4 ms·

If that was true, shouldn't they also see significant differences during night/day, and holidays?

by Chris2048 6y ago

marcosdumay 6y ago
Those cause dispersed changes on people's distribution on all directions around the clock, so no, they are probably not detectable. Also the sibling comparing with the Earth's mass isn't in a good direction, because the Earth's mass is mostly stable, and the clock only suffers disturbances from changes... But if the story is true, it should have an easy time detecting vulcanism.
- mikewarot 6y ago
  Tides cause clocks to change time... even the tail end of electromechanical clocks were precise enough to detect it. It was detected in 1929 with the Shortt–Synchronome clock See section 3 of this: https://hgss.copernicus.org/articles/11/215/2020/ https://hgss.copernicus.org/articles/11/215/2020/
  - fsh 6y ago
    This is because Shortt clocks are pendulum clocks whose oscillation period directly depends on the gravitational acceleration. Quarz oscillators and atomic clocks are not affected by this.
    - mikewarot 6y ago
      It took a while to sink in that gravity isn't anywhere near as simple as I usually think it is. I understand Gravitational Acceleration at the surface of the earth is on the order of 9.81m/sec^2 I also understand that if you take clocks to the mountains, they run slightly faster. What would happen if you put a clock at the bottom of a mine shaft a few miles down? (In an air-conditioned room, of course) How does the math work?
      - fsh 6y ago
        A pendulum is a mechanical oscillator in which gravity provides the restoring force. For small oscillation amplitudes, the oscillation frequency of a pendulum of length l is 1/(2pi)sqrt(g/l), where g is the gravitational acceleration. Pendulum clocks are therefore directly affected by the local strength of gravity. This is usually not a problem because g is fairly stable and only the most advanced pendulum clocks were able to resolve some slight variations due to the tides. Other clock oscillators, such as balance wheels or quartz crystals do not show this effect. The reason why atomic clocks run differently at different heights is due to General Relativity. In GR, there is no such thing as a gravitational force. Instead, the presence of mass changes the shape of space and time itself. In the limit of small masses and low velocities, the resulting equations of motion exactly reproduce the ones from Newtonian mechanics. On top of that, it turns out that a difference in gravitational potential between two locations also results in a slightly different progression of time. In the classical limit, the gravitational time dilation is given by the the gravitational potential divided by the speed of light squared. Not too far from the surface of the earth this comes out to approximately 1E-16/m (9.8 m/s^2/(3E8 m/s)^2). The best optical clocks are stable on the level of 1E-18, so they can resolve a high difference on the cm level due to the gravitational time dilation.
        mikewarot 6y ago
        What would be the dilation at the center of the earth? Classical gravity there would be zero.
        fsh 6y ago
        The gravitational force would indeed be zero and a pendulum clock would not work. But the gravitational redshift between to locations is given by the potential difference, not by the local force. The potential difference between the surface and center of the earth can be calculated using Gauss' law and comes out to be 1/2gR, where g is the gravitational acceleration at the surface and R is the earth radius. As a result, a clock at the center of the earth would run slower than one at the surface by 3.5E-10.
- Chris2048 6y ago
  I would have though mass would be centred at commercial areas during day; and I would also guess that peoples cars would be more mass-significant than themselves, so higher traffic during day would also be significant. During holidays, people would be more likely to go on vacation, and/or the streets less busy.
  - marcosdumay 6y ago
    Oh, if you place the clock somewhere where commercial areas are mostly on one side of it and residential areas are mostly on the other side, people movement will influence it. But most cities are not distributed that way, people cancel each other.
macintux 6y ago
It's possible they had already measured and accounted for those variations.
- Chris2048 6y ago
  They modelled the distribution of cars/people during holidays (if that's even possible), but then omitted one particular use case?

If that was true, shouldn't they also see significant differences during night/day, and holidays?

Those cause dispersed changes on people's distribution on all directions around the clock, so no, they are probably not detectable. Also the sibling comparing with the Earth's mass isn't in a good direction, because the Earth's mass is mostly stable, and the clock only suffers disturbances from changes... But if the story is true, it should have an easy time detecting vulcanism.

mikewarot 6y ago
Tides cause clocks to change time... even the tail end of electromechanical clocks were precise enough to detect it. It was detected in 1929 with the Shortt–Synchronome clock See section 3 of this: https://hgss.copernicus.org/articles/11/215/2020/ https://hgss.copernicus.org/articles/11/215/2020/
- fsh 6y ago
  This is because Shortt clocks are pendulum clocks whose oscillation period directly depends on the gravitational acceleration. Quarz oscillators and atomic clocks are not affected by this.
  - mikewarot 6y ago
    It took a while to sink in that gravity isn't anywhere near as simple as I usually think it is. I understand Gravitational Acceleration at the surface of the earth is on the order of 9.81m/sec^2 I also understand that if you take clocks to the mountains, they run slightly faster. What would happen if you put a clock at the bottom of a mine shaft a few miles down? (In an air-conditioned room, of course) How does the math work?
    - fsh 6y ago
      A pendulum is a mechanical oscillator in which gravity provides the restoring force. For small oscillation amplitudes, the oscillation frequency of a pendulum of length l is 1/(2pi)sqrt(g/l), where g is the gravitational acceleration. Pendulum clocks are therefore directly affected by the local strength of gravity. This is usually not a problem because g is fairly stable and only the most advanced pendulum clocks were able to resolve some slight variations due to the tides. Other clock oscillators, such as balance wheels or quartz crystals do not show this effect. The reason why atomic clocks run differently at different heights is due to General Relativity. In GR, there is no such thing as a gravitational force. Instead, the presence of mass changes the shape of space and time itself. In the limit of small masses and low velocities, the resulting equations of motion exactly reproduce the ones from Newtonian mechanics. On top of that, it turns out that a difference in gravitational potential between two locations also results in a slightly different progression of time. In the classical limit, the gravitational time dilation is given by the the gravitational potential divided by the speed of light squared. Not too far from the surface of the earth this comes out to approximately 1E-16/m (9.8 m/s^2/(3E8 m/s)^2). The best optical clocks are stable on the level of 1E-18, so they can resolve a high difference on the cm level due to the gravitational time dilation.
      - mikewarot 6y ago
        What would be the dilation at the center of the earth? Classical gravity there would be zero.
        fsh 6y ago
        The gravitational force would indeed be zero and a pendulum clock would not work. But the gravitational redshift between to locations is given by the potential difference, not by the local force. The potential difference between the surface and center of the earth can be calculated using Gauss' law and comes out to be 1/2gR, where g is the gravitational acceleration at the surface and R is the earth radius. As a result, a clock at the center of the earth would run slower than one at the surface by 3.5E-10.
Chris2048 6y ago
I would have though mass would be centred at commercial areas during day; and I would also guess that peoples cars would be more mass-significant than themselves, so higher traffic during day would also be significant. During holidays, people would be more likely to go on vacation, and/or the streets less busy.
- marcosdumay 6y ago
  Oh, if you place the clock somewhere where commercial areas are mostly on one side of it and residential areas are mostly on the other side, people movement will influence it. But most cities are not distributed that way, people cancel each other.

macintux 6y ago

It's possible they had already measured and accounted for those variations.

Chris2048 6y ago
They modelled the distribution of cars/people during holidays (if that's even possible), but then omitted one particular use case?