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Old 08-27-2011, 03:50 AM
Ian Moone Ian Moone is offline
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Join Date: Aug 2011
Posts: 108
Default Re: What is TIME exactly?

Ok so here it is,

Einsteins 21 equation proof for special relativity.



Diagram One above is a schematic of the M-M test. It was conducted on the basis that if an ether existed, the earth would be moving "through" it. Hence there would be a relative velocity between earth and the fluid of space.

It was reasoned that by splitting a beam of light (F) into two parts; sending one out and back in line with the direction of the earth's orbital path, (to mirror A) from Half silvered mirror (G) and glass plate (D); and recombining the two beams in the interferometer (E) one should be able to detect a shift in the phases of the two beams relative to one another.

This shift could accurately be predicted by knowing the velocity of light (c)
And the velocity (Ve) of Earth through orbital space. Their reasoning was as follows (refer diag. 1, diag. 2a, daig, 2b):

Assuming:

c2 = a2 + b2C = velocity of light = velocity from G to B by fixed extra-terrestrial observer
S = distance GA = GB
T1 = go-return time in-line (GA - AG)
T2 = go return time at right angles (GB-BG)
T = .5 t T2
V1= apparent velocity from g to B by earth observer.

Then the time (T1) is determined by:[s/(c-ve)] + [s/(c+ve))] = t1 which reduces to:

(Eq.1) 2sc/(c2 - ve2) = t1

Also, the time (t2) is determined by first solving for (v1) in terms of ( c ) and (Ve) using the Pythagorean Theorem (c2 = a2 + b2)…. Or, in this instance: (G to B)2 = (G to M)2 + (M to B)2

By substitution, c2 = ve2 + v12

Hence:

(Eq.2) v1= (c2 - ve2).5

Now, solving for the time (t) - which is the same over GM, GB, MB - of the GB trip by substituting s/t = v1 in (Eq.2) , one obtains:

(Eq.3) s/t = (c2 - ve2).5

rearranging:

(Eq.3) t = s/(c2 - ve2).5

Substituting: t = .5t2

Gives: t2/2=s/(c2 - ve2).5

Or:

(Eq.4) t2= 2s /(c2 - ve2).5

by comparing the ratio of the in-line go-return time (t1) to the right angle go-return time (t2) one obtains:

(Eq.5) t1/t2 =[2sc / (c2 - ve2).5 / 2s

which reduces to:

(Eq. 5.) t1/t2 = (1- ve2 / c2 ) - .5

Now then, if the light source is at rest with respect to the other, one sees:

(Eq 6.) ve = 0

Hence:

(Eq 7.) t1/t2 = 1/ (1 -0).5 = 1/1 = 1

Such a ratio as (Eq. 7) shows is exactly what each successive try of the linear M - M test has obtained…. (notice: Linear not angular!). Lorentz and Fitzgerald knew there had to be an ether; so they developed their well known transforms - an act which was in essence a way of saying, there has to be an ether…we'll adjust our observed results by a factor which will bring our hypothetical expectations and our test results into accord….
Their whole transform was based on the existence of ether space! Their transform, in essence said that length shortened, mass flattened, and time dilated as a body moved through the ether.

Einstein came along in 1905 saying the Mitchellson Morley test showed the velocity of light to be a universal constant to the observer. Seizing upon this and the Lorentz-Fitzgerald transforms, Einstein was able to formulate his Special Relativity which resulted in the now famous E = Mc2 …the derivation of which follows:

Starting with (Eq.5) t1/t2 = (1- ve2 / c2 ) - .5

The Lorentz-Fitzgerald transform factor for (Eq.5) becomes (1- ve2 / c2 ) - .5
(to bring t2= t1) giving t1/t2 an observed value of (1).

Assuming Lorentz and Fitzgerald's supposition to be correct one should look at mass-in-motion as the observer on the mass see's it versus mass-in-motion as the universal observer sees it,…

Let m1 = mass as it appears to the riding observer
Let v1 = velocity as detected by rider
Let m2 = mass as universal observer sees it
Let v2 = velocity as universal observer sees it
Then it follows (from Lorentz and Fitzgerald) that:

(Eq. 9) m1 v1 not = m2 v2

So - to equate the two products. Lorentz and Fitzgerald devised their transform factor (1- ve2 / c2 ) - .5 which would bring m1 v1 = m2 v2 to either observer,… yielding the following extension

(Eq. 10) m1s1/t1 Not = m2s2/t1

or,…

(Eq. 10) m1s1 Not = m2s2

then, by substitution of the transform factor s2 = s1(1- ve2 / c2 ) - .5(assuming time is reference) into (Eq. 10.) one obtains: m1s1 = m2s1(1- ve2 / c2 ) - .5
which reduces to:
(Eq. 11) m1 = m2 (1- ve2 / c2 ) - .5

To re evaluate this relative change in mass, one should investigate the expanded form of the transform factor (1- ve2 / c2 ) - .5 (which transforms t1=t2) It is of the general binomial type:

(Eq. 12) (1- b) -a

Hence it can be expressed as the sum of an infinite series:

(Eq. 13) 1 + ab = a(a+1)b2 /2! + a(a+1)(a+2)b3/3! + …etc

where b2 is less than 1

So - setting a = .5 and b = ve2 / c2

One obtains:

(Eq. 14) 1 + (ve2 / 2c2) + (3v4/8c4) + (5v6/16c6) + etc…

For low velocities in the order of .25c and less than the evaluation of (1- ve2 / c2 ) - .5
Is closely approximated by, the first two elements of (Eq. 14):

(Eq. 15) (1- ve2 / c2 ) - .5= 1+ve2 /2c2

so (Eq. 11) becomes:

(Eq. 16.) m2= m1(1+ ve2 / c2)…where ve less than .25c

developing further,… m2= m1 + m1 ve2 /2c2

(Eq. 17) m2 - m1 = .5 m1 ve2 /2c2

remembering energy (E) is represented by:

(Eq. 18) E = .5mv2…( where ve less than .25c)

One can substitute (Eq. 18) into (Eq. 17) giving…

(Eq. 19) m2 - m1 = E/c2…(assuming ve = v)

Representing the change in mass (m2 - m1) by M gives:

(Eq. 20) M = E/ c2

Or, in the more familiar form using the general (m) for (M):

(Eq. 21) E = m c2

(Note, however, that (Eq. 14) should be used for the greatest accuracy - especially where ve is greater than .25c)


Next we have Einsteins Fine Structure Constant Alpha.

Alpha = E^2/hc

where

The amount of time dilation or gravitational red-shifting of the electron in its ground state compared to the masses of the electron and proton are defined by the universally measured constant called "alpha."

The relationship between the "virtual" and "actual" velocity, meaning distance to time, of the electron is "c."

The relationship of mass/energy to time, meaning gravity, is hidden within Planck's Constant "h."

The relationship of electrical charge "e" to time and gravity is found in the "alpha" definition.

Attempting to produce a complete system of universal science based only on the triumvirate of "measured constants" e, c, and h, has proven to be insufficient and incomplete.

It turns out that a minimum of four constants are needed to define all the properties of time and space.

So

Starting at Equation 5 where velocity of light C^ 2 is introduced

(Eq.5) t1/t2 = (1- ve2 / c2 ) - .5

and

Substituting Alpha for C

We get

(Eq 5) t1/t2 = (1-ve2/Alpha^2) - .5

Fleshing out Alpha squared

we get

(Eq 6) t1/t2 = (1-ve2/
E^2/hc x E^2/hc) - .5

Resolve from here.

Cheers!
__________________
Madness takes its toll - please have exact change handy!

The primary manifestation of Time is Change

Ee does NOT equal Em Cee Squared!

M = Δ T

Last edited by Ian Moone : 08-27-2011 at 03:55 AM.
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