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Question Number 45646 by rahul 19 last updated on 15/Oct/18

According to relativistic theory, E^2 = p^2 c^2 +m_0 ^2 c^4 where m_0 is rest mass. For photon E= pc... (m_0 =0 for photon) For electron E=mc^2 ... Unlike photon ,Why p^2 c^2 is neglected in case of electron ?

$${According}\:{to}\:{relativistic}\:{theory}, \\ $$$${E}^{\mathrm{2}} =\:{p}^{\mathrm{2}} {c}^{\mathrm{2}} +{m}_{\mathrm{0}} ^{\mathrm{2}} {c}^{\mathrm{4}} \:{where}\:{m}_{\mathrm{0}} \:{is}\:{rest}\:{mass}. \\ $$$${For}\:{photon}\:{E}=\:{pc}... \\ $$$$\left({m}_{\mathrm{0}} =\mathrm{0}\:{for}\:{photon}\right) \\ $$$${For}\:{electron}\:{E}={mc}^{\mathrm{2}} ... \\ $$$${Unlike}\:{photon}\:,{Why}\:{p}^{\mathrm{2}} {c}^{\mathrm{2}} \:{is}\:{neglected}\:\: \\ $$$${in}\:{case}\:{of}\:{electron}\:? \\ $$

Commented by hassentimol last updated on 15/Oct/18

For photons, the mass is equal to 0. So, 0^2 ×c^2 =0, so this part is neglected for that reason. The mass of an electron, is not null indeed, however, as it is a very small number, the number become smaller while being squared thought it is multiplied by the square of the celerity. That′s why it is neglected and must be an aproximate answer and not an exact one.

$$ \\ $$$$ \\ $$$$\mathrm{For}\:\mathrm{photons},\:\mathrm{the}\:\mathrm{mass}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{0}. \\ $$$$\mathrm{So},\:\mathrm{0}^{\mathrm{2}} ×\mathrm{c}^{\mathrm{2}} =\mathrm{0},\:\mathrm{so}\:\mathrm{this}\:\mathrm{part}\:\mathrm{is}\:\mathrm{neglected}\:\mathrm{for}\:\mathrm{that}\:\mathrm{reason}. \\ $$$$ \\ $$$$\mathrm{The}\:\mathrm{mass}\:\mathrm{of}\:\mathrm{an}\:\mathrm{electron},\:\mathrm{is}\:\mathrm{not}\:\mathrm{null}\:\mathrm{indeed}, \\ $$$$\mathrm{however},\:\mathrm{as}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{very}\:\mathrm{small}\:\mathrm{number}, \\ $$$$\mathrm{the}\:\mathrm{number}\:\mathrm{become}\:\mathrm{smaller}\:\mathrm{while}\:\mathrm{being} \\ $$$$\mathrm{squared}\:\mathrm{thought}\:\mathrm{it}\:\mathrm{is}\:\mathrm{multiplied}\:\mathrm{by}\:\mathrm{the}\:\mathrm{square} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{celerity}. \\ $$$$\mathrm{That}'\mathrm{s}\:\mathrm{why}\:\mathrm{it}\:\mathrm{is}\:\mathrm{neglected}\:\mathrm{and}\:\mathrm{must}\:\mathrm{be}\:\mathrm{an} \\ $$$$\mathrm{aproximate}\:\mathrm{answer}\:\mathrm{and}\:\mathrm{not}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{one}. \\ $$

Commented by rahul 19 last updated on 15/Oct/18

thanks sir.

$${thanks}\:{sir}. \\ $$

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