The drag coefficient and drag force has been vital instruments in my own approach to the unification of gravity and cosmology with hopeful progress in quantum mechanics under semi-classical approaches. The drag coefficient was to refresh our memory directly proportional to the drag force:

[math]f = \frac{2F}{\rho v^2\ A} = \frac{A_b}{A_f} \frac{B}{R^2_e}[/math]

and the drag force itself is

[math]F = \frac{1}{2} \rho v^2\ f\ A[/math]

Now, since temperature and pressure are fundamentally related to gravity under the unification of Friedmann's equations obtained from Einstein's field equations, we should explore the physics of the drag force and its dimensionless proportionality [math]f[/math] with it. Certain equations from previous work will be important for such progress, a recent one is that the cosmological constant is more like a impetus thrust rather than associated to strictly a vacuum energy associated to zero point fields:

[math]- \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 [/math]

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

[math]- \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} [/math]

And of course, if we want to entertain the drag force relationship with thermodynamics, we also need to understand it within how the temperature relates to classical gravity:

[math]F = 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}[/math]

[math]N = \frac{A}{\ell^2} = \frac{Ac^3}{\hbar G}T = \frac{\hbar a}{2 \pi c\ k_B}[/math]

So, to find the drag force, we must find it as an application to all the physics underlying the equation above: Doing that is easy, we just plug its definition in so that in a loose way, all the terms are equal to each other

[math]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}[/math]

[math]N = \frac{A}{\ell^2} = \frac{Ac^3}{\hbar G}[/math]

In which again, the temperature is

[math]T = \frac{\hbar a}{2 \pi c\ k_B}[/math]

This will take us into ''new territory'' but first we have to justify the terms in such a way that it is consistent. One immediate consequence is the mass flow of a collection of particles, in which we collect the identity of a heat transfer which fortuitously it was already investigated in my paper yet to be properly updated and properly proof read:

An elementary form of the continuity equation for mass in hydrodynamics is (which is strictly for dynamics involving the cross section only, just as found in the entropic gravity relationships):

[math]\rho_1 v_1 \cdot A_1 = \rho_2 v_2 \cdot A_2[/math]

Such that a mass flow rate is given as

[math]\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A[/math]

with [math]\dot{V}[/math] is the volume flow rate and [math]\mathbf{j}[/math] is the mass flux.

This equation though is only true for a surface distribution of the flow. Only dimensionless equations are physically important in physics, which is why Bernoulli's equation is important for fluid mechanics for cosmological purposes -

[math](\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + \frac{P}{\rho})E = C_{vac}[/math]

And we will work on this later. But first of all, we can also identify for the work, a dynamic pressure [math]q = \frac{1}{2} \rho v^2[/math]. The force equation derived will present this quantity:

[math]F = \frac{A_b B}{Re^2_L} \cdot q\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}[/math]

(where we treat time as a true operator and observable in contradiction to quantum mechanics as an avenue to why the relativity theories and those of quantum mechanics seem at odds with each other).

This means it related the fluid pressure weighted by the back area with the dynamic pressure as proportional to the Bejan number and inversely related to the Reynolds number as related to the stress energy:

[math]P = \frac{F}{A_b} = \frac{B}{Re^2_L} \cdot q\ = \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00}[/math]

While there is so much to explore, the last obvious relationship that popped up to me was the mass flow identification which relates to the back area:

[math]F = \int\ \frac{1}{2}\frac{A_b B}{Re^2_L} \cdot \rho v\ = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00}[/math]

In which we refresh our memory on the following relationships -

[math]\dot{m} = \rho \cdot \dot{V} = \rho v \cdot A = \mathbf{j} \cdot A[/math]

If we just concentrate on the second equality of the equation, you will notice my derivations lead to:

[math]F = \frac{A_bB}{Re^2_L} \cdot q = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho v \cdot A_b = \int\ \frac{1}{2}\frac{B}{Re^2_L} \cdot \rho \dot{V} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b[/math]

The standard definition of the mass flow follows the limit where a velocity term has been set to unity for simplicity

[math]\dot{m} = \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t} = \frac{dm}{dt}[/math]

[math]F = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m} = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}[/math]

The total pressure is defined as

[math]P + q = P_0[/math]

In such a case, [math]P[/math] is defined as the static pressure. Recall also that [math]\mathbf{j}[/math] is the flux of [math]q[/math] and is the dynamic component of the pressure. Retrieving the dynamic component, by subtracting the static component from both sides gives:

[math]q = P_0 – P[/math]

Let's just cover very quickly the dimensions in which we have a set a velocity term to unity for [math]v = c = 1[/math] :

[math]F \cdot c \cdot t = mc^2[/math]

Since mass cannot move at light speed, we now deduct that [math]v<c[/math] for the first terms:

[math]F \cdot v \cdot t = mc^2[/math]

Which is already an established equation in relativity. Plugging in the terms is pretty simple, it just yields for the mass flow:

[math]F \cdot v = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m}c^2 = c^2 \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \lim_{\Delta t \rightarrow 0} \frac{\Delta m}{\Delta t}[/math]

With this in mind, we now have an argument for the drag under the entropic gravity case, in which the flow of mass will be translated as a heat flow or flux of heat and has already been stated, its form satisfied the cross section:

[math]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}[/math]

[math]N = \frac{A}{\ell^2} = \frac{Ac^3}{\hbar G}[/math]

[math]T = \frac{\hbar a}{2 \pi c\ k_B}[/math]

From here we can do loads of ''new'' stuff - we can rewrite the mass with the derivative directly allowing us to define the mass flux. We can even divide off area terms for the fluid mechanics in which we find the specific ratio for back and front areas as defined from the dimensionless coefficient:

[math]f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{R^2_e}[/math]

From it we find the first area term related to the front area:

[math]F = \frac{1}{2} \rho v^2\ f\ A_f = 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}[/math]

Which means the pressure is found through a simple division of the front area giving us also the important ratio from fluid mechanics:

[math]P = \frac{1}{2} \rho v^2\ f = \rho_A \frac{2 \pi c\ k_BT}{\hbar} = \rho_A \frac{4 \pi c}{\hbar} \frac{E}{N} = \rho_A \frac{4 \pi c^2}{\hbar} \frac{M}{N} = 4 \pi \frac{A_b}{A_f}\ \frac{ GMm}{r^4} = \rho_A\ \frac{GM}{r^2}[/math]

With [math]\rho_A[/math] being the area density term. Notice also, the configuration of the dimensionless particle number can also be expressed using the same physics we have been applying:

[math]N = \frac{A_b}{A_f}\frac{1}{\ell^2} = \frac{A_b}{A_f}\frac{c^3}{\hbar G}[/math]

Now let's implicate the mass flux. To do that we go back to the force equation and take the derivative with respect to time :

[math]\dot{F} = \frac{1}{2} \rho v^2\ f\ A_f = \dot{m}\frac{2 \pi c\ k_BT}{\hbar} = \dot{m}\frac{4 \pi c}{\hbar} \frac{E}{N} = \dot{m}\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \dot{m}\frac{A_b}{A_f}\frac{4 \pi GM}{r^2} = \dot{m}\frac{GM}{r^2}[/math]

The acceleration term arises from the equation, but we will come to that in the next post.

Force over time, is just how the force changes over time - but it can viewed a number of different ways, for instance, the impulse is

[math]I = F \Delta t[/math]

Solving for the force we get

[math]F = \frac{I}{(t_1 - t_2)}[/math]

and so it's derivative looks like

[math]\dot{F} = \frac{\dot{I}}{(t_1 - t_2)} = \frac{d^2I}{d(t^2_1 - t^2_2)}[/math]

Alternatively we can view it as being related to the relativistic expression [math]Fvt[/math] by derivation:

[math]F \cdot v \cdot vt = F \cdot v \cdot vt = mc^2 \cdot v[/math]

Dividing through by [math]v^2[/math] gives an inverse Doppler shift [math](\frac{v^2}{c^2})^{-1}[/math] as:

[math]F \cdot t = F \cdot t = (\frac{c^2}{v^2})\ m \cdot v[/math]

And twice the derivative yields:

[math]\dot{F} = (\frac{c^2}{v^2})\ \dot{m} \cdot \frac{dv}{dt} = (\frac{v^2}{c^2})^{-1}\ \dot{m}a [/math]

As far as I know, this equation is unique.

**Edited by Dubbelosix, 23 June 2019 - 10:55 AM.**