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

  • Thread starter goldust
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  • #1
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E = mc^2 results in c^2 = E / m, and division by 0 is undefined.
 

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  • #2
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Oops, my bad. Should be in the Special & General Relativity section.
 
  • #4
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.
 
  • #5
russ_watters
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E also equals FD: so how can force or distance ever be zero?
 
  • #6
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.
 
  • #7
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##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?
 
  • #8
ZapperZ
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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.
 
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  • #9
Dale
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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.
 

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