https://quantizedbla...XETnsMAp8sq_Nfk

# My Favorite Equations I Have Derived Over The Years.

### #18

Posted 13 June 2019 - 08:05 PM

### #19

Posted 16 June 2019 - 11:17 AM

Taking the concepts further:

Rearranging the standard deviation, in which we treated space and time as operators,

[math]<\frac{2F}{\rho v^2 }>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,t} [/math]

[math]= \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>[/math]

for the force, gives us (keep in mind we remove the standard deviation not for the usual [x,p] notation but instead strictly for [x,t]:

[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]

The Pressure is obtained from further rearranging:

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

[math] = \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00} [/math]

Which removes the squared form of the metric.

**Edited by Dubbelosix, 16 June 2019 - 01:29 PM.**

### #20

Posted 16 June 2019 - 01:33 PM

Just for dimensional purposes, there could also be a related charge under the stress energy tensor:

[math]Q = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})\ T_{00} [/math]

These are all just different idea's we can investigated, but it would indicate that the principal number as a discrete variable could have applications to various charges, just as it had during my investigation into the Weinberg formula for mass charges.

### #21

Posted 16 June 2019 - 05:00 PM

Rearranging the standard deviation, in which we treated space and time as operators,

[math]<\frac{2F}{\rho v^2 }>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,t} [/math]

[math]= \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>[/math]

for the force, gives us (keep in mind we remove the standard deviation not for the usual [x,p] notation but instead strictly for [x,t]:

[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]

The Pressure is obtained from further rearranging:

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

[math] = \frac{1}{2n^2\lambda}(x + y + z - \sigma_{x,t})\ T_{00} [/math]

Which removes the dimensions from the metric. Let's move on:

An elementary form of the continuity equation for mass in hydrodynamics is:

[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 -

[math]\frac{v^2}{2c^2} + \frac{\Psi}{\psi} + \frac{P}{\rho} = C[/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]

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]

And we will finish there for the day.

**Edited by Dubbelosix, 16 June 2019 - 11:08 PM.**

### #22

Posted 16 June 2019 - 06:13 PM

I said that would be enough for one night, but I realized I should have said something about the mass flow which features in the article in my blogs. The standard definition of the mass flow follows the limit

[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]

### #23

Posted 16 June 2019 - 07:00 PM

When I get an idea... I just can't put the proverbial pen down.

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]

This term is pretty much identical to the equation which describes an expanding or contracting sphere under the mathematical relationships describing sonoluminscence, which in a previous blog post was derived as:

[math]\frac{1}{R^2}\frac{P - (P_0 - P(t))}{\rho}(1 - (\frac{T_0}{T})^3) = \frac{\ddot{R}}{R} + \frac{3}{2}(\frac{\dot{R}}{R})^2 + \frac{4 \nu \dot{R}}{V} + \frac{2S}{\rho V} + \frac{e^2}{6 \pi mV} \frac{\dddot{R}}{R} + \frac{1}{V}(\frac{e}{m})\frac{\partial \dot{U}}{\partial R}[/math]

with

[math]P_{\infty} = P_0 + P(t)[/math]

Just using the solution for the dynamic pressure it explains the force also defined under:

[math]F = \frac{A_bB}{Re^2_L} \cdot (P_0 - P)[/math]

With the dynamic pressure retrieved as:

[math]q = \frac{F}{A_b} = \frac{B}{Re^2_L} \cdot (P_0 - P)[/math]

### #24

Posted 16 June 2019 - 11:30 PM

A post to clarify the dimensions

The force of drag was associated to the back area, and from previous relationships we introduced from mainstream physics, we ended up being able to describe it through a series of equivalences, which ended up with a metric term (in operator notation) and a stress energy tensor.

[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 = \frac{1}{2n^2}(x + y + z - \sigma_{x,t})^2\ T_{00} [/math]

[math]\int F \cdot dx = \int F \cdot cdt = \int \frac{1}{2} \frac{B}{R^2_e} \cdot dmc^2[/math]

and so the mass flow ~ but in such a case, we must take the speed of light as a velocity term on the force where now we set [math]c=1[/math] again

[math] F \cdot v = \int \frac{1}{2} \frac{B}{R^2_e} \cdot \frac{dm}{dt} = \int \frac{1}{2} \frac{B}{R^2_e} \cdot \dot{m}[/math]

So a few different ways to look at the correcting factor.

**Edited by Dubbelosix, 16 June 2019 - 11:30 PM.**

### #25

Posted 16 June 2019 - 11:56 PM

If we recall what the dragging coefficient is first of all

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

What kind of systems does the drag coefficient apply to? It applies to all kinds of sizes, but in theory it could be applied to an electron since it has been found, not to be a pointlike particle as we have been told in literature for many years, but has instead been carefully measured to be rather a spherical distribution of charge. For spherical systems, the drag coefficient is 0.47.- since we could in theory say that most particles are approximately spherical it wouldn't be too difficult to see that this number would only differ slightly depending on the structure of certain particles - for instance, a proton may appear spherical, but it's distribution inside is not entirely spherical which may mean it variates slightly in principle for different particle systems.

Retrieving the ratio of back to front area's for particular terms in the master equation, also called the wet and back area's, we can get the pressure flow:

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

The first equality then has a strict relationship to the pressure flow related to the drag coefficient

[math]P = \frac{1}{2}\frac{A_b}{A_f}\frac{B}{Re^2_L} \cdot q = f \cdot q[/math]

where again, we defined the dynamic/total pressure as a simple rearranging of the mainstream formula:

[math]q = (P_0 - P)[/math]

### #26

Posted 17 June 2019 - 12:02 AM

But that was just the first term, we will notice the drag coefficient is proportional to other terms from the master equation:

[math]P = f \cdot q = \frac{1}{2}f \cdot \rho v = \frac{1}{2}f \cdot \mathbf{j} = \frac{1}{2n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00} [/math]

Or as twice the pressure flow related to simplified terms on the rest of the equalities:

[math]2P = f \cdot \rho v = f \cdot \mathbf{j} = \frac{1}{n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00} = \frac{\hat{s}^2}{n^2A_f}\ T_{00} [/math]

These are nice simple relationships, but we can study these drag relationships now under the stress energy tensor more clearly for the future.

**Edited by Dubbelosix, 17 June 2019 - 12:46 AM.**

### #27

Posted 17 June 2019 - 01:04 AM

[math]2P = f \cdot \rho v = f \cdot \mathbf{j} = \frac{1}{n^2A_f}(x + y + z - \sigma_{x,t})^2\ T_{00} = \frac{\hat{s}^2}{n^2A_f}\ T_{00} [/math]

Now, let's write the stress energy tensor under the relativistic form of

[math]T^{00} = g^{00}(\rho + P_0 - P) = (\rho_m + \frac{P_0 - P}{c^2})v^0v^0 = (\rho_m + \frac{q}{c^2})v^0v^0[/math]

where [math]\rho[/math] is an energy density and [math]\rho_m[/math] is the mass density and while reminding ourselves that [math]q[/math] is the total pressure.

[math]2P = f \cdot \rho v = f \cdot \mathbf{j} = \frac{g^{00}\hat{s}^2}{n^2A_f}\ (\rho + P_0 - P) = \frac{\hat{s}^2}{n^2A_f}\ (\rho_m + \frac{q}{c^2})v^0v^0 [/math]

**Edited by Dubbelosix, 17 June 2019 - 01:08 AM.**

### #28

Posted 17 June 2019 - 06:35 AM

An elementary form of the continuity equation for mass in hydrodynamics is:

[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]

Using those relationships, we were able to create a drag force:

[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}[/math]

[math]= \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \mathbf{j} \cdot A_b = \int\ \frac{1}{2}\frac{ B}{Re^2_L} \cdot \dot{m}[/math]

And mass flow applies to the drag force

[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 continuity equation has many applications - for this case, let's say we rewrite it for the pressure: For instance, from it we can divide through by the energy density to get,

[math]\frac{P_1}{\rho_1} \cdot A_1 v_1 = \frac{P_2}{\rho_2} \cdot A_2 v_2[/math]

This too is a continuity equation related to the Bernoulli principle, which is stated concisely in a dimensionless form as

[math]\frac{v^2}{2c^2} + \frac{P_1}{\rho_1} = \frac{v^2_2}{2c^2} + \frac{P_1}{\rho_1}[/math]

To solve the velocities we obtain:

[math]v_1 = A_2 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}[/math]

[math]v_2 = A_1 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}[/math]

Before we go into an application involving the Friedmann equation, which is derived directly from fluid mechanics, we require to first investigate 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, often seen as a correction term to Einstein's formula.

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 [math]\Delta p[/math] we then have

[math]\Delta p =\frac{v}{2c^2}E[/math]

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

[math]2\Delta p=\frac{v}{c^2}E[/math]

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

[math]p′ = mv−2\Delta p = (m − \frac{E}{c^2})v[/math]

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.Thought I wrote about the possibilities of using this to explain the cosmological constant, my original approach ended up being wrong, so I will try and fix that later... but first, these corrections will also apply to the solved velocities spoke about not long ago:

[math]v_1 = A_2 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}[/math]

[math]v_2 = A_1 \sqrt{\frac{2 \Delta p}{\rho(A^2_1 - A^2_2)}}[/math]

Later, the equation we seek to unify with the notion of the cosmological constant as the dark energy will be of the form

:[math]\Delta p v = \frac{v^2}{2c^2}E[/math]

(though my previous formula's on this subject was wrong and I stated this clear in my previous posts, but it is something I am heading towards fixing).

But for now, the definition

[math]2\Delta p = \frac{v}{c^2}E[/math]

can be found in the numerator inside of the radical on the right hand side... To solve for it is a bit of a headache, but we want to solve it for the velocity terms and its coefficients:

[math]2 \Delta p = v^2(A^2_1 \rho - A^2_2 \rho)[/math]

(will show at the end how to solve this).

These solutions are for the right-momentum energy lost from the system.

**Edited by Dubbelosix, 17 June 2019 - 07:22 AM.**

### #29

Posted 17 June 2019 - 06:37 AM

Of course, I am not finished here, but it takes so much time to write this stuff out, I am taking a few hours break.

### #30

Posted 17 June 2019 - 07:12 AM

I solved that last equation wrong, I have now fixed it, the dimensions were not adding up and I realized what I did wrong. Then i did it wrong again, this time I have the right formula... my algebra needs a good brush up.

**Edited by Dubbelosix, 17 June 2019 - 07:19 AM.**

### #31

Posted 17 June 2019 - 07:34 AM

ok, so something is still not adding up, my last equation keeps coming up with the wrong dimensions! So we have

[math]2 \Delta p = v^2(A^2_1 \rho - A^2_2 \rho)[/math]

has units of

[math]mass \times\ velocity = \frac{mass }{position^3} \cdot\ velocity^2 \times\ area^2 - \frac{mass }{position^3} \cdot\ velocity^2 \times\ area^2[/math]

Since area has dimensions of position to the power of two, we can rewrite this

[math]mass \times\ velocity = \frac{mass }{position} \cdot\ velocity^2 - \frac{mass }{position} \dot\ velocity^2[/math]

But the dimensions are not so far away, it looks like there is a correction coefficient meant to be on the left hand side, but will work it out soon, this is the best part of physics and I hope people are learning how to do this stuff at the same time.

**Edited by Dubbelosix, 17 June 2019 - 07:51 AM.**

### #32

Posted 17 June 2019 - 07:44 AM

It appears that what I am missing on the left hand side to match the right required one extra factor of

[math]\frac{velocity}{position}[/math]

since [math]\frac{1}{time} = \frac{velocity}{position}[/math]

So there is an inverse time component on the left hand side missing.

### #33

Posted 17 June 2019 - 07:44 AM

I wonder how this has dropped out so easily, will need to check the derivation a few times.

Edit: Just a quick look through, I am suspecting this has to do with a derivative replacing the difference operator.

**Edited by Dubbelosix, 17 June 2019 - 07:50 AM.**

### #34

Posted 17 June 2019 - 07:54 AM

So if it has to do with the derivative of the formula instead of the change operator, then the formula I am probably seeking for should probably take the form:

[math]2 \dot{p} = v^2(A^2_1 \rho - A^2_2 \rho) = \rho v^2(A^2_1 - A^2_2)[/math]

And this is why dimensional analysis is very important - don't just rely on algebra, always check the units!

**Edited by Dubbelosix, 17 June 2019 - 03:19 PM.**