Re: Relativistic aberration

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

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? ? ?
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.
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