Recent content by PhysicsRock

Thank you very much, this helped a lot!

I am currently working on my bachelor thesis and ran into an issue regarding citation. I found some lecture notes from MIT which are quite useful (https://ocw.mit.edu/courses/8-333-statistical-mechanics-i-statistical-mechanics-of-particles-fall-2013/pages/lecture-notes/). Obviously, I'll have to...

Alright, thank you!

My solution is $$ Z = e^{-2\beta E_0} + 4 e^{-\beta (E_0 + E_1)} + e^{-2\beta E_1}, $$ since Fermions have non-zero spin and there are four options for distributing spins (assuming only ##\pm \frac{1}{2}##) among ##E_0## and ##E_1## such that ##E_\text{tot} = E_0 + E_1## and one option each...

In a homework problem, I had to derive the relationship ##R_{\mu\nu} = \pm K g_{\mu\nu}## on a surface, i.e. a ##2##-dimensional submanifold of ##\mathbb{R}^3##. Here, ##K## is the Gaussian curvature. I think I managed to do that, but from my derivation I don't see why this result is restricted...

The way I understand it is that if a point ##p \in M## lies within a chart ##(U,\varphi)## and ##\varphi(p) \in \partial\mathbb{H}^n## then ##p## is considered to be a boundary point. The set of all such ##p## is then called the boundary of ##M##. However, what I don't understand is why we...

In differential geometry, we typically define the boundary ##\partial M## of a manifold ##M## as all ##p \in M## for which there exists a chart ##(U,\varphi), p \in U## such that ##\varphi(p) \in \partial\mathbb{H}^n := \{ x \in \mathbb{R}^n : x^n = 0 \}##. Consequently, we also demand that...

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.