Re: The Physics Behind 'Contractions'.

On Fri, 02 Jan 2009 21:24:41 +0100, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxx> wrote:

Dr. Henri Wilson wrote:
On Thu, 01 Jan 2009 23:00:04 +0100, "Paul B. Andersen"
<paul.b.andersen@xxxxxxxxxxxxxxx> wrote:

So in the context of this:
http://home.c2i.net/pb_andersen/pdf/Stellar_aberration.pdf
your statement:
<<
This is totally inadequate. You haven't stressed the importance of the
telescope's orientation when the readings are taken.
>>
is indeed a joke.

What you write in the following is utterly irrelevant to the paper
you commented, but I will comment on it anyway.

I see that I have misunderstood your example below.
So I will snip my previous comments and write new ones.

For simplicity, let's assume that radius of the orbit is 1AU.
(You seem to have made that assumption.)

Yes

Let me explain again..... read carefully.
If a space telescope is sent into a circular ecliptic orbit and is accurately
aligned with a star that lies right on the ecliptic polar axis and if that
telescope spins on a similarly aligned axis once per its orbit period, the
image of that star will remain exactly in the viewing centre.

But to keep the star exactly in the viewing centre, the pointing of
the telescope must be moved along a circle synchronous with it's
orbital motion, the radius of this circle must be v/c radians where
v = 30km/s is the orbital speed of the telescope.
v/c = 10^-4 radians = 20.5".

If I understand you right (you haven't explained it well), you think
this could be achieved by letting the telescope spin (once per year!)
around an axis which is 20.5" off its optical axis.

Possible in principle, hardly in practice. And active steering would be
necessary, even if only tiny adjustments now and then would be needed.
In any case, it is doable.

It is no more difficult that steering the HST.

If all the other stars in view happened to be located at exacty the same
distance from Earth as the central star then they will appear to rotate around
the central star in parallel circles, the radius of which is just their
distance from the central star. In other words, aberration is the same for all
objects and doesn't affect their relative motions.

Sure. But only if the telescope is steered as explained above.

However, since they lie at vastly different distances, the sizes of those
circles will vary due to parallax....
and so will the positions of the CENTRES of those circles.
The offset of those centres can be used to estimate distances, using parallax.

The radii of the circles may be at least many minutes of arc, and their
centres offset by less than 1" (mostly considerably less).
It would be impossible to measure precisely!

Probably...that's why I repeated the experiment below but with the telescope
NOT rotating at all. See my 'flywheel experiment' in my other post to Jerry.

If instead the same telescope is pointed exactly parallel to the ecliptic polar
axis,

You mean that it is pointed such that an infinitely distant star at
the ecliptic pole would always be at the centre.

that's it

then a star right on that axis will appear to move in a circle whose
radius subtends an angle 1AU/D.

Right.

The whole fields of view will rotate and the
other stars in the window will also follow circular paths (in the rotating
frame) with the same kind of radii as above (1AU/Dx).

Right.
And since all the parallax circles will have the same phase,
the stars will follow circular paths where the centres
is offset by the parallax.

Impossible to measure!

Not so. At 10 LYs the angle is about 0.3"

This is basically how parallax is measured.

If the same vertically pointing telescope DOES NOT SPIN as it orbits, then the
whole fields DOES NOT rotate around the window either. Rather it 'wobbles' as a
whole due to aberration with each star moving in an additional small circle
(subtending 1AU/Dx) due entirely to parallax. CMIIW

You mean if the telescope doesn't rotate and it's axis always is
pointing in the same direction (like a gyroscope would do).

Yes.

Since the parallax always is 90 degrees out of phase with the aberration
(as you so correctly pointed out), all the stars are moving in circles
with radius _very_ slightly greater than 20.5".

Look at this figure :
p
<--|
|
|a
|
|
|
C

Spin it around C.
The star will appear to be at the arrowhead.
p = parallax << 1", a = aberration = 20.5"
If p is as big as 1" (bigger than any star), the radius = 20.52"

The difference in radius will be minute for any realistic value of p.

It isn't the difference in radius that matters. It is the movement of the
circle centre that indicates the relative distance of the various stars.

The situation is obviously much more complicated when the telescope is pointed
away from the ecliptic pole.

But what is your point with all this?
You cannot seriously propose that parallax could be measured this way?

How do you propose to measure it? I know Hipparcos uses a very ingenious method
but I think it still boils down to assessing differences in relative star
positions over six months.

If you are referring to my paper:
Not really.
It is correct that I in my paper explained how a star at the ecliptic pole
would move due to stellar aberration - along circles with radii 20.5",
but it should be fairly simple to understand that a star at the ecliptic
will appear to move back and forth along a line which is 2 x 20.5".
In the general case, it will appear to move along an ellipse where
the semi-major axis always is 20.5", and the semi-minor axis is
sin(phi)x20.5", where phi is the ecliptic latitude.

That is true if the star is kept right in the viewing centre and the telescope
angle is plotted over the year. The Earth's tilt has to be taken into account.
The plot produces an ellipse.

Why do you think the Earth's tilt matters?
(See at the bottom. Was that why?)

Ignoring precession and nutation, the Earth's axis always point
in the same direction; the Earth is a mighty gyroscope.
You can fix the telescope to the ground, and it will point in
the same 'absolute' direction once a day, throughout the year.

Bradley had his telescope fixed to a chimney (with screw adjustment
for accurate measurement). He observed the star gamma Draconis, which
passes close to zenith (London), and which is near the ecliptic
north pole. The star will drift through the viewing field once
a day (night), and during one year the trace will shift back and
forth during by 41".
Bradley measured 40". This was in 1725!
Bradley tried to measure the parallax, but realized that the phase
was wrong. The observed 40" was a mystery for some time, but he
figured it out in 1728, and used it to measure the speed of light.

To make the telescope always point in the same direction,
you must obviously compensate for Earth rotation, and the
rotation of the telescope must be done around an axis which
is parallel to the Earth axis. Was this what you meant?
This is of course routine, all telescopes are driven that way.



Henri Wilson. ASTC,BSc,DSc(T)

www.users.bigpond.com/hewn/index.htm.

......
.

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  • Re: The Physics Behind Contractions.
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    (sci.physics.relativity)
  • Re: The Physics Behind Contractions.
    ... a telescope which is stationary in that frame must be pointed to see the star. ... (A suspicion creeps in - have you still not understood stellar aberration? ... circles will vary due to parallax.... ...
    (sci.physics.relativity)