# How can the core of the Earth be younger than the crust due to gravitational time dilation if the crust experience more force than the core?

Gravitational force decreases with depth under the surface of Earth, and at centre, the surrounding forces cancel each other out effectively making it zero and hence a body at the center of gravity of the Earth (which would be very close to the centre of the Earth, the core) would experience weightlessness. Now the famous study in 2016 (https://arxiv.org/abs/1604.05507) says that the core is 2.5 years younger than the crust even though the former was formed first, due to gravitational time dilation causing the core to age slower even though the gravity acting on the core should be lower than what is on the crust. I understand that I might be confusing myself between two scientific principles but I want to understand both and know where my understanding is wrong.

• Gravitational time dilation is due to the gravitational potential, not the field strength. See physics.stackexchange.com/q/276522/123208 Dec 16, 2019 at 10:10
• I think this is a great question, very interesting as it really brings out a difference between verbal language and physics notation that bears on the subjective aspect of time, although I'd say "newer" rather than "younger", partly to associate it more clearly with such bouncing cosmologies as Poplawski's "cosmology with torsion" (which can be visualized with arrows of time pointing inward to the center of a spherical volume), and partly to to avoid further complications thru biology, what with there being little or no evidence that the earth is animated. Dec 16, 2019 at 13:00

Assuming the current understanding of Earths composition and formation , according to GR this is true up to the exact number of days/ years. I remember Feynman making a similar remark as an example of application of relativity. Let's say you are born in the center of a massive body and moved to the surface. Eventually as time is running more slowly for you, you would overtake your parents in age given enough time and relativity to work.

• Not to knock Feynman, but there would've probably been some problems with heat that would've left "your" humanity much in question by the time "you" would've steamed thru some vent into the surface. Which is itself pretty interesting, given the usual association between thermodynamics and time. So I've upvoted it as a "substantially" good answer (if not a perfect one), even while I'm lacking the competence to compare it with its only current competitor, especially with PM2ring's comment having been deleted before I saw it. Dec 16, 2019 at 13:12
• @Edouard For what it's worth, as of now there aren't any deleted comments on this question or its answers.
– rob
Dec 16, 2019 at 13:44
• I don't think this is an extct analogy Feynman but he did definitely mention the core would be younger than the surface. Dec 16, 2019 at 17:21

The confusion comes from a common misinterpretation that "time dilation is caused by gravity". In fact, the relation is the opposite. Time dilation, as as part of spacetime curvature, causes gravitational attraction. In a nutshell, while moving forward in time, things shift to where time runs slower, so (in a simplified view) gravity is defined by the gradient of the time dilation. This gradient is zero at the center of the Earth, so the gravitational attraction there is zero despite the time dilation being maximal.

Another common misconception is a misinterpretation of the Brinkhoff theorem, a relativistic version of the Newton shell theorem. The Newton theorem says that gravity caused by an empty massive spherical shell is zero inside, so gravity there is defined only by masses present inside. However, in General Relativity, spacetime inside such a shell s defined by both the masses inside and by the shell. For example, if the time dilation at the (inner) surface of the shell is such that time there runs twice slower than at infinity, then the time dilation inside the shell is the same everywhere and, due to the zero gradient, there is no gravitational force anywhere inside despite all things there agng twice slower.

• Why would time dilation be grethest in the center? Dec 16, 2019 at 17:42
• @Schaurberger Time dilation is a part of spacetime curvature created by mass (or more precisely by stress-energy-momentum). It is the highest where the concentration of mass is maximal. Since the center is surrounded by mass from all sides, the spacetime curvature (including the time dilation) there is the greatest. However, as explained in the answer, the gravitational attraction is defined not by time dilation, but by its gradient (derivative along a direction, e.g. radial), which is the greatest at the surface and zero at the center. Dec 16, 2019 at 20:21
• Fc= Fs'(1 − ∆Φ/ c 2 ), this is the explic relation between time (frequency) dilation and gravitational potential. There is no pressure or momentum here. Change in potential gives the change in the rate clocks tick. Dec 16, 2019 at 21:47
• @Schaurberger What you are saying is correct, but there is no contradiction. The potential and time dilation are directly related and (up to a simple conversion formula) have the same physical meaning. In turn, stress-energy (or, simply saying, mass) is what causes the time dilation (or equivalently the potential) to exist. While the Schwarzschild solution describes spacetime in vacuum (e.g. outside of the Earth), the initial source of the curvature is the mass (stress-energy) of the Earth, so the curvature, time dilation, and potential all have peak values at the center. Dec 17, 2019 at 4:36

Possible answer: Gravity inside a spherical body with radius $$a$$ and constant density $$\rho$$ obeys the metric $$d{{s}^{2}}=\left( 1-(1/3)\rho a^2 \right)d{{t}^{2}}-\frac{d{{r}^{2}}}{1-(1/3)\rho r^2}-{{r}^{2}}(d{{\theta }^{2}}+{{\sin }^{2}}\theta \,d{{\phi }^{2}}),$$ when choosing units of measurement convert the light velocity $$c$$ and the gravitational constant $$G$$ into $$c=8\pi G =1$$ Cambridge Open Engage. This approach is based on the assumption that distributed matter excites a negative vacuum pressure $$p = -(1/3)\rho$$ (see also Rom. Rep. Phys. 72, 113 (2020)). In standard units, this metric has the form $$d{{s}^{2}}=\left( 1-\frac{8\pi G}{3c^2}\rho a^2 \right)d{{t}^{2}}-\frac{d{{r}^{2}}}{1-\frac{8\pi G}{3c^2}\rho r^2}-{{r}^{2}}(d{{\theta }^{2}}+{{\sin }^{2}}\theta \,d{{\phi }^{2}}).$$ The flow of time inside the sphere is constant and coincides with one on the surface. Time flows more slowly than at a distance from the sphere in $$\left( 1-\frac{8\pi G}{3c^2}\rho a^2 \right)^{1/2}$$ times. The Earth consists of several regions, the density in which differs significantly. We consider the core density to be approximately $$\rho_{co}=11000 kg/m^3$$ and the density of the outer region, which includes the mantle and crust, $$\rho_{out}=4400 kg/m^3$$. In this case, the gravitational field inside the Earth will be the sum of the fields of two spheres with $$\rho_1=\rho_{out}, a_1=a_E=6378 km$$ and $$\rho_2=\rho_{co}-\rho_{out}=6600 kg/m^3, a_2=a_{co}=3488 km$$. The effect of time dilation in the center of the Earth will be the total of 2 spheres. The relative difference between the time in the absence of gravity and in the center of the Earth will be $$8.05*10^{-10}$$. The course of time in the center will differ little from the course of time in the entire region of the core. In the region of the mantle, it will begin to approach the surface. The relative difference between the clock readings in the center of the Earth and on the surface is $$1.09*10^{-10}$$.

According to a paper at the following link: https://ui.adsabs.harvard.edu/abs/2007AGUFM.V33A1161P/abstract the earth's crust is 2.77X10^22 kg. The mass of the core, from information obtained on the structure of the earth article on Wikipedia (https://en.wikipedia.org/wiki/Structure_of_the_Earth#Core) is 2.15X10^24 kg, approximately. (I'll let you calculate this number and check my work. You'll want the radii of the inner and outer core, the formula for volume of a sphere, and the lower limits on density of the inner and outer core, then sum the total masses of the inner and outer core together). Even if my math is off by just a little bit, the point is that the earth's core is more massive than the crust by at least 2 orders of magnitude. This means that the gravitational force near the core will be quite a bit stronger than at the surface of the crust, so yes, the crust would be older than the core, as time dilation from the higher gravitational field, means that less time will have passed for the core, than for the crust.

I hope this helps, and also, if I didn't format the math correctly, could someone be so kind as to fix my post so that the math is coded correctly? Thanks.

• So is the gravitational force on the surface of the core higher than the gravitational force on the crust? Would the distance matter more than the change in mass? Dec 16, 2019 at 10:38
• I haven't taken the difference in composition into consideration but still, on the surface it wouldn't just be the crust that contributes to the mass but Earth as a whole. I did some calculations by using the universal equation for gravitation, taking the mass of the Earth and radius of Earth on one side and mass of the core and radius of the core on the other and arrived at a conclusion that the force of gravity at the surface of the core would be around 10 times greater than that of the force on Earth's surface. Dec 16, 2019 at 11:02
• This answer is wrong on several levels: 1.) Inside a spherical mass distribution, famously, the gravitational acceleration increases linearly with distance to the core. Gauss's theorem, physics 101. 2.) How do you get the idea of only calculating the effect of the crust when standing on the crust? That's cherrypicking if I've ever seen some. 3.) As @PM2Ring's comment correct points out gravitational time dilation has to do with the potential, not with the force. Dec 16, 2019 at 11:10
• I went back over the paper and saw their reasoning. I was wrong with my answer. I would recommend checking out the book Gravity by J.B. Hartle for more information, which can be found here: archive.org/details/… Dec 16, 2019 at 14:04
• In that case, you should either correct your answer, or delete it. Dec 17, 2019 at 14:06