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#11
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A possible solution to the dark energy, cosmic acceleration and horizon problems
In article , Nicolaas Vroom writes:
Cosmology is now a data-driven science; I do not completely understand this sentence. It just means that there is now a huge amount of cosmological data, while not long ago this wasn't the case. As Malcolm Longair was told when he started studying cosmology in the 1960s, "There are only two-and-one-half facts in cosmology". 25 years later, Longair brought it up to 9 facts. Now, there are many, many more. |
#12
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A possible solution to the dark energy, cosmic acceleration and horizon problems
On Mar 26, 3:08 pm, Phillip Helbig---undress to reply
wrote: In article , Dave Rutherford writes: Hubble's law, in terms of the PROPER DISTANCE, holds exactly. But this is not observable. How can you claim that it holds exactly if it's not observable? Theory. Within the context of classical cosmology, proper distance is not directly observable, but nevertheless one can say something about it. Are you claiming that Hubble's proper distance-redshift law c*z = H_0*d_0, where d_0 is the (present) proper distance and z is cosmological redshift, holds exactly for all z? If so, I can show you that it doesn't. Look at http://www.softcom.net/users/der555/diagram.pdf The distance modulus derived from Hubble's proper distance-redshift law (the curve labeled "Hubble Law" in the diagram) does not hold for large z. If you mean does my curve pass through every data point in the data set, no. But the concordance model doesn't do that either. If you plot my model's m-z curve and the concordance model's m-z curve (\Omega_m = 0.31, \Omega_\Lambda = 0.69) against the data, you'll see that they almost exactly coincide. One needs a quantitative comparison of goodness-of-fit. Can you please explain to me what "a quantitative comparison" means relative to "goodness-of-fit"? |
#13
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A possible solution to the dark energy, cosmic acceleration and horizon problems
In article , Dave Rutherford
writes: Are you claiming that Hubble's proper distance-redshift law c*z = H_0*d_0, where d_0 is the (present) proper distance and z is cosmological redshift, holds exactly for all z? This holds for all z: the velocity, defined as the derivative of the proper distance (which has a very precise meaning in GR and cosmology) with respect to time, evaluated now, is proportional to the proper distance. Whether or not this is called "Hubble's Law" depends on the author. Note that this statement contains no physics: it is the only velocity-distance relation compatible with a homogeneous and isotropic universe. One needs a quantitative comparison of goodness-of-fit. Can you please explain to me what "a quantitative comparison" means relative to "goodness-of-fit"? Calculate the probability that your model is correct vs. the probability that the conventional model is correct, given the data. Alas, the details are too involved for a newsgroup post, but this is really standard stuff. |
#14
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A possible solution to the dark energy, cosmic acceleration and horizon problems
On Mar 27, 2:55 am, Phillip Helbig---undress to reply
wrote: In article , Dave Rutherford writes: Are you claiming that Hubble's proper distance-redshift law c*z = H_0*d_0, where d_0 is the (present) proper distance and z is cosmological redshift, holds exactly for all z? This holds for all z: the velocity, defined as the derivative of the proper distance (which has a very precise meaning in GR and cosmology) with respect to time, evaluated now, is proportional to the proper distance. Whether or not this is called "Hubble's Law" depends on the author. Note that this statement contains no physics: it is the only velocity-distance relation compatible with a homogeneous and isotropic universe. The velocity-distance part of my velocity-redshift-distance law (equation (14) in my article), v = c*ln(1 + z) = H_0*d_0 (14) is identical to the Hubble velocity-distance law v = H_0*d_0 |
#15
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A possible solution to the dark energy, cosmic acceleration and horizon problems
"Jonathan Thornburg [remove -animal to reply]"
schreef in bericht ... "Phillip Helbig---undress to reply" wrote Cosmology is now a data-driven science; Nicolaas Vroom wrote: I do not completely understand this sentence. IMO Cosmology (Astronomy) was always a data-driven science. IMO any science is data-driven i.e. is based on observations. Phillip's phrase is a commonly-used one today. The basic meaning is this: 30 years ago (say) there was only a small amount of cosmological data, and that data often had relatively large error bars. In contrast, today there's a HUGE amount of cosmological data, with big parts having quite small error bars. And more cosmological data is arriving each year. So, doing state-of-the-art cosmology today involves a lot of sophisticated modelling of many disparate data sets. This is what's being referred to in the phrase "data-driven science". I fully agree that today cosmology is based on much more and more accurate data (observations) than 30 years ago. In fact much more visible data is available now. Implying also that matter we could not detect 30 years ago now can be detected. This means that galaxies are much larger than previous thought. The same for the visible mass. The two last remarks because this makes the necesity of dark matter less obvious. When galaxies are more massive as previous thought also the amount of dark matter required to keep the galaxies clustered is reduced. The same for the dark matter to explain gravitational lenses. On the other hand if there is cosmic acceleration and if galaxies are more massive the amount of energy to explain that increases. Nicolaas Vroom http://users.pandora.be/nicvroom/ |
#16
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A possible solution to the dark energy, cosmic acceleration and horizon problems
On Mar 26, 3:08 pm, Phillip Helbig---undress to reply
wrote: In article , Dave Rutherford If you mean does my curve pass through every data point in the data set, no. But the concordance model doesn't do that either. If you plot my model's m-z curve and the concordance model's m-z curve (\Omega_m = 0.31, \Omega_\Lambda = 0.69) against the data, you'll see that they almost exactly coincide. One needs a quantitative comparison of goodness-of-fit. Are you referring to the reduced chi-square goodness-of-fit test? If so my reduced chi-square (according to Gnuplot) is 0.98, and the reduced chi-square of the concordance model is 1.06. So mine is magnitude 0.02 away from 1.0, and the concordance model's is magnitude 0.06 away from 1.0. If the best fitting model, between the two, is the one with a reduced chi-square that's closest to 1.0, then I guess I win that test. |
#17
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A possible solution to the dark energy, cosmic acceleration and horizon problems
On Mar 25, 3:16*pm, Dave Rutherford wrote:
[...] The relation between apparent magnitude and redshift is linear only at low redshift. The relation between the distance modulus (apparent magnitude minus absolute magnitude) and redshift, in my model, is linear only at low redshift. I don't see your point. I'm going to pick this point to restart a discussion we were in the middle of within sci.physics when you appear to have abandoned it to go post in here. You very much seem to be interested in straddling the line between using your own manifold + coordinate system, and wholesale borrowing of concepts that apply only on FRW manifolds. For example, your usage of the scale factor is inconsistent with homogeneity and isotropy of space which is explicit in the construction of the FRW manifold. But then you assert the usage of the factors of (1+z) that arise from the simultaneous time dilation and physical expansion of the universe that are limited to FRW manifolds. I showed in Figure 3 that my ideas match observation (Riess et al. 2007) very well. Quantitatively? *Better than conventional cosmology? If you mean does my curve pass through every data point in the data set, no. But the concordance model doesn't do that either. If you plot my model's m-z curve and the concordance model's m-z curve (\Omega_m = 0.31, \Omega_\Lambda = 0.69) against the data, you'll see that they almost exactly coincide. It helps a LOT when you borrow the exact (1+z)^2 proportionality factor from the concordance model. Sure, your overall functional form is different but the power series expansions match exactly (iirc) at 0th and 1st order, and reasonably close at 2nd order. When the errors of your formula are well within the random errors of observation, it undercuts the otherwise impressive looking picture's significance. [...] |
#18
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A possible solution to the dark energy, cosmic acceleration and horizon problems
On Mar 27, 3:55*am, Phillip Helbig---undress to reply
wrote: In article , Dave Rutherford writes: Are you claiming that Hubble's proper distance-redshift law c*z = H_0*d_0, where d_0 is the (present) proper distance and z is cosmological redshift, holds exactly for all z? This holds for all z: the velocity, defined as the derivative of the proper distance (which has a very precise meaning in GR and cosmology) with respect to time, evaluated now, is proportional to the proper distance. *Whether or not this is called "Hubble's Law" depends on the author. *Note that this statement contains no physics: it is the only velocity-distance relation compatible with a homogeneous and isotropic universe. One needs a quantitative comparison of goodness-of-fit. Can you please explain to me what "a quantitative comparison" means relative to "goodness-of-fit"? Calculate the probability that your model is correct vs. the probability that the conventional model is correct, given the data. *Alas, the details are too involved for a newsgroup post, but this is really standard stuff. Put the nitty-gritty of the statistics aside for a moment. Take the respective distance/luminosity formulas from his model and the concordance model. Taylor expand about z=0, keeping terms to O(z^2). Notice that they are rather similar. Now recognize that the overwhelming majority of the cosmological data relevant here is situated between z = 0 and z = 1.5. That's why his plot is so close. |
#19
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A possible solution to the dark energy, cosmic acceleration and horizon problems
On Mar 30, 10:37 pm, Eric Gisse wrote:
On Mar 25, 3:16 pm, Dave Rutherford wrote: The relation between the distance modulus (apparent magnitude minus absolute magnitude) and redshift, in my model, is linear only at low redshift. I don't see your point. You very much seem to be interested in straddling the line between using your own manifold + coordinate system, and wholesale borrowing of concepts that apply only on FRW manifolds. I'm using a Euclidean (flat) spacetime coordinate system. For example, your usage of the scale factor My r is really only related to the conventional (dimensionless) scale factor S when used as a ratio, as in r_0/r = S_0/S = 1+z, where r_0 and S_0 are the present values of r and S, respectively. Quantitatively? Better than conventional cosmology? If you mean does my curve pass through every data point in the data set, no. But the concordance model doesn't do that either. If you plot my model's m-z curve and the concordance model's m-z curve (\Omega_m = 0.31, \Omega_\Lambda = 0.69) against the data, you'll see that they almost exactly coincide. It helps a LOT when you borrow the exact (1+z)^2 proportionality factor from the concordance model. There's no (1+z)^2 in my article. Sure, your overall functional form is different but the power series expansions match exactly (iirc) at 0th and 1st order, and reasonably close at 2nd order. When the errors of your formula are well within the random errors of observation, it undercuts the otherwise impressive looking picture's significance. What errors? |
#20
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A possible solution to the dark energy, cosmic acceleration and horizon problems
On Apr 1, 12:04*am, Dave Rutherford wrote:
On Mar 30, 10:37 pm, Eric Gisse wrote: On Mar 25, 3:16 pm, Dave Rutherford wrote: The relation between the distance modulus (apparent magnitude minus absolute magnitude) and redshift, in my model, is linear only at low redshift. I don't see your point. You very much seem to be interested in straddling the line between using your own manifold + coordinate system, and wholesale borrowing of concepts that apply only on FRW manifolds. I'm using a Euclidean (flat) spacetime coordinate system. Bad start, because the universe likes to behave as if it were Lorentzian. But I'll ignore that and focus on my point : you are not using a FRW manifold. Regardless, Euclidean spacetime is *static*. No expansion, time dilation, or anything. Ignoring that, and pretending space has a logarithmic spiral structure (which negates your claim of Euclid...), you do not get to use the same distance modulus from FRW manifolds that specifically takes into account the fact that there are simultaneous time dilation and expansion effects. That's why the overall distance modulus has a factor of (1+z)^2, and why yours can't have that. Its' what distinguishes big bang cosmologies from the many other choices. For example, your usage of the scale factor My r is really only related to the conventional (dimensionless) scale factor S when used as a ratio, as in r_0/r = S_0/S = 1+z, where r_0 and S_0 are the present values of r and S, respectively. Well you were calling it a 'scale factor' explicitly. Quantitatively? *Better than conventional cosmology? If you mean does my curve pass through every data point in the data set, no. But the concordance model doesn't do that either. If you plot my model's m-z curve and the concordance model's m-z curve (\Omega_m = 0.31, \Omega_\Lambda = 0.69) against the data, you'll see that they almost exactly coincide. It helps a LOT when you borrow the exact (1+z)^2 proportionality factor from the concordance model. There's no (1+z)^2 in my article. Meant to say (1 + z^2), the power series expansion of your result is pretty close to the concordance cosmology result. Sure, your overall functional form is different but the power series expansions match exactly (iirc) at 0th and 1st order, and reasonably close at 2nd order. When the errors of your formula are well within the random errors of observation, it undercuts the otherwise impressive looking picture's significance. What errors? Your formula is indistinguishable from the concordance cosmology result within the errors of the data. |
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