
#1
Jun210, 02:11 PM

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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
Jun210, 02:22 PM

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PF Gold
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#3
Jun210, 04:49 PM

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#4
Jun210, 09:58 PM

P: 1,568

Atomic clock and LatitudeThis paper describes why all clocks at sea level tick at the same rate. 



#5
Jun210, 10:04 PM

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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
Jun210, 10:10 PM

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See here 



#7
Jun210, 10:15 PM

P: 15,325

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
Jun210, 10:43 PM

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#9
Jun210, 11:11 PM

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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
Jun210, 11:15 PM

P: 1,568

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



#12
Jun410, 01:07 PM

P: 1,568

[tex]\frac{d\tau}{dt}=\sqrt{1r_sr/\rho}\sqrt{1\frac{sin^2\theta}{1r_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. 



#13
Jun410, 01:57 PM

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