# B Gravity and Electromagnetism

1. Sep 19, 2016

### kent davidge

(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?

2. Sep 19, 2016

### Staff: Mentor

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.

3. Sep 19, 2016

### kent davidge

Thank you. How does the electron magnetic moment is related to its spin? And do photons and gravitons have a magnetic moment?

4. Sep 19, 2016

### Staff: Mentor

5. Sep 19, 2016

### kent davidge

6. Sep 20, 2016

### Staff: Mentor

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

Maybe @vanhees71 can help?

7. Sep 20, 2016

### vanhees71

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.

8. Sep 21, 2016

### kent davidge

Thank you.

9. Sep 22, 2016

### haushofer

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".

10. Sep 22, 2016

### vanhees71

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.