Is $E=mc^2$ true for all frame of references?
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?
1 answer
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.
0 comment threads
2 comment threads