Re: Einstein cheated with his fraudulent derivation of Lorentz transforms

Den 05.02.2025 23:51, skrev rhertz:

On Wed, 5 Feb 2025 12:29:56 +0000, Paul.B.Andersen wrote: 

Den 04.02.2025 22:54, skrev rhertz:

>
Now, considering that he started with Galilean transform x' = x-vt
>

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!!"

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

You used exactly those words back in 2020,
and as quoted above, you repeated that x' = x - vt
is the Galilean transform in §3.

>
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

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.

>
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

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

>
So the Galilean transform is: ξ = x - vt

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

And x' is a coordinate in k ? :-D

ξ(x') = x' in the moving frame. I attached a graphic to clarify this
but, with your "dog vision" you missed it.

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

 x' = x - vt is the well known Galilean transform, along with τ = t' = t.

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?
This is fun, isn't it?  :-D
There is only one meaning of x', right?
https://paulba.no/pdf/Dilbert.pdf
--
Paul
https://paulba.no/