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Say two photons moving beside each other through space at speed of light, since it carries energy it can bend spacetime. So both photons will exert gravitational force on each other at speed of light however so slightly, say strings theory is right that gravition really do exist. I'm thinking if the force carrier of gravity moving at speed could exchange between the two photons? Of course I'm aware this whole stuff makes little sense so what's am I missing here?

user6760
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Here is a lower level explanation:

Two photon physics , i.e. photon photon interactions exist and even gamma gamma colliders are proposed for experiments. Here is a two photon Feynman diagram, lowest order:

photon photon

A Feynman diagram (box diagram) for photon–photon scattering, one photon scatters from the transient vacuum charge fluctuations of the other

With electromagnetic couplings at the vertices, this diagram when calculated gives a very low probability due to the 1/137 value of the coupling constant Only at high energies due to the functional form under the integrals can a measurable probability be predicted.

Suppose that one has quantized gravity , and the hypothetical graviton exists. The corresponding coupling is 6*10^-39 . So the equivalent gravitational interaction diagram would be about 10^35 times weaker than the corresponding electromagnetic one, just at each vertex. Thus one is talking of infinitessimally small gravitational effects, not attainable in our laboratories. In cosmological models such energies can be hypothesized and in the primordial soup of the Big Bang it will be part of the gravitational interactions that bounds the universe but these are theoretical propositions.

anna v
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  • What does the corresponding Feynman diagram look like? Are the virtual electrons and positrons replaced by gravitons? If so, and if gravitons do exist (a very big if), the graviton and anti-graviton are the same so how are they incorporated in the diagram? And besides, gravitons are force mediators, just like the photons are which (correct me if I'm wrong) can't come together (couple) in a vortex of a Feynman diagram. – Deschele Schilder Jun 08 '19 at 12:59
  • @descheleschilder the virtual loop of massive particles will interact with the photon on one side and radiate graviton on the other, with very low probabilities. – anna v Jun 12 '19 at 06:27
  • But this does not represent a two-photon interaction mediated by gravitons. – Deschele Schilder Jun 12 '19 at 06:36
  • Onephoton breaks in to a loop of particleuntiparticle whch radites gravitons. It may be that in the final quatized theroy there s a photon graviton vertex. i.e. tthat the photon since it has an energy mometum vector can interact directly scattering off a graviton. I do not know – anna v Jun 12 '19 at 06:47
  • So the only two (external) photon reaction is one mediated by a massive virtual particle loop? – Deschele Schilder Jun 12 '19 at 11:08
  • @descheleschilder if there is no coupling "photon graviton" . It will depend on the final theory of quantization of gravity – anna v Jun 12 '19 at 19:13
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    @descheleschilder I found this paper about gravitons https://arxiv.org/pdf/gr-qc/0607045.pdf . there seemsto be a three graviton vertex, and possibly the analogoys of compton, photon in graviton out – anna v Jun 14 '19 at 13:11
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Photons don't move at all. What changes is the probability to find an excited photon state at a spacetime point. If gravity were just another field, then the existence of a photon state in one spacetime point would create excited graviton states around it. This is the usual formulation of quantum field theory on a preordained background.

Unfortunately gravity doesn't quantize this way. Gravity is the distortion of the background metric itself, rather than an excitation of a field that lives on the background metric.

I do not believe that string theory has achieved the required manifest background independence, yet, but I am willing to let a string theorist correct me on this. In absence of this string theory can still use a perturbative approach to the problem that one can hope gives the correct results without requiring a background independent mathematical formulation.

Having said that, the coupling between single photons and gravitons is incredibly weak until we get to extremely high energies. If the Planck scale exists (a big if), then we will have basically no way of producing photon energies high enough to see the production of high energy gravitons in our accelerator facilities anytime soon. At this point the question if gravitons exist is therefor experimentally not tractable, leaving little else but guesswork for theoreticians.

CuriousOne
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    I do not understand your statement Photons don't move at all, how does this comply with SR? – Courage Dec 28 '15 at 08:16
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    These are quantum mechanical photons, not quasi-classical light particles. In QM a photon is a field state, like the s- or p-state is a state of hydrogen. One can therefor say that "a photon has been measured in spacetime point A and another photon has been measured in point B". In classical theory one would then draw a straight line and say that "the photon" has been moving along this line. In QM one can't do this. Any attempt to detect the photon along the line would change the outcome of the experiment. It is therefor better to stick to photons as initial and final states without path. – CuriousOne Dec 28 '15 at 08:23
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Because in my view we have a coupled photon-graviton.

The Photon – Graviton pair (coupled) has the same speed and frequency, and the photon energy divided by the graviton energy is the electromagnetic energy divided by the gravitational energy, the electromagnetic force divided by the gravitational force.

Adrian Ferent
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