caracal Posted March 27, 2016 Author Posted March 27, 2016 Matter in neutron stars: The observations of pulsars and binary system neutron star may be able to prove or disprove this cosmological model. there are few thousand known pulsars currently. The gravitational time dilation inside non-rotating neutron star would be: [latex] ( \frac{\tau}{t_{inf}}) = \sqrt{ 1 - (\frac{1}{c^2}) ( \frac{GMr^2}{R^2} - \frac{3GM}{R})} [/latex] and you can see straight from this equation that the time dilation in the center is 3 times larger than on the surface. The real neutron star is rotating, so it has Kerr metric ? or different metric, and it is not currently known what kind of equation of state does the matter have inside the neutron star. If neutron star has mass m = 2 * mass of a sun and diameter 15 km, then its surface should have time dilation (1 - 0.1045) = (1 - D) According to particle contraction law what is the basis of this cosmological speculative model, that depends on the proper time of the matter, the matter in the neutron star should gain transformation difference relative to ordinary matter with no time dilation such that: [latex] D = 0.1045 [/latex] [latex] k = H = 6.93*10^{-11} [1/year] = 2.20 *10^{-18} [1/s] [/latex] [latex] L_{rel}=e^{HDt} = exp(0.1045* 6.93*10^{-11} [1/year] * t )= exp(7.240*10^{-12}[1/year]\cdot t[years]) [/latex] [latex] \approx( 1+7.240*10^{-12}[1/year]\cdot t[years])) [/latex] This should affect to the Newton's second law such that: 1. The gravitational acceleration of neutrons star in binary system should increase by factor 1/L 2. The moment of inertia of a pulsar should decrease by factor 1/L 3. The inertial mass of neutron star should decrease and since neutron star have more time dilation in the center than on its surface, maybe 3 times according to non-rotating case, then the effect may be approximately 3 times larger in the center of the neutron star. The older neutron star, the more time the matter has had time dilation and therefore the more it should have these 3 changes. Also the pulsation frequency of pulsar should decrease and its radius should grow - if the pulsar had homogenous time dilation then the decrease rate would be factor of 1/L and the increase of radius would be by factor L of course this is so only if i havent done any mistake.
Mordred Posted March 28, 2016 Posted March 28, 2016 (edited) Kerr metric is appropriate for a rotating body. Although there are artifacts of metrics in any metric. So far I'm impressed by the amount of diligence you've shown. Recognizing your still working on the model I've chosen not to interfere. For the equations of state inside a star the best reference I can think of is "physics of the interstellar medium" It is a textbook, it is extremely detailed in details such as temperature absorbtion of different elements etc Not my area of expertise. On this forum probably the best member I can think of is Sensei. He is far more adept at nuclear processes than I My suggestion is post a seperate thread on internal star processes. Sometimes in order to find the details on a forum a new thread on specific aspects will get the attention of the right people. (Key note I may not agree with your idea, but I refuse to interfere with your diligent study) Edited March 28, 2016 by Mordred
caracal Posted March 28, 2016 Author Posted March 28, 2016 ok, thanks:) I came cross into one interesting thing that is related to quasars - according to few studies - quasars with different distances do not show signs of cosmological time dilation in their power spectrum. (On time dilation in quasar light curve M. R. S. Hawkins 1988) I am not sure if this has some natural explanation. In this model there are two new changes that should happen in quasars in cosmological time scale: 1 Black holes should grow at rate exp( Ht ) = exp (6.93* 10^-2 [1/Byears] * t [byears] ) relative to contracting observer (also their gravitation constant also changes such that G= L^2 and in cross interactions with ordinary matter G = L(?), and their mass decreases M=1/L relative to contracting observer) 2 the matter in quasar jet, being ultrarelativistic or relativistic, should gain transformation difference at rate if they travel long distances
caracal Posted April 1, 2016 Author Posted April 1, 2016 (edited) Problem: Rotating black hole that has different inertial mass than gravitational mass Kerr metric describes rotating black hole: [latex] c^2 d\tau ^2 = \left ( 1- \frac{r_s r}{\rho ^2} \right )c^2 dt^2 - \frac{\rho ^2}{\Delta }dr^2 - \rho ^2 d\theta ^2 - \left ( r^2 + \alpha ^2 + \frac{r_s r\alpha ^2}{\rho ^2}sin^2\theta \right )sin^2\theta d\phi ^2 + \frac{2r_s \alpha sin^2 \theta }{\rho ^2}c dt d\phi [/latex] ,where [latex]\alpha = \frac{J}{Mc} [/latex] [latex]\rho ^2 = r^2 + \alpha ^2cos^2\theta [/latex] [latex] \Delta = r^2 + r_s r + \alpha ^2 [/latex] Einstein's theory of gravitation has a postulate that gravitational mass is equal to inertial mass. In this cosmological model however the matter that has transformation difference relative to observer, appears to behave as if it had different inertial mass than gravitational mass: [latex] m_{inertia} \neq m_{gravitation} [/latex] for black holes that has transformation difference relative to ordinary matter according to particle contraction law, the apparent difference between the inertial and gravitational mass is: [latex] \frac{m_{inertial}}{m_{gravitational}}= e^{-6.93 * 10^{-11} \cdot T[years]} \approx (1 - 6.93*10^{-11} \cdot T[years]) [/latex] ,where T is the age of the black hole The problem is - how does Kerr metric change, if the black hole has different inertial mass than gravitational mass? At least this change affects to inertial momentum, and therefore to all properties that depend on it, [latex] (\alpha , \rho ,\Delta ) [/latex] The different inertial momentum affects to : - the form of ergosphere - frame dragging - Penrose mechanism / Penrose process Edited April 1, 2016 by caracal
Mordred Posted April 7, 2016 Posted April 7, 2016 (edited) One of the key aspects to understand expansion is to understand the potential and kinetic energy aspects with conservation of energy. For this we can detail using Newtons laws. [latex] F=\frac{GMm}{r^2}[/latex] Mass density we will use [latex]\rho[/latex] which is the mass per unit volume. Now assume a field of test particles. Motion and mass currently unimportant. One of the aspects of the shell theorem in Newtons laws is the test particle will only notice a force from the center of mass. In a homogeneous and isotropic distribution any test particle or CoM can be used. As we're dealing with test particles we just need the mass relation. [latex]M=\frac{4\pi\rho^3}{3}[/latex] So [latex]E_p=-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}[/latex] Kinetic energy is [latex]E_k=1/2m\dot{r}^2[/latex] [latex]U=E_k+E_p[/latex] U is just a dimensionless constant to equate total energy must be set as a constant value. So the above translates to [latex]U=\frac{1}{2}m\dot{r}^2-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}[/latex] Now with the vector relation of the radius to length we can denote the scale factor. [latex]\overrightarrow{r}=a(t)\overrightarrow{x}[/latex] Where a is a function of time. This leads to [latex]U=\frac{1}{2}m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2 m[/latex] Multiply each side by [latex]2/ma^2x^2[/latex] Leads to [latex](\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}[/latex] [latex]kc^2=-2U/mx^2[/latex] [latex]k=-2U/mc^2x^2[/latex] K is the curvature constant. Please note the above doesn't care about what mass or momentum the test particles have except as a consequence of the above relations. [/quote[ Edited April 8, 2016 by Mordred
caracal Posted April 8, 2016 Author Posted April 8, 2016 (edited) i answer first to this and later to your other writings. Now my question is How does your contraction work and go from this[latex](\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}[/latex]To this equation[latex] \left ( \frac{\dot{a}}{a} \right )^2 =\left ( \frac{8 \pi G}{3} \right )\left ( \frac{\rho_{r,0}}{a^4} \right )+k^2 +\Lambda [/latex]The terms [latex](\frac{\rho_{r,0}}{a^4})+k^2 +\Lambda [/latex]Describe the total density value. Though you replaced the density of matter evolution with k^2What happened to the portion[latex]\frac{kc^2}{a^2}[/latex] You are mixing here two things. the k i use here has nothing to do with the curvature of the space, it is a constant factor parameter in the transformation factor function L, if L is decreasing exponential function of co-contracting time: [latex] L(t') = e^{-k(t'-t'_0)} [/latex] Contracting observer observes all cosmological distances grow simultaneusly such that [latex] R(t') = R_0 / L = R_0 \cdot e^{k(t'-t'_0)} [/latex] I could select ofher symbol instead of k, say p : [latex] L(t') = e^{-p(t'-t'_0)} [/latex] If i add the curvature in the equation, i should take into account that in the viewpoint of contracting observer, the curvature radius of the space R appear to increase (similarly as all other distances) such that: [latex] R_{curv}(t) = R_0 / L(t') = R_0 \cdot e^{p(t'-t'_0)} [/latex] and i think the curvature k is inverse of the curvature radius: [latex] k = 1/R [/latex] ,when [latex] k(t) = k_0 \cdot e^{-p(t'-t'_0)}[/latex] the term p^2 comes from that in the viewpoint of contracting observer, the space appears to expand more, and if the contraction of matter is exponential function of co-contracting time, the solution for vacuum universe is [latex](\frac{\dot{a}}{a})^2= p^2 [/latex] And then i claim a theorem that i can just add this vacuum solution term p^2 coming from contraction of matter to the ordinary Friedmann-equation In Lambda-CMD universe: so the Friedmann equation would be now: [latex]\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2 e^{-p(t'-t'_0)}}{a^2} + p^2 + \Lambda [/latex] Or, if you define the lambda and k differently: [latex] H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{ke^{-p(t'-t'_0)}}{a^2} + p^2 + \frac{\Lambda}{3} [/latex] This is the 1st. friedmann equation only if the transformation factor function is exponential [latex] L(t) = e^{-k(t'-t'_0)} [/latex] This is also equation for space that has ordinary matter, radiation, curvature, cosmological constant - and this new particle contraction / matter contraction - phenomenon, assuming that nearly all matter has same proper time (postulate is that particle contraction depends on proper time), same transformation factor L and also same transformation factor change rate dL/dt ( which is not the case in compact stellar objects, relativistic particles and generally in all matter that has had different time dilation history. Due to different time dilation histories of the matter, there should be so called differently transformed matter in the universe. ) Edited April 8, 2016 by caracal
Mordred Posted April 9, 2016 Posted April 9, 2016 (edited) No I understood what you were using k for. My point was to show you lost some key terms in the above equations without accounting for those lost terms. The curvature term is a consequence of the metrics I posted not an arbitrary choice. You have a different matter contraction rate you will need to account for a different curvature rate. Otherwise I don't see how you will get accurate geodesic relations to o observation. The curvature term is a key aspect of worldlines and lightcones seen in the Universe. The problem I have looking over your equations is its becoming more and more apparent that you fitting your terms into those equations without studying how the equations were derived in the first place. The equations I posted are specifically detailing potential and kinetic energy relations in an evolving volume. Simply slapping your terms into the later derivitaves is not the same as applying the needed changes at the original derivatives. With the above what you SHOULD be looking at is. With ypur model what is the potential and kinetic energy relations needed to maintain conservation of energy. One of the key aspects to understand expansion is to understand the potential and kinetic energy aspects with conservation of energy. For this we can detail using Newtons laws. [latex] F=\frac{GMm}{r^2}[/latex] Mass density we will use [latex]\rho[/latex] which is the mass per unit volume. Now assume a field of test particles. Motion and mass currently unimportant. One of the aspects of the shell theorem in Newtons laws is the test particle will only notice a force from the center of mass. In a homogeneous and isotropic distribution any test particle or CoM can be used. As we're dealing with test particles we just need the mass relation. [latex]M=\frac{4\pi\rho^3}{3}[/latex] So [latex]E_p=-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}[/latex] Kinetic energy is [latex]E_k=1/2m\dot{r}^2[/latex] [latex]U=E_k+E_p[/latex] U is just a dimensionless constant to equate total energy must be set as a constant value. So the above translates to [latex]U=\frac{1}{2}m\dot{r}^2-\frac{GMm}{r^2}=-\frac{4\pi G\rho^3 m}{3}[/latex] Now with the vector relation of the radius to length we can denote the scale factor. [latex]\overrightarrow{r}=a(t)\overrightarrow{x}[/latex] Where a is a function of time. This leads to [latex]U=\frac{1}{2}m\dot{a}^2x^2-\frac{4\pi}{3}G\rho a^2x^2 m[/latex] Multiply each side by [latex]2/ma^2x^2[/latex] Leads to [latex](\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}[/latex] [latex]kc^2=-2U/mx^2[/latex] [latex]k=-2U/mc^2x^2[/latex] K is the curvature constant. Please note the above doesn't care about what mass or momentum the test particles have except as a consequence of the above relations. I screwed up the original post lol. Ah well. This section and metrics doesn't give a hoot what the particles are. These metrics work if you have nothing but radiation or a matter only universe. Or even just a scalar field. [latex] H^2=(\frac{\dot{a}}{a})^2=\frac{8\pi G}{3}\rho-\frac{ke^{-p(t'-t'_0)}}{a^2} + p^2 + \frac{\Lambda}{3} [/latex] Work your model through all the equations in my quoted post before slapping the terms you added to your quoted equation. This last equation change makes no sense. You already have the time evolution with the scale factor. You've arbitrarily changed the RHS of the equation without any change to the LHS. Essentially what you've done is Take the RHS side which describes kinetic and potential energy relations and slapped a time component into it. When the time correlations are on the LHS of the equation. The right hand side is how the kinetic and potential energy evolves with (expansion or contraction) as a function of time on the LHS. For example I should be able to arrive at your conclusion that [latex]G_{12}=\sqrt{G_1}{G_2}[/latex] But then according to your pdf the scale factor a and the Hubble constant do have the standard relations either. Then you state [latex] T_\mu\nu =\frac{1}{L^4}[/latex] Which is complete garbage [latex] T_{\mu\nu}[/latex] is a tensor matrix each diagonal component has a specific function. The three pressure terms involve different pressure influences, such as vorticity. You've randomly slapped your terms into both the Ricci tensor which again is a 4*4 matrix each component of that matrix has a specific function. Each coordinate in those matrix'es has different formula applications. Then there is also [latex] G{\mu\nu}[/latex]. Which you state transforms as 1. Yet your entire model involves a different geometry relation. Using coordinates [latex]x^0,x^1,x^2,x^3= ct,r,\phi,\theta[/latex] [latex]G_{00}=1[/latex] [latex]G_{11}=-R(t)/(1-kr^2)[/latex] [latex]G_{22}=-R^2(t)r^2[/latex] [latex]G_{33}=-R^2(t)r^2sin^2\theta[/latex] So with all the transformation rules you described with [latex]L_x[/latex] How can you possibly state the metric tensor remains unchanged ? (This is just one example of modifying formulas without understanding the formulas and how they are derived. Here is another. [latex]\frac{f^\prime}{f}=L[/latex] Makes absolutely zero mathematical sense. look at the units itself. Frequency is measured in Hertz SI units is s^-1. Length has SI unit. metre. one of the rules of dimensional analysis is the LHS must equal in units to the RHS. On the left hand side the left over units is 1. No units because your using the same units in the numerator and denominator. On the right hand hand side you have a metre. So unless your intention is that L is a dimensionless parameter not a unit measuring change in distance this equation doesn't work. It's plain WRONG. However given that you've applied L everywhere being dimensionless is the only possibility Particularly with everything else you transform in your HUGE list. from below how can you possibly state the transform for f=L when both energy and wavelength also change by L??::::::???? 2: Transformed observer that has transformed by factor L observes ordinary photon to have following properties: wavelength: l/l = 1/L frequency: f/f = L Energy: E/E = L Momentum: p/p = L Volume: V/V = 1/L^3 Cross-section area: A/A = 1/L^2 Electric field: E/E = L Magnetic field: B/B = L now the Lorentz transformation follows specific rules. Yet you don't account for this. etc etc... [latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{x}=\frac{x-vt}{\sqrt{1-v^2/c^2}}[/latex] [latex]\acute{y}=y[/latex] [latex]\acute{z}=z[/latex] So lets take equation [latex]\acute{t}=\frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}[/latex] how do I apply your transformation list to the above. Normal method is substitution. So if I only read your pdf and applied your transformations. (I'll change the prime symbol to your transformation as simply t. To avoid confusion. [latex]\acute{t}=\frac{(t/t^t=L)-(v/v^t=L)x/c^2}{\sqrt{1-(v/v^t=L)^2/c^2}}[/latex] [latex]\acute{t}=\frac{L-Lx/c^2}{\sqrt{1-L^2/c^2}}[/latex] Why doesn't that work lol. Could it possibly be all your transformation rules (=L) oh I didn't notice c_o/c =1 rule. [latex]\acute{t}=\frac{L-Lx/1^2}{\sqrt{1-L^2/1^2}}[/latex] [latex]\acute{t}=\frac{L-Lx}{\sqrt{1-L^2}}[/latex] Edited April 9, 2016 by Mordred
caracal Posted April 9, 2016 Author Posted April 9, 2016 i am in hurry, but i make some statements here: -the key idea in the cosmological model is that 1. when particles contracts, the macroscopical objects will contract isotropically, that is why the contracting observer - like us cannot measure or observe whether he has been contracted or not 2. But matter does not contract isotropically in cosmological length scales - all concentrations of matter contracts towards their center. The matter kind of fragmentates, the concentration centers become all the time smaller and smaller. This causes that contracting observer measures that the universe is expanding in cosmological length scales. I call this as "apparent expansion" or "distance expansion" IF there were very little matter density in the universe, and its gravity was neglible then there should be only two cosmological phenomena: 1. Ordinary expansion of the space that is remnant of the big band 2. Apparent expansion of the space (that is actually an observation illusion and caused by the contraction of the meter stick of the observer) and i think then the scale factor that is observed by contracting observer should be then a= a0K * 1/L where K is some constant that describes the ordinary expansion and 1/L is so called "distance expansion" Other thing i say that these transformation equations that are on that table, such as f'/f = 1/L Are valid only when comparing two physical entities that are otherwise equal, say for example two stars, or two protons that would be otherwise equal but the other is perfectly contracted isotropically relative to the other. in other words, when the other is perfectly isotropical and homogenous transformation of the other. For example if the system of two stars contracts such that the both stars undergo isotropical contraction by factor L<1, but their distance remains fixed - then the whole system does not anymore contract isotropically - and for example the transformation equation for the distance between the star is not longer R'/R = L , while the radii of the stars still follow the equation r'/r = L thanks, i have to think about those issues you wrote.
Mordred Posted April 9, 2016 Posted April 9, 2016 (edited) I understand what your trying to do. The problem is the math your using is incorrect. If you change [latex]T/\acute{t}=L[/latex] to [latex] \acute{t}=t-L[/latex] for example but keep L dimensionless ( probably better to use a different symbol... to avoid confusion.) Now regardless of if I choose the LHS or the RHS of the second equation I must include the units. (Not to imply this relation is correct, as it certainly differs from the Lorentz transformation) I just randomly chose that as an example. Your formulas may make more sense. though quite frankly if all your after is compression of a volume the FLRW metric already has an appropriate formula without modification. [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 sinh,r/R &k=-1 \end {cases}[/latex] You just set k=-1. For the hidden commoving coordinates. then set k=0 for the flat metrics we observe in our universe. The above is simply the spatial component of the FLRW metric including curvature Now all you would need to focus on is why the compression portion differs from what we observe. (These equations don't care what Fluid your modelling) Matter, radiation or Lambda. Or any combination *** Please don't just jump in and try to modify the last equation until you understand how it was derived*** Particularly since it critically involves geodesic aspects. Edited April 9, 2016 by Mordred
caracal Posted April 30, 2016 Author Posted April 30, 2016 (edited) i think the 1st Friedmann equation in flat space actually is the one i first proposed in co-transforming coordinates? why? because: -in this coordinate system all matter in the universe has relative L=1 in some given time -the gravity works as usual because the relative L=1 for all matter -this new phenomenom what i call "distance expansion", if it is exponential function, then it just gives constant term to Friedmann equation the only difference is that not all matter contracts at equal rate (compact matter like black holes and neutron stars, and ultrarelativistic matter like cosmic rays and jets) but if i assume amount of this matter to be small the Friedmann 1 equation should be same as in Lambda-CMD universe, where lambda is substituted by this constant. but in non-flat space i think i have to change the Friedmann equation since the curvature radius increases like r'/r = 1/L = e^(pt) I think the transformation equation math works exactly as i described. The L is dimensionless factor, what i call "transformation factor" the equation for velocity is v'/v = 1 , therefore the lorentz gamma factor also changes like u'/u = 1 therefore the proportional lorentz transformation of infinitesimals dt and ds is also invariant equation for frequency is f/f = 1/L since the dimension of frequency is 1/t and t'/t = L you can derive these transformation equations from these nature laws: t'/t = L s'/s = L E'/E = 1/L lambda = s f = 1/t v = s/t p = h/lambda F = ma E = mc^2 p = mvu (u is lorentz gamma factor) F = G m1m2 /r^2 F = k Q1Q2 / r^2 F = q (E + v x B) c = 1/ sqrt ( e u ) i have to answer better but i thought to write about these two things. Edited April 30, 2016 by caracal
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