Q&A

# Post History

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1 answer  ·  posted 10mo ago by deleted user  ·  edited 8mo ago by deleted user

#2: Post edited by deleted user · 2022-02-21T01:42:09Z (8 months ago)
• 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 classical 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?
• 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: Initial revision by deleted user · 2021-12-10T07:58:06Z (10 months ago)
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 classical 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?