Re: Einstein cheated with his fraudulent derivation of Lorentz transforms

Am Mittwoch000005, 05.02.2025 um 23:51 schrieb rhertz:
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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
>
On the first page (page 5) Einstein defines the coordinate systems.
The "stationary system"  K(t,x,y,z) coordinates are Latin letters
The "moving system"      k(τ,ξ,η,ζ) coordinates are Greek letters
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So the Galilean transform is: ξ = x - vt

Sure, but that wasn't what Einstein wrote.

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

correct (without the insult, of course).

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

There wasn't a function named ξ in Einstein's text.
Because x' has a Latin letter 'x' in it, it was meant as a coordinate belonging to system K.
Therefore the equation ξ = x - vt was meant, while x' = x - vt was written.
x' should be capable of reflecting a light beam, hence a mirror should be placed there.
This would fit to the term 'If we place x' = x - vt...' but not to the equation.
That's why it is hard to say, what actually was meant with x'.

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

But that " τ = t' = t " wasn't, what Einstein had in mind.

 

You will _not_ find this anywhere in Einstein's paper.
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The x' is a point in the stationary system K, it is NOT
a coordinate in the moving system k.
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So x' = x - vt is a _moving_ point in K.
And since x' is moving with the speed v, it will be stationary
relative to k.

 relative to k.

No, the point x' MUST be stationary in K !!
If it is stationary in k, if would not move in respect to the emitter in the center of k and we had no use for the velocity v.
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TH