Kinetic Energy at (Very) High SpeedsWe 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

lookso different!Expand this relativistic Kinetic Energy equation using the

binomial expansion,( 1 + x ) ^{n}= 1 + n x + [ n (n - 1) / 2! ] x^{2}+( 1 - x )

^{n}= 1 - n x + [ n (n - 1) / 2! ] x^{2}+[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}andall 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 c

^{2}[ { 1 +^{1}/_{2}( v / c )^{2}} - 1 ]K = m c

^{2}[^{1}/_{2}( v / c )^{2}]K = (

^{1}/_{2}) m v^{2}just as it

must be(ta-dah!)

PowerSummaryReturn to ToC, Ch7, Work and Energy(c) Doug Davis, 2001; all rights reserved