Height of Cone Equals Acceleration as a Model for Gravity by Parker Matthew Davis Emmerson ©2009-2010

8/9/2019 Height of Cone Equals Acceleration as a Model for Gravity by Parker Matthew Davis Emmerson ©2009-2010

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Acceleration is Space-Time Distance and Time is theSpeed of Light - A Treatise on Metaphysics andPhysicsby Parker Matthew Davis Emmerson

The Second Derivative of the Height of the Cone Equals the Height of the Cone Itself

From, Lemma 1 and Lemma 6,

Lemma 1 The height of the cone can be caluclated in terms of r and q .

Proof.

q r = 2 p r - 2 p r 1

h = r 2 - r 1

2

q r = 2 p r - 2 p Hr^2 - h ^2 LSolving this equation we find that,

::h Ø -4 p r 2 q - r 2 q 2

2 p >, :h Ø4 p r 2 q - r 2 q 2

2 p >>Lemma 6

The initial radius can be calculated purely in terms of the angle q .Proof. From Lemma 1, the height of the cone has been solved in terms of the transformation. That expression for the heightdivided by the initial radius is set equal to the sine of b. Solving that equation yields an expression for b that includes r. Thisexpression for b is then set equal to the expression found from Lemma 5.

Sin @bD=h

r

=4 p r 2 q - r 2 q 2

r 2 p=

4 p r 2 q - r 2 q 2

4 p 2 r=

r H4 p - q Lq

4 p 2

b Ø ArcSin B4 p r q - r q 2

4 p 2 F= ArcSin BH4 p - q Lq

2 p FSolve B ArcSin B

4 p r q - r q 2

4 p 2 F== ArcSin BH4 p - q

Lq

2 p F, r F::r Ø

2 p H4 p - q Lq

H4 p - q Lq >>Thus, acceleration through of a distance through space - time is actually a distance through space - time.

Printed by Mathematica for Students



DBDB4 p Hr L2 q - Hr L2 q 2

2 p , qF, qF

-I4 p r 2 - 2 r 2 q M2

8 p I4 p r 2 q - r 2 q 2 M3 ê 2-

r 2

2 p 4 p r 2 q - r 2 q 2

DBDB4 p

2 p H4 p-q L q

H4 p-q L q

2

q -2 p H4 p-q L q

H4 p-q L q

2

q 2

2 p , qF, qF

-J 16 p 3

H4 p-q L 2-

4 p 2

4 p-q-

4 p 2 q

H4 p-q L 2 N2

8 p J16 p 3

4 p-q-

4 p 2 q

4 p-q N3 ê 2

+

32 p 3

H4 p-q L 3-

8 p 2

H4 p-q L 2-

8 p 2 q

H4 p-q L 3

4 p16 p 3

4 p-q-

4 p 2 q

4 p-q

Solve

B-J 16 p 3

H4 p-q L 2 -4 p 2

4 p-q-

4 p 2 q

H4 p-q L 2 N2

8 p J16 p 3

4 p-q-

4 p 2 q

4 p-q N3 ê 2

+

32 p 3

H4 p-q L 3 -8 p 2

H4 p-q L 2 -8 p 2 q

H4 p-q L 3

4 p 16 p 3

4 p-q-

4 p 2 q

4 p-q

==

4 p 2 p H4 p-q L q

H4 p-q L q

2

q -2 p H4 p-q L q

H4 p-q L q

2

q 2

2 p , qF

8<This proves Equality, because all parameters cancel out.

The height of the cone - the cone itself - is a model for acceleration and thus a description of gravity when q passes constantly

with "time" like a clock. The height of the cone equals the second derivative of the height of the cone and this is shown to be atrue equality.

Height of the Cone = Distance in Space - Time =4 p r 2 q - r 2 q 2

2 p.

Ifq

2 p= t, then as theta increases,

the height of the cone increases less per increment of increase in theta. In essence,the height decelerates as theta increases. The

absolute distance of the height accelerates as theta decreases.

Therefore, we are bold to inquire : Can the height be set equal to the second derivative of the height, and if so, whatposition of the variables does this signify in the cone? Can the first derivative of the innate velocity in the Lorentz transformationbe set equal to the height of the cone? Can the integral of the height with respect to time be set equal to the innate velocity? Andin all of this, since algebraically, it can be shown that r depends on theta, the initial radius being a function of theta in this system,but supposedly invariant based on the initial construction. How can theta change and not have r change? How do we account forthis in our use of calculus? I will address these questions.

2 Height of Cone Equals Acceleration as a Model for Gravity by Parker Matthew Davis Emmerson ©2009-2010.nb








Plot3D B-15 I4 p r 2 - 2 r 2 qM2 I8 p r q - 2 r q 2M2

32 p I4 p r 2 q - r 2 q 2M7 ê 2+

3 I4 p r 2 - 2 r 2 qM2 I8 p q - 2 q 2M16 p I4 p r 2 q - r 2 q 2 M5 ê 2

+

3 H8 p r - 4 r qL I4 p r 2 - 2 r 2 qM I8 p r q - 2 r q 2M4 p

I4 p r 2 q - r 2 q 2

M5 ê 2

-3 r 2 I8 p r q - 2 r q 2M2

8 p

I4 p r 2 q - r 2 q 2

M5 ê 2

-

H8 p r - 4 r qL2

4 p I4 p r 2 q - r 2 q 2M3 ê 2- H8 p - 4 qL I4 p r 2 - 2 r 2 qM

4 p I4 p r 2 q - r 2 q 2 M3 ê 2+

r 2 I8 p q - 2 q 2 M4 p I4 p r 2 q - r 2 q 2M3 ê 2

+

r I8 p r q - 2 r q 2Mp I4 p r 2 q - r 2 q 2M3 ê 2

-1

p 4 p r 2 q - r 2 q 2

, 8r, - 1, 1 <, 8q , - 2 p , 2 p <F

Height of Cone Equals Acceleration as a Model for Gravity by Parker Matthew Davis Emmerson ©2009-2010.nb 5






Solve B-J 16 p 3

H4 p-q L 2 -4 p 2

4 p-q-

4 p 2 q

H4 p-q L 2 N2

8 p J16 p 3

4 p-q-

4 p 2 q

4 p-q N3 ê 2+

32 p 3

H4 p-q L 3 -8 p 2

H4 p-q L 2 -8 p 2 q

H4 p-q L 3

4 p 16 p 3

4 p-q-

4 p 2 q

4 p-q

==

4 p 2 p H4 p-q L q

H4 p-q L q

2

q -2 p H4 p-q L q

H4 p-q L q

2

q 2

2 p , qF

8<Because space depends on time, acceleration and distance through space - time are the same thing. Now that I have

proven that distance through space - time is the same thing as acceleration when the angle theta passes constantly with time like aclock, I can be justified in setting the first derivative of phenomenal velocity (the innate velocity in the Lorentz transformation)equal to the height of the cone. I can also be justified in setting the integral of the height of the cone equal to the innate velocitywithin the Lorentz transformation.

16 p 3

4 p-q- 4 p 2 q

4 p-q

2 p ==

4 p 2 p

H4 p-q

Lq

H4 p-q L q

2

q -2 p

H4 p-q

Lq

H4 p-q L q

2

q 2

2 p

· 4 p

2 p H4 p-q L q

H4 p-q L q

2

q -2 p H4 p-q L q

H4 p-q L q

2

q 2

2 p „q = ·

16 p 3

4 p-q-

4 p 2 q

4 p-q

2 p „ q

· 16 p 3

4 p-q-

4 p 2 q

4 p-q

2 p „q

q

Solve B16 p 3

4 p-q-

4 p 2 q

4 p-q

2 p ã 1, qF

88<<

· 4 p Hr L2 q - Hr L2 q 2

2 p „q ==

r 2 H4 p - q Lq 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq

2 p F4 p H4 p - q Lq





RevolutionPlot3D Br 2 H4 p - q Lq 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq


, 8r, - 1, 1 <, 8q , - 2 p , 2 p <F

· 4 p Hr L2 q - Hr L2 q 2

2 p „ q =



=

DB4 p r 2 q - r 2 q 2

2 p , qF==

4 p r 2 - 2 r 2 q

4 p 4 p r 2 q - r 2 q 2

= Inst. Velocity

· 4 p Hr L2 q - Hr L

2 q 2

2 p „ q =

r2

H4 p - q Lq 4 p - q q3

ê2

- 2 p H4 p - q Lq + 8 p 2

ArcSin Bq


=

-I4 p r 2 - 2 r 2 qM I8 p r q - 2 r q 2M

8 p I4 p r 2 q - r 2 q 2M3 ê 2+

8 p r - 4 r q

4 p 4 p r 2 q - r 2 q 2

= Inst. Velocity

· · 4 p Hr L2 q - Hr L2 q 2

2 p „ q „ r

r r 2 H4 p - q Lq K4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin B q

2 p FO8 p H4 p - q Lq





RevolutionPlot3D Br r 2 H4 p - q Lq 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq


, 8r, - 1, 1 <, 8q , - 2 p , 2 p <F

Instantaneous Velocity HDerivative of the Height of the Cone L=Integral of Height of the coneH which is the integral of acceleration as being a distance in space time L=

r r 2 H4 p - q Lq K4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq


==

DB4 p r2

q - r2

q2

2 p, qF== 4 p r

2

- 2 r2

q

4 p 4 p r 2 q - r 2 q 2

,

because the height of the cone equals the second derivativeof the height of the cone.

Won' t Solve for Theta.

Solve B4 p r 2 - 2 r 2 q

4 p 4 p r 2 q - r 2 q 2

==

r r 2 H4 p - q Lq 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq

2 p F8 p

H4 p - q

Lq

, r F

::r Ø -4 J2 p H4 p - q Lq - q H4 p - q Lq N

q H- 4 p + q L 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq

2 p F>> Which can be shown through a couple of substitution examples and further visualized :





SphericalPlot3D B- J4 J2 p H4 p - q Lq - q H4 p - q Lq NNì q H- 4 p + q L 4 p - q q 3 ê 2 - 2 p 4 p - 2 p + p 2 - p 2 Sin @bD2

q + 8 p 2 ArcSin Bq

2 p F,

8q , - 2 p , 2 p

<,

8b , -p

ê2, p

ê2

<F

SphericalPlot3D B- J4 J2 p H4 p - q Lq - q H4 p - q Lq NNì q H- 4 p + q L 4 p - q 2 p + p 2 - p 2 Sin @bD2

3 ê 2

- 2 p 4 p - 2 p + p 2 - p 2 Sin @bD2 q +

8 p 2 ArcSin B2 p + p 2 - p 2 Sin @bD2

2 p F, 8q , - 2 p , 2 p <, 8b , - p ê2, p ê2<F









Instantaneous Velocity = Integral of Height of the cone with respect to thetaH which is the integral of acceleration as being a distance in space time L=

r r 2 H4 p - q Lq K4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq

2 p FO8 p

H4 p - q

Lq

==

DB4 p r 2 q - r 2 q 2

2 p, r, qF== - I4 p r 2 - 2 r 2 qM I8 p r q - 2 r q 2M

8 p I4 p r 2 q - r 2 q 2M3 ê 2+

8 p r - 4 r q

4 p 4 p r 2 q - r 2 q 2



==

-I4 p r 2 - 2 r 2 qM2

8 p I4 p r 2 q - r 2 q 2 M3 ê 2-

r 2

2 p 4 p r 2 q - r 2 q 2

· · 4 p Hr L2 q - Hr L

2 q 2

2 p„q „ r ==

r 2 H4 p - q Lq K4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq


ã

Instantaneous Velocity of the Height of the cone from derivatives with respect to r and q equals the integral of the height of the cone.

Solve Br 2 H4 p - q Lq 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq


==

-I4 p r 2 - 2 r 2 qM I8 p r q - 2 r q 2M

8 p I4 p r 2 q - r 2 q 2M3 ê 2+

8 p r - 4 r q

4 p 4 p r 2 q - r 2 q 2

, r F

::r Ø -

2 K2 p H4 p - q Lq - q H4 p - q Lq Oq H- 4 p + q L K4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin B q

2 p FO>>





RevolutionPlot3D B 2 J2 p H4 p - q Lq - q H4 p - q Lq Nq H- 4 p + q L 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq

2 p F, 8q , - 2 p , 2 p <F





SphericalPlot3D B2 2 p 4 p - 2 p + p 2 - p 2 Sin @bD2q - q H4 p - q Lq ì

q H- 4 p + q L 4 p - q 2 p + p 2 - p 2 Sin @bD23 ê 2

- 2 p H4 p - q Lq + 8 p 2 ArcSin Bq

2 p F,

8q , - 2 p , 2 p <, 8b , -p ê2, p ê2<F

Then we would state that the integral of the height of the cone equals the phenomenal velocity, because the height itself equals acceleration. We have shown that the second derivative of the height of the cone with respect to time equals the height of the cone itself. By analogy, the height of the cone is acceleration, thus its integral with respect to time is equal to phenomenalvelocity.





Solve B· 4 p

2 p H4 p-q L q

H4 p-q L q

2

q -2 p H4 p-q L q

H4 p-q L q

2

q 2

2 p „q ==

I,I

- 1.1294090667581471`*^18 q + 8.987551787368176`*^16 q 2 +

3.5481432270250993`*^18 Sin @bD2MMì - 12.566370614359172` q + q 2 + 39.47841760435743` Sin @bD2 , qF

99q Ø 2.99792 µ 10 8 ==Which tells us that time is the speed of light.

· · 4 p Hr L2 q - Hr L2 q 2

2 p „ q „ r ==

r 2

H4 p - q

Lq 4 p - q q 3 ê 2 - 2 p

H4 p - q

Lq + 8 p 2 ArcSin

Bq

2 p

F4 p H4 p - q Lq

=

I,I- 1.1294090667581471`*^18 q + 8.987551787368176`*^16 q 2 +

3.5481432270250993`*^18 Sin @bD2 MMì - 12.566370614359172` q + q 2 + 39.47841760435743` Sin @bD2

The above equation will only deliver solutions to beta.





Solve Br 2 H4 p - q Lq 4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin Bq


==

I,I- 1.1294090667581471`*^18 q +

8.987551787368176`*^16 q 2 + 3.5481432270250993`*^18 Sin

@b

D2

MMì - 12.566370614359172` q + q 2 + 39.47841760435743` Sin @bD2 , bF::b Ø - 1. ArcSin BKq - K1.5067 µ 10 98 q 2 - 3.59698 µ 10 97 q 3 +

2.86239 µ 10 96 q 4 - 7.59272 µ 10 94 q 5 - 1.59242 µ 10 79 q 6 + 4.22402 µ 10 77 q 7 +

2.21475 µ 10 61 q 8 - 5.87482 µ 10 59 q 9 - 4.15839 µ 10 83 q ArcSin B0.282095 q F2

+

6.61828 µ 10 82 q 2 ArcSin B0.282095 q F2

- 2.63333 µ 10 81 q 3 ArcSin B0.282095 q F2

+

1.84096 µ 10 65 q 4 ArcSin B0.282095 q F2

- 7.32494 µ 10 63 q 5 ArcSin B0.282095 q F2

+

2.86922 µ 10 68 ArcSin B0.282095 q F4

- 2.28325 µ 10 67 q ArcSin B0.282095 q F4

OOì K- K4.73344 µ 10 98 q 2 - 7.5335 µ 10 97 q 3 + 2.99748 µ 10 96 q 4 + 4.19107 µ 10 80 q 5 - 1.66758 µ 10 79 q 6 -

5.829 µ 10 62 q 7 + 2.31928 µ 10 61 q 8 - 1.3064 µ 10 84 q ArcSin B0.282095 q F2

+

1.0396 µ 10 83 q 2 ArcSin B0.282095 q F2

- 3.6339 µ 10 66 q 3 ArcSin B0.282095 q F2

+

2.89177 µ 10 65 q 4 ArcSin B0.282095 q F2

+ 9.01391 µ 10 68 ArcSin B0.282095 q F4

OOF>,

:b Ø ArcSin BKq - K1.5067 µ 10 98 q 2 - 3.59698 µ 10 97 q 3 + 2.86239 µ 10 96 q 4 -

7.59272 µ 10 94 q 5 - 1.59242 µ 10 79 q 6 + 4.22402 µ 10 77 q 7 + 2.21475 µ 10 61 q 8 -

5.87482 µ 10 59 q 9 - 4.15839 µ 10 83 q ArcSin B0.282095 q F2

+

6.61828 µ 10 82 q 2 ArcSin

B0.282095 q

F2

- 2.63333 µ 10 81 q 3 ArcSin

B0.282095 q

F2

+

1.84096 µ 10 65 q 4 ArcSin B0.282095 q F2

- 7.32494 µ 10 63 q 5 ArcSin B0.282095 q F2

+

2.86922 µ 10 68 ArcSin B0.282095 q F4

- 2.28325 µ 10 67 q ArcSin B0.282095 q F4

OOì K- K4.73344 µ 10 98 q 2 - 7.5335 µ 10 97 q 3 + 2.99748 µ 10 96 q 4 + 4.19107 µ 10 80 q 5 -

1.66758 µ 10 79 q 6 - 5.829 µ 10 62 q 7 + 2.31928 µ 10 61 q 8 - 1.3064 µ 10 84 q ArcSin B0.282095 q F2

+

1.0396 µ 10 83 q 2 ArcSin B0.282095 q F2

- 3.6339 µ 10 66 q 3 ArcSin B0.282095 q F2

+

2.89177 µ 10 65 q 4 ArcSin B0.282095 q F2

+ 9.01391 µ 10 68 ArcSin B0.282095 q F4

OOF>>If height of the cone equals acceleration and it is simply a distance through space - time, then the integral of the phenome-

nal velocity with respect to time will equal the height of the cone at certain solutions to the variables in the system.

b : = ArcSin BH4 p - q Lq

2 p F





‡ I,I- 1.1294090667581471`*^18 q +

8.987551787368176`*^16 q 2 + 3.5481432270250993`*^18 Sin @bD2 MMì - 12.566370614359172` q + q 2 + 39.47841760435743` Sin @bD2

„ q

q

The integral of the phenomenal velocity equals the integral of the height of the cone. Logically, therefore, the integral of theheight of the cone must be equal to the phenomenal velocity, and the integral of the phenomenal velocity may be set equal to theheight of the cone.

In[3]:= · 4 p Hr L2 q - Hr L2 q 2

2 p „ q

Out[3]=

r 2 H4 p - q Lq K4 p - q q 3 ê 2 - 2 p H4 p - q Lq + 8 p 2 ArcSin B q


· 4 p

2 p H4 p-q L q

H4 p-q L q

2

q -2 p H4 p-q L q

H4 p-q L q

2

q 2

2 p „q

q

In[5]:= Solve BI,I- 1.1294090667581471`*^18 q +

8.987551787368176`*^16 q 2 + 3.5481432270250993`*^18 Sin @bD2 MMì - 12.566370614359172` q + q 2 + 39.47841760435743` Sin @bD2 ã q , r F

Out[5]= 88<<In[6]:= Solve BI,I- 1.1294090667581471`*^18 q +

8.987551787368176`*^16 q 2 + 3.5481432270250993`*^18 Sin @bD2 MMì - 12.566370614359172` q + q 2 + 39.47841760435743` Sin @bD2

ã q , qFOut[6]= 99q Ø 2.99792 µ 10 8 ==



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Height of Cone Equals Acceleration as a Model for Gravity by Parker Matthew Davis Emmerson ©2009-2010