captcass Posted July 17, 2016 Author Share Posted July 17, 2016 Physics Expert said: "I do not understand this... when we move with respect to what?" When we consider motion, when we move in respect to other reference frames or they move in relation to us, the effects of relativity come into play. I am considering stationary reference frames where the only effects are due to the apparent differences in rates of time. Physics Expert said: "There is no meaningful notion of gravitational length contraction... sorry. You need to compare rulers to have length contraction. You need to describe a set-up for which one can do that in a meaningful way" We know that to maintain C space has to contract or expand when the rate of time changes. When we look up at a reference frame with a faster rte of time, the space in that reference frame has to also appear contracted to us. As I state in my paper, the reduced space is not in less meters, it is in a shortening of the length of a meter. When time slows and space expands, the increase is in the lngth of a meter, not the addition of meters. Physics Expert said: "I don't understand this statement. We model space-time as being curved irrespective of test bodies in motion. (But then you do have soem test bodies - your observers)" What I see is an evolving continuum. As reference frames within the continuum evolve forward at different rates of time the curvature of motion we see is manifested. In the paper I describe gravitational lensing as being produced by the up gradient side of the photon being updated in time sooner, and in faster time then the solar side. This applies to any particle moving across a time dilation gradient. As the Earth moves across the solar dilation gradient, for instance, the off solar side is being updated in time sooner than the solar side, evolving it forward in the continuum sooner than the solar side which causes the Earth's path to curve. As the off solar side is also in faster time, rotation is also induced. Of course this is simplistic and other effects influence the Earth's motion in the solar system such as the presence of other bodies, the Earth's velocity, etc. . I don't see a pre-existing curved space bodies move through, only an evolving continuum that manifests apparent curvature of motion. Physics Expert said: "However, the general notion of inflation is supported by observations of the CMBR. It is true however, that we are not at the stage of throwing particular models out." Once we accepted Hubble's law based upon the visual evidence, we began trying to make everything fit that theory, including the thermodynamics and the CMBR. Guth's attempt was impressive, but certainly not conclusive. And, of course, we still have no causality for why things began in the the first place, what set off the BB. Trying to make everything fit the inferences of Hubble's law has led us down a path to absurd conclusions and we are stuck in a rut. My theory gets us out of the rut. Link to comment Share on other sites More sharing options...
ajb Posted July 17, 2016 Share Posted July 17, 2016 I think I'm done... over to Mordred. Link to comment Share on other sites More sharing options...
captcass Posted July 17, 2016 Author Share Posted July 17, 2016 Does that mean you accept my answers as valid, or at least possibly valid? -1 Link to comment Share on other sites More sharing options...
ajb Posted July 17, 2016 Share Posted July 17, 2016 Does that mean you accept my answers as valid, or at least possibly valid? Not at all... it just means I have no plan to keep beating my head against a wall. It was fun, but I don't see that I nor you can gain anything from us continuing. Good luck with the 'physics'. Link to comment Share on other sites More sharing options...
StringJunky Posted July 17, 2016 Share Posted July 17, 2016 (edited) My theory gets us out of the rut. A theory is not such until it has widespread consensus that it's predictions match observation. At best, you have a hypothesis. Edited July 17, 2016 by StringJunky Link to comment Share on other sites More sharing options...
captcass Posted July 17, 2016 Author Share Posted July 17, 2016 Sorry 'bout that, I am having fun, too. As per the continuum, quantum physics tells us that nothing can be nailed down, position and momentum. I am not really here, is is just more probable I am here but I am in constant flux. Never, in any instant, can I be said to be anywhere. We perceive distance as though things are "there", but they are not really there, it is just more likely they are there and they are evolving forward in the continuum just like we are. If you could look at the universe from the side, it would have no depth. It is all an evolving energy field without even the thickness of a photon. It looks like we have space ahead of us we are moving into, but it is not "ahead" of us. It is just a perspective within the continuum. Each reference frame evolves forward at its own relative rate of time and the differences manifest as the perception of distance in both space and time.. Scientist says: "A theory is not such until it has widespread consensus that it's predictions match observation and can make predictions. At best, you have a hypothesis." What are the odds that my derivation of the Gravitational Equivalency Constant using the Earth's time dilation gradient works to derive the mass/energy of the dark matter of the Andromeda galaxy? Link to comment Share on other sites More sharing options...
Mordred Posted July 17, 2016 Share Posted July 17, 2016 (edited) your model isnt particularly testable for observation reality. Take for example GN z11. You believe your answer of 13.4 Gly is correct. Yet the professional community finds the distance as 32 Gly. So I asked you to show how your model accounts for this. You still haven't shown how your model accounts for this Hubble illusion. This is the part you keep denying. Quite frankly if you want to convince anyone we have a static vs an expanding universe. You need to mathematically show greater than c recessive velocity and explain why temperatures drop in a static universe. Until you do so Your model won't work After all the whole point of your model is to prove a static universe. You don't do this by assuming static distances ie 13.4 Gly is correct. When we measure 32 Gly and you ignore the observational evidence and use an assumed 13.4 Gly as being correct. Not without showing how the 32 Gly is wrong. Via the mathematics Quite frankly not accounting for mrasurement data beyond Hubble sphere isn't going to work. Edited July 17, 2016 by Mordred Link to comment Share on other sites More sharing options...
captcass Posted July 17, 2016 Author Share Posted July 17, 2016 The community finds it to be at 32 Gly because it is assuming an expanding universe. If it is not expanding, the distance is 13.4 Gly. The community assumes a faster than C recessive velocity because of the high red shift. I show the expansion isn't necessary to explain the high z. Thanks, guys, I'm off to work. I knew this would be a difficult sell. I could hardly believe it as it developed. Physicists like substantiality. But there is nothing substantial about the events in space/time. As Einstein said it is all an illusion. It is all just manifested in an evolving continuum. Link to comment Share on other sites More sharing options...
Mordred Posted July 17, 2016 Share Posted July 17, 2016 you still miss the point. You need to mathematically prove the 32 Gly as wrong. why do we measure that distance using both redshift and angular diameter distance? Show us in the math. Link to comment Share on other sites More sharing options...
captcass Posted July 17, 2016 Author Share Posted July 17, 2016 How do they derive the 32 Gly? They look at the apparent position and factor in the rate of expansion since 13.4 Gly ago. They don't see it at 32 Gly, They assume it is at 32 Gly based upon the expansion rate they think they see based on z. If the expansion isn't there, it is not where they think it is, it is at 13.4 Gly (not really, because we see it where it was 13.4 G years ago and it has a peculiar motion so it has moved some from the position we see.) It is not angular diameter distance at 32 Gly being used to determine that distance because we can't see it at 32 Gly. The angular diameter distance is used to determine its distance 13.4 Gy ago. Then expansion is factored in to determine the distance now, even though we can't see it at that distance. If I am correct about a nearly stable universe then my derivation works. It is not a matter of mathematically proving the 32 Gly is wrong. It is a matter of whether the universe is expanding or not, a conceptual difference. If I am wrong and z does reflect that expansion, then it is at 32 Gly. If I am correct and the universe is not expanding, then it is (was) at 13.4 Gly and my derivation works. I should add this to try to give you a perspective of my point of view. Because the universe is illusionary, it requires a perceiver. I see us (all life forms) as just different points of view for the same perceiver. We are all one in It. I see the continuum evolving around me, for me, so there is a continuity to events. When I move my arm, the continuum evolves in such a way that my arm ends up in the right place. Forgive me if this somehow violates forum rules or etiquette but I am going to tell a short story here so you can hopefully visualize the continuum I see. I own the world's first and only sea glass museum. I have spent many hours on our glass beaches (yes we have 3 glass beaches) collecting sea glass. I landed at the beach one day in my kayak and found 2 marbles right away right next to each other. They were just plain old marbles, but still, marbles are as rare as reds (1:5000).I talk to the Perceiver within me a lot because things like what follow happen all the time if I do. So I said, "Lord, thank you. I sure would like to find a red marble, though. I've got a blue, and I thank you, but I sure would like to find a red. Please, Lord?"So I went back picking and put that thought aside After about 2 hours of walking very slowly up and down the beach, looking for the gems among all the other glass, I was tired and it was time to go tide-wise, but there was just a little more beach to do, so I decided to just make a quick pass and see if I could spot anything special just lying there on top.Just before the end there is this HUGE red marble. Not just a regular little marble like the other 2 I found that day. It's 15/16ths of an inch. It is a beautiful blood red with a white swirl that forms A WAVE! (sea glass - remember?) When I bend down to pick up the marble, there are also 2 pieces of jewelry quality RED glass, one on either side of it! Reds are 1 in 5,000 pieces! Anyone care to calculate odds here? In my world, this is the Perceiver telling me the marble was not a co-incidence. I asked for red and got red. When I ask for something, if I get it, it is always much grander than I could have imagined. The reason I believe in the non-substantial evolving continuum is because I have been living in it for the last 42 years and can see it evolving around me, Sorry if I violated........ As per QM, neither the marble nor the red glass was there before I found it. What dimension did Einstein add to the 3 vector to get a 4 vector? The 4th vector is the ratio of what to what? The direction of curvature comes from the differences in what? Sorry, gotta go back to work. I'm out of time. Link to comment Share on other sites More sharing options...
captcass Posted July 18, 2016 Author Share Posted July 18, 2016 OK. Looks like I might have lost you guys. If I have, sorry ‘bout that and thank very much for trying to set me straight. I’m sorry if you think I wasted your time. So: Einstein’s 4 vector in determining μ is: μ = √(∆To/∆Tr + ∆X/X + ∆Y/Y + ∆Z/Z) ∆To/∆Tr is the relationship of change in the respective times in the inertial reference frame of the observer and the coordinate frame. The difference in the change in times is due to the different rates of time in the different frames. The differentials in time are determined by time dilation. This relates to the time dilation formula: To/Tr = √(1 – Ro/R) Which is the relationship my derivations are based upon. Ro/R, the relationship of distance (length) over distance (length) relates to To/Tr In the geodesics of Relativity, it is the ∆To/∆Tr relationship that determines the curvature of motion through space. As I stated earlier, I am not talking about this curvature in my paper (not directly anyway). What I am doing is taking a deeper look at the To/Tr relationship and how that shapes space in a different way by changing the length of a meter. We have two points of view being represented here: An expanding universe based upon the assumption that the Hubble shift is caused by recessive acceleration where the universe began as a minute concentrated singularity and will expand at an accelerating rate until it goes cold and dead. I’m sorry but this is just absurd. When the thermodynamics for this theory didn’t work out, we got Guth’s theory of inflation, which sounds nice, but is unproven. Just an idea of what might have happened that might make things work. And….it lets us exceed C! We look at the CMBR and say it is proof of a BB. But it is not what we thought it was. It isn’t homogenous. So we come up with unproven reasons why it is not homogenous. We make it fit like we used inflation to make the thermodynamics fit. So what causes this expansion? Dark Energy = “We don’t know”. The other point of view is a primarily stable universe, pre-Hubble assumptions. Z does not represent recessive velocity, but a relativistic increase in the length of a meter based upon To/Tr. Guth’s inflation can be discarded and the thermodynamics and the CMBR can be re-interpreted in relationship to a nearly stable universe. There is no Dark energy to consider. ∆To/∆Tr is the element in Einstein’s four-vector that determines the curvature of motion through space. Events in space evolve in the forward direction of time. What the heck does that mean? The forward direction of time curves into the Earth? That is what I was pondering when I came upon the concept of the lateral shift in the update: another, relativistic, forward direction of time, and yada yada as per my paper, that derives a Gravitational Equivalency Constant that relates the difference in rates of time per meter to the Newtonian force in Newtons that allows us to see the effects in time dragging everything along with it: a force in time, that lets us derive the mass/energy of the Andromeda galaxy. I know you folks are strongly committed to the BB and all that follows. It seems to be what we see. Relativity tells us that what we see is not what is there. With all the uncertainties and assumptions, can you truthfully say these theories are working? If you turn from the BB and all that to the stationary, relativistic, universe, i am describing, do my derivations work and explain the phenomena we see? Does this satisfy you that I know a thing or two about Relativity and Einstein’s field equations? Does it convince you I know the difference between relativistic effects and the effects I am proposing? C’mon, guys, don’t cut me off here! Give me a hand…… How about this? If you assume a stable universe like I describe, could the CMBR be due to interference, static, due to the interference of solar emanations and the reason it is not homogeneous is that solar density is not homogeneous? Is there another proposition you can think of for the CMBR for a primarily stable universe? Certainly there are other theories that fit a stable universe......I haven't checked it out, and will, but do you know anything that would fit? Can the thermodynamics work for such a universe if we discard inflation? Relatively, can the entropy of the CMBR be offset by the ordering of the CMBR's energy into mass? I see so many possibilities for lines in inquiry here.......Wanna play? Could solar emission interference result in the harmonic undertones of the CMBR? Sorry, I have a problem with verbosity compounded by free association........ Link to comment Share on other sites More sharing options...
Mordred Posted July 18, 2016 Share Posted July 18, 2016 (edited) no you haven't lost us. Your not doing the necessary steps that are required to properly test your theory with observational data. Sorry but your not. For reasons Ive already given. You seem to feel science states a certain distance based strictly upon redshift beyond Hubble sphere. Yet redshift is merely one rung on the cosmic distance scale. Other rungs include Tully-Fisher, Extragalactic parallax, Luminosity-distance D-q function angular diameter distance etc. You assume a static universe. Fine but the rest of the scientific community disagees with you. You need to prove them wrong. With the correct math. You dont have the correct math unless you can prove Hubbles law is wrong. distances beyond Hubbles sphere for greater than c recessive velocity has been confirmed by more than method. Other than just redshift. All you have to do is look for redshift calibration papers to see this. Yet you assume your static universe and wont show how greater than c recessive velocities are solved via your model other than assumption of a distance based on a static universe assumption Edited July 18, 2016 by Mordred Link to comment Share on other sites More sharing options...
captcass Posted July 18, 2016 Author Share Posted July 18, 2016 OK, thanks. What do you suggest? More derivations using other bodies to show it relates? I can't derive the 32 Gly distance as according to my theory it isn't there and the derivation of the 32 Gly is wrong as it is based on a non-existant expansion. How can I make you guys happy? Link to comment Share on other sites More sharing options...
Mordred Posted July 18, 2016 Share Posted July 18, 2016 (edited) You need to find a way to show how your model can account for the redshift values beyond Hubble sphere. There have been dozens of Galactoeccentric models that have used similar arguments that you have used. those models also stopped when v is less than c. Those even when peer reviewed didn't go anywhere. The scientific community essentially ignores them. They dont show what is needed. The math in them is more advanced than what you have. So if their papers didnt cut it..... The other problem is even photometric evidence shows an expanding universe without doing a single redshift calculation. For example did science stop at merely calculating the distance to Gn-z11 just using redshift? What other tests did they do? Why is it such a luminescence galaxy? What type of stars populate that galaxy? Most importantly what is the average plasma temperature. (this affects redshift wavelength). Hence posting Weins Displacement law earlier. Edited July 18, 2016 by Mordred Link to comment Share on other sites More sharing options...
captcass Posted July 18, 2016 Author Share Posted July 18, 2016 OK. Thanks. I'll get into that and I'll see where I get. Link to comment Share on other sites More sharing options...
Mordred Posted July 18, 2016 Share Posted July 18, 2016 (edited) very good now were moving forward. You can't assume a static measurement or a non static. example if 32 Gly is wrong. Then the age value of 400 years after BB is also wrong. This could very well mean the value 13.4 Gly is also wrong.... It would mean that all Lyman alpha forest data for that sector is also wrong. see where Im going on the dangers of assumptions? another common method is luminosity to distance measurements. You will eventually need to adress these as well as they involve redshift. To assist you. Historically speaking Scientist once assumed no signal can go farther than Hubble distance. Yet when we first dicovered objects lying outside Hubble distance the scientific community went up in an uproar. What tests were involved that changed their minds. (redshift error was first called into question) What tests did they need to perform to confirm it wasnt a redshift error? Edit there is another hickup I forgot to mention. For Gn-z11 is 32 Gly the commoving distance or the proper distance? piece of advise start with learning how this equation which includes GR and thermodynamics influence the light rays worldline. (All redshifted lightrays follow worldlines) In the FLRW metric the worldline is the line element ds^2 [latex]d{s^2}=-{c^2}d{t^2}+a{t^2}[d{r^2}+{S,k}{r^2}d\Omega^2][/latex] [latex]S\kappa,r= \begin{cases} R sin(r/R &(k=+1)\\ r &(k=0)\\ R sin(r/R) &(k=-1) \end {cases}[/latex] Note a redshift worldline is a null geodesic. Which is different from the normal infalling matter geodesics. I also recommend you graph your model vs the Standard model redshift. Distance vs redshift up to z= 1100. In proper distance not commoving distance. Which means you will need to convert the line element above to proper distance... THIS EQUATION CANNOT be used beyond Hubble sphere. [latex]1+Z=\frac{\lambda}{\lambda_o} or 1+Z=\frac{\lambda-\lambda_o}{\lambda_o}[/latex] I will post the correct equation beyond Hubble sphere. (When I work up how the first equation is derived. In Order to explain the new redshift equation beyond Hubble sphere.....) When I work up the two redshift equations. I hope you see your paper has the wrong equation being examined Edited July 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
captcass Posted July 19, 2016 Author Share Posted July 19, 2016 Thanks Mordred, I appreciate this. Been looking at the Weins Displacement law. A few questions; I ran across the limitations of the use of this formula when I began to get into this back when. As far as I can see, though, it works with the length of a meter, which doesn't have the limitations of light, which I why I went with meters. Would you agree with this thinking? If so, it seems all I need to adjust for is the Weins Displacement law, which I am assuming is accounted for in the Cosmological redshift formula I am not using, but looking to replace? When I look up components of the redshift I only get the three. Are there other components in the Cosmological redshift formula I would need to account for re the length of a meter? Please bear with me here. You are probably going to say, "AAAARRRGGGHH". I haven't been working with photons. My assumption is that if a meter lengthens, a photon within that meter will also lengthen by the same percentage. Thus I use meters. Thus it shouldn't matter what wavelength the photon is when it departs Gn-z11 or any other source. Does that mean I still have to account for the Weins Displacement law? I don't know how I can incorporate those factors into my approach, which is strictly visually relativistic based on meter length. The one common element I find is that I am using a null geodesic. See my problem here? I would get the CMBR z = 1100 at about 42.2 Gly Does that ring any bells for you? It could be less than this, which is just an approximation. A thought: A photon has energy E. When we say: E = hv we are not talking about a single sinusoidal wave with so many peaks. We are talking about the frequency of pulses of h energy from the emitting source. In a time dilation field like I am visualizing, as time slows between succeeding sets of frames, the timing of the pulses would also slow, decreasing the frequency. Link to comment Share on other sites More sharing options...
Mordred Posted July 19, 2016 Share Posted July 19, 2016 (edited) Weins Displacement law has more to do with the original emitter frequency. Which correlates back to hotter temperatures in the past. Anyways lets look at your questions above in more detail after. lets look at the corrections to the redshift formula. First we define a commoving field. This formula though it includes curvature (global) you can set for flat spacetime. A static universe is perfectly flat. [latex]ds^2=c^2dt^2 [\frac {dr^2}{1-kr^2}+r^2 (d\theta^2+sin^2\theta d\phi^2)][/latex] we write [latex](x^0,x^1,x^2,x^3)=(ct,r,\theta,\phi)[/latex] we set the above as [latex]g_{00}=1,g_{11}=-\frac{R^2(t)}{(1-kr^2)},g_{22}=-R^2 (t)r^2, g_{33}=-R^2 (t)r^2sin^2\theta [/latex] the geodesic equation of the above is [latex]\frac {du^\mu}{d\lambda}+\Gamma^\mu_{\alpha\beta}\mu^\alpha\mu^\beta=0 [/latex] if the particle is massive [latex]\lambda[/latex] can be taken as the proper time s. If it is a photon lambda becomes an affine parameter. So lets look at k=0. we set [latex]d\theta=d\phi=0 [/latex] this leads to [latex]ds^2=c^2t^2-R^2 (t)dr^2=c^2dt^2-dl^2=dt^2 (c^2-v^2)[/latex] where dl is the spatial distance and v=dl/dt is the particle velocity in this commoving frame. Assuming it to be a massive particle of mass "m" [latex]q=m (\frac {dl}{ds})c=(1-\frac {v^2}{c^2})^{\frac{1}{2}}[/latex] from the above a photon emitted at time [latex]t_1[/latex] with frequency [latex]v_1 [/latex] which is observed at point P at time [latex]t_0 [/latex] with frequency [latex]v_0[/latex] with the above equation we get [latex]1+z=\frac {R (t_0)}{R (t_1)}[/latex] Please note were still in commoving coordinates with a static background metric. [latex]z=\frac {v}{c}[/latex] is only true if v is small compared to c. from this we get the Linear portion of Hubbles law [latex]v=cz=c\frac{(t_0-t_1)\dot{R}t_1}{R(t_1)}[/latex] now the above correlation only holds true if v is small. When v is high we depart from the linear relation to Hubbles law. We start hitting the concave curved portion. The departures from the linear relation requires a taylor series expansion of R (t) with the present epoch for this we will also need H_0. note the above line element in the first equation does not use the cosmological constant aka dark energy. This above worked prior to the cosmological constant Now for the departure from the linear portion of Hubbles law. [latex] v=H_Od, v=cz [/latex] when v is small. To this end we expand R (t) about the present epoch t_0. [latex]R (t)=R[(t_0-t)]=R(t_0)-(t_0)-(t_0)\dot {R}(t_0)+\frac {1}{2}(t_0-t)^2\ddot{R}(t_0)...=R (t_0)[1-(t_0-t)H_o-\frac {1}{2}(t_0-t)q_0H^2_0...[/latex] with [latex]q_0=-\frac{\ddot{R}(t_0)R(t_0)}{\dot{R}^2(t_0)}[/latex] q_0 is the deceleration parameter. Sometimes called the acceleration parameter. now in the first circumstances when v is small. A light ray follows [latex]\int_{t_1}^{t^0} c (dt/R (t)=\int_0^{r_1}dr=r_1 [/latex] with the use of this equation and the previous equation we get [latex]r=\int^{t_0}_t=\int^{t_0}_t cdt/{(1-R (t_0)[1-(t_0-t)H_0-...]}[/latex] [latex]=cR^{-1}(t_0)[t_0-t+1/2 (t_0-t)^2H_0+...][/latex] here r is the coordinate radius of the galaxy under consideration. Solving the above gives.. [latex]t_0-t=\frac {1}{c}-\frac {1}{2}H_0l^2/c^2 [/latex] which leads to the new redshift equation [latex]z=\frac {H_0l^2}{c+\frac {1}{2}(1+q_0)H^2_0l^2/c^2+O (H^3_0l^3)}[/latex] Edited July 20, 2016 by Mordred Link to comment Share on other sites More sharing options...
captcass Posted July 19, 2016 Author Share Posted July 19, 2016 Thank you for that. Bed time for me so I'll look at it in the morning. So.....prepare to say, "ARRRRGGGHHHH". I worked it out on the basis of slowing the pulses and it comes out the same as with the length of a meter. I guess that figures as I use the differences in time to determine the length of a meter........ I do appreciate the work you did on that derivation and will look at it in the morning. Where does regular physics have z = 1100 occurring? I'm starting to wonder if my approach might explain just the relativistic visual perspective the normal science manifests. Headache time....... In my theory new space is always being created due to endless time dilation in the time vortices of the funadamental particles. I postulate this new space is absorbed in the spins of the particles, and the curvature of motion of events thru space. Perhaps instead it manifests in the expansion you see.......As time and therefore length (distance) are always relative, it might just always look like this is the way it works. Headache..... Link to comment Share on other sites More sharing options...
Mordred Posted July 19, 2016 Share Posted July 19, 2016 (edited) the CMB surface of last scattering is z=1100. However thats roughly 3 times the length of Hubbles radius. Its also referred to the cosmological event horizon (Observable universe which requires expansion in the commoving frame above). However to get an accurate distance you need the last formula I posted. In my theory new space is always being created due to endless time dilation in the time vortices of the funadamental particles. I postulate this new space is absorbed in the spins of the particles, and the curvature of motion of events thru space. Perhaps instead it manifests in the expansion you see.......As time and therefore length (distance) are always relative, it might just always look like this is the way it works. Headache..... this makes no sense sorry. Particularly since new space means new volume. Which means expansion. Remember your coordinates I really don't think you want to get into particle spin statistics. The terminology in the last quoted section is sheer gibberish. I strongly suggest studying how the equations of state of various particles influence expansion. Rather than make a fool of yourself that is... Oh by the way just so You can read a Peer review showing the first redshift equation as only accurate at close distances where v is much less than c. http://arxiv.org/abs/astro-ph/?9905116 "Distance measures in cosmology" David W. Hogg keep in mind Hogg's skipped numerous steps. He is also highly cited for other redshift works. One is trying to show cosmological redshift as gravitational redshift. (the method requires endless microsteps) the cosmological redshift with corrections is far more flexible and simpler. As I mentioned equations of state... here is another workup I did on another thread. [latex]DU=pdV[/latex]. First take the first law of thermodynamics. [latex]dU=dW=dQ[/latex] U is internal energy W =work. As we dont need heat transfer Q we write this as [latex]DW=Fdr=pdV[/latex] Which leads to [latex]dU=-pdV.[/latex]. Which is the first law of thermodynamics for an ideal gas. [latex]U=\rho V[/latex] [latex]\dot{U}=\dot{\rho}V+{\rho}\dot{V}=-p\dot{V}[/latex] [latex]V\propto r^3[/latex] [latex]\frac{\dot{V}}{V}=3\frac{\dot{r}}{r}[/latex] Which leads to [latex]\dot{\rho}=-3(\rho+p)\frac{\dot{r}}{r}[/latex] We will use the last formula for both radiation and matter. Assuming density of matter [latex]\rho=\frac{M}{\frac{4}{3}\pi r^3}[/latex] [latex]\rho=\frac{dp}{dr}\dot{r}=-3\rho \frac{\dot{r}}{r}[/latex] Using the above equation the pressure due to matter gives an Eos of Pressure=0. Which makes sense as matter doesn't exert a lot of kinetic energy/momentum. For radiation we will need some further formulas. Visualize a wavelength as a vibration on a string. [latex]L=\frac{N\lambda}{2}[/latex] As we're dealing with relativistic particles [latex]c=f\lambda=f\frac{2L}{N}[/latex] substitute [latex]f=\frac{n}{2L}c[/latex] into Plancks formula [latex]U=\hbar w=hf[/latex] [latex]U=\frac{Nhc}{2}\frac{1}{L}\propto V^{-\frac{1}{3}}[/latex] Using [latex]dU=-pdV[/latex] using [latex]p=-\frac{dU}{dV}=\frac{1}{3}\frac{U}{V}[/latex] As well as [latex]\rho=\frac{U}{V}[/latex] leads to [latex]p=1/3\rho[/latex] for ultra relativistic radiation. Those are examples of how the first law of thermodynamics fit within the equations of state. There is more intensive formulas involved. In particular the Bose-Einstein statistics and Fermi-Dirac statistics What the above correlates to is particle degrees of freedom. One can calculate how much influence any particle with known properties influence the temperature... pressure... expansion relations. Provided one knows the correct correlations to the Einstein field equations. (the above can and does affect geodesic equations. Which in turn can and does affect redshift to distance calculations) So to that end a sample of how to define a geodesic may be handy. For that Im going to cheat again and use a previous post... In the presence of matter or when matter is not too distant physical distances between two points change. For example an approximately static distribution of matter in region D. Can be replaced by tve equivalent mass [latex]M=\int_Dd^3x\rho(\overrightarrow{x})[/latex] concentrated at a point [latex]\overrightarrow{x}_0=M^{-1}\int_Dd^3x\overrightarrow{x}\rho(\overrightarrow{x})[/latex] Which we can choose to be at the origin [latex]\overrightarrow{x}=\overrightarrow{0}[/latex] Sources outside region D the following Newton potential at [latex]\overrightarrow{x}[/latex] [latex]\phi_N(\overrightarrow{x})=-G_N\frac{M}{r}[/latex] Where [latex] G_n=6.673*10^{-11}m^3/KG s^2[/latex] and [latex]r\equiv||\overrightarrow{x}||[/latex] According to Einsteins theory the physical distance of objects in the gravitational field of this mass distribution is described by the line element. [latex]ds^2=c^2(1+\frac{2\phi_N}{c^2})-\frac{dr^2}{1+2\phi_N/c^2}-r^2d\Omega^2[/latex] Where [latex]d\Omega^2=d\theta^2+sin^2(\theta)d\varphi^2[/latex] denotes the volume element of a 2d sphere [latex]\theta\in(0,\pi)[/latex] and [latex]\varphi\in(0,\pi)[/latex] are the two angles fully covering the sphere. The general relativistic form is. [latex]ds^2=g_{\mu\nu}(x)dx^\mu x^\nu[/latex] By comparing the last two equations we can find the static mass distribution in spherical coordinates. [latex](r,\theta\varphi)[/latex] [latex]G_{\mu\nu}=\begin{pmatrix}1+2\phi_N/c^2&0&0&0\\0&-(1+2\phi_N/c^2)^{-1}&0&0\\0&0&-r^2&0\\0&0&0&-r^2sin^2(\theta)\end{pmatrix}[/latex] Now that we have defined our static multi particle field. Our next step is to define the geodesic to include the principle of equivalence. Followed by General Covariance. Ok so now the Principle of Equivalence. You can google that term for more detail but in the same format as above [latex]m_i=m_g...m_i\frac{d^2\overrightarrow{x}}{dt^2}=m_g\overrightarrow{g}[/latex] [latex]\overrightarrow{g}-\bigtriangledown\phi_N[/latex] Denotes the gravitational field above. Now General Covariance. Which use the ds^2 line elements above and the Einstein tensor it follows that the line element above is invariant under general coordinate transformation(diffeomorphism) [latex]x\mu\rightarrow\tilde{x}^\mu(x)[/latex] Provided ds^2 is invariant [latex]ds^2=d\tilde{s}^2[/latex] an infinitesimal coordinate transformation [latex]d\tilde{x}^\mu=\frac{\partial\tilde{x}^\mu}{\partial x^\alpha}dx^\alpha[/latex] With the line element invariance [latex]\tilde{g}_{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g_{\alpha\beta}x[/latex] The inverse of the metric tensor transforms as [latex]\tilde{g}^{\mu\nu}(\tilde{x})=\frac{\partial\tilde{x}^\mu \partial\tilde{x}^\nu}{\partial x^\alpha\partial x^\beta} g^{\alpha\beta}x[/latex] In GR one introduces the notion of covariant vectors [latex]A_\mu[/latex] and contravariant [latex]A^\mu[/latex] which is related as [latex]A_\mu=G_{\mu\nu} A^\nu[/latex] conversely the inverse is [latex]A^\mu=G^{\mu\nu} A_\nu[/latex] the metric tensor can be defined as [latex]g^{\mu\rho}g_{\rho\nu}=\delta^\mu_\mu[/latex] where [latex]\delta^\mu_nu[/latex]=diag(1,1,1,1) which denotes the Kronecker delta. Finally we can start to look at geodesics. Let us consider a free falling observer. O who erects a special coordinate system such that particles move along trajectories [latex]\xi^\mu=\xi^\mu (t)=(\xi^0,x^i)[/latex] Specified by a non accelerated motion. Described as [latex]\frac{d^2\xi^\mu}{ds^2}[/latex] Where the line element ds=cdt such that [latex]ds^2=c^2dt^2=\eta_{\mu\nu}d\xi^\mu d\xi^\nu[/latex] Now assunme that the motion of O changes in such a way that it can be described by a coordinate transformation. [latex]d\xi^\mu=\frac{\partial\xi^\mu}{\partial x^\alpha}dx^\alpha, x^\mu=(ct,x^0)[/latex] This and the previous non accelerated equation imply that the observer O, will percieve an accelerated motion of particles governed by the Geodesic equation. [latex]\frac{d^2x^\mu}{ds^2}+\Gamma^\mu_{\alpha\beta}(x)\frac{dx^\alpha}{ds}\frac{dx^\beta}{ds}=0[/latex] Where the new line element is given by [latex]ds^2=g_{\mu\nu}(x)dx^\mu dx^\nu[/latex] and [latex] g_{\mu\nu}=\frac{\partial\xi^\alpha}{\partial\xi x^\mu}\frac{\partial\xi^\beta}{\partial x^\nu}\eta_{\alpha\beta}[/latex] and [latex]\Gamma^\mu_{\alpha\beta}=\frac{\partial x^\mu}{\partial\eta^\nu}\frac{\partial^2\xi^\nu}{\partial x^\alpha\partial x^\beta}[/latex] Denote the metric tensor and the affine Levi-Civita connection respectively. Starting to see how I mean when I say your 90 page paper lacks the required mathematical weight?????? 😉 Edited July 19, 2016 by Mordred 1 Link to comment Share on other sites More sharing options...
captcass Posted July 19, 2016 Author Share Posted July 19, 2016 Wow, thanks for this. You say "...Hogg's.....is also highly cited for other redshift works. One is trying to show cosmological redshift as gravitational redshift. (the method requires endless microsteps) the cosmological redshift with corrections is far more flexible and simpler." Are the "endless microsteps" similar to what I am doing but more detailed? Do you know where I can read that attempt? I had wanted to make my determination per meter but couldn't figure out how to do it. Too many microsteps.....just wondering if that was a similar attempt. Sorry about the gibberish. Like I said, I suffer from verbosity compounded by free association, especially if I have an extra glass of wine. I am going to do some more derivations for different bodies at different distances to see if what I am doing is consistent. If not, I'll trash this idea for the redshift. Trouble is, there is no way I can logically accept a universe that expands into a cold dead thing. This can only mean we are making a mistake about what the redshift indicates. There must be another relativisitic effect creating that impression. Thought I had it here, but if the other derivations don't work, I will have to assume I don't. As the diameter of the observable universe is thought to be 46 Gly and I came up with 42.2 in my z =1100 approximation, I am also going to try using 46 in a full derivation to see what I get. I really do appreciate all the time you've given me. I understand the science and reasoning. I used to embrace it. But the conclusions regarding the fate of the universe is just incomprehensible to me so I need to find another relativistic reason for z. I know I am not alone in this endeavor. Lots of people don't like the conclusions. I'll get back to you when I do more derivations. I know this thread is about the Hubble shift, but I would appreciate your ideas on my GEC derivation and the Andromeda derivation I use it for. Thanks. PS It did occur to me that the apparent creation of new space by time dilation would create an expanding universe. I believe this must be offset by another effect. For instance, as the update shifts through space it accelerates the rate of time in the frame it is updating, shortening the length of a meter, so the update appears to be shrinking the size of the universe, or it is absorbed in curvature. Or......? One more question. I have been searching to try to find out at what distance the visible spectrum would redshift to the extent there is no more visible light. Do you know that distance? Link to comment Share on other sites More sharing options...
Mordred Posted July 19, 2016 Share Posted July 19, 2016 (edited) Here is the paper. http://www.google.ca/url?q=http://ciencias.bogota.unal.edu.co/fileadmin/content/oan/documentos/maestria/documentos/Cosmologia.pdf&sa=U&ved=0ahUKEwi6he_E-f_NAhVS82MKHYbhDh4QFggbMAM&sig2=iydAcDLIS7F_DZDWkLB2ew&usg=AFQjCNFVBsrWlemsOxVudLQqjXhJTbuSrw here is a counter paper... http://www.google.ca/url?q=http://arxiv.org/pdf/0911.3536&sa=U&ved=0ahUKEwi6he_E-f_NAhVS82MKHYbhDh4QFggSMAA&sig2=mBEbBM8SOGCcLnwoqkmypg&usg=AFQjCNH9F3gn0P4YiZ_m0R0A3HxQklldGg I wouldn't let the fate of the universe bother you. The heat death fate is based on the assumption that the present dynamics will always hold true into the extremely far future. Science uses all forms of light not just visible light. However you can look at the wavelength range for visible light. Then calculate the redshift to find when visible light will no longer be visible. Edited July 19, 2016 by Mordred Link to comment Share on other sites More sharing options...
michel123456 Posted July 19, 2016 Share Posted July 19, 2016 Here is the paper. http://www.google.ca/url?q=http://ciencias.bogota.unal.edu.co/fileadmin/content/oan/documentos/maestria/documentos/Cosmologia.pdf&sa=U&ved=0ahUKEwi6he_E-f_NAhVS82MKHYbhDh4QFggbMAM&sig2=iydAcDLIS7F_DZDWkLB2ew&usg=AFQjCNFVBsrWlemsOxVudLQqjXhJTbuSrw here is a counter paper... http://www.google.ca/url?q=http://arxiv.org/pdf/0911.3536&sa=U&ved=0ahUKEwi6he_E-f_NAhVS82MKHYbhDh4QFggSMAA&sig2=mBEbBM8SOGCcLnwoqkmypg&usg=AFQjCNH9F3gn0P4YiZ_m0R0A3HxQklldGg I wouldn't let the fate of the universe bother you. The heat death fate is based on the assumption that the present dynamics will always hold true into the extremely far future. Science uses all forms of light not just visible light. However you can look at the wavelength range for visible light. Then calculate the redshift to find when visible light will no longer be visible. But then, other radiation that is not visible now will become visible, I suppose. It will not become an era of darkness. Or do I miss something? Link to comment Share on other sites More sharing options...
imatfaal Posted July 19, 2016 Share Posted July 19, 2016 But then, other radiation that is not visible now will become visible, I suppose. It will not become an era of darkness. Or do I miss something? Heat death is a proposed final stage when all useable mass - energy has been converted to heat and the universe is full of nothing but long wavelength radiation which is slowly stretched (like the CMBR was stretched from hot ultraviolet to very cold microwave). Nothing is hot enough, energetic enough to emit new short wavelength radiation. very boring very dull almost completely homogeneous - entropy has won and there are no structures, nothing above the almost lowest energy state; in fact nothing to do apart from sit back and hope one of the cyclical models were true. Link to comment Share on other sites More sharing options...
StringJunky Posted July 19, 2016 Share Posted July 19, 2016 Nothing is hot enough, energetic enough to emit new short wavelength radiation. very boring very dull almost completely homogeneous - entropy has won and there are no structures, nothing above the almost lowest energy state; in fact nothing to do apart from sit back and hope one of the cyclical models were true. The universe will flat-line. Link to comment Share on other sites More sharing options...
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