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Re: Re: How Einstein missed his opportunity to derive Lorentz in Point §3.

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Paul B. Andersen wrote:
> 
> Den 09.02.2025 00:40, skrev rhertz:
> > I was reading again the English version (1923) of his 1905 paper, and
> > got interested in a footnote from the editor:
> >
> > QUOTE:
> > 5The equations of the Lorentz transformation may be more simply deduced
> > directly from the condition that in virtue of those equations the
> > relation
> > x² + y² + z²  = c² t² shall have as its consequence the second relation
> > ξ² + η² + ζ²  = c²τ².
> > END QUOTE
> 
> Right.
> 
> A rather obvious consequence of the postulates of SR is that
> the speed of light is invariant c, that is c in all inertial
> frames of reference.
> 
> That is, the light will use the time t = L/c to travel the distance
> L = √(x²+y+z²) =>  c²t² = x²+y²+z² and thus c²τ² = ξ²+η²+ζ²
> 
> When Einstein in §3 of "Electrodynamics" writes
>    x² + y² + z² = c²t²
> and
>    ξ² + η² + ζ² = c²τ²
> it is a mathematical expression of
> "the speed of light in K is c" and
> "the speed of light in k is c"
> 
> You can find the derivation of the LT from this starting
> point in several textbooks, and on the net:
> 
> https://testbook.com/physics/derivation-of-lorentz-transformation
> https://www.vedantu.com/physics/derivation-of-lorentz-transformation
> https://people.iith.ac.in/kdmakwana/ep2017/lecture2.pdf  (see eqn. (33)
> https://tinyurl.com/26f3xgrp
> https://dacollege.org/uploads/stdmat/Physics-derivation-of-Lorentz-Transformation-Equation-dem4.pdf
> 
> And there are several other ways to derive the Lorentz transform,
> common for them all are that they start with the postulates
> of SR, usually in the form "speed of light is invariant c"-
> 
> February 7, Paul B. Andersen wrote
> | It is a _fact_ that the Lorentz transform follows from
> | the postulates of SR.
> |
> | This is proven many times by several physicists.
> | Einstein was the first to do it, but he did it in a rather
> | cumbersome way which, as you have demonstrated, may not
> | be simple to understand.
> | In the more than a century since Einstein did it, it has
> | been done many times in much more elegant ways which are
> | simpler to understand.
> 
> > ------------------------------------------------------------------------
> >
> > This is the derivation of Lorentz transforms using the above concepts:
> >
> >
> > ***********************************************
> >
> > He could have derived Lorentz transforms by simply postulating that as
> >
> > c² t² - x² - y² - z²  = 0
> >
> > and
> >
> > c² τ² - ξ² - η² - ζ²  = 0
> >
> > then
> >
> > c² τ² - ξ² - η² - ζ²  = c² t² - x² - y² - z²
> >
> > As y = η  and  z = ζ, this equation reduces to c² τ² - ξ²  = c² t² - x²
> 
> Perfect so far!
> 
> >
> > Assuming that ξ and x differs in a factor β from Galilean transform,
> > where β = 1, the new transform can be written as
> >
> > ξ  = β (x - vt)
> 
> This has nothing to do with the Galilean transform.
> You know the transform can be written on the form above.
> 
> Would you say:
> "Assuming that τ and t differs in a factor β from Galilean transform,
>   where β = 1, the new transform can be written as  τ = βt " ?
> 
> Obviously not, because you know the transform has nothing to do
> with the Galilean transform and has a different form.
> 
> It is however true that you by reasoning can find that
> the transform must be of the form  ξ  = β (x - vt).
> 
> But you have made no reasoning, you are using your knowledge
> of what the form of the transform should be to derive the transform.
> 
> But never mind, let it pass.
> It is true that the transform is of the form ξ = β (x - vt)
> 
> >
> > and
> >
> > x  = β (ξ + v τ)
> >
> > then
> >
> > x  = β² (x - vt) + β v τ
> >
> > β v τ = x (1 - β²) + β² v t
> >
> > τ = x (1 - β²)/βv + β  t
> >
> > τ = β [(1/β² - 1) x/v +  t]
> >
> > Replacing ξ and τ in ξ² = c² τ², it follows that
> >
> > β² (x - vt)² = c² β² [x (1/β² - 1) x/v +  t]²
> >
> > Factoring the above equation with a little help of algebra, it's
> > obtained
> >
> > [β² - c²β²/v² (1/β² - 1)] x² - 2 β² [v + c²/v (1/β² - 1)] xt = β² (c² -
> > v²) t²
> >
> > To verify
> >
> > x² + y² + z²  = c² t²
> >
> > It's required that, in the previous equation,
> >
> > [β² - c²β²/v² (1/β² - 1)] = 1
> >
> > [v + c²/v (1/β² - 1)] = 0
> >
> > β² (c² - v²) = c²
> >
> >  From the last equation,
> >
> > β²  = c²/(c² - v²)
> >
> > or
> >
> > β  = 1/√(1 - v²/c²) , which is the modern factor Gamma (γ).
> >
> > With this result of β  (γ), the transforms are
> >
> > ξ  = β (x - vt) = (x - vt)/√(1 - v²/c²)
> >
> > τ = β [(1/β² - 1) x/v +  t] = β (t - vx/c²) = (t - vx/c²)/√(1 - v²/c²)
> >


What is all dat stuff up there? i don't see 'any' numbers! It cannot be math if it ain't got any numbers...


Any why so many equal signs???? and so many half moons )))))(((((())))) 



come on, give me at least ONE number! wait, is dat a ONE i see??? o thought it was another letter L....




-- 
The Starmaker -- To question the unquestionable, ask the unaskable,
to think the unthinkable, mention the unmentionable, say the unsayable, 
and challenge the unchallengeable.

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How Einstein missed his opportunity to derive Lorentz  in Point §3. hertz778@gmail.com (rhertz) - 2025-02-08 23:40 +0000
  Re: How Einstein missed his opportunity to derive Lorentz  in Point §3. hertz778@gmail.com (rhertz) - 2025-02-09 18:40 +0000
    Re: How Einstein missed his opportunity to derive Lorentz  in Point §3. hertz778@gmail.com (rhertz) - 2025-02-10 21:50 +0000
  Re: How Einstein missed his opportunity to derive Lorentz in Point §3. "Paul B. Andersen" <relativity@paulba.no> - 2025-02-10 21:39 +0100
    Re: Re: How Einstein missed his opportunity to derive Lorentz in Point §3. The Starmaker <starmaker@ix.netcom.com> - 2025-02-10 15:10 -0800

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