DimaMazin Posted July 23, 2014 Share Posted July 23, 2014 factor of time=Ek / pv Can we use the factor anywhere at future? Link to comment Share on other sites More sharing options...
Bignose Posted July 23, 2014 Share Posted July 23, 2014 I'm going to use it now! Some 4 hours into the future from your post! I have no idea what you mean by factor of time, nor what the variables you chose [math]\frac{E_k}{pv}[/math] are supposed to mean. So, apart from my very facetious use of them above, your question is essentially meaningless without context. Link to comment Share on other sites More sharing options...
John Cuthber Posted July 23, 2014 Share Posted July 23, 2014 I could use it as a random number generator to choose my lottery numbers (if I ever played the lottery) Link to comment Share on other sites More sharing options...
swansont Posted July 23, 2014 Share Posted July 23, 2014 factor of time=Ek / pv Can we use the factor anywhere at future? Classically this value is 1/2. Constant in time. It does not have units if time, or of anything. 1 Link to comment Share on other sites More sharing options...
DimaMazin Posted July 24, 2014 Author Share Posted July 24, 2014 Classically this value is 1/2. Constant in time. It does not have units if time, or of anything. I have mistaken there. It should be then: ft = pv / Ek - 1 It is just like 1/gamma and doesn't need units, but is correct for light. time of observing object = time of observer * factor of time Link to comment Share on other sites More sharing options...
swansont Posted July 24, 2014 Share Posted July 24, 2014 I have mistaken there. It should be then: ft = pv / Ek - 1 It is just like 1/gamma and doesn't need units, but is correct for light. time of observing object = time of observer * factor of time How did you derive it? Link to comment Share on other sites More sharing options...
DimaMazin Posted July 24, 2014 Author Share Posted July 24, 2014 How did you derive it? Sorry, I did just lottery derivation, but maybe real derivation exists. -1 Link to comment Share on other sites More sharing options...
swansont Posted July 24, 2014 Share Posted July 24, 2014 Lottery derivation? Meaning you just picked it? Link to comment Share on other sites More sharing options...
Strange Posted July 24, 2014 Share Posted July 24, 2014 And lost. Link to comment Share on other sites More sharing options...
DimaMazin Posted July 25, 2014 Author Share Posted July 25, 2014 Lottery derivation? Meaning you just picked it? I was looking for communication of time with relation of energy and momentum therefore has found the formula.The formula is simple therefore it can be picked. And lost. Is the formula wrong? -1 Link to comment Share on other sites More sharing options...
Strange Posted July 25, 2014 Share Posted July 25, 2014 Is the formula wrong? Who knows. If you can't provide any justification for making it up, and you have no supporting evidence, then probably it is wrong. Or at best meaningless. Link to comment Share on other sites More sharing options...
swansont Posted July 25, 2014 Share Posted July 25, 2014 ft = 1 is simpler. But just as wrong. Link to comment Share on other sites More sharing options...
DimaMazin Posted July 26, 2014 Author Share Posted July 26, 2014 Who knows. If you can't provide any justification for making it up, and you have no supporting evidence, then probably it is wrong. Or at best meaningless. Let's check.For axemple v=0.8c gamma=5/3 then 1/gamma = 3/5 3/5= (5/3)*o.64c2m/[(2/3)mc2] - 1 3/5 = 3/5 My equation is completely right. ft = 1 is simpler. But just as wrong. Please use relativistic equations for definition of energy and of momentum. Link to comment Share on other sites More sharing options...
Klaynos Posted July 26, 2014 Share Posted July 26, 2014 Perhaps u should start with explaining what you are trying to do? Are you just trying to define 1/gamma? Link to comment Share on other sites More sharing options...
Strange Posted July 26, 2014 Share Posted July 26, 2014 Let's check.For axemple v=0.8c gamma=5/3 then 1/gamma = 3/5 3/5= (5/3)*o.64c2m/[(2/3)mc2] - 1 3/5 = 3/5 My equation is completely right. Please use relativistic equations for definition of energy and of momentum. I have no idea what you did there because I am not really sure what the symbols in your equation mean (I could guess some of them) and I don't know where the numbers in your example come from. So: 1. Why not explain what you have done 2. Showing that a single example works (if it does) is not a proof that the equation is correct. 3. If you can do it for a single value, then you it should be easy to show a general proof (substitute the variables, rather than numbers). 4. Even if you show the equation is equivalent to the Lorentz transform (if it is) doesn't make it meaningful. Link to comment Share on other sites More sharing options...
swansont Posted July 26, 2014 Share Posted July 26, 2014 My point was that being simple doesn't make it right. Showing that it works to get the right answer in one case is still ad-hoc. You need to derive it. Link to comment Share on other sites More sharing options...
DimaMazin Posted July 27, 2014 Author Share Posted July 27, 2014 And so I should prove that: gamma=Ek /(pv-Ek) Ek /(pv-Ek) = (gamma-1)mc2/[gamma*mv2-(gamma-1)mc2] = (gamma-1)c2/[gamma*v2-(gamma-1)c2]= =(gamma*c2-c2)/(v2gamma - gamma*c2+c2) gamma=c/(c2-v2)1/2 Ek /(pv-Ek) =[c3/(c2-v2)1/2-c2] / [ (v2c-c3)/(c2-v2)1/2 +c2]=[c2-c(c2-v2)1/2] / [v2-c2+c(c2-v2)1/2]= ={[(c2-v2)1/2-c]*c} / [c2-v2-c(c2-v2)1/2] = [(c2-v2)1/2 -c]c / [(c2-v2)1/2{(c2-v2)1/2-c}] = c/(c2-v2)1/2 Perhaps u should start with explaining what you are trying to do? Are you just trying to define 1/gamma? You can see in the equation a strange energy works Estrange=pv It works for speed and against increase of gamma.Maybe strange energy will be useful at future. Link to comment Share on other sites More sharing options...
Klaynos Posted July 27, 2014 Share Posted July 27, 2014 So, that doesn't really explain what you're trying to do. Link to comment Share on other sites More sharing options...
DimaMazin Posted July 31, 2014 Author Share Posted July 31, 2014 (edited) So, that doesn't really explain what you're trying to do. You can use known energy , it doesn't mean another energy doesn't exist. I already proved the equation. I have found unknown energy. The energy interacts with space. Edited July 31, 2014 by DimaMazin Link to comment Share on other sites More sharing options...
DimaMazin Posted September 20, 2014 Author Share Posted September 20, 2014 (edited) space energy=pv - Ek space energy=mc(c-(c2-v2)1/2) Edited September 20, 2014 by DimaMazin Link to comment Share on other sites More sharing options...
andrewcellini Posted September 20, 2014 Share Posted September 20, 2014 (edited) this all seems completely arbitrary. how did you come to these equations and terms and what do they mean? how are you approaching the problem you are trying to solve? you really aren't explaining anything by restating equations and coming up with new ones. Edited September 20, 2014 by andrewcellini Link to comment Share on other sites More sharing options...
DimaMazin Posted September 21, 2014 Author Share Posted September 21, 2014 (edited) this all seems completely arbitrary. how did you come to these equations and terms and what do they mean? how are you approaching the problem you are trying to solve? you really aren't explaining anything by restating equations and coming up with new ones. My equations have derivation,if you know algebra you can check them.With units you can define kind of physical phenomenon. Edited September 21, 2014 by DimaMazin Link to comment Share on other sites More sharing options...
Klaynos Posted September 21, 2014 Share Posted September 21, 2014 this all seems completely arbitrary. how did you come to these equations and terms and what do they mean? how are you approaching the problem you are trying to solve? you really aren't explaining anything by restating equations and coming up with new ones. I quite agree. The last post didn't help either. Some explanation is required, not just restating the same meaningless comments. Link to comment Share on other sites More sharing options...
DimaMazin Posted September 21, 2014 Author Share Posted September 21, 2014 (edited) I quite agree. The last post didn't help either. Some explanation is required, not just restating the same meaningless comments. Can you explain what does cause the phenomenons which are connected with gamma factor?And so you have only gamma=1/(1-v2/c2)1/2 Speed is a phenomenon,but any phenomenon can't create another phenomenon without interaction. What does interact with speed(v) for creation the phenomenons?There is only "c". But massive body can't have "c". Massive body has energies relative to us. Interaction of kinetic energy and of space energy causes the phenomenons. My equations show it. Edited September 21, 2014 by DimaMazin Link to comment Share on other sites More sharing options...
swansont Posted September 21, 2014 Share Posted September 21, 2014 What is "space energy"? Link to comment Share on other sites More sharing options...
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