Re: Cosmic Background Radiation

Dear George Dishman:

"George Dishman" <george@xxxxxxxxxxxxxxxxx> wrote in message
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N:dlzc D:aol T:com (dlzc) wrote:
Dear George Dishman:

"George Dishman" <george@xxxxxxxxxxxxxxxxx> wrote in message
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"N:dlzc D:aol T:com (dlzc)" <N: dlzc1 D:cox
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Dear jmetolius:

"jmetolius" <jmetolius@xxxxxxxxx> wrote in message
news:1148864950.517835.36650@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
....
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.

?

I took his statement to mean "how the astronaut
views the CMBR after he arrives on the planet".
The acceleration profile only controls the time it
takes to arrive at the planet, and yes the CMBR
will look differently on the trip, at different velocities.

The acceleration profile also affects his speed when
he reaches the planet which is very important, but
why I queried your response was simply that he said
"and it does matter" and you replied "Right. I never
said it did." which does not compute.

Not important, George. What he wrote is not what I "heard".
What he wrote is that the acceleration profile is important.
What I heard is that it is not important, as long as he winds up
on the planet at rest.

....
I'll leave you and George to it.

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

He believes he is agreeing with you. He has
not obtained isotropy with Earth as a starting
point... unless 300 km/sec is below his radar.

Yes, it is. Refer back to the original post:

jmetolius wrote:
If we imagine a planet in a particular far away
galaxy, where this galaxy has a considerable
relative recession to ourselves, and where
this relative velocity is almost entirely
^^^^^^
attributable to the cosmic expansion ( let's
say 20,000 km/sec).

The word "almost" indicates to me that he is
ignoring the anisotropy due to the planet's
motion, he is treating it as negligible
compered to the 20000km/s of the Hubble flow.

OK. I read it, but did not catch the copious use of "almost".

....
All of these uses of "almost" are to allow for the
small proper motion effects, it is the bulk flow that
he was puzzled about.

Agreed. Again, I am hearing imparied.

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.

No, this is clearly not true. There are an "infinite"
number of places in the Universe that say you
are wrong. They all obtain different *velocities*
for ours.

They all have different velocities relative to us at
which the CMBR is isotropic, but what I said is
still true, at _each_ of those infinite number of
points, there is only one velocity which makes
the CMBR isotropic.

But I'll be d*mned if I can
figure out a way to say it more clearly.

The trick is to say it in a way that shows that
what I said is "clearly not true", because I
can't figure out what you are objecting to.

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

A zero dipole moment occurs when all points equidistant in all
directions are receding at the same speed.

|v| = (distance x H) - |v_aniso| * cos(theta)cos(phi)
... might come closer with the value of zero for
theta and phi cleverly aligned with our own axis of
anisotropy.

Only in an Earth-centred coordinate system as
opposed to a co-moving system, but why does
that make what I said wrong?

Both the origin *and* "location of interest" (LOI) have to be
specified in order to obtain a *single value* for velocity at the
LOI.

I'm straining at gnats again, as you have pointed out. He was
interested in "almost", even though his formula was treated as
exact. Far enough from here, I guess it is "close enough".

David A. Smith


.

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