Mordred

Quite frankly if time didn't exist neither would spacetime. The only way time cannot exist is if there is literally nothing, not even a volume. In point of detail any absolutely empty volume would still have time. Even if it never changes. That's a measure of duration. Though quite frankly that's rather impossible to have a absolutely unchanging volume. So the concept of no time is also a meaningless concept. Time also measures duration not just change. Time isn't a force it's a measurement. It's not what I think, it's what GR allowed us to understand, and test. The tests of GR is incredible, they have even measured the time dilation between your feet and your head. It's used and tested everyday, for example particle accelerators and GPS satellites. Any cosmological distance measurements employs GR. It must.

The easiest way to think of time is the measured rate of change or duration. How we choose to model that rate can vary. Some models use "action" and geometry. GR uses geometry with vectors and scalars

This gets complex, I'll try to keep it simple. You look at a clock at rest with yourself at rest. Both you and the clock are in the same reference frame, you both being in the same gravitational potential and at rest. Now if we add a third observer who is also looking at the same clock, but is moving, who also has a clock that has been previously synchronized with the other clock. If the moving observer looks at the synchronized clock he is carrying he will see nothing unusual, however if he looks at the clock at rest, we will see it ticks at a different rate than the one he is carrying. The observer at rest will measure the same change in the clock that is moving. how you measure time depends on your reference frame (spacetime geometry) compared to the reference frame of the emitter. There is essentially a couple of reasons this occurs. light is the same velocity regardless of observer =c. So the velocity of measured signal (light pulses from the clock doesn't change) When you have relativistic effects occur two key affects occur simultaneously. Time dilation and length contraction. We model these two influences using time =ct. We map the spatial coordinates as x,y,Z. Hence 4d. Here time is treated coordinate with a vector component. Collectively we refer to this as spacetime. Reference frames being the coordinate condition of the observer and the coordinate difference of the emitter compared to the observer. if there is no difference in the coordinates the emitter and observer is in the same reference frame. Reference frames apply at all scales of measure, locally and globally in the case of modelling spacetime geometry of the Universe. As spacetime geometry also influences the light paths via null geodesics, we can essentially look for distortions in deep space images to measure potential spacetime geometry change. This is the function of the curvature term in the Cosmology FLRW metric, which is a simplified version of the Einstein field equations.

[latex]f=\frac{c+v_r}{c+v_s}f_o[/latex] Using this formula if the light path enters a gravity well there is time dilation. Ordinarily when it exits the gravity well the wavelength is restored. However if the background mass changes while the light is in the well, we can notice a difference. This forms the basis of the Sachs-Wolfe effect. Which is an application of using time dilation to our advantage. This technique is handy in CMB measurements as well as mapping possible gravitational lenses, and potential variations in the rate of expansion.

We know time varies depending on the observer. We have a solid good working knowledge of how time varies. However the speed of light is the same for all observers. Using the invariants of the speed of light as our baseline we can measure the amount of time dilation. By the formulas I posted above.... For gravitational time dilation this causes a change in frequency of light.

those equations use the invariance of the speed of light. In simple terms it measures how long it takes light to reach us. The amount of time. What did you think it meant? Or did you even bother reading it? If time did not exist at some mystical point, light from that point would never reach us. In point of detail we can use the above mathematics to measure the amount of gravity from gravity wells using the Sache Wolfe effect. https://en.m.wikipedia.org/wiki/Sachs%E2%80%93Wolfe_effect which takes advantage of gravitational time dilation to measure mass density using the gravitational redshift formula.

No you obviously don't have a good understanding of GR, because it does apply. I could show you mathematics but it would probably go over your head. Particularly if you didn't understand the basic equations I posted.

AH I see you have no knowledge of the basic relativity formulas. Well here is a paper that measures and tests relativity using Pulsars at various distances. http://www.google.ca/url?q=http://relativity.livingreviews.org/Articles/lrr-2003-5/download/lrr-2003-5Color.pdf&sa=U&ved=0ahUKEwjMnpGPhobMAhVO42MKHUJUB9MQFggkMAY&usg=AFQjCNFcdj07CFxj96F1IoobgMgdHVS0vw Point being relativity teaches us how observer influences occur due to gravitational potential and inertia There has been hundreds of tests as to its accuracy. Those mathematics above explain those observer to emitter influences, Coupled with the known influences upon redshift, we can determine the rate of time at a specific mass density. We know time isn't the same to all observers, yet if your in the same observer reference frame there is no time dilation. In order to have time dilation with mass density you require a gradient in mass distribution at a particular time slice. In cosmology this correlates to cosmic time, which employs a fundamental observer.

We measure rate of change by the following relativity rules. Lorentz transformation. First two postulates. 1) the results of movement in different frames must be identical 2) light travels by a constant speed c in a vacuum in all frames. Consider 2 linear axes x (moving with constant velocity and [latex]\acute{x}[/latex] (at rest) with x moving in constant velocity v in the positive [latex]\acute{x}[/latex] direction. Time increments measured as a coordinate as dt and [latex]d\acute{t}[/latex] using two identical clocks. Neither [latex]dt,d\acute{t}[/latex] or [latex]dx,d\acute{x}[/latex] are invariant. They do not obey postulate 1. A linear transformation between primed and unprimed coordinates above in space time ds between two events is [latex]ds^2=c^2t^2=c^2dt-dx^2=c^2\acute{t}^2-d\acute{x}^2[/latex] Invoking speed of light postulate 2. [latex]d\acute{x}=\gamma(dx-vdt), cd\acute{t}=\gamma cdt-\frac{dx}{c}[/latex] Where [latex]\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[/latex] Time dilation dt=proper time ds=line element since [latex]d\acute{t}^2=dt^2[/latex] is invariant. an observer at rest records consecutive clock ticks seperated by space time interval [latex]dt=d\acute{t}[/latex] she receives clock ticks from the x direction separated by the time interval dt and the space interval dx=vdt. [latex]dt=d\acute{t}^2=\sqrt{dt^2-\frac{dx^2}{c^2}}=\sqrt{1-(\frac{v}{c})^2}dt[/latex] so the two inertial coordinate systems are related by the lorentz transformation [latex]dt=\frac{d\acute{t}}{\sqrt{1-(\frac{v}{c})^2}}=\gamma d\acute{t}[/latex] So the time interval dt is longer than interval [latex]d\acute{t}[/latex] The above is what I would expect to see when one presents his own equation. The above isn't a full derivitave. Several missing steps. It was for another post. However it provides a better explanation of the Lorentz transformations than merely posting a formula. If your not using Lorentz then you need to define the coordinate transformation rules. Here is relativity of simultaneaty coordinate transformation in Lorentz. [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]

We see stars, moving, exploding, we can measure changes in plasma, such as the CMB. This is pointless, I'm recommending this thread to be locked, your obviously trolling.

I already supplied that, it's called observational evidence. We measure objects and events changing regardless of where we look with our most powerful telescopes. You need to supply evidence that your foolish conjecture has merit.

Your accusing that to be vaque balderdash? Did you not think your question better qualifies? If time did not exist there would be no change. We can observe events roughly 13 billion light years away and closer. What evidence do you have time doesn't exist elsewhere?

We know we see a rate of change when we observe events regardless of how far we look. We can measure those change of events. Those measurements correspond to relativity, precisely. That isn't based on religion but direct observational evidence. Where as your conjecture has none.

No spatial dimensions are used to measure geometric objects. There are 3 spatial dimensions one time dimension.

Not a bad link, it covers all the basic aspects. Without going too in-depth

Just to add some details. When you study particle physics in great detail. One comes to understand that particles are "Excitations in a field". So in point of detail from the classical formula I posted above, when you get into the quantum aspects, you start to learn that the number density of bosons and fermions can be calculated in any blackbody temperature. If you ever get into the Advanced formulas, the two main formulas is the Bose-Einstein statistics and the Fermi-Dirac statistics. These collectively can be combined into the Maxwell-Boltzmann statistics which is more practical in a blackbody where all known particles have dropped out of thermal equilibrium. Armed with the above and using particle physics one can then start calculating the % of elements that we can easily detect. For example, hydrogen, helium, lithium etc. These elements we can identify and measure with spectrography. Through These methods and numerous others we can TEST whether or not the number of particles are roughly the same. A reduction through nucleosynthesis and spectrography can be detectable. Another key evidence that there is no net outflow of energy, is that nucleosynthesis was able to predict the % of hydrogen, helium, lithium etc to a rather high degree of accuracy based upon the above prior to being able to measure those percentages. ( not too many people fully understand the significance of predicting those percentages as accurately as the LCDM model did) Even more incredible, is that the only method to average the number density of particles prior to being able to measure the CMB was by "Counting Stars" study a region and average that mass density to the Volume of the Universe. Yet even this seemingly inaccurate method allowed us to predict the correct values.

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.

Doesn't matter if the Universe is finite or infinite. The same gas laws are applicable to both. In the case of modelling our OBSERVABLE portion its finite. However there is no rule that states an infinite universe cannot increase in size either. The change in volume in the infinite case applies in the same way as the finite case. The distance from any measurements increases as the Volume increases. You can only measure a finite portion of an infinite quantity. However infinity can increase in size There is no limit to infinity. In terms of the ideal gas laws though we can't model infinite density too well. But compression not expansion.

The extra heat is distributed over a greater volume. Please study the following entry thermodynamic formula. [latex]pV=nRt[/latex] p=pressure V=volume N number of moles (particles) t =temperature. R is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant. Until you understand This equation there is no point in explaining how it works for an adiabatic fluid. Here is a half decent coverage. http://www.google.ca/url?q=http://authors.library.caltech.edu/25050/4/Chapter_03.pdf&sa=U&ved=0ahUKEwjJu-iE44DMAhUQ72MKHQB4D3AQFggRMAA&usg=AFQjCNGBdw9SjCUJw0FFlrOYC-hyANKbhg As the Volume increases BOTH the temperature and pressure drops The universe is modelled as an adiabatic and isentropic fluid. Meaning there is no net outflow of energy. The isentropic term refers to adiabatic and reversible.

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. 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. 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]

The majority of the mass in a solid is the binding energy of the electromagnetic force. Its so dominant it's a good approximation. It would be more accurate to think the field resistance. If you take the term binding energy as a field. Rather than two bodies. This will help as a field is modelled via coordinates in GR. So the relations to a particle from one coordinate to another will vary. Just a side note often people think of solids as having unique properties to a gas. The only real difference is density relations. It's perfectly valid to apply the ideal gas laws with a solid for example. However you have to be careful in the gas law definitions. For example you probably wouldn't treat a solid as an adiabatic fluid. An adiabatic fluid has no net flow of energy to its surrounding where a solid can lose temperature to its surroundings. So in your example if you want to model the interactions between two solids. You know from the above the total rest mass of each solid is due to total confinement energy/internal resistance to inertia. So mathematically in terms of fields its simpler to separate each solid and calculate each seperate. Then model the influence of one solid to another via the background field. This essentially is already being done via mass to mass relations between the two objects. A road can go anywhere you want it to. A photon follows a straight path, however a straight path can be curved Due to geometry change

It depends on the system. For example a vacuum has a pressure gradient near sources of mass. However on sufficiently large enough scales it uniform in distribution. Ie there is no pressure gradient on cosmological scales.

The conclusions above on just wavefunctions ignore numerous quantum numbers. For example spin correlations to velocity. - effects of charge to action. etc etc. yes you can successfully model interactions according to action this is one possible choice. Just judging from the scattering of references I don't get a good sense you fully understand the metrics involved behind each of the models you've referenced. Each model you mentioned have specified changes in relations. Quantum theory models the probability distribution. Yet relativity models more according to geometry change via vectors. (Which isn't specifically action... which has a more rigorous treatment). You can certainly adapt the four momentum to the Hamilton. However I seriously doubt the OP would understand the steps involved. Essentially the gist I get from the scattering of references you've provided is that your not truly familiar what each reference is designed to model in relation change. ( though like Klaynos, I'm having difficulty determing what your getting at)

Actually not bad. If you model spacetime as coordinate points. 4d. Time dilation and length contraction both occur. ( Time being the fourth coordinate.) So when you involve length contraction, from coordinate to coordinate the velocity doesn't change. What changes is the number of coordinates from a to b. (Sort of, this is where terms such as spacetime stretches etc come from). I'm going to cheat a bit on latex effort and use a previous post to help describe the above. Lorentz transformation. First two postulates. 1) the results of movement in different frames must be identical 2) light travels by a constant speed c in a vacuum in all frames. Consider 2 linear axes x (moving with constant velocity and [latex]\acute{x}[/latex] (at rest) with x moving in constant velocity v in the positive [latex]\acute{x}[/latex] direction. Time increments measured as a coordinate as dt and [latex]d\acute{t}[/latex] using two identical clocks. Neither [latex]dt,d\acute{t}[/latex] or [latex]dx,d\acute{x}[/latex] are invariant. They do not obey postulate 1. A linear transformation between primed and unprimed coordinates above in space time ds between two events is [latex]ds^2=c^2t^2=c^2dt-dx^2=c^2\acute{t}^2-d\acute{x}^2[/latex] Invoking speed of light postulate 2. [latex]d\acute{x}=\gamma(dx-vdt), cd\acute{t}=\gamma cdt-\frac{dx}{c}[/latex] Where [latex]\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}}[/latex] Time dilation dt=proper time ds=line element since [latex]d\acute{t}^2=dt^2[/latex] is invariant. an observer at rest records consecutive clock ticks seperated by space time interval [latex]dt=d\acute{t}[/latex] she receives clock ticks from the x direction separated by the time interval dt and the space interval dx=vdt. [latex]dt=d\acute{t}^2=\sqrt{dt^2-\frac{dx^2}{c^2}}=\sqrt{1-(\frac{v}{c})^2}dt[/latex] so the two inertial coordinate systems are related by the lorentz transformation [latex]dt=\frac{d\acute{t}}{\sqrt{1-(\frac{v}{c})^2}}=\gamma d\acute{t}[/latex] So the time interval dt is longer than interval [latex]d\acute{t}[/latex] The above is what I would expect to see when one presents his own equation. The above isn't a full derivitave. Several missing steps. It was for another post. However it provides a better explanation of the Lorentz transformations than merely posting a formula. If your not using Lorentz then you need to define the coordinate transformation rules. Here is relativity of simultaneaty coordinate transformation in Lorentz. [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] Now granted you may not understand the math above, but the velocity addition rules involve both length contraction and time dilation. They go hand in hand. In a sense you can't seperate one from the other.

Absolutely incorrect to relativity. Just judging from the above this post belongs in the Speculation forum. Your essentially stating it's this way without understanding time as a geometry vector. Although I can move it myself I would prefer a second opinion via a site Mod. Gravity is modelled via geometry relation, time is a coordinate ie a vector coordinate. Your above post shows that you don't understand basic relativity.