Re: Cosmic Background Radiation


N:dlzc D:aol T:com (dlzc) wrote:
Dear George Dishman:

"George Dishman" <george@xxxxxxxxxxxxxxxxx> wrote in message
news:e5f2pm$9cr$1@xxxxxxxxxxxxxxxxxxxxxxxx

"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.

?

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.

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.

*We* are at (distance x H) from a whole host of places, but *we*
do not see the CMBR as isotropic. So any "transferrence" of our
anisotropy to another locale, ends up with a similar anisotropy.

OK, I think it is a misunderstanding.

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.

Agreed. I misunderstood his intent/meaning. Isotropy
notwithstanding...

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.

An arbitrary starting point that the Earth is not located at at
two successive instants. We are moving, and not just by
expansion.

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.

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.

The inhabitants of this planet
will, of course, see themselves as almost 'at rest' relative to the
cosmic background radiation. They are almost entirely just moving with
the cosmic flow.

Again, "... almost 'at rest' ...", "... almost entirely just moving
...."

The astronaut may then have tea with an alien on that planet, who views
their backgrond radiation as (almost) not shfted at all.

And "... (almost) not shifted ..."

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.

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.

|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?

George

.

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