Comment by Ron V on December 11, 2011 at 1:47pm

Let us not forget, the full equation is E = mc^2*g*

 

Where g = 1 / Sqrt(1 - v^2 / c^2)

where m is the mass, c is the speed of light, and v is the velocity of the object.

When v = 0, it is just E = m * c^2, which is called the rest mass energy because the object is at rest (v = 0).

Note that if v is small compared to c, then E - m*c^2 = 1/2 * m * v^2. The energy minus the rest mass energy is equal to the old familiar kinetic energy...but only in the approximation that v/c 1.

Source(s):

Standard Theory of Relativity

Taylor Series approximation 1/Sqrt(1+v^2/c^2) ~ 1 + 1/2 * v^2/c^2 if v/c 1 is used.

Comment by Dave Gibbs on December 11, 2011 at 4:26pm

Excellent "haiku" answer he gave at the end in response to that cookie cutter false dilemma question.

Comment by KaraC on December 11, 2011 at 4:59pm

Actually it is more fully expressed by

E² = (p²c²) + (m²c⁴)

though physicists usually set c=1 and use

E² = p² + m²

but you do have to take care over units if you use that.

Comment by KaraC on December 11, 2011 at 5:00pm

The numbers below are superscript but the formatting got lost :-)

Comment by Ron V on December 11, 2011 at 5:42pm

Thanks, KaraC

There's a decent disussion about this at

http://www.physicsforums.com/archive/index.php/t-4096.html

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