I would like to know what happens with time dilation (relative to surface) at earth's center .

There is a way to calculate it?

Is time going faster at center of earth?

I've made other questions about this matter and the answers refers to:

$\Delta\Phi$ (difference in Newtonian gravitational potential between the locations) as directly related, but I think those equation can't be applied to this because were derived for the vecinity of a mass but not inside it.

Any clues? Thanks

  • $\begingroup$ May I suggest you spend some time reading about potentials in classical mechanics. Throwing off lines like "but I think those equation can't be applied to this because were derived for the vecinity of a mass but not inside it" does nothing to improve you reception here, and these issues are addressed in every textbook on the subject. Or at least ask questions about the subject (potential) you don't understand instead of speculating wildly. Please. $\endgroup$ May 19, 2011 at 15:43
  • $\begingroup$ @dmckee yes, you are right, and I do it whenever I can, I spend the time to go deeper, but there are a lot of interesting topics luckly and althought I am able to have the doubt, and even sometimes to understand an answer, but (life itself) give no so much time to study them all, so thanks to internet and to people like everyone here, one can know things that there is no other way to reach, same sense I help people in other areas I can dedicate more time, thanks $\endgroup$
    – HDE
    May 19, 2011 at 15:58

3 Answers 3


The rule I mentioned in another question, that the time dilation factor is $1+\Delta\Phi/c^2$, applies here. The derivation (found in various textbooks) depends only on the assumptions that fields are weak and matter is nonrelativistic, both of which are true for the Earth.

Modeling the Earth as a uniform-density sphere (not true, of course, but I don't care), we find that $g(r)=GMr/R^3$ where $R$ is the radius of the Earth. So $$ \Delta\Phi={GM\over R^3}\int_0^Rr\,dr={GM\over 2R}. $$ That means that $$ {\Delta\Phi\over c^2}={GM\over 2Rc^2}={1\over 4}{R_s\over R}. $$ Here $R_s=2GM/c^2$ is the Schwarzschild radius corresponding to the Earth's mass. Numerically, $R_s$ is about 9 mm, and $R$ is about 6400 km, so $\Delta\Phi/c^2=3\times 10^{-10}$.

The sign of the effect is that clocks tick slower when they're deeper in the potential well. That is, a clock at the Earth's surface ticks 1.0000000003 times faster than one at the center.

  • $\begingroup$ This seems a very simple approach, M/R define the ratio, a sort of "density", there is "an order of magnitude" (simply x10) of difference with Luboš Motl answer, but at least both answer said time inside earth would be slower than in the surface $\endgroup$
    – HDE
    May 19, 2011 at 15:46
  • 2
    $\begingroup$ Actually, the difference is a factor of 3 ($3\times 10^{-10}$ vs $10^{-9}$). The difference is because Lubos is calculating the potential difference between the center and infinity, whereas I calculate the difference between center and surface. That's precisely a factor of 3. $\endgroup$
    – Ted Bunn
    May 19, 2011 at 17:31
  • $\begingroup$ sorry I'd compared $10^{-10} vs 10^{-9}$, that's why I saw a x10 factor, now I see a 10/3 =3.3333.. factor $\endgroup$
    – HDE
    May 19, 2011 at 18:57
  • $\begingroup$ I thought if the Earth is a uniformly dense sphere the gravity at the center is zero. $\endgroup$ May 20, 2013 at 21:58
  • 3
    $\begingroup$ @BrandonEnright: Time dilation depends on the potential, not the field. $\endgroup$
    – user4552
    May 20, 2013 at 22:01

Dear HDE, it's not hard to estimate the gravitational potential at the Earth's center. Of course, it's smooth. Let me assume that the Earth's mass density is uniform which is an OK estimate - up to factors of two or so.

The gravitational acceleration at distance $R$ from the center is $GM/R^2$ if $R$ is greater than the Earth's radius $R_E$. However, for smaller values of $R$, you have to use Gauss' law $$\int d\vec S\cdot \vec g \sim GM_{inside}$$ and determine the total mass inside a smaller sphere. Because $M_{inside}$ goes like $R^3$ for $R<R_E$, and this $R^3$ is still divided by $R^2$ from $\int d\vec S$, it follows that the gravitational acceleration inside the Earth is pretty much proportional to $R$: $$ g(R) = g(R_E)\cdot \frac{R}{R_E} $$ In particular, the gravitational acceleration at the Earth's center is zero and near the center, a particle would oscillate like in a harmonic oscillator, $\vec F\sim -k\vec x$.

It's also trivial to calculate the extra decrease of the gravitational potential you get if you go from the surface to the center. On the surface, the gravitational potential is $-GM/R_E$, as you know, because the derivative of $-GM/R$ over $R$ gives the right acceleration. However, the potential is getting even more negative. If you integrate $g(R_E)\cdot R/R_E$ over $R$ from $0$ to $R_E$, you will get $g(R_E) R_E/2$. This has to be taken with the negative sign.

So the potential at the center, assuming uniformity, is $$ \Phi = -\frac{GM}{R_E} - g(R_E) \frac{R_E}{2} = -\frac 32 \frac{GM}{R_E} = -\frac 32 g(R_E) R_E $$ This gravitational potential determines the slowing of time, too. In SI units, $g(R_E)=10$ Newtons per meter and $R_E=6,378,000$. The product, with the $3/2$ factor added, is almost exactly $10^{8}$. Divide it by $c^2=10^{17}$ to get about $10^{-9}$ - the relative red shift from the center of the Earth to infinity.

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.

  • 5
    $\begingroup$ A small note: the question asks about the relative time dilation between the center and surface of the Earth, whereas the $\Phi$ you calculate is the potential of the center relative to infinity. That accounts for the factor of 3 difference between your answer and mine. $\endgroup$
    – Ted Bunn
    May 19, 2011 at 17:32

The CENTER of the earth will not have more gravity, but less. This is because half the mass will be "above" half "below" ( Regardless of orientation)...Sort of less g and in different direction/vectors. The thing is mass is not concentrated at a spot in the center with more and more g as one moves closer to the center. As you tunneled down, some of the mass, more and more would be behind you). More time dilation on surface... Where g is stronger. Time would not be slower at center of earth.

  • 4
    $\begingroup$ Time dilation depends on the potential, not the field. $\endgroup$
    – user4552
    May 20, 2013 at 22:00
  • 1
    $\begingroup$ While I understand you mean potential as distance from source, and your point that there is no dilation unless this potential is not the same for the two clocks. I do not think this is a point of confusion, and do not see why this comment is particularly relevant. Please explain. $\endgroup$
    – Tuesday
    May 21, 2013 at 15:57
  • $\begingroup$ the center of the earth has no gravitational attraction... so according to you, time should not pass there. $\endgroup$
    – udiboy1209
    Jul 7, 2013 at 6:40
  • $\begingroup$ @udiboy: No, that would mean that a Minkowski spacetime is equivalent to a black hole. Weird duality...(!) $\endgroup$ Sep 29, 2013 at 8:39

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