E=mc^2 Confusion (1 Viewer)

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"I beleive E=mc^2 works." But!
Photon (Packet of energy as described by Einstein in 1905, "Photoelectric effect") is mass less particle. So, the relation suggests that light should not have energy, though it has.

Is this wrong, if yes why?
 
It is nothing wrong. If a photon had a "nonzero resting mass", its "c speed traveling mass" would be infinite.

But it isn't.
 
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[itex]E={m_0}{c^2}[/itex] where [itex]m_0[/itex] is the rest mass. Since energy and mass are equivalent, when a particle moves it gains mass, even if its rest mass is zero, like a photon.

For an object in motion there is, where p is the momentum, [itex]{E^2}={{m_0}^2}{c^4}+{p^2}{c^2}[/itex], so the photon's energy (with it's zero rest mass) is [itex]E=pc[/itex]
 
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Jheriko said:
[itex]E={m_0}{c^2}[/itex] where [itex]m_0[/itex] is the rest mass.
No. [itex]E={m}{c^2}[/itex] where m is the relativistic mass.

You can understand that E is not [itex]{m_0}{c^2}[/itex], for a moving particle, from the correct equation you wrote then:

[itex]{E^2}={{m_0}^2}{c^4}+{p^2}{c^2}[/itex]

If it were [itex]E={m_0}{c^2}[/itex], substituting this value in the previous equation, you would have:

[itex] {{m_0}^2}{c^4} ={ {m_0}^2}{c^4} +{p^2}{c^2}[/itex]

that is: p = 0, which implies v = 0, so the particle is necessarily at rest.
 
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