Reference: UCI OpenCourseWare - GR 07 (Playlist )

Starting from 39:00 .

My Note: I've never been convinced of Mach's Principle. In fact, there has been many null results in testing Mach's principle. I don't know this has to be a thing in the first place.

Non-Cartitian coordinate system
Starting from 1:07:55 .

Starting from an example:

\xi^\alpha = \begin{cases}
x &=& r \sin \theta \cos \phi \\
y &=& r \sin \theta \sin \phi \\
z &=& r \cos \theta
\end{cases}
\\
x^\alpha = \begin{cases}
r &=& \sqrt{x^2 + y^2 + z^2} \\
\theta &=& \cos^{-1}(\frac{z}{r})\\
\phi &=& \tan^{-1}(\frac{y}{x})
\end{cases}
\\
\eta_{\alpha\beta} =\delta_{\alpha\beta}

We will try to calculate the metric in this case. The Jacobian:

\begin{align*}
\frac{\partial(x,y,z)}{\partial(r,\theta,\phi)} &= \left(\begin{array}{lll}
\frac{\partial x}{\partial r}&\frac{\partial x}{\partial \theta}&\frac{\partial x}{\partial \phi}\\
\frac{\partial y}{\partial r}&\frac{\partial y}{\partial \theta}&\frac{\partial y}{\partial \phi}\\
\frac{\partial z}{\partial r}&\frac{\partial z}{\partial \theta}&\frac{\partial z}{\partial \phi}\\
\end{array}\right)\\
&=\left(\begin{array}{lll}
\sin \theta \cos \phi & r \cos \theta \cos\phi & -r \sin \theta \sin \phi \\
\sin \theta \sin \phi & r \cos \theta \sin\phi & r \sin \theta \cos \phi \\
\cos \theta & -r \sin \theta \cos\phi & 0 \\
\end{array}\right)\\
g_{\mu\nu} &=
\eta_{\alpha\beta} \frac{\partial \xi^\alpha}{\partial x^\mu} \frac{\partial \xi^\beta}{\partial x^\nu}\\
&= \frac{\partial x}{\partial x^\mu}\frac{\partial x}{\partial x^\nu}+
\frac{\partial y}{\partial x^\mu}\frac{\partial y}{\partial x^\nu} +
\frac{\partial z}{\partial x^\mu}\frac{\partial z}{\partial x^\nu} \\
&= \left(
\begin{array}{lll}
1 & 0 & 0\\
0 & r^2 & 0 \\
0 & 0 & r^2 \sin^2\theta \\
\end{array}
\right)\\
\sqrt{\det g_{\mu\nu}} &= r^2 sin^\theta\\
ds^2 &= g_{\mu\nu}dx^\mu dx^\nu = dr^2 + r^2(d\theta^2 + sin^2\theta d\phi^2)
\end{align*}

The inverse is simple:

\begin{align*}
g^{\mu\nu} &= \left(
\begin{array}{lll}
1 & 0 & 0\\
0 & \frac{1}{r^2} & 0 \\
0 & 0 & \frac{1}{r^2 \sin^2\theta} \\
\end{array}
\right)\\
\Gamma^\lambda{}_{\mu\nu} &= -\frac{1}{2}g^{\mu\nu}(\partial g+\partial g-\partial g)\\
&= ...
\end{align*}

The important takeaway is that

\Gamma
is in fact not a tensor, for that it can be

0
in some coordinate systems but non-zero in others, even in the same planar space.