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Q&A

Is $E=mc^2$ true for all frame of references?

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Which one is correct? $$E=mc^2$$ or $$E^2=(mc^2)^2+(pc)^2$$

I mostly seen $$E=mc^2$$ from my childhood, and when I was learning problem solving in relativistic mechanics I had seen $$E^2=(mc^2)^2+(pc)^2$$ I know than if a frame at rest then momentum is $0$ hence $$E=mc^2$$ is true. But if a frame is not at rest then $E =mc^2$ true for that frame?

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Define your terms. (2 comments)
Rest energy and momentum (1 comment)

1 answer

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I will use $m_0$ for rest mass in the answer and $m$ for relativistic mass. The main equation is,

$$E^2=(m_0c^2)^2+(pc)^2$$ Here momentum is relativistic hence $p=\gamma m_0 v$

$$=(m_0c^2)^2+(\gamma m_0 v c)^2$$ $$=m_0^2c^4(1+\gamma^2 \beta^2)$$ $$=m_0^2c^4(1+\frac{1}{1-\beta^2} \cdot \beta^2)$$ $$=m_0^2c^4(\frac{1}{1-\beta^2})$$ $$E=m_0c^2\gamma$$ $$=mc^2$$

So energy-(relativistic)mass relation is true for all observer. But here mass is relativistic in mass-energy relation.

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