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Distance and the Hubble’s Constant 
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#19
Jul2208, 10:20 AM

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P: 521

Marcus: First and foremost:
[1] [tex] H = \frac{1}{t} = \sqrt{ \frac{2GM}{r^3}}[/tex] I was hoping to create a crude model of a matterdominated universe ranging from +1 billion years through to the presentday. My starting point was the quoted age of the universe, i.e. 13.7 billion years. The reciprocal of this leads to H=71km/s/mpc. This allows the Hubble radius to be defined as c/H=1.29E29m, which allows the associated volume to be calculated. As I have understood things, the only significance of the Hubble radius is that it is the point at which the expansion velocity exceeds the speed of light [c]. I selected a homogeneous density in the order of 9.5E27kg/m^3 from which I estimated the mass (M) within the Hubble radius volume, e.g. 8.58E52kg. This mass was also selected because it gives the measured value of H via equation (1). As a closed universe, I assumed the massdensity would not change over the period selected, i.e. matter dominated. From equation [1] I plotted H against time. See graph1.jpg in #13 for details. While H varies with time, I assumed that H was linear with distance for any given time. As such, I could plot recession velocity [v] at a given radius against time. By rearranging equation [1] you can calculate the radius corresponding to H and I assumed that this would allow me to produce a comparative plot of recessional velocity of a volume of space, expanding in time according to H. See graph2.jpg. What this graph suggested to me was that the edge of this volume of space, which today coincides with the Hubble radius, has expanded faster than light until now. If any of the assumptions of this model are valid, it seems to suggest that the universe was initially expanding faster than causality, i.e. ct. However, the model also suggests that as the expansion slows, new sections of the universe will appear within our future light cone, i.e. causality window. The way I was starting to consider the particle horizon is as follows: A photon emitted from an atomic transition at point X arrives on Earth after 13.7 billion years. While the distance travelled by the photon is ct, i.e. 13.7 billion lightyears, the distance between the Earth and point X is now much greater than 13.7 billion lightyears due to the subsequent expansion of the universe? Now based on my earlier model, if the photon was too far from the Earth, when emitted, the recessional velocity would be greater than [c] and the photon would still be chasing after us. However, there is also the issue that the recessional velocity appears to slow down over this vast amount of time. If I assume that the photon has travelled 13.7 billion lightyears, does the recessional velocity of point X give rise to the difference between (ct) and the particle horizon, such that: Particle horizon = ct + vt Where t=13.7 billion years, [c] is the speed of light and [v] is the variable recessional velocity of X with time? I accept this is probably too simplistic, but want to know if it basically explained the concept of receiving light from a point far in excess of ct. 


#20
Jul2208, 10:21 AM

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Nereid,



#21
Jul2208, 10:22 AM

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Krutsch, I am no expert, but here are some thoughts. 1) the bohr model is just a model that really doesn’t exist according to quantum theory. 2) I believe the velocity you are referring to, in the Bohr model, is orbital velocity, 3) I assume the velocity you are referring in connection to the Hubble effect is the recessional velocity. Hope this is of some help.



#22
Jul2308, 01:27 PM

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By luck, I came across an excellent analysis in this very forum on the subject of the particle horizon: http://www.physicsforums.com/showthread.php?t=23037
Footnote to PF admin or mentors: I am not that familiar with all the PF services, so is there a way to ‘weight’ a search on a contribution that has been approved into a library of accepted definitions. A search on ‘particle horizon’ returned 496 threads, which often contain dozens of individual posts, such that finding the thread above, posted in 2004, was highly unlikely. In post #7, Hellfire outlines a derivation of the particle horizon in a matterdominated universe, which I had been looking for. While I have only started to work through the Lineweaver reference given in #14, I am not sure that I agreed with the following descriptive definition of the particle horizon: As an additional issue, Hellfire’s post in the link above suggests that the particle horizon can be estimated as [tex] D(t_0) = 3 c t_0 [/tex]. If so, the estimate of particle horizon would come out at 3*c=1*13.7=41.1 billion lightyears. This differs from the Lineweaver estimate of 46GLYs, so does anybody know what expansion profile is being used in this estimate, which does not include inflation? 


#23
Jul2508, 09:49 PM

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#24
Jul2608, 12:11 PM

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Nereid: thanks for the paper, which I did quickly read. However, given my current A101 status in cosmology, I couldn’t always follow all the inferences being made. As such, I was not sure how this paper addressed the issue of H versus time. If possible, I would like to check some basic assumptions:
1) If on the very large scale the universe is flat, i.e. k=0, then the complexity of Riemann geometry can be reduced to Euclidean geometry. Equally, on this assumption, the FRW metric reduces to the flat spacetime metric of special relativity, where the complexity of GR curved spacetime is localised within much smaller regions of spacetime of higher gravitational potential, e.g. galaxies. 2) Today, H is estimated to be in the order of 71 km/s/mpc. This figure is primarily based on Cepheids (30mpc) and SN1A(400mpc) redshifts. From the aggregation of redshift measurements, we can determine recession velocity z=v/c, nonrelativistic case; from which H=v/d can be derived. 3) On the based of k=0, the Friedmann equation can be transposed to give the critical density of the universe, i.e. [1] [tex] \rho_c = \frac{3H^2}{8\pi G}[/tex] Based on H=71km/s/mpc, [tex]\rho_c=9.57*10^{27} kg/m^3 [/tex], which it is assumed is the total energy density of the universe, which theory is currently separating into 3 components, i.e. matter (4%), dark matter (23%) and dark energy (73%). 4) Matter plus dark matter (27%) constitutes the component that is slowing down the expansion of the universe due to gravity, whilst dark energy is a quantity that might be the cause of the universe expanding at an accelerated rate. If so, is equation [1] correct in that it only reflects (H) reducing with time based on gravitational attraction linked to the constant (G)? As a somewhat expansive question, can we plot the expansion factor (X) of the universe in conjunction with the falling value of (H) linked to gravity alone? One final point, this expansion property of space seems to have a threshold at which it can `overpower` the ‘internal forces’ of a given structure. Apparently, it is not big enough to overpower nuclear force holding together the atomic nuclei, nor the electromagnetic force between nuclei and electrons or even the gravitational force within a galaxy, but it is big enough to ‘overpower’ the weaker gravitational attraction between galaxies, which presumably helped form the largescale structures of the universe. Are there any papers that discuss this issue? As always, many thanks for your insights and the information provided. 


#25
Jul2608, 01:11 PM

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it does not mean that you have 4D flatness or that you can fit the picture to a Minkowski frameor even more radiclly to an Euclidean frame. It does NOT mean that the FRW "reduces to the flat spacetime metric of SR" people are often confused about this and get the mistaken notion that they can fit all or part of a Friedmann model universe onto an SR frame, in cases when this is inappropriate. the confusion partly arises from there being two different meanings of the word flat, namely spatial flat and 4D flat ========================= Nereid is doing a supercompetent response job on your questions, so I won't intrude any further. I only felt compelled to respond to that one point of yours because it presents a very real danger of confusion not only to you but to others here. the k=0 case is the spatial flat and typically the spatial infinite case (spatial flat can still be expanding like crazy and completely nonSR) the the k = 1 case is the spatial positive curvature and spatial finite caseit has spatial closure, but it does not need to eventually collapse, it can continue expanding indefinitely your formula for rhocrit looks fine. my figure for rhocrit, which I think is equivalent to yours, is 0.85 joules per cubic kilometer. You can see whether it is roughly equivalent or not by doing E=mc^2. I find it easier to remember in terms of joules and kilometers. nuff said, please pardon the intrusion 


#26
Jul2708, 03:10 AM

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Marcus, informed comments are no intrusion and the points you raise do home in on any aspect on which I am slightly confused about. I realise that there is still some debate over whether k=0 or is just very close to zero, but I wanted to just focus on the case of k=0.
[tex] c^2 \mathrm{d}\tau^2 =  c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1k r^2} + r^2 \mathrm{d}\theta^2 + r^2 \sin^2 \theta \, \mathrm{d}\phi^2 \right)[/tex] If I consider a radial path, I can reduce the complexity to: [tex] c^2 \mathrm{d}\tau^2 =  c^2 \mathrm{d}t^2 + {a(t)}^2 \left( \frac{\mathrm{d}r^2}{1k r^2}\right)[/tex] If I set k=0, the equation appears to reduce to: [tex] c^2 \mathrm{d}\tau^2 =  c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2[/tex] It was my understanding that today, a(t)=1, which seemed to imply that: [tex] c^2 \mathrm{d}\tau^2 =  c^2 \mathrm{d}t^2 + \mathrm{d}r^2[/tex] Which appears equivalent to SR spacetime metric at a given point in time. Many thanks 


#27
Jul2708, 06:04 AM

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P: 6,233

[tex] g =  dt^2 + dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right), [/tex] which does indeed look like Minkowski spacetime in spherical coordinates. Spacetime curvature, however, is calculated using second derivatives of the metric. At a particular instant of time, [itex]t_\mathrm{now}[/itex], [itex]a \left( t_\mathrm{now} \right) = 1[/itex] is chosen as a scaling convention, but [tex] \frac{d^2 a}{dt^2} \left( t_\mathrm{now} \right) \ne 0. [/tex] Consequently, even at the instant [itex]t_\mathrm{now}[/itex], spacetime curvature is nonzero, while the spacetime curvature of Minknoswki spacetime is zero at all spacetime events. [tex] dr^2 + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right) [/tex] Spatial curvature is zero, since spatial curvature is calculated using second derivatives of the spatial metric with respect to spatial coordinates. This can be easily seen after transforming to Cartesian coordinates. 


#28
Jul2708, 08:51 AM

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George: thanks for the feedback, but could I just press for a few more details? While, the Schwarzschild metric can be presented in a number of forms, I am use to the following form containing the ratio of the Schwarzschild radius [tex]R_s = 2GM/c^2[/tex] over radius [r], which reflects the gravitational field strength with distance.
[1] [tex]c^2 {d \tau}^{2} = \left(1  \frac{R_s}{r} \right) c^2 dt^2  \left(1\frac{R_s}{r}\right)^{1} dr^2  r^2 \left(d\theta^2 + \sin^2\theta \, d\varphi^2\right)[/tex] However, if we assume r>>Rs and we only consider a radial expansion along an equatorial plane, we could reduce this to: [2] [tex]c^2 {d \tau}^{2} = c^2 dt^2  dr^2[/tex] Which I am assuming is essentially equivalent to the Minkowski or SR spacetime metric. So, in word, this is saying in the absence of any significant mass density, spacetime is flat. As such, this was what I was referring to when I said: [3] [tex] c^2 \mathrm{d}\tau^2 =  c^2 \mathrm{d}t^2 + {a(t)}^2 \mathrm{d}r^2[/tex] Now the similarities between [2] and [3] are obvious except for the term a(t). 


#29
Jul2908, 08:24 AM

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#30
Jul2908, 10:25 AM

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