Relativistic segment in accelerated frame
From
Richard Hachel@21:1/5 to
All on Tue Dec 19 23:16:13 2023
We are in the Tau Ceti traveler's problem.
For example, we are in the last segment of one light year (there are
twelve).
We then ask the question:
“What will be the observable time required to cross this segment?”
Very strangely, the physicists, who have the real equation, since they
pose:
To=(x/c).sqrt(1+2c²/ax) know perfectly the observable time to go from
the earth to 11 al, and the time it takes to go from the earth to 12 al.
A fantastic confusion will then occur in their minds.
And they're going to say to themselves: "So we're going to do the subtraction, and it will be very simple and very practical."
Except that the equation above is only valid for a standing start.
To1 therefore concerns the journey from 0 to 11, and To2 the journey
from 0 to 12.
We cannot do (even if it seems so obvious) ΔTo=To2-To1
Carrots and turnips as I kill myself explaining to Paul B.Andersen, an excellent contributor, but who has not yet had the courage to get to the
bottom of things with me.
Scientific catastrophe is assured, and everyone is responsible for it
because everyone CAN only make it their own.
Everyone screams with laughter and starts singing as if mass had been
said, and Hachel was the perfect moron:
We will set ΔTo=To2-To1
Nope.
That's not how it works.
The good Lord (or the Big Bang for crazy scientists), he didn't create
his universe like that (which we now know is much bigger than 13al, and probably infinite).
Certainly, for clean times, that's how it works. It is clear that ΔTr=Tr2-Tr1 since the proper times
are measured by the same clock.
But not bad times. We forget that improper tenses are illusions, and
that illusions do not add up like that. A bit like if we wanted to add the square of 5 plus the square of 4 therefore equals the square of 9.
It doesn't work like that, my friends.
Now, let's give the real equation to calculate the observable time
between two segments (while knowing that it is abstract as Richard Verret rightly says on fsp and who, on this point, does not lie, because it IS abstract or even useless to calculate) .
In fact, there are two which give the same results, but I repeat, these equations are useless in RR, since the observable times have little
importance in this case: only the proper times count, and the real speeds,
much more useful.
We therefore have ΔTo=(Δx/c).sqrt(1+2c²/(a.[sqrt(x2)-sqrt(x1)]²)
but also if Vri=input speed=entry speed in the segment;
ΔTo= ΔTr.sqrt[1+(Vri+(1/2)a.ΔTr)²/c²]
Both equations give the same result. ΔTo =1.020 year
Note that ΔTo =1.020 year for this segment in both equations.
And not ΔTo =0.0029 year as physicists think.
Which gives an observable instantaneous speed of around 0.980c when
crossing Tau Ceti (and not Vo=0.997c as they say).
Note again that for a long time I gave the formula for finding the
observable instantaneous speed
in an accelerated frame of reference, and that this is not at all what
the physicists say either.
The equation is very simple: Vo/c=[1+c²/2ax]^(-1/2)
We then see their double error, one seeming to correct the other, but
still giving two false results.
Error on the mobile's own time, much too short.
Error on the instantaneous observable speed of the mobile, much too
high.
Thank you for your attention.
R.H.
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