Richard Hachel <r.hachel@tiscali.fr> wrote:
> Le 21/04/2022 à 17:01, Odd Bodkin a écrit :
>
>> A real scientist will not insist his idea is correct when it is discovered
>> that experimental data do not agree.
>
> C'est ce que je dis (That's what I say).

Except the results for proper time measured agree with physicists’ value,
and disagree with yours. This is already discovered in past measurements
and published. You perhaps need to catch up on the literature.

>
> R.H.
>

--
Odd Bodkin — Maker of fine toys, tools, tables

Le 21/04/2022 à 18:47, Odd Bodkin a écrit :
> Richard Hachel <r.hachel@tiscali.fr> wrote:
>> Le 21/04/2022 à 17:01, Odd Bodkin a écrit :
>>
>>> A real scientist will not insist his idea is correct when it is discovered
>>> that experimental data do not agree.
>>
>> C'est ce que je dis (That's what I say).
>
> Except the results for proper time measured agree with physicists’ value,
> and disagree with yours. This is already discovered in past measurements
> and published. You perhaps need to catch up on the literature.
>
>>
>> R.H.
>>

I don't see how charged and accelerated particles could give us any
information about their proper times.

On their impulse, on their speed, on their energy, etc... yes.

But how do you invalidate my equations, which are nevertheless very
precise and conceptually very well defensible?

You say "experience disproves your equations."

Which equations? By what experience?

There, I have total silence.

R.H.

On Thursday, 21 April 2022 at 18:47:20 UTC+2, bodk...@gmail.com wrote:
> Richard Hachel <r.ha...@tiscali.fr> wrote:
> > Le 21/04/2022 à 17:01, Odd Bodkin a écrit :
> >
> >> A real scientist will not insist his idea is correct when it is discovered
> >> that experimental data do not agree.
> >
> > C'est ce que je dis (That's what I say).
>
> Except the results for proper time measured agree with physicists’ value,

No, poor halfbrain, proper time measured is t'=t.
Anyone can check GPS.

Richard Hachel <r.hachel@tiscali.fr> wrote:
> Le 21/04/2022 à 18:47, Odd Bodkin a écrit :
>> Richard Hachel <r.hachel@tiscali.fr> wrote:
>>> Le 21/04/2022 à 17:01, Odd Bodkin a écrit :
>>>
>>>> A real scientist will not insist his idea is correct when it is discovered
>>>> that experimental data do not agree.
>>>
>>> C'est ce que je dis (That's what I say).
>>
>> Except the results for proper time measured agree with physicists’ value,
>> and disagree with yours. This is already discovered in past measurements
>> and published. You perhaps need to catch up on the literature.
>>
>>>
>>> R.H.
>>>
>
> I don't see how charged and accelerated particles could give us any
> information about their proper times.

Their decay rates are governed by their proper time.

>
> On their impulse, on their speed, on their energy, etc... yes.
>
> But how do you invalidate my equations,

With already-measured data, which you have not looked at, and I suspect you
have no idea even how to look them up.

> which are nevertheless very
> precise and conceptually very well defensible?
>
> You say "experience disproves your equations."
>
> Which equations?

Your equation for proper time.

Experiments already mentioned. Go look them up.

> By what experience?
>
> There, I have total silence.
>
> R.H.
>

--
Odd Bodkin -- maker of fine toys, tools, tables

Odd Bodkin <bodkinodd@gmail.com> wrote:
> Richard Hachel <r.hachel@tiscali.fr> wrote:
>> Le 21/04/2022 à 18:47, Odd Bodkin a écrit :
>>> Richard Hachel <r.hachel@tiscali.fr> wrote:
>>>> Le 21/04/2022 à 17:01, Odd Bodkin a écrit :
>>>>
>>>>> A real scientist will not insist his idea is correct when it is discovered
>>>>> that experimental data do not agree.
>>>>
>>>> C'est ce que je dis (That's what I say).
>>>
>>> Except the results for proper time measured agree with physicists’ value,
>>> and disagree with yours. This is already discovered in past measurements
>>> and published. You perhaps need to catch up on the literature.
>>>
>>>>
>>>> R.H.
>>>>
>>
>> I don't see how charged and accelerated particles could give us any
>> information about their proper times.
>
> Their decay rates are governed by their proper time.
>
>>
>> On their impulse, on their speed, on their energy, etc... yes.
>>
>> But how do you invalidate my equations,
>
> With already-measured data, which you have not looked at, and I suspect you
> have no idea even how to look them up.
>
>> which are nevertheless very
>> precise and conceptually very well defensible?
>>
>> You say "experience disproves your equations."
>>
>> Which equations?
>
> Your equation for proper time.
>
> Experiments already mentioned. Go look them up.
>
>> By what experience?
>>
>> There, I have total silence.
>>
>> R.H.
>>
>
>
>

This goes to what I told you earlier, Richard.

The test of ideas in physics is experiment. Not thought experiments, real
experiments.

The silliness of talking about Tau Ceti travelers is that no one has
traveled to Tau Ceti, and no one is going to travel to Tau Ceti in the
plannable future. It is therefore not accessible to a real measurement.
Without access to real measurement of proper time, the discussion will
devolve into “You say this, and I say that, and we have to choose, so we’ll
decide on the basis of beauty.” No. That is not science. Decisions about
which one to choose must be made with REAL DATA.

And so if there is a claim by relativity about how much proper time has
elapsed, really the question you have to ask is, “How could they have
possibly measured the proper time in real life to test that claim?” I
assure you, it HAS been measured. You are just unaware how, and you don’t
know how to research that either.

--
Odd Bodkin -- maker of fine toys, tools, tables

On Thursday, April 21, 2022 at 7:42:36 AM UTC-7, Richard Hachel wrote:
> There is something wrong with the way relativists calculate proper time.

Just the opposite: The scientific calculation is fine, but there is something
wrong with _your_ method. Proof:

Suppose in terms of your frame a traveler moves 1 light year at speed 0.6c,
and then he keeps going for another light year at 0.8c. This means that in your
frame he travels the first light year in 5/3 years, and he travels the second light
year in 5/4 years, for a total distance of 2 light years in a total time of 35/12
years. Right?

Now, you agree with special relativity that his elapsed proper times for those
segments are 4/3 years and 3/4 years, for a total of 25/12 years, which
is 2.0833 years. Right?

However, if another traveler went 2 light years for 35/12 years at the constant
speed, you agree with special relativity that his elapsed proper time would
be 2.1229 years. This is greater than 2.0833 years, and yet you claim that
these should be the same. That's why your beliefs are illogical and
self-contradictory.

Le 21/04/2022 à 20:54, Stan Fultoni a écrit :

> Suppose in terms of your frame a traveler moves 1 light year at speed 0.6c,
> and then he keeps going for another light year at 0.8c.

Yes.

> This means that in your
> frame he travels the first light year in 5/3 years, and he travels the second
> light
> year in 5/4 years, for a total distance of 2 light years in a total time of
> 35/12
> years. Right?

Yes. 2 years and 11 months.

>
> Now, you agree with special relativity that his elapsed proper times for those
> segments are 4/3 years and 3/4 years, for a total of 25/12 years, which
> is 2.0833 years. Right?

Tr=To.sqrt(1-v²/c²)

Tr(1)=5/3.sqrt(1-0.6²)
Tr(2)=5/4.sqrt(1-0.8²)

Tr = 4/3 + 3/4 = 25/12

2 years and 1 month.

Yes.

>
> However, if another traveler went 2 light years for 35/12 years at the constant
>
> speed, you agree with special relativity that his elapsed proper time would
> be 2.1229 years.

x= 2
To= 35/12

Vo=x/To=0.685714c Vr=Vo/sqrt(1-v²/c²)= 0.94208c

Tr=To.sqrt(1-v²/c²)=2.12295 ~ 2 years 1 month and 2 weeks.

Yes.

> This is greater than 2.0833 years, and yet you claim that
> these should be the same. That's why your beliefs are illogical and
> self-contradictory.

Yes.

Tr=x/Vr=2.1229

The two rockets arrive in the same terrestrial time, but not with proper
time.

However, what I say is neither illogical nor contradictory.

Simply, we are not in the same situation.

The three trips are different.

One is in constant speed of 0.6857c, the other uses two constant speeds
with very abrupt variation, and the third rocket (mine) is in constant
acceleration.

You seem to think (like the whole scientific class) that there is no
reason why we cannot equate trip 2 with trip 3 and say that it is the same
thing for Tr if the To are equal .

This is probably going too fast and relying on what is called a
misunderstood "a priori".

R.H.

On Thursday, April 21, 2022 at 3:02:44 PM UTC-7, Richard Hachel wrote:
> > Suppose in terms of your frame a traveler moves 1 light year at speed 0.6c,
> > and then he keeps going for another light year at 0.8c. This means that in your
> > frame he travels the first light year in 5/3 years, and he travels the second
> > light year in 5/4 years, for a total distance of 2 light years in a total time of
> > 35/12 years. Right?
>
> Yes.
>
> > Now, you agree with special relativity that his elapsed proper times for those
> > segments are 4/3 years and 3/4 years, for a total of 25/12 years, which
> > is 2.0833 years. Right?
>
> Yes.
> >
> > However, if another traveler went 2 light years for 35/12 years at the constant
> > speed, you agree with special relativity that his elapsed proper time would
> > be 2.1229 years.
>
> Yes.
>
> > This is greater than 2.0833 years, and yet you claim that
> > these should be the same. That's why your beliefs are illogical and
> > self-contradictory.
>
> Yes.

Right. So you agree that your beliefs are illogical and self-contradictory?

> The two rockets arrive in the same terrestrial time, but not with proper
> time.

Wait, your claim is that they have the same elapsed proper time, and yet
you also agree that they have different elapsed proper time, so your beliefs
are illogical and self-contradictory, right?

> The three trips are different.

There are two trips, one with constant speed and one with changing speed.
Yes, those two are different, and they have different elapsed proper times.
The problem is that you agree they have different proper times, but you also
claim they have the same proper times. Your claims are self-contradictory
and therefore false. Agreed?

> One is in constant speed of 0.6857c, the other uses two constant speeds
> with very abrupt variation...

Right, and they have different elapsed proper times, given by integrating the
quantity dtau = sqrt(dt^2 = dx^2) along the respective paths.

> and the third rocket (mine) is in constant acceleration.

The point is that your beliefs are already self-contradictory with
the simple example of two paths. There's no need to consider examples
where the speed changes twice, or three times, or 100 times, because your
beliefs are already shown to be self-contradictory and illogical with a rocket
that changes speed just once. Right?

Le 22/04/2022 à 01:46, Stan Fultoni a écrit :
>
> The point is that your beliefs are already self-contradictory with
> the simple example of two paths. There's no need to consider examples
> where the speed changes twice, or three times, or 100 times, because your
> beliefs are already shown to be self-contradictory and illogical with a rocket
> that changes speed just once. Right?

No, that's not what I said.

Care must be taken not to distort the words.

We cannot say "the two proper times of the two rockets are the same and
the two proper times of the two same rockets are not the same". It is
obviously contradictory, and, if you read me carefully, you will see that
I never said such contradictions.

I said clearly that the two proper times are not the same but that the
time observable by the terrestrial exterior observer is the same.

To put it better, he sees the two rockets leaving at the same time and
arriving at the same time.

But at the level of the two rockets, the proper times are not the same.

On the other hand, in the example of the uniformly accelerated reference
frame, not only the observable times are the same (To1=To2) but the proper
times are the same (Tr1=Tr2).

I know it's hard to understand and it completely confuses the scientist
who is not (but then not at all) used to my "fantasies".

I think we will certainly have to talk about all this again, because there
are extraordinary but very important things to understand.

R.H.

On Thursday, April 21, 2022 at 5:48:35 PM UTC-7, Richard Hachel wrote:
> I said clearly that the two proper times are not the same...

Right, they are not the same, and the correct proper time for the rocket
that travels one light year at 0.6c and one light year at 0.8c is the value
given by adding up the proper times for those two intervals.

Now we consider a rocket that moves from (x0,t0) to (x1,t1) at the
constant speed (x1-x0)/(t1-t0), and then it moves from (x1,t1) to (x2,t2)
at the constant speed (x2-x1)/(t2-t1), and so on, up to the final
destination at (x9,t9). What is the total elapsed proper time for
that rocket?

According to science, taking c=1 for convenience, the answer is to
add up the proper times of the nine individual segments, so it is

sqrt[(t2-t1)^2 - (x2-x1)^2]
+ sqrt[(t3-t2)^2 - (x3-x2)^2]
+ ...
+ sqrt[(t9-t8)^2 - (x9-x8)^2]

Do you agree that this is the total elapsed time for that rocket?

And if we consider a rocket following any path whatsoever, do
you agree that the total elapsed time is the integral of sqrt(dt^2 - dx^2)
along that path?

Le 22/04/2022 à 03:57, Stan Fultoni a écrit :
> Right, they are not the same, and the correct proper time for the rocket
> that travels one light year at 0.6c and one light year at 0.8c is the value
> given by adding up the proper times for those two intervals.
>
> Now we consider a rocket that moves from (x0,t0) to (x1,t1) at the
> constant speed (x1-x0)/(t1-t0), and then it moves from (x1,t1) to (x2,t2)
> at the constant speed (x2-x1)/(t2-t1), and so on, up to the final
> destination at (x9,t9). What is the total elapsed proper time for
> that rocket?
>
> According to science, taking c=1 for convenience, the answer is to
> add up the proper times of the nine individual segments, so it is
>
> sqrt[(t2-t1)^2 - (x2-x1)^2]
> + sqrt[(t3-t2)^2 - (x3-x2)^2]
> + ...
> + sqrt[(t9-t8)^2 - (x9-x8)^2]
>
> Do you agree that this is the total elapsed time for that rocket?
>
> And if we consider a rocket following any path whatsoever, do
> you agree that the total elapsed time is the integral of sqrt(dt^2 - dx^2)
> along that path?

You are tenacious, and that is an excellent quality for a man and a
scientist.
I really appreciate.
I will answer your question where you share, this time, the journey in
nine parts.

In the meantime, I invite you to reflect on one thing: the integration
used by scientists apparently correct mathematically is it physically
correct?

Is in there not a blunder?

R.H.

Le 22/04/2022 à 03:57, Stan Fultoni a écrit :
> Right, they are not the same, and the correct proper time for the rocket
> that travels one light year at 0.6c and one light year at 0.8c is the value
> given by adding up the proper times for those two intervals.
>
> Now we consider a rocket that moves from (x0,t0) to (x1,t1) at the
> constant speed (x1-x0)/(t1-t0), and then it moves from (x1,t1) to (x2,t2)
> at the constant speed (x2-x1)/(t2-t1), and so on, up to the final
> destination at (x9,t9). What is the total elapsed proper time for
> that rocket?
>
> According to science, taking c=1 for convenience, the answer is to
> add up the proper times of the nine individual segments, so it is
>
> sqrt[(t2-t1)^2 - (x2-x1)^2]
> + sqrt[(t3-t2)^2 - (x3-x2)^2]
> + ...
> + sqrt[(t9-t8)^2 - (x9-x8)^2]
>
> Do you agree that this is the total elapsed time for that rocket?
>
> And if we consider a rocket following any path whatsoever, do
> you agree that the total elapsed time is the integral of sqrt(dt^2 - dx^2)
> along that path?

(ΔTr)²=(ΔTo)²-(Δx/c)²

Yes.

But how do you integrate?

Warning, x is not a constant!!!

R.H.

On Friday, April 22, 2022 at 5:25:32 AM UTC-7, Richard Hachel wrote:
> > If we consider a rocket following any path whatsoever, do you
> > agree that the total elapsed time is the integral of sqrt(dt^2 - dx^2)
> > along that path?
>
> Yes.

Great!

> But how do you integrate?

To integrate sqrt(dt^2 - dx^2) along a given path you must specify
the path, such as by specifying x as a function of t. You already
agreed that, for a path of constant proper acceleration a, starting
from x=0,t=0, we have t = (x/c)sqrt[1 + 2c^2/(ax)]. For convenience
let's take units so c=1, and then this relation can be written as
t^2 = x^2 + 2x/a = (x + 1/a)^2 - 1/a^2. Differentiating and dividing
through by 2 gives tdt = (x + 1/a)dx, so we have dx = [t/(x + 1/a)]dt, and
hence dx = [t/sqrt(t^2 + 1/a^2)]dt. With this we can express the integrand
as sqrt[dt^2 - dx^2] = dt/sqrt[1 + (at)^2], which we can integrate to
give tau = arcsinh(at)/a, which gives (for your original question) the
result 3.14 years. Agreed?

Le 22/04/2022 à 21:58, Stan Fultoni a écrit :
> On Friday, April 22, 2022 at 5:25:32 AM UTC-7, Richard Hachel wrote:
>> > If we consider a rocket following any path whatsoever, do you
>> > agree that the total elapsed time is the integral of sqrt(dt^2 - dx^2)
>> > along that path?
>>
>> Yes.
>
> Great!
>
>> But how do you integrate?
>
> To integrate sqrt(dt^2 - dx^2) along a given path you must specify
> the path, such as by specifying x as a function of t. You already
> agreed that, for a path of constant proper acceleration a, starting
> from x=0,t=0, we have t = (x/c)sqrt[1 + 2c^2/(ax)]. For convenience
> let's take units so c=1, and then this relation can be written as
> t^2 = x^2 + 2x/a = (x + 1/a)^2 - 1/a^2. Differentiating and dividing
> through by 2 gives tdt = (x + 1/a)dx, so we have dx = [t/(x + 1/a)]dt, and
> hence dx = [t/sqrt(t^2 + 1/a^2)]dt. With this we can express the integrand
> as sqrt[dt^2 - dx^2] = dt/sqrt[1 + (at)^2], which we can integrate to
> give tau = arcsinh(at)/a, which gives (for your original question) the
> result 3.14 years. Agreed?

We have

To = (x/c)sqrt[1 + 2c²/(ax)]

Tr = sqrt (2x/a)

What must be done is to show, by the correct integration, that the two
values ​​are bijective and related.

How do we achieve Tr=sqrt(2x/a) : correct answer?

And not to :

Tr = (c/a)arcsinh(at/c)

Where is relativistic error in the derivation?

Is it possible (and I think so) that a carrot replaced a turnip in the
equations?

R.H.

On Friday, April 22, 2022 at 5:25:32 AM UTC-7, Richard Hachel wrote:
>>> > If we consider a rocket following any path whatsoever, do you
>>> > agree that the total elapsed time is the integral of sqrt(dt^2 - dx^2)
>>> > along that path?
>>>
>>> Yes, but how do you integrate?
>>
>> To integrate sqrt(dt^2 - dx^2) along a given path you must specify
>> the path, such as by specifying x as a function of t. You already
>> agreed that, for a path of constant proper acceleration a, starting
>> from x=0,t=0, we have t = (x/c)sqrt[1 + 2c^2/(ax)]. For convenience
>> let's take units so c=1, and then this relation can be written as
>> t^2 = x^2 + 2x/a = (x + 1/a)^2 - 1/a^2. Differentiating and dividing
>> through by 2 gives tdt = (x + 1/a)dx, so we have dx = [t/(x + 1/a)]dt, and
>> hence dx = [t/sqrt(t^2 + 1/a^2)]dt. With this we can express the integrand
>> as sqrt[dt^2 - dx^2] = dt/sqrt[1 + (at)^2], which we can integrate to
>> give tau = arcsinh(at)/a, which gives (for your original question) the
>> result 3.14 years. Agreed?
>
> We have To = (x/c)sqrt[1 + 2c²/(ax)]

If you are using the symbol To to denote the coordinate time t, then yes.

> Tr = sqrt (2x/a)

No, that is the elapsed proper time for a path that moves at constant
speed from the beginning event to the ending event. Your rocker is
not moving at constant speed, it is beginning at zero speed and
undergoing constant proper acceleration of "a" throughout the
journey. By integrating sqrt(dt^2 - dx^2) along the path we get
the correct result arcsinh(at)/a for the total elapsed proper time.

This is why I told you before that you were claiming two self-
contradictory things, because you claimed that the elapsed time for
any path between two given events is simply sqrt[delta t - delta x],
but then we considered some simple examples like splitting a journey
into two parts with different speeds, and you agreed that they gave
different answers than moving at constant speed between the same two
events. In fact, you denied ever claiming otherwise. But now you
are again claiming the same two self-contradictory things. You say
they are equal and you admit they are not equal. That is irrational.

You've agreed that the elapsed time is given by integrating the
quantity sqrt(dt^2 - dx^2) along the journey, and I've given you
(see above) the simple algebra that shows this gives the result
arcsinh(at)/a for constant proper acceleration a. So that's the
answer. Your self-contradictory claim that the elapsed proper
time is always sqrt(2x/a) has been shown to be logically untenable.
Agreed?

Le 23/04/2022 à 16:56, Stan Fultoni a écrit :
> On Friday, April 22, 2022 at 5:25:32 AM UTC-7, Richard Hachel wrote:

>> We have To = (x/c)sqrt[1 + 2c²/(ax)]
>
> If you are using the symbol To to denote the coordinate time t, then yes.

Merci.

R.H.

Le 23/04/2022 à 16:56, Stan Fultoni a écrit :

> Your self-contradictory claim that the elapsed proper
> time is always sqrt(2x/a) has been shown to be logically untenable.
> Agreed?

No.

R.H.

On Saturday, April 23, 2022 at 8:58:34 AM UTC-7, Richard Hachel wrote:
> Le 23/04/2022 à 16:56, Stan Fultoni a écrit :
> > Your self-contradictory claim that the elapsed proper
> > time is always sqrt(2x/a) has been shown to be logically untenable.
> > Agreed?
>
> No.

So, you're simply declaring intellectual bankruptcy? Very strange.

Le 23/04/2022 à 18:31, Stan Fultoni a écrit :
> On Saturday, April 23, 2022 at 8:58:34 AM UTC-7, Richard Hachel wrote:
>> Le 23/04/2022 à 16:56, Stan Fultoni a écrit :
>> > Your self-contradictory claim that the elapsed proper
>> > time is always sqrt(2x/a) has been shown to be logically untenable.
>> > Agreed?
>>
>> No.
>
> So, you're simply declaring intellectual bankruptcy? Very strange.

Yes, very strange.

R.H.