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What is the error when the classical exp...

What is the error when the classical expression for the kinetic energy is substituted for the relativistic expression? Calculate for `u_(1) = 0.1c`, for `u_(2) = 0. 9c` and for `u_(3) = 0. 99c`.

Text Solution

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The correct Answer is:
0.8%; 69%; 92%.
The relative error is
`epsi=(K_("rel")-K_("el"))/(K_("rel"))=1-(m_(0)u^(2))/(2(mc^(2)-m_(0)c^(2)))`
Denoting `beta = u//c` and noting that `m_(0)//m = sqrt(1-beta^(2))`, we obtain
`e=1-(beta^(2)sqrt(1-beta^(2)))/(2(1-sqrt(1-beta^(2))))=1-(beta^(2)sqrt(1-beta^(2))(1+sqrt(1-beta^(2))))/(2(1-1+beta^(2)))=`
`=1/2(1+beta^(2)-sqrt(1-beta^(2)))`
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