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Hubble deep field question



 
 
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  #21  
Old January 31st 09, 12:41 AM posted to sci.astro
Androcles[_8_]
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Posts: 1,135
Default Hubble deep field question


"Steve Willner" wrote in message
...
In article ,
"Craig Franck" writes:
I'm reading "Chasing Hubble's Shadows" by Jeff Kanipe, and he
states on page 140 that "blobjects" at redshift 6 that were
13,000 light years across would appear 0.2 seconds of arc in
size.


Let's check that with Ned Wright's cosmology calculator:
http://www.astro.ucla.edu/~wright/CosmoCalc.html

What we want is the "angular size distance." The calculator gives
1.2 Gpc, and the next line of the output gives the angular scale
directly. One fifth of an arcsecond at z=6 is about 1.2 kpc or about
4000 light years, so 13000 light years would be closer to 0.6 arcsec.

An interesting property of the angular size distance is that it
reaches a maximum around z=1.5 (for the current best estimate of
cosmological parameters). Beyond that, objects are magnified, and
the angular size distance decreases. Of course object surface
brightnesses drop drastically as they are magnified, which is why
distant objects are hard to detect.

But that's assuming the light left when the object was 12.7
billion light years away.


It's assuming a specific cosmological model, which will also give a
light travel time. According to the calculator, the light left the
object 12.7 Gyr ago, but the object was much closer to us then than
it is now.

I had thought that the objects would have been much closer when
the light first left and it took 12.7 billion years to reach us because
of cosmic expansion,


Yes, that's right.

which would not have made the objects look smaller.


I'm not sure how you figure that. Look at the explanations of the
angular size distance linked from Ned's calculator. In general,
there's a complicated relationship between angular size distance and
other distances.

At redshift 6 they would be traveling at about 0.9c,


More like 0.96 if you are thinking of Doppler shift, but it's not
best to think of cosmological redshift that way. See Ned's
explanatory material.

but how would you figure out how far away they were when the light
first left from that?


As noted on Ned's calculator, there are many different distances.
What you probably mean is the "proper distance" (which Ned calls
"co-moving radial distance"), which is now 27.5 G-light-year. When
the light was emitted, the scale factor of the Universe was 7 times
smaller than now (1+z), so the proper distance then was about 3.9
G-light-year. Unless I'm confused, but I don't think so.



You certainly are confused.
1 Parsec = 3.08568025 × 10^16 meters

"The parsec ("parallax of one arcsecond", symbol pc) is a unit of length,
equal to just under 31 trillion kilometres (about 19 trillionmiles), or
about 3.26 light-years. The parsec is used in astronomy.


http://en.wikipedia.org/wiki/Parsec











  #22  
Old January 31st 09, 11:48 PM posted to sci.astro
John Polasek
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Posts: 95
Default Hubble deep field question

On Thu, 22 Jan 2009 18:07:22 -0800 (PST), Craig
wrote:

On Jan 22, 5:55*pm, "Craig Franck" wrote:
I'm reading "Chasing Hubble's Shadows" by Jeff Kanipe, and he
states on page 140 that "blobjects" at redshift 6 that were
13,000 light years across would appear 0.2 seconds of arc in
size. But that's assuming the light left when the object was 12.7
billion light years away.

I had thought that the objects would have been much closer when
the light first left and it took 12.7 billion years to reach us because
of cosmic expansion, which would not have made the objects look
smaller.

...

Under the standard cosmologies, this is true. For example, see
http://en.wikipedia.org/wiki/Angular_diameter_distance
Beyond a redshift of about 1-2 objects of the same physical size start
to grow *larger* in apparent angular size.

Craig

The author most likely did a straight Doppler calculation. He may have
assumed that with z = 6, recession might be at 6/7 of c
at range 6/7 of 12.7Gly (= 10.9Gly). Then the angle works out to
13000 LY/10.9Gly = 1.193e-6 radians or .246 seconds of arc.

The cosomological distances derived with Ned's abstruse formulas leave
a lot to be desired. For example, setting z = 6 in his calculator
gives two values for distance that are imcommensurable:
For z = 6 and H0 =71
Angular size distance is 4.2Gly
and
Luminosity disance is 207Gly
which is 50 times larger.
That needs some explanation.
HIs light travel time of 10.418 is close to my 10.9Gy.
John Polasek
  #23  
Old February 2nd 09, 07:56 PM posted to sci.astro
Steve Willner
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Posts: 1,172
Default Hubble deep field question

In article ,
John Polasek writes:
The cosomological distances derived with Ned's abstruse formulas


Ned's calculator uses standard cosmological formulas. Ned's
contribution is the calculator. (Ned has, of course, done quite a
bit of work testing cosmological models.)

leave
a lot to be desired. For example, setting z = 6 in his calculator
gives two values for distance that are imcommensurable:
For z = 6 and H0 =71
Angular size distance is 4.2Gly
and
Luminosity disance is 207Gly
which is 50 times larger.
That needs some explanation.


As I mentioned, you'll find the explanation linked from the
calculator. However, the smaller angular size distance is exactly
what's expected from cosmic magnification.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
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  #25  
Old February 3rd 09, 07:41 PM posted to sci.astro
Steve Willner
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Posts: 1,172
Default Hubble deep field question

In article ,
John Polasek writes:
Still, if we're only 13.7 Gy old, how to see 207 Gly?


Take a look at the definition of luminosity distance, which is what
the 207 Gly is. And by the way, that's for an open Universe; the
concordance model is a flat Universe. Luminosity distance for z=6 is
192 Gly in that one.

To another poster: whatever it is I might be confused about, it isn't
the definition of parsec. Ned's calculator gives distances in both
parsecs and light-years.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)
  #27  
Old February 10th 09, 10:35 PM posted to sci.astro
Steve Willner
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Posts: 1,172
Default Hubble deep field question

In article ,
John Polasek writes:
I have striven mightily to make sense of the luminosity distance and
it smacks of necromancy, in, for one thing, tolerating 2 right
answers, 4GLy and 200GLy for the same z (IIUC).


Where do you find two different answers? Are you looking at two
different cosmological models? Luminosity distance is certainly
model dependent, but for any (standard) cosmological model, a
specified redshift corresponds to one and only one luminosity
distance. In particular, it's the distance that has the light from
an object at that redshift spreading over an area of 4*pi*D^2.

By the way, how can one take the log of parsecs which are meters?


If you are referring to proper distance (or "Hubble Law distance"),
the equation involves ln(1+z); z is dimensionless.

Luminosity distance sounds like a principal ingredient in the claim of
early acceleration, defined briefly as "stars too dim for their z" as
far as I can determine.


If you want to determine how bright a star should appear to us, given
its absolute magnitude and redshift, luminosity distance has to go
into the calculation. That should be obvious from the definition.
The point of the supernova results, which I think is what you are
referring to, is that luminosity distances from some cosmological
models agree with the SN data while distances from other cosmologies
don't. That rules out the latter cosmologies as accurate
representations of the Universe we live in, unless of course the SN
data or interpretation are wrong in some way.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)
  #28  
Old February 25th 10, 04:15 PM posted to sci.astro
John Polasek
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Posts: 95
Default Hubble deep field question

On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzc D:aol T:com \(dlzc\)"
wrote:

Dear Craig Franck:

"Craig Franck" wrote in message
...
I'm reading "Chasing Hubble's Shadows" by Jeff
Kanipe, and he states on page 140 that "blobjects"
at redshift 6 that were 13,000 light years across
would appear 0.2 seconds of arc in size. But that's
assuming the light left when the object was 12.7
billion light years away.


No,, that assumes the objects look to be 12.7 Gly away "now".

I had thought that the objects would have been
much closer when the light first left and it took
12.7 billion years to reach us because of cosmic
expansion, which would not have made the
objects look smaller.


... and hence further away, and with the correct intensity.

At redshift 6 they would be traveling at about
0.9c,


No. A Z of 1 is "receeding" at c.


No, at red shift 6, they would be traveling at c*6/7 =0.857c, because
in fact v/c = z/1+z.
You are making the common error of assuming that v/c =z.
This is a Doppler effect, in which v(z) is widely misinterpreted, even
on NASA's site. I am preparing a paper on this topic.

http://www.astro.ucla.edu/~wright/cosmo_01.htm
... the text below the little sketch with the text "You're
receding"

but how would you figure out how far away they
were when the light first left from that? I thought it
would be 1/6 * 12.7bly, but that's assuming the
author is wrong, which may not be warranted.


I think you are trying to get to a number that has little
meaning.

David A. Smith

John Polasek
  #29  
Old February 26th 10, 09:12 PM posted to sci.astro
dlzc
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Posts: 1,426
Default Hubble deep field question

Dear John Polasek
:

On Feb 25, 8:15*am, John Polasek wrote:
On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)"

....
"Craig Franck" wrote in message
...

....
At redshift 6 they would be traveling at about
0.9c,


No. *A Z of 1 is "receeding" at c.


No, at red shift 6, they would be traveling at
c*6/7 =0.857c, because in fact v/c = z/1+z.
You are making the common error of assuming
that v/c =z. This is a Doppler effect, in which v(z)
is widely misinterpreted, even on NASA's site. I
am preparing a paper on this topic.


*If* you accept our most successful theory, that even approaches
describing the cosmology we see, you are correct. If you ignore that
wealth of knowledge and success, a classical Doppler shift is as I
described it.

Now let me add back in the rest of the quote that was edited out
before now:
QUOTE
No. A Z of 1 is "receeding" at c.

http://www.astro.ucla.edu/~wright/cosmo_01.htm
.... the text below the little sketch with the text "You're
receding"
END QUOTE

David A. Smith
 




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