large. That's absurd: the initial inertial observers say the trailing rocket ends up back at the starting position, where those observers are ... they get run over by the rocket, even though it's got flame coming out its nozzle in their direction the whole trip. Here are a few data points: ("d" is the position of the curve for the people on the trailing rocket, and "d_new" is the position of the curve for the initial inertial observers.)

For t = 1 year, v = 0.762, gamma = 1.54, d = 0.434 and d_new = 0.281.
For t = 2 years, v = 0.964, gamma = 3.762 d = 1.325 and d_new = 0.352.
For t = 3 years, v = 0.995, gamma = 10.068 d = 2.309 and d_new = 0.229.
For t = 10 years, v = 1- , gamma = 11013 d = 9.307 and d_new = 0.00085

Thanks, but no thanks. I'm sticking with my curves in post #282 for the initial inertial observers. You can continue to maintain that the bottom curve in my chart is the view of the people on the trailing rocket, but then you get a view by the initial inertial observers which is absurd (and very painful for them).

 

I said no such thing. In the accelerating frame of the lower rocket, the rocket would always be at itself at all times, so the plot would be a flat line at x=0.
Don't worry, I'm going to plot that one as well.

The numbers for the black curve exactly match the curve in your chart in 282. They are your numbers. They also match the text values provided in this post, including the gamma values:
https://www.thenakedscientists.com/forum/index.php?topic=86264.msg708175#msg708175

If a mistake was made, then you made them. What I did (with both our numbers) is to take the inertial frame of the last event (at t=3) and use the inertial reference frame of the lower rocket at that event, and plotted the exact same data points in that reference frame (the red curve in each picture).
It should show a ship slowing down over exactly 3 years at 1g, an identical curve but mirror image.

Your numbers do not exhibit this property, meaning either
1) you don't mind your numbers being self-contradictory.
2) You are in denial of the validity of Lorentz Transforms (never know with you, since you are in denial of so much of SR)
3) You believe physics is different from one frame to the next (denial of 1st postulate), and a constant 1g acceleration to a halt in S' would actually look like that

I'm pretty much betting on option 1. Sure, the numbers show blatant inconsistencies, but that doesn't deter Mike, even though the use of corrected numbers doesn't effect your argument about the string not breaking. It's wrong for different reasons, which is why I never harped on them being wrong.


I have solid evidence that you don't believe your assertions. If you did, you would have the confidence to answer the questions I put to you.
The fact that you evade them (at least 6 times and counting) shows that you know it shows obvious contradictions in your assertions. You seem not to mind contradictions in general, but apparently these lead to ones that actually make you uncomfortable. 7th asking on some of these:

1) Your chart applies the LCE to the string between the ships (let's say a rod being pushed by the trailing ship). The upper ship's trajectory, instead of being a function of the formula for position over time at constant acceleration, instead, according to you, supposedly tracks the forward end of the rod, which is everywhere moving at the velocity of the trailing ship that is pushing it. If it is not over its entire length moving at the same speed as the trailing ship to which it is attached, then the LCE would not yield its actual length in S, and the end of the rod wouldn't be where you are computing/plotting it. If the entire rod is moving at the same speed as the trailing ship to which it is attached, then it cannot be attached to the leading rocket as well, which is (by your own admission above to one of the questions I asked) not moving at the same velocity as the trailing ship. So which is it? Is the string not moving at the same velocity as the trailing ship to which it is attached, or is it not moving at the same velocity as the leading ship to which it is also attached? Final option: Run away, because I don't actually believe the assertions behind which I stand are mathematically consistent.

2) Suppose there is a large number of ships taking off all at once, one every 10 light years. You know the series of ships is there, but you don't know where you are in the list. There might be an unbounded number of ships in the line. Now how far away from the launchpad does our ship get after 2 months?

Technically, it can be done with a roller coaster track, which wiggles and curves randomly. A series of roller-coaster cars are equally separated and attached by string between them (at midpoints of each, so contraction of cars won't change string lengths). They all commence identical constant proper acceleration relative to the track and unless new cars are added, the length contraction of the strings must break at least one of them. It's Bell's paradox, except a bit more complicated since it involves more than one spatial dimension. If the track is circular, it is Ehrenfest scenario, which also says that the string breaks.

3) The first derivative of velocity is (coordinate) acceleration. True or false or run-away? So far, 6 votes for the 3rd option and counting.

 

My (Mike's) comments in red. Halc's comments are in black.

In your (Halc's) post # 294, you (Halc) quoted me (Mike) as saying (in my post #293):

"Refer to the chart in my post #282. You are (incorrectly) saying that the bottom curve in that chart is according to the people in the trailing rocket, NOT the initial inertial observers."

Then you (Halc) said (in your post #294):

"I said no such thing."

But you DID say exactly that! In your post #283, I (Mike) had said:

"v = tanh(A * t) ."

You (Halc) then said (in your post #283):

"Corrected version is trivially different. v = tanh(A * τ), where τ is proper time. You've been using t (coordinate time) for this calculation."

 

According to the person in the trailing rocket, at each instant "tau" of his life he knows how far he has traveled since he ignited his rocket. I.e., he knows how far he is from the initial inertial observers at that instant in his life. Call that function D_rocket_person(tau).

And according to the initial inertial observers (who are perpetually stationary at the launch site), at each instant "t" of THEIR lives, they know how far the rocket is from them. Call that function D_initial_inertial_person(t).

We are arguing about which of these two functions is the lowest curve on the charts I've uploaded (in post #282). The answer is that in post #282, the bottom curve is D_initial_inertial_person(t). The other function, D_rocket_person(tau), is the one I described (but didn't plot) in post #293, and which gives an absurd result.

The chart I posted in post #282 IS the correct view of the initial inertial observers, including the lowest curve. The lowest curve on that chart is NOT the view of the person in the trailing rocket. It it WERE the view of the person in the trailing rocket, that would imply that the view of the initial inertial observers for the lowest curve is the one I described, but didn't plot, which I showed to be absurd ... the initial inertial observers would say the trailing rocket ends up coming back to the starting point, even though its rocket blast is still coming out in the direction of the initial inertial observers ... obviously a nonsensical result that must be rejected.

 

You are being entirely inconsistent. You indicate in your 'proof' that a person 'must conclude' this and that. Your proof depends on this obligation concerning beliefs. It implies that a person is obligated to always use the inertial frame in which he is stationary at a given moment. According to any person, that person thus cannot have traveled anywhere, else he would be at a nonzero displacement from where he currently is. So D_rocket_person(tau) = 0 at all values of tau, by your insistence on this obligation. He apparently is obligated to believe that it is everything else that travels, never himself.

I on the other hand stick to special relativity, the first postulate being that one inertial frame is just as good as any other, and that any person is free to use/believe any one of them that he might find most pragmatic at a given moment. Hence 'according to some person' has no meaning if the free choices of that person are not known, so I would never use such wording.

v(t) = tanh(A * asinh(At)) gives the correct velocity relative to frame S of an object accelerating from a stop in S at constant proper acceleration A (natural units) after t natural units of time. My code computes this, and also computes it your way. Both ways compute velocity (one incorrectly) of the one ship relative to S, not relative to the ship, which would be trivially zero at all times, and not 'according to person X' because the beliefs of person X have not been specified.

That would then be relative to that inertial observer, not 'according to the guy in the ship'. The guy on the ship might ask how far away the initial inertial observer is, a different question, and that would have a frame dependent answer. Hence I always specify a frame instead of referencing the personal beliefs of some occupant of a hypothetical scenario populated by potential morons.

Agree (despite your weird wording). That's what my code computes, both the way the physics text say, and also the way you say. It produces the black curve in both pictures (which is just a plot of the computer output). Those curves are not identical. The 'your way' numbers exactly match those you posted.

There is no upper curve on either chart. There is the one black curve, and it is what you call D_initial_inertial_person(t) in both charts, one computed per the physics texts, and one computed your way. The 'your way' one matches the curve you posted. The post says that the plot is relative to frame S. The code also says this. The red curve in each picture is not relative to S. It uses the same grid, but is a plot of the exact same 31 data points in S', also an inertial frame.

So you continue to assert, despite all the contradictions that result from it. You ignore those contradictions, but you've not refuted a one of them. You also chose 'run away' as the answer to each of the questions designed to see if you have any confidence in your supposed belief in your correctness. I said in the prior post that I had evidence that you very much know that you're slinging BS. I applied the test, and you confirmed what I've suspected all along. You will now re-profess your belief, which is exactly what I would expect from a slinger of BS. A mere moron would question his absurd beliefs when the contradictions were identified, or would at least attempt to refute them and not just assert that the contradictions are permissible when convenient, but absurd when that's what's convenient.

 

Halc, I'm not "baiting" you here. I've realized that the "rapidity" (or "velocity parameter") ISN'T

theta = A * t,

where "t" is the age of the initial inertial observer (IIO), as I have always been assuming. Instead, it's

theta = A * tau,

where "tau" is the age of the people on the trailing rocket. And with that change, I haven't been able to figure out how to proceed. Your equations in your last post seem to indicate you DO know how to proceed, so I could use some help.

 

Where tau is the elapsed proper time of the rocket since it has been stationary in the frame relative to which the rapidity is being computed. The people's ages could be anything. It's a nit. I know you cannot let go of the need to put humans everywhere to make things meaningful. Yes, rapidity is calculated that way. If there is a grid of little sign posts marking the light years or km, stationary in S, then that computes how frequently they pass by the rocket as measured by the rocket clock.

I don't think I bothered to compute rapidity in any of the stuff I've posted recently since none of the figures are relative to the non-inertial frame of any rocket. I can perhaps help if I know what you're trying to achieve. The one equation in my last post computes velocity relative to S after time t in S. If you look at the code I posted, that value is computed and output, but not what I plotted. I only plotted the distance values. The code uses simplified formulas that would need to be more generalized if proper acceleration was anything but 1, or if non-natural units are used.

The curve I plotted isn't visually a whole lot different than the one you plotted, but due to the large difference in gamma values, your plot of the lead ship (according to your assertions) would be effected by the string contracting less quickly. It would still go backwards and faster than light with sufficiently large initial separation. You still have the end of the string moving at a very different velocity than the lead ship to which it is supposed to be attached.