The Maths of General Relativity (3/8) - Geodesics

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Recommends user285989 · 2021-01 · Physics Stack Exchange · on “Trajectory By Christoffel Symbols”

So, basically Christoffel Symbols gives us the components of the vector that represents the rate of change of the basis vectors in a manifold (surface). And this equation below tells the rate of change of the vector components of a velocity vector summarized by a given object in a manifold where the Christoffel Symbols != 0, $$\frac{dv^a}{dt} = -\Gamma_{mn}^a v^m v^n$$ Let me be clear with simple terms, $v^m$ and $v^n$ are vector components respectively such that $v^1$ is the x-component, $v^2$ is the y-component of the velocity vector simultaneously. And $\Gamma$ with indices is the Christoffel Symbol, if you are interested with its derivation I will leave a link below for a video that expl

Recommends Guillermo Abad Lopéz · 2022-07 · Physics Stack Exchange · on “Trying to intuitively understand Christoffel symbols”

To gain a visual intuitionI would recomend cheking this Youtube video from science click (3rd video in the series of GR): [this video]

Recommends The Rizzler · 2024-10 · Physics Stack Exchange · on “Are 4-vectors really vectors in general relativity?”

After watching this video [this video] about geodesics, I am now confused about where 4-vectors fit in general relativity. I already checked this question [LINK] but it didn't help much. I am still learning the formalities of differential geometry. Let me know if I misunderstood anything. Let $M$ be a 4D spacetime manifold and $(U,\phi)$ be one of its coordinate charts where $U\subseteq M$ is an open patch of the manifold. A trajectory of an apple in $U$ is parametrized by a smooth map $\lambda:\mathbb R\to U$ . I think $$x^\alpha(\tau)=\phi(\lambda(\tau))=(cx^0(\tau),x^1(\tau),x^2(\tau),x^3(\tau))$$ is what one calls "position 4-vector" but it is a point, not a vector, despite its name. A "veloc