How do photons split into electrons and positrons?

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DuckAmuck
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I'm confused about how photons are able to split into electrons and positrons. Learning about four dimensional vectors, but it's still not clear how this happens.
The photon will have a vector of P=(E,p), where p^2=E^2, so P^2=0, since photons don't have mass.
It must be that P = P1 + P2, where P1 and P2 are electron and positron.
P1 = (E1,p1), where E1^2 = p1^2 + m^2.
So (E,p) = (E1,p1) + (E2, p2) = (E1+E2,p1+p2)
So (E,p)^2 = 0 = m^2 + m^2 + E1*E2 - p1*p2
So is the product of p1 and p2 really equal to m^2 + m^2 + E1*E2?!
That would suggest that either p1 or p2 is greater than it's corresponding energy, which would make mass negative.
What am I missing?
 
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Photons don't have mass, but they DO have momentum! ;) This is sneaky, but it comes from Special Relativistic equations. In fact, this is measurable, if you shine a light on a surface, the light will actually apply a force on that surface! F=dp/dt

check out some of these formulas:
http://en.wikipedia.org/wiki/Photon#Physical_properties

The equation for relativistic momentum is
[URL]http://upload.wikimedia.org/math/d/2/d/d2dec44ba56c41a31b4d334b144b51d6.png[/URL]
where
[URL]http://upload.wikimedia.org/math/9/c/3/9c3f2777ac6cb5f4c9c1edc647c68311.png[/URL]

If v=c, such is the case for a photon, the gamma term goes to infinity, and when multiplied with zero, well, the result is not easily determinable, but it often equals a real number, such as is the case here with photons.
 
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You are missing that for pair production to occur, another particle (usually an atomic nucleus) must be nearby to contribute momentum. Pair production does not occur in vacuum.
 
Yup. Four-momentum is not conserved in photon->electron+positron. You need another body to pick up the difference.