# Relativistic energy-momentum relation derivation

Relativistic energy momentum relation:

From Einstein mass energy relation

E = mc^{2 } (1)

Also from variation of mass with velocity relation

m = m_{0}/(1 – v^{2}/c^{2})^{1/2} (2)

Where m_{0} is the rest mass of the object

Put value of m in equation (1) and then square both sides, we get

E^{2}= m_{0}^{2}c^{4}/(1 – v^{2}/c^{2}) (3)

As momentum is given by

p = mv

Put equation (2) and square

p^{2} = m_{0}^{2}v^{2}/(1 – v^{2}/c^{2})^{ }

Multiply both sides by c^{2}

p^{2}c^{2} = m_{0}^{2}v^{2} c^{2}/(1 – v^{2}/c^{2}) (4)

Subtract equation (4) from (3) and solve, we get

E^{2} – p^{2}c^{2} = m_{0}^{2}c^{4}

Or E = (p^{2}c^{2 }+^{ }m_{0}^{2}c^{4})

This is Relativistic energy momentum relation

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