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From a little research done it seems as though they would create subatomic particles (maybe electrons?)If so, would photon collisions technically create matter? Is this a widely thought of theory?

fdafda12
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  • See http://physics.stackexchange.com/questions/159320/if-two-photons-collide-does-the-resulting-particle-have-zero-velocity/159326#159326 It is called Delbrück scattering and is a higher order effect in quantum electrodynamics. – Jon Jan 29 '15 at 14:16

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Take a look at this question. The OP stated that he wanted the energy of the photons to be so low pair production wasn't possible. I bolded these two words because, yes, it is possible for two colliding photons to create particles, in this case specifically electrons and their antiparticles, positrons (which are like positively charged electrons). Note that their charges are opposite, so charge is conserved. You could also create muons and antimuons, and so on.

You seem to be bothered by the fact that we just "created" matter, as electrons have a mass of roughly $511 \mathrm{keV}$, which is about $9.1 \times 10^{-31}$ kg. However, recall that while photons indeed have no rest mass (which makes them move at exactly the speed of light), they do indeed carry momentum. So, Einstein's formula $E=mc^2$ is nice, but in it's full form it should be $$ E^2=m_0^2c^4 + pc^2 $$ where $m_0$ is the rest mass and $p$ is the momentum.

So, photons carry energy. And from the equation you can see that mass is equal to energy, so they can be converted into each other. One electron-positron pair may just as well collide and produce 2 (or more) photons. This doesn't violate any fundamental physical principles like energy conservation, conservation of momentum etc. And if it's not allowed, it's bound to happen eventually, as Murphy's law states. Although that isn't exactly particle physics.

You could look into the Wiki article on pair production, Two-photon physics or check out the Feynman lectures; volume 3 eventually has something on this topic.

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John Doe
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  • I understand what you're saying but the equation for momentum is mv^2 and since protons do have no mass (or rest mass, but it shouldn't make a difference in this scenario) technically momentum should be equal to 0, giving them no energy even with the modified equation. I also understand that in a state of superconductivity, photons may act as though they have mass. Would this lead to their momentum? – fdafda12 Jan 29 '15 at 19:15
  • Actually for photons, p=hv/c. So the photon being mass less would make no difference...but why is the formula for photons different than momentum for everything else? Or does it end up being the same thing if the equation were really done out, and if something were going the speed of light then the mass would be irrelevant? – fdafda12 Jan 29 '15 at 19:26
  • Basically, the "common" equations for momentum ($mv^2$ etc.) are only useful for non-relativistic cases (that is, Newtonian physics). They are perfectly applicable there because the velocities are so low compared to c. However, as you move closer to c you start to all those weird relativistic effects like time dilation, mass getting bigger etc. However, this is quite a difficult topic to speak about in 600 cahracter answer. If you are interested in this topic, you should try an introductory quantum physics book or online course. I'm not the best teacher, unfortunately :) – John Doe Jan 30 '15 at 22:49