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 International Skeptics Forum Merged: Why the James Webb Telescope rewrites/doesn't the laws of Physics/Redshifts

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 16th November 2022, 08:09 AM #241 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by steenkh There is nothing new in the question of the Hubble constant. Is the paper offering tired light as a solution? This one? https://arxiv.org/abs/2211.04492 Far from it. But as far as nothing new, I can predict lookback times using a Hubble's constant that is actually constant.
 16th November 2022, 08:35 AM #242 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger I was unsure about the highlighted part, and it bothered me that Cappi placed equation (7) after equation (6) when he was so careful to place all of the other equations that you needed before performing the calculation implied by (6) before equation (6). I have now figured out how to calculate the lookback distance according to equation (6) without needing equation (7), and will explain in a few days when I have time to write it up. That's an interesting observation. Comparing the paper to his JavaScript implementation you can see that lookback time somewhat stands alone: Code: ``` // Lookback time lookback = timeint(1., z1) * Tnorm // Angular distance DA = DL / (z1 * z1) // Lunghezza in primi corrispondente a 1 Mpc. angle = (60. / rad) * z1 * z1 / DL; // Length in Mpc corresponding to 1 degree on the sky R = rad * DL / (z1 * z1);``` I look forward to your write up.
 16th November 2022, 10:33 AM #243 jonesdave116 Philosopher     Join Date: Feb 2015 Posts: 5,601 Originally Posted by Mike Helland Of course it does. That's why we're talking about it. Nope, you are talking about a way of describing redshift to explain the evidence it provides for accelerated expansion. The ISW and BAO observations have nothing to do with redshift. So, forget the redshift, you need to deal with them as a minimum before even bothering with redshift. __________________ “There is in every village a torch - the teacher; and an extinguisher - the priest.” - Victor Hugo “Never argue with an idiot. They will only bring you down to their level and beat you with experience.” - George Carlin
 16th November 2022, 11:00 AM #244 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by jonesdave116 Nope, you are talking about a way of describing redshift to explain the evidence it provides for accelerated expansion. The discrepancy between expecting a matter dominated model (omegaM=1) and what was observed was significant. Do you disagree?
 16th November 2022, 09:05 PM #245 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland The discrepancy between expecting a matter dominated model (omegaM=1) and what was observed was significant. Do you disagree? One would hope to see a discrepancy, since ΩM=1 is not at all consistent with mainstream cosmology's estimates of that parameter. Are you telling us that Helland physics expects ΩM=1? Or does Helland physics expect ΩM=0, as you told us you assumed when deriving your Helland equation for a relationship between red shift and distance?
 16th November 2022, 09:38 PM #246 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger One would hope to see a discrepancy, since ΩM=1 is not at all consistent with mainstream cosmology's estimates of that parameter. Are you telling us that Helland physics expects ΩM=1? Or does Helland physics expect ΩM=0, as you told us you assumed when deriving your Helland equation for a relationship between red shift and distance? Mainstream is that its about 30% matter and 70% dark energy. Do you disagree?
 17th November 2022, 10:50 AM #247 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by W.D.Clinger Mike Helland's equation for calculating distance from red shift does not involve any of the FLRW parameters, which was an extremely strong clue that the Helland equation is not based upon mainstream physics. In a recent post, however, Mike Helland has said his equation assumes a universe devoid of matter, which establishes the value Mike Helland's equation assumes for one of the FLRW parameters.... Helland's equation isd = (z/(1+z)) c/H0 Originally Posted by W.D.Clinger (For some reason, Mike Helland stubbornly persists in writing a more complicated equivalent of that equation, perhaps because the more complicated way of writing the equation makes it harder to compare against mainstream equations such as Cappi's.) Originally Posted by W.D.Clinger If you are assuming matter doesn't affect redshifts, then you are assuming Helland physics rather than mainstream physics. That is part of why your Helland equation for the relationship between red shift and distance is far less complicated than the mainstream equations for that relationship. (The other reason your equation is less complicated is that it either doesn't take model parameters into account, or it assumes those parameters have whatever fixed values you think they have in Helland physics.) Originally Posted by Mike Helland My equation is derived from inverting redshift formulas. That's all. It could be that it matches a universe where matter doesn't affect redshift is just a bizarre coincidence. Originally Posted by Mike Helland I can predict lookback times using a Hubble's constant that is actually constant. Originally Posted by Mike Helland The discrepancy between expecting a matter dominated model (omegaM=1) and what was observed was significant. Do you disagree? Originally Posted by W.D.Clinger One would hope to see a discrepancy, since ΩM=1 is not at all consistent with mainstream cosmology's estimates of that parameter. Are you telling us that Helland physics expects ΩM=1? Or does Helland physics expect ΩM=0, as you told us you assumed when deriving your Helland equation for a relationship between red shift and distance? Originally Posted by Mike Helland Mainstream is that its about 30% matter and 70% dark energy. Do you disagree? Why don't you quit dancing around and just tell us the values of the parameters that are wired into your Helland equation? You have insisted throughout that your Helland equation is based upon mainstream physics. But the mainstream equations that relate distance to red shift are considerably more complicated than your Helland equation, partly because that relationship is model-dependent. Unlike mainstream equations, your Helland equation mentions no model parameters apart from H0. That means your Helland equation implicitly assumes certain values for the other model parameters on which mainstream equations depend. What parameter values did you assume when you formulated your Helland equation?
 17th November 2022, 11:26 AM #248 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger Why don't you quit dancing around and just tell us the values of the parameters that are wired into your Helland equation? You have insisted throughout that your Helland equation is based upon mainstream physics. But the mainstream equations that relate distance to red shift are considerably more complicated than your Helland equation, partly because that relationship is model-dependent. Unlike mainstream equations, your Helland equation mentions no model parameters apart from H0. That means your Helland equation implicitly assumes certain values for the other model parameters on which mainstream equations depend. What parameter values did you assume when you formulated your Helland equation? The equation is derived by simply inverting the redshift equations. Apparently no one else thought of that. You know what the values of the cosmological parameters are.
 17th November 2022, 06:07 PM #249 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland The equation is derived by simply inverting the redshift equations. Apparently no one else thought of that. So you believez = d / ((c/H0) - d)is a mainstream physics equation? Citation needed! According to your Helland equationd = (z / (1+z)) c/H0a red shift of z=1 corresponds to a distance of about 7 billion light years. According to mainstream physics, a red shift of z=1 corresponds to a proper distance of about 10 billion light years, or 7.7 billion light years if you naively multiply the lookback time by the speed of light. Even with the most charitable interpretation of your Helland equation, it's about 10% off from the best estimates of mainstream physics.
 18th November 2022, 11:17 AM #250 jonesdave116 Philosopher     Join Date: Feb 2015 Posts: 5,601 Originally Posted by Mike Helland The discrepancy between expecting a matter dominated model (omegaM=1) and what was observed was significant. Do you disagree? Sorry? The EVIDENCE says the universe is accelerating in its expansion. And those observations do not all include redshift. As mentioned, the ISW and BAO observations strongly support an accelerated expansion. The former, at least, was predicted. So, faffing around with redshift equations is not going to make those predicted observations disappear, is it? Just as the CMB is not going to disappear, regardless of crank claims about what JWST has shown. __________________ “There is in every village a torch - the teacher; and an extinguisher - the priest.” - Victor Hugo “Never argue with an idiot. They will only bring you down to their level and beat you with experience.” - George Carlin
 18th November 2022, 12:26 PM #251 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger So you believez = d / ((c/H0) - d)is a mainstream physics equation? Citation needed! No. This is a mainstream physics equation 1+z=Eemit/Eobs. I just inverted it, so the range of redshifts is a percentage, which is what that integral is doing anyways. Quote: According to your Helland equationd = (z / (1+z)) c/H0a red shift of z=1 corresponds to a distance of about 7 billion light years. According to mainstream physics, a red shift of z=1 corresponds to a proper distance of about 10 billion light years, or 7.7 billion light years if you naively multiply the lookback time by the speed of light. Even with the most charitable interpretation of your Helland equation, it's about 10% off from the best estimates of mainstream physics. And according to LCDM, H0 should be 68, but we measure it to be 74. Mainstream physics is famously in tension with observation.
 18th November 2022, 02:13 PM #252 jonesdave116 Philosopher     Join Date: Feb 2015 Posts: 5,601 Originally Posted by Mike Helland No. This is a mainstream physics equation 1+z=Eemit/Eobs. Nope, that is an approximation, and only valid at small z. In other words <<1. Quote: And according to LCDM, H0 should be 68, but we measure it to be 74. Mainstream physics is famously in tension with observation. No, we have varying measurements of it, depending on whether the measurement are 'close' or distant. Which could imply new physics. Yippee! What it does not imply is that the universe is not expanding. Use whatever value you can find, and it leads to the same conclusion. People who want static universes need not only to overturn the redshift evidence, they need to overturn, and explain at least as well, the other observations, which show that we are in a universe which is accelerating in its expansion. Nobody can. Which is why nobody is even bothering anymore. __________________ “There is in every village a torch - the teacher; and an extinguisher - the priest.” - Victor Hugo “Never argue with an idiot. They will only bring you down to their level and beat you with experience.” - George Carlin
 18th November 2022, 04:43 PM #253 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by jonesdave116 Nope, that is an approximation, and only valid at small z. In other words <<1. Nope, that is how redshift z is quantified. Your thinking of the redshift-distance relationship. If you try to use with d = cz / H0, you will find this equation to be an approximation. However, by inverting the redshift function, you are multiplying c/H0 by a percentage (rather than to infinity). Which is what the lookback integral is doing too.
 18th November 2022, 06:58 PM #254 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger One would hope to see a discrepancy, since ΩM=1 is not at all consistent with mainstream cosmology's estimates of that parameter. Are you telling us that Helland physics expects ΩM=1? Or does Helland physics expect ΩM=0, as you told us you assumed when deriving your Helland equation for a relationship between red shift and distance? Prior to 1998, it was thought that ΩM=1. Since that was younger than the oldest stars, that couldn't be right. Using supernovae data they figured there needs to be the right amount of dark energy to negate the effects of matter. That ends up being ΩM=~0.3 and ΩΛ=~0.7. Redshifts in Helland physics (your name) aren't affected by gravity, so dark energy isn't necessary to counteract its affects.
 18th November 2022, 08:08 PM #255 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by W.D.Clinger Helland's equation isd = (z/(1+z)) c/H0 Originally Posted by Mike Helland The equation is derived by simply inverting the redshift equations. Apparently no one else thought of that. Originally Posted by W.D.Clinger So you believez = d / ((c/H0) - d)is a mainstream physics equation? Citation needed! Originally Posted by Mike Helland No. This is a mainstream physics equation 1+z=Eemit/Eobs. I just inverted it, so the range of redshifts is a percentage, which is what that integral is doing anyways. No. That "1+z=Eemit/Eobs" does not mention d at all, so it cannot be inverted to obtain an equation for d. The only equation that could possibly be inverted to obtain your Helland equationd = (z/(1+z)) c/H0isz = d / ((c/H0) - d)or some algebraic equivalent of that equation. Unsurprisingly, you have been unable to find a citation for any such equation in the mainstream literature. That refutes your claim that your Helland equation was derived by inverting a mainstream equation. Originally Posted by Mike Helland Prior to 1998, it was thought that ΩM=1. Since that was younger than the oldest stars, that couldn't be right. It takes a special level of cluelessness to think ΩM=1, taken in isolation, tells you anything about the age of stars. To infer anything about the age of stars, you have to combine that parameter with other observations and with the details of a mathematical model. I'm away from my library at the moment, but when I get back I will give you citations showing that mainstream cosmology of the 1970s was already estimating ΩM to be considerably less than 1. ETA: This spoiler contains the promised citations and quotations, obtained from an online search instead of my library. In 1976, P J E Peebles used A Cosmic Virial Theorem to estimate0.05 < ΩM < 0.5without proclaiming a great deal of confidence in that estimate. In 1983, that estimate was refined to ΩM ~ 0.2 plus or minus 0.1, which became the standard estimate until the modern mainstream estimate of ΩM ~ 0.3 became even more definitively established some 25 years ago (in 1998, as Mike Helland observed). As explained by James G. Bartlett and Alain Blanchard in 1995: Quote: Since its inception (Peebles 1976a,b), the so-called 'cosmic virial theorem' (CVT) has been considered one of the most reliable indicators of the cosmic mean matter density, and for good reason.... The theorem...was conscripted immediately (Peebles 1976a,b), but not with convincing results until the completion of the first CfA redshift survey (Davis & Peebles 1983)...Davis and Peebles concluded that Ω ~ 0.2....Such a low value of the mean density affronts the theorist's preference for a flat Universe.... Originally Posted by Mike Helland Redshifts in Helland physics (your name) aren't affected by gravity, so dark energy isn't necessary to counteract its affects. In mainstream physics, redshifts are affected by gravity, and the mainstream equations for estimating distance from redshift therefore have to take gravity into account. That is yet another proof that your mainstream equation for distance as a function of red shift is incompatible with mainstream physics. Last edited by W.D.Clinger; 18th November 2022 at 09:27 PM. Reason: added ETA and its spoiler
 19th November 2022, 01:59 AM #256 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger No. That "1+z=Eemit/Eobs" does not mention d at all I never said it did. Quote: The only equation that could possibly be inverted to obtain your Helland equationd = (z/(1+z)) c/H0isz = d / ((c/H0) - d)or some algebraic equivalent of that equation. Unsurprisingly, you have been unable to find a citation for any such equation in the mainstream literature. That refutes your claim that your Helland equation was derived by inverting a mainstream equation. You're confusing equations for redshift and equations for a redshift-distance relationship. My equation for blueshift is just the equation for redshift inverted. My equation for blueshift-distance relatoinship is a conjecture. If d = zc/H0 was what was initially thought to be redshift-distance relationship, but we discovered it was inaccurate over most of its domain, then I conjectured d = -bc/H0. It's not easy to compare two equations that have different variables, but since 1+b=1/(1+z), then b=1/(1+z)-1. Quote: It takes a special level of cluelessness to think ΩM=1, taken in isolation, tells you anything about the age of stars. Here's what Nobel prize winner Riess says about his discovery: "Coupled with the theoretical expectation of Ω m ∼ 1, the low expansion age implied by these measurements, still a few Gyr younger than the oldest stars, set off another “Hubble tension” until the discovery of cosmic acceleration amended the composition and recent expansion history and the age of the Universe grew comfortably higher." Last edited by Mike Helland; 19th November 2022 at 03:08 AM.
 19th November 2022, 04:34 AM #258 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger What you did say is that you obtained your Helland equation involving d by inverting that equation, which does not even mention d. False. You keep saying "Helland equation" but you keep confusing two different things. 1 + z = Eemit / Eobs That's a mainstream equation for redshift. Invert it. What do you have? 1/(1 + z) = Eobs / Eemit You can call that the Helland blueshift equation if you'd like. 1+b =1/(1+z) That's how the two values are related. Now I find this interesting, because while the range of z for valid redshifts is 0
 19th November 2022, 05:13 AM #259 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland Let's look at the lookback time integral. That's Cappi's equation (7). Originally Posted by Mike Helland t(z)=1/H0 * an integral As z goes to infinity, t approaches 1/H0. It will never get larger than that. So the integral must go from 0 to a limit of 1 (rather than infinity). I could just do this t(z)=1/H0 * -b. Yes, someone who is truly ignorant of calculus might do that, thinking your Helland equation for t(z) is consistent with Cappi's equation (7). But your Helland equation for t(z) isn't consistent with Cappi's equation (7). Guessing that the integral is always equal to -b is not at all the same as computing the value of the integral. By the way, the value of the integral in Cappi's equation (7) depends upon several model parameters, including ΩM, ΩR, and ΩΛ. Your Helland equation for t(z) doesn't mention any of those parameters, so it is freaking obvious that your Helland equation for t(z) is not consistent with Cappi's equation (7). Originally Posted by Mike Helland You can call that the blueshift-lookback time relationship if you want. If you want, you can call it the Helland equation for the blueshift-lookback time relationship. But you cannot truthfully say your Helland equation for whatever you want to call it is consistent with mainstream equations such as Cappi's equation (7). Originally Posted by Mike Helland I can also propose d = -bc / H0, since d=ct. You are the ultimate authority on Helland physics, so you can propose whatever idiocies you like. But you cannot truthfully say your proposed idiocies are consistent with mainstream physics. Last edited by W.D.Clinger; 19th November 2022 at 05:23 AM. Reason: corrected tags so Cappi's equation will be displayed
 19th November 2022, 05:25 AM #260 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger But you cannot truthfully say your Helland equation for whatever you want to call it is consistent with mainstream equations such as Cappi's equation (7). You've dubbed one of my equations as the Helland equation. And then another one of my equations as the Helland equation. And then you talk about them interchangeably. I only derived 1/(1+z)=Eobs / Eemit. I derived the blueshift equation by inverting a mainstream redshift equation. Those equations are not distance or time relationships. Those I invented. I modified the broke d=cz/H0 by replacing z with -b: d=-bc/H0. That wasn't derived. That was just conjectured. So the different equations came about in different ways. Quote: You are the ultimate authority on Helland physics, so you can propose whatever idiocies you like. Stop being so rude. Quote: But you cannot truthfully say your proposed idiocies are consistent with mainstream physics. How many times are you going to insult me before someone removes your posts? I didn't say all my equations are consistent with mainstream physics.
 19th November 2022, 05:46 AM #261 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland I only derived 1/(1+z)=Eobs / Eemit. I derived the blueshift equation by inverting a mainstream redshift equation. Those equations are not distance or time relationships. Those I invented. I modified the broke d=cz/H0 by replacing z with -b: d=-bc/H0. That wasn't derived. That was just conjectured. So the different equations came about in different ways. Thank you. So far as I know, this is the first time you have actually admitted that yourd=(z/(1+z))(c/H0)equation was your own invention/conjecture. Originally Posted by Mike Helland I didn't say all my equations are consistent with mainstream physics. The d=(z/(1+z))(c/H0) equation you invented/conjectured was the equation you were using to calculate the distances that, according to you, were a problem for mainstream physics. If there was a problem with the distances you calculated using an equation you invented/conjectured, that is a problem with Helland physics, not a problem for mainstream physics. To suggest that the failures of Helland physics are a problem for mainstream physics is dishonest.
 19th November 2022, 05:53 AM #262 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger Thank you. So far as I know, this is the first time you have actually admitted that yourd=(z/(1+z))(c/H0)equation was your own invention/conjecture. I said it straight away. http://www.internationalskeptics.com...&postcount=126 Quote: We see now the situation has reversed. Not only does energy and frequency rise when blueshift increases, but all the redshifting happens when -1 < b < 0. All of z > 0 now fits into -1 < b < 0. Does that affect anything? Distance The z-distance relation is: d = zc / H0 z << 1 This is only considered valid for very small values of z. At z = 10, the distance would be 140 billion light years. And since z can go up to infinity, the sky is the limit. How about when redshifts are represented as negative blueshifts (-b) instead of z? Perhaps: d = -bc / H0 I ask, what can we do with -b? I conjectured a distance relation with it.
 19th November 2022, 06:19 AM #263 jonesdave116 Philosopher     Join Date: Feb 2015 Posts: 5,601 Originally Posted by Mike Helland Nope, that is how redshift z is quantified. Your thinking of the redshift-distance relationship. If you try to use with d = cz / H0, you will find this equation to be an approximation. However, by inverting the redshift function, you are multiplying c/H0 by a percentage (rather than to infinity). Which is what the lookback integral is doing too. Gibberish. __________________ “There is in every village a torch - the teacher; and an extinguisher - the priest.” - Victor Hugo “Never argue with an idiot. They will only bring you down to their level and beat you with experience.” - George Carlin
 19th November 2022, 06:52 AM #264 jonesdave116 Philosopher     Join Date: Feb 2015 Posts: 5,601 Originally Posted by Mike Helland No. This is a mainstream physics equation 1+z=Eemit/Eobs. Really? Link to where you found this please. I think you'll find that you have got that bass ackwards. You end up with negative numbers. Try; z = Eobs/Eemit -1. __________________ “There is in every village a torch - the teacher; and an extinguisher - the priest.” - Victor Hugo “Never argue with an idiot. They will only bring you down to their level and beat you with experience.” - George Carlin
 19th November 2022, 07:13 AM #265 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by jonesdave116 Really? Link to where you found this please. I think you'll find that you have got that bass ackwards. You end up with negative numbers. https://en.wikipedia.org/wiki/Redshift If the photon is emitted with 1 eV, and observed with 0.5 eV, 1/0.5-1=1. You would only get negative numbers of emit < observe, but that's blueshift. Quote: Try; z = Eobs/Eemit -1. That's the inverted formula, which I've been calling b for blueshift to keep separate. b = Eobs/Eemit -1 If the photon is emitted with 1 eV, and observed with 0.5 eV, 0.5/1-1=-0.5. So z=1 is equal to b=-0.5. 1+b=1/(1+z) A CMB photon has a z=1100 and b=-1100/1101=-0.999091. CMB lookback time is -b/H0. Last edited by Mike Helland; 19th November 2022 at 07:36 AM. Reason: z=2 to z=1
 19th November 2022, 09:37 AM #269 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger And by the way, I suspect the calculators you've been using will automatically calculate that curvature parameter from the values you specify for ΩM, ΩR, ΩΛ, and H0. I'm doing this by hand. The calculator code I'm using is Cappi's, referenced in the first paragraph: "This short introduction is extracted from a paper on cosmological models with a dark energy component (Cappi 2001). For an on–line implementation of cosmological formulae see www.bo.astro.it/∼cappi/cosmotools." Code: `Omega_k = 1. - Omega_M - Omega_L;` Last edited by Mike Helland; 19th November 2022 at 09:40 AM.
 21st November 2022, 07:41 AM #270 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger I'll check my calculation later when I have more time. The integral is greatly simplified by assuming ΩM = ΩR = ΩΛ = 0, but I might have made a mistake while tracing the E(z) of equation (7) back through its defining equation (4) and the definitions preceding equation (4). In particular, I need to check whether the other density parameters really imply Ωk=1 using Cappi's definitions. As you can imagine, I haven't spent a whole lot of time contemplating the quantitative relationship between redshift and distance in a completely empty universe that, even so, has enough curvature to achieve the critical density. Does it though? This calculator shows that k<0 (while Ωk=1). http://icosmos.co.uk/index.html
 21st November 2022, 09:16 PM #271 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland Originally Posted by W.D.Clinger I'll check my calculation later when I have more time. The integral is greatly simplified by assuming ΩM = ΩR = ΩΛ = 0, but I might have made a mistake while tracing the E(z) of equation (7) back through its defining equation (4) and the definitions preceding equation (4). In particular, I need to check whether the other density parameters really imply Ωk=1 using Cappi's definitions. As you can imagine, I haven't spent a whole lot of time contemplating the quantitative relationship between redshift and distance in a completely empty universe that, even so, has enough curvature to achieve the critical density. Does it though? This calculator shows that k<0 (while Ωk=1). http://icosmos.co.uk/index.html k < 0 indicates negative curvature, and Ωk=1 indicates a great deal of curvature. As for the numerical value of k, there are two different conventions. I haven't looked at that specific calculator, but k is often taken to be a discrete parameter constrained to have one of these three values:k=0 means flat space* k=-1 means space has negative curvature k=+1 means space has positive curvature That is the convention used in Cappi's paper. With that convention, the scale factor a(t) is the radius of curvature, and is often written R(t) instead of a(t). I normally use the other common convention, in which a(t) is a dimensionless scale factor expressed as a fraction of the present-day scale, and k is the Gaussian radius of spatial curvature at the present day. The fact that Cappi's convention is different from the one I tend to use might cause confusion if you're trying to compare my notes on the Friedmann equations for flat space with Cappi's. *In many of my recent posts, I have written "flat spacetime" when I meant "flat space". Sorry about that.
 21st November 2022, 09:51 PM #272 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger k < 0 indicates negative curvature, and Ωk=1 indicates a great deal of curvature. The first part is true. The second part isn't. Ω indicates ratios in cosmology.
 22nd November 2022, 06:28 AM #273 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland Originally Posted by W.D.Clinger k < 0 indicates negative curvature, and Ωk=1 indicates a great deal of curvature. The first part is true. The second part is true as well. Originally Posted by Mike Helland The second part isn't. Ω indicates ratios in cosmology. Ωk is the curvature parameter, defined as the fraction of the critical density by which the sum of the matter, radiation, and dark energy parameters falls short of the critical density. When those other density parameters add up to 1, Ωk=0 and there is no curvature. When all of those other critical densities are 0, Ωk=1, implying a great deal of curvature. Taking H0=70 km/s/Mpc, Ωk=1 implies the radius of curvature is approximately 140 billion light years. With k=-1, that's a negative curvature, meaning space does not close upon itself, but opens upon itself. Ωk=1 is hundreds of times more curvature than would be consistent with the Planck mission's observations: Originally Posted by Wikipedia Final results of the Planck mission, released in 2018 show the cosmological curvature parameter, 1 – Ω = ΩK = –K c˛/a˛H˛, to be 0.0007±0.0019, consistent with a flat universe.[18] (i.e. positive curvature: K = +1, Ωκ < 0, Ω > 1, negative curvature: K = −1, Ωκ > 0, Ω < 1, zero curvature: K = 0, Ωκ = 0, Ω = 1).
 22nd November 2022, 07:10 AM #274 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger Ωk is the curvature parameter, defined as the fraction of the critical density by which the sum of the matter, radiation, and dark energy parameters falls short of the critical density. Right, so when ΩK=1, then it falls short of the critical density by 100%. You said: Quote: As you can imagine, I haven't spent a whole lot of time contemplating the quantitative relationship between redshift and distance in a completely empty universe that, even so, has enough curvature to achieve the critical density. So we agree it doesn't achieve critical density? Quote: 1 – Ω = ΩK http://hyperphysics.phy-astr.gsu.edu...denpar.html#c1 "It is customary to express the density as a fraction of the density required for the critical condition with the parameter Ω = ρ/ρcritical so that Ω = 1 represents the condition of critical density. " p = 0, so Ω = 0, and ΩK = 1.
 22nd November 2022, 09:09 AM #275 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland Originally Posted by W.D.Clinger Ωk is the curvature parameter, defined as the fraction of the critical density by which the sum of the matter, radiation, and dark energy parameters falls short of the critical density. Right, so when ΩK=1, then it falls short of the critical density by 100%. Right, assuming the antecedent of the highlighted pronoun is the sum of the matter, radiation, and dark energy parameters. Originally Posted by Mike Helland You said: Quote: As you can imagine, I haven't spent a whole lot of time contemplating the quantitative relationship between redshift and distance in a completely empty universe that, even so, has enough curvature to achieve the critical density. So we agree it doesn't achieve critical density? Right, assuming the highlighted pronoun refers to the sum of the matter, radiation, and dark energy parameters. My highlighted phrase was a poor joke, intended to highlight your mistake in this post: Originally Posted by Mike Helland Originally Posted by W.D.Clinger And then, three days later, you claimed that equation calculates results that are "exactly equal to LCDM in a default state": Right. Default as in before you add stuff to it, like matter, dark matter, and dark energy, and give it curvature. The highlighted phrase told me that, when you wrote that, you thought the calculator you were using defaulted to parameters corresponding to a completely empty universe, devoid of matter and devoid of energy, with no curvature. You were unaware that your calculator automatically supplied enough curvature to make up for the failure of the ordinary density parameters to add up to the critical density.
 22nd November 2022, 09:14 AM #276 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger Right, assuming the antecedent of the highlighted pronoun is the sum of the matter, radiation, and dark energy parameters. Right, assuming the highlighted pronoun refers to the sum of the matter, radiation, and dark energy parameters. My highlighted phrase was a poor joke, intended to highlight your mistake in this post: The highlighted phrase told me that, when you wrote that, you thought the calculator you were using defaulted to parameters corresponding to a completely empty universe, devoid of matter and devoid of energy, with no curvature. You were unaware that your calculator automatically supplied enough curvature to make up for the failure of the ordinary density parameters to add up to the critical density. Ha, I see. Yeah, it has negative curvature by default. On a related note, I have a lot of money... if you count having negative money as a lot.
 22nd November 2022, 09:28 AM #277 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland Ha, I see. Yeah, it has negative curvature by default. On a related note, I have a lot of money... if you count having negative money as a lot. You didn't really have to convince us you don't understand the concept of curvature in geometry, but I agree about your remarks being good for a laugh. Originally Posted by Wikipedia For example, a sphere of radius r has Gaussian curvature 1/r2 everywhere, and a flat plane and a cylinder have Gaussian curvature zero everywhere. The Gaussian curvature can also be negative, as in the case of a hyperboloid or the inside of a torus. ETA: Here's a picture. Last edited by W.D.Clinger; 22nd November 2022 at 09:30 AM. Reason: added ETA
 22nd November 2022, 12:11 PM #278 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger You didn't really have to convince us you don't understand the concept of curvature in geometry, but I agree about your remarks being good for a laugh. ETA: Here's a picture. So, an empty FLRW universe only expands, and has an open geometry. Fill it up with just the right amount of matter (and/or radiation) the expansion and gravity will be in balance, with is flat. And if it has more matter than that, gravity will overpower expansion and it is closed. But you could add repulsive dark energy to push it back to flat or open. So did you figure out how that lookback integral reduces? H0 or 2H0.
 22nd November 2022, 01:22 PM #279 W.D.Clinger Illuminator     Join Date: Oct 2009 Posts: 4,731 Originally Posted by Mike Helland So, an empty FLRW universe only expands, and has an open geometry. Not exactly. You apparently do not understand that the FLRW models are an entire family of mathematical models. One of the FLRW models is for an empty universe with flat space (zero curvature). That's because the general theory of relativity reduces to special relativity. Because you were playing around with a calculator whose models and parameters you did not understand, you told it to assume a completely empty universe with a positive Hubble constant. (I don't know whether you specified the Hubble constant yourself or let the calculator choose a default positive value for the Hubble constant, but I also don't much care how you screwed up.) With those assumptions, special relativity doesn't apply because of the positive Hubble constant. To accommodate the Hubble constant you specified (whether explicitly or implicitly), the calculator had to assume a lot of negative curvature to account for how you could have that Hubble constant despite empty space. Originally Posted by Mike Helland Fill it up with just the right amount of matter (and/or radiation) the expansion and gravity will be in balance, with is flat. If the matter, radiation, and dark energy densities add up to the critical density, then space will be flat. Note, however, that flat space is entirely consistent with expansion, even with an accelerating expansion, and that is the conclusion to which we are led by observations that imply realistic values for the density parameters. ETA: By the way, the distinction between flat and curved is not the same as the distinction between open and closed. Originally Posted by Mike Helland And if it has more matter than that, gravity will overpower expansion and it is closed. Well, if it has more matter etc than the critical density, then spacetime is closed in the sense that gravity will eventually overpower expansion, leading to a Big Crunch. But even that conclusion is based upon the FLRW models, and mainstream cosmologists are willing to consider effects that are not considered by the FLRW models. Originally Posted by Mike Helland But you could add repulsive dark energy to push it back to flat or open. Which is, at the moment, the mainstream explanation for why the universe is flat to within measurement error... ...instead of being hundreds of times more curved than would be consistent with measurements, which is the scenario you hilariously described as "LCDM in a default state". It took me a while to understand what you were smoking when you wrote that. By the way, I am not suggesting you were being ridiculous on purpose. Originally Posted by Mike Helland So did you figure out how that lookback integral reduces? H0 or 2H0. Within the next couple of days, I hope to finish my calculations and write them up for public consumption at this forum. Inspired by your "LCDM in a default state", I will do the math for your "LCDM in a default state" as well as for a variety of other simple models that are just as far out of touch with reality, mainly because I know how to compute closed form solutions for the integrals that come up when calculating with those simple models. Then I will use numerical integration to do the math for a couple of more realistic models. I prefer to present all of those calculations within a single long post. Last edited by W.D.Clinger; 22nd November 2022 at 01:27 PM. Reason: added ETA
 22nd November 2022, 01:43 PM #280 Mike Helland Master Poster   Join Date: Nov 2020 Posts: 2,998 Originally Posted by W.D.Clinger Within the next couple of days, I hope to finish my calculations and write them up for public consumption at this forum. Inspired by your "LCDM in a default state", I will do the math for your "LCDM in a default state" as well as for a variety of other simple models that are just as far out of touch with reality, mainly because I know how to compute closed form solutions for the integrals that come up when calculating with those simple models. Then I will use numerical integration to do the math for a couple of more realistic models. I prefer to present all of those calculations within a single long post. I look forward to it. The first parameters I compared to my distance relationship were ΩM=0.3 and ΩΛ=0.7. The blue line shows the difference between them. Considering I just went "what about quantifying redshift as negative blueshift" and use that in the distance relationship, I was pretty shocked it was actually that close. So you can imagine that finding they match exactly when you remove the effects of gravity and dark energy was completely not what I was expecting to find.

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