Le 26/12/2021 à 16:30, Thomas 'PointedEars' Lahn a écrit :
> Richard Hachel wrote:
>
>> When we study the theory of relativity, we abandon the idea of
>> ​​absolute space and time.
>
> Correct. We *must* abandon that idea because it is inconsistent with the
> assumption that comes from observation: that light only travels at one speed
> in vacuum.
>
>> Space, for example, is no longer seen as something that has a rigid
>> metric.
>
> In a way, that is correct.
>
>> When I put the equation y'= y and z' = z, I see that there is all the same
>> laterally in the direction of displacement, a certain form of rigidity, of
>> absolute metric.
>
> Not “laterally in the direction of displacement”, but only _perpendicular_
> to the direction of _motion_. Whereas then the direction of motion is
> defined to be parallel to the x-axis. (We must then *choose* the x-axis to
> be this way, and the y and z axis perpendicular to the x-axis and to each
> other.)
>
> If the motion is not parallel to the x-axis, then y' = y and z' = z are NOT
> correct statements.
>
> The identity transformations y' = y and z' = z are not definitions, but
> observable consequences. The same as the transformation x' = γ (x − v t)
> is. Together with t' = γ (t − v/c² x) – also observable – they define
> the
> Lorentz transformation *for motion parallel to the x-axis*.

I totally agree with what you are saying.

R.H.

On 12/26/2021 12:11 PM, Richard Hachel wrote:
> Le 26/12/2021 à 16:25, Michael Moroney a écrit :
>> On 12/25/2021 4:04 PM, Richard Hachel wrote:

>>> Exercice E=(x,y,z,To,t)
>>> (when observers is crossing t'=t=0) ---> E=(12,0,9,-15,0) in R what
>>> is E in R'? R.H.
>
>> Standard notation, please.
>
> I refuse.

Expect to be ignored, then.

It's like if someone writes in French in an English speaking newsgroup,
despite the fact his English is just fine. As deliberate obfuscation.

(wait, you already do that!)
>
>> I have no desire to try to translate to your hokey system and back.
>
>> Why do you claim -vt = sqrt(x²+y²+z²) ?
>
> I did not say that.

I used simple algebra to derive that from your
x' = [x+sqrt(x²+y²+z²)]/sqrt(1-v²/c²) and x'= γ(x-vt) Lorentz transform.

Thomas 'PointedEars' Lahn wrote:

> Richard Hachel wrote:
>> Except that in the second, since t "= t '= t = 0
>
> But is not. That two coordinate systems coincide in *one* point by
> definition does NOT mean that the temporal coordinates of events will
> always be zero, or the same. The very point of a coordinate
> *transformation* is that they are not:
>
> t t'
> ^ :
> : :
> : :'-._ e
> +---:-----'+
> : : :: x'
> : : _:.:--''
> ::__..--'' :
> ------------__.+'---------+-----> x
> __..--'' ::
> --'' : :
> : :
> : :

Now, less sleepy than before, I realize that I drew that wrong. It should
be more like this:

t t'
^ :
: : e
+-----:----_.+
: ..--'' ':
: : ' :
: . ' : x'
: : '__.:--''
::__..--'' :
------------__.+'-----------+---> x
__..--'' ::
--'' : :
: :
: :

[The lines for the coordinates must always be parallel to the respective
coordinate axes. Also, I drew event farther out and further in the future
now so that you can see it better.]

You can clearly see that the t'-coordinate of the event e is smaller than
the t-coordinate for the same event. Likewise for the x'- and x-
coordinates, as the Lorentz transformation tells us (e.g. substitute x ≔ 2,
t ≔ 2, v ≔ 0.6 c and c ≔ 1; you get x' = 1 and t' = 1).

PointedEars
--
Q: What happens when electrons lose their energy?
A: They get Bohr'ed.

(from: WolframAlpha)

On Sunday, December 26, 2021 at 12:12:00 PM UTC-5, Richard Hachel wrote:
> Le 26/12/2021 à 16:25, Michael Moroney a écrit :
> > On 12/25/2021 4:04 PM, Richard Hachel wrote:

> > Standard notation, please.
> I refuse.

Then get out of here.
If you come here, its up to *you* to use the appropriate language; that's respect.
If you don't use the appropriate language, you won't be understood.

> > Why do you claim -vt = sqrt(x²+y²+z²) ?
> I did not say that.

You *claimed* that, in the sense, you implied it. Google kept a record.

Richard Hachel wrote:

> Le 18/12/2021 à 22:08, Paul Alsing a écrit :
>>> > I say : x'= [x + sqrt (x² + y² + z²) v/c] / sqrt (1-v²/c²) y'= y z'=
>>> > z t'= t = 0
>
>> https://math.ucr.edu/home/baez/crackpot.html
>
> Look for the truth of the equations. Don't worry about men's brawls.

equations are nothing. In physics *inequations* are what physics is all
about.