# Surface and center of the Earth clocks?

1. Dec 31, 2011

### Powd

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?

2. Dec 31, 2011

### pervect

Staff Emeritus
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.

3. Dec 31, 2011

### Powd

Thank you pervect. How about from 1/2 the distance to the center vs. the center of the Earth? Which would experience more proper time?

4. Dec 31, 2011

### beamrider

Supposing earth has a constant density, ro, everywhere inside, then:

g(r) = (4/3)*K*pi*ro*r

An hypothetical pendulum suspended somewhere inside the earth at the depth R-r will swing with the period:

T(r)=2*pi*sqrt(L/|g(r)|)*ug(r)

where:

T = pendulum period
ug(r) = versor of g(r)
L = the length of pendulum string

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.

Proper time does not exist. You always need a reference, like for position or speed.

Last edited: Dec 31, 2011
5. Dec 31, 2011

### Powd

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?

6. Dec 31, 2011

### beamrider

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.

7. Dec 31, 2011

### Powd

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.

8. Jan 1, 2012

### beamrider

g(R) = (4/3)*K*pi*ro*R

g(R/2) = (4/3)*K*pi*ro*(R/2)

g(0) = 0

At half the way between earth surface and center, the acceleration g is half the g on the surface.

9. Jan 1, 2012

### Powd

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.

10. Jan 1, 2012

### Staff: Mentor

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.

Completely incorrect. Not only does proper time exist, it is frame invariant and does not require the specification of a reference.

Last edited: Jan 1, 2012
11. Jan 1, 2012

### 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.

Last edited: Jan 1, 2012
12. Jan 1, 2012

### beamrider

Instead of talking so much it is much more easy to help that guy, who asked the question, about proper time, with a formula.

Proper time expression in General Relativity can be found here:
http://en.wikipedia.org/wiki/Proper_time
(SEE: Example 4: The Schwarzschild solution — time on the Earth)

13. Jan 1, 2012

### Staff: Mentor

Note that the formula given in Example 4 of the Wikipedia link posted by beamrider applies only outside the earth. It is for the exterior metric, not the interior metric.

14. Jan 1, 2012

### Powd

Thanks I know all try and help.

15. Jan 1, 2012

### George Jones

Staff Emeritus
Like

16. Jan 1, 2012

### beamrider

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.)

17. Jan 1, 2012

### Staff: Mentor

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

I refer you back to my previous post.

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.

18. Jan 1, 2012

### beamrider

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.

Last edited: Jan 1, 2012
19. Jan 1, 2012

### umbeke

So, according to you it does not matter that gravitational forces cancels in a region of space.
What matters for GR is that forces exist there, even if they cancels each other.

20. Jan 1, 2012

### Staff: Mentor

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

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.