andrew s
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I suspect I did not understand that either. In comoving coordinates there is no time dilation. Any way we go in circles. We are done. Regards Andrew -
I have read and reread your concerns about the integration and I don't understand it. It seems metaphysical. All the integration is doing is integrating the instantaneous comoving speed of light c/a(t) over a time interval. It's just a integral version of distance = speed x time. That different approximations and approaches lead to the same order of distance so that makes me secure the integration is fine. If you look at my analytic solution for the matter dominated Universe you get 13.6 x 3 = 40.8 Glyrs The paper I linked to got 42.6 Glyrs using a numerical integration The pros get 46.5 Glyrs So given the approximations that looks solid to me. Regards Andrew
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I'll let others judge who is right. Regards Andrew
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Let's take a simpler model than LCDM. A flat matter dominated Universe then a(t) = (t/tn) where tn is the time since the BB I.e. now. (Normally written as t0) The r = ctn^2/3 Int( 1/t^2/3) from t = 0 to tn doing the integration gives r = 3ctn An approximation to the full calculation using LCDM is given here which you may find accessible. It uses a simple polynomial approximation to the values of a(t). This is as I have repeated is nonsense unless the tr= te. If you solve it you get z= 0. No red shift the source and detector are next to each other. Regards Andrew PS for a radiation dominated universe you get a(t) = (t/tn)^1/2 and the integration is again trivial as I said. PPS I am happy now I full understand the standard approach is correct so will waste no more time on it. Thanks for prompting me to get to grips with it.
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This post is mainly for my benefit so @kurdewiusz don't feel obliged to respond. The size of the observable Universe is the furthest distance a light ray can have got to us from at the current time I.e. Now. It is also called the particle horizon. This is the light of the CMB arriving with a red shift of z= 1100. One issue is to be clear what coordinate system we are using in the discussion Cosmological distances are most simply discussed in terms of comoving distance and time. That is a coordinate system which moves with the Hubble flow. Using the rubber band analogy consider the band have a set of tick marks at 1mm intervals at time t = 0 when the CMB was released. As the band expands objects on the band stay at the same "comoving" locations but move relative to a metre ruler layed next to the band. The ruler measures the local distance as in special relativity. A set of rulers layed out along the band measures what we call proper distance if done Now. As the band expands to Now the ticks expand from 1mm to 1m as measured by our ruler. To convert from rubber band coordinates to metres we need the number of meters per tick. This is the scale factor "a(t)". At t = 0 a(0) = 0.001 corresponding to a red shift of z ~ 1100. At t = Now a(now) = 1 (confusingly a(Now) is labled a0 and and t(Now) as t0. So how far can a light ray go in a radial direction from t = 0 to t = Now ? The standard metric in the LCDM model in the radial direction is just ds^2 = -c^2 dt^2 + a(t)^2 dr^2 Note r is the comoving coordinate distance For a light ray ds = 0 so we get cdt = a(t)dr or dr/dt = c/a(t) But dr/dr is just the speed of light in the comoving coordinates. SR tell us it is constant in local coordinates (our rulers) but it varies in comoving coordinates. Not very surprising as when a is small there are more ticks per metre than when it has stretched. To get the required distance we just integrate the equation dr/dt = c/a(t) from t = 0 to t = Now r = int(c/a(t))dt from t = 0 to t = Now (sorry can't do an integration symbol) Note a(t) changes smoothly as the Universe expands so the integration is not an issue. To get the proper distance D you need to use D = a(Now).r which as we have chosen a(Now) = 1 gives D = r. However, this only applies to Now Regards Andrew
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That's not how understand it. The red shift only depends on the scale factor at the times of emission and reception I.e at te and tr. However the size calculation depends on the scale factor for all times that the light is moving between its emission and reception I.e such that te =< t <= tr . I will expand on this tomorrow as its late now. Regards Andrew
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No it means there is zero time and distance between the time of emission and reception. That is the source and defector are next to each other. Totally irrelevant to the detection of the CMB. Regards Andrew
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That is not my formula. I have never posted that. It's yours. Regards Andrew
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As I said above for the calculation of the size of the Universe tr = Now (that's the time we receive the light) and by convention a(Now) = 1 te = 0 ( that is the time of emission of the CMB we observe now) a(0) ~ 0.001 "I remind you, that you need tr=te for a(tr)=a(te) for a=1." This is your error tr can't equal te. The CMB we see now was emitted some 13.6 Gyr ago and we observe it Now. We observe (1+z) = 1101 so using (1+z) = a(tr)/a(te) which for observations Now gives (1+z) = 1/a(te). With a(te) ~ 0.001 I have been consistent on this throughout. Unless you accept your error there is no point continuing this discussion. Regards Andrew
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No. I pointed out that te was the time of emission. It is a variable as it depends on what you are observing. For the CMB its zero but for Andromeda it will Now - 2.4 Myrs For GN -z11 it will be 430 Myrs Regards Andrew
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Correct. The CMB is the furthest back we can see at a red shift of about 1100 now. So for an observation now we have a(te) ~ 0.001, te = 0 and a(tr) =1, tr =Now Where a(te) is as I calculated before. Regards Andrew PS here is a plot of a(t) based on the Plank18 final mission data from here.
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Ok I lied. te is the time light is emitted. We can observer today light from the CMB emitted at te ~ 0 through any value up to now te ~ 13.6 Gyr since the Big Bang. For light we see today te only depends on how faraway it was when emitted. It certainly is not a constant. The end. Regards Andrew PS I lied about not responding again nothing else! PPS "If you don't change the definition of te or tr in a=1, you will never get the equality a(te)=a(tr), but you need it in a=1." No no no. For observations now a(tr) = 1, a(te) depends on when the light was emitted. a(te) = a(tr) if and only if they occur simultaneously.
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At te =0 a = 0 Just look at the scale on the right hand scale of the y axis in the Figure 1 that I posted. The CMB we see today has a red shift of about 1100 so (Z+1) = 1101 = a(tr=13.6 Gyr)/a(te) = 1/a(te) a(te) = 1/1101 ~ 0.001 and t ~ 0 I don't think I changed anything tr is always the time received and te the time emitted. This is my last reply. Good evening. Regards Andrew
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Absolutely if you emit a light beam now and detect it now + deltat then it will have travelled c×deltat a very short distance in a very short time and to zeroth order a(tr) = a(te) = 1 but to first order a(tr) = a(te) + deltat.da(te)/dt I have done my best to explain this to you. a(tr) = a(te) if and only if te = tr I.e they are at the same time. If the time are different the they are not equal. You seem to be stuck on a basic fact of calculus. I have tried my best. Good luck. I am retiring from the discussion as I don't seem able to help you. Regards Andrew
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A final try. I have tried to show normalised values can be greater than one e.g. my age at 25. I think you are hung up on the term normalised. A better term would be scaled. I scaled my age to that at age 12. This led to my scaled age at 0.001 being less than 1 and at age 25 greater than one. That's all ther is to it. Make a cosmic measurement now (say x meters) and its scale factor is defined as 1 i.e. (x/x = 1). If the same measurement had been made in the past the actual measurement in meters would be smaller (say y m) the scale factor would be such that y/x < 1. If made in the future you would get the longer distance (say z m) and the scale factor would be z/x> 1 Say your light ray was emitted at the early time te and received at the later time tr then (1+z) = a(tr)/a(te) = (z/x)/(y/x) = z/y > 1 Regards Andrew
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Ok one very last attempt. Note a(tr) and a(te) are the normalised scale factors at the time the light is received and emitted respectively. Looking at my height about age 12 for a small positive absolute increase in time my hight would be 5 + dh ft say and my scaled hight would be 1 + (dh/5) Looking at a change in hight about age 25 for a similar absolute change my hight would change to 6 + dh and my scaled hight would change to 1.2 + (dh/5) I have done my best not sure I can make it any clearer. Regards Andrew
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Sorry that totally wrong. Let's take a simple example my hight. When born I was about 1.7 ft tall at 12yrs I was 5 ft and at 25yrs 6 ft. Now I decide to scale myself about age 12 so the scale the a0(12) = 5/5 = 1. At birth my scaled hight was a(0.001) = 7/5 = 0.34 and at 25 yrs a(25) = 6/5 = 1.2. Yes a0 is fixed but "a" varies continuously about a0 = 1 As I age I will srunk but the Universe is not expected to. Regards Andrew
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It is always normalised about 1. In the region about now. It is continuous on 0 to infinity. To make matters concrete let's fix now to epoch 2000 and assume t = 13.6 Gyr then. Then a0 = a(13.6Gyr) =1. Now consider light from andromeda arriving in the year 2000 tr = 13.6 Gyr a(tr) = 1 It was emitted ~2.5 Myr prior to 2000 so te = 13.6 Gyr - 2.5 Myr when a(tr) = 1 - every small amount say 0.999999 (made up number) So (1+ z) = 1/0.999999 ~ 1 i.e we would not detect any red shift due to expansion but only peculiar velocity in fact it's blue shifted as its moving towards us. So to answer your question yes these points have physical meaning but the effect are very small until you get to large distances. Indeed Hubble' 1929 data showed significant scatter due to the relatively close proximity of the galaxies he measured and their peculiar motion. Regards Andrew
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No. a(te) is the scale factor for when the light was emitted I.e. in our past. a(tr) is the scale factor at the time it is received. Consider a light ray emitted when t ~ 1Gly I.e. te = 1 Gly (about 40 Glyrs away) a(te) ~ 0.01. It is received today tr ~ 13.6 Gly so a(tr) = 1 so (1+z) = 1/0.01 = 100 Now consider a light ray emitted today te = 13.6 Gyr about 10 Glys away it will reach us at tr ~ 25 Gyr and a(tr) ~ 2 so (1+z) = 2/1 = 2 I have taken the numbers very approximately from the diagram. Regards Andrew
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a0 = 1 always that is the scale is fixed so that the scale factor is normalised to one now. At future epochs a will be greater than 1 as the universe expands and was less than 1 in the past but a0 I.e. a at t = 13.6Gyr will remain 1. That's why you need the full equation for (1+z) = a(tr)/a(te) when doing sums about redshifts seen in the past or future not the one only valid for now. The key is that "a" is a normalised scale. Cosmologists chose to normalise it to 1 at the current time to make their sum with data colleted now simpler. Regards Andrew
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Ok. In the past t ~ 0 Gyr a ~ 0 (see attached image from the paper). By about 1 Gyr a~ 0.2. At 13.6 Gyr (now) a = 1 in the future at about 25 Gly a = 2 . It will not get to 1100 until 60 or more Gyr. So you have it in reverse so its the other way round. Regards Andrew PS It's confusing the a is sometimes labelled a0 for now. In all the above t=0 is at the start of the current expansion/ big bang and a0 refers to t ~ 13.6 Gyr
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Your equation (z+1) = 1/a only holds for the current now. In general the equation is (z+1) = a(tr)/a(te) where a(tr) is the scale factor at the the time the light is observed and a(te) the scale factor at the time of emission. In an expanding Universe a(tr) > a(te) and so no problems arise. Figure 1 of the Expanding Confusions paper show a plot with a -> infinity. Its equation 23 is the one I quote but with R in place of a. Hope this helps. Regards Andrew
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Should Science be a *practical* thing?
andrew s replied to Macavity's topic in Physics, Space Science and Theories
I always enjoyed practical science. Pops, bangs, unintended explosions and the odd rogue projectile enlivened the day. As did a near blind chemistry teacher spending some time trying to put a rubber bung into the wrong end of a test tube. He also created a flame thrower to melt a students bag. Having struck a match to light a Bunsen burner he could not get the rubber tube on the gas tap. As the flame reached his fingers he turned on the gas and lit it with the obvious result. He was a brilliant chemist who help develop the first epoxy resins. He could also be easily distracted into his favourite topic fireworks and suitable demonstrations followed. He did nearly axasphyxate us demonstrating oxygen would burn in ammonia gas as well as vice versa. (Health and Safety had yet to ..) The one area I did find a tad tedious was organic chemistry. Boiling up various clear liquids to produce another one with not quite melting or boiling point was somewhat dull. One trick was to get in with the stores personnel where a request for a sample of the target chemical could pull said melting or boiling point into the correct range. 😊 Regards Andrew PS recalling my days at Cambridge College of Arts and Technology reminded me of the wonder of nature. The class was a mix of 11 plus failures (including me) doing A levels and foreign students. One day the teacher had obviously lost the students attention. It had stated to snow and half the class had never seen it before. To his credit we stopped and went outside. Brilliant. -
Should Science be a *practical* thing?
andrew s replied to Macavity's topic in Physics, Space Science and Theories
Me too earth, fire, air and water. 😊 Regards Andrew -
Should Science be a *practical* thing?
andrew s replied to Macavity's topic in Physics, Space Science and Theories
This paper has a good discussion on the difficulties in doing the theoretical calculations. This diagram from the link shows the difference from two approaches Time and more experiments and theoretical number crunching will tell. Regards Andrew
