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From: hitlong@yahoo.com (gharnagel)
Newsgroups: sci.physics.relativity
Subject: Re: Oh my God!
Date: Sat, 28 Sep 2024 03:09:14 +0000
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On Sat, 28 Sep 2024 1:21:59 +0000, Richard Hachel wrote:
>
> Le 28/09/2024 à 03:10, hitlong@yahoo.com (gharnagel) a écrit :
> >
> > On Sat, 28 Sep 2024 0:23:18 +0000, Richard Hachel wrote:
> > >
> > > > > Vo=Vr/sqrt(1+Vr²/c²)
> > > > >
> > > > > If Vr---> ∞  then Vo=c
> > > >
> > > > It doesn't take a degree in mathematics to know that as Vr---> ∞,
> > > > Vo in your equation approaches zero, not c.
> > >
> > > In France, our experts in mathematics seem to say that
> > > Vo=Vr/sqrt(1+Vr²/c²) tends towards c.
> >
> > I can only see what I see, and I doubt you are being honest.
> >
> > Don't quote yourself as an "expert" in mathematics because you're
> > not.  Neither are you an expert in relativity.
>
>  Pffffff...
>
>  Well...
>
>  You say that the equation does not tend to c if Vr is infinite.
>
> Let's take Vr very large, example Vr=1000c
>
> Let's set Vo=Vr/sqrt(1+Vr²/c²)
>
> Vo=1000c/sqrt(1+1000²)
>
> Vo=(1000/1000.0005)c=0.9999995c
>
> And the higher you go, the more it tends to c.
>
> R.H.

I'm sorry, I didn't recognize the Vr in the numerator as
being the same as that in the denominator.

As I see now, the equation is actually quite similar to the
equation for tachyon momentum: p = mv/sqrt(v^2/c^2 - 1),
which follows from extending SR into the new domain of v > c.
So, it seems, that you are supporting tachyons while trying
to deny them :-)