Re: Einstein cheated with his fraudulent derivation of Lorentz transforms

On Wed, 5 Feb 2025 20:55:56 +0000, Paul B. Andersen wrote:

Den 04.02.2025 20:10, skrev rhertz:

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< big snip - let's start at the beginning >
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I'll try to explain systematically the steps that he did in the first
part of §3:
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1) RIGHT:  He used the equation 1/2 (τ₀ + τ₂) = τ₁ for times registered
in the moving frame k. No problem here, as long τ is the time MEASURED
in the moving frame.
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2) WRONG: Immediately after, he WRONGFULLY mixed perceptions of time (as
derived in §2 for the STATIONARY FRAME K, FORCING a relationship between
t, τ and x' = x - vt, which is A FALLACIOUS ASSERTION. The he came with
a modified equation from 1) as follows:

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In §1 light is emitted from a point A at the time tA,
reflected from a mirror at point B at the time tB,
and is back at the point A at the time tA'.
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The light uses the same time forth and back, so
   tB - tA = tA' - tB
or
   (tA + tA')/2 = tB
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Note that all the times are in the same stationary system.
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In §3 the light is emitted at an event with the coordinates
(0,0,0,t) in K, it is reflected at an event with coordinates
(x',0,0,tB) in K, and is back at the event with the coordinates
(0,0,0,t') in K.
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Since the the point x' is moving with the speed v in K,
tB = t + x'/(c-v)   (the light must catch up with x')
t' = t + x'/(c-v) + x'/(c+v)    (t + time forth + time back)
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The point is to find the function: τ(x,y,z,t)
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Now we have three instances of this function with three different
sets of coordinates:
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  τ₀(0,0,0,t),   τ₁(x',0,0,tB) and  τ₂(0,0,0,t')
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Knowing that (τ₀ + τ₂)/2 = τ₁ gives the equation:
>

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1/2 [τ(0,0,0,t) + τ(0,0,0,t + x'/(c - v) + x'/(c + v)] = τ[(x',0,0,t +
x'/(c-v)]
Simplyfing it by eliminating y=0 and z=0, this equation changes (with no
errors) to
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1/2 [τ(0,t) + τ(0,t + x'/(c - v) + x'/(c + v)] = τ[(x',t + x'/(c-v)]

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OK.
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Using the following relationships (developed in §2 for PERCEPTION in
frame K)

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What is  "PERCEPTION in frame K" ?
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Δτ₁ = x´/(c – v)

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Here is your giant blunder!
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x'/(c - v) is a time in the the stationary system K.

Innoble animal! I used Δτ₁=  x´/(c–v) and  Δτ₂ = x´/(c–v)+ x´/(c+v) ONLY
TO SIMPLIFY THE EQUATION
1/2 [τ(0,t) + τ(0,t + x'/(c-v) + x'/(c+v)] = τ[(x',t + x'/(c-v)]
before applying Taylor. If you didn't like it, you should have changed
them to Δt₁  and  Δt₂. This would have changed NOTHING in the following
analysis!
Now, simplified, it has the form
1/2 [τ(0,t) + τ(0,t + Δt₂)] = τ[(x',t + Δt₁]
Happy now?

So you could have written: ΔtB = x'/(c - v)
Setting ΔtB =  Δτ₁ is equivalent to setting τ = t
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So we have τ(x,y,z,t) = t or   τ = t
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This and the rest is NONSENSE.
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Δτ₂ = x´/(c – v)+ x´/(c + v)

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Good grief!
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<snip>

As you said: "The point is to find the function: τ(x,y,z,t)"
What is missing to add is that such function τ represents the time in
the moving frame k, which has to verify the equation
1/2 (τ₀ + τ₂) = τ₁
Now, you have TWO TIME VALUES to consider for each instance 0,1 and 2:
Time τ₀: Ray of light depart from ξ(0)=0, in the moving frame k. Set
τ₀=0.
Time τ₁: Ray of light reaches ξ(x')=x', in the moving frame k.
τ₁ = [ξ(x')-ξ(0)]/c = x'/c
Time τ₂: Reflected ray of light reaches ξ(0)=0, in the moving frame k.
τ₂ = τ₁ + [ξ(x')-ξ(0)]/c = τ₁ + x'/c = 2 x'/c
These values, in the moving frame k, satisfy 1/2 (τ₀ + τ₂) = τ₁
Now, as PERCEIVED from the stationary frame K:
Time t₀: Ray of light OBSERVED to depart from ξ(0)=0, from the frame K.
Set t₀=0.
Time t₁: Ray of light OBSERVED reaching ξ(x')=x', from the frame K.
t₁ = [ξ(x')-ξ(0)]/(c-v) = x'/(c-v)
Time t₂: Reflected ray of light IS OBSERVED reaching ξ(0)=0, from the
frame K.
t₂=t₁ + [ξ(x')-ξ(0)]/(c+v) = t₁ + x'/(c+v) = x'/(c-v) + x'/(c+v) =
2cx'/(c²-v²)
These values, PERCEIVED FROM THE STATIONARY FRAME K, don't satisfy
1/2(t₀+t₂)=t₁
Einstein FORCED/FUDGED a function τ(x,y,z,t) by mixing results from the
moving frame k and the stationary frame K. This wrongful decision to
obtain Lorentz AT ANY COST is EXPOSED in the FRAUDULENT MANIPULATION of
Taylor expansion, in particular by forcing that x' be ZERO instead of
the infinitesimal dx' and by making dx' to appear, magically, to obtain
the equation:
1/2 [(1/(c - v) + 1/(c + v)] ∂τ/∂t = ∂τ/∂x' + 1/(c-v) ∂τ/∂t
There is no place for ∂τ/∂x', which was INSERTED in a fraudulent way.
Otherwise, the equation would have been  ∂τ/∂t = 0, meaning that  τ = t,
which is THE REAL RESULT.
You are becoming more and more stupid as you age. Retire and do
gardening.