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B Gravity and Electromagnetism

  1. Sep 19, 2016 #1
    (I'm sorry for my poor English.) In GR the explanation for the bending of light by gravity is that gravity is a curvature in space (and time) and thus light follows the curved space. I was reading about the (undiscovered) graviton. It would have spin 2. Does it mean a graviton would interact with a magnetic field? If so, can we speak of bending of light as a photon-graviton interaction, like electrons (spin 1/2) interact with magnetic fields?
     
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  3. Sep 19, 2016 #2

    DrClaude

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    No. Having spin angular momentum has no relation with interaction with a magnetic field. The electron interacts with a magnetic field because it has a magnetic moment.

    No. The photon doesn't "bend," it goes in a straight line in a curved space-time.
     
  4. Sep 19, 2016 #3
    Thank you. How does the electron magnetic moment is related to its spin? And do photons and gravitons have a magnetic moment?
     
  5. Sep 19, 2016 #4

    DrClaude

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  6. Sep 19, 2016 #5
  7. Sep 20, 2016 #6

    DrClaude

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    I don't know that it can be "derived." There may be something coming from QFT.

    Maybe @vanhees71 can help?
     
  8. Sep 20, 2016 #7

    vanhees71

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    The relation between spin and magnetic moment comes from minimal coupling of the electromagnetic field. For Dirac spinors, e.g., you start from the free-field Lagrangian
    $$\mathcal{L}=\overline{\psi} (\mathrm{i} \gamma^{\mu} \partial_{\mu}-m) \psi,$$
    and substitute
    $$\partial_{\mu} \rightarrow D_{\mu}=\partial_{\mu} + \mathrm{i} q A_{\mu},$$
    where ##A_{\mu}## is the electromagnetic field. This leads to the magnetic moment of the electron with the correct tree-level gyrofactor of 2.
     
  9. Sep 21, 2016 #8
    Thank you.
     
  10. Sep 22, 2016 #9

    haushofer

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    Just to add: this can also be derived non-relativistically, which shows that the gyrofactor of 2 is not a relativistic effect, as sometimes is claimed (unlike the Darwin-term and the spin-orbit coupling, which are relativistic effects). See e.g. papers by Levy-Leblond. The idea is basically to write down the Dirac equation, but demand that every spinor component obeys the Schrodinger equation instead of the Klein-Gordon equation, resulting in a "nonrelativistic Clifford algebra".
     
  11. Sep 22, 2016 #10

    vanhees71

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    Well, in the non-relativistic case it's not so convincing, because you have to write
    $$\hat{\vec{p}}^2=(\vec{\sigma} \cdot \hat{\vec{p}})^2,$$
    and then introduce the minimal coupling in this way ##\vec{\sigma} \cdot \hat{\vec{p}} \rightarrow \vec{\sigma}(\hat{\vec{p}}-\mathrm{i} q \hat{\vec{A}})## and then square. This is just an ad-hoc description, leading to the correct gyro factor. Why one cannot simply put the minimal substitution without introducing the Pauli matrices is not clear. In th Dirac case it's a unique procedure.
     
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