# Ticking clocks

1. May 2, 2007

### Johnny R

How fast would a clock tick at the center of the earth compared to one on the surface?

2. May 2, 2007

### Janus

Staff Emeritus
Slower. I know this answer surprises some people as there is zero gravitational force at the center of the Earth and they think the clock should run faster. What they miss is that Gravitational time dilation is due to a difference in gravitational potential and not gravitational force. It is related to how much energy it would take to move the clock from one potential to another. Since it takes energy to move a clock from the center of the Earth to the surface (just like it takes energy to move a clock from the surface of the Earth to deep space.), the clock at the center is at a lower gravitational potential than the surface, and runs slower.

3. May 2, 2007

### Johnny R

How does the clock "know" this?

My reason for asking this question is that as the clock moves from the center toward the surface the energy required to move further increases proportionally and when it reaches the surface it requires even more energy to continue to move away from the earth.

4. May 2, 2007

### George Jones

Staff Emeritus
First, consider a baseball that is thrown upwards. As it rises, the baseball gains potential energy and loses kinetic energy.

Suppose you have two identical clocks. Dig a hole straight to the centre of the Earth, leave one clock at the centre of the Earth, and hoist yourself back to the surface of the Earth, where you left the other clock.

With one eye, look in a telescope that is trained on the clock at the centre of the Earth. With the other eye, watch the clock on the surface beside you. You will see the second hand on the clock at the centre move more slowly than the second hand on the clock beside you.

Why? Because, like the baseball, photons, as they rise from the centre to the surface, lose kinetic energy. Unlike the baseball, photons always move at the speed of light, so they can't lose kinetic energy by slowing down. Photon energy is proportional to frequency, so the frequency of the light decreases as it rises.

However, frequency is like ticks of a clock. So the image of central clock that you see is slower, because the photons lost energy as the rose from the centre to your eye.

Last edited: Dec 18, 2008
5. May 2, 2007

### pervect

Staff Emeritus
The clock doesn't have to know this.

A clock always ticks at 1 second per second. It is actually the process of comparing the clocks that is the tricky part. It is tricky because the geometry of space-time is curved.

Sorry if this isn't too clear, but I started to write a much longer response and I thought it was just as unclear, but much longer.

You may or may not appreciate the fact that one can only directly compare two clocks when one is at the same point in space-time. If the clocks are not at the same location, one has to go through the comparison process in considerable detail, working out the paths of all the signals involved. It is this sort of analysis that is needed here, but it's rather long.

6. May 2, 2007

### MeJennifer

Pervect, are you saying that there is a unique, and thus observer independent way, to determine the difference in clock rates at two distinct spacetime locations assuming the spacetime is curved?

7. May 2, 2007

### cesiumfrog

MeJennifer, the question is with respect to positions on/in the Earth, so I don't understand why you're introducing any suggestion of ambiguity? It's not "assuming the spacetime is curved", it's "assuming the curved spacetime is static, asymptotically flat, topologically trivial, and much more". It's not suggesting an "observer independent" comparison either, as specific observers worldlines (that of one on the surface, and of one at the center) and all other necessary specifics are unambiguously implied.

8. May 3, 2007

### pervect

Staff Emeritus
No, in general there isn't a unique way. But in the case of a static geometry (one that doesn't vary with time), there is a conventional way - a signal that shares the same property of "not varying with time" as the geometry (usually light is convenient, though one could also use carrier pigeons as long as they had the necessary properties of not varying with time) are used to compare the two different clocks.

9. May 3, 2007

### MeJennifer

Then the circumnavigation of the surface clock (assuming it is not positioned on the axis of rotation) is not a factor in this determination?

I am asking this since there are already difficulties in determining clock rate differences in the simpler case of a rotating disk in special relativity.

With regard to a static geometry, do you mean a static spacetime as opposed to a stationary spacetime? A rotating point mass is a stationary but not a static spacetime right?

Last edited: May 3, 2007
10. May 3, 2007

### DaveC426913

This has some interesting implications. I had always wondered if there is a measurable way of telling the difference between being gravitationally neutral and gravitationally unaffected.

Unless I'm wrong, a guy at the centre of the Earth cannot perform any tests to see if he's in a gravitational field. But he can look out of his field and see the world around him moving faster. He knows conclusively that he is in a grav field.

So, if we can look out of our Solar system towards the edge of the universe we see things moving much more rapdily than we expect. Can we hypothesize that our corner of the universe is at the bottom of a very large gravity well?

11. May 3, 2007

### Mentz114

George Jones says -
I don't follow this - can you explain further ? If we turned off the lights and there were no photons, would the clock still be running slower ?

12. May 3, 2007

### MeJennifer

I cannot agree with the interpretation that a photon's frequency somehow changes between the two events.

What seems to me the case is that the measured frequency of the photon is: fproper * clockrateemitter / clockrateabsorber

Am I wrong?