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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.(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. The photon doesn't "bend," it goes in a straight line in a curved space-time.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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Thank you. How does the electron magnetic moment is related to its spin? And do photons and gravitons have a magnetic moment?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.

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DrClaude

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Have a look at https://en.wikipedia.org/wiki/Electron_magnetic_momentThank you. How does the electron magnetic moment is related to its spin?

It is related to the fact that the electron has a charge.

No. We've discussed this before:And do photons and gravitons have a magnetic moment?

https://www.physicsforums.com/threads/photon-in-a-magnetic-field.884424

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Can you show me a derivation for the spin from the magnetic moment?Have a look at https://en.wikipedia.org/wiki/Electron_magnetic_moment

It is related to the fact that the electron has a charge.

No. We've discussed this before:

https://www.physicsforums.com/threads/photon-in-a-magnetic-field.884424

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DrClaude

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

Maybe @vanhees71 can help?

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$$\mathcal{L}=\overline{\psi} (\mathrm{i} \gamma^{\mu} \partial_{\mu}-m) \psi,$$

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.

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Thank you.

$$\mathcal{L}=\overline{\psi} (\mathrm{i} \gamma^{\mu} \partial_{\mu}-m) \psi,$$

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

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".

$$\mathcal{L}=\overline{\psi} (\mathrm{i} \gamma^{\mu} \partial_{\mu}-m) \psi,$$

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.

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

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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