I understand mathematically why they don’t, but I was hoping someone could provide a physical interpretation to this. Is there a physical consequence of this fact?
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Qmechanic
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Spencer Kraisler
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2Does this answer your question? What is a Christoffel symbol? – Miyase Feb 06 '23 at 05:57
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1With `mathematically' I suppose you know from the fibre bundle perspective it's a spacetime one-form and a Lie algebra curvature? – Guliano Feb 06 '23 at 06:06
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@Miyase that does answer my question! Should I deleted this post, or leave it? – Spencer Kraisler Feb 06 '23 at 06:23
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@Guliano I only understand them from the context of affine connections defined on manifolds. – Spencer Kraisler Feb 06 '23 at 06:23
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1@SpencerKraisler You can delete your question, but it's not necessary. It was already closed as duplicate, which is enough. – Miyase Feb 06 '23 at 12:14
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So the Christoffel symbols are a set of indexed scalar fields derived from your coordinates that can, like all sets of indexed scalar fields, be assembled into a tensor field.
The problem is, this derivation yields different tensor fields depending on the coordinate fields you start with.
I was in cond-mat so all of this is rusty to me but my memory is that the counterexamples are pretty simple. For example you start with flat 2D space, coordinate fields $u_1=x(p), u_2= y(p)$, your Christoffel symbols are zero. So if you assemble a tensor out of it, it is the zero tensor, and the zero tensor is the same in all coordinate systems. Then you switch to the equally valid coordinate $u_{1'}=x^3(p) = [x(p)]^3$ suddenly $\Gamma^{1'}_{1'1'}$ I think is nonzero and hence the resulting tensor isn't the zero tensor? Something like that.
CR Drost
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