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# 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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#### 2 comment threads

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

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

Rest energy and momentum
gmcgath‭ wrote 10 months ago:

I'm very rusty on this topic, so I'm offering a comment rather than an answer. Feel free to correct anything that looks dumb here.

In special relativity, there is no such thing as a "rest" frame of reference. All non-accelerated frames of reference have equal standing. But the classic Einstein equation refers to rest energy, i.e., the potential energy that can be obtained by converting mass. The more general form of the equation includes the energy of momentum (in a given frame of reference) as well, so it includes kinetic as well as potential energy. Kinetic energy obviously does depend on the reference frame. Both equations are correct; it depends on what you're trying to do. This page may be helpful: http://www.phys.ufl.edu/~acosta/phy2061/lectures/Relativity4.pdf

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