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Paul B. Andersen wanted to know...

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  Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-07-31 14:33 +0000
    Re: Paul B. Andersen wanted to know... The Starmaker <starmaker@ix.netcom.com> - 2022-07-31 10:57 -0700
    Re: Paul B. Andersen wanted to know... "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-07-31 19:59 +0200
      Re: Paul B. Andersen wanted to know... Maciej Wozniak <maluwozniak@gmail.com> - 2022-07-31 12:00 -0700
      Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-07-31 22:18 +0000
        Re: Paul B. Andersen wanted to know... "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-08-01 19:21 +0200
          Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-08-01 18:46 +0000
            Re: Paul B. Andersen wanted to know... "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-08-02 14:40 +0200
              Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-08-02 14:58 +0000
                Re: Paul B. Andersen wanted to know... "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-08-02 22:35 +0200
                  Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-08-02 21:11 +0000
                    Re: Paul B. Andersen wanted to know... "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-08-03 11:24 +0200
                      Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-08-03 12:46 +0000
                        Re: Paul B. Andersen wanted to know... "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-08-04 11:46 +0200
                          Re: Paul B. Andersen wanted to know... Maciej Wozniak <maluwozniak@gmail.com> - 2022-08-04 02:49 -0700
                            Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-08-04 11:27 +0000
                              Re: Paul B. Andersen wanted to know... Maciej Wozniak <maluwozniak@gmail.com> - 2022-08-04 05:02 -0700
                              Re: Paul B. Andersen wanted to know... Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-04 06:58 -0700
                                Re: Paul B. Andersen wanted to know... Maciej Wozniak <maluwozniak@gmail.com> - 2022-08-04 07:40 -0700
                                Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-08-04 14:51 +0000
                                  Re: Paul B. Andersen wanted to know... Maciej Wozniak <maluwozniak@gmail.com> - 2022-08-04 07:55 -0700
                                  Re: Paul B. Andersen wanted to know... Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-04 10:06 -0700
                                    Re: Paul B. Andersen wanted to know... Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-04 22:17 -0700
                                Re: Paul B. Andersen wanted to know... JanPB <filmart@gmail.com> - 2022-08-09 01:00 -0700
                                  Re: Paul B. Andersen wanted to know... JanPB <filmart@gmail.com> - 2022-08-09 01:02 -0700
                                  Re: Paul B. Andersen wanted to know... Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-09 02:03 -0700
                                    Re: Paul B. Andersen wanted to know... nospam@de-ster.demon.nl (J. J. Lodder) - 2022-08-09 11:49 +0200
                                      Re: Paul B. Andersen wanted to know... Prokaryotic Capase Homolog <prokaryotic.caspase.homolog@gmail.com> - 2022-08-09 04:25 -0700
                                        Re: Paul B. Andersen wanted to know... nospam@de-ster.demon.nl (J. J. Lodder) - 2022-08-09 14:23 +0200
                                      Re: Paul B. Andersen wanted to know... JanPB <filmart@gmail.com> - 2022-08-09 06:10 -0700
                                    Re: Paul B. Andersen wanted to know... JanPB <filmart@gmail.com> - 2022-08-09 06:03 -0700
                                  Re: Paul B. Andersen wanted to know... Juan Di pasqua <toed@iavdattg.ev> - 2022-08-09 10:57 +0000
                                    Re: Paul B. Andersen wanted to know... JanPB <filmart@gmail.com> - 2022-08-09 06:04 -0700
                          Re: Paul B. Andersen wanted to know... Richard Hachel <r.hachel@tiscali.fr> - 2022-08-04 11:11 +0000
                            Re: Paul B. Andersen wanted to know... "Paul B. Andersen" <paul.b.andersen@paulba.no> - 2022-08-05 08:48 +0200
                              Re: Paul B. Andersen wanted to know... The Starmaker <starmaker@ix.netcom.com> - 2022-08-05 10:23 -0700
                                Re: Paul B. Andersen wanted to know... The Starmaker <starmaker@ix.netcom.com> - 2022-08-08 21:53 -0700
      Re: Paul B. Andersen wanted to know... Neal Giordano <bqwc@gcfikwuf.ja> - 2022-08-01 14:33 +0000
      Re: Paul B. Andersen wanted to know... RichD <r_delaney2001@yahoo.com> - 2022-08-08 18:35 -0700
        Re: Paul B. Andersen wanted to know... Buck Baggio <toed@iavdattg.ev> - 2022-08-10 12:52 +0000

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#588970 — Paul B. Andersen wanted to know...

Paul B. Andersen wanted to know...

Please! Please! STOP!!!

R.H. 

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#588981

Richard Hachel wrote:
> 
> Paul B. Andersen wanted to know...
> 
> Please! Please! STOP!!!
> 
> R.H.


Fear not. Paul B. Andersen is just a collection of  chemical scums on
the surface of the earth with illusions of grandeur. 



-- 
The Starmaker -- To question the unquestionable, ask the unaskable,
 to think the unthinkable, mention the unmentionable, say the unsayable,
and challenge
 the unchallengeable.

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#588982

Den 31.07.2022 16:33, skrev Richard Hachel:
> Paul B. Andersen wanted to know...
> 
> Please! Please! STOP!!!
> 
> R.H.

The subject line is wrong.

It should be:
Richard Hachel wanted to know how can a traveler who observes
for nine years the earth returning towards him at an apparent
speed of 4c, can he see it evolve on 7.2al?

The answer is:
The traveller can't observe the earth returning towards
him at an apparent speed of 4c.

What he _can_ visually observe is this:

On the way out he will see the Earth clock run slow
by sqrt((1-0.8)/(1+0.8)) = 1/3.
So when he turns around when his clock shows 9 years,
he will see the Earth clock shows 3 years.

On the way back he will see the Earth clock run fast
by sqrt((1+0.8)/(1-0.8)) = 3.
So when he is back after another 9 years on his clock,
he will see the Earth clock showing (3+3*9) years = 30 years.

Keep asking what you want to know, Richard.
I will always answer a nice guy like you.

-- 
Paul

https://paulba.no/

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#588986

On Sunday, 31 July 2022 at 20:00:00 UTC+2, Paul B. Andersen wrote:
> Den 31.07.2022 16:33, skrev Richard Hachel: 
> > Paul B. Andersen wanted to know... 
> > 
> > Please! Please! STOP!!! 
> > 
> > R.H.
> The subject line is wrong. 
> 
> It should be: 
> Richard Hachel wanted to know how can a traveler who observes 
> for nine years the earth returning towards him at an apparent 
> speed of 4c, can he see it evolve on 7.2al? 
> 
> The answer is: 
> The traveller can't observe the earth returning towards 
> him at an apparent speed of 4c. 
> 
> What he _can_ visually observe is this: 
> 
> On the way out he will see the Earth clock run slow 
> by sqrt((1-0.8)/(1+0.8)) = 1/3. 
> So when he turns around when his clock shows 9 years, 
> he will see the Earth clock shows 3 years. 
> 

Fortunately we have GPS now, so we can be
absolutely sure that the relativistic tales
have nothing in common with the real clocks
and real measurements.

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#588990

Le 31/07/2022 à 19:59, "Paul B. Andersen" a écrit :
> 
> 
> Den 31.07.2022 16:33, skrev Richard Hachel:
>> Paul B. Andersen wanted to know...
>> 
>> Please! Please! STOP!!!
>> 
>> R.H.
> 
> The subject line is wrong.
> 
> It should be:
> Richard Hachel wanted to know how can a traveler who observes
> for nine years the earth returning towards him at an apparent
> speed of 4c, can he see it evolve on 7.2al?

 Yes !!! 

 Oh, thank God.

Paul has UNDERSTOOD what I am saying.

This is already partly miraculous in the sense that he, like me, fully 
understands the PROBLEM.

 Vo=0.8c
 Vapp=4c
 t'= 9 years.



On ne peut resoudre correctement un problème que si l'on a claurement 
conscience de ce que l'on est en train de faire. 

Et ici, je sais ce que je fais. 

Paul COMPREND(miracle un homme est capable de me suivre) ce que je dis et 
comprend que le temps propre est bien de 9 ans. 

Neuf ans pendant lesquels il va voir la terre revenir sur lui à une 
vitesse de 4c dans son téléscope. 

C'est d'une beauté et d'une logique infinie. 

Mais cette beauté, je l'explique depuis des années snas que personne 
n'ait le déclic, tellement ça parait a priori contre-intuitif quand on 
n'a pas l'habitude de la SR.

Pourtant, je vous supplie de me croire, si vous avez ce déclic, c'est 
d'une prodigieuse évidence et d'une grande beauté conceptuelle.

Et du même coup, c'est fini, il n'y a plus de "cranks", plus 
d'incompréhension, plus de paradoxe possible. 

Tout est clair comme un grand ciel bleu dans l'esprit du physicien 
relativiste. 

Mais il y a un petit effort à faire, comme une décharge électrique 
extraordinaire dans l'esprit. 

Et une fois cette lumière éclatante dans l'esprit, le physicien ne 
pourra plus jamais retourner en arrière.  

La mauvaise maison relativiste actuelle ne pourra jamais être 
reconstruite. 

Le paradoxe de Langevin n'existe plus. 


 You can only solve a problem correctly if you are clearly aware of what 
you are doing.

And here I know what I'm doing.

Paul UNDERSTANDS (miracle a man is able to follow me) what I say and 
understands that the proper time is 9 years.

Nine years during which he will see the earth return to him at a speed of 
4c in his telescope.

It is of infinite beauty and logic.

But I've been explaining this beauty for years without anyone having the 
click, so a priori it seems counter-intuitive when you're not used to SR.

However, I beg you to believe me, if you have this trigger, it is 
prodigiously obvious and of great conceptual beauty.

And at the same time, it's over, there are no more "cranks", no more 
incomprehension, no more possible paradox.

Everything is clear as a big blue sky in the mind of the relativistic 
physicist.

But there is a little effort to be made, like an extraordinary electric 
shock in the mind.

And once that light shines in the mind, the physicist can never go back.

The current relativistic bad house can never be rebuilt.

Langevin's paradox no longer exists.

> 
> The answer is:
> The traveller can't observe the earth returning towards
> him at an apparent speed of 4c.


It is obvious that the traveler who has turned and is returning to earth 
SEES in his telescope the earth returning to him at 4c, AS, on the other 
side (total and perfect covariance) the brother sees the rocket returning 
to earth at 4c.

I don't understand why someone is stuck on this.

> 
> What he _can_ visually observe is this:
> 
> On the way out he will see the Earth clock run slow
> by sqrt((1-0.8)/(1+0.8)) = 1/3.
> So when he turns around when his clock shows 9 years,
> he will see the Earth clock shows 3 years.
> 
> On the way back he will see the Earth clock run fast
> by sqrt((1+0.8)/(1-0.8)) = 3.
> So when he is back after another 9 years on his clock,
> he will see the Earth clock showing (3+3*9) years = 30 years.

Yes, that is absolutely correct.

I see that you have understood perfectly what I have been saying for years 
about times.

It's exactly that.

But what you don't understand is the relativistic effect on distances.

There is complete covariance in the universe.

When I see the rocket coming back towards me (let's say it is thirty 
meters long), I see a rocket which apparently measures 90 meters.

Everyone agrees on that.

But the point of the reasoning is to consider that it is symmetrical and 
reciprocal; that it is of an absolute covariance.

In the rocket's reference frame, it is the earth-aphelion distance which, 
just after the half-turn, will appear to be multiplied by 3.

It's counter intuitive.

But I've been explaining all this for years, without anyone wanting to 
make the effort to understand how it really works.

> Keep asking what you want to know, Richard.
> I will always answer a nice guy like you.

 Thank you for that, and the reverse is also true. I have a lot of respect 
and pleasure to read the things you explain.

BUT that's not what I'm asking you (and the prospective reader): I've been 
begging you for years to understand what I'm trying to tell you, because 
it's so worth having this snap in the mind.

D=Vapp.Tr

Vapp (apparent speed)=4c
Tr (proper time)=9 years
D (distance)= 36ly

And not D=7.2!!! 

R.H. 

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#589044

Den 01.08.2022 00:18, skrev Richard Hachel:
> Le 31/07/2022 à 19:59, "Paul B. Andersen" a écrit :
>>

> 
> Vo=0.8c
> Vapp=4c
> t'= 9 years.

>>
>> What he _can_ visually observe is this:
>>
>> On the way out he will see the Earth clock run slow
>> by sqrt((1-0.8)/(1+0.8)) = 1/3.
>> So when he turns around when his clock shows 9 years,
>> he will see the Earth clock shows 3 years.
>>
>> On the way back he will see the Earth clock run fast
>> by sqrt((1+0.8)/(1-0.8)) = 3.
>> So when he is back after another 9 years on his clock,
>> he will see the Earth clock showing (3+3*9) years = 30 years.
> 
> Yes, that is absolutely correct.

Indeed.
When the traveller just has turned around,
the distance to Earth in his rest frame is D.

Since the traveller observes a Doppler shift = 3,
he can calculate that the speed of the Earth must be:
v = c⋅(3²+1)/(3²-1) = 0.8c

Since the traveller will be back on Earth after T = 9 years,
we have:
D = v⋅T = 7.2 light years

Case closed.

> 
> D=Vapp.Tr
> 
> Vapp (apparent speed)=4c
> Tr (proper time)=9 years
> D (distance)= 36ly
> 
> And not D=7.2!!!
> R.H.
> 



-- 
Paul

https://paulba.no/

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#589046

Le 01/08/2022 à 19:21, "Paul B. Andersen" a écrit :
> 
> 
> Den 01.08.2022 00:18, skrev Richard Hachel:
>> Le 31/07/2022 à 19:59, "Paul B. Andersen" a écrit :
>>>
> 
>> 
>> Vo=0.8c
>> Vapp=4c
>> t'= 9 years.
> 
>>>
>>> What he _can_ visually observe is this:
>>>
>>> On the way out he will see the Earth clock run slow
>>> by sqrt((1-0.8)/(1+0.8)) = 1/3.
>>> So when he turns around when his clock shows 9 years,
>>> he will see the Earth clock shows 3 years.
>>>
>>> On the way back he will see the Earth clock run fast
>>> by sqrt((1+0.8)/(1-0.8)) = 3.
>>> So when he is back after another 9 years on his clock,
>>> he will see the Earth clock showing (3+3*9) years = 30 years.
>> 
>> Yes, that is absolutely correct.
> 
> Indeed.
> When the traveller just has turned around,
> the distance to Earth in his rest frame is D.
> 
> Since the traveller observes a Doppler shift = 3,
> he can calculate that the speed of the Earth must be:
> v = c⋅(3²+1)/(3²-1) = 0.8c
> 
> Since the traveller will be back on Earth after T = 9 years,
> we have:
> D = v⋅T = 7.2 light years
> 
> Case closed.

 
What's tragic is that you don't understand what I'm telling you.


I've given you one of the most beautiful theories ever, and you don't want 
to bother to figure it out.

There, frankly, it all becomes grotesque.

We are no longer in science, but in dogma.

"We don't want Dr. Richard Hachel to reign over us."

This is no longer science, but madness.

I can explain, re-explain, and explain again: "WE DON'T WANT YOU, RATHER 
FREE BARABBAS".

It's in dogma and religion, there.

Possibly in sociology.

Not in science.

If we were really into science, we would wonder why a traveler who has 
just made his U-turn, who is going to observe for 9 years an earth 
returning towards him at 0.8c, that is to say at 4c of apparent speed, can 
see this event other than with Hachel's equations, that is to say those 
that I give and give again here.

It is absolutely obvious, and you have not yet had the click in your mind, 
that 4*9 is 36, and that this fantastic zoom effect does exist, as on the 
other side, by covariance, the terrestrial observer sees the rocket three 
times longer than it is.

I beg you one last time to understand what I am telling you.

You have the intellectual capacities, but I beg you to believe that the 
intellectual effort must be made (if possible with strong coffee).

I promise you, on my honor, two things.

A huge slap and a few minutes of "disorientation" if this happens to you.

But also the certainty that the house that will collapse before you cannot 
be rebuilt forever.

It's not yet the case.

When you write what you write to answer me, I see that you still haven't 
understood what I'm saying, nor had the fantastic click that can occur in 
a human mind.

This click goes through the very counter-intuitive notion of total 
covariance, the support of universal anisochrony in all equations and not 
only those dealing with the Doppler effect on electromagnetic waves.

<http://news2.nemoweb.net/jntp?pmC8rWDkCvHQHwM3ENDGJT5-RjU@jntp/Data.Media:1>

R.H. 

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#589086

Den 01.08.2022 20:46, skrev Richard Hachel:
> Le 01/08/2022 à 19:21, "Paul B. Andersen" a écrit :
>>
>>
>> Den 01.08.2022 00:18, skrev Richard Hachel:
>>> Le 31/07/2022 à 19:59, "Paul B. Andersen" a écrit :
>>>>
>>
>>>
>>> Vo=0.8c
>>> Vapp=4c
>>> t'= 9 years.
>>
>>>>
>>>> What he _can_ visually observe is this:
>>>>
>>>> On the way out he will see the Earth clock run slow
>>>> by sqrt((1-0.8)/(1+0.8)) = 1/3.
>>>> So when he turns around when his clock shows 9 years,
>>>> he will see the Earth clock shows 3 years.
>>>>
>>>> On the way back he will see the Earth clock run fast
>>>> by sqrt((1+0.8)/(1-0.8)) = 3.
>>>> So when he is back after another 9 years on his clock,
>>>> he will see the Earth clock showing (3+3*9) years = 30 years.
>>>
>>> Yes, that is absolutely correct.
>>
>> Indeed.
>> When the traveller just has turned around,
>> the distance to Earth in his rest frame is D.
>>
>> Since the traveller observes a Doppler shift = 3,
>> he can calculate that the speed of the Earth must be:
>> v = c⋅(3²+1)/(3²-1) = 0.8c
>>
>> Since the traveller will be back on Earth after T = 9 years,
>> we have:
>> D = v⋅T = 7.2 light years
>>
>> Case closed.
> 
> 
> What's tragic is that you don't understand what I'm telling you.
> 
>

I do understand that what you are telling me is nonsense.

But you don't understand what I just told you.

Please point out what is wrong in the calculation above,
repeated below:

When the traveller just has turned around,
the distance to Earth in his rest frame is D.

Since the traveller observes a Doppler shift = 3,
he can calculate that the speed of the Earth must be:
v = c⋅(3²+1)/(3²-1) = 0.8c

Since the traveller will be back on Earth after T = 9 years,
we have:
D = v⋅T = 7.2 light years

>>> 
>>> D=Vapp.Tr
>>> 
>>> Vapp (apparent speed)=4c
>>> Tr (proper time)=9 years
>>> D (distance)= 36ly
>>> 
>>> And not D=7.2!!!

How ridiculous!

-- 
Paul

https://paulba.no/

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#589095

Le 02/08/2022 à 14:40, "Paul B. Andersen" a écrit :
> 
> 
> Den 01.08.2022 20:46, skrev Richard Hachel:
>> Le 01/08/2022 à 19:21, "Paul B. Andersen" a écrit :
>>>
>>>
>>> Den 01.08.2022 00:18, skrev Richard Hachel:
>>>> Le 31/07/2022 à 19:59, "Paul B. Andersen" a écrit :
>>>>>
>>>
>>>>
>>>> Vo=0.8c
>>>> Vapp=4c
>>>> t'= 9 years.
>>>
>>>>>
>>>>> What he _can_ visually observe is this:
>>>>>
>>>>> On the way out he will see the Earth clock run slow
>>>>> by sqrt((1-0.8)/(1+0.8)) = 1/3.
>>>>> So when he turns around when his clock shows 9 years,
>>>>> he will see the Earth clock shows 3 years.
>>>>>
>>>>> On the way back he will see the Earth clock run fast
>>>>> by sqrt((1+0.8)/(1-0.8)) = 3.
>>>>> So when he is back after another 9 years on his clock,
>>>>> he will see the Earth clock showing (3+3*9) years = 30 years.
>>>>
>>>> Yes, that is absolutely correct.
>>>
>>> Indeed.
>>> When the traveller just has turned around,
>>> the distance to Earth in his rest frame is D.
>>>
>>> Since the traveller observes a Doppler shift = 3,
>>> he can calculate that the speed of the Earth must be:
>>> v = c⋅(3²+1)/(3²-1) = 0.8c
>>>
>>> Since the traveller will be back on Earth after T = 9 years,
>>> we have:
>>> D = v⋅T = 7.2 light years
>>>
>>> Case closed.
>> 
>> 
>> What's tragic is that you don't understand what I'm telling you.
>> 
>>
> 
> I do understand that what you are telling me is nonsense.
> 
> But you don't understand what I just told you.
> 
> Please point out what is wrong in the calculation above,
> repeated below:
> 
> When the traveller just has turned around,
> the distance to Earth in his rest frame is D.
> 
> Since the traveller observes a Doppler shift = 3,
> he can calculate that the speed of the Earth must be:
> v = c⋅(3²+1)/(3²-1) = 0.8c

 Vapp=Vo/(1-Vo/c)

 Vapp=0.8c/(1-0.8)

 Vapp=4c

 Tr=To*sqrt(1-v²/c²)

 Tr=15*sqrt(1-0.64)

 Tr=9 years

 x = 36 ly (AND NOT 7.2)

 
> 
> Since the traveller will be back on Earth after T = 9 years,
> we have:
> D = v⋅T = 7.2 light years
> 
>>>> 
>>>> D=Vapp.Tr
>>>> 
>>>> Vapp (apparent speed)=4c
>>>> Tr (proper time)=9 years
>>>> D (distance)= 36ly
>>>> 
>>>> And not D=7.2!!!
> 
> How ridiculous!

Sir, sir, I beg you to understand that when you speak to me, I fully 
understand what you are saying to me.

But I beg you to understand something else, which is that when it's me 
speaking, you don't understand anything I'm telling you.

I beg you on my knees to understand this.

There are very good contributors on this forum, but they don't understand 
anything I'm trying to explain to them.

I repeat, for 9 years, the traveler will see the earth come back on him, 
and with an apparent speed of 4c.

This is what he SEE.

Any other conclusion is absurd or dishonest, and part not of truth but of 
dogma.

I've spent thousands of hours studying this theory, and it's a bit 
disgusting to pass myself off as a moron who hasn't understood anything, 
when precisely no one in the world has gone further than me in the 
complete understanding of this geometry.

I beg you to make the effort to understand it.

Then I allow you to spit on it.

But I beg you, first, to understand what beauty and what scientific 
evidence you are spitting on, without realizing it.

R.H. 

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#589125

Den 02.08.2022 16:58, skrev Richard Hachel:
> Le 02/08/2022 à 14:40, "Paul B. Andersen" a écrit :
>>
>>
>> Please point out what is wrong in the calculation above,
>> repeated below:
>>
>> When the traveller just has turned around,
>> the distance to Earth in his rest frame is D.
>>
>> Since the traveller observes a Doppler shift = 3,
>> he can calculate that the speed of the Earth must be:
>> v = c⋅(3²+1)/(3²-1) = 0.8c
> 
> Vapp=Vo/(1-Vo/c)
> 
> Vapp=0.8c/(1-0.8)
> 
> Vapp=4c
> 
> Tr=To*sqrt(1-v²/c²)
> 
> Tr=15*sqrt(1-0.64)
> 
> Tr=9 years
> 
> x = 36 ly (AND NOT 7.2)
> 
> 
>>
>> Since the traveller will be back on Earth after T = 9 years,
>> we have:
>> D = v⋅T = 7.2 light years

Are you not going to answer?

Repeat:
Please point out what is wrong in the calculation above,
repeated below:

When the traveller just has turned around,
the distance to Earth in his rest frame is D.

Since the traveller observes a Doppler shift = 3,
he can calculate that the speed of the Earth must be:
v = c⋅(3²+1)/(3²-1) = 0.8c

Since the traveller will be back on Earth after T = 9 years,
we have:
D = v⋅T = 7.2 light years


> Sir, sir, I beg you to understand that when you speak to me, I fully 
> understand what you are saying to me.

If you understand it and still insist it is wrong,
why can't you point out what's wrong?

Is the speed of the Earth relative to the traveller
different from what you defined it to be, namely 0.8c?

Is the duration of the traveller's way back different
from 9 years on his own clock, as you defined it to be?

Is the distance different from (speed X time)?


-- 
Paul

https://paulba.no/

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#589126

Le 02/08/2022 à 22:35, "Paul B. Andersen" a écrit :
> Are you not going to answer?
> 
> Repeat:
> Please point out what is wrong in the calculation above,
> repeated below:
> 
> When the traveller just has turned around,
> the distance to Earth in his rest frame is D.
> 
> Since the traveller observes a Doppler shift = 3,
> he can calculate that the speed of the Earth must be:
> v = c⋅(3²+1)/(3²-1) = 0.8c
> 
> Since the traveller will be back on Earth after T = 9 years,
> we have:
> D = v⋅T = 7.2 light years
> 
> 
>> Sir, sir, I beg you to understand that when you speak to me, I fully 
>> understand what you are saying to me.
> 
> If you understand it and still insist it is wrong,
> why can't you point out what's wrong?
> 
> Is the speed of the Earth relative to the traveller
> different from what you defined it to be, namely 0.8c?
> 
> Is the duration of the traveller's way back different
> from 9 years on his own clock, as you defined it to be?
> 
> Is the distance different from (speed X time)?

Things have to be very clear between us and between other readers.

What I am telling you is very important, and is part of one of the 
greatest controversies in the history of science.

I'm not doing this for fun, and I know a few people here are serious, and 
really want to understand.

So I resume:

When the traveler turns around, he has just traveled 12 ly in the earth 
reference frame, and he will again travel 12 ly in the earth reference 
frame.

On that everyone agrees (even Maciej, I think).

It will do so in 15 years of observable time in the terrestrial reference 
frame To=15.

But with an apparent time of Tapp=3 (in the terrestrial reference frame).

And with a proper time of Tr (real time; proper time)=9.

On that everyone relativist agrees.

Where we no longer agree, you and I, is on what happens at the level of 
length elasticity.

You think there is a contraction of lengths, and that in the frame of the 
rocket L'=L.sqrt(1-v²/c²)

This is what is wrong.

This is why if we pass in apparent speeds (what we actually see in 
telescopes) the theory of relativity becomes absurd and contradictory.

This is where the error of the relativists lies, and this is where the 
famous paradox was born.

At the time, physicists had written an article called "one hundred authors 
against Einstein".

They understood that something was unclear.

They were right, those hundred physicists.

It's not that the theory is wrong, it's that there's something no one 
understands yet if they don't have the click.

It took me years to one day have this fantastic click.

As for the apparent speed of the earth returning to the rocket (in the 
reference frame of the rocket), it is 4c.

The covariance also applies on the other side.

For the earth, it is the rocket which returns to 4c of apparent speed 
(Vo=0.8c).

I don't understand your blockage on this.

The subject sees, for 9 years (this is his own time, his real time), a 
rocket which returns to him at 0.8c, but it is obvious that in his 
telescope, he sees it return to Vapp=4c.

It's this confusion between apparent speed (what he actually sees) and 
traditional Vo speed that makes you lose track of logic.

He actually sees something (which is apparent speed, I grant, but that's 
what he sees).

And for 9 years of his own time.

It is infinitely logical to think that covariance on speeds, on times, 
also works on distances and lengths (this is the step that you have so 
much trouble taking a priori, but yet once understood, we only see that).

Just as the terrestrial observer sees the distances and lengths of the 
rocket's reference frame multiplied by three (and a rocket that measures 
90 meters instead of 30), the observer in the rocket sees, ON THE FRONT of 
the rocket , an earth that returns to him at 4c, for 9 years, and from 
three times further than in the terrestrial reference frame.

That is to say D=12ly in R, and D'=36ly in R'.

I can't explain it better.

Afterwards, it's up to you to make the effort to understand why I say that 
I have internal perfection (the beauty of the theory) and external 
perfection (I explain all the experiments and even the instantaneous 
quantum effect) .

R.H. 

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#589156

Den 02.08.2022 23:11, skrev Richard Hachel:
> Le 02/08/2022 à 22:35, "Paul B. Andersen" a écrit :
>> Are you not going to answer?
>>
>> Repeat:
>> Please point out what is wrong in the calculation above,
>> repeated below:
>>
>> When the traveller just has turned around,
>> the distance to Earth in his rest frame is D.
>>
>> Since the traveller observes a Doppler shift = 3,
>> he can calculate that the speed of the Earth must be:
>> v = c⋅(3²+1)/(3²-1) = 0.8c
>>
>> Since the traveller will be back on Earth after T = 9 years,
>> we have:
>> D = v⋅T = 7.2 light years
>>
>>
>>> Sir, sir, I beg you to understand that when you speak to me, I fully 
>>> understand what you are saying to me.
>>
>> If you understand it and still insist it is wrong,
>> why can't you point out what's wrong?
>>
>> Is the speed of the Earth relative to the traveller
>> different from what you defined it to be, namely 0.8c?
>>
>> Is the duration of the traveller's way back different
>> from 9 years on his own clock, as you defined it to be?
>>
>> Is the distance different from (speed X time)?
> 
> Things have to be very clear between us and between other readers.
> 
> What I am telling you is very important, and is part of one of the 
> greatest controversies in the history of science.
> 
> I'm not doing this for fun, and I know a few people here are serious, 
> and really want to understand.

Please answer the questions above, and do not try to escape
by talking about something else.

The scenario in question, defined by you is:

Immediately after his turnaround the traveller is inertial
and is travelling towards the Earth.
His speed is 0.8c in the Earth frame, which means that
the speed of the Earth in the traveller's rest frame is v = 0.8c.
We know that the traveller will use the time T = 9 years
on his clock to get to the Earth.

That means that the distance to the Earth in the travellers rest
frame must be:
D = v⋅T = 7.2 light years

You claim that this is wrong and that the answer should be 36 ly.

So please explain what's wrong in the above by answering the questions
below:

Q1:
Is the speed of the Earth in the travellers rest frame = 0.8c?

Q2:
Is the traveller's proper time between turnaround and
back at Earth = 9 years?

Q3:
Is distance = speed X time?

If you have answered NO on any of these questions,
please explain what the correct answer should be, and how this
answer can make the distance to be 36 light years.

-- 
Paul

https://paulba.no/

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#589167

Le 03/08/2022 à 11:24, "Paul B. Andersen" a écrit :

> Q1:
> Is the speed of the Earth in the travellers rest frame = 0.8c?

 Yes.

> Q2:
> Is the traveller's proper time between turnaround and
> back at Earth = 9 years?

 Obviously.

> 
> Q3:
> Is distance = speed X time?

 Yes. 

 C'est cela que je dis, et ce que je fais, et c'est cela que vous ne 
parvenez pas à comprendre.

 La distance doit être celle mesurée par l'observateur lui-même (et pas 
un autre placé dans un autre référentiel).

 La distance est D.

 Le temps doit être mesuré par l'observateur, c'est à dire par son 
temps propre (et pas celui d'un autre).

 Ce temps est Tr.

 Attention! Et c'est là que la formidable claque dans la figure arrive 
(préparez-vous, tenez-vous au fauteuil). 

 La vitesse mesurée DOIT être la vitesse apparente. 

 L'observateur VOIT une terre revenir vers lui selon une vitesse 
apparente.

 L'équation correcte, je vous SUPPLIE de faire attention au concept est 
forcément D=Tr.Vapp

 Je vous supplie de comprendre ça.

 JE VOUS SUPPLIE DE VOUS METTRE EN PLACE D'UN HOMME QUI VOUT LA TERRE 
REVENIR SUR LUI PENDANT 9 ans, 
et avec constamment une vitesse apparente 4c.

 Je vous supplie de poser alors l'équation D=Tr.Vapp comme IL FAUT le 
faire.

R.H.  

 

 

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#589238

Den 03.08.2022 14:46, skrev Richard Hachel:
> Le 03/08/2022 à 11:24, "Paul B. Andersen" a écrit :
> 
>> The scenario in question, defined by you is:
>> 
>> Immediately after his turnaround the traveller is inertial
>> and is travelling towards the Earth.
>> His speed is 0.8c in the Earth frame, which means that
>> the speed of the Earth in the traveller's rest frame is v = 0.8c.
>> We know that the traveller will use the time T = 9 years
>> on his clock to get to the Earth.
>> 
>> That means that the distance to the Earth in the travellers rest
>> frame must be:
>> D = v⋅T = 7.2 light years
>> 
>> You claim that this is wrong and that the answer should be 36 ly.
>> 
>> So please explain what's wrong in the above by answering the questions
>> below:

>> Q1:
>> Is the speed of the Earth in the travellers rest frame = 0.8c?
> 
> Yes.
> 
>> Q2:
>> Is the traveller's proper time between turnaround and
>> back at Earth = 9 years?
> 
> Obviously.
> 
>>
>> Q3:
>> Is distance = speed X time?
> 
> Yes.

OK.
So you agree that immediately after the turnaround
the distance to the Earth in the traveller's rest frame is
D = v⋅T = 7.2 light years.

Thanks for finally admitting that you were wrong when
you claimed that this distance was 36 light years.

-- 
Paul

https://paulba.no/

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#589239

On Thursday, 4 August 2022 at 11:46:47 UTC+2, Paul B. Andersen wrote:

> >> Is distance = speed X time? 
> > 
> > Yes.
> OK. 

A common sense prejudice, refuted by your bunch
of idiots  with the discovery of inflation.

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#589243

Le 04/08/2022 à 11:49, Maciej Wozniak a écrit :
> On Thursday, 4 August 2022 at 11:46:47 UTC+2, Paul B. Andersen wrote:
> 
>> >> Is distance = speed X time? 
>> > 
>> > Yes.
>> OK. 
> 
> A common sense prejudice, refuted by your bunch
> of idiots  with the discovery of inflation.

Maciej...

Science (the real one) needs you here.

I try to explain that when a person sees an object coming towards them at 
0.8c, they actually see this object coming at 4c, because of the 
longitudinal Doppler effect.

When I say that, everyone agrees with me, ie 100% of the people who post 
or who would come to post.

I'm trying to say that if this happens for 9 years, then the person will 
see the object move 36 ly.

Posters of sci.physics.relativity have a hard time realizing this, and all 
of them tell me that x=9*0.8=7.2

This is a huge blunder which nevertheless exists in them in an incredible 
way, and if you can support me by saying that you do not understand why 
they are making this blunder, I will be very grateful to you.

It's very important what I'm asking you.

Thank you for not "missing" me when I need you.

R.H.

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#589245

On Thursday, 4 August 2022 at 13:27:09 UTC+2, Richard Hachel wrote:
> Le 04/08/2022 à 11:49, Maciej Wozniak a écrit : 
> > On Thursday, 4 August 2022 at 11:46:47 UTC+2, Paul B. Andersen wrote: 
> > 
> >> >> Is distance = speed X time? 
> >> > 
> >> > Yes. 
> >> OK. 
> > 
> > A common sense prejudice, refuted by your bunch 
> > of idiots with the discovery of inflation.
> Maciej... 
> 
> Science (the real one) needs you here. 
> 
> I try to explain that when a person sees

And -  do you have a SLIGHEST CLUE why it
has to be a person? 
That's because a person has a neural network 
called brain, and it's trained in the skill of claiming.

"Seeing", "observing" and so - is a social 
process from outside of your tale. 

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#589250

On Thursday, August 4, 2022 at 6:27:09 AM UTC-5, Richard Hachel wrote:

> I try to explain that when a person sees an object coming towards them at 
> 0.8c, they actually see this object coming at 4c, because of the 
> longitudinal Doppler effect. 

You need to carefully distinguish between what you *see* versus
what you *measure/observe*.

This is one of my favorite animations for a cube in transverse motion:
https://en.wikipedia.org/wiki/Special_relativity#Measurement_versus_visual_appearance

I think that I ought to create one for near-longitudinal motions. 

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#589252

On Thursday, 4 August 2022 at 15:58:26 UTC+2, prokaryotic.c...@gmail.com wrote:
> On Thursday, August 4, 2022 at 6:27:09 AM UTC-5, Richard Hachel wrote: 
> 
> > I try to explain that when a person sees an object coming towards them at 
> > 0.8c, they actually see this object coming at 4c, because of the 
> > longitudinal Doppler effect.
> You need to carefully distinguish between what you *see* versus 
> what you *measure/observe*. 

> 
> This is one of my favorite animations for a cube in transverse motion: 
> https://en.wikipedia.org/wiki/Special_relativity#Measurement_versus_visual_appearance 

In the meantime in the real world, of course,
forbidden by your bunch of idiots gPS and
TAI keep measuring t'=t, just like all serious 
clocks always did.

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#589253

Le 04/08/2022 à 15:58, Prokaryotic Capase Homolog a écrit :
> On Thursday, August 4, 2022 at 6:27:09 AM UTC-5, Richard Hachel wrote:
> 
>> I try to explain that when a person sees an object coming towards them at 
>> 0.8c, they actually see this object coming at 4c, because of the 
>> longitudinal Doppler effect. 
> 
> You need to carefully distinguish between what you *see* versus
> what you *measure/observe*.

Your answer is very violent.

You take your opponent for a fool, and you reverse the roles.

It's obviously a little "disgusting".

I remind you that it is ME, who asks the other speakers to be very careful 
about what they say.

They are the ones who confuse or use in a completely wrong way the notions 
of apparent, real and observable speed.

I've been explaining the differences between these speeds, and how to use 
them, for decades.

Sir, sir, once again I beg you not to be silly and arrogant.

In the problem we are discussing, apparent velocities MUST be used, not 
common observable velocities.

When you see an object coming back towards you, for 9 years, with an 
apparent speed of 4c,
it is absolutely clear that you are looking at a course of 36 par.

This is what the true theory of relativity says, and this is what the 
Lorentz transformations show.

Everything else is your fantasies and your arrogance of the type: "Doctor 
Hachel is shit, we are smarter and more educated than him".

Just that behavior.

That is rediculous.

R.H.

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