Kinetic Energy at (Very) High Speeds
We have just defined Kinetic Energy as
K = (1/2) m v 2 But, Einstein's Theory of Relativity defines Kinetic Energy as
How can these be reconciled? They look so different!
Expand this relativistic Kinetic Energy equation using the binomial expansion,
( 1 + x )n = 1 + n x + [ n (n - 1) / 2! ] x2 + 
( 1 - x )n = 1 - n x + [ n (n - 1) / 2! ] x2 +
[1 - ( v / c )2 ]-1/2 = ?
That is, use x = ( v / c )2 and n = - 1/2
[1 - ( v / c )2 ]-1/2 = = 1 - ( - 1/2 ) ( v / c )2 + [ ( - 1/2)( - 3/2) / 2! ] [( v / c )2 ]2 +
[1 + ( - v / c ) ]-1/2 = 1 + 1/2 ( v / c )2 + [ ( 3/8) ( v / c )4 +
And for v << c, we have ( v / c )4 << ( v / c )2 so ( v / c )4 and all other higher-order terms may be ignored.
That is, for v << c,
[1 - ( v / c )2 ]-1/2 = 1 + 1/2 ( v / c ) 2 So our relativistic KE equation becomes
K = m c2 [ { 1 + 1/2 ( v / c ) 2 } - 1 ]
K = m c2 [ 1/2 ( v / c ) 2 ]
K = (1/2) m v2
just as it must be (ta-dah!)
Power Summary Return to ToC, Ch7, Work and Energy (c) Doug Davis, 2001; all rights reserved
