Phi for All Posted July 18 Posted July 18 ! Moderator Note Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos. Videos and pictures should be accompanied by enough text to set the tone for the discussion, and should not be posted alone. Please present your topic here, and give us enough of an opener to steer us in the direction you'd like the discussion to take.
Mordred Posted July 18 Posted July 18 That being said the first mistake of that link is that recessive velocity is not an inertial velocity and does not equate to the gravitational redshift formula. It is a relation involving separation distance and not spacetime curvature involving time dilation relations. The treatment is simply incorrect
Nicram Posted July 18 Author Posted July 18 (edited) 1 hour ago, Mordred said: That being said the first mistake of that link is that recessive velocity is not an inertial velocity and does not equate to the gravitational redshift formula. It is a relation involving separation distance and not spacetime curvature involving time dilation relations. The treatment is simply incorrect The first mistake you make is thinking that spacetime is curved along the scale factor's time axis. The most fundamental description of spacetime is given by the metric tensor. For the current intergalactic space it has completely different values from the one that was describing the early, dense-energy universe. Physical link between the former tensor and the present one does not exist. There is no time curvature across the universe history, because the past is completely gone with the former values of the metric tensor. Spacetime in its form described by the tensor exists only for the time of its metric. That's my point of view and also the reason why the changing rate of expansion, given by a'(t)/a(t) is not causing the curvature of spacetime. That's also the reason why we can compare GR and SR (velocity vs redshift as well as scale f. vs time) on the same plot. That's because a single point on GR curve (ΛCDM) represents the inertial snapshot of the universe. If we're considering only snapshots, that are the single points on the curves, these velocities are equal up to now. Curves are made of these points, so these velocities are always equal (up to now). This makes cosmological redshift and Doppler redshift equal. 1 hour ago, Phi for All said: ! Moderator Note Links, pictures and videos in posts should be relevant to the discussion, and members should be able to participate in the discussion without clicking any links or watching any videos. Videos and pictures should be accompanied by enough text to set the tone for the discussion, and should not be posted alone. Please present your topic here, and give us enough of an opener to steer us in the direction you'd like the discussion to take. I'm begging you to make an exception this time. I'm already mentally exhausted by proving Doppler's accuracy for CMB redshift and recession velocity. Edited July 18 by Nicram
iNow Posted July 18 Posted July 18 An exception to presenting your key points here? What a ridiculous request
Nicram Posted July 18 Author Posted July 18 1 hour ago, Mordred said: The treatment is simply incorrect Please, try to correct the inconsistency I'm showing in a different way. Just now, iNow said: An exception to presenting your key points here? What a ridiculous request And yet it is.
Phi for All Posted July 18 Posted July 18 20 minutes ago, Nicram said: I'm begging you to make an exception this time. I'm already mentally exhausted by proving Doppler's accuracy for CMB redshift and recession velocity. ! Moderator Note No exceptions. And you've proven nothing here. Give us an overview and a direction for discussion, Already Mentally Exhausted.
Nicram Posted July 18 Author Posted July 18 Just now, Phi for All said: ! Moderator Note No exceptions. And you've proven nothing here. Give us an overview and a direction for discussion, Already Mentally Exhausted. These are on astronomy stackexchange. My request to you - If I'm wrong, try to correct the inconsistency I'm showing in a different way. -2
Mordred Posted July 18 Posted July 18 (edited) 34 minutes ago, Nicram said: The first mistake you make is thinking that spacetime is curved along the scale factor's time axis. No that's is totally incorrect and not what is described by the scale factor of the FLRW metric to begin with. The curvature terms of the FLRW metric differs from GR in the sense that the curvature term is determined by the critical density formula relations. The GR solution for the FLRW metric is strictly in the Newtonian weak field limit. Where the stress energy momentum term that tells spacetime how to curve is essentially zero with the only term being the energy density in the T_00 component. The rest of your post can readily be shown erroneous by simply knowing that due to expansion the recessive velocity can be greater than c. That occurs beyond the Hubble horizon. In point of detail at z=1100 the recessive velocity is 2.3 c. So your graphs are in error as they do not show this detail. Due to the equations of state the rate of expansion is also not consistent the Hubble parameter decreases in time even though expansion described by recessive velocity is accelerating. This can readily be seen by the calculator in my signature with all the above. (That calculator only required 5 formulas ) Edited July 18 by Mordred
Nicram Posted July 18 Author Posted July 18 (edited) 8 minutes ago, Mordred said: The GR solution for the FLRW metric is strictly in the Newtonian weak field limit. Where the stress energy momentum term that tells spacetime how to curve is essentially zero with the only term being the energy density in the T_00 component. Doesn't it make the spacetime flat? Because GR solution for FLRW is used to describe it, and also your description is a description of flat space-time. I can't quote you anymore using the Quote button, so I'll use italics. The rest of your post can readily be shown erroneous by simply knowing that due to expansion the recessive velocity can be greater than c. That occurs beyond the Hubble horizon. In point of detail at z=1100 the recessive velocity is 2.3 c. That's a statement based on assumption that superluminal velocities are correct. It's like saying, that I can't be correct, because I'm wrong, but without an explanation. I'm showing that these velocities are incorrect. So your graphs are in error as they do not show this detail. - same thing. Due to the equations of state the rate of expansion is also not consistent the Hubble parameter decreases in time even though expansion described by recessive velocity is accelerating. I'm not sure I can see this inconsistency, because I haven't changed the shape of the curves. Does the increasing Hubble constant is the effect of changing the time direction by me? Also, is decreasing Hubble constant proved by observations? Please, try to correct the inconsistency I'm showing in a different way. I can't post anymore today. I've reached the maximum number of posts I can make per day. Edited July 18 by Nicram
Mordred Posted July 18 Posted July 18 (edited) Not necessarily, a critically dense universe is one where the actual density perfectly matches the critical density. If the actual density is greater than the critical density then you have positive curvature for the FLRW metric. If the actual density is less than the critical density then the curvature term is negative curvature. However with combination of the Cosmological Principle the mass density of the overall universe (global metric is uniformly distributed) the 2 terms describing this is homogeneous and isotropic. Meaning no preferred location and no preferred direction. If you apply Newtons shell theorem to this mass distribution gravity equals zero.... You have no center of mass term nor any directional vector field. So you do not have any divergence term such as the gradient term of gravity. These details is why your gravitational redshift formula does not work regardless of your using the relativistic Doppler shift. There is no gravitational nor inertial mass term to begin with The recessive velocity is due to the separation distance described by Hubbles law : the greater the separation distance the greater the recessive velocity. That is certainly not an inertial velocity that is described by kinematics as per GR. see below for the key relations of the FLRW metric the proper time component for a commoving observer being \{c^2d{t^2}\} of the lime element (worldline) next equation below FLRW Metric equations \[d{s^2}=-{c^2}d{t^2}+a({t^2})[d{r^2}+{S,k}{(r)^2}d\Omega^2]\] \[S\kappa(r)= \begin{cases} R sin(r/R &(k=+1)\\ r &(k=0)\\ R sin(r/R) &(k=-1) \end {cases}\] \[\rho_{crit} = \frac{3c^2H^2}{8\pi G}\] \[H^2=(\frac{\dot{a}}{a})^2=\frac{8 \pi G}{3}\rho+\frac{\Lambda}{3}-\frac{k}{a^2}\] setting \[T^{\mu\nu}_\nu=0\] gives the energy stress mometum tensor as \[T^{\mu\nu}=pg^{\mu\nu}+(p=\rho)U^\mu U^\nu)\] \[T^{\mu\nu}_\nu\sim\frac{d}{dt}(\rho a^3)+p(\frac{d}{dt}(a^3)=0\] which describes the conservation of energy of a perfect fluid in commoving coordinates describes by the scale factor a with curvature term K=0. the related GR solution the the above will be the Newton approximation. \[G_{\mu\nu}=\eta_{\mu\nu}+H_{\mu\nu}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}\] Thermodynamics Tds=DU+pDV Adiabatic and isentropic fluid (closed system) equation of state \[w=\frac{\rho}{p}\sim p=\omega\rho\] \[\frac{d}{d}(\rho a^3)=-p\frac{d}{dt}(a^3)=-3H\omega(\rho a^3)\] as radiation equation of state is \[p_R=\rho_R/3\equiv \omega=1/3 \] radiation density in thermal equilibrium is therefore \[\rho_R=\frac{\pi^2}{30}{g_{*S}=\sum_{i=bosons}gi(\frac{T_i}{T})^3+\frac{7}{8}\sum_{i=fermions}gi(\frac{T_i}{T})}^3 \] \[S=\frac{2\pi^2}{45}g_{*s}(at)^3=constant\] temperature scales inversely to the scale factor giving \[T=T_O(1+z)\] with the density evolution of radiation, matter and Lambda given as a function of z \[H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}\] this last equation shows how the equations of state evolve due to expansion and how that affects the Hubble parameter as a function of redshift. This form is the professionally used form that you rarely see in introductory textbooks as its at a more advanced level of understanding. This will further equate to any luminosity distance formulas as well as affect how we determine conformal time A more detailed analysis here including the curvature term k Cosmological Principle implies \[d\tau^2=g_{\mu\nu}dx^\mu dx^\nu=dt^2-a^2t{\frac{dr^2}{1-kr^2}+r^2d\theta^2+r^2\sin^2\theta d\varphi^2}\] the Freidmann equations read \[(\frac{\dot{a}}{a})^2+\frac{k}{a^2}=\frac{8\pi G}{3}\rho\] for \[\rho=\sum^i\rho_i=\rho_m+\rho_{rad}+\rho_\Lambda\] \[2\frac{\ddot{a}}{a}+(\frac{\dot{a}}{a})^2+\frac{k}{a^2}=-8\pi Gp\] for \[p=\sum^ip_i=P_{rad}+p_\Lambda\] with conservation of the energy momentum stress tensor \[T^{\mu\nu}_\nu=0\] \[\dot{p}a^3=\frac{d}{dt}[a^3(\rho+p)]\Rightarrow \frac{d}{dt}(\rho a^3)=-p\frac{d}{dt}a^3\] \[p=\omega\rho\] given w=0 \(\rho\propto a^{-3}\) for matter, radiation P=1/3 \(\rho\propto{-3}\), Lambda w=-1.\(p=-\rho\) for k=0 \[H_o^2=\frac{k}a^2_O=\frac{8\pi G}{3}(\rho^0+\rho_{rad}^0\\rho_\Lambda)\] dividing by \(H^2_0\) and \(P^0_{crit}=\frac{3H^2_0}{8\pi G}\) gives \[1=-\frac{k}{h_0^2a^2_0}+\Omega^o_m+\Omega^0_{rad}=\Omega^0_\Lambda\] \[\Omega_k^0=-\frac{k}{h^2_0a^2_0}\Rightarrow 1=\Omega_k^0+\Omega^0_{rad}+\Omega^0_\Lambda\] densities can be written as \[\rho_{rad}=\rho^0_{rad}(\frac{a_o}{a})^4=\frac{3}{8\pi G}H_0^2\Omega^0_{rad}(\frac{a_o}{a})^4\] \[\Omega_m=\rho^0_m(\frac{a_o}{a})^3=\frac{3}{8\pi G}H_0^2\Omega^0_{rad}(\frac{a_o}{a})^3\] \[\rho_\Lambda=\rho_\Lambda^0=\frac{3}{8\pi G}H_o^2\Omega^0_\Lambda\] \[-\frac{k}{a^2}=\overbrace{-\frac{k}{a^2_0H_o^2}}^{\Omega^0_k}H^2_0(\frac{a_o}{a})^2\] with \(1+z=\frac{a_0}{a}\) densities according to scale factor as functions of redshift. \[\rho_{rad}=\frac{3}{8\pi g}H^2_o\Omega^0_{rad}(\frac{a_o}{a}^4=\frac{3}{8\pi G}H^2_0\Omega^0_{rad}(1+z)^4\] \[\rho_m=\frac{3}{8\pi g}H^2_o\Omega^0_m(\frac{a_o}{a}^3=\frac{3}{8\pi G}H^2_0\Omega^0_m(1+z)^3\] \[\rho_\Lambda=\frac{3}{8\pi G}H_0^2\Omega^0_\Lambda\] \[H^2=H_o^2[\Omega^2_{rad}(1+z)^4+\Omega_m^0(1+z)^3+\Omega_k^0(1+z)^2+\Omega_\Lambda^0]\] the Hubble parameter can be written as \[H=\frac{d}{dt}ln(\frac{a(t)}{a_0}=\frac{d}{dt}ln(\frac{1}{1+z})=\frac{-1}{1+z}\frac{dz}{dt}\] look back time given as \[t=\int^{t(a)}_0\frac{d\acute{a}}{\acute{\dot{a}}}\] \[\frac{dt}{dz}=H_0^{-1}\frac{-1}{1+z}\frac{1}{[\Omega_{rad}(1+z^4)+\Omega^0_m(1=z0^3+\Omega^0_k(1+z)^2+\Omega_\Lambda^0]^{1/2}}\] \[t_0-t=h_1\int^z_0\frac{\acute{dz}}{(1+\acute{z})[\Omega^0_{rad}(1+\acute{z})^4+\Omega^0_m(1+\acute{z})^3=\Omega^0_k(1+\acute{z})^2+\Omega^0_\Lambda]^{1/2}}\] here is the curvature term from the critical density relation \[\Omega_k^0=-\frac{k}{h^2_0a^2_0}\] note the logarithmic functions here \[H=\frac{d}{dt}ln(\frac{a(t)}{a_0}=\frac{d}{dt}ln(\frac{1}{1+z})=\frac{-1}{1+z}\frac{dz}{dt}\] in point of detail the cosmological redshift is a logarithmic function Edited July 18 by Mordred 1
Mordred Posted July 18 Posted July 18 (edited) Lastly observational evidence shows that our universe is very close to being critically dense with a very slight curvature term. In essence the global spacetime metric is close to flat. K=0 this is the correct term for our universe according to observational evidence. This was determined via CMB lack of distortions that curvature would cause. By the way there is a professional paper by Bunn and Hoggs that tried to equate cosmological redshift as Doppler shift however it required using a continous series of infinitisimals to get it to work. Here is the paper https://arxiv.org/abs/0808.1081 As you hit your 5 post first day limit so will check if you reply tomorrow. Edited July 18 by Mordred
Mordred Posted July 18 Posted July 18 (edited) Just an aside many will miss the significance of this equation \[H=\frac{d}{dt}ln(\frac{a(t)}{a_0}=\frac{d}{dt}ln(\frac{1}{1+z})=\frac{-1}{1+z}\frac{dz}{dt}\] The natural log function has a term being the "natural log function" \[log_e(x)\] where the irrational number is 2.71828 this is identical to the e-folds term e used by inflationary theories. This relation carries on into the expansion rate today as well. As you can see both the scale factor and cosmological redshift are Natural logarithmic functions defined by number of e-folds. Aka your accelerating separation distance Recall I highlighted proper time being \{c^2(dt)\} that's a second order term. Velocity and equations involving strictly velocity relation such as v/c is a first order relation. While we see from the above relation Hubble constant, Z and scale factor are second order. (Acceleration in this case) Now consider Hubble law \[v_{recessive}=H_O d\] The recessive velocity is shown to be accelerating it's not an inertial velocity but is oft described as a peculiar velocity based on separating distance. The accelerating expansion. While the value of the Hubble constant is decreasing in time shown by the relation above of Hubble parameter to redshift relation in the previous post. Edited July 18 by Mordred
Nicram Posted July 19 Author Posted July 19 @Mordred This is the biggest avalanche of equations that has ever hit me I must be extremely brave or stupid for not running away from it. My mental powers are too small and lifespan too short to address all the aspects of Friedmann equations used in ΛCDM. However, I hope I can deal with the crucial one, that combines the recession velocity, Hubble parameter and a large distance. Seems like Hubble's Law. My reply is in the attached pictures. Aside from that, how do we know for sure, that dz/dt is increasing slower (maybe also with the decreasing second derivative?) than z+1? Do observations confirm it? Aside from that, did you figure out how to fix the inconsistency? 19 hours ago, Mordred said: By the way there is a professional paper by Bunn and Hoggs that tried to equate cosmological redshift as Doppler shift however it required using a continous series of infinitisimals to get it to work. Here is the paper https://arxiv.org/abs/0808.1081 Love it ❣️
Mordred Posted July 19 Posted July 19 (edited) The first graph is correct The details you are missing in the correct graph which your attempted graph did not reproduce lies in the relations I posted above. Which describes how the commoving coordinates EVOLVE over proper time. This is what is missing from your graph not the graph on top. Now lets understand the correct graph ( the top graph showing the distinctions between Doppler, SR and GR) Notice the cutoff in SR where velocities approach c ? This detail highlights that The Lorentz transformation laws have an effective cutoff when v=c described by its light cone its Global metric is Minkowskii. The FLRW metric using GR graph applies with regards a commoving (conformal Observer). The commoving observer is also inertial as well as the The object being observed. It is the class of observer as well as the coordinate choice and evolution over time that gives the distinction between SR treatment under Minkowskii which The FLRW metric is only locally equivalent and not globally equivalent. The weak field Newton limit directly relates to this detail as it employs the Minkowskii tensor but the FRLW metric has key distinctions on how it evolves over time. Hence the use of a commoving observer. The Doppler relations given in the top graph do not include the Gamma factor and its influence on the coordinate changes. The SR version does Now I cannot stress the Hubble parameter ( value used evolves over time) that is this key relation \[H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}\] So if your applying strictly Hubble's law and the SR transformations your ignoring the coordinate transformations as well as the observer in accordance to the coordinates. Hubble's Law \[V_{recess}=H_o D\] That's why your graph doesn't work it doesn't take in the observer type nor the coordinate transformations itself described via the acceleration equations that your graph doesn't include. The proper time relations provided in the Lineweaver paper. This equation is the line element but does not include how the FLRW metric it describes evolves over time. \[d{s^2}=-{c^2}d{t^2}+a({t^2})[d{r^2}+{S,k}{(r)^2}d\Omega^2]\] \[S\kappa(r)= \begin{cases} R sin(r/R &(k=+1)\\ r &(k=0)\\ R sin(r/R) &(k=-1) \end {cases}\] \[\rho_{crit} = \frac{3c^2H^2}{8\pi G}\] to include how the above changes over time you need the FLRW metric acceleration equations. So you cannot ignore the expansion history and expect to get the correct results as that is an essential component as to why the metric works the way it does. It is also why the Cosmological redshift equation is strictly Doppler shift. The geometry itself changes over time due to thermodynamic evolution Edited July 19 by Mordred 1
Nicram Posted July 19 Author Posted July 19 10 minutes ago, Mordred said: So you cannot ignore the expansion history and expect to get the correct results as that is essential an essential component as to why the metric works the way it does. Sorry, but where is expansion history stored, so you can build your universe from the different parts of its history? 16 minutes ago, Mordred said: It is also why the Cosmological redshift equation is strictly Doppler shift. The geometry itself changes over time due to thermodynamic evolution Nice to agree with you for the first time.
Mordred Posted July 19 Posted July 19 (edited) 26 minutes ago, Nicram said: Sorry, but where is expansion history stored, so you can build your universe from the different parts of its history? Thats provided above under the thermodynamic sections by myself using the equations of state I also included the acceleration equation above. \[H^2=(\frac{\dot{a}}{a})^2=\frac{8 \pi G}{3}\rho+\frac{\Lambda}{3}-\frac{k}{a^2}\] it can also be described as \[\frac{\ddot{a}}{a}=-4\pi G(\acute{\rho}+3 \acute{p})\] as per wiki on the equations of state. The double overdot is the second order time derivative describing the acceleration term of the scale factor ( the equations you used in your graph did not involve acceleration) they used the first order relations of velocity. In the previous equation the single overdot is the time derivative for the velocity term of the scale factor. If you look again at the equations of state in the Lineweaver paper the single overdot used in the equations on the right hand side correspond by the same time derivatives. Everything I posted above in yesterdays post directly applies these equations of state... I simply showed how they were involved https://en.wikipedia.org/wiki/Equation_of_state_(cosmology) using those relations which the calculator in my signature includes from z=1100 to present \[{\scriptsize\begin{array}{|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|r|} \hline z&Scale (a)&S&T (Gyr)&R (Gly)&D_{now} (Gly)&V_{gen}/c&V_{now}/c&H(t)&Temp(K)&rho(kg/m^3)&OmegaM&OmegaL&OmegaR&OmegaT \\ \hline 1.09e+3&9.17e-4&1.09e+3&3.71e-4&6.25e-4&4.53e+1&2.12e+1&3.13e+0&1.56e+6&4.59e-18&2.97e+3&7.56e-1&1.29e-9&2.44e-1&1.00e+0\\ \hline 5.41e+2&1.84e-3&5.42e+2&1.18e-3&1.91e-3&4.47e+1&1.40e+1&3.09e+0&5.13e+5&4.94e-19&1.48e+3&8.62e-1&1.20e-8&1.38e-1&1.00e+0\\ \hline 2.68e+2&3.71e-3&2.69e+2&3.59e-3&5.64e-3&4.38e+1&9.51e+0&3.03e+0&1.73e+5&5.64e-20&7.34e+2&9.26e-1&1.05e-7&7.36e-2&1.00e+0\\ \hline 1.33e+2&7.47e-3&1.34e+2&1.07e-2&1.64e-2&4.25e+1&6.58e+0&2.94e+0&5.95e+4&6.66e-21&3.65e+2&9.62e-1&8.89e-7&3.80e-2&1.00e+0\\ \hline 6.55e+1&1.50e-2&6.65e+1&3.11e-2&4.73e-2&4.07e+1&4.59e+0&2.81e+0&2.07e+4&8.01e-22&1.81e+2&9.81e-1&7.39e-6&1.92e-2&1.00e+0\\ \hline 3.20e+1&3.03e-2&3.30e+1&8.98e-2&1.36e-1&3.80e+1&3.22e+0&2.63e+0&7.20e+3&9.73e-23&9.00e+1&9.90e-1&6.09e-5&9.65e-3&1.00e+0\\ \hline 1.54e+1&6.09e-2&1.64e+1&2.58e-1&3.89e-1&3.43e+1&2.27e+0&2.37e+0&2.52e+3&1.19e-23&4.47e+1&9.95e-1&4.99e-4&4.82e-3&1.00e+0\\ \hline 7.15e+0&1.23e-1&8.15e+0&7.39e-1&1.11e+0&2.89e+1&1.60e+0&2.00e+0&8.81e+2&1.46e-24&2.22e+1&9.94e-1&4.06e-3&2.39e-3&1.00e+0\\ \hline 3.05e+0&2.47e-1&4.05e+0&2.10e+0&3.12e+0&2.14e+1&1.14e+0&1.48e+0&3.13e+2&1.84e-25&1.10e+1&9.67e-1&3.22e-2&1.16e-3&1.00e+0\\ \hline 1.01e+0&4.97e-1&2.01e+0&5.80e+0&8.05e+0&1.12e+1&8.92e-1&7.73e-1&1.22e+2&2.77e-26&5.49e+0&7.86e-1&2.14e-1&4.67e-4&1.00e+0\\ \hline 0.00e+0&1.00e+0&1.00e+0&1.38e+1&1.45e+1&0.00e+0&1.00e+0&0.00e+0&6.77e+1&8.60e-27&2.73e+0&3.11e-1&6.89e-1&9.18e-5&1.00e+0\\ \hline \end{array}}\] Edited July 19 by Mordred
Nicram Posted July 19 Author Posted July 19 @Mordred These equation are perfectly correct, until you start to integrate the scale factor (its inverse to be precise) and I've explained why in the second picture with lots and lots of text using all my powers. Where are the former values of the scale factor stored, because you need them to perform the integration.
Mordred Posted July 19 Posted July 19 Where did you show that ? you need to include here if you recall. secondly we cross posted the calculator in in signature puts it all together (see above and notice the Stretch term S=1+z compared to Z. Thirdly if you included the acceleration equations why did you ask where I had described it in yesterdays post when 90 percent of that post directly applies the acceleration equation relations ? where is that included in your first post link ? ! Moderator Note you do need to bring them here as its a forum requirement that you agreed upon when you agreed to our forum policies edit forgot to add I set the calculator above on our cross post using the Planck Dataset+BAO 2018 and further set to GyR /Gly
Nicram Posted July 19 Author Posted July 19 @Mordred Our misunderstanding is epic. Where is the spacetime that stores the former values of the scale factor, so you can add their reciprocals using integration to build a distance in this way? Did you open my second picture and read the text?
Mordred Posted July 19 Posted July 19 (edited) The first three equations when you solve them They are power law that I showed further solved to provide the natural logarithmic functions. \[d{s^2}=-{c^2}d{t^2}+a({t^2})[d{r^2}+{S,k}{(r)^2}d\Omega^2]\] \[S\kappa(r)= \begin{cases} R sin(r/R &(k=+1)\\ r &(k=0)\\ R sin(r/R) &(k=-1) \end {cases}\] \[\rho_{crit} = \frac{3c^2H^2}{8\pi G}\] \[H^2=(\frac{\dot{a}}{a})^2=\frac{8 \pi G}{3}\rho+\frac{\Lambda}{3}-\frac{k}{a^2}\] spacetime doesn't store anything how it evolves is the thermodynamic processes given by the equations of state. I would never ever describe spacetime as storing anything as it is a Metric its simply the arena where the standard model of particles reside. I certainly hope you don't consider spacetime as some fabric like entity and recognize that spacetime fabric is just an expression used to explain curvature to laymen. Where we choose to keep track of numbers and values and relations is our problem not the universes. It is the distribution and their momentum terms at any given point in time that determines what occurs and when. though if you noticed its oft easier to use the natural log functions with regards to the scale factor which I also provided Edited July 19 by Mordred
Nicram Posted July 19 Author Posted July 19 @Mordred I call it the downvote out of a personal grudge. Why not downvoting all of my replies now, to show a truly scientific attitude. -3
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