Recent content by PhysicsRock

Does that make a difference? Probably easy to check, but I know myself very well and I tend to make the worst mistakes with the easiest problems.

This is a very helpful insight that I've actually never seen. Unfortunately, I haven't been able to take a proper differential geometry class so far, all my current knowledge comes from a very short and not quite detailed basic introduction in last semesters electrodynamics lecture. All the...

To my understanding, an orientation can be expressed by choosing a no-where vanishing top form, say ##\eta := f(x^1,...,x^n) dx^1 \wedge ... \wedge dx^n## with ##f \neq 0## everywhere on some manifold ##M##, which is ##\mathbb{H}^n := \{ x \in \mathbb{R}^n : x^n \geq 0 \}## here specifically. To...

This is actually brilliant, because Stokes by itself works without a metric, but I'm still supposed to cover how one may appear when considering the Hodge-Star, which is included here. Amazing, thank you!

Great, thank you!

Thank you. I'll check that out.

True, my mistake. I guess I'll ask my professor if a proof would do the job, assuming we already know the theorem and just wanted to confirm it's true, not build it from scratch / a certain point we can assume to be true.

To me, a proof means starting with the theorem and ending somewhere else, a true statement would be good. A derivation would do the same, but the other way around. Start somewhere, do some math-magic and finish off with the theorem.

Hello everyone, as part of my bachelor studies, I need to attend a seminar with the aim to prepare a presentation of about an hour on a certain topic. I have chosen the presentation about the generalized Stokes theorem, i.e. $$ \int_M d\omega = \int_{\partial M} \omega. $$ After hours of...

Yes, of course. That makes it pretty clear. Would've been nice to see this mentioned somewhere, instead of just doing it and not even addressing the trick. Thank you for your help!

How exactly do the fact that ##\nabla## is a differential operator and the doubling of the last term go together?

I was / am trying to derive the energy shift resulting from the normal Zeeman-Effect by coupling the Hamiltonian to the external field ##\vec{A}##, that carries the information about the field ##\vec{B}## via ##\vec{B} = \nabla \times \vec{A}##. Let ##q = -e## be the charge of the electron and...

No, although this is also interesting to see. The question was more or less in regard to the example I gave earlier with the scalar field action / Lagrangian. If such an assignment does come up, I obviously wouldn't wanna write everything out. I thought the property ##dx^i \wedge dx^j \wedge...

Great, thank you.

Yes, of course. The question was not about how the Levi-Civita worked, but rather if there was some additional ##-1##, i.e. ## dx^i \wedge dx^j \wedge dx^k = (-1)^\text{something} \cdot \varepsilon^{ijk} dx^1 \wedge dx^2 \wedge dx^3##. However, your answer seems to indicate that the minus sign...