# Proving The Coherence Of The Drag Formula

13 replies to this topic

### #1 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 01 July 2019 - 06:20 PM

Summery

Three equations of interest to me derived coherently 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:

$Thrust >\ Drag$

or much greater than the drag

$Thrust >>\ Drag$

We also deducted the law of gravitational mass and inertial mass was also equivalent to the drag of the mass, extending the Einstein equivalence:

$m_d = m_g = m_i$

Instead of a negative mass, the thrust appears to be thermodynamically-related to a pressure difference as found from previous independent speculations from Bernoulli's equations when using a correction factor to Einstein's equivalence principle.

Extending the Investigation for the Compton Wavelength

We will use the master equation linking drag to the thermodynamic entropy:

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

And remind ourselves of the corrections we introduced, one by Lorentz (which included the dispersion of an acoutsic gravitational wave in the modified theory of mine) and the other I derived for the extension into three dimensions in which the realization was that the index of refraction had to be quantized, revealing that the principal quantum number plays the identical role:

$1 + \frac{A_b}{A_f} \frac{B}{Re^2_L} = \frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}$

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{1}{\rho v^2 A} \cdot \rho v^2 f A = -f$

And that

$f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

Moving forward, the pressure was defined from a Planck unit investigation which led to a derivation involving the wavelength:

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

Solving for the wavelength we have, and obtain the Doppler shift:

$P = \frac{\rho}{4 \pi \rho_A} (\frac{v^2}{c^2})\ f = \frac{\rho}{4 \pi \rho_A}\ (1 - \frac{1}{\gamma^2}) = \lambda^{-1}$

And the inverse of the force equation is:

$F^{-1} = \frac{2}{\rho v^2\ f\ A} = \frac{\hbar}{2 \pi mc}(\frac{1}{k_BT}) = \frac{r^2}{GMm} = \frac{r^2}{\hbar c}$

If either the force or the density increases, likewise the wavelength becomes larger! Since the force is proportionally inverse to the kinetic energy, This reserves a specific rule in physics in which short wavelegths carry more energy than longer wavelengths. Suppose we take one factor of $r^2$ for the Schwarzchild radius, we would have

$r^2_s = \frac{G^2M^2}{c^4}$

we obtain a simple equation

$F^{-1} = \frac{G}{c^4}(\frac{M_e}{M_P})(\frac{m_e}{m_P}) = \frac{G}{c^4}\ \frac{Gm^2}{\hbar c} = \alpha_G \ (\frac{G}{c^4})$

In which $\frac{G}{c^4}$ has units of the inverse of the upper limit of the gravitational force.

Edited by Dubbelosix, 01 July 2019 - 06:27 PM.

### #2 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 02 July 2019 - 03:51 AM

In base units, the SI base length can be associated to the unit of a metre - similarly, the Planck length as a base to the force is expressable as

$\frac{F}{F_P} = \frac{(\frac{M}{M_P})(\frac{m}{m_P})}{(\frac{r}{\ell_P})^2}$

This demonstrates further coherence:

$F = \frac{G}{c^4}\frac{(\frac{M_e}{M_P})(\frac{m_e}{m_P})}{(\frac{r}{\ell_P})^2} = \frac{G}{c^4}\ \frac{(\frac{Gm^2}{\hbar c})}{(\frac{r}{\ell_P})^2} = \alpha_G \ \frac{(\frac{G}{c^4})}{(\frac{r}{\ell_P})^2}$

This demonstrates further coherence in the formula's.

### #3 ralfcis

ralfcis

Creating

• Members
• 1121 posts

Posted 02 July 2019 - 08:37 AM

I've lived long enough to know when people over-sell something using complex but meaningless explanations and terminology, they're always frauds.

### #4 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 02 July 2019 - 10:29 AM

I've lived long enough to know when people over-sell something using complex but meaningless explanations and terminology, they're always frauds.

lol... are you speaking about yourself? Or reflecting on the fact other people have spent literally years and years doing this stuff and you cannot? Either way troll, it takes one snap of my fingers and either you go or I do.....

### #5 ralfcis

ralfcis

Creating

• Members
• 1121 posts

Posted 02 July 2019 - 11:21 AM

Snap away, I'm pretty sure you'll find another home soon.

### #6 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 02 July 2019 - 11:28 AM

Snap away, I'm pretty sure you'll find another home soon.

Mods, you decide. Him or me. Take your pick.

### #7 ralfcis

ralfcis

Creating

• Members
• 1121 posts

Posted 02 July 2019 - 12:26 PM

006 I don't care if you stay or go but I don't understand how you're allowed to post outside the rubber room forums. Even so, one of the best features of this site is that you're allowed to do so.

### #8 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 02 July 2019 - 12:42 PM

You don't understand?

I started posting on forums when I was 16, mostly as a futuristic thinker... and my math was not entirely brilliant. I have learned so much more since, to the point objections now would show that someone like me cannot progress. I have told you plenty of times, if you stop thinking about your own past and put in some proper effort, you could do the same... but it's like speaking to a brick wall. Then you attack people who have learned, where you think you are already learned.

### #9 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 02 July 2019 - 12:42 PM

As I demonstrated further coherence:

$F = \frac{G}{c^4}\frac{(\frac{M_e}{M_P})(\frac{m_e}{m_P})}{(\frac{r}{\ell_P})^2} = \frac{G}{c^4}\ \frac{(\frac{Gm^2}{\hbar c})}{(\frac{r}{\ell_P})^2} = \alpha_G \ \frac{(\frac{G}{c^4})}{(\frac{r}{\ell_P})^2}$

And this further demonstrates further coherence in the formula's.

$F\ r^2 = \alpha_G \ (\frac{G}{c^4})\ \ell^2_P$

or

$\Delta E \Delta \ell \geq \alpha_G \ (\frac{G}{c^4})\ \ell^2_P$

$\frac{G}{c^2} = \frac{mass}{length}$

why? Let's take a look at dimensional analysis:

$F = \frac{m_P\ell_P}{t^2_P} = G (\frac{M^2_P}{\ell^2_P})$

or

$F_P = \frac{E_P}{\ell_P} = \frac{\hbar}{\ell_P t_P} = \frac{c^4}{G}$

$r_G = \frac{r_s}{2} = \frac{Gm}{c^2}$

Rearrange

$\frac{r_G}{m} = \frac{r_s}{2m} = \frac{G}{c^2}$

Edited by Dubbelosix, 30 July 2019 - 09:14 AM.

### #10 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 02 July 2019 - 07:15 PM

From Previous Work on the Equivalence Principle and the Relativity of Heat

To satisfy both the Ott and Einstein (et al.) transformations for both warmer and cooler bodies with respect to relativistic motion.

$\Delta \mathbf{S} = \frac{\Delta S}{k_B} = 2T \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2T}{\sqrt{1 - \frac{v^2}{c^2}}}$

In which the Ott covariant transformation is

$T = T_0 \sqrt{1 - \frac{v^2}{c^2}} = \gamma T$

And the suggestion by Einstein and Planck was

$T = \frac{2T_0}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{2T}{\gamma}$

par a factor of two that we have picked up through this derivation. The equation will be universal in understanding the relationship between the two transformation laws:

$T’ = \gamma T$

$T’ = \frac{T}{\gamma}$

There was not much dispute for many years until Ott showed that a relativistically moving body can appear hotter. I came to [eventually] realize from my own investigation, that both Einstein and Ott where right and in fact temperature is not alone subjected to the relativistic motion of systems and observers at rest, or the converse, but is also subject to redshift and blueshift which means it [also] depends on the direction of motion, so I concluded the two laws for cooler and hotter descriptions of relativistic bodies depend on additional relativistic premises and so both laws should hold in different circumstances.

I also notice these transformation laws linked like this together could may be implemented into the equation I developed earlier on for the power emitted by an accelerated black hole with a mass, charge and angular momentum:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3} \frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2} = \frac{2}{3} \frac{r_g}{mc}\frac{1}{(1 - \frac{R}{r})^2}(\frac{dp}{dt})^2$

If temperature is covariant, is entropy covariant? I don't think so, I think the laws have to follow the contraction of the volume of the system, just as we applied to the black hole to satisfy the Penrose theorem of rotated spheres in relativity. The temperature in such a case, has to transform in the same way as the four volume. It seems fair to do this, since entropy can be either dimensionless or have units of the $k_B$ (the Boltzmann constant) which is well known to be a constant.

We may see how the change in entropy becomes dependent on the only covariant object which can lead to a variation in the temperature due to Lorentz contractions:

$\Delta \mathbf{S} = \log_2(\frac{V_2}{V_1}) = \log_2(2) = 1$

Or as a more compact argument,

$\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})$

Which reminds me of a four-dimensional entropy equation which measured the temperature through a similar ratio. In such cases, a ratio like

$\frac{V_0}{V} = \frac{T_0}{T}$

is something I suspect can be formed. The equation I suggested not long ago for the transformation of the volume lead to a change in the entropy:

$\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})$

On my blog a while back I derived the following (with some adjustments for clarity here):

$\Delta S = Nk_B \log_2(\frac{T_2}{T_1})$

So it stands to reason the identity may hold to satisfy a covariant temperature following the same transformation laws as the four volume.

$\Delta S = Nk_B(\log_2(\frac{V_2}{V_1}) = \log_2 (\frac{T_2}{T_1}))$

The only two invariant quantities in this equation is entropy and the Boltzmann constant. The covariant temperature transforms in exactly the same way as the volume and should be approximated under the kinetic energy of the equipartition theorem.

Ok, the suggested equations I had given was:

$\Delta \mathbf{S} = 2k_BT \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2k_BT}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma^{-1}k_BT$

how do you transform between Ott's law and Einstein's? Well, you can see the above as already satisfying Einstein's version. In which case, to get Ott's law you just multiply through by $\gamma^{-2}$ giving:

$\Delta \mathbf{S} = 2k_BT \sqrt{1 - \frac{v^2}{c^2}} = \gamma^{-2} \frac{2k_BT}{\sqrt{1 - \frac{v^2}{c^2}}} = \gamma k_BT$

Really is as simple as that. From Lorentz contractions, the transverse and longitudinal direction in time is given as:

$t_l = \frac{2L \sqrt{1 - \frac{v^2}{c^2}}}{c} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2L}{c}\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = t_t$

Extending the Investigation for the Compton Wavelength

We will use the master equation linking drag to the thermodynamic entropy:

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

Melding the Coherence of Thermodynamics with Gravitational Drag

Moving forward, the pressure was defined from a Planck unit investigation which led to a derivation involving the wavelength:

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

Solving for the wavelength we have, and obtain the Doppler shift:

$P = \frac{\rho}{4 \pi \rho_A} (\frac{v^2}{c^2})\ f = \frac{\rho}{4 \pi \rho_A}\ (1 - \frac{1}{\gamma^2}) = \lambda^{-1}$

And the inverse of the force equation is:

$F^{-1} = \frac{2}{\rho v^2\ f\ A} = \frac{\hbar}{2 \pi mc}(\frac{1}{k_BT}) = \frac{r^2}{GMm} = \frac{r^2}{\hbar c}$

If either the force or the density increases, likewise the wavelength becomes larger! Since the force is proportionally inverse to the kinetic energy, This reserves a specific rule in physics in which short wavelegths carry more energy than longer wavelengths. The formula linking all of this together, would be through the Doppler shift, which we have already stated using the identity in previous posts:

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

So for example if

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

then

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

and its inverse form is

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

While reminding ourselves that the Ott covariant transformation with a squared notation is

$T = T_0 (1 - \frac{v^2}{c^2}) = \gamma^2 T$

And the suggestion by Einstein and Planck was

$T = \frac{2T_0}{(1 - \frac{v^2}{c^2})} = \frac{2T}{\gamma^2}$

... We'll finish here as these will become useful to bridge all the equations together.

### #11 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 03 July 2019 - 07:54 AM

As I demonstrated further coherence:

$F = \frac{G}{c^4}\frac{(\frac{M_e}{M_P})(\frac{m_e}{m_P})}{(\frac{r}{\ell_P})^2} = \frac{G}{c^4}\ \frac{(\frac{Gm^2}{\hbar c})}{(\frac{r}{\ell_P})^2} = \alpha_G \ \frac{(\frac{G}{c^4})}{(\frac{r}{\ell_P})^2}$

And this further demonstrates further coherence in the formula's.

$F\ r^2 = \alpha_G \ (\frac{G}{c^4})\ \ell^2_P$

or

$\Delta E \Delta \ell \geq \alpha_G \ (\frac{G}{c^4})\ \ell^2_P$

$\frac{G}{c^2} = \frac{mass}{length}$

why? Let's take a look at dimensional analysis:

$F = \frac{m_P\ell_P}{t^2_P} = G (\frac{M^2_P}{\ell^2_P})$

or

$F_P = \frac{E_P}{\ell_P} = \frac{\hbar}{\ell_P t_P} = \frac{c^4}{G}$

$r_G = \frac{r^2_s}{2} = \frac{Gm}{c^2}$

Rearrange

$\frac{r_G}{m} = \frac{r^2_s}{2m} = \frac{G}{c^2}$

The so-called ''uncertainty'' which is really about what we can extract from the system, not purely by statistics or even random processes, but a lack of information specifically, I should have added the uncertainty relationship has a proportionality to the force combined by extra uncertainties ~

$\Delta F \Delta \ell_P\ c\Delta t \propto \Delta E \Delta \ell \geq \alpha_G \ (\frac{G}{c^4})\ \ell^2_P$

### #12 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 03 July 2019 - 06:22 PM

The so-called ''uncertainty'' which is really about what we can extract from the system, not purely by statistics or even random processes, but a lack of information specifically, I should have added the uncertainty relationship has a proportionality to the force combined by extra uncertainties ~

$\Delta F \Delta \ell_P\ c\Delta t \propto \Delta E \Delta \ell \geq \alpha_G \ (\frac{G}{c^4})\ \ell^2_P$

Since I derived this idea independently that there can be three uncertainty relationships, I decided to look for literature to back this up. It does in fact happen in physics I was glad to learn:

Under a canonical representation it is given as

$\Delta p \Delta q \Delta r \geq (\frac{\hbar}{2})^{\frac{3}{2}}$

ref:

https://arxiv.org/pdf/1407.0083.pdf

When you see something raised to the power of three over two, is the same as writing it as

$\hbar \sqrt{\hbar}$

It should be further noted, that the length term can be interpreted as a spacetime uncertainty relationship under the guise of:

$\Delta F\ [\Delta \ell_P\, \Delta t] \propto [\Delta (mv), \Delta \ell] \geq \alpha_G \ (\frac{G}{c^5})\ \ell^2_P$

To obtain the energy from the canonical relationship requires that we take a dimensional look from Maupertius' principle for action, in which the canonical relationship is a momentum and position

$\hbar = \int p \cdot dq = mass \cdot velocity \cdot length$

such that a correcting factor is

$\Delta p \approx \frac{1}{\ell_P}(\frac{\hbar}{2})^{\frac{3}{2}}$

And so that a distribution of the velocity gives us:

$\Delta E \approx \Delta pv^2 \approx \frac{c}{\ell_P}(\frac{\hbar}{2})^{\frac{3}{2}}$

In which 1 Planck time per Planck length is only equal to a speed of light.

Edited by Dubbelosix, 30 July 2019 - 09:11 AM.

### #13 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 04 July 2019 - 08:09 AM

It is possible to saturate the bound with the correct dimensions:

$\hbar = \int p \cdot dq = mass \cdot velocity \cdot length$

such that a correcting factor is

$\Delta p \Delta q \geq \frac{1}{\ell_P}(\frac{\hbar}{2})^{\frac{3}{2}}$

$\Delta p \Delta q \geq \sqrt{2}\frac{1}{\ell_P}(\frac{\hbar}{2})^{\frac{3}{2}}$

In the paper referenced, apparently it is called the ''Schrodinger triple'' ~ but I do not like the use of these canonical relationships as they are not entirely clear so we will rewrite it as I have recommended:

$[\dot{p}, \dot{q}] \geq \sqrt{2}\ [\frac{1}{t_P}\frac{1}{\ell_P}]\ (\frac{\hbar}{2})^{\frac{3}{2}} = \sqrt{2}\ (\frac{<\frac{\hbar}{t_P}\frac{1}{\ell_P}> - <\frac{\hbar}{\ell_P}\frac{1}{t_P}>}{2} )^{\frac{3}{2}}$

under this notation, we notice the saturation point is in fact an inverse Planck space-time uncertainty as $(\Delta t \Delta \ell)^{-1} \geq [\frac{1}{t_P}\frac{1}{\ell_P}]$

Edited by Dubbelosix, 04 July 2019 - 08:09 AM.

### #14 Dubbelosix

Dubbelosix

Creating

• Members
• 3434 posts

Posted 04 July 2019 - 08:21 AM

$[\dot{p}, \dot{q}] \geq \sqrt{2}\ [\frac{1}{t_P}\frac{1}{\ell_P}]\ (\frac{\hbar}{2})^{\frac{3}{2}} = \sqrt{2}\ (\frac{<\frac{\hbar}{t_P}\frac{1}{\ell_P}> - <\frac{\hbar}{\ell_P}\frac{1}{t_P}>}{2} )^{\frac{3}{2}}$

If we denote the numerator as simply $a$ and denominator as $b$ then we have

$\sqrt{2}\ (\frac{<\frac{\hbar}{t_P}\frac{1}{\ell_P}> - <\frac{\hbar}{\ell_P}\frac{1}{t_P}>}{2} )^{\frac{3}{2}}$

is the same as

$\sqrt{2}(\frac{a}{2b}) \sqrt{\frac{a}{2b}}$