Does Gravitational Time Dilation Affect How We Measure Time on Earth?

[Ibix]

And to be even more precise, you need to define a notion of simultaneity between the Earth's surface and its center, so you know what "since the formation of the earth" actually means along each worldline.

Indeed. But it's about 0.02 light seconds to the centre of the Earth and the planet formed around 1.3×10¹⁷s ago, so it's a fairly minor issue in this case.

 

[David Lewis]

… a clock at the centre would have ticked less than one at the surface since the formation of the earth

Would a uranium deposit near the surface have more fission products than one deeper down?

 

[Ibix]

Would a uranium deposit near the surface have more fission products than one deeper down?

If the Earth were geologically inactive and a uranium deposit could be relied upon to stay at the depth it was at, yes. But that's not the case so, without detailed calculations of the predicted time dilation compared to the precision of a decay clock and a consultation with a geologist, I'd have to go with "maybe".

It would not surprise me if someone had taken an atomic clock down a mine, which would be the same test you are trying to do, I think. A visit to Google might turn up something. In fact I seem to recall from the faster than light neutrino incident that clocks at CERN and in Italy were synchronised by slow clock transport, which must have involved accounting for gravitational time dilation effects from leaving the CERN tunnel.

 

[FactChecker]

Would a uranium deposit near the surface have more fission products than one deeper down?

Yes, but from @Ibix 's post #31, the difference would be tiny. The difference is in time itself and all physical processes change accordingly.

 

[jbriggs444]

Yes, but from @Ibix 's post #31, the difference would be tiny. The difference is in time itself and all physical processes change accordingly.

In post #31, the figure that is presented represents how much discrepancy a different choice of synchronization convention could introduce when comparing an elapsed time measured over there with an elapsed time measured over here.

It does not tell you how much difference there will be in the elapsed proper times. It only tells you how ambiguous the endpoints are.

 

[David Lewis]

How seriously should I take the following claims?

"Young at heart

Plugging this difference into the equations of relativity gives a time dilation factor of around 3 x 10^-10, meaning every second at the Earth’s centre ticks this much slower than it does on the surface. But since the Earth is around four billion years old, the cumulative effect of this time dilation adds up to a difference of around a year and a half.

These calculations assume a uniform density throughout the Earth, which we know isn’t accurate since the core is denser than the mantle. Using a more realistic model of Earth’s density, Uggerhøj’s team found the difference in age is actually around two-and-a-half years."
Read more: https://www.newscientist.com/articl...f-years-younger-than-its-crust/#ixzz60sNRAxfU

 

[PeterDonis]

How seriously should I take the following claims?

The calculations look ok to me.

 

[PhDnotForMe]

There is not. We can say there is, but that won't make it so. (I think that was Abraham Lincoln)

So gravitational potential does NOT cancel out is what you are saying

 

[jbriggs444]

So gravitational potential does NOT cancel out is what you are saying

Correct. For three reasons.

1. Gravitational potential is not defined in a space-time containing black holes.

2. If we hand-wave that away and talk about some appropriate heuristic, gravitational potential is not additive in general relativity.

3. If we hand-wave that away and consider an approximation, the gravitational potentials from two black holes have the same sign. The resulting potential is increased rather than cancelled.

 

[FactChecker]

I think that the cause of confusion in the original post is in thinking of a balanced gravitational force as a zero gravitational potential. The total gravitational force is zero at exactly one point. The gravitational potential is determined by the gravitational difference between one point and other points. So the two are quite different.

 

[Ibix]

The calculations look ok to me.

I'm badly confused at the moment.

The Newtonian gravitational potential inside a sphere of constant density is $GMr^2/2c^2r_g^3$ where $r_g$ is the radius of the sphere. Plugging in the numbers for the Earth leads directly to the naive 3×10^-10 figure for time dilation quoted in the New Scientist article (edit: the "naive" is nothing to do with relativity - rather that the Earth isn't constant density).

However, assuming that we use coordinates where $\partial_t$ is parallel to the timelike Killing vector, shouldn't this be (approximately) equal to $1-\sqrt{|g_{tt}(0)/g_{tt}(r_g)|}$? My reasoning is that the interval between the $t=T$ and $t=T+dt$ planes along a worldline parallel to $\partial_t$ at $r$ is $\sqrt{|g_{tt}(r)|}$ and we are interested in the ratio of these intervals for two such clocks (MTW equation 25.25 seems to agree). The problem I have is that for the interior Schwarzschild solution,$$g_{tt}=\frac 14\left(3\sqrt{1-\frac{R_S}{r_g}}-\sqrt{1-\frac{R_Sr^2}{r_g^3}}\right)^2$$(see my Insight article and the wiki article linked therein, which seem to concur with Schutz p267 and MTW p610) and hence the ratio of tick rates between clocks at $r=0$ and $r=r_g$ is $2/3$. Which is clearly silly - quite apart from the magnitude it's independent of the mass and would apply to a cricket ball as well as to the Earth.

Can't see what I'm missing...

 

[A.T.]

... hence the ratio of tick rates between clocks at $r=0$ and $r=r_g$ is $2/3$. Which is clearly silly - quite apart from the magnitude it's independent of the mass and would apply to a cricket ball as well as to the Earth.

The Newtonian potential at the center of a uniform sphere is 3/2 of the potential at the surface, also independently of the sphere's mass.

 

[PeterDonis]

I'm badly confused at the moment.

The actual paper [1] might help to clarify what's going on. Equations (1) and (2) in the paper are the Newtonian approximations to the potential $\Phi$ in the exterior and interior of a spherically symmetric mass of constant density. Equation (2), in particular, is the Newtonian approximation to the $g_{tt}$ you wrote down, if we take $\Phi = 1 - \sqrt{g_{tt}}$. Note that we have $\Phi(0) = \frac{3}{2} \Phi(r_g)$ (the paper uses $R$ for what this thread has been calling $r_g$), which agrees with the 2/3 ratio you mention. Note also that, for this idealized case, this result indeed does not depend on the mass; that's because all of the mass dependence is hidden inside the factors that are the same for $\Phi(0)$ and $\Phi(r_g)$. Varying $M$ and/or $r_g$ will change the density of the object, but as long as the density is uniform inside, and as long as the weak field approximation is still valid (basically, as long as $M / r_g$ is small enough), this variation will not change the ratio of $\Phi(0)$ to $\Phi(r_g)$, although it will change the absolute values of both of them.

The paper then derives the difference in time dilation from the difference in $\Phi$, in equations (4) and (5). In this approximation we can use the difference in $\Phi$ instead of having to take ratios. The more realistic calculation integrates the actual density profile inward from $r_g$ to get $\Phi(0)$.

[1] https://arxiv.org/abs/1604.05507

 

[Herman Trivilino]

Would it be possible for the high man to read a newspaper that, from the low man's standpoint, hasn't been printed yet?

It would not be possible for the low man to determine anything from the newspaper about the future at his own location. News can't travel FTL.

 

[David Lewis]

… each successive sunrise should occur even earlier and earlier for the high man. The upshot would be that he would rack up sunrises faster than the low man over a long enough period. That simply does not make any sense.

If enough time elapses, wouldn't eventually low man's calendar read Monday, and high man's calendar read Tuesday?

 

[David Lewis]

... each successive sunrise should occur even earlier and earlier for the high man. The upshot would be that he would rack up sunrises faster than the low man over a long enough period. That simply does not make any sense.

Please elaborate. If we wait long enough, high man's calendar will show Tuesday, and low man's will show Monday, if I understand correctly.

 

[jbriggs444]

Please elaborate. If we wait long enough, high man's calendar will show Tuesday, and low man's will show Monday, if I understand correctly.

Yes. If both men maintain their calendars by marking off each day after 86,400 seconds have elapsed then one will find that he is crossing off Tuesday when the other is crossing off Monday. So what? There is nothing particularly strange about that.

By using their own wristwatches as their source of calendar time, both men have divorced themselves from the calendar maintained by the newspaper publisher.

 

[pervect]

If enough time elapses, wouldn't eventually low man's calendar read Monday, and high man's calendar read Tuesday?

Not according to current conventions, no. Coordinate time and proper time are different, though closely related, entities, and that the calendar is based on coordinate time, not proper time. Note that the coordinate time is generally regarded as a convention. A less modern example of the conventional nature of the calendar is the difference between the Gregorian calendar and the Julian calendar. Usually there isn't much ambiguity anymore about what convention (calendar) to use, though this wasn't always the case. In ages past, I've read that different social agencies would use different time conventions (I don't have a reference for this handy,unfortunately), and historians would report important dates via several different conventions, such as who was in office at the time

Our modern realization of coordinate time is TAI, International Atomic TIme. In the modern system, it is recongizned that physical clocks, that keep proper time, tick at a different rate than coordinate clocks, which keep coordinate time. The difference in rate between coordinate clocks and the physical clocks is commonly referred to as "gravitational time dilation", the title of this thread.

TAI time started out by being an average of all clocks on Earth from participating institutions, but when the accuracy of our clocks became high enough, the averaging procedure was changed to recognize and account for the fact that clocks at different altitude tick at different rates. So the rate is adjusted by altitude, first - then the average is taken.

The coordinate time is based on the concept of the reference clock for the coordinate system being at sea level. This is a good enough system for now. The issue of the effect of solar and lunar tides and how that affects the very concept of "sea level" isn't currently important enough to be an issue, but may become an issue in the future as our timekeeping standards improve. The current paradigm is that all clocks at "sea level" tick at the same rate, and that sea level can be regarded as something static, indepenent of time, rather than something dynamic, dependent on time.

It may not be obvious at first glance how or why all clocks at sea level (ignoring tides) tick at the same rate, but references such as Wiki and (MTW's Gravitation) will support this claim.

A competing system, that avoids the whole sea level issue, puts the reference clock not at "sea level", but at the center of the earth. Unfortunately, this means that clocks on the Earth's surface don't keep coordinate time when this system used. This is barycentric coordinate time, TCB. It, or similar systems derived from it, are sometimes used for astronomy. This is an oversimplified overview, but it wouldn't be helpful to go into more detail at this point, and frankly I don't recall all the details offhand anymore.https://en.wikipedia.org/w/index.php?title=International_Atomic_Time&oldid=917191735

In the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation, and the combined TAI scale therefore corresponded to an average of the altitudes of the various clocks. Starting from Julian Date 2443144.5 (1 January 1977 00:00:00), corrections were applied to the output of all participating clocks, so that TAI would correspond to proper time at mean sea level (the geoid).

 

[David Lewis]

A competing system... puts the reference clock… at the center of the earth. Unfortunately, this means that clocks on the Earth's surface don't keep coordinate time when this system used.

That makes sense. If calendars measured their respective proper times, and there were a 4.54 billion-year-old habitable lab at the center of the Earth, would the lab calendar, translated to Gregorian, show approximately March 2017? (October 2019 minus 2.5 years).

 

[Ibix]

That makes sense. If calendars measured their respective proper times, and there were a 4.54 billion-year-old habitable lab at the center of the Earth, would the lab calendar, translated to Gregorian, show approximately March 2017? (October 2019 minus 2.5 years).

Yes.

 

[pervect]

That makes sense. If calendars measured their respective proper times, and there were a 4.54 billion-year-old habitable lab at the center of the Earth, would the lab calendar, translated to Gregorian, show approximately March 2017? (October 2019 minus 2.5 years).

My understanding is that the current date at time at the center of the Earth would be the same as it is at the surface, i.e. October 2019 as of the time of this post. By "at the same time", I mean by use of the Einstein clock synchronization convention, though since the one-way light travel time is about 21 milliseconds, the details of the clock synchronization are not particularly important.

However, if two SI standardized clocks were started at an idealized event representing "the creation of the Earth", one clock in the center, and one on the surface, each clock would read a different number of seconds "now" (using the same Einstein convention as mentioned previously). If I haven't made a sign error, the clock at the center of the Earth would read more seconds than the one on the surface.

 

[jbriggs444]

However, if two SI standardized clocks were started at an idealized event representing "the creation of the Earth", one clock in the center, and one on the surface, each clock would read a different number of seconds "now" (using the same Einstein convention as mentioned previously). If I haven't made a sign error, the clock at the center of the Earth would read more seconds than the one on the surface.

Yes, if the two clocks are started "simultaneously" and then stopped "simultaneously" later on, they will read a different number of seconds.

However, the one at the center of the Earth reads fewer seconds. It runs "slow" relative to the clock at the surface.

 

[pervect]

Yes, if the two clocks are started "simultaneously" and then stopped "simultaneously" later on, they will read a different number of seconds.

However, the one at the center of the Earth reads fewer seconds. It runs "slow" relative to the clock at the surface.

You're probably right - I'll try to work it out. We can write:

$$d\tau^2 \approx 1 + 2(\Phi - \Phi_0) dt^2 / c^2$$

where $\Phi$ is the gravitational potential, and $\Phi_0$ is the gravitational potential at sea level, see for instance https://arxiv.org/abs/gr-qc/9508043 which uses slightly different symbols, but the idea is the same.

This metric makes $dt = d\tau$ at sea level. We don't need the other terms in the metric, since nothing is moving - dx,dy,dz are all zero.

In the exterior region $\Phi = -GmM/r$. While the paper doesn't give an expression for $\Phi$ in the interior region, we know that $\Phi$ should still have it's minimum (most negative) value at the center of the earth, and increase monotonically as one moves away from the center, reaching a maximum of 0 at infinity.

Taking the square root of both sides, we expect that for a fixed dt (the assumption made here), $d\tau$ will be larger for large $\Phi$ and smaller for small $\Phi$.

 

[David Lewis]

… one will find that he is crossing off Tuesday when the other is crossing off Monday. So what? There is nothing particularly strange about that.

If they were born on the same day, one would have to celebrate his birthday on a different day.

 

[jbriggs444]

If they were born on the same day, one would have to celebrate his birthday on a different day.

I still do not see the problem. If they choose to use different calendars then the dates on the two calendars may not match. So what?

 

[PeterDonis]

If they were born on the same day, one would have to celebrate his birthday on a different day.

That depends on what you mean by "birthday" and "a different day". Is your birthday the point at which you have physically aged another year? Or the point at which another year has elapsed by some other process--for example, the Earth going around the Sun? And which definition do you use to determine the calendar day?

None of this has anything to do with physics; it's just human conventions.

 

[Janus]

If they were born on the same day, one would have to celebrate his birthday on a different day.

To add to what @PeterDonis has already said. It depends on how they count "days"
If our person at the center of the Earth had a "peep hole" drilled through the Earth so that he could count how many times the Sun passed over the opening and counted the time between each passing as a "day", then he would celebrate his birthday on the same day as the one on the surface, regardless of whether or not an atomic clock beside him counted off the same number of seconds as the one on the surface. His atomic clock would just record a slightly different length of time for each "day" than the clock on the surface does.