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

[goldust]

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

 

[goldust]

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

 

[Dale]

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

Done.

 

[Nugatory]

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]

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

 

[Psychosmurf]

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.

 

[ModestyKing]

$E=mc^2$ 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]

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

Please read the Relativity FAQ subforum.

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

Zz.

 

[Dale]

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.