[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.

  • $\begingroup$ I guess it's saying something about extremizibg proper time. But it makes more sense to say that you experience pseudo-forces because you are accelerating, as EM forces are pushing you. $\endgroup$
    – innisfree
    Apr 20, 2016 at 9:43
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    $\begingroup$ Related: physics.stackexchange.com/q/3009/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 20, 2016 at 10:50

4 Answers 4


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.

  • $\begingroup$ A quibble: you're maximizing, not minimizing, the proper time along your worldline. Any deviation from a geodesic incurs time dilation and results in a shorter proper time (c.f. the twin paradox). $\endgroup$ Apr 21, 2016 at 0:31
  • $\begingroup$ The twin paradox has nothing to do with minimization/maximization. Each of the twins rides along its own geodesic (if we state the problem as to remove the problem of the infinite acceleration required by one twin to turn back; it is possible to do so in such a way that the paradox is still a paradox, for example by adding a third twin to the picture), so they're both really minimizing/maximizing their own proper time. $\endgroup$ Apr 21, 2016 at 8:10
  • $\begingroup$ As for the minimization/maximization, as you can see I wrote the full action, with a $-mc^{2}$ in front of proper time, which is necessary in order to get the correct energy, say, for a (special)-relativistic particle, and the equations of motion in presence of external forces. The functional is thus minimized. I am really minimizing proper time times $-mc^{2}$, which you can interpret as a physical definition of the "distance between today and tomorrow" (non-intuitively, just because it naturally appears in the equations). $\endgroup$ Apr 21, 2016 at 8:16
  • $\begingroup$ We always define things as to strictly minimize the action and have a positive-definite energy. $\endgroup$ Apr 21, 2016 at 8:21
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    $\begingroup$ In the twin paradox, only one twin rides a geodesic; the other rides a timelike curve that is not a geodesic (although some sub-segments of the curve may be coincident with a geodesic, i.e. parts of the trip where the travelling twin is not accelerating). $\endgroup$ Apr 21, 2016 at 16:21

What GR says is correct: the straight line between, say, London today and London tomorrow is not the curve that spends all the time between in London: whether it actually passes through the centre of the Earth I'm not sure, and it depends on how fast you are moving as well as where you are.

The caveat is that the straight line (geodesic) not the shortest path, it's the longest (there is no shortest path) and the length is proper time.

This not inconsistent with you being able to take other paths: you can, but they are not extrema of length and therefore you experience acceleration on the path: the acceleration which is currently sticking you to the ground, for instance.

  • $\begingroup$ Can you explain in which sense it's the longest path? $\endgroup$ Apr 20, 2016 at 20:53
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    $\begingroup$ @romkyns It has the greatest proper time, or in other words you would experience more time on it than on any other path. This is why, for instance, the twin 'paradox' works: the twin that experiences acceleration also experiences less time than the one who does not, $\endgroup$
    – user107153
    Apr 20, 2016 at 22:15
  • $\begingroup$ Each of the twins rides along a different geodesic, so the twin paradox is not a good example in this case. They both maximize/minimize their proper time, given different spatial velocities. One can set up the problem in such a way that no acceleration is needed (the point of the paradox is not the acceleration required for one twin to turn back). $\endgroup$ Apr 21, 2016 at 7:59
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    $\begingroup$ @GiorgioComitini At least one of the twins does not travel along a geodesic in the usual version, as they turn around. In flat spacetime (without any odd topology) then there is a unique geodesic between any two events, so the twin paradox can't happen at all if they travel on geodesics. $\endgroup$
    – user107153
    Apr 21, 2016 at 8:46
  • $\begingroup$ @tfb Read my comments to my answer, there I explained in what sense my statement was to be read. $\endgroup$ Apr 21, 2016 at 17:55

It makes sense as a "visual" description.

In GR, free particles with mass move on time-like geodesics. A common description of geodesics are such curves that locally minimalize path length, but this desciption comes from Riemannian geometry, not Lorentzian geometry, which GR is. In Lorentzian geometry, timelike geodesics are those that locally maximalize proper time.

The reason the quote sounds so nonsensical, is that in GR time is also curved, and the geodesics move through space-time, not just space. If the ground was not beneath your feet, you'd fall through the center of the earth, as time would pass, hence you could say that the "path with greatest proper time between today and tomorrow leads through the center of the earth".

But there is a ground beneath your feet, the ground exerts EM force on you that makes you deviate from this geodesic, since you are no longer a "free particle".

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    $\begingroup$ Quibble: Pauli exclusion is not "EM" force. $\endgroup$
    – JDługosz
    Apr 21, 2016 at 10:24
  • $\begingroup$ @JDługosz Yeah, but that's not the immediate cause. The thing preventing you from falling through the ground is static electromagnetic repulsion. Sure, that repulsion wouldn't exist if PEP didn't limit "effective size" of fermions, but that's going too far IMO :D And of course, if the pressure you exert on ground was large enough, PEP would be the only thing blocking your way, but that certainly isn't something the human body would survive... $\endgroup$
    – Luaan
    Apr 22, 2016 at 7:59
  • $\begingroup$ I don't think so. Electric repulsion and attraction cancel out and don't form a barrier. It's the Exchange Interaction that prevent matter from passing through each other. $\endgroup$
    – JDługosz
    Apr 22, 2016 at 12:43

That quote requires some modification for it to make sense:

"General Relativity basically says that the reason I am sticking to the floor is that the path of maximal aging between 'here now' and 'here tomorrow' is through Earth's center."

  • $\begingroup$ How will you end up "here" if you are able to fall through to the center of the Earth? $\endgroup$
    – JDługosz
    Apr 21, 2016 at 10:22
  • $\begingroup$ @JDługosz If you avoided hitting anything you would return to the same point in your 'orbit' after a brief visit to the antipodes. $\endgroup$
    – richardb
    Apr 21, 2016 at 13:33
  • $\begingroup$ I don't see how that's relevant, since the effect is for all future times and not related to the orbital period. $\endgroup$
    – JDługosz
    Apr 21, 2016 at 14:55

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