Re: Relativistic aberration

Den 02.08.2024 23:17, skrev Richard Hachel:

 We will now move on to numerical applications.

Yes, Let's do that!

 We have an observer on Earth who has just noticed the explosion of a supernova in the sky and who sends its coordinates to Paul B Andersen. Paul B Andersen takes out his telescope, and at the indicated location, finds the supernova. He notes E = (12000, 9000.0, -15000.0).

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

He then asks for spatial confirmation from the commander of a rocket
which crosses the Earth on the x-axis, but since he knows the laws of aberration of the position of the stars, Paul, thanks to an ultra-fast computer, sends the correct coordinates in the rocket's reference frame.

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)

So he sends: position in x':
x'=(x+Vo.To)/sqrt(1-Vo²/c²) (attention negative To)
Or again, which is the same thing:
x'=[x+sqrt(x²+y²+z²)Vo/c]/sqrt(1-Vo²/c²)
 <http://news2.nemoweb.net/jntp?aGJtGFi-pcZdeYKlbLrP7fJkFGw@jntp/Data.Media:1>
 x'=40000
Or for the rocket: E'=(40000,9000,0,-41000,0)

Interesting. The rocket and Earth are colocated and receive
the same light, but on Earth the star is seen as it goes supernova,
while the rocket will see the star 6000 years before it goes supernova.
Richard, you live in a very weird world! :-D

We will note that always, always, always, for two joint observers, t=t'=0.
Which translated means "they see the same universe present" but not with the same spatial coordinates.
 R.H.

--
Paul
https://paulba.no/