Why covariant derivative is partial derivative in SR?

Question

I'm new in these topics and i've been confuse at some relations between the limit of SR for GR.

In cartesian coordinates, basis do not change, so \begin{equation}\Gamma^{\mu}_{\alpha\beta}=0 \quad \quad(1)\end{equation} and thus $V^{\alpha};_{\beta}=V^{\alpha},_{\beta}$.

I've seen that polar coordinates changes point to point: \begin{equation}|\vec{e_{\theta}}|^{2}=r^{2}\quad\quad(1.1),\end{equation} so polar coordinates has non constant basis and for these cordinates, equation (1) isn't true, as same than the covariant derivative isn't the partial derivative for that cordinate system, that is \begin{equation}V^{\alpha};\beta\ne V^{\alpha},\beta\quad (1.2)\end{equation}

My book Schutz define a locally inertial frame as a frame in which everything is locally like SR, and formally by a theorem we see that

Choose any point $\mathcal{P}$ of the manifold, a coordinate system can be found whose origin is at $\mathcal{P}$ and which the metric is Minkowski metric (2)

He also says

In particular, we say that the derivatives of the basis vectors of a locally inertial coordinate system are zero at $\mathcal{P}$' (3)

Why ???, the coordinate system is arbitrary, can't be polar or spherical ?, i understand that locally the inertial frame has behaviour as SR but in SR i can have polar, spherical and some other's coordinate system's that derivative of basis are not equals to zero.

This definition mentioned in (3) leads he to conclude that \begin{equation} V^{\alpha};\beta=V^{\alpha},\beta \quad \mbox{ at } \mathcal{P} \mbox{ in this frame } \quad \quad (4)\end{equation}

Why ?? again this frame is in SR, okay, but this frame can assume a lot of coordinate systems that derivative of basis isn't necessary equals to zero, but if i consider this point $\mathcal{P}$ as a instantaneous measure, the derivative of non constant basis will be zero, but how $V^{\alpha},\beta$ exists in instantaneous measure ??

than he follows with these conclusions \begin{equation}g_{\alpha\beta;\gamma}=g_{\alpha\beta,\gamma}=0\quad \mbox{ at } \mathcal{P}\quad \quad (5)\end{equation} because the last equation is a tensor equation, so is valid

\begin{equation}g_{\alpha\beta;\gamma}=0\quad \mbox{in any basis }\quad (6) \end{equation}

this will be valid in any basis in this SR frame ?? at SR frame near point $\mathcal{P}$, or in all frames ??.

All these doubts lead me to ask myself if is SR defined only for Cartesian coordinates, that is, when we talk about SR is supposed to assume only Cartesian coordinates ??.

Sorry about a ton of question, i'm lost in these doubts a few days ago

Answer 1

Why ???, the coordinate system is arbitrary, can't be polar or spherical ?, i understand that locally the inertial frame has behaviour as SR but in SR i can have polar, spherical and some other's coordinate system's that derivative of basis are not equals to zero.

All of those statements of Schultz revolve around the idea of a coordinate system whose origin is $\mathcal P$, and in which $\Gamma^i_{jk}(\mathcal P)=0$; a polar coordinate chart (of a smooth manifold) cannot even include the coordinate origin, so such coordinate systems are not suitable for this kind of analysis.

I admit, Schultz is perhaps not so good at making this distinction. What you say is essentially true - polar coordinates are as good as any other coordinates - with the exception that they are poorly behaved at the origin.