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Electromagnetism can be modeled as a $U(1)$-principal bundle over Minkowski spacetime. The strength of the electromagnetic field is given by the 2-tensor $F_{\mu\nu}$. In differential geometry this is the curvature tensor, and so when $F_{\mu\nu}$ is nonzero there is some curvature present in the bundle. Is there any interpretation of this curvature similar to how one interprets gravity and curvature?

This question asks something similar but is largely focused on the mathematics, whereas I am curious about the physical interpretation of the curvature.

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  • $\begingroup$ The $F$ satisfies Maxwell’s equations. That’s ALL physics. Surely your question isn’t just this? $\endgroup$
    – peek-a-boo
    Jan 4 at 11:22
  • $\begingroup$ @peek-a-boo Yes that is not what I am asking. I am trying to visualize the $U(1)$-bundle and get an intuitive and geometric picture of how the curvature can relate to the electromagnetic field. $\endgroup$
    – CBBAM
    Jan 4 at 11:26
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    $\begingroup$ The curvature is the field strenght. What else is there to understand here? This is what allows mathematicians to understand gauge theory in terms of differential geometry. $\endgroup$
    – Mauricio
    Jan 4 at 11:45
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    $\begingroup$ The Bernstein & Philipps article discussed in this post develops a nice geometrical picture of that curvature. Nonetheless: the real analogy is not very geometrical in a visual sense. The book of Nakahara is the way to go. $\endgroup$
    – Kurt G.
    Jan 4 at 12:59

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I am not sure that it is exactly what do you want to see, but let me try.

Differential Geometry Object Electromagnetism
local triviliazation $U_i\times G$ gauge
local form of connection $\omega_i=A_{\mu}^i dx^{\mu}$ 4-potential in a chosen gauge
connection $\nabla$ electromagnetic (EM) field
local form of connection curvature $\Omega_i$ EM tensor in a chosen gauge
connection curvature tensor $K$ EM tensor
Bianchi identity $D\Omega_i=0$ Faraday part of Maxwell equations
bundle space $P(\mathbb{R}^4\times S^1)$ Phase space of physical system
bundle base space $B(\mathbb{R}^4)$ Space-time
structural bundle group $G(U(1))$ "Rotation" group in charge space

Updatde: I understand the OP desire, but my knowledge ends here. Nevertheless, I would like to cite B. Hatfield (opening remarks to "Feynman lectures on gravitation"):

Presently we have a geometrical intepretation of classical gauge theories such as electrodynamics and Yang-Mills. The vector potentials are connection coffecients on a principle fiber bundle where the structure group is gauge group. The field strength are the curvatures associated with the connectiones. The chargeed matter that the fields couple to are associated vector bundles. .... While it can be arguted that the geometric intepretation of gauge fields has not helped us solve QED or QCD, it has certainly led to many useful insights into the topological aspects of these theories (e.g. the Gribov ambiguity, instantonts, the vacuum angle and toplologically inequivalent vacuums) ...

From my point of view, the geometrical interpretation is useful in QFT but in classical theory it seems that it cannot give any more.

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    $\begingroup$ Nice Wu-Yang dictionary! $\endgroup$
    – Mauricio
    Jan 4 at 14:11
  • $\begingroup$ Thanks, this is very nice but I'm looking for something a bit more visual/intuitive as I'm okay with the technical definitions. Maybe such a thing does not exist. $\endgroup$
    – CBBAM
    Jan 4 at 21:02
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    $\begingroup$ @CBBAM to me you are looking for a mathematical abstraction that is different from what people understand from this kind of analogies. The electromagnetic field is the physical interpretation, the connection is the mathematical one, not the other way around. If I show you the equations of EM in mathematical setting, you could use a table like this to see that these could be interpreted as force fields. When you say that you want the physical interpretation I think you are looking for some surface in some mathematical space not a physical one. $\endgroup$
    – Mauricio
    Jan 8 at 14:00
  • $\begingroup$ @Mauricio Thank you, I think you're right that I have this backwards. $\endgroup$
    – CBBAM
    Jan 9 at 8:31

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