Mordred Posted November 7 Posted November 7 (edited) the geometry is changing your photons must travel through spacetime. as the universe expands the distance your photons must travel increases. This leads to gravitational redshift which is Z. https://en.wikipedia.org/wiki/Gravitational_redshift That was the FLRW metric above the scale factor changes as your photons are trying to reach the observer. Here is the geometry again assume the flat case as its the easiest case \[d{s^2}=-{c^2}d{t^2}+a({t^2})[d{r^2}+{S,k}{(r)^2}d\Omega^2]\] \[S\kappa(r)= \begin{cases} R sin(r/R &(k=+1)\\ r &(k=0)\\ R sin(r/R) &(k=-1) \end {cases}\] if you want the line integral for flat the total interval along the Worldline (null geodesic photon path is) \[\delta S=\int^b_a=\sqrt{ds}=\int^b_a\sqrt{-cdt)^2+(dx)^2+(dy)^2+(dz)^2}\] so using the Radius of the universe and how it changes the distance becomes \[(dL)^2=(\frac{dr}{\sqrt{1-r^2}{/R^2}})^2+(r d\phi)^2\] so your photons must travel further the last 2 equations do not include scale factor changes but include the same relation to determine scale factor by \[a(t)\frac{r}{R}\] Edited November 7 by Mordred
kawiusz Posted November 7 Author Posted November 7 Mister, what exactly is moving with the relative velocity v<c in my universe with a single observer and a single photon that he emitted?
Mordred Posted November 7 Posted November 7 (edited) The blooming radius of the Observable universe pick any object or galaxy at the furthest you can examine and measure it how else do you think physics measures expansion pick any three or more galaxies and measure how they change distance from each other. I'm honestly beginning to feel your just trolling as no one an be that blind either that or you really need to study basic cosmology either that or I'm talking to some kid between ages 10 to 15. That has no prior knowledge of physics if thats true on the last part then we can help teach you the basics Edited November 7 by Mordred
kawiusz Posted November 7 Author Posted November 7 Is the blooming radius of the observable universe shorter than the distance traveled by my only photon emitted by my only observer? If it's not, then it's relative velocity can't be less than c. Is it? Don't edit your answers after you were replied.
Mordred Posted November 7 Posted November 7 (edited) The universe is smaller in the past when the photon gets emitted during its travel the universe expands how are you missing that detail ? the Universe is at its largest today and smaller in the past. Do you understand that part ? that is literally what expanding means Edited November 7 by Mordred
kawiusz Posted November 7 Author Posted November 7 Do you understand the definition of the radius of the observable universe? Has it ever increased with the velocity v<c? Don't edit your answers after you were replied.
Mordred Posted November 7 Posted November 7 (edited) I already answered that read back prior to Hubble radius v<c after Hubble radius v>c Hubble radius The Observable universe if bigger than the Hubble radius by a significant amount. Hubble radius is defined as the radius where the age of the universe times c. So if our universe is 13.8 Gly the Hubble radius is only 13.8 Gly our observable universe is however 46.3 Gly in radius https://en.wikipedia.org/wiki/Hubble_volume equation from above link \[R_H=\frac{c}{H_0}\] the SR transformations will give ds^=0 at the above Hubble radius our universe is larger precisely what I described in my first couple of posts https://en.wikipedia.org/wiki/Observable_universe The reason why the universe exceeds the Hubble horizon is the following \[H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}\] \(H_0\) is not constant that is todays value but in the past its higher and at its highest at BB. Those equations of state I mentioned Edited November 7 by Mordred
kawiusz Posted November 7 Author Posted November 7 Your answer to my question should be: No, the observable universe radius has never increased with the velocity v<c. So I repeat my previous question: Mister, what exactly is moving with the relative velocity v<c in my universe with a single observer and a single photon that he emitted?
Mordred Posted November 7 Posted November 7 I'm describing as measured from the Commoving observer using recessive velocity MISTER.
kawiusz Posted November 7 Author Posted November 7 (edited) There is no other comoving observer in my simulation, mister. There must be at least two to be comoving. Edited November 7 by kawiusz
Mordred Posted November 7 Posted November 7 THEN define your reference frame mathematically and post the relevant details for discussion as you are required to have a geometry that should be a cinch. observer is measured by you it is commoving as everything in our universe is a commoving coordinate. So you better get to work on your observer. As it doesn't fall under SR nor GR unless your in some absolute frame of reference in which case this thread will belong under Eather based and isn't main stream physics.
kawiusz Posted November 7 Author Posted November 7 (edited) Can't you seriously imagine a universe with a single observer with his own reference frame, that is also the only reference frame in this universe? Edited November 7 by kawiusz
Mordred Posted November 7 Posted November 7 As I stated that Observer could be you but you are a commoving observer under the equations I posted. Those work just as well expressing you but invariance requires any observer for proper velocity relations for the four momentum. All part of GR requirements also required for SO(3.1) Poincare group = spacetime metric. Simply arguing your the observer so it shouldn't matter doesn't work when the very coordinates your located at are commoving with the universe. hence you would need a different geometry with a different flow of any measurements you take of any particles or objects around you unless you are moving with the coordinates ie fixed coordinate.
kawiusz Posted November 7 Author Posted November 7 And these comoving coordinates manifest themeselves in the expanding distance between the observer and the photon that he emitted. "hence you would need a different geometry with a different flow of any measurements you take of any particles or objects around you unless you are moving with the coordinates ie fixed coordinate." My geometry is expressed by the same expanding distance and the expanding wavelenght of the photon.
Mordred Posted November 7 Posted November 7 (edited) Then you better show your equations if it deviates from GR you have your work cut out for you and believe me I'll be able to tell. If you don't understand the EFE and how it applies to the FLRW metric then you really don't understand its true flexibility. Every equation I posted you can be the observer. Even the only one why is the recessive velocity important is simple, velocity as shown by the Lorentz transformations directly apply to how we measure time so using recessive velocity is how we factor in the time time component vs the space space components. Using GR relations I will show how the FLRW metric fits with GR. but first here is an interesting trick simply take \[v_{recessive}=H_oD\] and to get an accurate recessive velocity all the way out to the cosmological event horizon do this substitution \[v_{recessive}=(H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}D\] Now this shows that the rate of change in distance to the Cosmological event horizon is accelerating and that the Hubble value for expansion is also not constant over time. The above substitution calculates how H changes as a function of cosmological redshift. That applies the equations of state and includes both equations of the FLRW metric its geometry previously shown. The second term is the acceleration equation for how radiation and matter energy densities evolve over time expansion relations. That's the portion under the square root including cosmological term. now under GR the above relations would give the following including all others I have already posted such as the FLRW metric. take the EFE (Einstein field equation) which is needed for its field treatments of multipoint coordinates. Any coordinate can be an observer including yourself \[G_{\alpha\beta}=\frac{8\pi G}{c^4}T_{\alpha\beta}\] \(T_{\alpha\beta}\) being the stress energy momentum tensor. \[ds^2=g_{\alpha}{\beta}dx^\alpha dx^\beta\] where \(g_{\alpha\beta}\) is the metric tensor as this is an orthogonal matrix above the non vanishing elements can be given in matrix form for the FLRW metric as below for the metric \[g_{\alpha\beta}=\begin{pmatrix}1&0&0&0\\0&-\frac{a^2}{1-kr^2}&0&0\\0&0&-a^2r^2&0\\0&0&0&a^2r^2\sin^2\theta\end{pmatrix}\] for the stress energy momentum tensor \(T_{00}=\rho c^2,,,T_{11}=\frac{Pa^2}{1-kr^2}\) the left hand side of the Einstein field equation becomes \[G_{00}=3(a)^{-2}(\dot{a}^2+kc^2)\] \[G_{11}=-c^{-2}(a \ddot{a}+\dot{a}^2+k)(1-kr^2)-1\] using above the time evolution of the cosmic scale factor then becomes \[\frac{a}{a}^2+\frac{kc^2}{a^2}=\frac{8\pi G}{3}G\rho\] \[2\frac{\ddot{a}}{a}+\frac{\dot{a}}{a}^2+\frac{kc^2}{a^2}=\frac{8\pi}{3}P\] where \(\rho\) is the energy density and P is the pressure. The overdot 's above the scale factor terms are the velocity for single dot with two dots its the acceleration term. This is shows why we use velocity and acceleration the choice of observer is irrelevant its obviously practical to more often than not treat yourself as the observer. The above also shows that the FLRW metric is a GR solution and its generalized relations. They already include any SR application but under GR field treatment which is better suited for spacetime curvature. Spacetime under GR always include the equations of momentum given by \[E^2=\sqrt{(pc)^2+(m_oc^2)^2}\] which is the full equation for \(e=m_oc^2\) called the energy momentum relation for previous. the above shows a mathematical proof that the substitution below is valid and how its applied. SR works for the first term but only at very close range and it degrades in accuracy due to equation 2 below. Equation two is a product of those relations above including how radiation, matter and the cosmological constant evolve over time in energy density and pressure relations \[v_{recessive}=H_oD\] \[v_{recessive}=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}D\] That should give you a good overview of why commoving coordinates are an essential aspect to an expanding universe. Its the influence of our matter/energy content and how they affect expansion. Yes the above is complex but once you understand it. It is absolutely remarkable how flexible the above is in describing how the scale factor evolves over time and why the affine connection for proper time or cosmic time is tied to the scale factor as given as \(a(t)\). Edited November 7 by Mordred
kawiusz Posted November 7 Author Posted November 7 (edited) You probably forgot to throw in the multiplication table. "Then you better show your equations if it deviates from GR you have your work cut out for you and believe me I'll be able to tell." Every possible and arbitrary but monotonically increasing function of the scale factor function will have the same result for my simulated observer and I gave the explanation. That's why my explicit form of [math]a(t)[/math] is meaningless and makes no difference. Edited November 7 by kawiusz
Mordred Posted November 7 Posted November 7 (edited) Not a chance mate your equations are essentially useless as they do not describe how our universe evolves over time. Nor have you really anything particularly useful for physics scale factors are a dime a dozen in numerous theories Every major theory will have some form of scale factor. If your scale factor doesn't include any other related metrics specifically under a geometry treatment applicable to the system its describing then its essentially of little use as a replacement. Generating scale factor simulations is nothing new to physics and they can be widely varied. If your equations don't conform to observational evidence it's insufficient as proof. Edited November 7 by Mordred
kawiusz Posted November 7 Author Posted November 7 (edited) I'm not your mate, pal. I can get the most accurate, explicit form of the equation for a(t) calculated from the Friedmann eq. in Lambda CDM with the most accurate density parameters and I will have the same result for my simulated observer, but you simply don't get it. Don't edit your answers after you were replied. Edited November 7 by kawiusz -1
Mordred Posted November 7 Posted November 7 (edited) So you claim yet have only shown a single equation. Prove it mathematically here go ahead I challenge you to put your mathematics where your mouth is. Instead of claiming mathematically prove your claims. Start by proving it will work under SR first After all we still have to prove it will work in curved spacetime. Come on pit your mathematics skills and your single equation under examination that it will work under GR. I would love to see that but I already know you can't Edited November 7 by Mordred
kawiusz Posted November 7 Author Posted November 7 Because I have a total number equal zero of my own equations to prove. You can't prove the arbitrary scale factor function and the Friedmann's a(t) is already perfectly proved. You totally can't read with comprehension. What institute gave you your degrees?
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