# Gravity and Electromagnetism

(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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DrClaude
Mentor
(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?
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.

If so, can we speak of bending of light as a photon-graviton interaction, like electrons (spin 1/2) interact with magnetic fields?
No. The photon doesn't "bend," it goes in a straight line in a curved space-time.

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

DrClaude
Mentor
Can you show me a derivation for the spin from the magnetic moment?
I don't know that it can be "derived." There may be something coming from QFT.

Maybe @vanhees71 can help?

vanhees71
Gold Member
2019 Award
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.

kent davidge and bhobba
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.
Thank you.

haushofer
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.
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".

kent davidge
vanhees71
$$\hat{\vec{p}}^2=(\vec{\sigma} \cdot \hat{\vec{p}})^2,$$