Den 01.08.2024 23:03, skrev Richard Hachel:
Le 01/08/2024 à 22:05, "Paul.B.Andersen" a écrit :
Den 01.08.2024 13:39, skrev Richard Hachel:
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The event that occurred at the supernova located at 15,000 ly has just been recorded by the Earth observer.
He notes E = (x, y, z, To, t)
Let x = 12000, y = 9000, z = 0, To = -15000, t = 0.
This is how I write in Hachel notation.
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A very stupid notation!
Giving the position of a star in Cartesian coordinates
would obviously include the distance to the star, and then
in addition giving the distance to the star in light years
or years is redundant.
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Astronomers give the position to the star as an angle
in a spherical coordinate system. The angle is given as
observed from the Sun.
The distance is given in parsecs (or ly) and will be determined
by parallax or other method.
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The Earth is orbiting the Sun, which will mean that
an observer on the Earth will see the direction to
the star varying during the year. The star will appear
to move along an ellipse with major axis 40.98 arcsecs.
The direction to the star observed from the Sun will be in
the centre of the ellipse.
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https://paulba.no/pdf/Stellar_aberration.pdf
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An observer crosses the Earth at this moment at a few thousand kilometers, his speed is 0.8c (his rocket has very advanced technology, and he is heading on the Ox axis).
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Along the x-axis?
So his velocity is 37⁰ from the direction to the star.
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It is clear, at the risk of being absurd, and of not having understood anything at all about the theory of relativity that the events will be simultaneous for the two observers who will be conjoined at this moment.
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Their watch will mark t = 0 (since they are triggered at this moment).
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We want to know what the coordinates of the supernova event will be in the new reference frame of the rocket, and what the commander will write on his on-board carent.
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Due to aberration he will see the supernova in a very
different direction than the person on Earth.
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https://paulba.no/pdf/AberrationDoppler.pdf
equation (19)
The rocket will see the supernova
an angle 12.7⁰ from the x-axis.
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Let E'=(x,y,z,To,0) since not yet knowing the Poincaré-Lorentz transformations, we can only guess the other four coordinates.
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But you, Paul, who knows the TL and who knows a little about SR, can you give the five new coordinates in RH mode?
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The aberration above is from the LT.
It is obviously meaningless to transform the distance,
there is no way the rocket man can measure it.
It is measured by other means.
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Don't be afraid to make mistakes, we are here to learn and understand, and even if you make a mistake, it will allow us to go further.
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If you don't make a mistake, it's because you have already understood this point perfectly.
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Did you have a point?
We don't care.
I give this notation because it is the best.
It isn't so different from what astronomers use.
The position of the star is given as two angles and a distance.
The angles are Right ascension (RA) and Declination(DEC).
The former is equivalent to longitude and the latter to latitude.
The distance is given in Parsecs or light years.
It is of course a trivial matter to convert these three coordinates
to a Cartesian frame of reference.
Your system have the transit time of the light from the star
as a fourth coordinate, which is redundant because it is given
by the distance.
I didn't fail to notice that your √(x² + y² + z²) = -To⋅c.
So you could remove the To from your system.
When it comes to the time of the observation t, it would be
very inconvenient if the stars in the star catalogues were
observed at different times given in the catalogue, so the
data are given as they would be at the same time.
The current standard is Epoch J2000 (January 1, 2000)
That means that if a star is observed at another time,
the data must be calculated to what they were at Epoch J2000.
(The angular and radial velocity will normally be known).
Now to the real reason why this system is better than yours.
When you know the RA and DEC of the star, you know where
to point the telescope! (Corrected for stellar aberration and parallax.)
You can now buy amateur telescopes where you can enter the RA and DEC
of an astronomical object, and the computer will know where to point
the telescope on a rotating Earth, and even track the object.
You do not have to know the distance.
And when you observe a star, the direction of your telescope
give you the RA and DEC.
Afterwards, everyone can take their own.
I note a simple event E at a time t=0.
At this time, a rocket crosses me.
We call this a joint event.
The two observers are therefore at the place and time A=(0,0,0,0,0).
It is very simple.
x,y,z, are the metric coordinates.
To is the time when the phenomenon occurred in referential measurement.
No. To is a distance.
The distance to the star is usually measured by parallax.
There is no other way to determine To than to set To = -d/c
That you change the unit from ly to y doesn't change the fact
that -To⋅c = √(x² + y² + z²). To is redundant.
t is the time when the phenomenon actually, and live, occurred (which is a bit disconcerting, I admit, if we refuse universal anisochrony). It is obviously, if we understand correctly, the time at which the phenomenon is perceived (live-direct at Hachel).
This notation, I repeat, seems to me to be the best, although it requires a compromise to accept it.
If I write x = 12000, y = 9000, z = 0, To = -15000, t = 0. I have all the necessary coordinates for the observer in the rocket.
The transformations necessary to find the five coordinates for the guy in the rocket, it's up to you to give them to me. It's very easy.
To transform the coordinates to the rocket frame can only mean
that the observer will see the star in the direction
to the point x', y', z' where √(x² + y² + z²) = -To⋅c
I will repeat what I wrote in my previously post, which you obviously
haven't read:
If the rocket is moving along the x-axis,
its velocity is 37⁰ from the direction to the star.
Due to aberration the rocket observer will see the supernova
in a very different direction than the person on Earth.
https://paulba.no/pdf/AberrationDoppler.pdfequation (19)
The rocket will see the supernova
an angle 12.7⁰ from the x-axis.
(From Earth the supernova is seen in the direction 37⁰ from the x-axis)
Here is how an observer moving at 0.8c will observe the universe:
The stars in front of him will be much closer to each other and
much brighter.
This phenomenon is called "beaming".
https://paulba.no/div/Beaming.pdfThe following answer most of your questions:
https://math.ucr.edu/home/baez/physics/Relativity/SR/Spaceship/spaceship.htmlI would like you to give them to me to show me (even if you don't believe it) that you have perfectly understood how I reason, and what I'm talking about. The first three coordinates are in ly, it's a metric unit. The other two in years, these are two time units.
Be careful, I gave To = -15000 years.
This means that in reference time, this happened already 15000 years ago.
Why are you stating the bleeding obvious? :-D
I gave t = 0. This is the time coordinate of the image reception (in live reception, this is what is a bit strange to understand, and why To is different from t in Hachel due to the geometry used).
There you go, do a little test and try to give me E'=(x',y',z',To',t').
Don't be afraid to make a mistake, it doesn't matter if you make a mistake.
I just want to know if you understand my notation, and my relativistic geometry because it is the basis of the reasoning.
R.H.
I will not repeat what I have written yet another time,
so if you don't read it (you seem never to do) don't respond!
-- Paulhttps://paulba.no/
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