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At what values of the ratio of the kinetic energy to rest energy can the velocity of a particle be calculated from the classical formula with the relative error less than epsilon=0.010? |
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Answer» Solution :Relativistically `(T)/(m_0c^2)=((1)/(sqrt(1-beta^2))-1)=1/2beta^2+3/8beta^4` So `beta_(rel)^2~~(2T)/(m_0c^2)-3/4(beta_(rel)^2)~~(2T)/(m_0c^2)-3/4((2T)/(m_0c^2))^2` Thus `-beta_(rel)=[(2T)/(m_0c^2)-3(T^2)/(m_0^2c^4)]^(1//2)=sqrt((2T)/(m_0c^2))(1-3/4(T)/(m_0c^2))` But CLASSICALLY, `beta_(CL)=sqrt((2T)/(m_0c^2))` so `(beta_(rel)-beta_(cl))/(beta_(cl))=3/4(T)/(m_0c^2)=epsilon` Hence if `(T)/(m_0c^2)lt4/3epsilon` the velocity `beta` is GIVEN by the classical formula with an error less than `epsilon`. |
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