Gravitational Time Dilation

  • #51
pervect
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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.
 
  • #52
jbriggs444
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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.
 
  • #53
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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##.
 
  • #54
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… 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.
 
  • #55
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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?
 
  • #56
PeterDonis
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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.
 
  • #57
Janus
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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.
 

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