# Time dilation without curvature at the center of mass, how is that possible?

What is the general relativity explanation for why objects at the center of the Earth are weightless?

And John Rennie's answer where he says:

"When $r = 0$ the Christoffel symbol $\Gamma_{tt}^r$ is zero and that means the radial four-acceleration is zero and that means you're weightless."

And this Question:

Gravitational time dilation at the earth's center

And Luboš Motl's answer where he says:

"If you spend 1 billion years at the center of the Earth, your twin brother outside the gravitational field will get 1 billion and one years older. If you wish, you may interpret it by saying that it's healthy to live at the center of the Earth. Good luck."

So what the two are saying is that at the center of mass:

• there is time dilation
• there is no curvature

I read and understand what they are writing about the Christoffel symbol

$$\Gamma_{tt}^r= \frac{r}{2R^6}\left[2M^2r^2+MR^3\left(3\sqrt{1-\frac{2Mr^2}{R^3}}\sqrt{1-\frac{2M}{R}}-1\right)\right]$$

And the potential at the center, assuming uniformity, is

$$\Phi = -\frac{GM}{R_E} - g(R_E) \frac{R_E}{2} = -\frac 32 \frac{GM}{R_E} = -\frac 32 g(R_E) R_E$$

And that

"This gravitational potential determines the slowing of time, too."

And somehow I just cannot understand how it is possible to have time dilation without curvature. I thought that time dilation is caused by (in GR, not SR) curvature, which is the effect of gravity, the gravitational potential.

I don't understand how there can be gravitational potential without curvature, I thought the only effect gravity has is the curvature of spacetime.

1. Can somebody please explain how (with math and with simple way too) how that is possible together? How can there be gravitational potential and time dilation without curvature?

2. Isn't gravity's only effect the curvature? Is there another effect with what gravity can create potential (so without creating curvature) and so create time dilation?

• "I thought the only effect gravity has is the curvature of spacetime. " - it's not that gravity curves spacetime, spacetime curvature is gravity. – Alfred Centauri Dec 14 '16 at 22:19
• why did I get a downvote on this? – Árpád Szendrei Dec 15 '16 at 20:04
• Can somebody please give me a formula for the curvature at the center? So I can see in the math that it is not zero? – Árpád Szendrei Dec 15 '16 at 20:38

Time dilation is not a local property. That is, you can't just take a point in space and say what it's time dilation is. Time dilation is always relative to some other point in space.

Suppose we take two points $A$ and $B$, then time dilation means that for the coordinate time $t$ we have:

$$\frac{dt}{d\tau}(\text{at point A}) \ne \frac{dt}{d\tau}(\text{at point B})$$

and the time dilation isn't determined by what happens at $A$ or what happens at $B$, but by what happens in between $A$ and $B$. In particular it's possible for spacetime to be flat at point $A$ and flat at point $B$, and there can still be time dilation if the spacetime is curved in between points $A$ and $B$.

To take a concrete example lets consider a thin spherical shell of mass $M$ and radius $D$. Outside the shell the metric is simply the Schwarzschild metric, so if we go out to $r=\infty$ spacetime is flat.

Inside the shell the metric is:

$$ds^2 = -\left(1 - \frac{2GM}{c^2D}\right) c^2dt^2 + \frac{dr^2}{1 - \frac{2GM}{c^2D}}$$

where for simplicity I'll leaving off the angular bit since we dealing only with spherically symmetric systems. At first glance this looks curved, but we can define new time and radial coordinates:

\begin{align} T &= \sqrt{1 - \frac{2GM}{c^2D}}\,t \\ R &= \frac{r}{\sqrt{1 - \frac{2GM}{c^2D}}} \end{align}

and using these coordinates the metric becomes:

$$ds^2 = -c^2dT^2 + dR^2$$

and this is just the Minkowski metric in polar coordinates. So spacetime is flat inside the spherical shell, which should come as no surprise since it's just Newton's shell theorem.

However even though spacetime is flat inside the shell, the flat time coordinate $T$ is not the same as the time coordinate $t$ outside the shell at $r=\infty$. They differ by a factor of $\sqrt{1-2GM/c^2D}$, and that means the time inside the shell is dilated i.e. runs slower, than the time at infinity:

$$\frac{dT}{dt} = \sqrt{1-\frac{2GM}{c^2D}}$$

This gives you an example of two points, both in flat spacetime, that have a relative time dilation. This difference in the time coordinates is due to the spacetime curvature caused by the shell in between the points inside and outside the shell.

• wow! thank you! best answer, you are correct, but #1 it is in contradiction with other answers, where they say "there is curvature at the center of mass". So is there curvature at the center of mass or not? (from your answer I guess there is NOT) – Árpád Szendrei Dec 15 '16 at 17:59
• @ÁrpádSzendrei this is for a spherical shell. The Riemann tensor is zero inside a spherical shell however it is not zero inside a solid sphere like the Earth. – John Rennie Dec 15 '16 at 18:01
• #2 Now I understand that time dilation is caused by the 'difference' caused by the curvature INBETWEEN the points although BOTH points are in flat spacetime LOCALLY correct? (Which is logical because the shell changes the way of causality inbetween the two points, so according to SR there has to be time dilation) What confused me was, that there was weightlessness at the center point. And I thought that weightlessness is caused by ZERO CURVATURE. Now what I do not have an answer for yet, is in the point of center of mass case, if there is CURVATURE, then what causes weightlessness? – Árpád Szendrei Dec 15 '16 at 18:02
• @ÁrpádSzendrei a stationary particle is weightless if all the Christoffel symbols $\Gamma^\alpha{}_{tt}$ are zero. However this does not mean the Riemann tensor has to be zero as well. It's possible for the Riemann tensor to be non-zero, i.e. there is curvature, and the Christoffel symbols to be zero. This is what happens at the centre of the Earth. – John Rennie Dec 15 '16 at 18:06
• @ÁrpádSzendrei: see Getting started self-studying general relativity – John Rennie Dec 15 '16 at 18:18

Equation (1) in John Rennie's answer is

$$ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\sqrt{1-\frac{2Mr^2}{R^3}}\right]^2dt^2 + \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)} + r^2 d\Omega^2 \tag{1}$$

which is, we are told, the Schwarzschild interior metric for a spherical body of uniform density. At the center, where $$r=0$$, this line element becomes

$$ds^2 = -\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\right]^2dt^2 + \frac{dr^2}{\left(1-\frac{2Mr^2}{R^3}\right)}$$

Assuming one remains at $$r=0$$, we have that

$$d\tau = \left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\right]\,dt$$

so, a clock at the center runs slower than a clock 'at infinity' by a factor of

$$\frac{d\tau}{dt}=\left[\frac{3}{2}\sqrt{1-\frac{2M}{R}} - \frac{1}{2}\right]$$

So, there is time dilation at the center according to this line element. Note that for $$M \rightarrow 0$$, the clock at the center runs at the same rate as the clock at infinity.

It's not clear to me what you mean by "there is no curvature". Spacetime curvature is manifest in geodesic deviation which is to say that, if there's curvature, two nearby geodesics that are initially parallel will not remain parallel.

And that is the case here. A particle that remains at $$r=0$$ is on a geodesic. A nearby freely falling particle at $$r=\epsilon$$, initially at rest with respect to the other particle, will fall towards the other particle which means there is curvature.

What I do not understand is #1 is the particle at rest at the center really on a geodesic, then why is it not moving in space (if it is not moving in space, that means it is still 'moving' in the time dimension at speed c, I understand that)

A timelike geodesic $$x^\mu(\tau)$$ satisfies the geodesic equation

$$\frac{d^2x^\mu}{d\tau^2}=-\Gamma^\mu_{\;\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau}$$

Consider the world line of a (massive) particle given by $$x'^\mu(\tau)=(t,0,0,0)$$. Since $$\frac{d^2x'^\mu}{d\tau^2}=0$$, then $$x'^\mu(\tau)$$ is a geodesic if $$\Gamma^\mu_{\;tt}=\frac{1}{2}g^{\mu\nu}(g_{\nu t,t}+g_{\nu t,t}-g_{tt,\nu})|_{r=0}=0$$

By inspection, the metric depends only on the coordinate $$r$$ and so we need to evaluate

$$\Gamma^r_{\;tt}=-\frac{1}{2}g^{\mu\nu}g_{tt,r}|_{r=0}$$

But, in John Rennie's answer, this evaluates to zero thus a world line given by $$(t,0,0,0)$$ is a geodesic; a particle that remains at $$r=0$$ is in free fall.

• Both great answers. I like this one because this tells me that :" A particle that remains at r=0r=0 is on a geodesic. A nearby freely falling particle at r=ϵr=ϵ, initially at rest with respect to the other particle, will fall towards the other particle which means there is curvature." So do I understand correctly, that BOTH the particle at r=0 at rest and the other nearby particle freely falling, they are BOTH on geodesics? On different geodesics but as the nearby particle starts moving toward the one in the center, THAT really means that there is curvature AROUND the center right? – Árpád Szendrei Dec 15 '16 at 17:24
• And because the two geodesics do not remain parallel (because they start moving toward each other), that means there is curvature AROUND the center right? What I do not understand is #1 is the particle at rest at the center really on a geodesic, then why is it not moving in space (if it is not moving in space, that means it is still 'moving' in the time dimension at speed c, I understand that)? So how can a geodesic mean being at rest in space? Does this mean that the center point is a special geodesic, different from allother geodesics (cause particles on other geodesics will move in space)? – Árpád Szendrei Dec 15 '16 at 17:29
• #2 do I understand correctly, that you are talking about the curvature AROUND the center? Because if you choose the two particles, being BOTH at the center (so two photons, that do not obey the Pauli exclusion), they will remain parallel right? – Árpád Szendrei Dec 15 '16 at 17:33
• Thank you for the edit! Can you please give me a formula for the curvature at the center? So I can see in the math that it is not zero? – Árpád Szendrei Dec 15 '16 at 20:39

I'd like to add a comment to Alfred's correct answer and all the correct answers you cite.

Something to emphasize here, and which I believe is causing problems for your understanding, is that weightlessness and curvature are about different terms in the Taylor series for the metric (and other geometric properties) of spacetime when we expand these objects in geodesic (Riemann normal) co-ordinates. Weightlessness is a first order notion: you can understand it by expanding the Taylor series to first order. Curvature, on the other hand, is a second order or (as it is often called) a tidal notion.

At any point in spacetime (curvature present or not), we can always make it look Minkowskian if we restrict ourselves to a small enough neighborhood of that point: that's just part and parcel of the manifold as the geometric object describing spacetime in GTR. And it is in this restricted view that we define weightlessness: there is always has a momentarily comoving inertial frame (a particular choice of basis for the tangent space) wherein an observer is weightless; but in this case when people say things are weightless at the Earth's center, what they are emphasizing is that this comoving frame is stationary relative to the Earth's center, i.e. relative to the stuff that makes the Earth up.

The curvature / geodesic deviation is about how the true picture deviates from this localized one as we shift over length / timescales greater than the ones whereover the first order picture, given by the linear tangent space, holds good. So these tidal effects only show up in the second order terms of the relevant Taylor series.

As for any Taylor series, I hope that you can understand that the first and second order co-efficients can be set wholly independently.

• Great answer thank you. So do I understand correctly, that John Rennie says only that at the center point, you are weightless (that is caused by the first order of the metric), but what he says DOES NOT mean that there is no curvature? Then what I do not understand is, where he says: "where the $\Gamma^\mu_{\alpha\beta}$ are called [Christoffel symbols][5] (of the second kind) and they depend on the spacetime curvature. ". – Árpád Szendrei Dec 15 '16 at 17:39
• So then he says "$$\Gamma_{tt}^r= \frac{r}{2R^6}\left[2M^2r^2+MR^3\left(3\sqrt{1-\frac{2Mr^2}{R^3}}\sqrt{1-\frac{2M}{R}}-1\right)\right]$$ Yet another fiendishly complicated expression, but note that the whole thing is multiplied by $r/2R^6$ and that means if $r = 0$ the whole complicated expression is just zero." So he says that the Christoffel symbols are 0. So if the Christoffel symbols are 0, and they mean the curvature itself, then is he saying that the curvature is 0 at r=0 or not? – Árpád Szendrei Dec 15 '16 at 17:40
• @ÁrpádSzendrei the curvature is described by the Riemann tensor not by the Christoffel symbols (though they can be derived from each other). The Riemann tensor is not zero at the centre of the sphere. – John Rennie Dec 15 '16 at 17:44
• Thank you! So if there is curvature at the center, then what causes weightlessness? – Árpád Szendrei Dec 15 '16 at 18:50
• @ÁrpádSzendrei, in a locally inertial frame at $r=0$, the Riemann tensor is $$R^\alpha_{\;\beta\mu\nu}=\frac{1}{2}g^{\alpha\sigma}(g_{\sigma\nu,\beta\mu}-g_{\sigma\mu,\beta\nu}+g_{\beta\mu,\sigma\nu}-g_{\beta\nu,\sigma\mu})$$ evaluated at $r=0$. Note that this involves 2nd and mixed partials of the metric. Since the metric depends on the coordinate $r$ alone, only the $,rr$ derivatives are non-zero. – Alfred Centauri Dec 15 '16 at 22:14