Liste des Groupes | Revenir à p relativity |
Den 10.02.2025 00:18, skrev rhertz:Paul, I think that you are a lame and pathetic IMBECILE beyond>On Sat, 8 Feb 2025 19:54:58 +0000, Paul.B.Andersen wrote:
>>
I have now read your attachment, and I didn't panic.
I was however mildly shocked by the extent of your confusion.
>
What I found was this:
Δτ₁ = x'/(c-v)
Δτ₂ = x'/(c-v)+x'/(c+v)
τ₁ = τ₀ + Δτ₁
τ₂ = τ₀ + Δτ₂
>
I won't bother to yet again explain why this is nonsense,
because you are unable to learn.
>
>>>
You keep behaving as a relativistic cornered rat.
>
When Einstein wrote x'=x-vt, it's CLEAR that x' is a "point at rest" in
the moving frame k (he wrote so), and does the Galilean transform for a
point P(x,y,z,t) that exists only in K. It's crystal clear that x' IS
not a FIXED POINT in K but it's moving away from the origin of K at v
speed.
It is irrelevant that the x' is moving in K.
The coordinates x' is the position of the mirror in K
at the time of reflection.
>
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>
You wrote:
Δτ₁ = x'/(c-v)
Δτ₂ = x'/(c-v)+x'/(c+v)
τ₁ = τ₀ + Δτ₁
τ₂ = τ₀ + Δτ₂
>
If you really don't understand why this is ridiculous,
I will have to explain:
>
The function: τ₁ = τ[(x', x'/(c-v)] means that
the temporal coordinate of k: τ₁ is a function of the event
with the coordinates in K: x = x' and t = x'/(c-v).
>
The function: τ₂ = τ[(0, x'/(c-v) + x'/(c+v)] means that
the temporal coordinate of k: τ₂ is a function of the event
with the coordinates in K: x = 0 and t = x'/(c-v) + x'/(c+v).
>
But you write:
τ₁ = τ₀ + x'/(c-v)
τ₂ = τ₀ + x'/(c-v) + x'/(c+v)
>
That is;
If we have an event with coordinates (ξ,τ) in k
and (x,t) in K, you write τ = t.
>
How is it possible to fail to understand of ridiculous this is?
>
>>>
Analyze the attached figure that I've, and try to deny that he didn't
use Galilean transforms.
No figure can save your _giant_ blunder!
>
You have thoroughly demonstrated that you are not competent to
evaluate Einstein's derivation of the LT.
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