If two photons traveling in opposite directions along the same line collide, will the resulting particle have a velocity of zero relative to the rest of time space in the instant of the collision?
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1$\begingroup$ Photons are bosons, so they can't technically collide per se. $\endgroup$– Hritik NarayanCommented Jan 14, 2015 at 6:45
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6$\begingroup$ They can interact via a fermion loop. It's very very rare, but possible (and has even been experimentally measured I think). $\endgroup$– David ZCommented Jan 14, 2015 at 6:50
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4 Answers
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If two photons traveling in opposite directions along the same line collide will the resulting particle have a velocity of zero relative to the rest of time space in the instant of the collision?
Photons are quantum mechanical particles. In the microscopic dimensions where quantum mechanical particles interact there are Nature's rules that dominate these dimensions, though they are usually insignificant in macroscopic dimensions.
One of these rules is the Heisenberg uncertainty principle, HUP,: one cannot define the location of a particle and the momentum of a particle with accuracy better than:
where $\hbar =6.62606957(29)×10^{−34}$ Joule second a very small number that is why it is effectively zero in macroscopic dimensions.
Thus two photons even with the same energy will not collide at a point.
Going into the mathematics of it, photon-photon interactions are very very weak, since there is no first-order interaction between two photons, but they have to go through a particle loop. In addition, momentum conservation requires two particles out.
A Feynman diagram (box diagram) for photon-photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other
Feynman diagrams have one to one correspondence with calculable integrals that will give the probability for a given interaction.
A photon carries energy, two photons have an invariant mass. In their center of mass system, depending on the energy available from each, the output can be again two photons, or if there exists energy enough to generate massive particles, there will exist a quantum mechanical probability for the interaction. They are proposing high energy photon colliders, gamma gamma colliders.
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$\begingroup$ This is known as Delbrück scattering en.wikipedia.org/wiki/Delbr%C3%BCck_scattering and being a four vertexes effect is really small. $\endgroup$– JonCommented Jan 14, 2015 at 8:28
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$\begingroup$ you mean... that if the trajectory that led them towards each other was one in which these photons initiate their cycle in tight spherical formation around a vacuum, where every photon that forms the sphere travels outward in a circular trajectory away from each other at the same instant with equal momentum and where the collective trajectory of them all traces a vortex sphere, where each pair of photon meet head-on twice; 1)furthest from the initial photon sphere and 2) again at the surface of the photon sphere, and collision point 2 is high in energy matter will be created? $\endgroup$ Commented Jan 14, 2015 at 8:33
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$\begingroup$ Vortices and stuff belong to macroscopic physics. Quantum mechanical calculations are represented by Feynman diagrams and have nothing to do with vortices and "photon spheres". The photon is an elementary particle, i.e. a point particle , no spheres. Quantum mechanical rules tell us that given enough energy to be able to create paraticles, there will be a probability that at the scatering they will be created, which is what the gamma gamma colliders are aiming at studying. $\endgroup$– anna vCommented Jan 14, 2015 at 8:44
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$\begingroup$ Sorry Anna i did not intend to use the terms vortex and photon sphere to describe a specific phenomena but rather theoretical points along the trajectory of the photons im describing. i'll attempt to represent it graphically $\endgroup$ Commented Jan 14, 2015 at 9:01
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$\begingroup$ @JuanBautista, are imagining that the photons have well defined trajectories? $\endgroup$ Commented Jan 14, 2015 at 13:14
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Generally no, because velocity is not a conserved quantity. It is momentum that is conserved in all interactions. For photons, the magnitude of momentum is simply $$ p = \frac{E}{c} = \frac{h\nu}{c} = \frac{h}{\lambda}, $$ so photons with different energies/frequencies/wavelengths will have different momenta. If the total momentum is nonzero before the collision, it will be nonzero after.
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$\begingroup$ Interesting, so theoretically in a vacuum if the two photons part from a single and exclusive gravitational center by one same force in opposite directions along the elliptical trajectory of one of the poles of the dipole moment of a ring through which's center (the same as the gravitational center) only one photon fits we could expect that the photons would meet once in a low energy collision point at the outer rim of the ring and again at the high energy point of the center of the ring where only one could pass would be the exception to your "generally no" answer to my question? $\endgroup$ Commented Dec 25, 2015 at 22:00
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Note that the collision probability can be strongly enhanced in some nonlinear materials (such as a Kerr medium). As stated above, vacuum is a very weakly nonlinear material.
The resulting 'velocity' for the photon would be its momentum $\mathbf{k}$, the rule being that, if no loss occurs in the material, momentum and energy must be conserved: $\mathbf{k}_{1}+\mathbf{k}_{2} = \mathbf{k}'_{1}+\mathbf{k}'_{2}$. But these are vectorial quantities and $\mathbf{k}_{1}+\mathbf{k}_{2} = \mathbf{0}$ does not imply $\mathbf{k}'_{1} = \mathbf{0}$ and $\mathbf{k}'_{2} = \mathbf{0}$. Many solutions are possible, these are studied by nonlinear optics. Moreover, energy conservation puts further restrictions.
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Photos are bosons. They can occupy the same quantum states and the same place in space, so they can't actually "collide". If two photons traveling in opposite directions along the same line collide, they just pass right through each other.
I think you were probably thinking of pair production. It's a completely different process in which two energetic photons interact with nucleus and create a particle-antiparticle pair.
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$\begingroup$ Inaccurate, as the other answers show. $\endgroup$ Commented Jan 14, 2015 at 12:56
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1$\begingroup$ To clarify: the fact that two photons can occupy the same state doesn't stop them interacting. Maxwell's equations are linear, so in any regime where Maxwell's equations apply, photons will rarely interact. Nonetheless, that doesn't make it impossible. $\endgroup$ Commented Jan 14, 2015 at 14:25