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The young centre of the Earth (2016) (arxiv.org)
121 points by ungerik on Nov 5, 2019 | hide | past | favorite | 71 comments



How is it possible that the core is several years younger if the core and the surface have gone around the sun the same number of times?


Time isn't universal, it doesn't tick for everybody at the same rate, in fact every particle has its own unique clock.

One thing that changes how your clock flows is being in a gravity well. We have clocks which can measure it over just a few feet of height difference. Gravity isn't so much a "force" but the bending of space and time. Clocks moving differently is the result.


> We have clocks which can measure it over just a few feet of height difference

Wait, seriously? Do you have a link??


Measuring the height of a mountain with a portable atomic clock (1000m accuracy) https://www.nature.com/articles/s41567-017-0042-3

NIST measuring speed differences < 10 m/s and height < 1 m with a pair of atomic clocks https://science.sciencemag.org/content/329/5999/1630.long


Tests of gravitational time dilation are also within the reach range of enthusiastic amateurs using second-hand atomic clocks:

Part 4 (starting pdf page 44) of http://web.stanford.edu/group/scpnt/pnt/PNT18/presentation_f... (slides)

Or if you prefer non-PDF:

http://leapsecond.com/great2005/

http://leapsecond.com/great2016a/


Amazing. They observe fractional frequency changes of around 4.1E-17. Thanks for sharing the links!


Couldn't see that number in the info at the link above, is that from the full article?


Yeah: "The two measurements consist of approximately 100,000 s of low-height data and 40,000 s of high-height data, and the clocks exhibit (Fig. 3) a fractional frequency change of (4.1 ± 1.6) × 10^−17."


Mass, gravity and general relativity. It’s a fascinating and very counter-intuitive at some moments, such as this.

Same goes for people that stay on the moon, when they go back to earth their watches are slightly different than those on earth. GPS satellites need to be synced with earth clock periodically as well for this reason.


The GPS satellites have multiple general relativity effects on their definition of time. Since they're moving so fast, the internal clocks on the satellites move slower than Earth (about 7 microseconds a day), while the decreased gravity on the satellites causes the internal clocks to speed up compared to Earth (45 microseconds a day). Net effect is a speed up of 38 microseconds.

Also, the on-board chips are pre-programmed to take into account the relativistic effects and compensate when doing their calculations.


In addition to the other excellent comments about general relativity, a "year" is only valid in expressing the time it takes to go around the Sun right NOW, not the entire existence of the Earth. The Earth's orbit has shifted throughout the lifetime of the solar system. A year circa 1000000 B.C. isn't the same as the year circa recorded history if we are considering it one revolution. so that number wouldn't represent the number of times it's gone around the Sun even excluding relativistic effects.


> The Earth's orbit has shifted throughout the lifetime of the solar system.

I'm by no means qualified to state this as fact but also I imagine that as we travel through space and hit random stuff in space (meteors, particles from stars, comet trails etc) we ever so slightly change our speed outside of the gravitational influences of other bodies, it would probably be incredibly small fractions of a second over centuries but it's something.

I wonder how much of an effect something like an extinction event asteroid on the speed of the earth both around the sun and on its rotation on its axis.


https://en.wikipedia.org/wiki/Gravitational_time_dilation

Now a detail that really makes things more interesting is that gravity is not constant upon the surface ( https://svs.gsfc.nasa.gov/11234 ), due to mass variations. So with that, it would be interesting to work out the age difference as a surface map. Though would need to historically adjust for tectonic shifts and other effects upon the surface mass. Of course the difference won't be as large, but still, I'd dare say that a day or so, would be more than viable upon the scale of things.


The core and the surface have gone around the Sun the same number of times, but a clock at the core would record less elapsed time for each orbit around the Sun than a clock at the surface.


General Relativity


Parent comment is making a joke about “year” as a unit measuring how many times you go around the sun. Less time has passed for the core per relativity, but is it not the same number of years?

The more official definition of years defines it in terms of days, which break down further to SI’s rigorously defined second as the base unit of time. But it’s still weird to think about.


Unless the second is defined on a Hubble flow co-moving observer, it's ill-defined enough to be ambiguous by at least 2 years out of 4 billion.


Yes, you wouldn't be able to measure it accurately, but it's nonetheless a static amount of time unrelated to the Earth's orbital period.

I'm not an astrophysicist, but I'd guess the Earth's orbital period isn't quite the same as when the planet first formed either. So using today's defined measurement of "years", the planet's age and the number of times it's been around the sun might not match up at the surface either. The impact that formed the moon must have changed our velocity a bit, right? And the sun has been losing mass (into energy via fusion) since it first formed.


That explains how it can be ...does the conversion... 75 megaseconds younger, but the whole of the earth modulo ejected material like spacecraft should be the same number of _years_ old.



We no longer measure time as laps around the sun.


because spacetime is curved by mass the matter in the centres of the Earth and Sun computes at a slower rate than matter at their surfaces so that from our point of view the centres of the Earth and Sun are younger because their clocks have ticked fewer times compared to their surfaces


Universe is not heliocentric


Here is a more accessible explanation for those (like me) who are unfamiliar with the maths: https://www.sciencealert.com/earth-s-core-is-2-5-years-young...


From that article:

> If you want an even more stark illustration of gravity’s effect on time, take the Sun - the team calculates that the Sun’s core is around 40,000 years younger than its surface.

Wow!


> The pedagogical value of this discussion is to show students that any number or observation, no matter who brought it forward, must be critically examined.

Excellent, I hope this value is also encouraged more in schools now than when I was there.


What's the deepest core sample; 10 km? I assume isotope dating is no where near accurate enough to see the difference.


10 kms would not show any effect at all. The difference in gravity would be too low and mixing in the crust high.


12 km to be precise, of the earth’s 6000 odd km radius. That’s all we could accomplish in entire human history.

https://www.smithsonianmag.com/smithsonian-institution/ask-s...


Even if it were accurate enough, I think plate tectonics moves the crust around enough such that one would have a hard time finding a piece of rock that has been more or less always in the same depth. And I have no idea how static the core is, maybe even it undergoes some mixing process over long enough periods of time.


If there are any such stable deep-crust locations, they're most likely located under a continental location rather than oceanic, if my understanding is correct. Ocean basins reform at a higher rate (and hence: are more-recently-exposed crust) than continental plates.

The very oldest regions are two locations, presently in South Africa and Australia, previously joined, dating back several billions of years.

The Jack Hills region in Australia has been dated to 4.4 billions of years (the Earth itself is 4.5 billion):

https://www.nationalgeographic.com/news/2014/2/140224-oldest...

A region presently under Greenland was dated to 3.8 billion years, in 2007:

https://www.newscientist.com/article/dn11438-oldest-chunk-of...

The Nuvvuagittuq Greenstone Belt, near Hudson Bay, Canada, has been dated to between 3.7 and 4.3 billion years old:

https://en.wikipedia.org/wiki/Nuvvuagittuq_Greenstone_Belt

All of these are surface rock. My thought is that drilling from these locations might find very old subsurface structures as well. Though whether any relativistic time dilation could be observed is hard to say.


It seems like it would be the opposite:

- If you are smack dab in the middle of the Earth, although you'd be mashed, gravitational forces should be equal in all directions and therefore cancel out to zero.

- Ergo, if you are on the surface of the Earth, gravity should make your clocks run slower than in the middle.

What am I missing that makes this untrue?


The gravitational forces at the center do cancel out, but the important term for this effect is not the sum of forces but the gravitational potential, essentially the energy gained/lost by moving from deep space to the given location.

Force is the derivative of this potential, so the abscence of forces in the center indicates an extremum.


It's not the gravitational force that determines the relative passage of the time but the gravitational potential, which can be much higher despite not experiencing any force.


Oh, interesting. Thanks for the clarification!


What about the center of a black hole? Is it younger than the surface?


The question is slightly ill-formed. I understand why you'd ask it, but there are a couple of possible answers, none of which work directly.

- No, because a black hole has nothing underneath the event horizon. (This is true from the perspective of anyone outside it, at least to a first approximation. Gravity is affected by curvature, and whatever might be in the centre cannot be the cause of the event horizon, except historically, but the event horizon is stable in itself. It's a self-sustaining cascade of space-time.)

- Mu, because once you pass the event horizon the extreme warping of space-time turns "inwards" into "future", making the centre of the hole a point in time, not space. This makes it difficult to determine what the question is asking.

- Maybe, because singularities probably don't exist. There should be something in the centre, which might evolve over time. It might just be a faster-than-light cascade of space-time, though. We still can't answer your question, because we don't know how to accurately model the 'singularity'. (As anything other than a singularity, and that's essentially just the math giving up.)


> self-sustaining cascade of space-time

I am in awe of these words and fail to understand them in equal parts.


The way I usually explain it is that gravity is a form of energy, and that energy has mass, and that at the event horizon the energy of the gravity is exactly sufficient to maintain the gravity itself, basically becoming a self-powering loop of gravitational energy.


> the extreme warping of space-time turns "inwards" into "future", making the centre of the hole a point in time, not space.

Assuming an exact black hole solution to the Einstein Field Equations (EFEs), what's happening is that any observer sufficiently close to the singularity will inevitably collide with the singularity. One can then define the horizon the outer boundary at which the future light cone attached at every point ends in the singularity. In order to avoid the singularity in the future one would need a causal cone broader than that of the light cone. The boundary is only a surface in the abstract; it's not material, you can't touch it, nothing bounces off it, and the no-drama conjecture says that if you were to float through the horizon you won't immediately notice you've done so.

Outside the horizon as defined above, some part of the future light cone will have the singularity in it, however as one choose an observer at a point further and further away from the black hole, the singularity occupies a smaller and smaller part of the future light cone for that observer at that point, and the observer would have to aim ever more carefully to collide with it. (The observer simply has to reach a point at the horizon).

> a black hole has nothing underneath the event horizon

That's true of vacuum solutions of the EFEs which contain black holes, but it's not true in general. The Vaiyda metric is an exact solution which equips a spherically symmetrical central mass with spherically symmetric non-interacting dust of incoming and/or outgoing radiation: a Vaiyda black hole can have some incoming radiation within the horizon.

Astrophysical black holes are more complicated since the matter outside them is generally far from spherically symmetrical. However, within the horizon will be some of the cosmic microwave background's (CMB's) photon gas, and whatever else reaches the horizon.

But, you're right that something must be deep inside an astrophysical black hole, and the something is perhaps not a classical singularity, but rather some configuration that preserves the quantum mechanical properties of the field-content that reached the horizon.

> might evolve over time

If there's infalling matter, it must evolve over time, as the mass will increase, and likely so will the angular momentum, and charges are likely to go back-and-forth as one cannot guarantee that only chargless matter (e.g. photons) will fall in, nor that if we let charged matter fall in, it will always be charge-neutralized (e.g., no ions allowed!). That's true of a classical singularity, and it will be even more true for any ultradense non-singularity structure that preserves QFT information.

> faster-than-light cascade of space-time

What's that?

> Gravity is affected by curvature

In General Relativity, gravity is curvature.

> whatever might be in the centre cannot be the cause of the event horizon

Mass-energy generates the metric. In the case of the Schwarzschild black hole solution to the EFEs, the central mass (which is all the mass and energy in the vacuum Schwarzschild spacetime) is static and concentrated at one point. However, a non-rotating star, or a slowly-rotating object like the moon, generates a very close approximation of the Schwarzschild solution. Astrophysical black holes generate an approximation of the vacuum Schwarzschild black hole spacetime, but the Kerr metric is a better approximation for rotating black holes, and there is plenty of matter in the universe outside black holes. The true metric of the entire universe is much much more complicated than our (useful!) approximations, but nevertheless it is determined by the scatter of mass-energy throughout the spacetime. The true metric almost certianly describes regions which behave almost indistinguishably from the interiors of Schwarzschild (and Kerr, etc.) black holes.

Another way of looking at this: as one throws matter into a black hole, the mass of the black hole increases, and so therefore does the diameter of the black hole. If you allow matter to cross into the black hole and collide with the singularity (or whatever replaces it), how can you maintain the italicized part of the quote above?

> the event horizon is stable in itself

If you allow mass to increase by letting dust and gas and cosmic microwaves reach the horizon, how do you keep the horizon stable?

Moreover, if you merge two black holes, how do you keep both black holes' horizons stable?

> It's a self-sustaining

Given Hawking radiation then a black hole that's a solution to the Einstien Field Equations and equipped with one or more quantum fields filling the whole spacetime where the energy-density of the quantum fields is below a critical threshold will shrink. In the limit of quantum vacuum outside, this shrinking becomes complete evaporation: the horizon simply is not found in the future, because the mass in the whole spacetime has reconfigured from concentration at a central point to diffusion away from the central point. Since mass-energy generates the metric, the metric ceases to be Schwarzschild (or whatever black hole metric had existed prior to final evaporation).

In our universe, there is lots of mass-energy outside black holes, and even though the metric expansion of space will make the CMB energy-density very small, it will still be higher than the critical threshold for stellar mass and larger black holes for a lonnnnnng time. But this isn't the black holes self-sustaining: they would evaporate, except for the CMB dust falling onto them, so it's the CMB (and its neutrino equivalent) that's doing the sustaining.


Sigh... I wrote the above while sleep-deprived, and it came out garbled. Thanks for the corrections.


That is a bit more difficult to reckon.

A distant observer watching their unfortunate friend falling into a black hole will see their friends clock tick slower and slower such that they will never, from the outside, be seen to cross the event horizon, the passage of time will slow towards zero. From the outside, time does not appear to pass at all inside a black hole.

The unfortunate traveller, however, won't even notice the crossing on a bigger black hole.


Our universe is more complicated than e.g. a vacuum Schwarzschild black hole spacetime.

For starters, there are probably lots of black holes in our universe (and even in our galaxy). What happens if the unfortunate friend falling into the black hole is being watched from an orbit close to another black hole?

(General Relativity has the somewhat frustrating property that adding a black hole to a black hole spacetime does not obey the principle of linear superposition, so things can get quite messy if the two black holes in the paragraph above are close together. One can get even kinkier, for instance by considering a stellar-mass black hole orbiting a supermassive black hole, and kinkier still if both of them rotate with unaligned rotational axes and/or with opposite spins.)

Next, the metric expansion of space means that an isolated black hole would be more like a Schwarzschild-de Sitter solution. There we have the problem that two widely-separated observers can be highly cosmologically redshifted with respect to each other even if neither is anywhere near a black hole. One can of course also consider a collapsing universe in which two observers are blueshifted with respect to one another, and one could consider what that blueshift would do to observations by A of B if B were falling into a black hole in the collapsing universe.

Moreover, the presence of matter can make a mess of things. One can contrive an arrangement of matter near to a black hole which is sufficiently dense as to offset the gravitational time dilation of someone near the horizon.

Finally, there's kinematics: we can subject the unfortunate friend and the observer to a relative ultraboost wherein the doppler blueshift undoes the gravitational redshift.

A relatively ultraboosted Schwarzschild black hole would look a little odd to the relatively ultraboosted observer: https://en.wikipedia.org/wiki/Aichelburg%E2%80%93Sexl_ultrab... Things get weirder when you add other black holes, stars, dust, and so on to the Aichelburg-Sexl picture.

> From the outside, time does not appear to pass at all inside a black hole

From outside the event horizon of a static black hole there is no way to tell, even if one uses tricks like the above to do away with the objection that in general time appears to come to an effective standstill infinitesimally outside the horizon.

However, if the infaller brings in significant mass or angular momentum, we would expect the horizon to reflect the internal configuration change, since it's the internals that generate the horizon in the first place. This is a dynamical rather than a static (up to tiny perturbation) black hole.

This is extremely fiendishly difficult, and is run into in practice in terms of trying to work out exactly what the full theory says about the final merger of two black holes. Roughly speaking, the gravitational interactions between two merging black holes is best described by a metric wherein there is an effective third (and sometimes fourth) body.


Black holes are a singularity. They have an even horizon but it's not really a surface of an object it's just a place in space where certain conditions occur. What happens underneath that horizon (and in the singularity) isn't solidly known.


The "center" of a black hole is not a place in space. It's a moment of time (that is to the future of all other moments inside the horizon).


Not disagreeing that there's a point of no return wherein an infaller will inevitably collide with the singularity, just whether it's really correct to say that makes the singularity a moment of time. For starters, different infallers can arrive at the singularity at different times (even ones that arrive at the horizon at the same time can be relatively accelerated, resulting in having different interval lengths).

Also not sure it makes sense for non-eternal BHs. For a BH that forms by gravitational collapse and eventually completely evaporates, surely the singularity has a timelike worldline? Some "lucky" too-early infallers (a cosmic neutrino, say) might pass through the centre collapsing progenitor before the horizon forms, while too-late infallers might pass through the region after final evaporation (assuming no horizon-equipped remnant). The "just-miss" is depends on where in spacetime the coincidence happens, and any reasonable slicing or threading would attribute the miss to being at the wrong moment in time.

Maybe easiest to think about that if the singularity is not always at the spatial origin of a system of coordinates. For instance, if SN1987a's remnant contains a black hole, that black hole is surely moving around the galaxy with the luminous matter "now", but before the final collapse there was neither horizon nor singularity.

I'm also perhaps unreasonably superstitious about the slogan's survival of arbitrary parameterizations, e.g., one can do an affine parameter on even an eternal BH's singularity and have what looks like a decent proper time for it.

Perhaps another way of putting it is that the usual Carter-Penrose BH diagrams showing long horizontal wavy line segments for the singularity are misleading because of exaggerations caused by the particular conformal chart it uses, kinda like how the Earth's very-near-polar regions are not that big even though they look that way on Mercator charts (those are also conformal, and you get really different results in the near-polar regions in other conformal projections even as closely related as a transverse Mercator/Gauss conformal projection, https://en.wikipedia.org/wiki/Map_projection#/media/File:Usg... versus https://commons.wikimedia.org/wiki/File:Usgs_map_traverse_me... ).

Just thinking aloud in the wee hours, and maybe having given too little weight (pardon the pun) to the qualification "inside the horizon" just before your close paren. There's a lot of rigour that can be hidden in those three words.


> maybe having given too little weight (pardon the pun) to the qualification "inside the horizon" just before your close paren

That qualifier is indeed crucial.


> Not disagreeing that there's a point of no return wherein an infaller will inevitably collide with the singularity, just whether it's really correct to say that makes the singularity a moment of time.

A more technically correct term would be "spacelike line". (It's a line and not a 3-surface because it is the locus with r = 0.) But I think "moment of time" is easier to understand for an audience that is likely not to be familiar with GR technical terms.

> different infallers can arrive at the singularity at different times

If we're going to that level of detail, then there is no invariant way to specify "at the same time" inside the horizon anyway, since the spacetime is not stationary there so there is no timelike Killing vector field and no set of spacelike surfaces picked out by the spacetime geometry as surfaces of "constant time". So "at different times" and "at the same time" inside the horizon are just a matter of coordinate choice and have no invariant physical meaning either way. But again, that's probably too much detail for people who are not familiar with GR technicalities.

> For a BH that forms by gravitational collapse and eventually completely evaporates, surely the singularity has a timelike worldline?

Not for the case where the final black hole is a Schwarzschild hole, no. The locus r = 0 is a timelike line at the center of the collapsing matter until the collapsing matter's surface reaches r = 0; up to that point r = 0 is not a singularity and is a perfectly ordinary location inside a blob of collapsing matter. But once the collapsing matter's surface reaches r = 0 and forms the singularity, the locus r = 0 becomes a spacelike line, and that's the spacelike line that anything falling in after the collapse has finished will end up hitting.

> Maybe easiest to think about that if the singularity is not always at the spatial origin of a system of coordinates

It's not at the spatial origin of any system of coordinates. That's impossible for a spacelike line.

> if SN1987a's remnant contains a black hole, that black hole is surely moving around the galaxy with the luminous matter "now"

It depends on what you claim is "moving". You can certainly pick out a "world tube" in the external spacetime (say by taking a timelike 3-surface formed by the dynamics of a surface of constant r just outside the horizon) and say that "whatever is inside the tube" is moving around the galaxy, because that's what the world tube describes. But you can't say that the singularity is "a point" that is "somewhere" inside that world tube. The singularity doesn't have a spatial location, and spacetime inside the hole's horizon simply doesn't work the way the ordinary language that we use to describe objects moving in space through time assumes that things work.

> the usual Carter-Penrose BH diagrams showing long horizontal wavy line segments for the singularity are misleading because of exaggerations caused by the particular conformal chart it uses

They are not misleading about the fact that the singularity is spacelike. And if you look at a Penrose diagram of a model like the 1939 Oppenheimer-Snyder model of collapse of a spherically symmetric star to a black hole, you will see that the line marking r = 0 has a "corner" at the upper left of the diagram where it changes from timelike to spacelike; that "corner" is where the singularity begins.


This only works because the core is denser than the surface.

(hint: imagine the limit case where the earth were a hollow sphere)


They calculated the 1.5 years assuming a homogenous Earth (equal density throughout), so this is not true. The 2.5 years is calculated using a more realistic density distribution for the Earth.


> imagine the limit case where the earth were a hollow sphere

If it were a hollow sphere with the same total mass as our actual Earth, as measured from the outside, then clocks everywhere inside the hollow sphere would run slow relative to clocks just on the outer surface of the sphere, by roughly the same amount as clocks at the center of the actual Earth. The "clock rate" in the interior of the sphere's material would decrease a lot faster, per unit distance down from the outer surface, than it does in our actual Earth, because the material of the hollow sphere would have to be a lot denser than the material of our actual Earth, in order to have the same total mass.


What is the force of gravity at any point in such a sphere? So what does GR have to say?


The force of gravity is 0. But the potential of gravity is higher than on the outside, so clocks run slower.


> the potential of gravity is higher than on the outside, so clocks run slower

Correction: the potential of gravity is lower inside the sphere than outside, so clocks run slower.


Does the same apply to the Sun? How old is the centre of the Sun compared to its surface?


There's a little bit of smoke-and-mirrors going on here. Yes, the same effect happens on/in the sun (in fact it's even more extreme because the sun's gravitational field is much stronger than earth's). But no, it's not quite correct to say that the center of the sun -- or of the earth for that matter -- is "older" than the surface because the material in both bodies moves around. If you put a clock at the center and another synchronized clock at the surface and left them running for a few billion years and then compared them, you'd see a difference. But of course there are no such clocks. The only "clock" is the material that exists in both locations, and in all cases this material moves around over time. On earth there is some stability because heavier material sinks towards the center and more or less stays there, while lighter material floats to the surface and again more or less stays there. But the sun is made almost entirely of hydrogen and helium and that material is constantly being moved around due to convection.


> it's not quite correct to say that the center of the sun -- or of the earth for that matter -- is "older" than the surface

Correction: "younger", not "older".


There is no convection in the core of the sun. It is plausible that material has mostly stayed there since the formation of the sun.


What form(s) of motion are there, if any, according to theory?

(I'll presume ... direct observations ... are limited.)


Heh, you're right. TIL.


”If you want an even more stark illustration of gravity’s effect on time, take the Sun - the team calculates that the Sun’s core is around 40,000 years younger than its surface.”

https://www.sciencealert.com/earth-s-core-is-2-5-years-young...


> Comments: Published version, apart from minor editing (e.g. corrections of 'center' to 'centre')

> Journal reference: Eur.J.Phys. vol. 37, 035602 (2016)

If Brexit occurs, Britain will no longer be a European nation, and therefore the European Journal of Physics will no longer be obliged to accept "centre" as the standard spelling of "center", right?


European Physical Society which produces this journal is unrelated to EU, national physical societies have its membership directly without involvement of their countries' governments. IOP Publishing which publishes this journal is based in UK. "IOP" in "IOP Publishing" is Institute of Physics which is also based in UK.

Equating being "a European nation" with EU membership is surely a trolling.


We will most certainly still be a European nation. Just not a member of the European Union. We are still located on the European continent. Its spelled centre.


“Some years ago I opined that London was not really an English city any more. Since then, virtually all my friends from abroad have confirmed my observation So there must be some truth in it.” -- John Cleese

In that sense, perhaps London remains a European city while still the capital of a not really European nation.


The UK will no longer be a European Union country but it will definitely keep being European. And I guess Ireland has a "centre" too.


Centre is the standard for all English countries except the US, no? We use centre in Canada


This was the work of spelling reformer Noah Webster[1]. He was quite influential in the United States, but unsurprisingly the British Empire was less enthused by his early 19th century linguistic innovations.

[1] https://en.wikipedia.org/wiki/Noah_Webster


It's all a wibbly-wobbly, timey-wimey big BIG ball of string.


There's a "yo momma" joke just waiting to be made here...


According to General relativity your mom is actually more attractive compared to other mothers...




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