Re: Relativistic aberration

Den 03.08.2024 23:40, skrev Richard Hachel:

Le 03/08/2024 à 22:28, "Paul.B.Andersen" a écrit :

 E(x,y,z,t) = (12000 ly, 9000 ly, 0 ly, -15000.0 y)
t is when the light left the telescope, now = 0 y
 Let us assume that, in accordance with convention,
the x-axis is pointing toward the vernal point Epoch J200,
and the x-y plane is the ecliptic plane. The z-axis will
then point toward the ecliptic north pole.
 The RA is then arctan(9000/12000) = 37⁰
The DEC = 0⁰
The distance d =  √(x² + y² + z²) = 15000 ly
 Since the rocket is moving along the x-axis'
the angle velocity - (direction to star) = 37⁰,
the RA in the rocket frame will due to aberration be 12.7⁰
the DEC = 0.
Since the rocket and the Earth are colocated at the time of reception,
they will obviously receive the same light which was emitted from
the star 15000 years ago.
That means that the distance in the Rocket frame must be 15000 ly.
>
Simple geometry will give:
x' = 15000⋅cos(12.7⁰) ly = 14633 ly
y' = 15000⋅sin(12.7⁰) ly =  3297 ly
z' = 0 ly
t' = -15000/c year = -15000 year
>
E '= (14633 ly, 3297 ly, 0 ly, -15000 y)

 ? ? ?
<http://news2.nemoweb.net/jntp?n6nnyNLQR1tXDC_uShX3k3bxE5g@jntp/Data.Media:1>
  But what are you talking about? ? ?
 You're talking nonsense!!!
 Your thing IS nonsense!
 How do you want the object to be at the same distance in both frames of reference? ? ?

This is about what the observers SEE in their telescopes.
They are co-located when they receive the light from the supernova,
so they obviously both see the star as it goes supernova.
Since earth observers have _measured_ (via parallax ?) the distance
to the star to be 15000 ly, and the speed of light is c in the
Earth frame, the earth observer can deduce that the light must
have left the star 15000 years ago.
The rocket observer will see in his telescope the star
as it goes supernova in the direction RA =  12.7⁰ and DEC = 0⁰.
He has no possibility of knowing the distance to the star or
when it went supernova, but the Earth observer know that
since they were co-located at the reception, they both have
received the light that was emitted 15000 years ago.
That means that since the speed of light is c in the rocket frame,
the distance in the rocket frame must be 15000 ly.
But the rocket observer can't know this without being told.
But we can calculate that the coordinates of the star as
it went supernova must have been:
E '= (x' = 14633 ly, y' = 3297 ly, z' 0 ly, t' = -15000 y)
Case closed.

 All this is sad to cry and you show EXACTLY what I have been saying for years, namely that physicists do not understand anything at all about the theory of relativity, and use mathematics in a completely ridiculous and anarchic way!
 But this is nonsense, Paul!!!
 You practice a stupid rotation, and we can clearly see all the stupidity of the Minkowski space-time block, stupid and abstract.
 PAUL, PAUL, PAUL, I beg you to understand something!
 There is NO rotation, there is NO change in y, nor change in z.
 Poincaré was right and his geometry is magnificent, and we must take up its numerical applications again.
 y'=y=9ly
z'=z=0ly
 This is dramatically simple.
 x=12 ly
x'=40 ly
 To=15 ly
To'=41ly
 t'=t=0
 There is a relativistic translation on the x-axis.
 NOTHING MORE.
 This produces a ROTATION OF THE AXIS OF VIEW, but NOT of the star!!!
 But damn it, if you don't understand that, you who are one of the best posters of relativity, we are in a terrible mess, and we will never progress.
 R.H.
 <http://news2.nemoweb.net/?DataID=n6nnyNLQR1tXDC_uShX3k3bxE5g@jntp>

You and Archimedes Plutonium have a lot in common.
Neither of you qualify as a crank.
Too nonsensical.
:-D
--
Paul
https://paulba.no/