Einstein's potential energy equation

What happened to the 1/2 in the transformation of classical potential energy equation to Einstein's potential energy equation. Is it dropped because giving off all rest energy would require annhilation of a particle pair thereof 1/2mv2+1/2mv2=1mv2 gets rid of the half?
 
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krab

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Re: question

Originally posted by einsteinian77
What happened to the 1/2 in the transformation of classical potential energy equation to Einstein's potential energy equation. Is it dropped because giving off all rest energy would require annhilation of a particle pair thereof 1/2mv2+1/2mv2=1mv2 gets rid of the half?
What do you mean by "Einstein's potential energy equation"? I don't understand the question. Do you mean to compare mc^2 with (1/2)mv^2? Apples and oranges?
 
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Potential energy doesn't equal 1/2mv^2(that's kinetic energy) it equals mgh

I think you are asking about the energy E = mc^2 equation relating to classical systems.

First E = mc^2 is only for stationary objects. The actual equation is E = mc^2/[1-((v^2)/(c^2))]^1/2(you see now why when discussing it in general people assume v = 0)

Einstein's equation deals with the total energy of an object at any speed. and the derivation can be found in most Modern Physics books( a convineince for me because I don't remember it off the top of my head)
 

mathman

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kinetic energy from Einstein

E=mc2/(1-(v/c)2)1/2
where m is rest mass.
Expand Lorentz term in power series:
(1-(v/c)2)-1/2=1+(v/c)2/2+...
Net result:
E=mc2+mv2/2+...
The first 2 terms are the energy due to rest mass and the kinetic energy.
 
whats so difficult about my question krab all i was asking was how come there is no half in the rest energy equation.
 

krab

Science Advisor
896
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Well, then it's been competently answered by Mathman and VBPhysics.
 

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