Gravitational time dilation near Earth

[PeterDonis]

In SI units, the above formula becomes

The first one you gave is correct.

 

[DanMP]

Optic fibres keep the FCC (or whatever your local regulatory body is) off your back and keep relatively rapidly varying atmospheric conditions from ...

What means FCC?

Do you know their method? How they compared optical clock rates?

 

[Ibix]

What means FCC?

Federal Communications Commission. They object to unlicensed radio signals in the US - similar bodies exist elsewhere. There are plenty of ways of not annoying them - using light signals in fibres is one.

Do you know their method? How they compared optical clock rates?

No. The classic experiment is the Pound-Rebka experiment, which is a really clever way of comparing the frequency of light emitted at the top of a tower to the frequency received at the bottom. The details of that aren't paywalled and are easily Googleable.

 

[DanMP]

The first one you gave is correct.

Thank you!

So, we have:
$$
\phi(r_E, r_S, v_E, v_S) = 1 - \frac{GM_E}{c^2 r_E} - \frac{GM_S}{c^2 r_S} - \frac{v_E^2}{2 c^2} - \frac{v_S^2}{2 c^2}
$$

where: $r_E=x$, $r_S=R-x$ (R=Sun-Earth distance), $v_E=\omega x$, $v_S=\omega (R-x)$ ($\omega$ = Earth's angular speed around the Sun)

So, we can write:
$$
\phi(r_E, r_S, v_E, v_S) = \phi(x)= 1 - \frac{GM_E}{x c^2} - \frac{GM_S}{(R-x) c^2} - \frac{(\omega x)^2}{2 c^2} - \frac{(\omega (R-x))^2}{2 c^2}
$$
and
$$
\frac{d \phi(x)} {dx} = \frac{GM_E}{x^2 c^2} - \frac{GM_S}{(R-x)^2 c^2} - \frac{\omega^2 x}{c^2} + \frac{\omega^2 (R-x)}{c^2}
$$

We need $\frac{d \phi(x)} {dx} = 0$

so
$$
\frac{GM_E}{x^2 c^2} - \frac{GM_S}{(R-x)^2 c^2} - \frac{\omega^2 x}{c^2} + \frac{\omega^2 (R-x)}{c^2}=0
$$
or
$$
\frac{GM_E}{x^2} - \frac{GM_S}{(R-x)^2} - \omega^2 x+ \omega^2 (R-x)=0
$$
or
$$
\frac{GM_S}{(R-x)^2} + \omega^2 x = \frac{GM_E}{x^2} + \omega^2 (R-x)
$$

so x is the place where gravitational and centrifugal forces acting on the clock cancel each other ... Interesting.

The value is: 1,349,811,584 m

 

[PeterDonis]

x is the place where gravitational and centrifugal forces acting on the clock cancel each other

Yes, that is to be expected since, in this approximation, the force is the gradient of the potential, and the potential is what determines the clock rate. So the maximum of the potential, which is where the maximum clock rate will be, is also where the gradient is zero and therefore the net force is zero.

 

[DanMP]

...So the maximum of the potential, which is where the maximum clock rate will be, is also where the gradient is zero and therefore the net force is zero.

How about the opposite side of the Earth? There is a place there where the 4 forces cancel each other but, beyond it, the clock rate will continue to increase.

Is there a round the Earth transition surface? It is like the rim of the Earth's gravitational well?

What do you think about the optical clocks spiraling upwards from the Earth experiment I suggested? I have reasons to say that such an experiment is very important for relativity. I can't reveal the reasons, because personal theories are forbidden here (not the same in the other forum ... where I intend to post them in the near future), but we may find that the transition will be anything but smooth.

 

[Dale]

I can't reveal the reasons, because personal theories are forbidden here (not the same in the other forum ... where I intend to post them in the near future),

We thank you for respecting our rules here.

This thread is now closed. Best of luck in your academic pursuits.

 

[PeterDonis]

How about the opposite side of the Earth?

As a bit of thread cleanup, I should clarify that my statements were only referring to the potential in between the Sun and the Earth. The maximum of the potential between those bodies is only a local maximum, not a global maximum.