- #1

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

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- Thread starter goldust
- Start date

- #1

- 89

- 1

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

- #2

- 89

- 1

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

- #3

- 32,774

- 9,883

Done.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

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- #7

- 142

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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

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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

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

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