My Favorite Equations I Have Derived Over The Years.

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#35 Dubbelosix

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Posted 17 June 2019 - 08:08 PM

Ok so let;s get into the cosmological fluid equation, the Friedmann equation. The most simplest form of the Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{4 \pi G}{3c^2}\rho$

One correction I made I applied in the past, inspired through the work of Motz and Kraft was that the equation need not be taken as a conservation equation - this is after all at odds with relativity in which energy is ill-defined under the Noether theorem, the distribution of a time variable indicates we get

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = -\frac{4 \pi G}{3c^2}\dot{\rho}$

Motz' quite rightly stated that the equation need not be conserved, just as this previous equation indicates, as he said it was an ''unfounded assumption by Friedmann to conserve the energy in such a way) - and general relativity hints at this since not only is energy ill-defined, it is ill-defined because time itself is ill-defined (as it states there is no cosmological constant) under the Wheeler de Witt equation, in which diffeomorphism invariance is not a true time evolution of the theory. Further more, the Wheeler de Witt equation is also often taken as a statement of a conserved universe in which no changes happen inside of it, the consequence of it having no time derivative. But logic should be indicating to us that  many phase transitions have occurred and so the general statement should be taken with a pinch of salt.

Using the non-conserved form of the Friedmann equation, we can introduce the so-called continuity equation, but this is a misnomer if the derivative on the density term implies a change inside of the universe in a non-conserved way. The misnamed continuity equation takes the form:

$\dot{\rho}= -3H(\rho + P)$

where we take all terms here as the energy density and pressure. $H$ is the usual Hubble constant, but we should read this that density is not fixed, but dynamic component. From previous posts I have shown that the total pressure would imply a correction to this relativity term:

$\dot{\rho}= -3H[\rho + (P_0 - P)] = -3H[\rho + q]$

You can plug this into this into the non-conserved form of the Friedmann equation as

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = -\frac{4 \pi G}{c^2}H[\rho + (P_0 - P)]$

$= \frac{4 \pi G}{c^2}H[\rho - (P_0 - P)] = \kappa [\dot{\rho} + H(P_0 - P) = \kappa [\dot{\rho} + H(P_0 - P) = \kappa [\dot{\rho} + \dot{q}]$

$\kappa = \frac{4 \pi G}{3}$

Also an important quantity, the density parameter is equivalent to

$\Omega \equiv \frac{P}{\rho} = \frac{8 \pi G \rho}{3H^2}$

Later we will come back to this as it will play a role in the Bernoulli's principle. We must also consider for this following work the following work the presence of a second correction term for the total pressure of the fluid, as we had defined earlier as

$q = (P_0 - P)$

Which is a correction term, for the ''continuity equation'' a misnomer if a third derivative actually implies a diabatic, non-conserved solution for the Friedmann equation:

$\dot{\rho}= -3H(\rho + (P_0 - P)) = 3H((\rho - (P_0 - P)) = - 3H((\rho - (P_0 - P)) = 3(\dot{\rho} - \dot{q})$

the last set of terms in this equation has also featured in investigations into practically similar expanding and contracting models for sonoluminscence models - this is no accident of course since both those solutions are also derived from fluid mechanics.Plugging the main corrections into the non-conserved form of the Friedmann equation yields:

$\frac{\dot{R}}{R} \frac{\ddot{R}}{R} = \frac{4 \pi G H}{c^2}(\rho - (P_0 - P))$

$= \frac{4 \pi G }{c^2}(\dot{\rho} - \dot{q})$

I have made previous corrections terms, such as introduction a zero point energy term for a pre-big bang state that existded at very near absolute zero to attempt to formulate a semi-classical model. The Friedmann equation, even in principle, cannot predict why spacetime should be flat - let me quickly explain. The density parameter, $\Omega$, is defined as the ratio of the actual (or observed) density $\rho$ to the critical density$\rho_c$ of the Friedmann universe. The relation between the actual density and the critical density determines the overall geometry of the universe; when they are equal, the geometry of the universe is flat (Euclidean). Appealing to authority, Susskind himself and Lisi Garret have expressed their opinions, that the universe is not flat - a way to articulate why this is , is because the universe would have to be infinitely large and this isn't really how we think about the universe in a scientific way, because it would take the universe an infinite steps to do so - but steps involve numbers and infinities are not numbers by definition. At any moment in time, no matter what time, the universe could be measured being finite in value always.

In a similar notion, the only way a black hole can become flat, would that the radius

$R \rightarrow 0$

Meaning the curvature of an expanding sphere must disappear

$K \rightarrow 0$

But the universe can only approach these values, but never reach it perfectly.

As wiki states itself: ''In earlier models, which did not include a cosmological constant term, critical density was initially defined as the watershed point between an expanding and a contracting Universe. To date, the critical density is estimated to be approximately five atoms (of monatomic hydrogen) per cubic metre, whereas the average density of ordinary matter in the Universe is believed to be 0.2–0.25 atoms per cubic meter.''

I have become a hard disbeliever against dark matter, too much mounting evidence against it over the years - I could go into loads of reasons why, but I will spare the reader. I also believe that dark energy has been misunderstood, but we will get to that later. The entire basis of the density parameter to the actual observed ratio of matter in the universe, is too many magnitudes off, which should have been taken seriously that the Friedmann equation may itself be flawed fundamentally.

Next we will get into Bernoulli's principle and how the work above will come together to explain a source of the cosmological constant and how it applies to dark energy due to a weakening of gravity in the universe - the cosmological constant has often been believed that it is not really a constant, but there are some intriguing reasons why we may have to consider new alternatives to explain this in a more rationally-mechanical way due to pressure, perhaps thermodynamic in nature.This time around i plan to write the correct formula, in whcih we retain that dark energy remains constant, but the acceleration entertained by a weakening of gravity over galactic scales, only giving the impression dark energy is soley due to it - basically, the weakening of gravity only appears to make the constant change, when it is not. The gravitational binding of the universe would arise in my picture as an effect of gravitation become weaker as it gets larger and so only appears to accelerates. Anyway, back to this tomorrow.

Edited by Dubbelosix, 18 June 2019 - 03:55 PM.

#36 Dubbelosix

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Posted 18 June 2019 - 09:30 AM

There is a density parameter which can be used in Bernoulli's principle to apply for cosmological applications. But before
we jump into that, the best way to see how Bernoulli's principle can explain the cosmological constant is literally by
writing the constant as a constant energy density in the style of conservative forces:

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

Now, let's not move to quickly, I want to now define the so-called ''density parameter'' for the Friedmann equation ~

$\Omega \equiv \frac{P}{\rho} = \frac{8 \pi G \rho}{3H^2}$

And the relative formula for the density parameters Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{2c^2}(\rho + P)$

So we have a number of things we can do here; first, let's plug in the density parameter and then divide through by the energy density:

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}\frac{P}{\rho}$

So with this equation and the other what do we do now?

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

We now plug in the co-called ''critical density'' for both formula's - but ''critical density'' is a misnomer in the absence of dark matter and the proper interpretation of dark energy:

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}\Omega = \frac{8 \pi G}{3c^2}\frac{8 \pi G \rho}{3H^2}$

$(\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\Omega)E = (\frac{v}{2c^2} + \frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{8 \pi G \rho}{3H^2})E = \mathbf{C}_{vac}$

I will be introducing once again the Leonard extension of the Einstein equivalence to explain why a universe expands rather than contracts, but these formula's are a really good starting point for new invetsigation, my last attempts were a bit clumsy, this  is a much better approach.

Edited by Dubbelosix, 23 June 2019 - 03:27 AM.

#37 Dubbelosix

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Posted 18 June 2019 - 01:49 PM

Ok, so we spoke about $C$ being a constant, today we know it is related to energy and the main main reason why

this Freidmann expansion melding into Bernoulli's principle was not a ''an assumption'' as was challenged by someone who followed the work, because he seemed to be unaware that we work with fluid dynamics when working with an expanding universe, which relies on different pressures through various complicated parameters. In other words, the two have stronger relationships to each other than most would like to agree on, or have been unaware of.

To present the laws spoken in previous posts, a general application from the last equation would be to distribute an energy, such that the constant on the right hand side, is a constant of energy. We will leave out the right hand side until I can try and simplify it further, the real equation we should concentrate will be of the form:

$(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

where now C is no longer ''just a constant'' but is define under the units of energy defined through the vacuum contribution.

This model, finally is becoming a unified equation for the universe.... but there is one final thing to do... and that will be the last highlight of the thread, an implication no less, of a driving force which we may consider as an interpretation of the cosmological constant. To do so, we require the work of Fritz Rohrlich who demonstrated that the force on a moving system (as in opposition to one at rest) is important when considering the mass-energy equivalence.

$\frac{1}{\rho}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^2}\Omega = \frac{8 \pi G}{2c^2}\frac{8 \pi G \rho}{3H^2} = \frac{64 \pi^2 G^2 \rho}{3c^2}$

It's also interesting we can find the classical upper inverse limit of the gravitational force (but with an additional gravitational constant term), which implies an inverse of the stress energy tensor on the left hand side:

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{2c^4}\frac{8 \pi G \rho}{3H^2} = \frac{64 \pi^2 G^2 \rho}{3c^4}$

$T_{00}^{-1}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2} = \frac{64 \pi^2 G^2 }{3c^4}\ \rho = \frac{64 \pi^2 G^2 }{3c^4}\ \rho$

$\frac{\ddot{R}}{R} = \frac{4 \pi G}{3c^2}\rho$

The model he proposed considered in a frame that moves with velocity to the left, the driving force moving to the left is redshifted, while the driving force moving to the right is blueshifted. The blue light carries more momentum than the red light, so that the momentum of the light in the moving frame is not balanced - in other words, a cosmological constant is interpreted here-on-in as a non-balanced force since the energy is carrying some net momentum to the right.

The object has not changed its velocity before or after the emission, however, in this frame it has lost some right-momentum to the energy driving it in a particular direction. The only way it could have lost momentum is by losing mass - this may be also a statement of non-conservation and is not only quintessentially tied to Poincaré's radiation paradox, it also solves it.

So the right-moving energy carries extra momentum $\Delta p$ we then have

$\Delta p =\frac{v}{2c^2}E$

The left-moving energy will carry a little less momentum, by the same quantity $\Delta p$ such that the total right-momentum in the energy is twice the value of $\Delta p$. This is the right-momentum energy lost from the system (universe)

$2\Delta p=\frac{v}{c^2}E$

$2\Delta p v =\frac{v^2}{c^2}E$

In which $\frac{v^2}{c^2}$ is the Doppler shift, more appropriately given in the squared form, rather than the form Leonard had given, because it allows us to write the metric in a relativistic way.

The momentum of the universe moving in the directional frame after the emission is reduced by the amount of

$p′ = mv−2\Delta p = (m − \frac{E}{c^2})v$

So the change in the universes mass is equal to the total energy lost divided by the speed of light squared - the big implication here is that any emission of energy can be carried in a two-step process in which energy used by the universe is converted to mass, while the emission of an energy is accompanied by a loss of the mass in the universe. To understand this pressure difference, we now include Leonards correction to the mass energy equivalence in terms of the first term in the paranthesis: To do that, we remind ourselves first of the formula we want

$\Delta p v =\frac{v^2}{2c^2}E$

and then we plug in:

$\Delta p v + (\frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

The physics of the first term, will not only explain why something expands rather than contracts, but it will also provide under its own definition, a redshift term - meaning things are moving outwards, not inwards, (ie. the universe is expanding, not contracting due to pressure differences) and because we have used a conservative form, it means it does not rely on the gravitational field soley.

Edited by Dubbelosix, 23 June 2019 - 03:27 AM.

#38 Dubbelosix

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Posted 18 June 2019 - 02:30 PM

So to finish up, I will write the last equation in a different way, for the last time then leave you with a link describing most of the theory in a nutshell:

$(\frac{v^2}{2c^2} + \Psi + (\frac{T_0}{T})\frac{P}{\rho})E = \Delta p v + (\frac{\Psi}{\psi} + (\frac{T_0}{T})\frac{P}{\rho})E = \mathbf{C}_{vac}$

Without the aether drag threads, this was finally what was intended in my investigations, I won't lie, it was hard to put together. I hope people have learned something, even if my theory is not the right one.

https://quantizedbla...ological-Theory

#39 Dubbelosix

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Posted 18 June 2019 - 03:59 PM

Numerical mistake here? Anyone spot it? I accidently forgot to take into consideration $3H$

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2} = \frac{64 \pi^2 G^2 \rho}{3c^4}$

$T_{00}^{-1}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega = \frac{8 \pi G}{2c^4}\frac{8 \pi G \rho}{3H^2} = \frac{64 \pi^2 G^2 }{3c^4}\ \rho = \frac{64 \pi^2 G^2 }{3c^4}\ \rho$

$\frac{\ddot{R}}{R} = \frac{4 \pi G}{3c^2}\rho$

So let's fix that part....

Edited by Dubbelosix, 12 August 2019 - 09:17 AM.

#40 Dubbelosix

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Posted 19 June 2019 - 12:22 AM

The relative formula for the density parameters Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}(\rho + P)$

Let's divide through by the relative density and pressure

$\frac{1}{\rho + P}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}$

But we need not take such a relativistic case, because this will drop out naturally when we take the non-conserved form of the Friedmann equation. We shall get to that soon.

I also entertained

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega$

Where we remind ourselves that the misnamed ''critical density'' is:

$\Omega = \frac{\rho}{P} = \frac{8 \pi G}{3H^2} \rho$

To retrieve the upper limit of the classical gravitational force.

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

The continuity equation is

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

and the respective energy density is

$\dot{\rho}c^2 = -H(\rho c^2 + P)$

So, we've been through this a few times, the non-conserved form of the Friedmann equation requires one extra derivative which removes one such factor of the Hubble ''constant'' (or in the last term as explained not long ago, it gives us the continuity which is strictly defined under the density and pressure parameters) - we also distribute the factor of three from the denominator:

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{3c^4}\frac{8 \pi G}{H^2} \dot{\rho}$

This gives by plugging in the continuity:

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{3c^4}\frac{8 \pi G}{H^2} \frac{\dot{R}}{R}(\rho + P)$

Which neatly gives a new interpretation/form of the expansion of a universe, in diabatic from.

Edited by Dubbelosix, 30 July 2019 - 07:09 AM.

#41 Dubbelosix

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Posted 19 June 2019 - 12:30 AM

Oops, before I leave, I meant to distribute through an inverse factor of $8 \pi G$ to neaten it up without over-simplifying by combining components:

$\frac{3}{8 \pi G \rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{1}{c^4}\dot{\Omega} = \frac{8 \pi G}{2c^4}\frac{ \rho}{H} = \frac{8 \pi G}{2c^4}\frac{1}{H^2} \frac{\dot{R}}{R}(\rho + P)$

Edited by Dubbelosix, 19 June 2019 - 12:31 AM.

#42 Dubbelosix

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Posted 19 June 2019 - 04:51 AM

I should have seen this early, a more elegant solution is, we rearrange the last equation for the Hubble constant:

$\frac{3H^2}{8 \pi G \rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^4} \frac{\dot{R}}{R}(\rho + P)$

Just because it fits in nicely without the inverse term of the same constant spoke about.

Edited by Dubbelosix, 20 June 2019 - 10:00 AM.

#43 Dubbelosix

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Posted 20 June 2019 - 09:32 AM

The first line implies f = 1/n² -1.

The second line says 3/n² +f = 1, so substituting for f, we have 3/n² +1/n² -1 = 1

So rearranging, we have 4/n²= 2; so 2/n² = 1, hence n² = 2 and so n = √2.

Now we can solve for f, using the first line: f = 1/2 - 1 = -1/2.

n = √2; f = -1/2.

Because you noticed this, it will be written up into my paper I plan to have published at the end of the year, rest assured, a link will be provided soon to the entire derivations, including the original investigation into the transition of a black hole by discrete quantum processes.

#44 Dubbelosix

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Posted 23 July 2019 - 03:13 AM

Been away but been doing lots more work on this, pages in fact, which also implements what Exchemist noticed on the first page. Will be back in due time to write the final formula's.

#45 Dubbelosix

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Posted 25 July 2019 - 03:41 AM

I've still been busting away with  numerical values, so just for now, a classical representation of drag can be made from Newtons approximated equations in the low energy limit.

The three equations that implemented the dragging coefficient was:

$F = \frac{1}{2} \rho v^2\ f\ A = m\frac{2 \pi c\ k_BT}{\hbar} = m\frac{4 \pi c}{\hbar} \frac{E}{N} = m\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \frac{4 \pi GMm}{A} = G \frac{Mm}{r^2}$

$\frac{1}{f}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - \frac{A_b}{A_f} \cdot \rho v^2 = - \frac{A_b}{A_f} \cdot T^{00}$

$- \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - g^{00}\ \frac{\Lambda c^4}{8 \pi G} = T^{00}$

The first was a drag equation unified with thermodynamic gravity in two dimensions. The second equation was transcended into a three dimensional format and the third explained the cosmological constant as a thrust. The thrust in this sense, had to be greater than the drag of gravity: It does represent a small quantity of the the universal energy in one direction favored from another - this impetus has been formulated under the areal-thermodynamic viewpoint of gravity from the first equation. The true form of the equation does contain a sign convention:

$F_{21} = \frac{1}{2} \rho v^2\ f\ \vec{A} = - G \frac{M_1m_2}{r^2_{[12]}}\frac{r^2_2 - r^2_1}{|r^2_2 - r^2_2|}$

From ordinary classical mechanics in which the sign appears from the vector form of the equation. The drag coefficient may be interpreted as a thrust force in such a form:

$F_{21} = \frac{1}{2} \rho v^2\ \vec{A} = - G \frac{M_1m_2}{f\ r^2_{[12]}}\frac{r^2_2 - r^2_1}{|r^2_2 - r^2_2|}$

Because again, the negative of the drag is a thrust equation. We vcan apply a circular integration similar to the Gauss law while dropping the unecessary factors:

$\oint_R\ F \cdot dr = 2 \pi \rho v^2\ \vec{V} = - 4 \pi G \frac{Mm}{f\ r}$

In the low energy limit approximation it becomes with a standard mass (of the sun) we have a gravitational drag on the earth (denoted by the small mass term as);

$\oint_r\ F_{drag} \cdot dr = 2 \pi \rho v^2\ f \vec{V} = - 4 \pi G\frac{M_{sun}m}{ r} = -4 \pi m \phi$

Division of $c^2$ gives us the Doppler shift:

$\frac{1}{c^2} \oint_r\ F_{drag} \cdot dr = 2 \pi \rho (\frac{v^2}{c^2})\ f \vec{V} = - 4 \pi G\frac{M_{sun}m}{rc^2} = -4 \pi m (\frac{\phi}{c^2})$

Which can also be written as:

$\frac{1}{c^2} \oint_r\ F_{drag} \cdot dr = 2 \pi \rho (1 - \frac{1}{\gamma^2})\ f \vec{V} = - 4 \pi G\frac{M_{sun}m}{rc^2} = -4 \pi m (\frac{\phi}{c^2})$

Reference:

Solar mass - Wikipedia

I was able to continue even further with the new equations:

Continuing, we also have,

$\frac{1}{c^2} \oint F_d \cdot dr = 2 \pi \rho (1 - \frac{1}{\gamma^2})fV = -4 \pi G (\frac{M_{sun}m}{rc^2}) = -4 \pi m (\frac{\phi}{c^2})$

-this is similar to the apsidol motion inverse cube law, similar but not quite the same, with Newton defining a correction factor of

$F = G \frac{Mm}{r^2} + B\frac{Mm}{r^3}$ with $B$ having units of a constant but with a length. Expending to first order we have:

$\frac{1}{c^2} \oint_{r} F_d \cdot dr = 2 \pi \mu (\frac{\phi}{c^2})\ f = -4 \pi G(\frac{Mm}{rc^2}) \rightarrow -4 \pi m(\frac{\phi_1 - \phi_2}{c^2}) = - \frac{4 \pi G}{c^2} (\frac{Mm}{r_1r_2}|r_2 - r_1|)$

with

$\Delta \phi = \phi_1 - \phi_2 = \frac{m}{r_1r_2}|r_2 - r_1|$

Edited by Dubbelosix, 25 July 2019 - 03:42 AM.

#46 Dubbelosix

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Posted 25 July 2019 - 03:46 AM

The drag force is also related to the stress energy and quipartition laws due to the velocity pressure as:

$\oint_r F \cdot dr = \frac{3}{2}k_BT\ f = \frac{1}{2}mv^2\ f = \frac{1}{2}\rho v^2\ fV$

and so

$\epsilon_d = \oint_V T_{00}\ f\ dV$

Edited by Dubbelosix, 25 July 2019 - 03:50 AM.

#47 Dubbelosix

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Posted 29 July 2019 - 07:05 AM

The relative formula for the density parameters Friedmann equation is

$\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}(\rho + P)$

Let's divide through by the relative density and pressure

$\frac{1}{\rho + P}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^2}$

But we need not take such a relativistic case, because this will drop out naturally when we take the non-conserved form of the Friedmann equation. We shall get to that soon.

I also entertained

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\Omega$

Where we remind ourselves that the misnamed ''critical density'' is:

$\Omega = \frac{\rho}{P} = \frac{8 \pi G}{3H^2} \rho$

To retrieve the upper limit of the classical gravitational force.

$\frac{1}{\rho c^2}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^4}\Omega = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{3H^2}$

The continuity equation is

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

and the respective energy density is

$\dot{\rho}c^2 = -H(\rho c^2 + P)$

So, we've been through this a few times, the non-conserved form of the Friedmann equation requires one extra derivative which removes one such factor of the Hubble ''constant'' (or in the last term as explained not long ago, it gives us the continuity which is strictly defined under the density and pressure parameters) - we also distribute the factor of three from the denominator:

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{3c^4}\frac{8 \pi G}{H^2} \dot{\rho}$

This gives by plugging in the continuity:

$\dot{\rho} = -3H(\rho + \frac{P}{c^2})$

$\frac{3}{\rho c^2}\frac{\dot{R}}{R}\frac{\ddot{R}}{R} = \frac{8 \pi G}{3c^4}\dot{\Omega} = \frac{8 \pi G}{3c^4}\frac{8 \pi G \rho}{H} = \frac{8 \pi G}{2c^4}\frac{8 \pi G}{H^2} \frac{\dot{R}}{R}(\rho + P)$

Which neatly gives a new interpretation/form of the expansion of a universe, in diabatic from.

An alternative diabatic form of the thermal state of equation, I get reformulated a gas law under the guise of a Gibbs-Helmholtz equation in the form:

$f\ \oint_V\ (\frac{\dot{\rho}}{n} + \dot{P}(\frac{1}{n})) \frac{d^2V}{dt^2} = 6 k_BT\ (\frac{\ddot{R}}{R}\frac{\dot{R}}{R})$

And I further state that the drag coefficent is better understood as being:

$f = \pm \frac{2F}{\rho v^2\ A}$

So that the negative of the force can be a drag or a thrust. The conclusions from the first equation (the equation of state), the universe expands due to thermal contribution acting a pressure-density related flux. The non-conserved form that it is in, may have implications which understanding Bernoulli'ds modified equation for cosmology - if the pressure is negative then it is possible for the the drag coefficient also to be either positive or negative.

Edited by Dubbelosix, 29 July 2019 - 07:12 AM.

#48 Dubbelosix

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Posted 30 July 2019 - 09:01 AM

The pressure formula from the fundamental principles of entropic gravity from my own derivations found that:

$\frac{1}{c^2} (\frac{P}{\rho_A}) = \frac{1}{c^2}\frac{2P}{8 \pi^2 \rho_A} = \frac{\rho}{8 \pi^2 \rho_A} (\frac{v^2}{c^2})\ f = \frac{mc}{2 \pi \hbar} = (2 \pi \lambda)^{-1}$

So what is the significance of the Doppler term $\frac{v^2}{c^2}$. Ok, special relativity is inherently tied to the Gamma function, which itself describes the variations of velocity to that of the speed of light. Following algebra hacks I wrote up a while ago,

More Identities and Useful Hacks

We find that in a simplified expression, the Doppler effect is

$\frac{v^2}{c^2} = 1 - \frac{1}{\gamma^2}$

When solving for $\gamma$ we get a very nice set of solutions:

$\gamma = i\frac{c\sqrt{v^2 - c^2}}{v^2 - c^2}$

and

$\gamma = -i\frac{c\sqrt{v^2 - c^2}}{v^2 - c^2}$

Where we notice that for this to be true, $v \ne c$. You could go further and state that $v \ne - c$ but this is where pure algebra reaches a limit because there is no such thing as a negative speed of light.

This brings me to two articles I have read in the past:

https://arxiv.org/abs/1403.0389

And I am now considering that the second solution for $-c$ may be part of the imaginary number attached to a generic mass term for the Tachyon models. If there is a way to coherently express this in some other language I will try the next few days.

Edited by Dubbelosix, 30 July 2019 - 09:03 AM.

#49 ralfcis

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Posted 02 August 2019 - 08:07 AM

"but this is where pure algebra reaches a limit because there is no such thing as a negative speed of light. "

Yup  if c=x/t algebra just doesn't know what to do with -c = -x/t because that would mean light is going in the opposite direction and no one knows what that means because only tachyons can do that and there are no tachyons in a flashlight if you turn it away from your face. But why would any normal person not want the light to shine in their face because everything is much brighter that way. Also, your equation for Y, why would anyone needlessly introduce i when you can just take the sqrt of c2-v2 instead of v2-c2. The sign of c wouldn't matter anyways because c is squared.You are just spouting useless nonsense. Take your meds unless equation porn is self-medication, then, continue.

Edited by ralfcis, 02 August 2019 - 08:13 AM.

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#50 OceanBreeze

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Posted 02 August 2019 - 10:57 AM

"but this is where pure algebra reaches a limit because there is no such thing as a negative speed of light. "

Yup  if c=x/t algebra just doesn't know what to do with -c = -x/t because that would mean light is going in the opposite direction and no one knows what that means because only tachyons can do that and there are no tachyons in a flashlight if you turn it away from your face. But why would any normal person not want the light to shine in their face because everything is much brighter that way. Also, your equation for Y, why would anyone needlessly introduce i when you can just take the sqrt of c2-v2 instead of v2-c2. The sign of c wouldn't matter anyways because c is squared.You are just spouting useless nonsense. Take your meds unless equation porn is self-medication, then, continue.

You definitely don't want those negative tachyons shining in your eyes. Be careful how you hold that flashlight!

I thought my derivation was much more straight forward:

The derivation of   $Z=\sqrt { \frac { 1+v/c }{ 1-v/c } }$ comes about from these considerations:

1. The clock on Alice’s spaceship is running slower, as seen by Bob the earth twin, according to the familiar Lorentz transform. It will take $\frac{1}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ seconds before the next tick of the spaceship’s clock and the next frame of video to be sent (assuming one frame per second for simplicity)
2. The spaceship is also moving away from the earth, in this example, so Bob will not receive that next video frame until the signal, travelling at the speed of light, has travelled back to him over the distance covered by the spaceship between video frames. So, we have the ship moving away at v, and the time between frames is $\frac{1}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ seconds.

Therefore the distance the spaceship covers between video frames as measured by Bob  is

$\frac{v}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$, as distance is just velocity multiplied by time.

Since the video signal travels at the speed of light c, it covers this distance in time

$\frac{v}{c \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ since time is just distance divided by velocity.

So now we can calculate the total time between Bob seeing one video frame and the next if the frames are generated one second apart according to the clock on Alice’s ship as:

$\frac{1}{ \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ + $\frac{v}{c \sqrt{1 - \frac { v^{2} }{ c^{2}} }}$ =   $Z=\sqrt { \frac { 1+v/c }{ 1-v/c } }$

This is the change in frequency (Doppler shift) which is 0.5 for a halving of the frequency on the way out and 2 or a doubling of the frequency on the way back in. (In octaves, it is a decrease or an increase of one octave)

#51 exchemist

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Posted 02 August 2019 - 11:33 AM

Yes it makes no sense to “solve” Dubbelsox’s equation for gamma, since that equation is just a rearranged version of the definition of gamma in the first place. Gamma is just the Lorentz factor. It is quite obviously NOT imaginary, so the intrusion of i shows that he has gone off the rails, as usual.