A student takes a science exam, and a question is about the theory of
relativity.
Unfortunately, he forgot about Lorentz's transformations.
He is then asked to give the coordinates of a space rocket by changing the
frame of reference.
Do not panic!

He knows that y'= y and z'= z.

He remembers Hachel's equation:
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)

With that, he can get by.

For To', just set To'=sqrt(x'²+y²+z²)/c

He is saved.

Thanks Hachel.

R.H.

Richard "Hachel" Lengrand (M.D.) wrote:
> A student takes a science exam, and a question is about the theory of
> relativity.
> Unfortunately, he forgot about Lorentz's transformations.

No problem, he can derive them easily from first principles.

> He is then asked to give the coordinates of a space rocket by changing
> the frame of reference.
> Do not panic!
>
> He knows that y'= y and z'= z.
>
> He remembers Hachel's equation:
> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)

Why would he remind such an atrocious bullshit which is not
linear (so in contradiction with Newton laws of motion) while
the correct equation is far simpler?

> With that, he can get by.
>
> For To', just set To'=sqrt(x'²+y²+z²)/c

A real student in physics wouldn't use a silly equation in
contradiction with Newton's laws.

> He is saved.

He is dead in the water, he failed miserably. He won't pass.

Maybe he should consider a career in medicine, standard there
are clearly lower.

On Saturday, 8 January 2022 at 15:45:45 UTC+1, Python wrote:
> Richard "Hachel" Lengrand (M.D.) wrote:
> > A student takes a science exam, and a question is about the theory of
> > relativity.
> > Unfortunately, he forgot about Lorentz's transformations.
> No problem, he can derive them easily from first principles.

And, of course, in the meantime in the real world forbidden
by your moronic religion TAI, using sane methods of
synchronization - keep measuring t'=t, just like all serious
clocks always did.

Maciej Wozniak wrote:
> On Saturday, 8 January 2022 at 15:45:45 UTC+1, Python wrote:
>> Richard "Hachel" Lengrand (M.D.) wrote:
>>> A student takes a science exam, and a question is about the theory of
>>> relativity.
>>> Unfortunately, he forgot about Lorentz's transformations.
>> No problem, he can derive them easily from first principles.
>
> And, of course, in the meantime in the real world ...

In the real world you are a wannabe "information engineer" with
0 knowledge in physics, maths or engineering yelling on Usenet
without convincing anyone with your idiotic rant. Meanwhile
real engineering is done using principles from Galileo, Newton,
Laplace, Einstein and Quantum Mechanics and your rant is just
a annoying, marginally entertaining, pile of crap with 0 consequence.

Richard Hachel wrote:

> A student takes a science exam, and a question is about the theory of
> relativity. Unfortunately, he forgot about Lorentz's transformations.
> He is then asked to give the coordinates of a space rocket by changing
> the frame of reference. Do not panic!

BEYOND RETARDED - Hospitalized Canadian Patient Diagnosed With 'Climate
Change' 🤣🤣 https://www.bitchute.com/video/f0njKXIKXwpb/

Le 08/01/2022 à 15:45, Python a écrit :
> Richard "Hachel" Lengrand (M.D.) wrote:
>> A student takes a science exam, and a question is about the theory of
>> relativity.
>> Unfortunately, he forgot about Lorentz's transformations.
>
> No problem, he can derive them easily from first principles.

Easily, easily, c'est toi qui le dit.

Il a quand même fallu que Poincaré s'y colle, lui le plus grand
mathématicien de l'époque,
et le dernier homme au monde à maitriser l'intégralité de la physique
et de la mathématique de son temps.

>
>> He is then asked to give the coordinates of a space rocket by changing
>> the frame of reference.
>> Do not panic!
>>
>> He knows that y'= y and z'= z.
>>
>> He remembers Hachel's equation:
>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>
> Why would he remind such an atrocious bullshit which is not
> linear (so in contradiction with Newton laws of motion) while
> the correct equation is far simpler?

But because I'm crazy.

You said it yourself.
>> With that, he can get by.
>>
>> For To', just set To'=sqrt(x'²+y²+z²)/c
>
> A real student in physics wouldn't use a silly equation in
> contradiction with Newton's laws.

And?
>
>> He is saved.
>
> He is dead in the water, he failed miserably. He won't pass.

Snifff...
> Maybe he should consider a career in medicine, standard there
> are clearly lower.

Ad hominem attack.

R.H.

On Saturday, 8 January 2022 at 16:26:41 UTC+1, Python wrote:
> Maciej Wozniak wrote:
> > On Saturday, 8 January 2022 at 15:45:45 UTC+1, Python wrote:
> >> Richard "Hachel" Lengrand (M.D.) wrote:
> >>> A student takes a science exam, and a question is about the theory of
> >>> relativity.
> >>> Unfortunately, he forgot about Lorentz's transformations.
> >> No problem, he can derive them easily from first principles.
> >
> > And, of course, in the meantime in the real world ...
>
> In the real world you are a wannabe "information engineer" with
> 0 knowledge in physics, maths or engineering yelling on Usenet
> without convincing anyone with your idiotic rant.

Have you already learnt the difference between a square
root of 7 and sqrt(7)?

Meanwhile
> real engineering is done using principles from Galileo, Newton,
> Laplace, Einstein

Keep dreaming, poor stinker; , in the meantime in the real world
forbidden by your moronic religion TAI, using sane methods of
synchronization - keep measuring t'=t, just like all serious
clocks always did.

Idiot crank Maciej Wozniak wrote:
> On Saturday, 8 January 2022 at 16:26:41 UTC+1, Python wrote:
>> Maciej Wozniak wrote:
>>> On Saturday, 8 January 2022 at 15:45:45 UTC+1, Python wrote:
>>>> Richard "Hachel" Lengrand (M.D.) wrote:
>>>>> A student takes a science exam, and a question is about the theory of
>>>>> relativity.
>>>>> Unfortunately, he forgot about Lorentz's transformations.
>>>> No problem, he can derive them easily from first principles.
>>>
>>> And, of course, in the meantime in the real world ...
>>
>> In the real world you are a wannabe "information engineer" with
>> 0 knowledge in physics, maths or engineering yelling on Usenet
>> without convincing anyone with your idiotic rant.
>
> Have you already learnt the difference between a square
> root of 7 and sqrt(7)?

Still fighting with the basics, Woz? Sigh...

Le 08/01/2022 à 15:45, Jean-Pierre Messager (Python) a écrit :

>> He remembers Hachel's equation:
>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)

> is not linear

It's bijective, and that's enough for me.

I am a great humble man, you know.

I'm not looking for numbers in silver, or mathematical signs in gold.

The coherence of nature is enough for me.

I leave scientific puritanism to others.

R.H.

Richard "Hachel" Lengrand (M.D.) wrote:
> Le 08/01/2022 à 15:45, Jean-Pierre Messager (Python) a écrit :
>
>>> He remembers Hachel's equation:
>>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>
>> is not linear
>
> It's bijective, and that's enough for me.

Could you prove it is even bijective ?

> I am a great humble man, you know.

No. You are an arrogant idiot.

> I'm not looking for numbers in silver, or mathematical signs in gold.

This is not a proper English sentence, it is meaningless.

> The coherence of nature is enough for me.

Newton's laws are part of this "coherence". Learn.

> I leave scientific puritanism to others.

This is not "puritanism", it is sticking with overwhelming experimental
evidence.

Le 08/01/2022 à 19:56, Python a écrit :

Bon, Jean-Pierre.

Maintenant, tu arrêtes, tu n'es pas gentil du tout.

C'est mal.

>>>> He remembers Hachel's equation:
>>>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>>
>>> is not linear
>>
>> It's bijective, and that's enough for me.
>
> Could you prove it is even bijective ?

Not only is it bijective, but it is reciprocal if you change v to -v.

Exemple :

E(R)=(12,9,0) (E)R'=(40,9,0) v=0.8c

x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)

x'= [12+sqrt(12²=9²+0²).0.8]/0.6 ---> x=40

Réciproque :

x= [x'-sqrt(x'²+y'²+z'²).v/c]/sqrt(1-v²/c²)

x= [40-sqrt(40²+9²+0²).0.8]/0.6 ---> x=12

Cette équation permet d'obtenir très facilement les coordonnées
spatiales et temporelles
dans R' d'un événement qui se passe dans R au moment du croisement O et
O'.

Evidemment, puisqu'on déclenche les montres à cet instant, on a t'=t=0.

Pour connaître l'heure de l'événement en temps local, il suffit de
poser
To=-d/c dans R, et To'=-d'/c dans R'.

Ici To=sqrt((12²+9²+0²)/c= -15 et To'=sqrt(40²+9²+0²)/c= -41

On obtient la même chose en passant par les TL(attention au signe de
To).

x'=(x-vTo)/sqrt(1-v²/c²)
y'=y
z'=z
To'=(To-xv/c²)/sqrt(1-v²/c²)

<http://news2.nemoweb.net/?DataID=o99IT4IvqyjD7HA01v15TsMcMEo@jntp>
R.H.

--
"Mais ne nous y trompons pas. Il n'y a pas que de la violence
avec des armes. Il y a des situations de violence".
Abbé Pierre.

Richard "Hachel" Lengrand, (M.D.) wrote:
> Le 08/01/2022 à 19:56, Python a écrit :
....
>>>>> He remembers Hachel's equation:
>>>>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>>>
>>>> is not linear
>>>
>>> It's bijective, and that's enough for me.
>>
>> Could you prove it is even bijective ?
>
> Not only is it bijective, but it is reciprocal if you change v to -v.

You don't know what bijective mean? Really? So
why do you use the word?

On Saturday, January 8, 2022 at 4:12:18 AM UTC-8, Richard Hachel wrote:
> He knows that y'= y and z'= z. He remembers Hachel's equation:
> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)

That equation is valid only for events that are light-like separated from the origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² - x² - y² - z², so to cover time-like intervals your equation would have to be
x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
where tau is the proper time along the interval. Writing the transformation equation that way is not advantageous.

On Saturday, 8 January 2022 at 22:17:39 UTC+1, Townes Olson wrote:
> On Saturday, January 8, 2022 at 4:12:18 AM UTC-8, Richard Hachel wrote:
> > He knows that y'= y and z'= z. He remembers Hachel's equation:
> > x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
> That equation is valid only for events that are light-like separated from the origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² - x² - y² - z², so to cover time-like intervals your equation would have to be
> x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
> where tau is the proper time along the interval. Writing the transformation equation that way is not advantageous.

In the meantime in the real world, however, forbidden
by your moronic religion TAI keep measuring t'=t, just
like all serious clocks always did.

On Saturday, January 8, 2022 at 1:02:13 PM UTC-8, Python wrote:
> Richard "Hachel" Lengrand, (M.D.) wrote:
> > Le 08/01/2022 à 19:56, Python a écrit :
> ...
> >>>>> He remembers Hachel's equation:
> >>>>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
> >>>
> >>>> is not linear
> >>>
> >>> It's bijective, and that's enough for me.
> >>
> >> Could you prove it is even bijective ?
> >
> > Not only is it bijective, but it is reciprocal if you change v to -v.
> You don't know what bijective mean? Really? So
> why do you use the word?

What quantities are you using you moron?

Le 08/01/2022 à 22:17, Townes Olson a écrit :
> On Saturday, January 8, 2022 at 4:12:18 AM UTC-8, Richard Hachel wrote:
>> He knows that y'= y and z'= z. He remembers Hachel's equation:
>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>
> That equation is valid only for events that are light-like separated from the
> origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² -
> x² - y² - z², so to cover time-like intervals your equation would have to be
> x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
> where tau is the proper time along the interval. Writing the transformation
> equation that way is not advantageous.

Ah! Un posteur qui réfléchit et qui ne vient pas faire du pugilat.
Je ne dis pas qu'il n'y en a pas sur ce site, car il s'en trouve. Mais il
n'y en a pas des dizaines.
Donc félicitations, monsieur.

Ah! A poster who thinks and does not come to fight.
I'm not saying there aren't any on this site, because there are. But there
aren't dozens of them.
So congratulations, sir.

Je ne sais pas à quoi correspond votre équation, je vous donne donc la
mienne puisque vous voulez aller plus loin et parler en fonction de t.

Je précise bien en fonction de t. C'est à dire le moment où
l'observateur perçoit l'événement.

Je différencie le temps local To par rapport au temps propre t de la
montre placée en O. Ce n'est pas la même chose.

I don't know what your equation is, so I'm giving you mine since you want
to go further and speak in terms of t.

I am precise in terms of t. That is to say the moment when the observer
perceives the event.

I differentiate the local time To compared to the proper time t of the
watch placed in O. It is not the same thing.

L'équation qui prend en compte t est la suivante :
x'=[x+({sqrt(x²+y²+z²)/c}-t).v/c]/sqrt(1-v²/c²)

Sa réciproque est :
x=[x'-({sqrt(x'²+y²+z²)/c}-t).v/c]/sqrt(1-v²/c²)

The equation which takes into account t is the following:
x'=[x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)

Its reciprocal is:
x = [x'- ({sqrt(x'²+y²+z²)/c}-t').v/c]/sqrt(1-v²/c²)

Prenons un exemple simple. Une fusée passe à proximité de la terre à
vitesse v=0.8c.
On lui assigne l'axe Ox.
Dans le référentiel terrestre une explosion cosmique à lieu (c'est à
dire est VISIBLE) au moment où les deux observateurs O et O' se croisent.
x=12ly, y=9ly z=0
On a alors l'événement cosmique E suivant E=(12,9,0,-15,0) selon les
coordonnées (x,y,z,To,t).
Plusieurs sur ce forum ont alors chercher à savoir (et y sont parvenus)
ce qui se passait dans le référentiel de la fusée. Je les en remercie.
Ils ont trouvé E=(40,9,0,-41,0) pour la fusée.
Ils ont donc reçu mes félicitations.
Let's take a simple example. A rocket passes near the earth at speed v =
0.8c.
We assign it the axis Ox.
In the terrestrial frame of reference a cosmic explosion takes place (ie
is VISIBLE) at the moment when the two observers O and O 'cross. x=12ly
y=9ly z=0
We then have the following cosmic event E E = (12,9,0,-15,0) according to
the coordinates (x, y, z, To, t).
Several on this forum then tried to find out (and succeeded in doing so)
what was happening in the rocket's repository. I thank them for that.
They found E=(40,9,0,-41,0) for the rocket.
So they received my congratulations.

Mais il faut aller plus loin, vous avez raison. Car ici je n'ai employé
que l'équation :
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
Cette équation n'est valable que pour tous les observateurs qui se
croisent en O à t=0.

Mais imaginons, que cet événement n'ai pas lieu au moment où la fusée
passe, mais six ans plus tard.
Il va donc me falloir employer une équation qui contient t (le moment où
la terre perçoit l'explosion cosmique) pendant que la fusée a
progressé, elle, en mouvement galiléen régulier de 0.8c.
Posons t=6 (six ans).
L'équation devient :
x'=[x+({sqrt(x²+y²+z²)/c}-t).v/c]/sqrt(1-v²/c²)
On a alors dans le référentiel terrestre E=(12,9,0,-9,6)
et dans le référentiel de la fusée E=(32,9,0,-31, 2.24154)

But we have to go further, you are right. Because here I only used the
equation:
x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
This equation is only valid for all the observers who intersect at O
​​at t=0.

But imagine, that this event does not take place when the rocket passes,
but six years later.
So I will have to use an equation which contains t (the moment when the
earth perceives the cosmic explosion) while the rocket has progressed in
regular Galilean motion of 0.8c.
Let t=6 (six years).
The equation becomes:
x '= [x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)
We then have in the terrestrial frame of reference E = (12,9,0, -9,6)
and in the frame of the rocket E = (32,9,0, -31, 2.24154)

Vérifions maintenant que la réciproque est correcte (sinon c'est
absurde).
Now let's check that the converse is correct (otherwise it's absurd).

x=[x'-({sqrt(x'²+y²+z²)/c}-t').v/c]/sqrt(1-v²/c²)

x=[32-({sqrt(32²+9²+0²)/1}-2.24154)*0.8]/0.6
x=[32-(33.24154-2.24154)*0.8]/0.6
x=[32-(31*0.8)]/0.6
x=12

R.H.

Le 08/01/2022 à 22:02, Python a écrit :
> Richard "Hachel" Lengrand, (M.D.) wrote:
>> Le 08/01/2022 à 19:56, Python a écrit :
> ...
>>>>>> He remembers Hachel's equation:
>>>>>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>>>>
>>>>> is not linear
>>>>
>>>> It's bijective, and that's enough for me.
>>>
>>> Could you prove it is even bijective ?
>>
>> Not only is it bijective, but it is reciprocal if you change v to -v.
>
> You don't know what bijective mean? Really? So
> why do you use the word?

Mais tu ne t'arrêtes donc jamais, toi?

R.H.

On Saturday, January 8, 2022 at 4:13:30 PM UTC-8, Richard Hachel wrote:
> >> He knows that y'= y and z'= z. He remembers Hachel's equation:
> >> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
> >
> > That equation is valid only for events that are light-like separated from the
> > origin, meaning that c²t² = x²+y²+z². In general we have tau^2 = c²t² -
> > x² - y² - z², so to cover time-like intervals your equation would have to be
> > x'= [x + sqrt(x²+y²+z²+tau^2)v/c] / sqrt(1-v²/c²) .
> > where tau is the proper time along the interval. Writing the transformation
> > equation that way is not advantageous.
>
> I don't know what your equation is...

It's the space component of the Lorentz transformation, written in a rather silly way, by replacing the time coordinate, t, with the expression sqrt(x²+y²+z²+tau^2). Again, writing the equation that way is not advantageous.

> The equation which takes into account t is the following:
> x'=[x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)

No, that is correct only if x=y=z=0, which is not generally the case. The simplest way of expressing the Lorentz transformation, in units with c=1, is just x'=(x-vt)g, t'=(t-vx)g where g=1/sqrt(1-v^2). That is all there is to it.

Le 09/01/2022 à 01:36, Townes Olson a écrit :

>> The equation which takes into account t is the following:
>> x'=[x+({sqrt(x²+y²+z²)/c}-t)v/c]/sqrt(1-v²/c²)
>
> No

LOL.

>that is correct only if x=y=z=0,

re-LOL

> which is not generally the case. The simplest way of expressing the Lorentz
> transformation, in units with c=1, is just x'=(x-vt)g, t'=(t-vx)g where
> g=1/sqrt(1-v^2). That is all there is to it.

No.

You do not read what I write correctly.

My rating is this in this case:

x'=γ(x-v.To)
y'=y
z'=z
To'=γ(To-xv/c²)

N.B. Attention To and To' are negative. Here To=-15 et and To'=-41.
To2=-9(-15+6) and To2'=-31

R.H.

Le 08/01/2022 à 22:02, Python a écrit :
> Richard "Hachel" Lengrand, (M.D.) wrote:
>> Le 08/01/2022 à 19:56, Python a écrit :
> ...
>>>>>> He remembers Hachel's equation:
>>>>>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>>>>
>>>>> is not linear
>>>>
>>>> It's bijective, and that's enough for me.
>>>
>>> Could you prove it is even bijective ?
>>
>> Not only is it bijective, but it is reciprocal if you change v to -v.
>
> You don't know what bijective mean? Really? So
> why do you use the word?

Yes, is the question.
why this fool use that words.
Because : "Il est fou".
jma

Jean-Michel Affoinez, Richard "Hachel" Lengrand's sock puppet, wrote:
> Le 08/01/2022 à 22:02, Python a écrit :
>> Richard "Hachel" Lengrand, (M.D.) wrote:
>>> Le 08/01/2022 à 19:56, Python a écrit :
>> ...
>>>>>>> He remembers Hachel's equation:
>>>>>>> x'= [x+sqrt(x²+y²+z²).v/c]/sqrt(1-v²/c²)
>>>>>
>>>>>> is not linear
>>>>>
>>>>> It's bijective, and that's enough for me.
>>>>
>>>> Could you prove it is even bijective ?
>>>
>>> Not only is it bijective, but it is reciprocal if you change v to -v.
>>
>> You don't know what bijective mean? Really? So
>> why do you use the word?
>
> Yes, is the question.
> why this fool use that words.
> Because : "Il est fou". jma

This is the best you can do when your crap is debuncked, Lengrand? Post
more nonsense using a sock puppet fake name?

On Sunday, 9 January 2022 at 15:35:33 UTC+1, Python wrote:

> This is the best you can do when your crap is debuncked, Lengrand? Post
> more nonsense using a sock puppet fake name?

So, is Euclidean geometry a valid geometry of our
universe? According to you and your idiot gurus, of
course.
Still no answer, poor stinker?

Maciej Wozniak wrote:
> On Sunday, 9 January 2022 at 15:35:33 UTC+1, Python wrote:
>
>
>> This is the best you can do when your crap is debuncked, Lengrand? Post
>> more nonsense using a sock puppet fake name?
>
> So, is Euclidean geometry a valid geometry of our
> universe? According to you and [snip profanity], of
> course.
> Still no answer, [snip profanity]?

You shouldn't ask questions that you don't understand what
they actually mean. You're only making a fool of yourself
and exposing what an idiot you are Maciej.

On Sunday, 9 January 2022 at 16:12:00 UTC+1, Python wrote:
> Maciej Wozniak wrote:
> > On Sunday, 9 January 2022 at 15:35:33 UTC+1, Python wrote:
> >
> >
> >> This is the best you can do when your crap is debuncked, Lengrand? Post
> >> more nonsense using a sock puppet fake name?
> >
> > So, is Euclidean geometry a valid geometry of our
> > universe? According to you and [snip profanity], of
> > course.
> > Still no answer, [snip profanity]?
>
> You shouldn't ask questions that

An opinion of a fanatic, lying stinker not even knowing
what a function is doesn't interest me, unfortunately.
So, is Euclidean geometry a valid geometry of our
universe? According to you and your idiot gurus, of
course.
Still no answer, poor stinker?

Le 09/01/2022 à 16:11, Python a écrit :
> Maciej Wozniak wrote:
>> On Sunday, 9 January 2022 at 15:35:33 UTC+1, Python wrote:
>>
>>
>>> This is the best you can do when your crap is debuncked, Lengrand? Post
>>> more nonsense using a sock puppet fake name?
>>
>> So, is Euclidean geometry a valid geometry of our
>> universe? According to you and [snip profanity], of
>> course.
>> Still no answer, [snip profanity]?
>
> You shouldn't ask questions that you don't understand what
> they actually mean. You're only making a fool of yourself
> and exposing what an idiot you are Maciej.

Ce qui m'étonnera toujours, c'est la connerie humaine, teintée de haine,
de violence, de censure,
d'appels à la délation et au "doute professionnel".

Cette saloperie humaine a existé.

On appelle cela la profondeur du serpent.

Des centaines de siècles attestent cela, couronnés par la sagesse des
sages.

Le serpent croit qu'il a une grosse bite (alors qu'il n'a même pas de
pattes).

Il se donne des noms, le serpent.

Il est fou.

On lui demande d'arrêter.

On lui dit : Jean-Pierre, maintenant, tu arrêtes. STOP!

Il continue.

Il est fou.

R.H.