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Den 03.08.2024 23:40, skrev Richard Hachel:Again common sense prejudices, refuted by yourLe 03/08/2024 à 22:28, "Paul.B.Andersen" a écrit :This is about what the observers SEE in their telescopes.>>
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
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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.
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The RA is then arctan(9000/12000) = 37⁰
The DEC = 0⁰
The distance d = √(x² + y² + z²) = 15000 ly
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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.
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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
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E '= (14633 ly, 3297 ly, 0 ly, -15000 y)
? ? ?
<http://news2.nemoweb.net/jntp?n6nnyNLQR1tXDC_uShX3k3bxE5g@jntp/Data.Media:1>
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But what are you talking about? ? ?
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You're talking nonsense!!!
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Your thing IS nonsense!
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How do you want the object to be at the same distance in both frames of reference? ? ?
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.
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