Why is E=pc with Light's Zero Rest Mass?

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Grim Shady
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I was reading the derivation of this equation on hyperphysics.com and after searching all over the internet it seems that eveywhere seems to just accept this equation is valid for light, even though a rest mass of 0 would make p equal to 0.

Sorry if I'm being stupid or if this has been asked a million times before but I can't find an explanation anywhere as to why light having zero rest mass gives the equation E = pc, despite p being proportional to the (zero) rest mass of light.
 
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For any particle, massive or massless, the relativistic relation between mass, energy, and spatial momentum is given by

[tex]E^2 - \left( cp \right)^2 = \left( mc^2 \right)^2 .[/tex]

Using [itex]m = 0[/itex] in this relation results in [itex]E = cp[/itex].
 


But surely if p = m0v/sqrt(1-v2/c2), that would make p equal to 0?
 


Yeah sorry my mistake, but that still would mean that p is an undefined value, despite it being in the equation George posted.
 


But you can't just ignore the equation for momentum and state that light has a momentum, p, purely for convenience.

Unless momentum can be defined as something else, other than the equation I posted.
 


Grim Shady said:
But you can't just ignore the equation for momentum and state that light has a momentum, p, purely for convenience.

Unless momentum can be defined as something else, other than the equation I posted.


E^2 = p^2 + m_0^2

is the general formula

E = gamma m_0

applies only to massive particles

where is your lack of understanding?
 


It was in the definition of momentum itself, I've found the answer the answer anyway in the de Broglie wavelength equation, which is something I knew already was just having trouble remembering it. Cheers for the help people.
 


I think he's saying that in the equation E=pc, p=mv, so that if m=0, then p=zero, then E=0.

Unless maybe saying m=0 means that relativistic mass is zero, not rest mass?