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Most virtual particles comes in matter-antimatter pair and are quickly being annihilated, I know we shouldn't think of virtual particle as real particle but I think as long as the virtual particle can avoid being destroyed it will exist as real particle instead. I can only think that virtual photon is constantly being absorbed and emitted by other virtual particle... else why virtual photon?

Also a complementary question: how bending spacetime (intense gravity like the black hole) can convert virtual particle into real particle(hawking radiation)?

user6760
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    A virtual photon is no more a particle than a real photon. All virtual particles are merely calculation devices in a perturbation series. – CuriousOne Jan 03 '16 at 04:29

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Here is whence the term "virtual particle" derives, the Feynman diagram:

repulsion

This diagram shows how to write the integral that will give the cross section for electron electron scattering or positron positron scatterng. The e lines are called real electrons because one can have a detector before and a detector after the interaction, and measure momenta and energies. The exchanged line is called virtual because it is within the integration, its variables spanning the phase space of integration and even though from quantum number conservation laws it get the label of a particle, in this case a photon, its mass is variable because of the integration and thus off mass shell (not zero).

With the above background:

Most virtual particles comes in matter-antimatter pair and are quickly being annihilated,

The heisenberg uncertainty principle allows one to have a model where within the uncertainty Feynman diagram loops with no input of real particles can be envisaged mathematically :

vacuum loop

The result of the integration ( Feynman diagrams are always about integration, and virtual particles exist only within Feynman diagrams in a mathematial space) will be zero momentum zero energy because zero was the input.

I know we shouldn't think of virtual particle as real particle but I think as long as the virtual particle can avoid being destroyed it will exist as real particle instead.

A virtual particle is a point in a function that will be integrated over, its "time of life" is within the integral and unless energy and momentum is supplied from outside, and real lines enter and leave the Feynman diagram, there is no "reality" as a particle.

Also a complementary question: how bending spacetime (intense gravity like the black hole) can convert virtual particle into real particle(hawking radiation)?

As I stress, virtual particles have a meaning only within Feynman diagrams. One has to write a Feynman diagram where there exists an exchange of energy and momentum with the black hole through a graviton. The graviton is the particle hypothesized to transfer gravitational interaction , but quantization of gravity is still part of on going research. An effective diagram can be thought of, where the whole of the black hole exchanges a graviton with one of the virtual loops, i.e. supplies energy and momentum , then a probability exists that one of the pairs in the loop is drawn into the black hole, while the other gets enough energy and momentum to escape. See also the answer here.

anna v
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  • Pardon me if I am off here (and I will even delete the comment) but don't virtual particle have meaning in the context of quantum 'foam'? Don't they play a role in the Casmir Effect? (I suppose the is a FD for the Casmir Effect...) – Jiminion Jan 05 '16 at 15:44
  • There is a Feynman diagram for any usage of "virtual" particle. It has no meaning outside of the prescribed intergrations dictated by the FD. The "foam" is the second figure above. Its value is zero unless there exist interactions with external lines ( i.e. real particles) – anna v Jan 05 '16 at 16:58