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
have previously described.

On Tuesday, July 4, 2023 Mike Fontenot attempted to write:
> [In terms of standard inertial coordinates x,t, consider two initially
> stationary objects along the x axis, one at x=0 and the other at x=D.
> At time t=0 they each begin to undergo equal constant proper
> acceleration "a".]

Okay, that's a well-defined sequence of events, and using geometric units such that c=1 for convenience, the trajectories of the two objects are

x1 = sqrt[(1/a)^2 + t^2] - 1/a
x2 = sqrt[(1/a)^2 + t^2] - 1/a + D

so we obviously have x2 - x1 = D for all t. This is not something you are free to disagree with, it is stipulated in the conditions.

Now, in order for two accelerating objects to maintain constant proper separation (Born rigid motion), such that the distance in terms of their instantaneously co-moving inertial coordinate systems is constant, the leading object would need to have slightly less constant proper acceleration than the trailing object. But the two objects you have described are stipulated to have equal proper accelerations, so the leading object is accelerating faster than it would need to for constant proper separation. Hence the proper separation is increasing, and the string breaks.

Of course, the easy way of seeing this is to simply note that the string contracts as it accelerates in terms of x,t, whereas the objects maintain constant separation D in terms of x,t, so the string must break. Bell's point is that even people who supposedly understand special relativity (like his friends at CERN) sometimes have to take a few minutes to figure this out, but you must understand that every one of them ultimately agreed that indeed the string would break, and those who got the wrong answer to Bell's little pop quiz initially were embarrassed. There is no doubt as to the correct answer. Bell believed people would be able to answer more quickly and easily if special relativity was taught differently.

On 7/4/23 12:17 PM, Bill wrote:
>
> [...] using geometric units such that c=1 for convenience, the trajectories of the two objects are
>
> x1 = sqrt[(1/a)^2 + t^2] - 1/a
> x2 = sqrt[(1/a)^2 + t^2] - 1/a + D
>

Those two equations aren't even dimensionally consistent.

On Tuesday, July 4, 2023 at 1:27:03 PM UTC-7, Mike Fontenot wrote:
> On 7/4/23 12:17 PM, Bill wrote:
> >
> > [...] using geometric units such that c=1 for convenience, the trajectories of the two objects are
> >
> > x1 = sqrt[(1/a)^2 + t^2] - 1/a
> > x2 = sqrt[(1/a)^2 + t^2] - 1/a + D
> >
> Those two equations aren't even dimensionally consistent.

Sure they are. As stated above, this is in geometric units, with c=1, so x and D and t have units of (say) meters, and "a" has units of meters^-1. I take it you are unacquainted with geometric units. You should learn about them... it will maked things much easier and more clear for you.

On Tuesday, July 4, 2023 at 1:37:56 PM UTC-7, Bill wrote:
> On Tuesday, July 4, 2023 at 1:27:03 PM UTC-7, Mike Fontenot wrote:
> > On 7/4/23 12:17 PM, Bill wrote:
> > >
> > > [...] using geometric units such that c=1 for convenience, the trajectories of the two objects are
> > >
> > > x1 = sqrt[(1/a)^2 + t^2] - 1/a
> > > x2 = sqrt[(1/a)^2 + t^2] - 1/a + D
> > >
> > Those two equations aren't even dimensionally consistent.
> Sure they are. As stated above, this is in geometric units, with c=1, so x and D and t have units of (say) meters, and "a" has units of meters^-1. I take it you are unacquainted with geometric units. You should learn about them... it will maked things much easier and more clear for you.

How is the speed of light geometric?

On Tuesday, July 4, 2023 at 3:27:03 PM UTC-5, Mike Fontenot wrote:
> On 7/4/23 12:17 PM, Bill wrote:
> >
> > [...] using geometric units such that c=1 for convenience, the trajectories of the two objects are
> >
> > x1 = sqrt[(1/a)^2 + t^2] - 1/a
> > x2 = sqrt[(1/a)^2 + t^2] - 1/a + D
> >
> Those two equations aren't even dimensionally consistent.

CAN'T YOU READ?
Bill *stated*, just above, that for convenience he was using geometric
units where c=1. The equations are a lot simpler, and if, at the end
of a series of computations, you actually need the c's, it is very simple
to add them back as necessary. It is a standard trick that reduces
the complexity of a calculation.

(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.

Note that the people on the spaceships would NOT agree that the above
constructed chart represents THEIR viewpoint. They say their separation
is constant. And the people in the trailing rocket say that the people
in the leading rocket are ageing faster that the people in the trailing
rocket, by a specific factor f(D, A).

On Tuesday, July 4, 2023 at 2:39:00 PM UTC-7, Mike Fontenot wrote:
> People on the spaceships will say... ... the conclusions of INERTIAL OBSERVERS...

Relativity isn't about what various "people will say", nor about "conclusions of inertial observers" (with or without caps), it is about the unique objective facts, which can of course be described in terms of whatever system of coordinates you like.

> That chart is wrong...

Again, you are mistaken. Remember, in terms of standard inertial coordinates x,t, there are two initially stationary objects along the x axis, one at x=0 and the other at x=D. At time t=0 they each begin to undergo equal constant proper acceleration "a". Using geometric units such that c=1 for convenience, the trajectories of the two objects are

x1 = sqrt[(1/a)^2 + t^2] - 1/a
x2 = sqrt[(1/a)^2 + t^2] - 1/a + D

so we obviously have x2 - x1 = D for all t. This is not something you are free to disagree with, it is stipulated in the conditions.

These two objects maintain constant coordinate separation in terms of x,t, but not constant proper separation. In order for two accelerating objects to maintain constant proper separation (Born rigid motion), such that the distance in terms of their instantaneously co-moving inertial coordinate systems is constant, the leading object would need to have slightly less constant proper acceleration than the trailing object. But the two objects you have described are stipulated to have equal proper accelerations, so the leading object is accelerating faster than it would need to for constant proper separation. Hence the proper separation is increasing, and the string breaks.

Of course, an easy way of seeing this is to simply note that the string contracts as it accelerates in terms of x,t, whereas the objects maintain constant separation D in terms of x,t, so the string must break. Bell's point is that even people who supposedly understand special relativity (like his friends at CERN) sometimes have to take a few minutes to figure this out, but you must understand that every one of them ultimately agreed that indeed the string would break, and those who got the wrong answer to Bell's little pop quiz initially were embarrassed. There is no doubt as to the correct answer. Bell believed people would be able to answer more quickly and easily if special relativity was taught differently.

The bottom line is that, just as everyone has been telling you, the string breaks. Do you understand this now? Please make the effort to actually reply to the substance of this message. Use your words. If something is unclear, go ahead and ask.

On 7/4/23 4:09 PM, Bill wrote:
>
> In order for two accelerating objects to maintain constant proper separation (Born rigid motion), such that the distance in terms of their instantaneously co-moving inertial coordinate systems is constant, the leading object would need to have slightly less constant proper acceleration than the trailing object.

Einstein disagreed with you. In his 1907 paper, he gives an example of
two rockets accelerating with the same constant acceleration, which
maintain a constant separation. He also showed that the leading
rocket's clock runs faster than the trailing rocket's clock, by the
factor f(L, A), where "L" is the constant separation, and "A" is their
acceleration.

He also used the equivalence principle to specify an gravitational
scenario with no motion, but with a gravitational field that doesn't
vary with distance along its direction. Two stationary clocks,
separated by distance "L" along that field, will tick at different
rates, with the one farther from the source of the field running faster
by the factor f(L, g), where "f" is the same functional form in both of
those two scenarios.

On Tuesday, July 4, 2023 at 10:08:53 AM UTC-7, Mike Fontenot wrote:
> 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) .

NO. This is wrong. You are misunderstanding what the length contraction
is and how it works.

Don't waste any more time on this until you understand how
special relativity works.

--
Jan

On Tuesday, July 4, 2023 at 3:32:39 PM UTC-7, Mike Fontenot wrote:
> > In order for two accelerating objects to maintain constant proper separation
> > (Born rigid motion), such that the distance in terms of their instantaneously
> > co-moving inertial coordinate systems is constant, the leading object would
> > need to have slightly less constant proper acceleration than the trailing object.
>
> Einstein disagreed with you. In his 1907 paper, he gives an example of
> two rockets accelerating with the same constant acceleration, which
> maintain a constant separation.

They maintain a constant *coordinate* separation, just as do in Bell's scenario, but they do not maintain constant *proper* separation. Also, I will point out that Einstein's 1907 survey article did not mention rockets, nor did it state that two clocks initially at rest in x,t and then undergoing equal acceleration beginning at a certain t would maintain constant proper separation. You are simply lying about this. If you would care to type whichever sentence from that paper you believe supports your erroneous claim, then please do so. Otherwise, please stop lying. Isn't that a fair request?

> He also showed...

Everything he said in that Jahrbuch article is correct, but everything YOU are saying is incorrect. Your characterizations and descriptions of what he said are all false, based on nothing but misreading and misunderstanding. Again, if YOU have something to say, please say it. I dispute that Einstein's paper contains any of the statements that you claim it contains, so please provide verbatim quotes and I'll be happy to explain what he was actually saying.

> He also used the equivalence principle...

Stop... you assured me you are only interested in special relativity, so do not try to talk about the equivalence principle. Again, Einstein's article is fine, but your posts are gibberish. Do you now agree that the trajectory of the objects in Bell's scenario are exactly what I told you there were, and that they do not maintain constant proper separation, and that therefore the string breaks? If not, what part don't you understand?

On 2023-07-04 17:08:47 +0000, Mike Fontenot said:

> 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:

Both curves are obtained the same way:

> The rapidity "theta" is

> theta(t) = A * t,

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

The acclerations of the two spaceships are the same, so the rapidities
are the same, too.

> The velocity is

> v(t) = tanh(theta[t]);

Consequently, the velocities are the same, too.

> The distance traveled is

> d(t) = ln(cosh[v{t}]) .

Both spaceships have travelled the same distance, just starting at
different positions.

> 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".

The upper curve is as correct as the lower one. Both are plotted according
to the same equations (except the starting point) so if one is wrong then
the equations are wrong and the other curve is wrong, too.

Mikko

On 7/5/23 1:15 AM, Mikko wrote:

> [...]
You are ignoring one indispensable thing: the famous length contraction
equation (LCE) says that any inertial observer MUST conclude that any
yardstick or other physical rod that is moving (in the direction of its
length) with respect to himself is shorter (according to him) than an
identical rod that is stationary with respect to himself. That requires
that the distance between the two curves must decrease as the
acceleration continues, according to the viewpoint of inertial observers
stationary with the rockets immediately before they are ignited.

On Wednesday, July 5, 2023 at 6:51:50 AM UTC-7, Mike Fontenot wrote:
> On 7/5/23 1:15 AM, Mikko wrote:
>
> > [...]
> You are ignoring one indispensable thing: the famous length contraction
> equation (LCE) says that any inertial observer MUST conclude that any
> yardstick or other physical rod that is moving (in the direction of its
> length) with respect to himself is shorter (according to him) than an
> identical rod that is stationary with respect to himself. That requires
> that the distance between the two curves must decrease as the
> acceleration continues, according to the viewpoint of inertial observers
> stationary with the rockets immediately before they are ignited.
Cranko,

Look at the title of your thread: the distance between the two rockets is CONSTANT. By the mere definition of the exercise. Otherwise, since the speed is increasing (due to acceleration), the distance would contract (in the frame of the launchpad).

On 2023-07-05 13:51:44 +0000, Mike Fontenot said:

> On 7/5/23 1:15 AM, Mikko wrote:

> You are ignoring one indispensable thing:

Irrelevant, as there is no error in my message that has been attributed
to that.

Mikko

JanPB wrote:
> On Tuesday, July 4, 2023 at 10:08:53 AM UTC-7, Mike Fontenot wrote:
>> 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) .
> NO. This is wrong. You are misunderstanding what the length contraction
> is and how it works.

you are a liar. You change your name and move to america. That's a lie in
itself, because it's not your real name.

*_Medvedev_names_how_Ukraine_conflict_could_‘end_in_days’_*
https://%72t.com/r%75%73%73ia/579250-medvedev-ukraine-conflict-end-days/

“If NATO, primarily the US *_and_its_vassals_*, had stopped supplying
weapons and munitions to Ukraine, the [Russian] special military operation
would have ended in mere months,” Medvedev told the news agency. It can
still end “in days,” however, if Washington and its allies stopped the
deliveries, he added.

you stupid country of electricians and vassals. Why buying cheap fuel and
gas from Russia, when you can pay 4 times more from capitalist corporation
of america, a defenseless shithole.

On Wednesday, July 5, 2023 at 10:03:19 AM UTC-7, Harris Theofilopoulos wrote:
> JanPB wrote:
> > On Tuesday, July 4, 2023 at 10:08:53 AM UTC-7, Mike Fontenot wrote:
> >> 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) .
> >
> > NO. This is wrong. You are misunderstanding what the length contraction
> > is and how it works.
>
> you are a liar. You change your name and move to america. That's a lie in
> itself, because it's not your real name.

Did you happen to work on the Steele dossier as well?

--
Jan

On 7/4/23 12:08 PM, Mike Fontenot wrote:
> To get a chart [...]

This is all nonsense. You attempt to use "length contraction" (i.e.
dividing by gamma(v)) in a situation where it simply does not apply.

In the inertial frame in which the rockets are initially at rest, their
separation remains constant. If they stop accelerating at equal proper
times, they will come to rest in some inertial frame moving wrt that
initial frame -- "length contraction" applies and unequivocally implies
the rockets are further apart in their final frame than they were in
their initial frame.

Give up. Your arguments are all bogus. Look up the actual calculations
using SR -- your attempt to use a shortcut (invoking the equivalence
principle) is just plain wrong.

If Einstein's 1907 paper actually says what you claim, then he was
wrong, too; I rather doubt that (but have no interest in revisiting
ancient history). For instance, two stationary clocks in a gravitational
field are NOT "equivalent" to Bell's spaceship paradox, because the
clocks have constant proper separation while the spaceships do not.

Tom Roberts

On Wednesday, July 5, 2023 at 6:51:50 AM UTC-7, Mike Fontenot wrote:
> > In order for two accelerating objects to maintain constant proper separation
> > (Born rigid motion), such that the distance in terms of their instantaneously
> > co-moving inertial coordinate systems is constant, the leading object would
> > need to have slightly less constant proper acceleration than the trailing object.
>
> Einstein disagreed with you. In his 1907 paper, he gives an example of
> two rockets accelerating with the same constant acceleration, which
> maintain a constant separation.

Einstein didn't specify in that paper whether the clocks (not rockets) maintained constant *coordinate* separation (as in Bell's scenario), or constant *proper* separation, but this ambiguity doesn't affect his result, because he is only considering the first-order effects of relativity of simultaneity and how this affects the comparative rates of the clocks based on the acceledrating temporal foliation. That's a first-order effect that Einstein evaluates over an incremental span of time, which doesn't yield the secular change in the proper length for equal accelerations over an extended period of time.

> He also showed...

Everything Einstein said in that Jahrbuch article is correct, but everything YOU are saying is incorrect. Your characterizations and descriptions of what he said are all false, based on nothing but misreading and misunderstanding. Again, if YOU have something to say, please say it. If you claim that Einstein said something other than I just described, point out the exact sentence.

> He also used the equivalence principle...

Stop... you assured me you're only interested in special relativity, and have no interest in understanding general relativity, so you have forfeited the right to invoke general relativity and the equivalence principle in your arguments. Remember, the equivalence principle does not mean what you think it means, and this has been carefully explained to you, but you disregarded the explanation on the grounds that you aren't interested in general relativity. Well, so be it, but then you can't turn around and make claims based on your misunderstaidng of the equivalence principle and general relativity.

On 7/5/23 8:51 AM, Mike Fontenot wrote:
> the famous length contraction
> equation (LCE) says that any inertial observer MUST conclude that any
> yardstick or other physical rod that is moving (in the direction of its
> length) with respect to himself is shorter (according to him) than an
> identical rod that is stationary with respect to himself.  That requires
> that the distance between the two curves must decrease as the
> acceleration continues, according to the viewpoint of inertial observers
> stationary with the rockets immediately before they are ignited.

You got it BACKWARDS.

The distance between the rockets remains constant in their initial
inertial frame, so if they stop accelerating simultaneously wrt that
frame, their proper distance must be LARGER than the value measured in
their initial frame.

[Of course if they have identical proper accelerations
as a function of their proper time, they will also have
identical accelerations relative to their initial frame,
so their separation wrt their initial frame is constant.]

Tom Roberts

Bill wrote:

> On Wednesday, July 5, 2023 at 6:51:50 AM UTC-7, Mike Fontenot wrote:
>> Einstein disagreed with you. In his 1907 paper, he gives an example of
>> two rockets accelerating with the same constant acceleration, which
>> maintain a constant separation.
>
> Einstein didn't specify in that paper whether the clocks (not rockets)
> maintained constant *coordinate* separation (as in Bell's scenario), or
> constant *proper* separation, but this ambiguity doesn't affect his
> result, because he is only considering the first-order effects of

I keep telling Zelensky, which is a zelenske, not to fight if you have no
weapons, but he won't listen to me. I think he banned RT.

also, I keep wondering how stupid these ukronazians are, thinking they
have a country, when the zelenske, a khazar goy, is selling it to
Blackrock etc 10 times, which is 1000%.

what the fuck are these 87% of the population, which are Russian, do? The
country of whom?

also, that shithole of a wannabe "country" was sold already 100% to the
*_capitalist_collective_west_*, *_before_2014_coup_* detate. *_And_* they
were ready to take over Russia too!!