Bei's Study Notes

General Relativity 06
Last updated: 2016-08-14 19:21:06 PDT.

Reference: UCI OpenCourseWare - GR 07 (Playlist)

Some remarks (Mach's Principle)

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.