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Old 07-June-2006, 06:09 AM
snowflakeuniverse snowflakeuniverse is offline
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Dimensional Analysis and Equivalence

This topic may or may not be presented to the class. It depends upon time. It describes the initial reason for my interest in theoretical physics.

Sometimes one only knows the units or dimensions of a relationship without knowing the theoretical basis for the relationship. Often by simply knowing the dimensions or units of a relationship it is possible to get close to the proper equation that does express the theoretical relationship. It will give a clue as to what to look for.

For example, lets say we forgot the formula that describes how far an object falls while experiencing a constant acceleration. We could derive the relationship using calculus, but we can also guess by using dimensional analysis.

We know that Acceleration has units of distance /time^2, D/T^2. We know the answer we want has units of distance and we are trying to determine how far the object falls per a measure of time T
We can guess the equation to be D = A T, but a dimensional check indicates that it does not work
D =? A T
D =? D/T^2 T = D/T is not correct but if the relationship was multiplied by T^2 dimensionally things would be correct.

D = A T^2. This is close, the correct form is D = 1/2 A T^2. But this gives a good idea of the value of dimensional analysis; it will establish the proper dimensional form of an equation that is only off by a constant factor, in this case 1/2.

Now lets look at the Equivalence of Gravitational and Inertial Forces


Drawing of two identical objects experiencing a 1 Newton or 1 pound force. One has the force applied directly by pushing on it, the other force is applied by Gravity using another mass near the mass.

Fg = Fi
1 Newton force applied by gravity = 1Newton force applied directly
g M1 M2/ D^2 = MA

Dimensionally, g is an experimentally derived term. It is not based upon a theoretical model. (Although those familiar with general relativity could disagree with this statement). The F = MA relationship can be directly produced without resorting to an experimentally derived term that carries units. Since the dimensional characteristics are known for inertial forces, and many of the terms for gravitational force are known, this raises the possibility to use dimensional analysis to help give some kind of clue as to what theoretical relationship would be that could establish the force of gravity with no constant that carries dimensional terms to make it work. So by using dimensional analysis it is hoped that the proper dimensional form a theoretical relationship will be found.

An investigation of the terms involved led me to think that measures of Distance and Time are fundamental, but mass is more complicated. It is usually defined by other terms that use units of distance and time. Solving the above relationship equating gravitational force to inertial force, without the ‘fudge factor g”, results in
M M /D^2 = M D/T^2
M = D^3/ T^2

If mass were described by dimensional measures of Distance cubed per time squared, then perhaps I could be on to something. The D^3 term made sense since mass occupies a volume but the T^2 term made no sense unless there was some kind of inherent acceleration or plane.

A dimensional check of the M = D^3/ T^2 led me to believe that I was on to something.

First off, the equations equaling inertial and gravitational forces are now equal with no gravitational constant that carries dimensional measures.

Fg = M M /D^2 = (D^3/ T^2) * (D^3/ T^2) / D^2 = D^4/T^4
Fi = MA = (D^3/ T^2) D/T^2 = D^4/T^4

This is not too surprising since the equation was solved for M to work

However this is

Energy = Force times distance
Energy = Mcc this is a much different form of equation derived directly from theoretical relationships.

Fi *D = MAD = D^5/T^4

E = mcc = D^5/T^4 The same!!


Then when I had the epiphany that “if nothing changed there would be no Time”, and allowed space itself change and found that the dimensional measure for space had units of

D^3 = kT^2 (See original derivation of Ratios of Time formulas), which is dimensionally the same as M = D^3/ T^2. All that was different was a scalar number k, which would correspond directly to the amount of mass or a volume of spacetime.

I knew I was on to something. It matched the dimensional form of the searched for theoretical relationship that would unit gravitational force with inertial force.

To me it made sense that the same dimensional relationship that defines mass should be the same dimensional relationship that defines spacetime.

I became real excited when I applied the relationship to check if celestial stability was preserved, as shown earlier.



Continued
Observational Confirmation