GORDON HERMA Posted January 15, 2006 Author Share Posted January 15, 2006 Well i got my presentation all worded, im sure it still lacks some more preceptive explanations. This is all i can come up with at the moment though. Ill put it in Quotes... Mass, and the force that bind it together are bound at the hip. You can not have one without the other. Einstein does not account for the force binding the mass together in a state of rest in his equation E=MC^2. If mass can not pass the speed of light, then there is always a portion that does not get “converted“. I think the reason for this is because there is a binding force already causing some energy within it at a state of rest. The more mass in the equation, the more energy that is “converted”. The more mass you have, the smaller the amount of force not used to sustain/bind the mass. In conclusion there is a binding force that is causing energy within the mass in a state of rest. This is being mistaken as converted mass in the original equation E=MC^2. Since that Energy cannot be converted because its already Energy , there is an amount of mass that should not be calculated in the totality or “E” equals “MC” squared. It would seem to me that you cant effect one without effecting the other or leave one out without leaving the other. The energy is there in the totality, not because it was converted, but because it was already energy. This is like a word with a silent “E” inside of Einstein’s equation. Here is the method… M=Mass A= Actual Energy E= Energy C= The constant of the speed of light O= “Phase/Oscillation” (I guess I am viewing it like light from the sun in an analogy, If you take a portion of light you have x-rays, visible light, ect all in that portion. Just at different wavelengths. I guess I am viewing mass (light) the same ways because of permeability/phase (wavelength) R= Resisted Binding Force/Mass (The amount that could not be converted because of phase/oscillation vs. strong force/energy being used in a state of rest.) The method and equation are as follows.. mc^2=e mc=o e-o=r rc^2=a a-r = True (mc^2 -mc)c^2 - (mc^2-mc) = True or…… (e-o)c^2 - (e-o) = True or……. a-(e-o) = True or ……. a-r = True First we must determine the oscillation or “phase” of the mass being used so we multiply the mass ”M” by the speed of light “C” because of its properties associated with relativity. “M*C=O” Secondly we must equalize both the amount being subtracted and the amount that is being subtracted from. So we do the equation “E”= MC^2 for both amounts. Then we subtract the oscillation or “phase” from the converted mass “E” This gives them both an oscillation or “phase” to be used as the same baseline. Third we convert the amount being subtracted from by multiplying the new amount by the speed of light squared “(E-O)C^2 = A“ Fourth, we subtract the first amount from the newly converted amount to give us the amount “True”, so that we are able to find the percentage. (E-O)C^2 - (E-O) = True Fifth we find the percentage that should be taken out of the equation by Multiplying “True” by 100, then dividing it by “A”. We then subtract that amount from 100% to give us the correct percentage. With this we can calculate the amount that is truly converted in the equation “E=MC^2” by subtracting that percentage from “E” Do.... (True*100)/A = % Then... (E*%)/100 = # Then... E-# = Einsteins Missing Resistance 8) Let me know what you think Gordon Herman.... Link to comment Share on other sites More sharing options...
insane_alien Posted January 15, 2006 Share Posted January 15, 2006 Maybe you should consult swansont or one of the other phyisicists from now on. I'm a chemist at heart and i won't be on much for a couple of days(Going to my cousins in england and they only have 18.8kbps dial-up shudder.). hope you find some answers. Link to comment Share on other sites More sharing options...
GORDON HERMA Posted January 15, 2006 Author Share Posted January 15, 2006 Thank you again alien, have a good journey. Gordon... Link to comment Share on other sites More sharing options...
swansont Posted January 15, 2006 Share Posted January 15, 2006 Einstein does not account for the force binding the mass together in a state of rest in his equation E=MC^2. Yes, he does. The mass of a bound system is less than that of a free system, by that exact relationship. Link to comment Share on other sites More sharing options...
GORDON HERMA Posted January 15, 2006 Author Share Posted January 15, 2006 Thanks for the reply I will look up the definitions of those systems, so i understand what you mean a bit better. What could acount for the slideing percentage based on the mass in the equation. Just seem like there is some other factor that keeps a portion of mass from converting into energy(annihilation). I dont know what it is though. Gordon.. Link to comment Share on other sites More sharing options...
GORDON HERMA Posted January 16, 2006 Author Share Posted January 16, 2006 I see what you mean and i understand the equations relationship, has this been used for mass in this way before? They did mention phase on and http://en.wikipedia.org/wiki/Closed_systembut still no mass relationship is mentioned on there. Can this be used as a resistance/mass relationship equation because of the ratio properties and the different states that phase that can give systems more or less permeability? I think alien mentioned there was not an equation used ,in an earlier post, regarding a mass/resistance relationship. The percentages definetly show a ratio between phase and mass. Gordon.. Link to comment Share on other sites More sharing options...
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