Re: Einstein cheated with his fraudulent derivation of Lorentz transforms

On Thu, 6 Feb 2025 18:48:16 +0000, Paul.B.Andersen wrote:

Den 05.02.2025 23:51, skrev rhertz:

On Wed, 5 Feb 2025 12:29:56 +0000, Paul.B.Andersen wrote:
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Den 04.02.2025 22:54, skrev rhertz:

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Now, considering that he started with Galilean transform x' = x-vt
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You can't have read §3 properly, you have only
scrutinised the text to find "x' = x − vt", and when you found
it, you got an orgasm, shouting:
"EINSTEIN USED GALILEAN TRANSFORM TO DERIVE LORENTZ WITHOUT ETHER!!"

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NO, innoble animal!. I didn't write that claim.

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You used exactly those words back in 2020,
and as quoted above, you repeated that x' = x - vt
is the Galilean transform in §3.
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See:
https://paulba.no/paper/Electrodynamics.pdf
Read §3
  Theory of the Transformation of Co-ordinates and
  Times from a Stationary System to another System in
  Uniform Motion of Translation Relatively to the Former

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The paragraph is about the transformation of the coordinates
of an event from one frame of reference to another frame
of reference which is moving relative to the first.
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On the first page (page 5) Einstein defines the coordinate systems.
The "stationary system"  K(x,y,z,t) coordinates are Latin letters
The "moving system"      k(ξ,η,ζ,τ) coordinates are Greek letters

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So:
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The coordinate transform  K-> k is generally:
ξ = ξ(x,y,z,t)
η = η(x,y,z,t)
ζ = ζ(x,y,z,t)
τ = τ(x,y,z,t)
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The coordinate transform k -> K is generally:
x = x(ξ,η,ζ,τ)
y = y(ξ,η,ζ,τ)
z = z(ξ,η,ζ,τ)
t = t(ξ,η,ζ,τ)
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The Galilean transform K-> k is:
ξ = x - vt
η = y
ζ = z
τ = t
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The Galilean transform k -> K is:
x = ξ + vτ
y = η
z = ζ
t = τ
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The Lorentz transform K-> k is:
ξ = (x - vt)/√(1−v²/c²)
η = y
ζ = z
τ = (t - (v/c²)x)/√(1−v²/c²)
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The Lorentz transform k-> K is:
x = (ξ + vτ)/√(1−v²/c²)
y = η
z = ζ
t = (τ + (v/c²)ξ)/√(1−v²/c²)
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So the Galilean transform is: ξ = x - vt

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NO, innoble animal! ξ is the name of the horizontal axis in the moving
frame k.

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And x' is a coordinate in k ? :-D
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ξ(x') = x' in the moving frame. I attached a graphic to clarify this
but, with your "dog vision" you missed it.

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What kind of transform is ξ(x') = x' ?
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Did you possibly mean: ξ = (x - vt)/√(1−v²/c²) ?
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But at this point we don't know this yet, we only know: ξ = ξ(x,y,z,t)
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x' = x - vt is the well known Galilean transform, along with τ = t' = t.

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So  τ is a coordinate in k, but ξ is not a coordinate in k,
because the coordinates in k are (x',η,ζ,τ), right?
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This is fun, isn't it?  :-D
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There is only one meaning of x', right?
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https://paulba.no/pdf/Dilbert.pdf

As Dono did, you're deviating the attention about the main topic, which
can be read in the title of this thread.
Using ξ(x') = x' is my way TO ENFORCE that x' is a point on the ξ axis.
I also used this way to express ξ(0) = 0, so I could write  the length
of the
"virtual rod r_AB" as  ξ(x') - ξ(0), being 0 and x' the points in the
frame k
where clocks, detectors and the mirror are located.
If you insist with formalisms, better make a forensic analysis of your
pagan
god Einstein, who was pretty good cheating, hacking and taking gullible
assholes like you like 100% IDIOTS.
In your case, and after more than 30 years behaving as a relativistic
sucker,
it's noticeable that you are tired of that shit. You think and write
like a
extremely worn out person, after so much time repeating the same
idiocies.
Accept it: if you EVER had a bit of wit, you've lost it. Do gardening
now, before you become the plant.
You haven't written a word about the graphic. Something to say, asshole?