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

• goldust
This is a consequence of the full relativistic energy formula.In summary, E = mc^2 is a special case of the more general energy formula E^2 = (m_0c^2)^2 + (pc)^2. The massless particles have m_0 equal to zero but still have momentum. This formula only applies in the rest frame of the particle and does not account for momentum. For more information, please refer to the Relativity FAQ subforum.

#### goldust

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.

goldust said:
Oops, my bad. Should be in the Special & General Relativity section.
Done.

goldust said:
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.

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.

Nugatory said:
##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?

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