Yet another Special Relativity question - Tilted Forum Project Discussion Community

vinaur · 04-12-2005, 05:17 AM

Can anyone explain how Einstein arrived at mc^2/sqrt(1-v^2/c^2) and E_0/sqrt(1-v^2/c^2) in his book?? The page that I'm refering to is right here. Thanks a bunch.

the420star · 04-12-2005, 12:07 PM

I think this is one of his assumptions... for the first thing... it goes to talk about how the limit of that function is limited as V approaches the speed of light. you know his whole deal that it would require infinate energy to go C. It apears he adjusted the previeous equation to have a limit, rather than be boundless.... that make any sense... i am sure it doest but i thought i would give it a shot

vinaur · 04-12-2005, 04:31 PM

If you're talking about the sqrt(1-v^2/c^2) then that is not really assumed. It comes from the Lorentz transformation, and trusting that it is correct, the speed of light is the limit. This denominator also appears in the Lorentz contraction and time dilation as the limiting factor.

I think you thought that m*c^2/sqrt(1-v^2/c^2) was derived from E=m*c^2 (at least that's what I thought you did), but actually it's the other way around. I can sort of understand how to get E=m*c^2 from the two equations listed in my first post, but what I really want to know is how to arrive at those two equations basically just from Lorentz contraction and I guess some sort of Maxwell's equation or theory.

Also, could someone check if my reasoning is correct for E=m*c^2?? Let G=sqrt(1-v^2/c^2). So here's my reasoning: First I think that the "Kinetic energy" equation that Einstein gives as m*c^2/G is in fact more like total energy equation. So then when the particle of mass M absorbs E_0 energy via radiation without the increase of velocity, his total energy becomes (E_0+Mc^2)/G, which can also be written as (E_0/c^2+M)*c^2/G. This particle now has the same amount of energy as a particle of mass m+M. Now we can set E_0/c^2+M=M+m, which reduces to E_0/c^2=m. Now multiply both sides by c^2 and you get E_0=m*c^2. This is for absorption of energy, but I suppose that the process can go both ways. Is this correct???

04-12-2005, 05:17 AM	#1 (permalink)
vinaur Insane	Yet another Special Relativity question Can anyone explain how Einstein arrived at mc^2/sqrt(1-v^2/c^2) and E_0/sqrt(1-v^2/c^2) in his book?? The page that I'm refering to is right here. Thanks a bunch. __________________ If you multiply that by infinity and take it to the depths of forever, you will, perhaps, get just a glimpse of what I am talking about. --Meet Joe Black--

04-12-2005, 12:07 PM	#2 (permalink)
the420star Psycho Location: Cow Country, CT	I think this is one of his assumptions... for the first thing... it goes to talk about how the limit of that function is limited as V approaches the speed of light. you know his whole deal that it would require infinate energy to go C. It apears he adjusted the previeous equation to have a limit, rather than be boundless.... that make any sense... i am sure it doest but i thought i would give it a shot __________________ No, they arnt breasts, they are personalities, because its ok to like a girl for her personalities.

04-12-2005, 04:31 PM	#3 (permalink)
vinaur Insane	If you're talking about the sqrt(1-v^2/c^2) then that is not really assumed. It comes from the Lorentz transformation, and trusting that it is correct, the speed of light is the limit. This denominator also appears in the Lorentz contraction and time dilation as the limiting factor. I think you thought that mc^2/sqrt(1-v^2/c^2) was derived from E=mc^2 (at least that's what I thought you did), but actually it's the other way around. I can sort of understand how to get E=mc^2 from the two equations listed in my first post, but what I really want to know is how to arrive at those two equations basically just from Lorentz contraction and I guess some sort of Maxwell's equation or theory. Also, could someone check if my reasoning is correct for E=mc^2?? Let G=sqrt(1-v^2/c^2). So here's my reasoning: First I think that the "Kinetic energy" equation that Einstein gives as mc^2/G is in fact more like total energy equation. So then when the particle of mass M absorbs E_0 energy via radiation without the increase of velocity, his total energy becomes (E_0+Mc^2)/G, which can also be written as (E_0/c^2+M)c^2/G. This particle now has the same amount of energy as a particle of mass m+M. Now we can set E_0/c^2+M=M+m, which reduces to E_0/c^2=m. Now multiply both sides by c^2 and you get E_0=mc^2. This is for absorption of energy, but I suppose that the process can go both ways. Is this correct??? __________________ If you multiply that by infinity and take it to the depths of forever, you will, perhaps, get just a glimpse of what I am talking about. --Meet Joe Black*--