general relativity - GR and my journey to the centre of the Earth

[General Relativity] basically says that the reason you are sticking to the floor right now is that the shortest distance between today and tomorrow is through the center of the Earth.

I love this, not the least because it sounds nonsensical.

(From an unassessed comment in the internet)

OK so I love this too, but is it a completely looney description or does it make any sense in which case I'm in for some serious enlightenment today since last time I checked light cones allowed me to move somewhat freely unless in significant proximity with a singularity.

Answer

That is awesome! And it makes complete sense too! (other than a possible misusage of the word "distance"). Let's have a look at the equations of motion of you in Earth's curved spacetime, assuming that your feet are not touching the ground:

$$\frac{\mathrm d^{2}x^{\mu}}{\mathrm ds^{2}}+\Gamma^{\mu}_{\nu\sigma}(x(s))\ \frac{\mathrm dx^{\nu}}{\mathrm ds}\frac{\mathrm dx^{\sigma}}{\mathrm ds}=0$$ where $x^{\mu}(s)$ is your world line, $s$ is some parameter,

$$\Gamma^{\mu}_{\nu\sigma}=\frac{1}{2}\ g^{\mu\tau}(\partial_{\nu}g_{\sigma\tau}+\partial_{\sigma}g_{\nu\tau}-\partial_{\tau}g_{\sigma\nu})$$ with $g^{\mu\tau}$ the inverse of the metric and $$g=\left( 1 - \frac{r_{s} r}{\rho^{2}} \right) c^{2}\, \mathrm dt^{2} - \frac{\rho^{2}}{\Delta} \mathrm dr^{2} - \rho^{2} \,\mathrm d\theta^{2}+ \\ - \left( r^{2} + \alpha^{2} + \frac{r_{s} r \alpha^{2}}{\rho^{2}} \sin^{2} \theta \right) \sin^{2} \theta \,\mathrm d\phi^{2} + \frac{2r_{s} r\alpha \sin^{2} \theta }{\rho^{2}} \, c \,\mathrm dt \, \mathrm d\phi$$ where $$r_{s}=\frac{2GM}{c^{2}}\ ,\quad\alpha=\frac{J}{Mc} \ ,\quad \rho^{2}=r^{2}+\alpha^{2}\cos^{2}\theta\ ,\quad \Delta=r^{2}-r_{s}r+\alpha^{2}$$ with $M$ and $J$ Earth's mass and angular momentum.

The equations of motion can be derived from the action functional

$$S[x(s)]=-mc\int_{a}^{b}\sqrt{g_{\mu\nu}(x(s))\,\frac{\mathrm dx^{\mu}}{\mathrm ds}\frac{\mathrm dx^{\nu}}{\mathrm ds}}\ \mathrm ds$$ where $m$ is your mass and, as gravity goes, it plays no role at all in how you fall to the ground. You find the equations of motion by minimizing S with respect to the curve $x(s)$, which amounts to minimizing the (proper) time you spend on your worldline, times $-mc^{2}$ (this is why you are minimizing rather than maximizing): $$\begin{align} S[x(\tau)]&=-mc^{2}\int_\textrm{today}^\textrm{tomorrow}\sqrt{g_{\mu\nu}(x(\tau))\,\frac{\mathrm dx^{\mu}}{\mathrm d\tau}\frac{\mathrm dx^{\nu}}{\mathrm d\tau}}\,\mathrm d\tau\\ &= \text{the distance between today and tomorrow}\,. \end{align}$$ As you'll fall in the direction that connects you to the center of the Earth, the shortest distance between today and tomorrow is indeed through the center of the Earth. The reason why you are sticking to the floor right now is really that the ground is preventing you from taking the shortest path from today to tomorrow, which passes through the center of the Earth.