Bell’s Spaceship Paradox. Does the string break?

So. Have we figured out whether SR is correct, or whether Mike_Fontenot is correct? I prefer the one without the 'Absurdities', but what do I know

 

Log in or Sign up to hide all adverts.

The bottom curve on my plot of the initial inertial observers' view of the position of the trailing rocket is given by the equation

X = tanh(A*t).

(It's shown in post #282).

I now know that to be wrong. I want see what the corrected curve is, and I want to know the equation of that new curve, and how the equation is derived.

Also, if we replace "t" with "tau" in the above equation, does that give the view of the people on the trailing rocket about how far they have traveled from the initial inertial observers?

 

Log in or Sign up to hide all adverts.

I made a mistake as well in the graphs of post 292. I switched the labels on the axes of all the pictures. Time should be horizontal (the way you do it), and X vertical. Too much habit doing it the other way.

No, that equation was (incorrectly) used to compute the velocity relative to S after time t in S.
Your equation for distance was the correct integration of the above incorrect velocity, yielding X(t) = log(cosh(t)).

All the correct equations can be found at https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity), which I posted in 294.

The simplified equations (natural units, A=1) is posted in 293, and is used in the code posted in 292.
I didn't personally derive any of this, letting the experts do that. I did get a derivation sent to me by somebody watching this topic but not posting ever, but I cannot follow it, so I defer to those who did. It contains an interesting alternate way to compute X(t)

In natural units,
X(t) = (1/A) (cosh(asinh(At) - 1) (as used in the code posted in 292)
X(t) = (1/A) (sqrt(A²t² + 1) - 1) (alternate method, produces same number)

Both of those are shown on the wiki site

In the accelerating frame of the rocket, the rocket is by definition stationary at all times and has traveled nowhere, but the launchpad recedes from the rocket over proper time tau.
The distance traveled by the launchpad depends on the choice of frame used to measure it.
If you put the grid of distance markers through the universe that the rocket sees go by, then that's using S as the choice of frame, so it would be 'according to S'. That distance is given by
X(τ) = 1/A cosh(τ) - 1 (Distance in S traveled by the rocket after rocket's proper time τ)

But if you want to do it in the rocket frame, the launchpad is much closer. Consider the CMIF of the rocket at tau=1.82 which happens after 3 years in S.
We will call this CMIF frame S', and we're asking how far away the launchpad is in S'. So I did a hand-modify of my earlier picture to add the worldline of the launchpad in green.

Please Register or Log in to view the hidden image!

Our rocket is the red worldline, and is stationary at time 3 in S'. Ignore the black curve, meaningless in S'.
The launch tool place 3 years ago in S', at the upper-left event. The green line is the launchpad worldline in S', and is nearly -0.7 behind the rocket in S' at the time the rocket is momentarily stationary in S'. Let's actually check that more accurately.

All these numbers come from post 292.
Position of launchpad at time 0 in S' : 2.162
Speed of launchpad in S' at any time: -0.9497
Current position of launchpad in S' at τ=1.8184: 2.162 - 3*0.9497 = -0.687

Not a neat formula, but I'm sure one can be constructed. I showed the way to get the number.

 

Log in or Sign up to hide all adverts.

Not sure where -0.9497 came from. The speed of the launchpad in S' should be -0.9487
so Current position of launchpad in S' at τ=1.8184: 2.162 - 3*0.9487 = -0.684, not -0.687

 

(I'll come back and respond to your latest post (#303) shortly, but I was halfway through writing a post when your latest one came in, and I need to finish that first ... here it is:

I keep coming back to this argument:

Consider the lowest curve (on the chart we've been discussing for so long), which we have been told is the view of the people on the rockets, NOT the view of the initial inertial observers as I believed before. If that is the case, we get the initial inertial observers view by dividing that curve by gamma (where gamma [a function of the velocity, which is the slope of the line] gets larger as time increases). I DID that exercise before, and the result is that the modified curve approaches the horizontal axis as "t" (the initial inertial observer's ages) get large.

Here is a more detailed discussion of that scenario:

An inertial observer (Tom) is present at the launch of the rocket (and colocated with the rocket at the instant of the launch). And suppose that at the instant of the launch, there is another rocket much farther back, behind Tom by a very large distance "L", which simultaneously (according to Tom, and to Dick [the person on the rocket at Tom's position], and to Harry [the person on the rocket at distance "L" behind Tom) starts its rocket.

Much later, the rear rocket (with Harry) gets to Tom. At that instant, Tom knows that the front rocket is VERY far away from him, and the speeds are VERY large, and gamma is VERY large. Tom knows the length contraction equation (LCE) of special relativity, and so he knows that, in HIS frame, the two rockets are now separated by the distance L/gamma. That means that the leading rocket, and the trailing rocket, and Tom, are all essentially at the same place. And it means that the front rocket, which was previously very far away from Tom, has backed up to Tom, even though its rocket has been pointed away from Tom the whole time, with its rocket blazing. Clearly, that is an absurd situation. So we must reject the contention that the lowest curve should be according to the people on the rockets, because it leads to an absurdity. So the curve must be the view of the initial inertial observers ... there is no alternative.

 

I wrote something to plot distance to lauchpad relative to the ever-changing CMIF of the rocket, as a function of ship time tau

Code:

// Compute distance to lauchpad from ship frame at ship time tau
#include <stdio.h>
#include <math.h>
#define ITTERS 36      // itterations
#define STEP 0.05
int main()
{
  int    itter;
  double ctime;        // Coordinate time of inertial frame
  double tau = STEP;   // Proper time of ship
  double pos;          // Computed position of ship in S
  double lpad;         // Computed position of lauchpad relative to ship
  double vel;          // Relative velocity between pad and ship

  puts(" tau   ctime   S pos  launchpad   v      gamma");
  for (itter = ITTERS; itter-- >= 0; tau += STEP) {
     ctime = sinh(tau);       // coordinate time
     vel = tanh(tau);         // speed as a function of proper time
     pos = cosh(tau) - 1.;    // distance traveled by ship
     lpad = pos - ctime*vel;
     printf("%.4f %.4f %.5f %.5f %.5f %7.4f\n",
            tau, ctime, pos, lpad, vel, 1 / sqrt(1 - vel*vel));
  }
}

This yields the following output, consistent with my numbers above:

Code:

 tau   ctime   S pos  launchpad   v      gamma
0.0500 0.0500 0.00125 -0.00125 0.04996  1.0013
0.1000 0.1002 0.00500 -0.00498 0.09967  1.0050
0.1500 0.1506 0.01127 -0.01115 0.14889  1.0113
0.2000 0.2013 0.02007 -0.01967 0.19738  1.0201
0.2500 0.2526 0.03141 -0.03046 0.24492  1.0314
0.3000 0.3045 0.04534 -0.04337 0.29131  1.0453
0.3500 0.3572 0.06188 -0.05827 0.33638  1.0619
0.4000 0.4108 0.08107 -0.07499 0.37995  1.0811
0.4500 0.4653 0.10297 -0.09336 0.42190  1.1030
0.5000 0.5211 0.12763 -0.11318 0.46212  1.1276
0.5500 0.5782 0.15510 -0.13428 0.50052  1.1551
0.6000 0.6367 0.18547 -0.15645 0.53705  1.1855
0.6500 0.6967 0.21879 -0.17952 0.57167  1.2188
0.7000 0.7586 0.25517 -0.20329 0.60437  1.2552
0.7500 0.8223 0.29468 -0.22761 0.63515  1.2947
0.8000 0.8881 0.33743 -0.25230 0.66404  1.3374
0.8500 0.9561 0.38353 -0.27721 0.69107  1.3835
0.9000 1.0265 0.43309 -0.30221 0.71630  1.4331
0.9500 1.0995 0.48623 -0.32715 0.73978  1.4862
1.0000 1.1752 0.54308 -0.35195 0.76159  1.5431
1.0500 1.2539 0.60379 -0.37648 0.78181  1.6038
1.1000 1.3356 0.66852 -0.40067 0.80050  1.6685
1.1500 1.4208 0.73741 -0.42443 0.81775  1.7374
1.2000 1.5095 0.81066 -0.44771 0.83365  1.8107
1.2500 1.6019 0.88842 -0.47046 0.84828  1.8884
1.3000 1.6984 0.97091 -0.49262 0.86172  1.9709
1.3500 1.7991 1.05833 -0.51417 0.87405  2.0583
1.4000 1.9043 1.15090 -0.53508 0.88535  2.1509
1.4500 2.0143 1.24884 -0.55533 0.89569  2.2488
1.5000 2.1293 1.35241 -0.57490 0.90515  2.3524
1.5500 2.2496 1.46186 -0.59380 0.91379  2.4619
1.6000 2.3756 1.57746 -0.61202 0.92167  2.5775
1.6500 2.5075 1.69951 -0.62956 0.92886  2.6995
1.7000 2.6456 1.82832 -0.64643 0.93541  2.8283
1.7500 2.7904 1.96419 -0.66264 0.94138  2.9642
1.8000 2.9422 2.10747 -0.67820 0.94681  3.1075
1.8500 3.1013 2.25853 -0.69311 0.95175  3.2585

Plot that, and you get 'the view according to the rocket'. No prior chart has computed this. They've all depicted inertial frames.

 

Sorry, I couldn't follow what you are doing. Why wouldn't the CMIF results be exactly the same as the "tau" (accelerating) results? I would think the collection of CMIF results at each instant precisely give the "tau" (accelerating) results.

Assuming you correctly get the curve according to the people on the rocket, I THINK my results show that when you convert that to what the initial inertial observers conclude (by dividing by gamma), you get an absurd result. That seems to imply that the curve you're starting with (which is the curve we've been discussing for the last month) can't be the view of the people on the rocket.

 

I THINK your computations say that at tau = 1.0, the distance of the curve above the horizontal axis is 0.35195. (I've switched the sign on your result, so I can plot it on my previous plot with the initial inertial observer view). But I plotted your value of 0.35195 at the horizontal location 1.0 on my old chart, which is the "t" value there ... maybe I should be converting your tau value to the corresponding "t" value, and plotting that. OR, maybe I just need to not try to compare the two curves in a quantitative way, but rather just in a qualitative way. They definitely look like they both have essentially the same shape.

But despite all that, I THINK when I take your curve, and then ask the question "What does the initial inertial observer say about that curve", he will get the same absurd result that I have described. First according to the length contraction equation (LCE), he has to divide all the distances by the value of gamma at that point, to get HIS view of what's going on. He'll say that, although the leading rocket DOES at first get far away from him (when gamma is small), it eventually "backs up" toward him (when gamma gets large), and ends up at his position (all the while with the rocket pointing away from him and spewing exhaust toward him). That's clearly an absurd situation, so something is wrong here.

 

One other comment: The weird things that happen that I have described don't show up until gamma is very large. So you need to extend your computations to larger tau to get to large gamma ... that's when the rocket will start moving back toward the initial inertial observer, according to the initial inertial observer (even though it's still pointing away from him, and still thrusting).

 

What absurdity exactly does this lead to? I'm not sure what you consider absurd. It it that the rocket accelerating away should be far away after quite some time, but also that it has 'come back' so to speak? Or is there a different thing you find absurd?

Because the CMIF picture is that of an inertial frame (which is what the CMIF part stands for). It being an inertial frame, inertial (unaccelerating) things like the green launchpad line are straight lines.
The rocket on the other hand is accelerating, so any attempt to graph using the rocket frame would result in a non-inertial picture with the rocket always at X=0 by definition, and the launchpad line being curved, which the data in post 306 indeed shows.

They do. That's how I generated post 306. Each data point was calculated in a different CMIF frame. The rocket is always at location zero in any of those frames at the time in which it is stationary in that particular frame. I didn't bother to draw a picture plotting the launchpad curve. No point in plotting the rocket since it follows the time axis.

It is relative to the accelerating frame of the rocket and has nothing to do with the choices of the people on the rocket. They're quite free to choose frames other than that of the rocket. I never use people's choices as a frame specification since nothing in theory of relativity obligates what they should be.

Don't agree with the logic, but I do agree that the curve in question does not depict motion relative to any rocket.

OK, by switching signs you're saying that the rocket is accelerating in the negative direction and leaving the lauchpad behind in a positive displacement from the rocket. That can be done, but inconsistent with the story up until now where we had the lone rocket accelerating forward in the +x direction.

OK, it fits on the graph better that way. So the launchpad is above the rocket, not below it.

Yes, we noted your unconventional habit of putting t horizontal. I expected this, and have tried to use this convention in my latest post where the one picture (of the CMIF S') did exactly that. I never drew a picture of the rocket PoV since it is confusing. For instance, the time axis only works at x=0 and at different x values, time scales differently, so a clock stationary at x=5 would not be in sync with the clock at x=0. There might be discontinuities in worldlines, stuff like that. Such is the nature of attempts to draw non-inertial frames. You rarely seem them drawn on physics sites, but you requested the data.

What curve are we talking about? The column labeled S-pos? That curve is the location of the rocket relative to S, an inertial frame, not something relative to the rocket. The curve relative to the rocket is the launchpad curve, the only object that moves in the rocket frame.
Yes, if you plot the S-pos number along with the ctime for the time value, you get the black curve posted in 303. But that curve, again, is relative to S, not relative to the rocket.

That's why I don't worry about what your inertial observers say. They're morons. I just graph where the rocket is relative to S and let the observers delude themselves if they choose.

You asked where stuff is relative to the accelerating rocket, and I wrote a program to illustrate it. Anything else needed?
You need to answer my questions above if you want to make any progress about your assertions about what people decide to say. That was the whole point of the questions, to move this forward instead of just repeating the same fallacious assertions over and over.

Don't they now... They look pretty weird right out of the gate to me.

Your story, not mine. Answer the questions if you have any faith at all in this assertion.

 

Taylor and Wheeler assumed that the rapidity "theta", which is related to the velocity "v" by the equation

v = tanh(theta),

is given by

theta = A * tau,

where "tau" is the age of the people who are accelerating at "A" ly/y/y.

I now believe that they were incorrect in that assumption, because it leads to the absurd result that I have described in my post #305. I now believe that the correct assumption is that

theta = A * t,

where "t" is the age of the initial inertial observers.

I believe that, because it avoids the absurd results, and gives results that are consistent in all respects.

So the chart given in post #282 is the correct chart for the initial inertial observers' view. And if you want the chart for the view of the people in the trailing rocket, you just multiply every point on every curve of the chart in post #282 by gamma.

 

Full denial of physics texts. Where in the text does T&W say that the first equation is an assumption?
As I recall, SR has only two assumptions, and that wasn't one of them. Seems it is only you making new ones.

You only asserted that you found something absurd. No specific absurdity was described, so your statement above is pretty empty.

Why hasn't this topic been long ago moved to Alternate Theories with all the other denialists?

 

This is a old and repeated quote which I present as evidence that being in the presence of a rocket that literally vanishes and appears a finite distance away constitutes an absurd situation.

I said that a ways back, and wanted to illustrate it.

So I put the lead ship 6 ly ahead of the trailing ship in S and had them both accelerate at 1, plotting 4 years as measured by S using your calculations. I then took only the data points of the lead ship and plotted them in a frame moving at -0.866c relative to S. I will call this inertial frame S", depicted below.
I placed the origin of S" exactly at the event where the ship is launched. It is inertial before that. I also added two observers stationary in S" the entire time since you seem to put a lot of stock in things only if there is an inertial observer watching something absurd occur in their presence.
Light blue lines are past light cones of certain observer events.

Please Register or Log in to view the hidden image!

So the black line is the worldline of the ship. Note that Bob at time -1.53 witnesses an accelerating ship appearing out of nowhere right in his presence. There was no ship nearby before that, but there is a similar one parked at a fast moving launchpad passed by him about 1.5 years ago. He can watch all three ships for the next ~3.2 years, at which point he sees two of them vanish without a trace, all of which is absurd by your admission.

Alice meanwhile sees only the approaching launchpad for several years, watches the launch before the pad gets to her, and suddenly sees two ships explode into existence by her, only to have in a few weeks, one of them, plus the one that just launched, just vanish without a trace. In fact, the two ships vanished from existence exactly in her presence, but the light from both of them only gets to her later, so they seem to actually appear, not vanish, at that event. Both ships are moving faster than light.

Observers aside, there very much exist 3 lead ships in this frame for a bit over 1.8 years between the event where new ships time travel into existence from a future event where the two ships vanish out of existence.

Now of course your selection bias will kick in and you'll somehow not see this as absurd when the same sort of thing was declared absurd in your 'proof', because by definition Mike cannot be wrong, no matter how many clowns get out of that car.