What is the relationship between latitude and the accuracy of an atomic clock?

In summary, atomic clocks record the same time at all latitudes due to the cancellation of the special relativity time dilation and the general relativity gravitational time dilation. This is because the Earth's surface is an equipotential, meaning there is no difference in the rate at which time flows at different points. However, there is a difference in time dilation at different altitudes, which is why clocks must be adjusted to account for this and the average time is calculated as the TAI. Furthermore, the difference in time dilation is a result of both special and general relativity effects, and can be calculated using the Schwarzschild or Kerr metric.
  • #1
Dmitry67
2,567
1
Atomic clock should record different time depending on their geo location: different latitude -> different rotation speed.

Question:
The most precise time on Earth - where is it valid? On what Latitude?
 
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  • #2
Dmitry67 said:
Atomic clock should record different time depending on their geo location: different latitude -> different rotation speed.

No, this is incorrect. This was one of Einstein's original predictions in SR, but by the time he figured out GR, he realized that there is no such effect. Since the surface of the Earth is an equipotential, there is no difference in the rate at which time flows at different points. Loosely, you can say that the SR time dilation cancels out with the gravitational time dilation.
 
  • #3
Dmitry67 said:
Atomic clock should record different time depending on their geo location: different latitude -> different rotation speed.

Question:
The most precise time on Earth - where is it valid? On what Latitude?
As bcrowell said since the Earth is not a sphere the SR and GR effects cancel such that time dilation is not a function of latitude. It is, however, a function of altitude.
 
  • #4
DaleSpam said:
As bcrowell said since the Earth is not a sphere the SR and GR effects cancel such that time dilation is not a function of latitude. It is, however, a function of altitude.

Indeed, the fact that the amount of time dilation is a function of altitude, coupled with the fact that most clocks are not situated at sea level forces the ongoing adjustment of the clocks such that the TAI is indeed the average over hundreds of readings of different clocks around the globe. This article describes the situation very well.
This paper describes why all clocks at sea level tick at the same rate.
 
  • #5
You guys seem to be concentrating on GR in the gravity well. But the OP is talking about SR - the equator is moving 1000mph faster than the poles.
 
  • #6
DaveC426913 said:
You guys seem to be concentrating on GR in the gravity well. But the OP is talking about SR - the equator is moving 1000mph faster than the poles.

True. Nevertheless, he's asked "Where is the most precise time? At what latitude?". The answer is that all latitudes are equipotential at sea level. All atomic clocks "show" the same time due to the fact that the "GR blueshift" exactly cancels out the "SR redshift". I put quotation marks because both effects are truly GR.
See here
 
  • #7
starthaus said:
True. Nevertheless, he's asked "Where is the most precise time? At what latitude?". The answer is that all latitudes are equipotential at sea level. All atomic clocks "show" the same time due to the fact that the "GR blueshift" exactly cancels out the "SR redshift". I put quotation marks because both effects are truly GR.
See here

Sorry for being dogged.

Are you in fact saying that
- there does indeed exist a real SR time dilation due to differential velocities at differing latitudes? (This effect would have nothing to do with equipotential at sea level.)
- but it is but is canceled out by some GR effect?
 
  • #8
DaveC426913 said:
Are you in fact saying that
- there does indeed exist a real SR time dilation due to differential velocities at differing latitudes? ...
- but it is but is canceled out by some GR effect?
Yes.

DaveC426913 said:
This effect would have nothing to do with equipotential at sea level.
It has everything to do with equipotential at sea level.
 
  • #9
DaveC426913 said:
Sorry for being dogged.

Are you in fact saying that
- there does indeed exist a real SR time dilation due to differential velocities at differing latitudes? (This effect would have nothing to do with equipotential at sea level.)
- but it is but is canceled out by some GR effect?

This is one way of looking at it, but I don't think it's the only way of looking at it. In GR these are not two separate effects.

In the frame that rotates with the earth, if two clocks are both at rest, then any frequency difference between them has to be explained as a gravitational redshift, which varies as [itex]e^{-\phi}[/itex]. There is no SR time dilation, because both clocks are at rest. If they run at the same frequency (as they will if they're both at sea level), then [itex]\phi_1=\phi_2[/itex]. You can actually use this kind of frequency mismatch to *define* the gravitational potential in GR.

In the frame where we see the Earth as rotating, there is both an SR time dilation and a gravitational time dilation.

The distinction between these two explanations is dependent on which frame you choose.

Note that in the Newtonian context, if you choose the inertial frame in which the Earth is seen to rotate, the Newtonian gravitational potential is *not* equal between different latitudes. This is similar to spinning a bucket full of water about a vertical axis; the parabolic surface of the water is not a Newtonian equipotential.
 
  • #10
DaveC426913 said:
Sorry for being dogged.

Are you in fact saying that
- there does indeed exist a real SR time dilation due to differential velocities at differing latitudes?

Yes.

- but it is but is canceled out by some GR effect?

Yes. With a small correction: both effects are part of one GR effect, the effect is derived from the Schwarzschild metric, the frequencies of two clocks situated at radiuses [tex]r_1[/tex] and [tex]r_2[/tex] respectively is expressed by the ratio:

[tex]\frac{f_2}{f_1}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}\sqrt{\frac{1-(r_1sin\theta_1\omega/(1-r_s/r_1))^2}{1-(r_2sin\theta_2\omega/(1-r_s/r_2))^2}}[/tex]

where [tex]r_s[/tex] is the Schwarzschild radius.The above is valid for a uniform density sphere.
Replace [tex]r_s[/tex]with the net gravitational potential [tex]\Phi[/tex] and you get the correct answer for the geoid.
 
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  • #11
Thank you
Very interesting indeed!
 
  • #12
Dmitry67 said:
Thank you
Very interesting indeed!

You are welcome. You can get the result of the paper I linked by using the Kerr (instead of the Schwarzschild) metric. Instead of filling pages with calculations, you get the result in one line:

[tex]\frac{d\tau}{dt}=\sqrt{1-r_sr/\rho}\sqrt{1-\frac{sin^2\theta}{1-r_rr/\rho}[(r^2+\alpha^2+r_sr\alpha^2/\rho^2)(\omega/c)^2-\frac{2r_sr\alpha}{\rho^2}(\omega/c)]}[/tex]

The problem in your OP is a perfect application for the Kerr solution.
 
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  • #13
DaveC426913 said:
Sorry for being dogged.

Are you in fact saying that
- there does indeed exist a real SR time dilation due to differential velocities at differing latitudes? ...

- but it is but is canceled out by some GR effect?

I agree with all the comments by bcrowell, DaleSpam and starthaus. The two effects cancel out. The increased time dilation due to increased velocity at the equator relative to a clock at one of the Poles, is canceled out by reduced gravitational time dilation due to the increased radius at the equator. (All clocks being at sea level).

DaveC426913 said:
... (This effect would have nothing to do with equipotential at sea level.)
...

The "effective potential" (due to velocity and gravitational effects) at sea level is the same everywhere on the globe and this is why all clocks at sea level run at the same rate. It is also why the Earth is oblate rather than perfectly spherical. (Sea water moves from a higher effective potential to a lower effective potential until the effective potential at sea level is the same everywhere.)
 

FAQ: What is the relationship between latitude and the accuracy of an atomic clock?

What is an atomic clock and how does it work?

An atomic clock is a highly accurate timekeeping device that uses the oscillation of atoms to measure time. It works by measuring the frequency of the transition between two energy levels of a specific atom, typically cesium or rubidium. This frequency is then used to keep track of time with extreme precision.

Why is an atomic clock important for measuring latitude?

An atomic clock is important for measuring latitude because it relies on precise timekeeping to determine the location of an object. By measuring the time it takes for a signal to travel from an atomic clock to a receiver and back, the receiver can calculate its distance from the atomic clock, allowing for accurate latitude measurements.

How does the accuracy of an atomic clock affect latitude measurements?

The accuracy of an atomic clock is crucial for latitude measurements because even the smallest error in timekeeping can result in significant errors in determining the location. An atomic clock's accuracy is typically within one second over millions of years, making it the most precise method for measuring time and latitude.

What is the relationship between latitude and the Earth's rotation?

Latitude is a measurement of how far north or south an object is from the Earth's equator. The Earth's rotation plays a significant role in determining latitude because it causes a difference in the amount of centrifugal force at different latitudes. This force affects the Earth's shape, resulting in different latitudes having slightly different distances from the Earth's center.

How does the Earth's rotation affect the accuracy of atomic clocks for measuring latitude?

The Earth's rotation has a minimal effect on the accuracy of atomic clocks for measuring latitude. While the Earth's rotation does impact the Earth's shape, the atomic clock's precision is not affected by this change. However, other factors such as atmospheric conditions and the position of the receiver can affect the accuracy of the measurement.

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