- #1
DaveC426913
Gold Member
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In explaining to a curious member on a another forum what [tex]E=mc^2[/tex] means, I finally came to understand it better myself.
The member wanted to know why c is squared. What sense does it make to square a velocity? I stated comparing it to the kinetic energy formula [tex]K=1/2mv^2[/tex] A 1 ton car moving at 40mph has four times as much energy as a 1 ton car moving at 20mph. That's why you square the velocity to calc the energy.
Einstein's formula is the same conversion, except that c has been substituted for v (since the resultant photons are all moving at c).
It's as simple as that. Take a mass, figure out what velocity it is moving at, square the velocity and you get the amount of energy.So, assuming my thoughts are correct, what happened to the [tex]1/2[/tex]? Einstein's formula doesn't contain it.
(I suspect it has something to do with the car transferring its energy to another mass a la Newtons Law, so you're only counting half the energy? But I'm not sure.)
D'OH! I just saw the 'Helpful Posts' at the bottom. This exact question has already been answered...
The member wanted to know why c is squared. What sense does it make to square a velocity? I stated comparing it to the kinetic energy formula [tex]K=1/2mv^2[/tex] A 1 ton car moving at 40mph has four times as much energy as a 1 ton car moving at 20mph. That's why you square the velocity to calc the energy.
Einstein's formula is the same conversion, except that c has been substituted for v (since the resultant photons are all moving at c).
It's as simple as that. Take a mass, figure out what velocity it is moving at, square the velocity and you get the amount of energy.So, assuming my thoughts are correct, what happened to the [tex]1/2[/tex]? Einstein's formula doesn't contain it.
(I suspect it has something to do with the car transferring its energy to another mass a la Newtons Law, so you're only counting half the energy? But I'm not sure.)
D'OH! I just saw the 'Helpful Posts' at the bottom. This exact question has already been answered...
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