Surface and center of the Earth clocks?

I am curious about time dilation of atomic clock positions on and in the Earth. If you had a long tube going through the Earth with one atomic clock on the surface of the Earth and one in the center of the Earth which would experience more proper time?

Signals from the clock in the center of the earth would be redshifted as they come out. The stationary observer on the surface would conclude from this that the clock in the center of the Earth ticked more slowly.

There should still be gravity towards the center of the Earth. That would be acceleration at so many feet per second similar to the surface acceleration but less. Proper time should exist everywhere. A difference in proper time should exist with a difference in g=a. r=0 should be just the position of least amount of proper time or is my logic incorrect?

In the center of an ideal spherical earth, isolated in space, there is no gravity.
If you move inside the earth 1 km left, right, up and down from its center, the maximum g you will encounter will be the same order of magnitude the g that characterizes the surface of an asteroid 2 km in diameter, which is quite low, negligible.

In the center of an ideal spherical earth, isolated in space, there is no gravity.
If you move inside the earth 1 km left, right, up and down from its center, the maximum g you will encounter will be the same order of magnitude the g that characterizes the surface of an asteroid 2 km in diameter, which is quite low, negligible.

Yes, I think you might be weightless at the center. But you took the statement out of context. From 1/2 way to the center of the Earth to the center of the Earth is there a difference in the proper time between those two points? At half way to the center there should be acceleration (gravity) until you are close to the center.

By logic in a gravity well there should be a red shift difference causing a difference between the proper time of 1/2 the distance to the center of the Earth and the center of the Earth. I thought Einstein said GR and SR were basically the same. I was thinking if a ship could instantly accelerate to 9.8 m/s/s and start deceleration evenly 8 thousand miles about the same as the Earths radius the Earth and the ship might have the same proper time. This is just a thought my thinking might be off.

As you come to the surface, from the center of the earth, the period T of the pendulum will vary between 0 and pi*sqrt(L/((1/3)*K*pi*ro*R))

This T is the so called classical time dilatation (not related to Einstein's time dilatation) measured with a pendulum.
As you see none of the T(r) with r=[0, R] can be called more proper time.

Completely irrelevant. A pendulum clock would not tick in orbit, but time still passes. A pendulum can only be used as a clock if the proper acceleration is known. If the proper acceleration is unknown, then the use of a different clock to measure the period of the pendulum can give you a measure of the proper acceleration.

Staff: Mentor

Powd, please disregard all of beamrider's comments. I cannot see any that are both correct and relevant.

In a static spacetime the observed gravitational redshift between two points is not a function of the difference in gravitational acceleration between the two points, but a function of the difference in gravitational potential. For a uniform density sphere, the acceleration is zero far from the sphere, maximum at the surface of the sphere, and zero at the center of the sphere, but the potential decreases monotonically towards the minimum at the center of the sphere.

where [itex]R[/itex] is the [itex]r[/itex] coordinate at the surface of the Earth.

If an observer on the Earth's surface uses a telescope to look down a tunnel to a clock at the Earth's centre, he will see his clock running faster than the clock at the Earth's centre.

Consider two dentical clocks, one moving around the Earth once a day on the Earth's surface at the equator ([itex]\theta = \pi/2[/itex]) and one at the Earth's centre. Both clocks have constant [itex]r[/itex] values, so [itex]dr=0[/itex] for both clocks, and, after factoring out a [itex]dt^2[/itex], the above equation becomes

where [itex]v=rd\phi/dt[/itex] is, approximately, the speed of something moving along a circular path. At the centre, [itex]v=r=0[/itex], and, on the surface, [itex]r = R[/itex] and [itex]v = 1.544 \times 10^{-6}[/itex], which is one Earth circumference in one day.

For earth interior you have to replace the mass m corresponding to R = 6356 km with a smaller mass corresponding to a smaller earth, for example an earth having R' = 6356 km / 2.
(At a depth h inside the earth everything behaves like you have a smaller earth with a ray, R = R-h. The spherical shell of depth h that is above the R-h surface produces no gravity inside.)

Staff: Mentor

For earth interior you have to replace the mass m corresponding to R = 6356 km with a smaller mass corresponding to a smaller earth, for example an earth having R' = 6356 km / 2.
(At a depth h inside the earth everything behaves like you have a smaller earth with a ray, R = R-h. The spherical shell of depth h that is above the R-h surface produces no gravity inside.)

While this is true, it is not relevant. You seem to not be able to distinguish between gravitational acceleration and gravitational potential.

In a static spacetime the observed gravitational redshift between two points is not a function of the difference in gravitational acceleration between the two points, but a function of the difference in gravitational potential. For a uniform density sphere, the acceleration is zero far from the sphere, maximum at the surface of the sphere, and zero at the center of the sphere, but the potential decreases monotonically towards the minimum at the center of the sphere.

I appreciate your enthusiasm, but you really need to be careful. You are adding more confusion than you are helping. Please, if you don't understand something ask questions, and only provide answers where you do understand something.

1) I came with a formula I have found in Wikipedia which has much more value than 1000 words.

What that m ( from example: Example 4: The Schwarzschild solution — time on the Earth, https://en.wikipedia.org/wiki/Proper_time ) should be for earth interior in order to make the formula valid inside our planet?

I proposed m as being the mass of an inside the earth sphere having the ray R' = R -h. You say this is wrong.

2) Anyway, there is a phrase at: https://en.wikipedia.org/wiki/Proper_time : "When standing on the north pole, we can assume dr = dteta = dfi = 0 (meaning that we are neither moving up or down or along the surface of the Earth)."
which implies that the formula should be also valid if one moves inside the earth as long as they speak about moving up or down. This word "down" should not be there as long as the formula it is not valid for moving down as you say.

Staff: Mentor

1) I came with a formula I have found in Wikipedia which has much more value than 1000 words.

What that m ( from example: Example 4: The Schwarzschild solution — time on the Earth, https://en.wikipedia.org/wiki/Proper_time ) should be for earth interior in order to make the formula valid inside our planet?

I proposed m as being the mass of an inside the earth sphere having the ray R' = R -h. You say this is wrong.

I would start with the interior metric suggested by George Jones in post 15.

2) Anyway, there is a phrase at: https://en.wikipedia.org/wiki/Proper_time : "When standing on the north pole, we can assume dr = dteta = dfi = 0 (meaning that we are neither moving up or down or along the surface of the Earth)."
which implies that the formula should be also valid if one moves inside the earth as long as they speak about moving up or down. This word "down" should not be there as long as the formula it is not valid for moving down as you say.

Why not? It says that you are NOT moving down, which is perfectly acceptable to say in this case. If you infer from this that the formula is valid inside the earth then you are making incorrect inferences.