Time dilation for two clocks at different altitudes on Earth

[zuz]

Physicists took two atomic clocks, placed one at sea level and the other on top of Mount Sunapee. After four days the clock on the mountain top was 4 nanoseconds ahead of the clock at sea level. They said that since the clock was further from the center of the earth, it moved faster, because their was less gravity. My question is...what about the mass of the mountain? It's farther away from the center sure, but there is more earth between it and the clock at sea level. Please explain.

 

[phinds]

Good question, but the mass of the mountain is essentially zero compared to the mass of the earth, and the gravitational time dilation is based on distance from the center of the earth, not local mass anomalies.

To make any sense of using a local mass anomalie for the time dilation, you would have to be measuring the time dilation of the sea level one with that of the one on the mountain, AND you would have to ignore the mass of the earth, which would make no sense.

 

[Dale]

My question is...what about the mass of the mountain?

The important thing is the gravitational potential. Yes, the mountain has some mass, but the top is still at a higher gravitational potential than the bottom.

 

[physics2222]

You can kind of treat Earth as an idealized point particle where the mass is concentrated at the center and the potential (metric) is a function of the radius R. There are mass anomalies that change the field but those are negligible compared to the change in field strength you would get at from the height at the top of mount Everest.

 

[zuz]

Good question, but the mass of the mountain is essentially zero compared to the mass of the earth, and the gravitational time dilation is based on distance from the center of the earth, not local mass anomalies.

To make any sense of using a local mass anomalie for the time dilation, you would have to be measuring the time dilation of the sea level one with that of the one on the mountain, AND you would have to ignore the mass of the earth, which would make no sense.

Thanks.

 

[PAllen]

If you imagine two places on earth such that the total work to raise a small body from the surface to very far from earth was greater at A than B, then a clock at A would run slower (equivalently, escape velocity at A is larger). However, the top of a mountain has slight less escape escape velocity than at its base, so the mountain top clock runs faster.

 

[PeterDonis]

They said

Can you give a specific reference?

 

[cianfa72]

If you imagine two places on earth such that the total work to raise a small body from the surface to very far from earth was greater at A than B, then a clock at A would run slower (equivalently, escape velocity at A is larger). However, the top of a mountain has slight less escape velocity than at its base, so the mountain top clock runs faster.

I'd add that, since the two clocks involved are fixed at two different location A and B, "clock run faster" is not the same as twin paradox's scenario where twins share the initial and final events of their own journeys through spacetime. Here, determining which clock runs faster, requires the definition of a physical process to use to compare them.

 

[A.T.]

If you imagine two places on earth such that the total work to raise a small body from the surface to very far from earth was greater at A than B, then a clock at A would run slower (equivalently, escape velocity at A is larger). However, the top of a mountain has slight less escape escape velocity than at its base, so the mountain top clock runs faster.

Why not simply consider the sign of net work needed to bring the small object from A to B?

 

[PAllen]

Why not simply consider the sign of net work needed to bring the small object from A to B?

You, could, of course. I was trying to realize the notion of potential in contrast to surface gravity, as a function of position.

 

[A.T.]

I'd add that, since the two clocks involved are fixed at two different location A and B, "clock run faster" is not the same as twin paradox's scenario where twins share the initial and final events of their own journeys through spacetime. Here, determining which clock runs faster, requires the definition of a physical process to use to compare them.

Two meetings are required if the clocks are in relative motion at some point. If their arrangement remains static permanently (in some frame where the gravitational potential is static too), then you can just continuously sent signals of a predefined frequency between them and look at the frequency shift. Based on that, both locations will agree about who's clock runs faster by how much, so there is no need for them to ever meet and sort this out.

 

[cianfa72]

Two meetings are required if the clocks are in relative motion at some point. If their arrangement remains static permanently (in some frame where the gravitational potential is static too), then you can just continuously sent signals of a predefined frequency between them and look at the frequency shift. Based on that, both locations will agree about who's clock runs faster by how much, so there is no need for them to ever meet and sort this out.

Exactly, I was referring to what you described (i.e. measuring frequency shift of a predefined frequency signal sent) as the physical process/procedure involved in determining which clock runs faster.

 

[A.T.]

Exactly, I was referring to what you described (i.e. measuring frequency shift of a predefined frequency signal sent) as the physical process/procedure involved in determining which clock runs faster.

My point was that, unlike with kinetic time dilation and no second meeting, here there is no ambiguity which stationary clock runs faster, over a sufficiently long period of time. Every physical process/procedure will have to agree with the frequency shift, when it's deviations from the stationary state are accounted for. So you actually don't have to define the exact comparison procedure, to tell which clock runs faster.

For example, if you bring the clocks together regularly for comparison, you can do it in a symerical fashion. You can also make the relative error from the transport arbitrarily small, by making the static intervals arbitrarily large compared to the transport intervals. In the limit of the error going to zero this will approach the frequency shift result during the static period.

 

[cianfa72]

For example, if you bring the clocks together regularly for comparison, you can do it in a symmetrical fashion. You can also make the relative error from the transport arbitrarily small, by making the static intervals arbitrarily large compared to the transport intervals. In the limit of the error going to zero this will approach the frequency shift result during the static period.

By static interval I believe you mean the segment of clock's worldline along a part of timelike static congruence. For instance, assuming a Schwarzschild spacetime in Schwarzschild coordinates, it is a segment/part of the timelike worldline followed by a body fixed at a given Schwarzschild radius $r_A$ and $\theta_A, \phi_A$.

Likewise clock's transport intervals are pieces of timelike curves joining the events to compare.

 

[Sagittarius A-Star]

it moved faster, because their was less gravity.

This is wrong. It is a common misconception. You have such "gravitational" time-dilation also between bow and stern of an accelerating rocket in flat spacetime.

As others asked: What is the reference you are citing?