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In general relativity, can the differential Bianchi identity be interpreted as "non existence of gravitomagnetic monopoles"?

This was suggested by Avantgarde's answer in What is the geometric interpretation of the Einstein tensor $R_{\mu \nu} - \frac{1}{2} g_{\mu \nu} R$

The idea seems intriguing and kind of makes sense from the point of view of gravitoelectromagnetism. But he compares the Riemann tensor $R_{\mu\nu\rho\sigma}$ to the gravitational field strength, whereas I thought the Christoffel symbols are a closest thing to "force field" in GR. So the analogy breaks down? Otherwise the arguments seem good to me.

So can someone help me? Thank you.

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  • $\begingroup$ Riemann tensor is indeed the field strength of gravity, just like the Faraday tensor is the field strength of electromagnetism. In this analogy, the Christoffel symbols (or more precisely the gamma symbols) are analogous to the gauge field. $\endgroup$ Commented Jan 3, 2022 at 19:32
  • $\begingroup$ Gravitomagnetism is real but gravity is so weak compared to electric force the effects are neglible unless we talk about the gravitomagnetic field of rotating neutron stars or rotating black holes. $\endgroup$
    – Jun Seo-He
    Commented Jan 3, 2022 at 19:55
  • $\begingroup$ So would you agree with Avantgardes arguments that Bianchi identity represents non existence of gravitomagnetic monopoles? $\endgroup$
    – Anssmirk
    Commented Jan 4, 2022 at 20:30
  • $\begingroup$ But if the metric is a generalization of the gravitational potential, then Christoffel symbols (being derivatives of the metric) would represent the force? The curvature tensor would represent derivatives of force, right? Thank you so much for your answers. $\endgroup$
    – Anssmirk
    Commented Jan 4, 2022 at 20:38

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