john hunter Posted March 23, 2010 Posted March 23, 2010 The redshift of light might not be related to scale factor like 1+ z = a(1)/a(2) as traditionally thought but as 1+z = [a(1)/a(2)]^2 ........................(1) This leads to a luminosity distance of c/H*(1+z)(sqrt(1+z) -1)..........(2) , with H half the traditional value, and gives a very good match to supernovae data without dark energy (for small z, d=c/2H*z so matches observations). The WMAP value for omeg(matter) of 0.25 comes about because omega = rho/rho(crit) and rho(crit) = 3H^2/8*pi*G and the value used for H is twice the real one. ---------------------------------------------------- The reason for (1) is not known for the purposes of this post, but it seems to be the answer to the dark energy problem. The details are in http://vixra.org/abs/1003.0222 ,including derivation of (2).
ajb Posted March 23, 2010 Posted March 23, 2010 (edited) The idea that the red shift is modified as you have stated seems very phenomenological. Can you offer some calculation to show that this is correct? To derive the red shift one usually considered null-geodesics in the FRW solutions. Interestingly if we have a simply connected, homogeneous, isotropic universe then we just about uniquely single out the FRW solutions. This is in fact independent of general relativity. The question is can you derive your red shift without completely destroying the standard model of cosmology? Edited March 23, 2010 by ajb corrected typo Arch2008 identified
ajb Posted March 23, 2010 Posted March 23, 2010 "derive your red shit"Hmmm...Freudian slip? Thanks, now corrected.
john hunter Posted March 23, 2010 Author Posted March 23, 2010 Dear ajb, the derivation of the luminosity distance (2) is in appendix A of the six page paper...link on first page. It uses the Robertson-Walker metric as you suggest. There are reasons but too speculative to convince anyone about why (1) should be true. But the evidence speaks for itself.....it looks like (1) must be true,....it's the simplest and probably only way to understand the dark energy phenomenon. The next problem for physics might be to puzzle out why (1) is true.... John H.
john hunter Posted March 23, 2010 Author Posted March 23, 2010 Ajb, A4 is the new relation that is proposed, same as formula 2. John H.
ajb Posted March 24, 2010 Posted March 24, 2010 You should be able to calculate the red shift based on the geometry of the space-time. I am fairly sure that the geometry (expansion) + null properties light rays pretty much fix the red shift. Am I wrong on this?
john hunter Posted March 24, 2010 Author Posted March 24, 2010 (edited) Dear ajb, It would be best just to suggest the new redshift - scalefactor relation 1+z = [a(0)/a(t)]^2 ........(1)....as this is the most certain part of the theory....it gives a great match to the data without dark energy. For why it's true we might need Tom Bakers help! The best that I can do is to suggest that the Robertson-Walker relation (A1) should be changed so that the c^2 becomes c^2a(t)^2 i.e the speed of light depends on the scale factor too. This would follow through to (A6) so the R.H.S. has an extra a(t). Then if we assume the scale factor changes the size of all objects, atoms etc too, then the 'apparent/measureable' dX can be divided again by a(t) leaving (A7) and so on unchanged... This would mean that the scale factor changes the size of objects, distance between all objects, all physical constants which contain length dimensions accoording to (for quantity Q).... Q(t)=QexpnHt, where n is the number of length dimensions in quantity Q This leaves the universe apparently static, but with one difference... if no time passes for a photon of light of energy E=hf, the energy is unchanged during travel, but as h(t)=hexp(2Ht) (for the observer), the frequency changes to conserve energy as 1+z = exp(2Ht) which is proportional to a(t)^2. So there is a way to get to (2)...... John H. Edited March 24, 2010 by john hunter
ajb Posted March 25, 2010 Posted March 25, 2010 I know people have thought about variable light speed cosmology. Not that I have much of an idea of exactly what has been done. You have to think about the preservation of local Lorentz invariance which has been tested to some high degree of accuracy. John Moffat I know proposed such a cosmology. I am worried about modifying the FRW metric, the cosmological principle just about fixed this. Then maybe the cosmological principle is too strict. One then also has to wonder what is the matter content of your theory? (Stick it in the field equations).
michel123456 Posted March 25, 2010 Posted March 25, 2010 The best that I can do is to suggest that the Robertson-Walker relation (A1) should be changed so that the c^2 becomes c^2a(t)^2 i.e the speed of light depends on the scale factor too. (...) This would mean that the scale factor changes the size of objects, distance between all objects, all physical constants which contain length dimensions (..) This leaves the universe apparently static, (...) I embrace those ideas. Go on John, I believe you are on the right path. I am sure you'll find on the way a natural explanation for gravity. The square you inserted in (2) comes from acceleration IMHO. A world living in a scale factor of constant rate is impossible to detect IMO, by exact analogy to the impossibility to detect absolute motion. Only if the rate is not constant you may be able to detect the scale factor. And I think it is exactly what is happening. And that gravity is the consequence of this increasing rate. I suppose from you post that you have noticed that a slight acceleration can give a simple explanation to the apparent expanding universe. Michel.
john hunter Posted March 25, 2010 Author Posted March 25, 2010 Dear ajb, Thanks, Prof Moffat has been contacted for his opinion..... Michel123456, It might be that the change of scale factor causes gravity as you mentioned, in fact if all quantities change according to the number of length dimensions in them, the total energy due to a mass (m) goes from mc^2 - GMm/R................to (mc^2 - GMm/R)exp2Ht Where M and R are the mass and radius (c/H) of the rest of the universe. If energy is conserved mc^2 - GMm/R = 0, this gives the value of G as rc^2/M, a nice easy way to solve the flatness problem. This means that the energy gained by a mass during the change is balanced by the loss in gravitational potential energy, i.e gravity is caused by the change of scale factor. (there need be no slowing to conserve energy, H is constant). John H. 1
michel123456 Posted March 25, 2010 Posted March 25, 2010 Please call me Michel. Merged post follows: Consecutive posts mergedI guess from your model you may deduce the value of H instead of inserting it.
Akhenaten2 Posted March 25, 2010 Posted March 25, 2010 (edited) If I was going to be seduced by this "rescaling" theory I should want to see its curve providing a better fit to observations than the Dark Energy Model it was intending to replace. But it does not. In fact - above a redshift of 0.7 the rescaled curve simply does not provide a good fit at all!! Furthermore since I predicted several effects in the early 1990's (before D.E. was discovered) that would beset the Voyager and Ulysses spacecraft as a result of an "energy" of this kind - which effects have been patently in evidence and are continuing (for the 2 voyager craft at least) - I'm inclined to vote in favour of Dark energy - at least for now. I am however, not convinced that the values of several so called "Constants" are actually absolute. Additionally - you might like to visit Astronomy Now On-Line. Edited March 25, 2010 by Akhenaten2
john hunter Posted March 25, 2010 Author Posted March 25, 2010 Michel, The universe is at critical density in this model. Gravity and the value of G are caused by the changing scale-factor G = 3H^2/8*pi*rho, we could get H from this but it is uncertain due to uncertainties in G and rho. H will be about 36km/s/Mpc or about 1x10^-18s-1, if the theory is on the right lines, it is a fundamental constant of nature. John H. Merged post follows: Consecutive posts mergedAkenaten, The match is pretty good up to z=1, in fact it's a better fit than dark energy for intermediate redshifts, but goes a bit high at higher redshifts. However the uncertainties in the higher redshift points is greater, and there are fewer points....so it's a good match. Also, importantly, dark energy has an extra variable parameter which helps it get a good fit...the value for omega(matter), their curve can be varied by choosing omega(matter) to be 0.4,0.3, 0.2 etc...the curve from the new relation (1) has no extra variable parameter. John H.
Akhenaten2 Posted March 25, 2010 Posted March 25, 2010 Fair Comment I'll give it more thought - also I've edited my previous posting - any comments?
michel123456 Posted March 26, 2010 Posted March 26, 2010 (edited) (...)G = 3H^2/8*pi*rho, we could get H from this but it is uncertain due to uncertainties in G and rho. H will be about 36km/s/Mpc or about 1x10^-18s-1, if the theory is on the right lines, it is a fundamental constant of nature. I was thinking about something else. I'll put a new thread about it. Sometime. BTW you value of H is about half the value mentionned on Wiki. "about 2.5×10−18 s−1 with an uncertainty of ± 15%.[6] NASA summarizes existing data to indicate a constant of 70.8 ± 1.6 (km/s)/Mpc if space is assumed to be flat, or 70.8 ± 4.0 (km/s)/Mpc otherwise.[7]" in http://en.wikipedia.org/wiki/Hubble's_law Edited March 26, 2010 by michel123456
ajb Posted March 26, 2010 Posted March 26, 2010 So, you want to think about a metric like [math]ds^{2} = R^{2}(t)\left(- c^{2}dt^{2} + \frac{d \rho^{2}}{1- k \rho^{2}} + \rho^{2}(d\theta^{2} + sin^{2}(\theta) d \phi^{2})\right)[/math]. So, such a metric would not satisfy the cosmological principle. Anyway, examining radial null motion we end up at [math]R^{2}(t)c^{2} dt^{2} = \frac{ R^{2}(t)}{1- k \rho^{2}} d\rho^{2}[/math]. Then assuming [math]R(t)[/math] is never zero we can simply remove it. [math]c^{2} dt^{2} = \frac{ 1}{1- k \rho^{2}} d\rho^{2}[/math]. The do the integrations as in the text books then [math]\frac{\Delta t_{0}}{\Delta t_{1}} = 1[/math] no red shift. So you could consider a metric like [math]ds^{2} = - R^{-2}(t) c^{2}dt^{2} + R^{2}(t)\left( \frac{d \rho^{2}}{1- k \rho^{2}} + \rho^{2}(d\theta^{2} + sin^{2}(\theta) d \phi^{2})\right)[/math]. Again this does not satisfy the cosmological principle. But now we get the sort of relation you are looking for.
john hunter Posted March 26, 2010 Author Posted March 26, 2010 (edited) Dear ajb, Yes the metric at the top of your last post might be a possibility, why do you say it wouldn't satisfy the cosmological principle? Wouldn't it satisfy the perfect cosmological principle as well? There would be no redshift as you say, from the change of scale factor in the usual way. If the speed of light changes, and in fact the interpretation could be that all lengths and physical constants which contain length dimensions change too....then the universe would appear static. The redshift arises as, for a photon of light E=hf....no time passes for the photon, so from conservation of energy, as it enters regions where h has increased, the frequency decreases. The change for each physical quantity would be Q(t)=Qexp(nHt) where n is the number of length dimensions in the quantity. This gives a symmetric/conformal change. Plancks constant varies as h(t)=hexp(2Ht), so f(t) = fexp(-2Ht) which gives the new redshift relation (1) John H. Edited March 26, 2010 by john hunter
michel123456 Posted March 26, 2010 Posted March 26, 2010 There would be no redshift as you say, from the change of scale factor in the usual way. If the speed of light changes, and in fact the interpretation could be that all lengths and physical constants which contain length dimensions change too....then the universe would appear static. Try with acceleration. The apparent constancy of speed of light introduces an observational delay, and this delay in relation with acceleration will give you H. Without acceleration the universe would appear static, as you said. The change of all physical constants simply means they are all linked together. When one change, the other change too. And the observator inside the system notices nothing. The only way to notice that something is going on may come from the change in the rate of change. Change of a change = change^2= something like acceleration. IMHO.
john hunter Posted March 29, 2010 Author Posted March 29, 2010 Michelo, the value of H is meant to be half the traditional value....that's the point...the cause of the dark energy mix-up! Ajb, still wondering why you said the ammended R-W metric violates the cosmological principle.. J.H.
ajb Posted March 30, 2010 Posted March 30, 2010 Michelo, the value of H is meant to be half the traditional value....that's the point...the cause of the dark energy mix-up! Ajb, still wondering why you said the ammended R-W metric violates the cosmological principle.. J.H. You want a space-time that is isotropic and homogeneous in space, but not time. This means you can foliate the space-time into space-like slices each themselves isotropic and homogeneous. Then we can take our space-time (at least locally?) to look like [math]R \times \Sigma[/math] where the homogeneity means that the 3-manifold [math]\Sigma[/math] is maximally symmetric. In fact this means that the entire space-time is maximally symmetric., i.e. has the maximum number of Killing vectors possible. we are then free to use comoving coordinates and put the metric into the form [math]g= - dt^{2} + a^{2}(t)\gamma_{ij}dx^{i}dx^{j}[/math]. So, maybe it is the case that your modified metric does satisfy the cosmological principle, but you are not in comoving coordinates. If so a coordinate transformation would render it in the form above. We are then back to the standard model of cosmology. (My gut feeling is as we have no cross terms dtdx, I think it cannot be rendered as the above, but please prove me wrong.) The other option is that it does not satisfy the cosmological principle.
john hunter Posted March 30, 2010 Author Posted March 30, 2010 Dear Ajb, "So, maybe it is the case that your modified metric does satisfy the cosmological principle, but you are not in comoving coordinates. If so a coordinate transformation would render it in the form above." This is getting a bit mathematical for me...but surely a co-ordinate transformation would remove the a(t)^2 term from the dx*dx terms too...as well as from the cdt term....so the g is apparently constant in time. From a more instinctive approach, there is no reason why any region should be different from any other with the modified metric...and since the universe is apparently static, in scale size (all distances including the size of atoms, people etc..varying with the scale factor in this model) it should satisfy the perfect cosmological principle too. John H.
ajb Posted March 31, 2010 Posted March 31, 2010 (edited) This is getting a bit mathematical for me... But modern cosmology is really a mathematical pursuit. If you want to make any real headway you will need to learn a little about general relativity, as well as many other things. but surely a co-ordinate transformation would remove the a(t)^2 term from the dx*dx terms too...as well as from the cdt term....so the g is apparently constant in time. You could consider a conformal transformation [math]g \rightarrow g a^{-2}(t)[/math] to remove the scale factor, but that is not quite what I was thinking of. See below. From a more instinctive approach, there is no reason why any region should be different from any other with the modified metric...and since the universe is apparently static, in scale size (all distances including the size of atoms, people etc..varying with the scale factor in this model) it should satisfy the perfect cosmological principle too. Well, the idea in cosmology is usually to approximate everything as a (near) perfect fluid. Today, this fluid would be very large collections of galaxies. Local gravitational effects will swamp the cosmological expansion. That is why when say studying the motion of the planets we don't need to worry about the expansion of the universe. On the scale of people and atoms the electromagnetic force dominates. On further thinking, I think you have just chosen "conformal time" and written the metric in these coordinates. You have chosen coordinate in which [math]dt = a(t) d\eta[/math] and we get your "modified term". So, it does satisfy the cosmological principle. However, you are really back to the standard model of cosmology, just not in "comoving coordinates". Conformal time is a very well known a used construction. (I should have spotted this straight away, my only defence is that it has been a while since I look at cosmology properly ) Edited March 31, 2010 by ajb
john hunter Posted April 1, 2010 Author Posted April 1, 2010 Dear Ajb, The metric suggested was the one you wrote on post 18, whether this is the same as the conformal time version, is not so clear...but it's good to know it probably satisfies the cosmological principle. The interpretation of the new metric is quite relevant.... The a(t) would effect all physical quantities and constants with length dimensions n, so that Q(t) = Q*a(t)^n, this would leave a metric which is apparently static, and the interpretation is that gravity doesn't change the scale factor (or rate of change of scale factor), but that the H is constant and the changing scale factor causes gravity (without slowing) to conserve energy. John Hunter.
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