A Model for the Propagation of Visible Light and Other Rays

The value of energy is a quantity that is dispersed or gained, regardless of time. Then the value of energy is independent from time.

For example: A candy bar will have a certain amount of energy within it. Say that, we also have a stick of dynamite with the same amount of energy as the candy bar. The energy contained within the candy bar will be dispersed slowly, with a long period of time, after consumption. But, the energy contained within the stick of dynamite will be dispersed rapidly, over a shorter period of time, after it is ignited. As one can clearly see from this example, time is not a factor for the quantity of energy contained within a system.

Now once more, we will derive the kinetic energy formula, hopefully from an approach that you will prefer.
Starting from this equation:
\(v=at\)
The equation above can be plotted on a Cartesian graph, as seen in the following image: https://drive.google.com/file/d/0B-vl7kjDEZ8HXzByUWtrSUd5TDA/view

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After multiplying the above equation by dt:
\([v=at]\cdot dt\)
Performing integration:
\(v\int\limits_0^{t_c}dt=a\int\limits_0^{t_c}t~dt\)
\(v_{max}t_c=\frac{1}{2}at^2_c\)
Where \(v_{max}t_c=s_{max}\)
\(s_{max}=\frac{1}{2}at^2_c\)
\([t_c=\frac{v_{max}}{a}]^2\)
\(t^2_c=\frac{v^2_{max}}{a^2}\)
After substituting the value of \(t^2_c\) into the equation for \(s_{max}\)
\(s_{max}=\frac{1}{2}a\frac{v^2_{max}}{a^2}\)
\(s_{max}=\frac{1}{2}\frac{v^2_{max}}{a}\)
After solving for a:
\(a=\frac{v^2_{max}}{2s_{max}}\)
After substituting the value of a into Newton's Second Law, F=ma:
\(F=m\frac{v^2_{max}}{2s_{max}}\)
\(F\cdot s_{max}=\frac{1}{2}mv^2_{max}\)
Where \(F\cdot s_{max}=E_k\)
Then:

\(E_k=\frac{1}{2}mv^2_{max}\)​

It is not a mere coincidence that we keep arriving to the same result for kinetic energy with all of these different approaches, by only using a precise language known as Mathematics. Are you really claiming that all of these proofs are wrong... and you are right? One has to be honest with the results, these proofs are not subjective.

As we keep stating, this kinetic energy proof is not part of the 77 pages within The Yaldon Particle Theory. Here will be a direct link to The Yaldon Particle Theory: https://drive.google.com/file/d/0B-vl7kjDEZ8HV2lZN1dId3JpcUU/preview

When one fully understands this model, then one will appreciate the work done in order to develop it. Please do not continue to derail the topic of this thread away from the subject matter contained within this book. Thank you.

 

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How can anything be "dispersed" or "gained", "regardless of time"? Dispersing and gaining are processes, and thus are per definition dependent on the passage of time. If the value of (net?) energy is independent of time, is cannot change, and one cannot talk about "dispersing" or "gaining".

Summary: take two objects with the same amount of energy. Look, they have the same amount of energy! Your example is a tautology, and contains no insight on the time-dependent nature of energy.

What does \(t\) stand for? Time, sure, but what time? Can I pick \(t=0\) to be tomorrow?

 

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I got an alert saying my complaint about this thread and yaldonTheory has been rejected ('sob').

And that I can point out the problems here if that's my bent. Well, bugger that.

But I will say this about potential energy and momentum: there is an equivalence relation since force is the first derivative of momentum and also the first derivative of potential. But this has been known for a long time, and does not imply that potential energy and momentum are the same thing physically.

I think the big Yal actually said force and potential are the same, or something. Obviously bollocks, the man is off his nut.

 

Let me assist you in pointing out the problems with the Yaldon 'arm waving conjecture'.

The problem is the yaldon 'wild ass conjecture' is wrong.

Bet you a nickel the yaldon 'team' consist of some guy living in his parents basement named Mr. Yaldon.

 

(FTY: There are two authors named by the text, "Menketh and Stephen Yalda", so it's a team of two.)

 

Well that's too bad! It means that I lost a nickel and there are 2 equally clueless people instead of just one.