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As it turns out, your intuition might be in the right path. As was recently noted and published by physicists of the caliber of Arkani Hamed and Freddy Cachazo, you can avoid the trouble of doing your Feynmann integrals off-shell, by replacing the vertices with on-shell particles of complex momenta and coordinates. There is a conjecture that this replacement ...


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Ever since Newton and the use of mathematics in physics, physics can be defined as a discipline where nature is modeled by mathematics. One should have clear in mind what nature means and what mathematics is. Nature we know by measurements and observations. Mathematics is a self consistent discipline with axioms, theorems and statements having absolute ...


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To understand this one shall take in quantum-mechanical approximation method namely perturbation theory into account. In perturbation theory, systems can go through intermediate virtual states which often have energies different from that of the initial and final states. This is because of time energy uncertainty principle. Consider an intermediate state ...


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No, the virtual photon is not a particle, since a virtual particle is what one calls the internal lines in a Feynman diagram, and there are no asymptotic particle states associated to these lines, so a virtual particle is not a particle in the usual (or any other rigorous) sense. Therefore, the question is non-sensical because it is not clear what an ...


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It seems to me that you just cannot tell the difference between a Bose condensate and nothing in this case. What will change if you add some photons or phonons with zero energy to the system? No characteristics of the system will change. So it seems to me we have no criterion to decide if there is a Bose condensate in this case, and what's more important, it ...


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Technically no, because if something went faster than c it would be classified as a Tachyon. If they did however, they would only be a Tachyon by definition, physically they would still be Photons.



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