Answer to Question #112880 in Mechanics | Relativity for DEEPAK SINGH

Question #112880
Prove the relation E^2-P^2C^2=m•^2C^4 where P is the momentum.
1
Expert's answer
2020-05-03T17:19:10-0400
E=mc21v2c2E2=m2c41v2c2=m2c4(1+v2c21v2c2)=m2c4(1+v2c2γ2)E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}\\E^2=\frac{m^2c^4}{1-\frac{v^2}{c^2}}=m^2c^4\left(1+\frac{\frac{v^2}{c^2}}{1-\frac{v^2}{c^2}}\right)\\=m^2c^4\left(1+\frac{v^2}{c^2}\gamma^2\right)

E2=m2c4+m2c2γ2v2=m2c4+p2c2E^2=m^2c^4+m^2c^2\gamma^2v^2=m^2c^4+p^2c^2


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