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[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.

Qmechanic
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Lloeki
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  • 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. – innisfree Apr 20 '16 at 09:43
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    Related: http://physics.stackexchange.com/q/3009/2451 and links therein. – Qmechanic Apr 20 '16 at 10:50

4 Answers4

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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.

  • 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). – Nathan Reed Apr 21 '16 at 00:31
  • 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. – Giorgio Comitini Apr 21 '16 at 08:10
  • 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). – Giorgio Comitini Apr 21 '16 at 08:16
  • We always define things as to strictly minimize the action and have a positive-definite energy. – Giorgio Comitini Apr 21 '16 at 08:21
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    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). – Nathan Reed Apr 21 '16 at 16:21
  • @NathanReed Let me elaborate on what I meant. The (mathematically) simpler set up for the twin paradox, as you know, is that which uses a piecewise geodesic for the traveling brother: one for the outward voyage and one for the return voyage. A piecewise geodesic of course is not strictly a geodesic, in the sense that it doesn't solve the geodesic (autoparallel) equation: its solutions must have continuous derivatives (i.e. velocities), which is not true in this case. Nonetheless, each piece of the curve is a geodesic and, as such, it minimizes proper time. (Continues) – Giorgio Comitini Apr 21 '16 at 17:14
  • The paradox itself lays in the fact that the twins age differently when traveling at different speeds. In order to check this, one uses the fact that the twins start and end at the same spacetime point, so that they can exchange information on the proper times they measured. Imagine though that at the traveling twin's turning point there was a third twin traveling on a curve that, after passing that point, reproduces the return trip of the second twin, just as if the two were running a relay race. (Continues) – Giorgio Comitini Apr 21 '16 at 17:28
  • This way, there is no need for the second twin to turn around and go back to the (spatially) starting point. In particular, each of the three twins can be taken to follow three different true (i.e. not piecewise) geodesics: the second one can just keep on going without turning back, while the third one can be taken to have been traveling at the same speed as he is after the encounter with the second one. As the world lines of the second and the third twins intersect, they can exchange information in such a way that the proper time of the second is known by the third. (Continues) – Giorgio Comitini Apr 21 '16 at 17:36
  • The basic information the third needs in order to know the proper time of the second is how fast the latter is moving with respect to his own frame of reference. The fact that the three geodesics intersect, together with the information that the three of them are are moving along geodesics, the proper time of the first between the second leaving and the third arriving and the proper time of the third twin between his encounter with the second and the first, given a basic knowledge of special relativity, is enough to calculate the proper time elapsed on the piecewise-geodesic curve. – Giorgio Comitini Apr 21 '16 at 17:43
  • Of course, we postulate (as always) that at the intersection of two world lines the observers can measure each other's velocities. This way, the first twin will find that his proper time was longer than that elapsed on the piecewise geodesic, the paradox is still a paradox, but we are only using geodesics in order to test it. What I meant when I said that the paradox has nothing to do with the maximization/minimization of proper time is basically the statement that the paradox can be formulated by using true geodesics only, removing the singularity at the turning point. – Giorgio Comitini Apr 21 '16 at 17:52
  • It is a subtlety after all. Of course what you are really measuring in the end is proper time on a non-truly geodesic curve, which of course is not maximum. – Giorgio Comitini Apr 21 '16 at 17:53
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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.

  • Can you explain in which sense it's the longest path? – Roman Starkov Apr 20 '16 at 20:53
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    @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, –  Apr 20 '16 at 22:15
  • 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). – Giorgio Comitini Apr 21 '16 at 07:59
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    @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. –  Apr 21 '16 at 08:46
  • @tfb Read my comments to my answer, there I explained in what sense my statement was to be read. – Giorgio Comitini Apr 21 '16 at 17:55
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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".

Bence Racskó
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    Quibble: Pauli exclusion is not "EM" force. – JDługosz Apr 21 '16 at 10:24
  • @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... – Luaan Apr 22 '16 at 07:59
  • 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. – JDługosz Apr 22 '16 at 12:43
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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."

Johannes
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  • How will you end up "here" if you are able to fall through to the center of the Earth? – JDługosz Apr 21 '16 at 10:22
  • @JDługosz If you avoided hitting anything you would return to the same point in your 'orbit' after a brief visit to the antipodes. – richardb Apr 21 '16 at 13:33
  • I don't see how that's relevant, since the effect is for all future times and not related to the orbital period. – JDługosz Apr 21 '16 at 14:55