1
$\begingroup$

The electromagnetic gauge field is $A + d\theta$, where $\theta \colon \mathbb{R}^n \to \mathbb{R}$ comes from a gauge function, $e^{i\theta(\vec x)} $. Let's set $A=0$. The curvature form is $0$ since the gauge field is closed (in other words $d(d\theta) = 0$).

My question is partially mathematical in nature. I am also curious about physical interpretation. $\theta(\vec x)$ appears to have no curvature, because of my reasoning (the connection it defines is a closed form). However... as a function it does not necessarily describe a flat manifold. And integral curves of the vector field created by the connection (for example, the connection coefficients $(\partial_{1} \theta(x_1)\ldots\partial_n \theta(x_n))$ after we dot $d\theta$ with a set of basis vectors) will have acceleration if the coefficient functions are not constants. So there's a set of contradictory observations for me about the role these calculations play in curvature. How do I combine these facts? Specifically, why am I calculating the curvature of the form is always zero when $\theta$ could be a curved surface (or $n$ dimensional generalization), and what does this have to do with not having a gauge force when the geodesics of that surface may curve?

Its possible I'm conflating a few different ideas and this is leading to my confusion, so any help straightening this out in my head would be really appreciated.

$\endgroup$

1 Answer 1

1
$\begingroup$

Well, one simple confusion is that $A$ is the gauge potential whilst $A+d\theta$ is a gauge transformation.

The mathematical and physical description of gauge potentials differ. Essentially, the physical gauge potential is the mathematical gauge potential pulled back to sime local section.

The usual framework for all this, in both maths and physics, is fibre bundles. However note terms differ. For example, the gauge group in physics is what is called the structure group in maths. However, this gets confusing as physicists also ise the term for a much larger related group and this is isomorphic (but not equal) to the gauge group used in maths.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.