# E = mc^2, so how can anything have zero mass?

E = mc^2 results in c^2 = E / m, and division by 0 is undefined.

Oops, my bad. Should be in the Special & General Relativity section.

Dale
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Oops, my bad. Should be in the Special & General Relativity section.
Done.

Nugatory
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E = mc^2 results in c^2 = E / m, and division by 0 is undefined.

##E=mc^2## is a special case of the more general ##E^2=(m_0{c}^2)^2+(pc)^2##. The massless particles have ##m_0## equal to zero but ##p## non-zero.

russ_watters
Mentor
E also equals FD: so how can force or distance ever be zero?

E = mc^2 only applies in the special case where the particle in question is in its rest frame. Photons do not have a rest frame.

##E=mc^2## is a special case of the more general ##E^2=(m_0{c}^2)^2+(pc)^2##. The massless particles have ##m_0## equal to zero but ##p## non-zero.

The 'p' in that is momentum, correct?
Yet momentum is m * v, mass times velocity. I understand that photons have a velocity of c, but their mass...? If you use the argument that it isn't talking about rest mass, but total mass, then isn't it sort of circular logic?

ZapperZ
Staff Emeritus
E = mc^2 results in c^2 = E / m, and division by 0 is undefined.

https://www.physicsforums.com/forumdisplay.php?f=210 [Broken]

Zz.

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Dale
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2021 Award
Yet momentum is m * v, mass times velocity.
Not in general. Photons have momentum but 0 mass, and for massive objects momentum is unbounded as v approaches c.