Re: Cosmic Background Radiation


"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox T:net@xxxxxxxxxx> wrote in
message news:Pzseg.4713$AB3.695@xxxxxxxxxxxxx
Dear jmetolius:

"jmetolius" <jmetolius@xxxxxxxxx> wrote in message
news:1148864950.517835.36650@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
David, you're not addressing the issue. What you
wrote was wrong.

No one is placing the Earth in a special position,

You keep referencing distance. DIstance is not an issue.

and it does matter
what the acceleration proifile of an astronaut leaving
Earth has, (as to deciding how he views the CMBR).

Right. I never said it did.

?

In the case of the thought experiment, the average
acceleration is determined by the fact that the
astronaut must land on the planet that has a
(distance x H) velocity away fom Earth, so it is
obvious

You still don't get it. The planet can be coincident with Earth, have
zero distance (briefly), and still be at rest wrt the *total* CMBR.

David, I'm not entirely following your point here
but perhaps what is said later may clarify.

that the final velocity of the astronaut relative to
Earth at the time of any observation must be
(distance x H).

We'll let George handle that one, since you don't like hearing it from me.

OK, what I hear the OP saying is that if you start
at _any_ point in the universe and in a state of
motion such that you observe no anisotropy in the
CMBR and move some distance in any direction then
your motion must change by (distance x H) relative
to your initial velocity in order to again observe
no anisotropy. That appears correct to me so I'm
not clear what you are objecting to.

This (distance x H) velocity relative to Earth
is the determining factor as to whether any
observer anywhere in the Universe
will see CMBR as uniform.

Which places the Earth in a special position, since all distance
measurements (from which you obtain a velocity) are wrt the Earth.

AFAICS, the OP's choice of Earth as a starting point
was completely arbitrary, he is not implying it is
specuial at all.

This determining factor of whether an observer
will see a uniform CMBR does not HAVE to be
a (distance x H) velocity away from EARTH.

And the velocity is not a function of distance from Earth either.

The change os speed needed to maintain an isotropic
CMBR is a function of the distance moved from the
(arbitrary) starting point.

It can be a (distance x H) velocity away from
ANY point in space where an observer at that
point sees the background as also uniform .

This is really going to hurt when you finally get it.

I'll leave you and George to it.

The OP and I don't disagree unless I am misreading
his posts.

In the meantime consider how many (distance x H) there are coincident at
any point, from all possible sources. All the velocities can't be the
same, yet you seek to make them all the same.

At any location in the universe, there is only one
velocity in which the CMBR has zero dipole moment.

George


.

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