KJW Posted November 25, 2023 Share Posted November 25, 2023 (edited) On 11/25/2023 at 4:48 AM, KJW said: On 11/25/2023 at 12:33 AM, Markus Hanke said: I also suspect (not sure though) that this would introduce off-diagonal terms into the metric? No, it doesn't. Only the time coordinates are involved in the coordinate transformation. If the original metric describes a flat three-dimensional space, then the transformed metric will be a scalar function multiple of the Minkowskian metric: g_{uv} = ƒ(t) η_{uv} Consider: (ds)² = (c dt)² – a(t)² ((dx)² + (dy)² + (dz)²) t = ƒ(t') ; x = x' ; y = y' ; z = z' dt = ∂t/∂t' dt' + ∂t/∂x' dx' + ∂t/∂y' dy' + ∂t/∂z' dz' = dƒ(t')/dt' dt' dx = ∂x/∂t' dt' + ∂x/∂x' dx' + ∂x/∂y' dy' + ∂x/∂z' dz' = dx' dy = ∂y/∂t' dt' + ∂y/∂x' dx' + ∂y/∂y' dy' + ∂y/∂z' dz' = dy' dz = ∂z/∂t' dt' + ∂z/∂x' dx' + ∂z/∂y' dy' + ∂z/∂z' dz' = dz' dƒ(t')/dt' = a(ƒ(t')) Solve for ƒ(t'), then let A(t') = dƒ(t')/dt' = a(ƒ(t')) Then: (ds)² = A(t')² ((c dt')² – (dx')² – (dy')² – (dz')²) .............................. For example, let a(t) = k t. Then: (ds)² = (c dt)² – (k t)² ((dx)² + (dy)² + (dz)²) dƒ(t')/dt' = k ƒ(t') ƒ(t') = exp(k t' + C) where C is an arbitrary constant. A(t') = k exp(k t' + C) (ds)² = (k exp(k t' + C))² ((c dt')² – (dx')² – (dy')² – (dz')²) Edited November 26, 2023 by KJW 1 Link to comment Share on other sites More sharing options...

KJW Posted November 27, 2023 Share Posted November 27, 2023 (edited) Continuing from my previous post: On 11/26/2023 at 9:47 AM, KJW said: (ds)² = (k exp(k t' + C))² ((c dt')² – (dx')² – (dy')² – (dz')²) This can be further developed by manipulating the arbitrary constant to obtain: (ds)² = exp(k (t' – t'_{0}))² ((c dt')² – (dx')² – (dy')² – (dz')²) where t'_{0} is an arbitrarily chosen value of t' at which the metric is locally Minkowskian On 11/26/2023 at 9:47 AM, KJW said: (ds)² = A(t')² ((c dt')² – (dx')² – (dy')² – (dz')²) For the general case: dƒ(t')/dt' = a(ƒ(t')) 1/a(ƒ(t')) dƒ(t')/dt' = 1 Let F(t) be such that: dF^{–1}(t)/dt = 1/a(t) Then: F^{–1}(ƒ(t')) = t' – t'_{0} where t'_{0} is an arbitrarily chosen value of t' ƒ(t') = F(t' – t'_{0}) dƒ(t')/dt' = dF(t' – t'_{0})/dt' = A(t' – t'_{0}) (ds)² = A(t' – t'_{0})² ((c dt')² – (dx')² – (dy')² – (dz')²) Edited November 27, 2023 by KJW 1 Link to comment Share on other sites More sharing options...

KJW Posted December 4, 2023 Share Posted December 4, 2023 (edited) If one has a conformally flat metric, the corresponding Friedmann-Lemaître-Robertson-Walker (FLRW) metric can be obtained from it by a coordinate transformation: (ds)² = A(t)² ((c dt)² – (dx)² – (dy)² – (dz)²) t = ƒ(t') ; x = x' ; y = y' ; z = z' dt = ∂t/∂t' dt' + ∂t/∂x' dx' + ∂t/∂y' dy' + ∂t/∂z' dz' = dƒ(t')/dt' dt' dx = ∂x/∂t' dt' + ∂x/∂x' dx' + ∂x/∂y' dy' + ∂x/∂z' dz' = dx' dy = ∂y/∂t' dt' + ∂y/∂x' dx' + ∂y/∂y' dy' + ∂y/∂z' dz' = dy' dz = ∂z/∂t' dt' + ∂z/∂x' dx' + ∂z/∂y' dy' + ∂z/∂z' dz' = dz' Note that the primed coordinates (t', x', y', z') and unprimed coordinates (t, x, y, z) have reversed roles compared to the earlier posts in this thread. dƒ(t')/dt' = 1/A(ƒ(t')) A(ƒ(t')) dƒ(t')/dt' = 1 Let F(t) be such that: dF^{–1}(t)/dt = A(t) Then: dF^{–1}(ƒ(t'))/dt' = A(ƒ(t')) dƒ(t')/dt' = 1 F^{–1}(ƒ(t')) = t' – t'_{0} where t'_{0} is an arbitrarily chosen value of t' ƒ(t') = F(t' – t'_{0}) Let a(t' – t'_{0}) = A(ƒ(t')) Then: dƒ(t')/dt' = 1/A(ƒ(t')) = dF(t' – t'_{0})/dt' = 1/a(t' – t'_{0}) And therefore: (ds)² = (c dt')² – a(t' – t'_{0})² ((dx')² + (dy')² + (dz')²) .............................. For example: (ds)² = exp(k t)² ((c dt)² – (dx)² – (dy)² – (dz)²) A(t) = exp(k t) 1/A(ƒ(t')) = dƒ(t')/dt' = exp(–k ƒ(t')) exp(k ƒ(t')) dƒ(t')/dt' = 1 (1/k) exp(k ƒ(t')) = t' – t'_{0} where t'_{0} is an arbitrarily chosen value of t' ƒ(t') = (1/k) ln(k (t' – t'_{0})) 1/a(t' – t'_{0}) = dƒ(t')/dt' = 1/(k (t' – t'_{0})) a(t' – t'_{0}) = k (t' – t'_{0}) Therefore: (ds)² = (c dt')² – (k (t' – t'_{0}))² ((dx')² + (dy')² + (dz')²) as would be expected from the earlier post. Edited December 4, 2023 by KJW 1 Link to comment Share on other sites More sharing options...

KJW Posted December 20, 2023 Share Posted December 20, 2023 (edited) (ds)² = A(t)² ((c dt)² – (dx)² – (dy)² – (dz)²) For the cosmological redshift, Z, the use of the conformally flat metric simplifies the calculation because the equation of a light-like trajectory in spacetime has the simple form of a straight line. Let the observer be at the origin of the coordinate system (t = 0 ; x = 0 ; y = 0 ; z = 0), and let the emitter of two light-pulses, an infinitesimal interval of time apart, be at x = X > 0 ; y = 0 ; z = 0. Then, the equation of the two light-pulses: t + x/c = 0 ; y = 0 ; z = 0 and: t + x/c – dt = 0 ; y = 0 ; z = 0 Thus, the emitter emitted the two light-pulses at: t = –X/c and: t = –X/c + dt These two light-pulses were observed at: t = 0 and: t = dt Then the cosmological redshift, Z: Z = (ds(t = 0) / ds(t = –X/c)) – 1 = (A(0) c dt / A(–X/c) c dt) – 1 And therefore: Z = (A(0) / A(–X/c)) – 1 Note that for a redshift, Z > 0, A(0) > A(–X/c), and for a blueshift, Z < 0, A(0) < A(–X/c) Also note that the x, y, and z coordinates of the conformally flat metric are the same as for the Friedmann-Lemaître-Robertson-Walker (FLRW) metric, and therefore the cosmological redshift, specified in terms of –X/c is unchanged for the FLRW metric. For example: (ds)² = exp(k t)² ((c dt)² – (dx)² – (dy)² – (dz)²) Z = exp(k X/c) – 1 Edited December 20, 2023 by KJW Link to comment Share on other sites More sharing options...

AbstractDreamer Posted April 26 Author Share Posted April 26 I not going to pretend to follow the math. Neither do I want to interrupt where the thread has gone, but I want to bring it back layman speak on my level. On 11/24/2023 at 6:48 PM, KJW said: The OP enquired as to why it is only space that expands and not time. The answer is that the difference between time expanding with space and time not expanding with space is just a coordinate transformation, which means that there is no physical difference. However, as you correctly point out, the time coordinate of the time-expanding metric doesn't correspond to anything, in particular, not a co-moving clock, whereas the spatial coordinates actually do correspond to the co-moving cosmological fluid. Earlier in this thread I did already mention that the FLRW metric is "orientated" where time does not expand with space. I had suspected that a transformation could orientate it differently such that time does expand with space. And I suspected at the other extreme we can have a solution where space does not expand at all and only time does. In another thread I asked about a variable "metric of time", and the "rate of flow of time", which was very difficult to conceptualise and it sort of ended there. My position is that l still maintain the validity of the interpretation that: non-relative time expansion/contraction is indistinguishable from space expansion/contraction. When you look up cosmological redshift in wiki there is no "Temporal Redshift" type. That is, redshift caused by an expanding temporal metric. It doesn't exist. Not a single reference, no studies, no papers. Why? Just because its too complex and abstract compared to space-expansion-only theory? I don't believe complexity is a reason for the entire physics community to shy away from such an interpretation. If it is valid, and no-one has researched into temporal redshift, it can only be because "space expansion" and its universal acceptance has blinded us to the truth that is only one alternative of other interpretations. -1 Link to comment Share on other sites More sharing options...

Markus Hanke Posted April 27 Share Posted April 27 7 hours ago, AbstractDreamer said: I had suspected that a transformation could orientate it differently such that time does expand with space. You are of course always free to pick different coordinates to describe the same spacetime - which is one of the central insights in GR. However, when you do this you also change the physical meaning of those coordinates. In the standard FLRW metric, the time coordinate is chosen such that it corresponds to a clock that is co-moving with the cosmological medium, meaning it fits well with our own physical clocks here on Earth, and thus the “phenomenology” of the metric corresponds to what we actually observe, without any need for complicated transformations. You are free to choose a coordinate system where eg tick rates aren’t constant, but then you need to be very careful how you relate the metric to real-world observations, since the t-coordinate no longer corresponds to Earth-bound clocks. Ultimately it is best to describe the spacetime in terms of geometric properties that are independent of coordinate choice; in the case of FLRW for example, we can say the spacetime is conformally flat, meaning during free fall angles are preserved, but not volumes. 7 hours ago, AbstractDreamer said: When you look up cosmological redshift in wiki there is no "Temporal Redshift" type. That is, redshift caused by an expanding temporal metric. It doesn't exist. Not a single reference, no studies, no papers. These aren’t different “theories”, but simply coordinate choices. You’re describing the same spacetime in different coordinates. KJW has given an example how a “time-only” expansion metric could look like. Ultimately you want to choose coordinates that make your calculations as simple as possible, and that’s often ones based on the cosmological medium. But in principle, the choice is yours, so long as they’re related by valid transformations. Link to comment Share on other sites More sharing options...

KJW Posted April 27 Share Posted April 27 (edited) 13 hours ago, AbstractDreamer said: And I suspected at the other extreme we can have a solution where space does not expand at all and only time does. No, it is not possible to coordinate-transform a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) to a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) In general, a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) has non-zero Ricci curvature, whereas a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) describes flat spacetime. Note that this metric can be transformed to the Minkowskian metric by the coordinate transformation: t' = t'(t) ; x' = x ; y' = y ; z' = z where t'(t) is a solution to the differential equation: dt'(t)/dt = α(t) Edited April 27 by KJW Link to comment Share on other sites More sharing options...

AbstractDreamer Posted April 27 Author Share Posted April 27 7 hours ago, Markus Hanke said: In the standard FLRW metric, the time coordinate is chosen such that it corresponds to a clock that is co-moving with the cosmological medium, meaning it fits well with our own physical clocks here on Earth, and thus the “phenomenology” of the metric corresponds to what we actually observe, without any need for complicated transformations. So by choosing such time coordinates, it also inherits the assumption that the cosmological medium of time is moving uniformly, everywhere and always (an Earth bound clock must tick at the same rate as the rest of the universe now and in the past and in the future). And yet, our observations are bounded to a infinitesimally small location of the universe (observations from our solar system compared to the size of the universe), and a very small period of time (150/13billion years). There is a problem here I cant quite put into words, so I will use bad analogies. Its like everyone being colour-blind and believing the universe is shades of grey. You can observe light wavelengths, but you cant observe the colour blue. It has no physical meaning. Its like believing gravity is a force before GR modelled spacetime curvature. Everything must fit what we observe (of course, to be empirically tested), but we don't acknowledge enough how severe our observations are restricted/limited. In many areas of science, where and when you perform an observation has no bearing on what is being tested. In THIS particular case of redshift, when and where you perform an observation is of paramount significance... and we are straight-jacketed into observations from our solar system location (where ever it is in the universe), and observations from our moment in time (a few hundred years). The limitations of our observations are significant relative to the field of study. 7 hours ago, Markus Hanke said: These aren’t different “theories”, but simply coordinate choices. You’re describing the same spacetime in different coordinates. KJW has given an example how a “time-only” expansion metric could look like. Ultimately you want to choose coordinates that make your calculations as simple as possible, and that’s often ones based on the cosmological medium. But in principle, the choice is yours, so long as they’re related by valid transformations. But space-expansion IS a theory, as is the more absurd temporal-expansion. The premise for the theories is from choice of coordinates. 2 hours ago, KJW said: No, it is not possible to coordinate-transform a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) to a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) Is this saying there is no transformation that will allow only time to expand and not space? What is the meaningful consequence of this? Link to comment Share on other sites More sharing options...

Mordred Posted April 27 Share Posted April 27 (edited) 3 hours ago, AbstractDreamer said: So by choosing such time coordinates, it also inherits the assumption that the cosmological medium of time is moving uniformly, everywhere and always (an Earth bound clock must tick at the same rate as the rest of the universe now and in the past and in the future). And yet, our observations are bounded to a infinitesimally small location of the universe (observations from our solar system compared to the size of the universe), and a very small period of time (150/13billion No there is no assumptions due to coordinate choice. You already know time dilation is a consequence of spacetime curvature or Relativistic inertia. The math and observational evidence shows us that there is no curvature term k=0. So where would you get time dilation ? This has already previously been mentioned. As massless particles travel at c we can ignore the inertial gamma factor. A higher density past the answer either. To go into greater detail if you take 3 time slices say time now, time at the CMB say z=1100. And a slice at say universe age 7 billion years old. If you describe the geometry of each slice. Each slice has a uniform mass distribution so no slice has a non uniform mass distribution to have a curvature term. Hint this is the real advantage of the scale factor a. No time slice has any change in geometry or curvature it's simply volume change between slices and density changes as a result of the ideal gas laws Edited April 27 by Mordred Link to comment Share on other sites More sharing options...

AbstractDreamer Posted April 27 Author Share Posted April 27 31 minutes ago, Mordred said: No there is no assumptions due to coordinate choice. I don't know the maths at all. But it seems fundamental to me that if you make a choice, you instantiate something. When something is instanced, things are set and other settings are rejected. When you reject other settings, there are fundamental consequences. These consequences are the assumptions. If time coordinates are chosen such that earth-bound clocks are comoving with the cosmological medium, that has consequences. The very choosing of those coordinate forbids a non-relative (non-gravitational) time dilation effect. That is why FLRW metric forbids temporal redshift, because it was chosen to be orientated that way. Am I wrong? 33 minutes ago, Mordred said: The math and observational evidence shows us that there is no curvature term k=0. So where would you get time dilation ? This has already previously been mentioned. As massless particles travel at c we can ignore the inertial gamma factor. The observational evidence in this case is dubious solely because of the narrow range of observation relative to the field of study. We've never made an observation of cosmological redshift from outside of our solar solar system, let alone from a distance where space-expansion or temporal-expansion is significant. We've never made an observation of cosmological redshift from a time in the past or the future, where spatial or temporal expansion is significant. I'm not saying any of this is possible. I'm just saying our sample range of observations is far too narrow to be confident to say our evidence is significant. We've taken a handful of stones from a beach, and assumed all beaches must be stoney. As for where would we get time dilation? Where do we get space-expansion? Dark energy? We can make up anything to fit the narrative. 36 minutes ago, Mordred said: A higher density past the answer either. To go into greater detail if you take 3 time slices say time now, time at the CMB say z=1100. And a slice at say universe age 7 billion years old. If you describe the geometry of each slice. Each slice has a uniform mass distribution so no slice has a non uniform mass distribution to have a curvature term. Hint this is the real advantage of the scale factor a. No time slice has any change in geometry or curvature it's simply volume change between slices and density changes as a result of the ideal gas laws Advantage for what purpose? Simplicity and accuracy to fit other observations are similarly limited in their scope? This again falls foul of confirmation bias. BTW, someone anonymous is downvoting all my threads. Not that I care about reputation, but being anonymous and not saying why I'm wrong feels like im being victimised and rather abusive. Link to comment Share on other sites More sharing options...

Mordred Posted April 27 Share Posted April 27 (edited) 2 hours ago, AbstractDreamer said: I don't know the maths at all. But it seems fundamental to me that if you make a choice, you instantiate something. When something is instanced, things are set and other settings are rejected. When you reject other settings, there are fundamental consequences. These consequences are the assumptions. If time coordinates are chosen such that earth-bound clocks are comoving with the cosmological medium, that has consequences. The very choosing of those coordinate forbids a non-relative (non-gravitational) time dilation effect. That is why FLRW metric forbids temporal redshift, because it was chosen to be orientated that way. Am I wrong? no coordinate choice affects the mass distribution. I could describe the universe in numerous different coordinate choices example Euclidean, spherical cylindrical etc without causing any difference. It is precisely why we use invariance. The mathematics is set up that way so that we do not have any coordinate choice dependency. 2 hours ago, AbstractDreamer said: As for where would we get time dilation? Where do we get space-expansion? Dark energy? We can make up anything to fit the narrative. you know full well GR fully describes time dilation the FLRW metric is a GR solution. We don't arbitrarily choose DM and DE as the full explanation those two terms are simply placeholders until we can determine the cause of each. We still can measure their effects through indirect evidence. 2 hours ago, AbstractDreamer said: BTW, someone anonymous is downvoting all my threads. Not that I care about reputation, but being anonymous and not saying why I'm wrong feels like im being victimised and rather abusive. I rarely give downvotes so its someone else. As far as sampling range is concerned, redshift is only one of many pieces of evidence of an expanding universe. In point of detail its not even close to the strongest evidence. Its the one most ppl are familiar with but the real evidence comes from our thermodynamic laws in regards to temperature and how it influences the SM model of particles via processes such as BB nucleosynthesis in regards to the CMB. One danger of trying to understand cosmology by rote instead of learning the math is that too often you get incorrect information. I will give an example if I looked up hydrogen and its temperature it could form with stability a google search will state 3000 kelvin. However if one knows how to use the Saha equations that would reveal that value equates to 75 % of the potential hydrogen. Hydrogen can start to form as low as 6000 kelvin=25% 4000 kelvin for 50 %. That is just one example. however knowing this one can study the metallicity of our universe evolution via hydrogen, lithium, deuterium etc. So I just described another piece of evidence for expansion. In other words were not restricted to redshift to determine if our universe is expanding . In point of detail we do not rely on redshift in cosmology it is too full of other influences such as gravitational redshift, transverse redshift, Integrated Sache-Wolfe effect, Doppler redshift. etc etc. We examine all pieces of possible evidence to confirm the accuracy of cosmological redshift. Nor do we use the generic formula everyone sees on google. https://en.wikipedia.org/wiki/Redshift this formula only works for nearby objects it loses accuracy as near as one MPC. The full formula includes the influence of the evolution history of matter, radiation and Lambda. details can be found here "Distance measures in cosmology" David W. Hogg https://arxiv.org/abs/astro-ph/9905116 side note the paper also applies to luminosity distance we also have a different formula for Luminosity distance than what one would google. \[H_O dl=(1+z)|\Omega_k|^{-1/2}sinn[\Omega_k^{1/2} \int^z_o\frac{d\acute{z}}{\sqrt{(1+\acute{z})^2\Omega_R+(1+\acute{z}\Omega_m-\acute{z})(2+\acute{z})\Omega_\Lambda}}]\] What this equation shows is that matter, radiation and Lambda density not only influences expansion rates it also influences redshift and luminosity as well as any curvature term k Edited April 27 by Mordred Link to comment Share on other sites More sharing options...

Mordred Posted April 27 Share Posted April 27 (edited) you won't find that equation in a textbook, textbooks only show the basic equations in math speak in this case you would usually see the first order equation this delves into the second order. just as most textbooks won't show the equation \[H_z=H_o\sqrt{\Omega_m(1+z)^3+\Omega_{rad}(1+z)^4+\Omega_{\Lambda}}\] this shows the expansion rate H varies over time (it will also help to better understand the first equation as well as the Hogg paper I posted. now as you mentioned DM and DE one line of research is Higgs being responsible. Sterile neutrinos (right hand are heavier than left hand neutrinos ) antimatter and matter neutrinos. so the calculated abundance could fall into range \[\Omega_pdmh^2=\frac{G^{3/2}T_0^3h^2}{H_0\sigma v}=\frac{3*10-{27} cm^3s^{-1}}{\sigma v}\] research is still on going. Just as the equation of state for the Higgs field may explain inflation as well as the cosmological constant. That should sufficiently show that what really goes on in the professional circles isn't something one can simply google at best that just gives hints Edited April 27 by Mordred Link to comment Share on other sites More sharing options...

Markus Hanke Posted April 28 Share Posted April 28 (edited) 15 hours ago, AbstractDreamer said: So by choosing such time coordinates, it also inherits the assumption that the cosmological medium of time is moving uniformly Well, the fundamental assumptions underlying this solution are homogeneity and isotropy - if you feed this kind of energy-momentum distribution into the field equations, you get as solution a spacetime that expands. You are free to choose yourself what kind of coordinate system you wish to use to describe this, but obviously it is smart to use a system where your intended calculations are easy. 15 hours ago, AbstractDreamer said: There is a problem here I cant quite put into words I understand what you are trying to say. The FLRW metric does rely on the cosmological principle, that’s an assumption we make - that on large scales the universe is homogenous and the same in all directions. Since there’s an observational horizon past which we can’t see, it’s possible at least in principle that perhaps one of these doesn’t actually hold. 15 hours ago, AbstractDreamer said: But space-expansion IS a theory, as is the more absurd temporal-expansion. The premise for the theories is from choice of coordinates. The underlying premise is really the laws of gravity, meaning Einstein’s equations. If you start off with a distribution of energy-momentum that interacts (approx) only gravitationally, then it’s actually difficult to avoid solutions that metrically expand in some way. FLRW is by no means a unique thing, it’s just a particular example of a large number of such solutions. This is not just due to coordinate choices. 15 hours ago, AbstractDreamer said: Is this saying there is no transformation that will allow only time to expand and not space? Indeed. 14 hours ago, AbstractDreamer said: But it seems fundamental to me that if you make a choice, you instantiate something. And you are correct - you need to pick some boundary conditions to solve the EFE, which in this case is the cosmological principle. But in GR, the choice of coordinate system has no physical consequences, so it’s not due to that. 14 hours ago, AbstractDreamer said: If time coordinates are chosen such that earth-bound clocks are comoving with the cosmological medium The coordinates are chosen such that they correspond to an observer co-moving with the medium; this seems to apply to Earth too, since we don’t observe anything different. We remain in our galaxy, which is part of a local cluster, which co-moves along with everything else. 14 hours ago, AbstractDreamer said: I'm just saying our sample range of observations is far too narrow to be confident to say our evidence is significant. In physics you are always restricted by the set of available data - our task is to find a model that best fits this currently available data. If the data set changes, then sometimes the model needs to change too. There’s many possible objections to the Lambda-CDM model, but honestly, right now there’s nothing else that fits all available data better. Let’s just consider this a work in progress. Physics would be boring if all the last words had been spoken already. Edited April 28 by Markus Hanke Link to comment Share on other sites More sharing options...

AbstractDreamer Posted April 29 Author Share Posted April 29 21 hours ago, Markus Hanke said: Well, the fundamental assumptions underlying this solution are homogeneity and isotropy - if you feed this kind of energy-momentum distribution into the field equations, you get as solution a spacetime that expands. You are free to choose yourself what kind of coordinate system you wish to use to describe this, but obviously it is smart to use a system where your intended calculations are easy. I understand what you are trying to say. The FLRW metric does rely on the cosmological principle, that’s an assumption we make - that on large scales the universe is homogenous and the same in all directions. Since there’s an observational horizon past which we can’t see, it’s possible at least in principle that perhaps one of these doesn’t actually hold. But homogeneity and isotropy in the cosmological principle is an assumption of spatial distribution of energy momentum. Choosing time coordinates for a solution to EFE such as FLRW, is an assumption of temporal distribution. Isotropy of time would mean there is no preferred direction of time, but all our observations of time show it does have a preferred direction - time goes forwards. Observationally, the universe is temporally anisotropic. Homogeneity of time would mean there is no preferred moment in time. The universe looks different at different coordinates in time - it was pure plasma very early on, and now it isn't. Similarly, observationally, the universe is temporally inhomogeneous. 21 hours ago, Markus Hanke said: The underlying premise is really the laws of gravity, meaning Einstein’s equations. If you start off with a distribution of energy-momentum that interacts (approx) only gravitationally, then it’s actually difficult to avoid solutions that metrically expand in some way. FLRW is by no means a unique thing, it’s just a particular example of a large number of such solutions. This is not just due to coordinate choices. Right, but FLRW is a particular solution that inherently forbids temporal expansion because of the choice of coordinates. Therefore it cannot be used to justify why all expansion is spatial. On 4/27/2024 at 3:24 PM, Mordred said: As far as sampling range is concerned, redshift is only one of many pieces of evidence of an expanding universe. In point of detail its not even close to the strongest evidence. Its the one most ppl are familiar with but the real evidence comes from our thermodynamic laws in regards to temperature and how it influences the SM model of particles via processes such as BB nucleosynthesis in regards to the CMB. My position is NOT that the universe is NOT expanding. My position is why all the expansion is attributed to spatial expansion and not temporal expansion. I suspect all the other evidence that supports metric expansion does not directly refute temporal expansion. Cosmological redshift does not refute temporal expansion. But if we use the same FLRW solution to interpret the evidence, then our conclusion will be constrained to the assumptions of the solution we chose. It is the solution that assumes all expansion is spatial, not the evidence. On 4/27/2024 at 3:24 PM, Mordred said: "Distance measures in cosmology" David W. Hogg https://arxiv.org/abs/astro-ph/9905116 I am really interested in distance measures in cosmology. In particular the margins of error, models and assumptions when interpreting observations. But will ask those questions another day. On 4/27/2024 at 11:07 AM, KJW said: No, it is not possible to coordinate-transform a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) to a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) This is piqued my interest. No transformation that allows time to expand and not space. Why does expansion have to be at least in part spatial? Why can expansion have no temporal component? What does this physically mean? Link to comment Share on other sites More sharing options...

Markus Hanke Posted April 29 Share Posted April 29 10 minutes ago, AbstractDreamer said: But homogeneity and isotropy in the cosmological principle is an assumption of spatial distribution of energy momentum. Yes. FLRW spacetime is a “dust solution” - a universe homogeneously and isotropically filled with energy-momentum that interacts only gravitationally. 10 minutes ago, AbstractDreamer said: Choosing time coordinates for a solution to EFE such as FLRW, is an assumption of temporal distribution. The choice of coordinate system is arbitrary, it represents no physical assumption. You are basically just picking an observer on whose point of view you base your labelling. You can take the ordinary FLRW metric (usually written in what is called Gauss coordinates) and just perform a valid coordinate transformation to arrive at a different point of view; this can be done directly, and has nothing to do with the field equations or the physics. For example, you could choose an observer that is accelerated at all times - you would get a metric that at first glance looks very different, but still describes the same spacetime. 28 minutes ago, AbstractDreamer said: Right, but FLRW is a particular solution that inherently forbids temporal expansion because of the choice of coordinates. As I said, the choice of coordinates is arbitrary. For example, if you were to base your coordinate system on a clock that is not comoving with the cosmological medium (eg one that is accelerated at all times, possibly non-uniformly), you would get a metric where both the time and space parts explicitly depend on the t-coordinate. So long as the coordinate transformation is a valid diffeomorphism, this is perfectly allowed, though probably an algebraic nightmare to actually work with. It’s important again to realise that this describes the same spacetime, just in terms of different coordinates. What is not possible though is to try and have only the time part expand - there’s no valid transformation that yields this. 37 minutes ago, AbstractDreamer said: My position is why all the expansion is attributed to spatial expansion and not temporal expansion. See above - you could “distribute” the expansion across both time and space parts of the metric by a suitable coordinate transformation, which has no physical consequences. It’s the same spacetime, you’d just label events in it differently. The question is why you would want to do this - it would greatly complicate most calculations relevant to us, since such coordinates wouldn’t straightforwardly correspond to our own clocks and rulers. But of course you can do this, if you really wanted to. Link to comment Share on other sites More sharing options...

Mordred Posted April 29 Share Posted April 29 (edited) 2 hours ago, AbstractDreamer said: My position is why all the expansion is attributed to spatial expansion and not temporal expansion. I beg to differ on this score the FLRW metric is a GR solution and in GR time has dimensionality of length via the Interval (ct). It is that relation that includes length contraction and time dilation. Whether or not its required depends on the spacetime geometry. The simple reason you only really need the spatial component is that observational evidence shows a flat spacetime geometry. That's not some arbitrary choice of the metric. That the findings of all observational evidence. We have very useful methods for seeking spacetime curvature terms at our disposal. One example is distortions curvature causes light paths to bend this leads to distortions. Those distortions are constantly looked for. They can also be useful such as boosting viewing distance by gravitational lensing. That's just one method of detecting spacetime curvature there are others. The point being the metric does factor in the time component simply by being a GR solution. It's simply not needed due to all observational evidence. As far as observer effects, we do indeed need to take those into consideration. The dipole anistrophy due to Earths motion through spacetime in relation to the object we are observing must be factored in. A clear example was the findings of the first Planck dataset that had a dipole anistrophy in its first dataset. That dataset didn't have the correct calibration. That led to all kinds of pop media and scrambling. The next dataset had eliminated that dipole as we then had a better understanding of Earths momentum. As well as other localized effects. There isn't any arbitrary choice made the FLRW metric is quite capable of dealing with curvature. It's simply not needed beyond the weak field limit. You really only need the Minkowsii metric for the weak field limit. In a flat curvature parallel beams will remain parallel. If you have positive curvature those beams will converge. They will diverge for negative curvature. The converging or diverging is detectable and quite apparent in spectography in particular....which makes hydrogen a particularly useful test for distortions in its spectrographic readings. In particular the 21 cm line. That is what spacetime geometry ddescribes. All major findings show miniscule at best curvature best fit of a global geometry is flat. So the FLRW metric follows GR in the appropriate manner described by GR for a flat geometry Edited April 29 by Mordred 1 Link to comment Share on other sites More sharing options...

KJW Posted April 29 Share Posted April 29 On 4/27/2024 at 8:07 PM, KJW said: No, it is not possible to coordinate-transform a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) to a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) In general, a metric of the form: (ds)² = c² (dt)² – a(t)² ((dx)² + (dy)² + (dz)²) has non-zero Ricci curvature, whereas a metric of the form: (ds)² = α(t)² c² (dt)² – ((dx)² + (dy)² + (dz)²) describes flat spacetime. Note that this metric can be transformed to the Minkowskian metric by the coordinate transformation: t' = t'(t) ; x' = x ; y' = y ; z' = z where t'(t) is a solution to the differential equation: dt'(t)/dt = α(t) On 4/27/2024 at 10:55 PM, AbstractDreamer said: Is this saying there is no transformation that will allow only time to expand and not space? Yes. The expanding time only metric is flat, whereas the FLRW metric is not flat, so there is no coordinate transformation between them. On 4/27/2024 at 10:55 PM, AbstractDreamer said: What is the meaningful consequence of this? 16 hours ago, AbstractDreamer said: This is piqued my interest. No transformation that allows time to expand and not space. Why does expansion have to be at least in part spatial? Why can expansion have no temporal component? What does this physically mean? In this thread, I demonstrated that an expanding space and time metric does not exist as distinct from an expanding space only metric. I did this by showing that an expanding space and time metric can be transformed to an expanding space only metric by a coordinate transformation, thus proving that they are physically the same. Mathematically however, the manifestly conformally flat form of the metric is useful for two reasons: 1, it indicates that the Weyl conformal tensor field is zero (the Weyl conformal tensor field represents pure gravitation, and therefore the only gravitation in the flat space FLRW spacetime is that which is intrinsic to the energy-momentum tensor field itself); and 2, it simplifies the equation of light-like geodesics (though this also assumes the homogeneity and isotropy of the FLRW spacetime¹). My most recent post in this thread demonstrated that there are limits to what coordinate transformations can do. By showing that an expanding time only metric can be transformed to a flat spacetime metric by a coordinate transformation, I established that it cannot be obtained from an expanding space only metric, which is not flat, by a coordinate transformation. It is worth noting that any metric of the form: (ds)² = T(t)² c² (dt)² – X(x)² (dx)² – Y(y)² (dy)² – Z(z)² (dz)² describes flat spacetime. The following coordinate transformation exists between this metric and the Minkowskian metric: t' = t'(t) ; x' = x'(x) ; y' = y'(y) ; z' = z'(z) where t'(t), x'(x), y'(y), and z'(z) are solutions to the differential equations: dt'(t)/dt = T(t) dx'(x)/dx = X(x) dy'(y)/dy = Y(y) dz'(z)/dz = Z(z) ¹ Although the equations for the light-like trajectories in the manifestly conformally flat form of the metric given earlier in this thread are indeed light-like trajectories, I did not prove that these are geodesics. To prove that they are geodesics, note that in the two-dimensional tx-spacetime, the light-like trajectory passing through a given point from a given direction is unique, and therefore must be geodesic. But to prove that in the four-dimensional spacetime, the light-like geodesic doesn't deviate from the two-dimensional tx-spacetime, the symmetry arising from homogeneity and isotropy can be invoked, requiring that the unique light-like geodesic remain in the two-dimensional tx-spacetime (symmetry breaking can only occur if there are multiple solutions). 1 Link to comment Share on other sites More sharing options...

AbstractDreamer Posted May 24 Author Share Posted May 24 (edited) On 4/29/2024 at 10:02 PM, KJW said: My most recent post in this thread demonstrated that there are limits to what coordinate transformations can do. By showing that an expanding time only metric can be transformed to a flat spacetime metric by a coordinate transformation, I established that it cannot be obtained from an expanding space only metric, which is not flat, by a coordinate transformation. I have been trying to understand this for some time but still fail. A flat spacetime metric cannot be obtained by coordinate transformation from an non-flat expanding space-only metric? How does this refute an expanding time-only metric? On 4/29/2024 at 6:56 AM, Mordred said: The simple reason you only really need the spatial component is that observational evidence shows a flat spacetime geometry. That's not some arbitrary choice of the metric. That the findings of all observational evidence. But if there was a temporal component, could it be observed as distinct from the spatial component? If you cant observe a distinction, then just as you can argue that all observational evidence says you only need a spatial component, the position that none of the observational evidence refute a temporally expanding component is equally as strong. And the important thing is that EFE suggests BOTH components form a single manifold. There's no observational reason why the temporal component is zero in the case of expansion. The only reason, as far as I can tell, is simplicity of calculations - which is a good reason but not one based on observation. Thought experiment: Say we observe two redshift galaxies at z=5. Let's say one galaxy is only spatially expanding away from us, and the other is both spatially and temporally expanding away from us, and all three locations (two galaxies and the observer) are on a spacetime plane that has observably flat geometry. Could you distinguish which is which? On 4/29/2024 at 5:32 AM, Markus Hanke said: you could “distribute” the expansion across both time and space parts of the metric by a suitable coordinate transformation, which has no physical consequences. It’s the same spacetime, you’d just label events in it differently. The question is why you would want to do this - it would greatly complicate most calculations relevant to us, since such coordinates wouldn’t straightforwardly correspond to our own clocks and rulers. But of course you can do this, if you really wanted to. I accept the argument why we would want to complicate the calculations. Like the geocentric theory of the solar system is valid if you choose that coordinate system, but it makes the calculations impossibly complicated, versus the Copernican model which simplifies things a lot. My question is what might we be missing when we simplify them. Just like the equivalence principle, there's no difference between being in a gravitational field or in a rocket that is being accelerated. For local calculation purposes they are physically equivalent. But there is a materialistic difference. An accelerated rocket is a far less stable environment than a gravitational field. Maybe locally there is no physical consequence of calculations that assume zero temporal expansion. But maybe in the bigger picture, or some grander theory there is a difference. Could the Crisis in Cosmology be partly due to the lamba-CDM modelling both spacetime geometry as too "flat" and expansion as spatial-only? Edited May 24 by AbstractDreamer Link to comment Share on other sites More sharing options...

Mordred Posted May 24 Share Posted May 24 It seems your confusion is thinking the FLRW metric for flat spacetime only includes the 3d portion for spatial components. It also includes the time component for 4d spacetime. However for flat spacetime you don't require the Gamma correction between coordinate time and proper time. It's literally no different Than GR for flat spacetime or Euclidean geometry the only difference is the commoving coordinates which is equated via the scale factor "a". The FLRW metric is literally a GR solution. Link to comment Share on other sites More sharing options...

AbstractDreamer Posted May 25 Author Share Posted May 25 17 hours ago, Mordred said: It seems your confusion is thinking the FLRW metric for flat spacetime only includes the 3d portion for spatial components. It also includes the time component for 4d spacetime. However for flat spacetime you don't require the Gamma correction between coordinate time and proper time. It's literally no different Than GR for flat spacetime or Euclidean geometry the only difference is the commoving coordinates which is equated via the scale factor "a". The FLRW metric is literally a GR solution. I do accept that the FLRW does include a time component. Perhaps I'm using the wrong words when I accuse the FLRW of having no temporal component. What I mean to say is FLRW has a temporal component that has a net zero value, and consequently all observed expansion must be spatial according to EFE. But where is the direct evidence that Cosmological spacetime is absolutely flat in the absence of a gravitational field, even if there is no evidence of local spacetime expansion, neither spatial nor temporal? The logical fallacy here is: Because gravity curves spacetime, local spacetime is not flat, therefore (fallacy) in the absence of gravity, non-local spacetime is also flat. If gravity can curve spacetime locally, why must spacetime be flat cosmologically? Link to comment Share on other sites More sharing options...

AbstractDreamer Posted May 25 Author Share Posted May 25 On 4/29/2024 at 6:56 AM, Mordred said: All major findings show miniscule at best curvature best fit of a global geometry is flat. How small is miniscule? Why does a globally flat geometry forbid an expansion of the temporal metric? Link to comment Share on other sites More sharing options...

Mordred Posted May 25 Share Posted May 25 2 hours ago, AbstractDreamer said: I do accept that the FLRW does include a time component. Perhaps I'm using the wrong words when I accuse the FLRW of having no temporal component. What I mean to say is FLRW has a temporal component that has a net zero value, and consequently all observed expansion must be spatial according to EFE. But where is the direct evidence that Cosmological spacetime is absolutely flat in the absence of a gravitational field, even if there is no evidence of local spacetime expansion, neither spatial nor temporal? The logical fallacy here is: Because gravity curves spacetime, local spacetime is not flat, therefore (fallacy) in the absence of gravity, non-local spacetime is also flat. If gravity can curve spacetime locally, why must spacetime be flat cosmologically? GR for starters. Ok we don't require to understand that gravity only occurs when you have a non uniform mass distribution. Newtons shell theorem is more than adequate. Take a point at (0,0,0) now have the same amount of mass density everywhere surrounding that point. The force exerted due to the mass density will be identical regardless of which direction you examine. So the sum of force is identical at every angle. (force is a vector). The net force is now zero at (0,0,0) so you have zero acceleration due to gravity. Hence gravity is zero for flat spacetime. The mass distribution is uniform. That's been understood and tested since the 16th century. Its no different for GR or the FLRW metric. Link to comment Share on other sites More sharing options...

KJW Posted May 25 Share Posted May 25 (edited) On 5/25/2024 at 12:29 AM, AbstractDreamer said: On 4/30/2024 at 7:02 AM, KJW said: My most recent post in this thread demonstrated that there are limits to what coordinate transformations can do. By showing that an expanding time only metric can be transformed to a flat spacetime metric by a coordinate transformation, I established that it cannot be obtained from an expanding space only metric, which is not flat, by a coordinate transformation. I have been trying to understand this for some time but still fail. A flat spacetime metric cannot be obtained by coordinate transformation from an non-flat expanding space-only metric? How does this refute an expanding time-only metric? A flat spacetime metric cannot be obtained from a non-flat spacetime metric by a coordinate transformation. A flat spacetime metric can only be obtained from a flat spacetime metric by a coordinate transformation. This is central to the mathematics of general relativity. In general, a tensor that is zero in one coordinate system is zero in every coordinate system, and curvature is a tensor, so a curvature of zero (ie flat) in one coordinate system is zero (flat) in every coordinate system. An FLRW metric of an expanding flat space is a non-flat spacetime (do not confuse space with spacetime). But I have shown that an expanding time only metric can be coordinate-transformed to a flat spacetime metric. Therefore, the expanding time only metric is a flat spacetime. And because it is a flat spacetime, it cannot be obtained from a non-flat spacetime such as an FLRW metric of an expanding flat space by a coordinate transformation. Edited May 25 by KJW 1 Link to comment Share on other sites More sharing options...

Markus Hanke Posted May 25 Share Posted May 25 On 5/24/2024 at 4:29 PM, AbstractDreamer said: A flat spacetime metric cannot be obtained by coordinate transformation from an non-flat expanding space-only metric? Such a coordinate transformation wouldn’t be a diffeomorphism, meaning you are dealing with two different spacetime geometries, and thus two completely different solutions to the EFE. IOW, an expanding-time-only metric would not just be an FLRW spacetime written in different coordinates. On 5/24/2024 at 4:29 PM, AbstractDreamer said: How does this refute an expanding time-only metric? A metric where only the time part is “expanding” describes a completely flat spacetime, only from the perspective of an observer that undergoes some form of accelerated motion. This is not a good description of our universe, since that contains matter and radiation, and thus can’t be a flat spacetime. On 5/24/2024 at 4:29 PM, AbstractDreamer said: But if there was a temporal component, could it be observed as distinct from the spatial component? So far as I can tell (someone correct me if I’m wrong), a metric where both the time and space parts are expanding could be the natural description of an observer who undergoes non-uniformly accelerated motion with respect to the cosmological medium, provided the time part has a suitable mathematical form (if not, it won’t correspond to any physical observer). For such an observer, measurements of both spatial distances and time durations would explicitly depend on when they are performed. He’d see an expanding universe, but also detect proper acceleration, which would play the role of the distinct time component. On 5/24/2024 at 4:29 PM, AbstractDreamer said: Let's say one galaxy is only spatially expanding away from us, and the other is both spatially and temporally expanding away from us, and all three locations (two galaxies and the observer) are on a spacetime plane that has observably flat geometry. The FLRW spacetime is not Riemann-flat, so I’m not sure how to answer this. Given the time component has the right form, the distinction would be in the presence of proper acceleration in the motion of the observer. On 5/24/2024 at 4:29 PM, AbstractDreamer said: My question is what might we be missing when we simplify them. The details of the spacetime geometry - curvature, geodesic structure etc - are independent of coordinate choices, so I don’t think we’re missing anything. It’s like having an electric charge - in the rest frame of that charge, you detect only an E-field. An observer who moves past that same charge sees both an E-field and a B-field. In both cases, you have the same electromagnetic field. It’s the same physical situation, seen from different vantage points. But no observer under these circumstances will ever see just a B-field alone, since that would be a physically different situation. The expanding space situation is similar - you can re-distribute the expansion among the components of the metric by a suitable coordinate transformation (same spacetime, different vantage point), but no observer will ever see a time-only expansion, since that would imply being in a flat spacetime. 4 hours ago, AbstractDreamer said: If gravity can curve spacetime locally, why must spacetime be flat cosmologically? FLRW spacetime is not flat, so I don’t quite understand what you’re trying to ask…? Link to comment Share on other sites More sharing options...

AbstractDreamer Posted May 28 Author Share Posted May 28 (edited) On 5/25/2024 at 4:10 PM, Markus Hanke said: FLRW spacetime is not flat, so I don’t quite understand what you’re trying to ask…? Is FLRW spacetime not flat because of curvature caused by mass, and the scale factor of expansion that changes over time? If we zero the non-flatness effect of gravity AND zero the non-flatness effect of the scale factor, is FLRW spacetime otherwise flat? In other words, if we take a period of time that is very small cosmologically, say 1 day, where the scale factor of expansion is constant; and if we remove all gravity from the universe. Would then the FLRW and the spatial expansion it describes be over a flat geometry? Edited May 28 by AbstractDreamer Link to comment Share on other sites More sharing options...

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