# Effort to get us all on the same page (balloon analogy)

by marcus
Tags: analogy, balloon, effort
 Astronomy Sci Advisor PF Gold P: 23,232 To begin to deal with distances in terms of this simple model I need to add up this cumulative sum of terms ts where s is the reciprocal scalefactor 1/a and run thru some range like [1,2] in steps of 0.1, say. Let's start it at s = 1.1 and go 1.2, 1.3,.... and at each value of s we will evaluate this formula atanh((.375136* s^3+1)^-.5) using http://web2.0calc.com to get ts the time (expressed in billions of years) when the light was emitted that we now see wave-lengthened by a factor of s. 1.14678............(starting with 1.1) 2.1893904284755 3.14075638766957 4.01193171154709 4.81239653369235 5.55030772362728 6.2327011396146 6.86566003274671 7.45445670103674 8.00367220130405 (this was for 2.0) 8.51729768096363 8.99882016696824 9.4512951759886 9.87740815857389 10.2795265021777 10.65974356945189 11.0199160299271 11.36169555160973 11.68655575226624 11.99581516649566 (the last one was for 3.0) So how to use this cumulative sum to get distances? Well suppose a galaxy's light comes in with a scaleup s=2.0 (wavelengths twice as long as when emitted.) The number from the list we use is the one for 1.9 namely 7.45445670103674, and that gets multiplied by the Δs, the step size, which is 0.1 In addition there are two other things to do: add (1+Δ/2)*t1 which is 1.05*1.2661864372681=1.329495759131505 and subtract (2-Δ/2)*t2 = 1.95*0.54921550026731=1.0709702255212545 The difference is 0.2585255336102505, so that's what gets added: 0.745445670103674+0.2585255336102505 = 1.0039712037139245 Finally multiply that by 16.3/1.5 and get 10.9098... Gly. Ned Wright's calculator says 10.901 Gly (with equivalent model parameters). So the accuracy is not so bad. that's the current distance: 10.91 billion lightyears. All those extra decimals are ridiculous but it is too much trouble to be rounding off all the time so I just take what the calculator gives and finally round it off to something sensible at the end. the corresponding thing for s=3 1.168655575226624+1.329495759131505-0.91231527197679=1.585836062381339 and then again finally multiply by 16.3/1.5, to get 17.23275...billion lightyears Ned Wright says 17.220, so again we are off by 1 in the fourth digit.
 Astronomy Sci Advisor PF Gold P: 23,232 I didn't get around to editing the previous post until after the deadline for changes and it needs some clarification. The key formula in the toy version cosmic model I'm working with is (16.3/1.5)*atanh((.375136* s^3+1)^-.5) This gives the time (expansion age in billions of years) at which the reciprocal scalefactor was a particular value s. Another way to say it: ts is the time (expressed in billions of years) when the light was emitted that we now see wave-lengthened by a factor of s. s can be thought of, if you like, as the "scale-up factor": because since the time ts, distances and wavelengths have been scaled up by a factor of s. Back at time t2 distances were 1/2 their present size, back at t9 distances were 1/9 their present size, t1 is the universe's present expansion age, and so on. I won't be using the traditional notation for s, which is "1+z", since it makes the formulas even messier than they are already. The Hubbletime parameter 16.3 billion years represents the cosmological constant (asymptotic distance growth rate.) The other Hubbletime parameter, giving present growth rate, is 13.9 Gy and since that most often appears in combination as (16.3/13.9)2-1=0.375136... I have, for simplicity, packaged it in that number. When I have to do a lot of calculating, e.g. numerical integration, I leave off the coefficient (16.3/1.5) and factor it in only at the end. It turns out that we can do a pretty good job of estimating the distances to sources at various scaleups by essentially just adding up a long string of arctanh values, with s advancing from 1 to s in steps of some smallish stepsize Δ. If we take Δ = 0.1, this amounts to: 1.05t1+ 0.1( t1.1+ t1.2+...+ ts-0.1) - (s-0.05)ts This is the bare bones of a numerical integration for c∫ s dt. The idea is that at each interval dt of time in the past, the light from an object travels a distance cdt and this gets scaled up by a the appropriate factor s. So at present the distance to the object is the sum of all those scaled-up segments and equals c∫ s dt. It looks messy but seems to work out all right. Here's a cumulative sum of atanh((.375136* s^3+1)^-.5) using http://web2.0calc.com 1.14678............(starting with 1.1) 2.1893904284755 3.14075638766957 4.01193171154709 4.81239653369235 5.55030772362728 6.2327011396146 6.86566003274671 7.45445670103674 8.00367220130405 (this was for 2.0) 8.51729768096363 8.99882016696824 9.4512951759886 9.87740815857389 10.2795265021777 10.65974356945189 11.0199160299271 11.36169555160973 11.68655575226624 11.99581516649566 (this was for 3.0) 12.29065686210409 12.57214523554962 12.84124042899438 13.09881073851359 13.34564332215593 13.58245346592667 13.80989262370986 14.02855541222707 14.23898571313266 14.44168201025018 (for 4.0) 14.63710206991032 14.82566705565418 15.00776515463345 15.18375478139281 15.3539674149637 15.51871011700947 15.67826777187171 15.83290508355766 15.98286835979816 16.12838710914517 (for 5.0) So suppose we evaluate 1.05t1+ 0.1( t1.1+ t1.2+...+ ts-0.1) - (s-0.05)ts to find the distance now to a source with s = 4 (its light comes in wavelengthened by a factor of 4). As long as we are using the stepsize Δ=0.1, the first terms is always 1.329495759131505, and the sum, multiplied by the stepsize, can be read off that list: 1.423898571313266. The term at the end, that gets subtracted, is 3.95*0.202696297=0.800650373 1.329495759 +1.423898571 - 0.800650373 = 1.95274396 And then at the end the whole thing gets multiplied by the cosmological constant term 16.3/1.5, to give 21.2198≈ 21.22 billion light years. Sorry about all the meaningless extra digits but it is too much trouble to be rounding off every time I take a result from the calculator, so I just round off at the end. Let's compare this with Wright's calculator. Well, Wright's says 21.204 Gly. So as usual we are OK for three significant figures and off in the 4th place. Notice that since this numerical summing procedure gives us the NOW distance to the source, all we need to do is multiply by the scalefactor a, or alternatively divide by the scaleup s, and we get the THEN distance---how far from our matter or galaxy the thing was when it emitted the light. So this primitive model already does quite a bit that one expects from serious cosmology calculators. Given a scalefactor a (or the reciprocal 1/a = s) it can give the corresponding expansion age---the time when the source galaxy emitted the light. And it can give the Now and Then distances to the source (proper distance, as if you could halt expansion at the given moment and measure directly). The THEN distance is essentially the ANGULAR SIZE distance (our model is spatial flat) so that's taken care of. It still might be nice to be able to calculate the HUBBLETIME corresponding to a given scalefactor a, or its reciprocal s. That is hour handle on the rate of expansion going on at the time the light was emitted. Notice that the light itself, when it arrives, tells us the scalefactor, or equivalently its reciprocal s, which we focus on here, so the other things we want to know should be calculated from s.
 Astronomy Sci Advisor PF Gold P: 23,232 It looks as if the Hubbletime Ys corresponding to a given scaleup s should be given in billions of years by: 16.3(0.375136s3+1)-.5=16.3/sqrt(.375136*s^3 + 1) so using the calculator let's try that for s=3 It gives Y3 = 4.8861403 ≈ 4.886 billion years. Let's see if I've made a mistake. Apparently not, Jorrie's calculator (with the corresponding parameters) gives 4.885 billion years. Remember that the two parameters we're using in the model's formulas, namely 16.3 and 0.375136, are just an equivalent form of the two Hubbletimes which determine two key expansion rates, now and in distant future. Ynow = 13.9 billion years Y∞ = 16.3 billion years The number .375136 is simply what (Y∞/Ynow)2 - 1 = (16.3/13.9)2 - 1 works out to be. You can think of 1.375136 as a ratio of two expansion rates, squared. It is simply (Hnow/H∞)2 so it tells you how much more the percentagewise expansion rate is now than it will be in the longterm future. When people talk about "acceleration" what they mean is what you see when the H expansion rate is declining only very slowly or is steady at some given value. As long as H is not declining too rapidly, if you watch a particular distance it will grow by increasing annual amounts as the principal grows. Not terribly dramatic, given the very low "interest rate" but there is acceleration in a literal sense. In the previous post we calculated that a galaxy we see with scaleup factor s = 4 (wavelengths quadrupled) is now at a distance of 21.22 Gly. How fast is that distance now growing? That is very simple to calculate. We just divide 21.22 Gly by 13.9 Gy. Dnow/Ynow = 21.22/13.9 = 1.53c. That means it is growing at 1.53 times the speed of light. Calculation easy with these quantities For comparison and a bit more practice, back in post #433 we found that the distance NOW for s=3 was 17.23275 billion lightyears, which means distance THEN was 17.23275/3=5.744 billion lightyears. However we just found that also for s=3 we have Hubbletime Ythen = 4.886 Gy. So for a s=3 galaxy, whose distance THEN at time light was emitted was 5.744 Gly, how fast was that distance then growing? Well obviously Dthen/Ythen = 5.744 Gly/4.886 Gy = 1.18 ly/y = 1.18c. The notation is still far from perfect, but I hope some of this is comprehensible ============== referring back to post #434 the last increment was 0.14551874934701 and to find the now distance to an s=5 source D5(now) one would take 4.95*0.14551874934701=0.720317809 off at the end, so it looks like 1.329495759 +1.598286836 - 0.720317809 = 2.207464786 which then gets multiplied finally by 16.3/1.5 to give 23.987784 billion lightyears. So unless I've made a mistake that's the distance now to an s=5 source. I'll compare with what Wright's says. It says 23.970 Gly. So we are still OK for three digits. D5(now) = 23.99 Gly D5(then) = 23.987784/5 ≈ 4.798 Gly Y1 = 13.9 Gy Y5 = 16.3/sqrt(.375136*5^3 + 1)= 2.35535 ≈ 2.355 Gy So we can say that for an s=5 galaxy, when the light we are now getting was emitted, the distance was expanding at a speed 4.798 Gly/2.355 Gy = 2.037 c, over twice the speed of light. The then and now speeds of expansion are given by: D5(then)/Y5 and D5(now)/Y1
P: 757
 Quote by marcus It looks as if the Hubbletime Ys corresponding to a given scaleup s should be given in billions of years by: 16.3(0.375136s3+1)-.5=16.3/sqrt(.375136*s^3 + 1)
It appears to me that your 'scaleup factor' (s = 1/a = z+1) is a useful one, since it does not go negative for future times, but just goes smaller than 1. I would caution against the terminology though, as it is too close to the conventional 'scalefactor' and may cause confusion. Maybe something like 'upscale ratio' or 'expansion ratio'? I would prefer 'upscale ratio of distances' rather than 'scaleup factor of wavelengths', so as to not also cause potential confusion with Doppler effects.

I'm working on a variant of the cosmo-calculator that will give a table for a range of z (or s?), with some useful values in the columns. Not quite there yet, but it looks practical.
 Astronomy Sci Advisor PF Gold P: 23,232 How about calling it "stretch factor"? Or "extension ratio"? Another idea, similar to something you suggested is to call the s number the "enlargement" because it is the ratio by which distances are enlarged during the time the light is on its way somewhat reminiscent of photographs being blown up. I'd welcome more suggestions. I'm glad to hear your new calculator is looking practical! I think the idea of generating tables for a range of z (or for s !!!) is a good one. Even fairly short tables with 5 to 10 lines can give someone extra perspective and intuition about how things are evolving. Just being able to compare two or three lines can be informative. After trying different things I do agree that the reciprocal scalefactor (whatever you call it and whether or not you subtract 1 from it) is the most useful handle on the situation.
 Astronomy Sci Advisor PF Gold P: 23,232 Here's one way to think about it: say we number the stages of expansion history according to how much distances have been enlarged since then. It means that earlier slices of spacetime have larger s numbers, which at first seems turned around, but in fact it's cleaner formula-wise to do the numbering backwards that way. To illustrate, suppose we are watching a galaxy as it was when distances were 1/4 of present size. We can denote that stage of expansion history by saying s = 4. While the light was on its way, distances and wavelengths have been enlarged by that factor. So that galaxy, as we see it, is in "slice 4" of expansion history. In that way of denoting stages of expansion, the present is s=1, because enlarging by a factor of 1 is the identity. In the simplified toy model, the expansion age ts associated with a stage s is given (in billions of years) by: $$t_s = \frac{16.3}{1.5}arctanh \left( \left( 0.375136 s^3 + 1\right)^{-1/2}\right)$$ And the corresponding Hubbletime at stage s, also in billions of years, is: $$Y_s = 16.3 \left( 0.375136 s^3 + 1\right)^{-1/2}$$ These are the two basic equations of the model--the other usual quantities such as distances and expansion speeds can be derived from these two. There is one peculiar thing to notice, which is that with expansion stages numbered this way, not only do we have the present tagged s = 1 but also future infinity is s = 0. So the eventual, or longterm value of the Hubbletime (a key parameter in the model) is: $$Y_0 = 16.3 \left( 0.375136 \times 0^3 + 1\right)^{-1/2} = 16.3 Gy$$ while the present Hubbletime is: $$Y_1 = 16.3 \left( 0.375136 \times 1^3 + 1\right)^{-1/2} = 13.9 Gy$$ I keep having to write this number 0.375136, which is kind of like a parameter of the system being the square ratio of our two Hubbletimes, less one. (Y0/Y1)2 - 1. So I will call that number capital Theta Θ. The two basic equations of the model are then: $$t_s = \frac{2}{3}Y_0 arctanh \left( \left( \Theta s^3 + 1\right)^{-1/2}\right)$$ $$Y_s = Y_0 \left( \Theta s^3 + 1\right)^{-1/2}$$ People who don't like greek letters should just remember it is a shorthand for an ordinary number ≈ 0.375 that essentially says something about the amount bigger current expansion rate is than the eventual longterm rate. (their ratio is about sqrt(1.375)
 Astronomy Sci Advisor PF Gold P: 23,232 I want to try out some terminology in part suggested to me by Jorrie's comments and which he might be puttng to use in another project.But I can try several ideas out, tentatively, in connection with this simple cosmic model. The main variable could be called the "stretch" because it is the factor by which distances from a past slice of spacetime are enlarged (and wavelengths too) between then and now. The idea is that if you start back to some earlier stage in expansion history and the enlargement of distances (and wavelengths) from then to now is a stretch factor of four (say S = 4) then the scale back then, relative to now is 1/4, or 0.25. So the stetch and scalefactor are reciprocals, like 4 and 1/4. It just turns out that the stretch is a convenient variable to run the model, or the calculator, on. You get the simplest formulas that way, of the various things I've tried. So I'm using S to stand for the stretch and the conventional letter a (= 1/S) to stand for the scale factor. The lineup of numerical information could (tentatively) go like this: Stretch---Scale factor---Expansion age---Hubble time---Distance now---Distance then and then again, this time showing the symbols that might be used to denote these quantities: Stretch (S=1/a)---Scale factor (a)---Expansion age (tS)---Hubble time (YS)---Distance now (DS[now])---Distance then (DS[then]) The idea is, we observe a galaxy and its light tells us the stretch factor S, say it is 4. Wavelengths 4 times what they were at the start of the trip. The galaxy is living back when distances were 1/4 present size. Then D4[now] tells us proper distance to the galaxy NOW, and D4[then] tells us distance back then, when light was emitted, from our matter (that became us) to the galaxy. If we want to know the SPEED of distance growth, you simply divide the distance by the Hubble time belonging to that slice. Back then when the distance was D4[then], it was growing at speed D4[then]/Y4. The present is denoted S=1 and the present Hubbletime is Y1 = 13.9 billion years. So the present distance is expanding at speed D4[now]/Y1. Today's distances, if you want to know what speed they are expanding, you just divide them by 13.9 billion years. So that's a provisional idea for a list of 6 related numbers that the model, or a calculator, can give you, that seems like enough to work with and get a picture of the expansion history from.
 Astronomy Sci Advisor PF Gold P: 23,232 I don't want to forget that simple model, although Jorrie has now put an excellent online tabulator on line. Here is the single-line formula calculator we were using, see post #434 http://web2.0calc.com Here is the calculator formula to compute the time given the stretch S: (16.3/1.5)atanh((.375136* S^3+1)^-.5) There's also a more complicated version for it given the scalefactor, but we probably won't use it. (16.3/1.5)atanh((((16.3/13.9)^2 -1)/a^3+1)^-.5) I just had a kind of exciting look into the future. I went to web2.0calc and put in exactly what I mentioned, namely (16.3/1.5)atanh((.375136* S^3+1)^-.5) And decided to see when distances would be 100 times what they are today which means scalefactor a=100 and reciprocal S = 1/a = 0.1 So I put 0.1 in place of S, in the formula and pressed = and it said that would happen in year 87.92 billion. So that is kind of cool. When expansion has been going on for about 88 billion years distances can be expected to be about 100 times what they are today. Let's try another. when will it be that distances are FIFTY times what they are today? Put in S=0.02 to the web2.0calc. Bingo. It says that will happen in year 76.625 billion. And just as a side comment with continuous compounding the Hubbletime 16.3 billion years corresponds to a doubling time of 11.3 billion years. That is the natural log(2) times 16.3. So it seems right that you go from scale 50 to scale 100 in something a little over 11 billion years.
 Astronomy Sci Advisor PF Gold P: 23,232 In post#434 I used a crude numerical integration of Sdt to find the distance now to a galaxy in the past in era S It turned out that when you rearrange an Sdt integration to make it easy to add up you get what LOOKS like a tdS integration (with extra terms at either end). this is just algebraic rearrangement. Then the steps can be of S rather than time and we can use the formula tS (16.3/1.5)*atanh((.375136* S^3+1)^-.5) which gives the time (expansion age in billions of years) when the reciprocal scalefactor (stretch) was a particular value S In the earlier post we had S advancing from 1 to S in steps of some smallish stepsize Δ. If we take Δ = 0.1, this amounts to: 1.05t1+ 0.1( t1.1+ t1.2+...+ ts-0.1) - (s-0.05)ts This is what the numerical integration for c∫ S dt boiled down to. The idea was that at each interval dt of time in the past, the light from an object travels a distance cdt and this gets scaled up by the appropriate factor S. So at present the distance to the object is the sum of all those scaled-up segments and equals c∫ S dt. ==================== So I decided to look into the future with the same technique and I found that if a galaxy is going to pass thru your forward lightcone at S=0.5, that is when distances are TWICE what they are today, then the distance NOW to it is 7.5 Gly. Where are the galaxies NOW which you could hit with a flash of light you send today and which arrives wavestretched to double length? They are 7.5 billion lightyears from here. It's like a time reverse image of the earlier game when we asked things about a galaxy whose light comes in wavestretched by a factor of 2, where was it when it emitted the light, where is it now etc. The numerical integration boiled down to: .55t.5+ 0.1( t.6+ t.7+...+ t.9) - .95t1.0 And I evaluated that and got 7.5 billion lightyears. So then we can say that when the signal we send arrives the distance to the target galaxy will be 15.0 billion light years. Because we know the expansion of scale between now and that time in the future.
 Astronomy Sci Advisor PF Gold P: 23,232 In the previous post I did a rough numerical integration based on toy model and got the estimate that a galaxy we can send a message to which will arrive when distances are TWICE today (namely an S=0.5 galaxy) is currently at distance 7.5 Gly and when the message arrives it will be at S=15 Gly. Now I can confirm that with the A20 calculator, which sees the shape of expansion history in the future as well as the past http://www.einsteins-theory-of-relat...oLean_A20.html I just make a small table running from present S=1 out to S=.1 in future, in steps of 0.1. So it covers the S=0.5 case but also gives me a little context (to help grow intuition/feel for the expansion process.) ===quote=== Hubble time now (Ynow) 13.9 Gy Change as desired (9 to 16 Gy) Hubble time at infinity (Yinf) 16.3 Gy Change as desired (larger than Ynow) Radiation and matter crossover (S_eq) 3350 Radiation influence (inverse: larger means less influence) Upper limit of Stretch range (S_upper) 1.0 S value at the top row of the table (equal or larger than 1) Lower limit of Stretch range (S_lower) 0.1 S value at the bottom row of table (S_lower smaller than S_upper) Step size (S_step) 0.1 Step size for output display (equal or larger than 0.01) Stretch (S) Scale (a) Time (Gy) T_Hubble (Gy) D_now (Gly) D_then (Gly) 1.000 1.000 13.756 13.900 0.000 0.000 0.900 1.111 15.250 14.444 -1.417 -1.575 0.800 1.250 16.981 14.929 -2.887 -3.608 0.700 1.429 19.004 15.342 -4.401 -6.287 0.600 1.667 21.396 15.677 -5.952 -9.921 0.500 2.000 24.279 15.930 -7.533 -15.066 0.400 2.500 27.856 16.108 -9.135 -22.839 0.300 3.333 32.507 16.218 -10.752 -35.840 0.200 5.000 39.097 16.275 -12.377 -61.886 0.100 10.000 50.388 16.297 -14.006 -140.059 ===endquote=== So the quick and dirty estimate I did earlier worked OK. For an S=.5 galaxy (where our message reaches when distances are TWICE) the present distance really is 7.5 and the distance then when message arrives really is 15 Gly. the minus signs have to do with the direction the light is going, from us to them. whereas in the past the distances have positive sign because the light is coming from them to us----itself a kind of nice feature. Also as an extra bonus the A20 tells me that the message that we send today (expansion age 13.75 billion years) will arrive when expansion age is 24.3 billion years. So it will take around 11 billion years to get there. That makes sense: when it arrives at destination the message will be 15 billion lightyears from us, and will have been traveling 11 billion years---you have to allow for some expansion of distances so naturally 15 > 11.
P: 757
 Quote by marcus One thing that could be highlighted, that we haven't discussed much here so far, is that if a massive particle or object is given a kick so that it has its own individual motion relative to the universe rest frame---the "Hubble flow"---CMB rest, it will gradually slow down relative to CMB rest and given enough time will REJOIN the "Hubble flow", or come approximately to a STOP relative to the ancient light. This is rather un-Newtonian and could be unintuitive to newcomers. It violates conventional conservation notions. But it is really basic to understanding so we should talk about it. It is analogous to the redshifting of light. the light loses energy and momentum as it travels across cosmological distances, although its speed doesn't change. With a massive object, the mass doesn't change but the speed does, so there is the same loss of energy and momentum.
Actually, there is a way in which the balloon analogy can make cosmic particle momentum decay intuitive. Simply consider a massive, frictionless particle that moves along the surface of the spherical balloon as a Kepler orbit around the center of a balloon. This particle must conserve angular momentum relative to the center of the balloon, i.e.

$L = r^2 m d\phi/dt$ = constant. Since $v = r d\phi/dt$, it means that for non-relativistic speeds, the particle speed scales with $1/r$.

If the balloon is being inflated, the particle must lose surface speed, just like a Kepler orbit that is losing orbital speed at larger radius. If the increase in balloon radius is kept up, the particle’s surface speed will eventually approach zero, as radius tends to infinity. In cosmology, this is usually described as 'joining the Hubble flow'.

The analogy seems to hold even for relativist particles. The relativistic Keplerian equation for the conservation of orbital angular momentum is:

$L = (1-v^2/c^2)^{-0.5} r^2 m d\phi/dt$ = constant (e.g. MTW eq. 25.18).

This simple scheme can be shown to reproduce the curves of figure 3.5 obtained by Davis (2004) [http://arxiv.org/abs/astro-ph/0402278] (with the exception of $v=c$).

I think that the $v=c$ case can also be handled by the analogy; the particle’s momentum must then be expressed in terms of the de Broglie wavelength.

Edit:
Relativistic de Broglie wavelength is given by: $\lambda=\gamma h/(m v)$, where $\gamma$ is the Lorentz factor.

If we write the angular momentum of the 'balloon particle' in terms of surface velocity, it is simply $L = \gamma r m v$. Taking $\gamma$ from the de Broglie wavelength, gives the conservation of angular momentum as

$L = r h / λ =$ constant, valid for photons and matter.
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 P: 26 I have no ability to understand a sphere with nothing off the surface. I'd rather have Plotinus's hypersphere. This is more or less a balloon analogy, with the spacetime universe on the surface as usual, but it has a centre and at least one extra dimension. Then we have the ability to include an extra dimension (or bundle of them) at each point in spacetime, since this would be represented as a connection to the centre point. Plotinus was not talking physics exactly, but his model seems more sophisticated than a 3D balloon representing a 2D spacetime. But I am way out my depths here.
Astronomy
PF Gold
P: 23,232
 Quote by PeterJ I have no ability to understand a sphere with nothing off the surface. I'd rather have Plotinus's hypersphere. This is more or less a balloon analogy, with the spacetime universe on the surface as usual, but it has a centre and at least one extra dimension. Then we have the ability to include an extra dimension (or bundle of them) at each point in spacetime, since this would be represented as a connection to the centre point. Plotinus was not talking physics exactly, but his model seems more sophisticated than a 3D balloon representing a 2D spacetime. But I am way out my depths here.
Instructive example! A mystic will postulate additional details (like the "edge" of the universe, or extra spatial dimensions) because they appeal to his imagination. Or even that they are "required" by his imagination.

the type of person we could call pragmatic or perhaps "Occamite" will avoid adding features which lack an operational meaning---i.e. some way to experience, even if very tenuous or indirect.

I would say try to think of the EXPERIENCE of being 2D and living in a 2D sphere. Don't picture the sphere as if you are a God, outside and looking from outside at the sphere. Using some new type of lightrays that travel in 3D rather than 2D. Think of a sphere as the experience of living in it. And also think of a hypersphere that way.

Let's say that you and your brothers discover a remarkable fact about the world namely that there is a special area K which you have determined experimentally which allows you to reliably predict the area of any triangle!
You just have to sum the angles, subtract π, and multiply by that area K!
this always turns out to give the area (if you take the trouble to measure the area carefully.

The rule used by Euclid, namely 1/2 the base times the height does not work for you, it is only approximately right for small area triangles and gets progressively wronger for larger ones.

That's part of what I mean by the experience. It would apply also to living in a hypersphere. It does not involve postulating an extra dimension which we don't experience and cannot access. It just involves experimenting with triangles and determining the value of the area K.

Circumnavigating is another aspect of the experience which you (as creature living in sphere or hypersphere) might have. You can think of various others.
 P: 26 No it's okay. I'm happy with my way of thinking about it. I'd say that the topography of the universe is unrepresentable as a visual model, and so within quite wide limits it would be a matter of personal preference how we do it. A Klein bottle or Necker cube would also be relevant images. I'm not sure what you mean about 'mystics' and the stuff that appeals to their imagination. It has nothing to do with imagination. If Plotinus is to believed he is trying to describe what he is seeing, and he was not seeing any edges, nor any inside or outside. Perhaps he is not to be believed, but his model does at least allow for the idea that distance is arbitrary, which seems to make it useful, and it's only one more dimension, making it more economical than string theory. He even adds the proviso 'it is as if'. I'm not suggesting that his description is 'true', just something to consider.
 Astronomy Sci Advisor PF Gold P: 23,232 New narrowed-down values of the cosmological parameters, coming out of the SPT (south pole telescope) http://arxiv.org/pdf/1210.7231v1.pdf Scroll to Table 3 on page 12 and look at the rightmost column which combines the most data: ΩΛ 0.7152 ± 0.0098 H0 69.62 ± 0.79 σ8 0.823 ± 0.015 zEQ 3301 ± 47 Perhaps the most remarkable thing is the tilt towards positive overall curvature, corresponding to a negative value of Ωk For that, see equation (21) on page 14 Ωk =−0.0059±0.0040. Basically they are saying that with high probability you are looking at a spatial finite slight positive curvature. The flattest it could be IOW is 0.0019, with Ωtotal = 1.0019 And a radius of curvature 14/sqrt(.0019) ≈ 320 billion LY. Plus they are saying Omega total COULD be as high as 1.0099 which would mean radius of curvature 14/sqrt(.0099) ≈ 140 billion LY. So the idea which is traditionally favored of perfect flatness and spatial infinite is hanging on by its 2 sigma fingernails. It is still "consistent" with the data at a 2 sigma level. But the hypersphere ( abbreviated S3 for the 3D analog of the 2D surface of a ball ) is looming realer and realer as a kind of ignored elephant in the room. It could still go away of course. We avert our eyes and hope it will have the politeness to do so. For Jorrie's A25 calculator the important parameters as estimated by the SPT report are current Hubble time = 14.0 billion years future Hubble time = 16.6 billion years matter radiation balance Seq = 3300 or with more precision put these into Google calculator: 1/(69.62 km/s per Mpc) 1/(69.62 km/s per Mpc)/.7152^.5
 P: 6 On the subject of space expansion.... If I built a spaceship with a machine (an engine if you will) on the tail end of the ship that eliminated all effects of gravity at the tail end of the ship , so as to counter act any gravitational pull on the tail end of the ship, would my spaceship thus be capable of traveling at light speed as I ride the expansion of space in what ever direction my ship is pointed? Or can I only travel away from the center of the universe? Would I instantly be moving at the same speed as the space is expanding the moment I turn on my spaceship, but feel no acceleration?
Astronomy
PF Gold
P: 23,232
 Quote by hagendaz On the subject of space expansion.... If I built a spaceship with a machine (an engine if you will) on the tail end of the ship that eliminated all effects of gravity at the tail end of the ship , so as to counter act any gravitational pull on the tail end of the ship, would my spaceship thus be capable of traveling at light speed as I ride the expansion of space in what ever direction my ship is pointed? Or can I only travel away from the center of the universe? Would I instantly be moving at the same speed as the space is expanding the moment I turn on my spaceship, but feel no acceleration?
The expansion of distances doesn't GO anywhere. No person or object approaches any destination. Simply put: Things that aren't held together by their own gravity or molecular forces just get farther apart.

There is no "center" that anyone can point to.

So you cannot "ride" the expansion of space in any direction. Since there is no center you cannot " travel away from the center" as you say, either.

Typical very largescale distances grow several times faster than the speed of light but nothing travels anywhere.

The balloon analogy is intended to illustrate those things to make them easy to visualize. You might try studying the brief animation movie of it, reading some of this thread, or the FAQs.

You could start your own thread with this question, since it does not fit in so well in this balloon analogy thread.

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