# Atomic clock and Latitude

1. ### Dmitry67

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?

2. ### bcrowell

6,096
Staff Emeritus
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.

### Staff: Mentor

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. ### starthaus

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. ### DaveC426913

16,439
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. ### starthaus

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. ### DaveC426913

16,439
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?

### Staff: Mentor

Yes.

It has everything to do with equipotential at sea level.

9. ### bcrowell

6,096
Staff Emeritus
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 $e^{-\phi}$. 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 $\phi_1=\phi_2$. 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. ### starthaus

Yes.

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 $$r_1$$ and $$r_2$$ respectively is expressed by the ratio:

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

where $$r_s$$ is the Schwarzschild radius.The above is valid for a uniform density sphere.
Replace $$r_s$$with the net gravitational potential $$\Phi$$ and you get the correct answer for the geoid.

Last edited: Jun 3, 2010
11. ### Dmitry67

Thank you
Very interesting indeed!

12. ### starthaus

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:

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

The problem in your OP is a perfect application for the Kerr solution.

Last edited: Jun 4, 2010
13. ### yuiop

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 cancelled out by reduced gravitational time dilation due to the increased radius at the equator. (All clocks being 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.)