 |
 | Home | Contact Us | FAQ | Search & Site Map | Link to Us |
 | Sign In | Join | Other 45 Sites in Network |
| HomeDiscussion GroupsSpace ScienceAstronomyAmateur AstronomySpace FlightSpace StationShuttleSpace HistorySpace PolicySETI SpaceKB.com Contact UsLink To UsSearch & Site Map |
Space Forum / Astronomy / February 2010Hubble deep field question
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. At redshift 6 they would be traveling at about 0.9c, 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.  SignatureCraig Franck craig.franck@verizon.net Cortland, NY > I'm reading "Chasing Hubble's Shadows" by Jeff Kanipe, and he > states on page 140 that "blobjects" at redshift 6 that were [quoted text clipped - 11 lines] > that? I thought it would be 1/6 * 12.7bly, but that's assuming the > author is wrong, which may not be warranted. Nobody sane would take seriously the so-called "Big Bang" creation theory, yet the apparent red shift of distant galaxies is perplexing. However, if one takes the Principle of Relativity seriously one is compelled to accept a variation in the speed of light that is source velocity dependent, and with a little study one can soon realise that what works for one star in a galaxy must work for them all. This diagram: http://www.androcles01.pwp.blueyonder.co.uk/Doolin'sStar.GIF is for one star, and one can quickly see that dT is greater than dt for most of the period. For a distant galaxy at the appropriate distance, ALL stars will show a red shift MOST of the time. So statistically, red shift is exactly what is to be expected for a galaxy without any recessional velocity. Attempting to gather light from a single star in a galaxy is an impossibility, even the closest appears as a blur. http://antwrp.gsfc.nasa.gov/apod/ap090109.html In other words if your theory is that the speed of light is fixed and what you see isn't illusion then you'll forever run into ever- increasing conundrums such as objects being 13,000 light years across! Dear Craig Franck: > 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. 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 On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzc D:aol T:com \(dlzc\)" <dlzc1@cox.net> wrote: >Dear Craig Franck: > >"Craig Franck" <craig.franck@verizon.net> wrote in message >snip > >No. A Z of 1 is "receeding" at c. I beg to differ. This is a common and grievous error, if we attribute redshift to straight Doppler. The laboratory measurement is L'/L = 1 + z Letting v/c = b (for beta), doppler causes the frequency reduction f'/f = 1 - b from which the wavelength is given by the inverse L'/L = 1/1-b = 1 + b + b^2 + b^3 which is not 1 + z z is not v/c. A little algebra shows that recession velocity is given by c*z/1+z, so for z = 1, the recession is c/2. I think this has been lost sight of in the current fiction that space stretches wavelengths by 1+z, a concept which panders to Einsteins space-time. I see the graphs that routinely present v/c>1 with velocities equal to 2 and 3c. It isn't true, and this is the reason, which should come as a great relief, since viewing >c is impossible. >http://www.astro.ucla.edu/~wright/cosmo_01.htm >... the text below the little sketch with the text "You're >receding" > >David A. Smith John Polasek > On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzc D:aol T:com \(dlzc\)" > <dlzc1@cox.net> wrote: [quoted text clipped - 30 lines] >> > John Polasek Either the galaxies are speeding up or the light from them is slowing down. The former requires dork energy, the latter a better understanding of light's (and human) behaviour. "Hubble had confused two different kinds of Cepheid variable stars used for calibrating distances" says it all. http://www.androcles01.pwp.blueyonder.co.uk/Doolin'sStar.GIF >Either the galaxies are speeding up or the light from them is slowing down. >The former requires dork energy, the latter a better understanding of >light's (and human) behaviour. If the value of c changes it changes the value of the fine structure constant and that affects emission and absorption lines in spectra. No such change seems to be visible. Bud >>Either the galaxies are speeding up or the light from them is slowing >>down. [quoted text clipped - 6 lines] > > Bud Ok. Changing the value of pi affects the ratio of the circumference of a circle to its diameter. No such change seems to be visible. Change the value of the fine structure constant so that you affect emission and absorption lines in spectra, then publish your results. f.cking dork! Dear John Polasek: > On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" > [quoted text clipped - 8 lines] > I beg to differ. This is a common and grievous > error, if we attribute redshift to straight Doppler. The Universe has no unique center, therefore the expansion is not from an explosion in space. So there is no theoretical support to your "manipulations". David A. Smith Dear John Polasek: On Jan 23, 9:51 pm, John Polasek <jpola...@cfl.rr.com> wrote: > On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" > [quoted text clipped - 8 lines] > I beg to differ. This is a common and grievous > error, if we attribute redshift to straight Doppler. The Universe has no unique center, therefore the expansion is not from an explosion in space. So there is no theoretical support to your "manipulations". David A. Smith ================================ David A. Smith has no brain, therefore what he says has no meaning. So there is no factual support to your "rambling rants". >Dear John Polasek: > [quoted text clipped - 16 lines] > >David A. Smith I did not discuss 'center'. My statement treats of a line joining the observer and his star- there's no explosion. I derived the velocity for you as a function of z, but you neglected to examine it. No center you say, OK, but then space has no substance either, that could stretch by 1+z, ostensibly to fend off superluminal problems. But these problems stem from erroneous use of the illegal expression cz leading to 2c, 3c, which are not impossible, just that there's nothing to see past 1c and we need to see redshift. The red shift is from Doppler alone, but a whole legend has developed from this 1+z stretch fixation. The paper below argues for a replacement of z with z/1+z but from purely mathematical consideration of algebraic poles and the like. http://arxiv.org/abs/ 0710.1887. This notion is correct as I have shown in the prior note. I repeat, it is a serious error with serious consequences John Polasek Dear John Polasek: > >> On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" > [quoted text clipped - 15 lines] > > I did not discuss 'center'. But you did invoke straight classical Doppler shift, with the inherent energies required to blast masses away from us at significant percentages of the speed of light. > My statement treats of a line joining the > observer and his star- there's no explosion. Sure, if you completely ignore what your "model" requires. ... > I repeat, it is a serious error with serious > consequences You are right. So why don't you face those consequences? David A. Smith >Dear John Polasek: > [quoted text clipped - 19 lines] > >But you did invoke straight classical Doppler shift, which, I think you will agree, is merely the most plausible *explanation* for a universally acknowledged phenomenon, redshift, and which involves mere kinematics. >with the inherent >energies required to blast masses away from us at significant >percentages of the speed of light. You seem to have accepted the responsibility of providing an explanation for Doppler redshift, but I resent your making me a party to this hypothetical detonation. In any case, this dynamite thing is just an obfuscatory distraction, which in no way undermines my assertion that z/1+z is the proper distance modulus, not z. You stated that recession velocity at z =1 is c, and tolerated a reported measurement of z = 6 making v = 6c, without considering how you could see the redshift. Of course with D = z/1+z you get v = c*6/7 and we are back in real physics. >> My statement treats of a line joining the >> observer and his star- there's no explosion. [quoted text clipped - 4 lines] >> I repeat, it is a serious error with serious >> consequences The inherent error lies in failing to recognize that while the Doppler effect has a transparent explanation as a frequency phenomenon, it is *measured* as a wavelength phenomenon and then persisting to employ wavelength algebra through thick and thin despite the glaring incongruities. >You are right. So why don't you face those consequences? You better modify all those z*c expressions to zc/1+z. >David A. Smith John Polasek Dear John Polasek: > >> >> On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" > [quoted text clipped - 22 lines] > acknowledged phenomenon, redshift, and which > involves mere kinematics. It is the first baby step to be taken in exploring a cosmology. Essentially, it requires that we are the center of the Universe. Again. No, I don't find that "plausible". David A. Smith >Dear John Polasek: > [quoted text clipped - 32 lines] > >David A. Smith You are confusing redshift with the Hubble precept that the radial velocity is proportional to range (an extrapolation by others, as Hubble didnt go that far). At first glance that makes us the center, as you seem to presume, but that is easily debunked. If I see a star with redshift it is inescapable that the perceived frequency will be reduced by the factor 1 - v/c where v is the separation rate between source and observer. The wavelength will therefore be increased in the factor 1/1-v/c, observed in the lab as 1 + z. z is not v/c but that seems to be the popular 'take'. So space stretches by 1+z. That's puerile. There's no 'it' to stretch. John Polasek Dear John Polasek: > >> >> >> On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" > [quoted text clipped - 33 lines] > that the radial velocity is proportional to range (an > extrapolation by others, as Hubble didnt go that far). He actually did, but Hubble is not the problem... > At first glance that makes us the center, as you > seem to presume, but that is easily debunked. Actually required by your mathematics. You simply say "well that is puerile", but you do not follow your mathematics to their inescapable conclusion. David A. Smith >Dear John Polasek: > [quoted text clipped - 42 lines] > >Actually required by your mathematics. If you would please be specific, what is required? >You simply say "well that is >puerile", but you do not follow your mathematics to their inescapable >conclusion. What inescapable conclusion? Be specific. This has all the earmarks of a debate that could continue far into the spring. Just for starters, accept or reject: f' = f(1-dS/cddt) where S is the separation. >David A. Smith Dear John Polasek: > >> >> >> >> On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" > [quoted text clipped - 42 lines] > > If you would please be specific, what is required? Motion in a Euclidian Universe, with all objects visible moving away from us, with speed (more or less) proportional to distance. > > You simply say "well that is puerile", but you > > do not follow your mathematics to their > > inescapable conclusion. > > What inescapable conclusion? Be specific. I have been. You do not acknowledge the cosmology required by your "simple solution". > This has all the earmarks of a debate that > could continue far into the spring. > Just for starters, accept or reject: > f' = f(1-dS/cddt) > where S is the separation. No, the "debate" ends now. David A. Smith >Dear John Polasek: > [quoted text clipped - 66 lines] > >David A. Smith You must think my logic requires a Hubble-expanding universe to have a center, (why didn't you say so?). Not at all. There is little science that can be extracted from Hubbles constant, since cosmology cannot put a definite value on it, nor is it possible to implicate it as a possible cause of expansion (for which cosmology presently has no raison). But since it and z are about the only real data available, we analyze the Hubble principle: There is no center, as follows, using stars A(us), B and X: A sees X with vector distance and its colinear velocity = AX Clearly, B equally sees a different distance and velocity BX which, given large separation of A and B, the vectors mismatch by a wide angle yet BX can claim legitimacy also. How? the 3d leg AB. So A can view B and get redshift range and velocity = AB. And B can view A and get BA. Simply combine: Now it's obvious that AB + BX = AX so B is like A, and v.v. BA + AX = BX so A is like B. There's only one conclusion: they are separating without the semblance of a center. So to cover this contretemps, cosmology uses the anomalous coefficient a, which for us now is a0, and its value, fortuitously is unity: We are just the right size now! In fact that's just what H is: H = 100%/age of our star 13Gyr. So, what do you think, is f' = f(1-dS/cdt) or not? John Polasek On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzc D:aol T:com \(dlzc\)" <dlzc1@cox.net> wrote: >Dear Craig Franck: > [quoted text clipped - 19 lines] > >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 [quoted text clipped - 9 lines] > >David A. Smith John Polasek Dear John Polasek > On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" ... ... > >> At redshift 6 they would be traveling at about > >> 0.9c, [quoted text clipped - 7 lines] > 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 Dear John Polasek On Feb 25, 8:15 am, John Polasek <jpola...@cfl.rr.com> wrote: > On Thu, 22 Jan 2009 17:00:01 -0700, "N:dlzcD:aol T:com \(dlzc\)" ... ... > >> At redshift 6 they would be traveling at about > >> 0.9c, [quoted text clipped - 7 lines] > is widely misinterpreted, even on NASA's site. I > am preparing a paper on this topic. *If* you accept our most successful theory, ==================================== Your most successful theory is a pile of ignorant sh.t and you have your head up your bigoted arse. It takes a real moron to claim 1000 * 1.3 seconds is 1.02 km/sec as you have, Smiffy. Why don't you just shut the f.ck up and piss off, you clueless cretin? > Dear John Polasek > [quoted text clipped - 22 lines] > is 1.02 km/sec as you have, Smiffy. Why don't you > just shut the f.ck up and piss off, you clueless cretin? Look at the equation again. 1000 * 1.3 * 1.02 = what? I realize units were never your strong suite. David A. Smith On Feb 26, 4:04 pm, "Androcles" <Headmas...@Hogwarts.physics_u> wrote: > "dlzc" <dl...@cox.net> wrote in message > [quoted text clipped - 25 lines] > is 1.02 km/sec as you have, Smiffy. Why don't you > just shut the f.ck up and piss off, you clueless cretin? Look at the equation again. 1000 * 1.3 * 1.02 = what? I realize units were never your strong suite. David A. Smith ===================================================== Smiffy wrote: Let's see... hmmm so the fact that the CMBR is blocked by objects, including the Moon, and it is redshifted by more than 1000... http://blogs.nature.com/news/blog/2010/01/ ... down to "Antarctica 2010: Hunting for galaxy clusters" 1000 * 1.3 seconds * 1.02 km/sec. Looks like we'd see the CMBR in addition to the "CMBR shadow" with a very large offset. Note that the CMBR is black body and the Moon is not, so the signal is detectable. So lame insults, and a requirement for others to do your work for you was your response before. Yes, that is what I rememebered... David A. Smith ================================================= Too senile to recall your own post, Smiffy? It takes a real moron to claim 1000 * 1.3 seconds is 1.02 km/sec as you have, Smiffy. Why don't you just shut the f.ck up and piss off, you clueless cretin? > I'm reading "Chasing Hubble's Shadows" by Jeff Kanipe, and he > states on page 140 that "blobjects" at redshift 6 that were [quoted text clipped - 6 lines] > 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 >> I'm reading "Chasing Hubble's Shadows" by Jeff Kanipe, and he >> states on page 140 that "blobjects" at redshift 6 that were [quoted text clipped - 14 lines] > >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 > 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 [quoted text clipped - 5 lines] > 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. SignatureSteve Willner Phone 617-495-7123 swillner@cfa.harvard.edu 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.) >> The cosomological distances derived with Ned's abstruse formulas > [quoted text clipped - 15 lines] >calculator. However, the smaller angular size distance is exactly >what's expected from cosmic magnification. Still, if we're only 13.7 Gy old, how to see 207 Gly? John Polasek > 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. SignatureSteve Willner Phone 617-495-7123 swillner@cfa.harvard.edu 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.) >> 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. 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). By the way, how can one take the log of parsecs which are meters? 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. But there are plots of deltaM that only peak slightly at z =.5 (too dim) and then slump away sharply at z>1 IIRC. This is discouraging. John Polasek > 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. SignatureSteve Willner Phone 617-495-7123 swillner@cfa.harvard.edu 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.) > 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. SignatureSteve Willner Phone 617-495-7123 swillner@cfa.harvard.edu 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.) >> I'm reading "Chasing Hubble's Shadows" by Jeff Kanipe, and he >> states on page 140 that "blobjects" at redshift 6 that were [quoted text clipped - 52 lines] > 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 |
Reply to this Message |
 Sign In
 Join
 My Latest Posts My Monitored Threads My Blog My Photo Gallery My Profile My Homepage Start New Thread Enable EMail Alerts Rate this Thread
|
©2010 Advenet LLC Privacy Policy - Terms of Use
This website includes both content owned or controlled by Advenet as well as content owned or controlled by third parties.