# Principia Of Gravity Revised For Drag

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### #1 Dubbelosix

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Posted 19 March 2020 - 11:59 AM

It should be noted that the references at the end should be explored first to understand some dynamics and essentials of the investigation of the following work. When reading Newtons Principia, I was surprised to learn that he had objectively denied gravity as a drag phenomenon and in recent light of evidence discovered in which a star was found to drag space and time around with it, I felt the need to revisit his laws and implement the drag coefficient for the new physics. It has been an essential argument of mine for a while now that the supermassive black holes at the centers of most typical spiral galaxies will play the role of a spacetime polarization in which dark matter effects results from the same binding energy they harbour with trillions of solar masses.

To start off, Newtons law is:

$F = \frac{Gm^2}{R^2} = \frac{mv^2}{R}$

He went off to find a unification to prove that Keplers law for planetary motion around the sun was justified. To do this he took the rotation velocity

$v_{rot} = \frac{2 \pi R}{t}$

And he plugged it in giving

$G(\frac{m}{R})^2 = \frac{m(2 \pi R/t)^2}{R}$

By rearranging he found that Keplers third law was obtained as

$\frac{R^3}{t^2} = \frac{Gm_{sun}}{4 \pi^2} = \frac{\mu_{sun}}{4 \pi^2}$

It is interesting that he wholeheartedly objected to the drag interpretation of gravity since all the planets part from one, rotates in the direction of the spin of the sun. That rule even extended to spiral galaxies in which there is no typical spiral galaxy that rotated in a direction opposite to the black hole they harbour. This was no coincidence in my eyes and strongly suggested to me that the thing we call dark matter was in fact produced from a more local effect inside of the galaxies themselves as opposed to some fundamental field rendering it superfluous and excessive. I will continue from the drag formula I created to predict a heuristic approach to explaining the drag dynamics of the supermassive black hole, from it the binding energy is found as

$E_{binding} = F_{drag} \cdot \int dR = \frac{1}{2}f(\rho + 3P)\ \int dV_{smbh}$

The drag in this case changes linearly to the volume of the supermassive black hole and I also invited to relativistic correction of $3P$ where $f$ is the gravitational drag coefficient. We can rewrite the last equation as

$E_{binding} = m_{drag}g \cdot \int dR$

In which we define

$g \equiv \frac{m_1}{m_2}\frac{v^2}{R} = \frac{Gm}{R^2}$

Where $Gm$ is the gravitational parameter denoted as $\mu$ and has dimensions of $Rv^2 = Gm$.

where $g$ is the usual gravitational acceleration, by solving for the drag coefficient I get

$f \equiv 2g(\frac{R}{v^2}\frac{V_2}{V_1}) = \frac{4 \pi^2 R^2}{v2t^2} = \frac{4 \pi^2 R}{t^2}\frac{R}{v^2} = \frac{4 \pi^2 R}{t^2}g^{-1}$

Distributing the factor of acceleration, we get

$f(\frac{Gm}{R^2}) = \frac{4 \pi^2R}{t^2}$

and by distributing the radius squared we obtain Keplers third law under the drag coefficient interpretation of gravity:

$f \cdot Gm = \frac{4 \pi^2R^3}{t^2}$

From here I wanted to get a little bit more creative, and found that by plugging in the rotational velocity into the gravitational acceleration defined near the beginning of the work I could obtain:

$g \equiv (\frac{m_1}{m_2})\frac{4 \pi^2 R^2}{t^2 R}$

Multiplying through by a mass term yields Newtons force equation with a twist, mind the pun, because we obtain a quantity known as the rotational inertia $mR^2$

$F_{drag} \equiv mg = (\frac{m_1}{m_2})\frac{4 \pi^2 mR^2}{t^2 R}$

and can be rearranged for the rotational energy associated to the moment of inertia with a double time derivative

$F_{drag} \cdot \int dR = 4 \pi^2 (\frac{m_1}{m_2}) \ddot{I_{rot}}$

This subtle unification of gravity drag into Newtons formula will provide a toy model, or starting point towards a revised Newtonian law that satisfies the new experimental evidence.

References

https://en.wikipedia...zeau_experiment

https://en.wikipedia...Reynolds_number

https://en.wikipedia...rag_coefficient

Edited by Dubbelosix, 27 March 2020 - 07:16 AM.

### #2 Dubbelosix

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Posted 19 March 2020 - 04:18 PM

As I stated before, the idea of a gravity drag I came to learn was even taken seriously by NASA, as you will find in the following link

https://www.grc.nasa...ket/termvr.html

It has almost all the essential ingredients within my heuristic approach, but contains an essential feature if the model is to be taken seriously, and that is, the notion of weight. Defining weight is critical under a drag theory ad yet weight is remains ill-defined in general relativity but not ill-defined under Newtonian mechanics. If I was to modify the rough equations set in the link to satisfy my own approach, it not only requires the weight but also the relativistic correction and would feature as:

$F_{drag} = \frac{1}{2} f (\rho + 3P) \frac{v^2}{R} \int dV - W$

In the work I provided, we remove area definitions for the volumetric interpretation, something equally important within the idea of the Ricci flow discussed in my posts a few weeks back. The work in NASA's page suggests the frontal area, a definition found within fluid dynamics respectively from the same type of model satisfying the Fresnel drag. Further more, the fluid equations describing this I suspect to be equally important for cosmological purposes since the most important of cosmological equations are derived from using the ideal laws of fluid motion. Tomorrow, if I find time, I will define the weight under an inspection of the classical Newtonian physics and then try interpret it all under general relativity. It is not a lack of general relativity being incapable of describing weight, it is just that gravity was never a fundamental force as Newton had erroneously believed and this somewhat influenced Einstein directly in the lack of definition within general relativity.

Edited by Dubbelosix, 27 March 2020 - 06:25 AM.

### #3 Dubbelosix

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Posted 19 March 2020 - 06:21 PM

I thought bugger it, I will complete the classical definition while I can since I cannot promise I will be here tomorrow for sure as I may be tied up with house work. Weight is defined using the 'the most hated integral' notation (a personal beast against me since I know infinities do not exist in nature) from the following equation by Newton: To add before we cover the equation, there is a large assumption of Newton that gravity is an instantaneous force which can be easily challenged at a later date... but for now, the weight is defined through the potential energy equation ~

$PE = \int_{r=R}^{R = \infty} G(\frac{m}{R})^2\ dr$

$= Gm^2 \int_{R}^{\infty}\ \frac{dr}{r^2} = G\frac{m^2}{R} = G(\frac{m}{R})^2 R = mgr = W \cdot R$

The drag formula I derive is

$F_{drag} = \frac{1}{2}f\ \rho \frac{v^2}{R} \int dV - mg$

Edited by Dubbelosix, 27 March 2020 - 06:25 AM.

### #4 Dubbelosix

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Posted 20 March 2020 - 08:25 AM

Yes, but more pertinent is that dark matter effects cannot be observed for the first four billion years of the universe... this logically tells me that the supermassive black holes where not supermassive at all and had to feed for four billion years before their effects became noticable. Why the field theory of dark matter remains as part of the standard model in face of this confounds me.

### #5 Dubbelosix

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Posted 26 March 2020 - 08:12 AM

One major consequence discovered a while back by myself, was that Einsteins weak equivalence explaining that gravitational mass being equal to the inertia of the system would further be extended to mean it was also related to the drag exerted on the mass. In simple terms the inertia is a resistance of a piece of matter to be moved from rest into motion. This resistance is very small and is actually due to the gravitational field locking a piece of matter in its place of rest, the extended principle follows:

$M_{gravitational}g = M_{inertial}g\ (is\ itself\ equal\ to)\ M_{drag}g$

This extension explains what inertia is, and was something Einstein could not fully explain outside of saying it had something to do with the energy content of the system. Here by explaining it as a phenomenon of drag, sufficiently throws away a purely local effect for the influence of gravity on the surrounding system acting upon it.

Edited by Dubbelosix, 26 March 2020 - 08:13 AM.

### #6 Dubbelosix

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Posted 26 March 2020 - 03:27 PM

Moving on, the idea now is to transpose the drag formula

$F_{drag} = \frac{1}{2}f\ \rho \frac{v^2}{R} \int dV - mg$

into the language of general relativity. To do this isn't as difficult as one might make it out to be. First of all, we must realize that the force as defined by Newton, is actually a pseudo force as defined through general relativity. Let's explain this with some detail...

The Christoffel symbol can be 'loosely' thought of as being analogous to a force in Newtons equations (where mass has been set to 1 to denote that it is a constant in this formulation):

$\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}$

Newtonian formulation of this acceleration is

$F = -\frac{\partial \phi}{\partial x}$

However as mentioned, the gravitational force is not actually a true definition of a force as we come to expect say, in the proposed fundamental fields of nature, which are inherently complex (when quantum gravity is not) and that requires quantization of field particles acting as mediators of the force (something which gravity is expected to use to form the unification theory in the opinions of many scientists). It's actually a crucial component of many theories, most notably string theory.

Gravity is a pseudo force and can be understood in the following (neat) and (concise and short) way:

$\frac{d^2x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\nu \lambda} \frac{dx^{\nu}}{d \tau} \frac{dx^{\lambda}}{d\tau} = 0$

where

$\Gamma^{\mu}_{\nu \lambda} = \frac{\partial x^{\mu}}{\partial \eta^{a}}\frac{\partial^2 \eta^a}{\partial x^{\nu}\partial x^{\lambda}}$

or more compactly

$\Gamma^{\mu}_{\nu \lambda} = J^{\mu}_{a} \partial_{\nu} J^{a}_{\lambda} = J^{\mu}_{a} \partial_{\lambda} J^{a}_{\nu} \equiv J^{\nu}_{a} J^{a}_{\nu \lambda}$

which represents a pseudo force for gravity which makes it in the same league as the Coriolis and the Centrifugal forces.

Now, in general relativity, there are a number of ways you can write the force, there is such a thing as the three force but its character leaves out essential dynamics which explains why curvature should even exist. In general, we wish to speak of the four force:

$\mathbf{F} = \frac{d \mathbf{p}}{dt}$

The bolded characters just means we deal with the four force and four momentum. But when transposing an equation into general relativity, we like to use the correct notation such as replacing any acceleration terms (ie $g$) for the Christoffel symbol $\Gamma$. It is not always defined like this, some times authors make the Christoffel symbol dimensionless... but in our case we will define it as the acceleration as it too defines the gravitational pseudo field. So let us have this post for now sink in... before we start modifying the drag equation.

Edited by Dubbelosix, 27 March 2020 - 06:25 AM.

### #7 Dubbelosix

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Posted 26 March 2020 - 04:30 PM

We defined the acceleration from general relativity under the Christoffel symbol as

$\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}$

we define the metric here as a velocity term, meaning we have to redefine or correct the dimensions found in the derivative on the RHS. One way would be to put in a factor of $c$, but alternatively, we may just define $g_{00}$ as having units of velocity, and of course, the rate of change of velocity to that of time, is acceleration, so we rewrite

$\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial t}$

Plugging this into the last term, which clearly contains the acceleration from the definition of weight in classical physics means that the half is not attached to one term as found in the NASA page, but is in fact intuitively attached to both terms,

$E_{drag} = F_{drag} \cdot \int dR$

$= \frac{1}{2} (f\ \rho v^2 \int dV - m \frac{\partial g_{00}}{\partial t} \cdot R)$

(note, the time time components areb usually reserved for the stress energy, so we explicitely solved for this)

Also in general relativity, the density multiplied by the factor of $v^2$ yields the stress energy tensor of the form $T_{00}$. We may also plug this in and define the drag energy:

$E_{drag} = F_{drag} \cdot \int dR$

$= \frac{1}{2}(f\ T_{00} \int dV - m\ \frac{\partial g_{00}}{\partial t} \cdot R)$

Now, there are other ways we can approach this, for instance, we can just plug in the Christoffel symbol, but if we do this, we can define the weight say, under the four force definition, allthewhile keeping the stress energy tensor in the first term on the RHS, but in this instance, we wane away from using the time time components for a better array for the matrices which define such four force. While I said the four force was

$\mathbf{F} = \frac{d \mathbf{p}}{dt}$

we would get a slightly different version of what might be written in your standard textbook for the four force, only because we have defined the units of the Christoffel symbol with that of acceleration and massively simplifies the standard notation giving:

$F_{\mu \nu} = g_{\mu \nu} \frac{dp}{dt} + m \Gamma_{\mu \nu}$

now, when the weight is redefined with the Christoffel symbol, we find it produces an extra term:

$E_{\mu \nu} = F_{\mu \nu} \cdot \int dR$

$= \frac{1}{2}(f\ T_{\mu \nu} \int dV - g_{\mu \nu} \frac{dp}{dt} + m \Gamma_{\mu \nu} \cdot R)$

These are just two preliminary ways we could have achieved a modified equation for the drag under general relativity.

Edited by Dubbelosix, 27 March 2020 - 07:24 AM.

### #8 Dubbelosix

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Posted 26 March 2020 - 04:39 PM

Since this last post has defined the drag formula under general relativity, we have reached our goal. Any further investigation into drag will be done so in a new post, hope you enjoyed this!

edited...

Or maybe not, because all the while we have assumed that we define weight under the four force definition. It may be that this assumption is wrong and I may need to investigate further what we might actually mean when it comes to weight itself. It is only that it seemed natural to do this, when we have a drag equation with one part being the drag force and the other the weight defined from the NASA files. For instance, a following post will concur that we may need to define weight more properly. I will look into this very soon.

Edited by Dubbelosix, 26 March 2020 - 04:44 PM.

### #9 Dubbelosix

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Posted 26 March 2020 - 04:54 PM

So with a quick read on stackechange, it has not made this all the more easier to grasp, but it seems I may have been on the right track, since they do have to use the four acceleration definition. Their final result however defines the force getting back the Newtonian limit as

'' $F=ma=\frac{GMm}{r^2}\frac{1}{\sqrt{1-\frac{2GM}{c^2r}}}$

As $c\rightarrow \infty$ we recover the Newtonian definition, but nobody bothers phrasing it in these terms.''

Mind you, they did start with the Schwarschild metric, which led to this final result to recover Newtons equations, the denominator has featured in many papers I have read and can have many uses. What is pertinent here though is that they recognise that the definitions must require the four acceleration, or in my approach, more directly the four force. It would be interesting to gather their definition into the drag formula since the denominator is for redshift analysis. At least in below, Lubos Motl does mention that weight is not obsolete, but as I have stated before, it is very ill-defined in Einstein's general relativity to the point that such questions on its nature are hardly ever spoke about

### #10 Dubbelosix

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Posted 26 March 2020 - 04:58 PM

On a different site, I find an answer more similar to my approach, it is said very simply that:

''In GR, the "weight" of an object is the magnitude of the object's 4-acceleration multiplied by its rest mass. The magnitude of the 4-acceleration of an object stationary on the Earth's surface is g; the magnitude of the 4-acceleration of an object stationary at an altitude above the Earth equal to the Earth's radius is g/4.

So in this sense, we did indeed take the objects four acceleration multiplied by its mass term, but the result was that I defined the acceleration with a Christoffel symbol as having those exact dimensions. After reading this link, I feel a bit more confident in this heuristic approach.

### #11 Dubbelosix

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Posted 26 March 2020 - 06:21 PM

I think weight can be simplified as volume/lambdamax.

That was how I got my Planck density as simply Planck temp

Four force weight isnt weight its mass

what is lambda max? A thermal wavelength, if it is, a volume divided by a wavelength is not a Planck density or Planck temperature.

### #12 Dubbelosix

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Posted 26 March 2020 - 07:01 PM

It's more purely mathematical when $\lambda$ is used to define entropy, these days we are more accustomed in physics to use an ordinary notation for the entropy of $S$. Either way, I still don't know how this let you reach any Planck definitions, let alone Planck units are way over-rated and poorly understood. If you want to clarify the ''work'' you may want to start a thread on it rather than discussing it here. It's normal rules not to hijack a thread unless your work is any way related to the work of the OP that is being discussed.

Edited by Dubbelosix, 27 March 2020 - 07:27 AM.

### #13 Dubbelosix

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Posted 26 March 2020 - 10:52 PM

Right anyway, back on topic. A question I raised to myself tonight was the interpretation of any inverse solution to the stress energy tensor in regards to the drag coefficient, ie. $T_{\mu \nu}$~$\frac{1}{f}$ and it seems a bit off to even find a true interpretation for the inverse because it would mean as the source field gets larger, the drag becomes smaller. What I did find in previous investigations was an inverse solution except for a small detail, the drag coefficient took on a negative value. I was quick to assume the following:

The cosmological constant is more like a impetus thrust rather than associated to strictly a vacuum energy associated to zero point fields:

$- \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = \frac{\Lambda c^2}{8 \pi G} = \rho_{vac} v^2$

Which relates the cosmological constant as a ''thrust.'' And the second replacement for the stress energy:

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

In such an interpretation, as the stress energy became larger, the drag on the universe became smaller, which would indicate an acceleration feature. How fast it accelerates is a matter of opinion, if it accelerates at all. This inverse solution will be the next thing I will investigate because I find the whole thing a bit of an ordeal. The hardest thing in physics, is not always the making of an equation, but how to interpret it.

Edited by Dubbelosix, 27 March 2020 - 02:04 AM.