Strange question about time dilation

[Bart]

Suppose that at the top of Mt. Everest precision clock has been installed, and the same second clock was installed at the bottom of the mountain, 8 km below the summit.

Before installing, clocks were carefully synchronized.

The question is: what difference of time these clocks will show per day (24h), according to the theory of relativity?

[granpa]

its not a strange question at all.

 

http://www.iopb.res.in/~phatak/relativity/node75.html

 

A direct measurement of gravitational time dilation requires measurement of time delays between two clocks which are placed at different gravitational potentials. An example would be two clocks, one placed in a satelite and another on the surface of Earth. Let us compute the difference between two clocks, one placed on Earth and another at a distance of twice the radius of Earth

 

That is, in a day the two clocks will show a time difference of about 30 microseconds in one day. Such small time differences can be observed in atomic clocks.

 

http://www.physicsforums.com/showthread.php?t=58359

 

Multiply the time as measured by a distant observer by sqrt(1-(2GM/Rc^2)).

This is independent of any velocity induced dilation.

:biggrin:

 

escape velocity of earth = 10 km/s (actually its 11.18 kilometers per second)

 

at that velocity gamma = 1.000,000,000,556,325,2

 

gamma calculator

 

http://www.1728.org/reltivty.htm

Edited October 27, 2011 by granpa

[Bart]

its not a strange question at all.

 

http://www.iopb.res....ity/node75.html

 

A direct measurement of gravitational time dilation requires measurement of time delays between two clocks which are placed at different gravitational potentials. An example would be two clocks, one placed in a satelite and another on the surface of Earth. Let us compute the difference between two clocks, one placed on Earth and another at a distance of twice the radius of Earth

 

That is, in a day the two clocks will show a time difference of about 30 microseconds in one day. Such small time differences can be observed in atomic clocks.

 

http://www.physicsfo...ead.php?t=58359

 

 

 

escape velocity of earth = 10 km/s (actually its 11.18 kilometers per second)

 

at that velocity gamma = 1.000,000,000,556,325,2

 

gamma calculator

 

http://www.1728.org/reltivty.htm

 

Thanks for your response. But this question is strange, because I forgot to add that the two clocks are timed remotely from a radio transmitter 1 Ghz, installed on the slope.

[granpa]

what do you mean by timed remotely?

 

you mean you measure how much time each records?

[Bart]

what do you mean by timed remotely?

 

you mean you measure how much time each records?

 

 

Clocked remotely means that the clocks are driven by pulses from 1 GHz radio transmitter, and not their own oscillators.

[granpa]

then whats the question?

[Bart]

then whats the question?

 

The question is: what difference of time these clocks will show per day (24h), according to the theory of relativity?

 

In my understanding the clocks will always indicate the same time, regardless of their distance and period of time measurement, day, year, 10 years ... But what about the theory of relativity?

[swansont]

Near the earth's surface (where you can assume g is constant) the difference is gh/c^2, so clocks or signals will differ in frequency by 8.7 x 10^-13, which gives a timing difference of about 75 ns/day; the clock on top of the mountain will run faster.

 

That is, in a day the two clocks will show a time difference of about 30 microseconds in one day. Such small time differences can be observed in atomic clocks.

For atomic clocks, 30 microseconds a day is huge. Atomic clocks can be remotely measured to better than 1 ns/day, and the limitation is in the signal, not the clocks.

 

Clocked remotely means that the clocks are driven by pulses from 1 GHz radio transmitter, and not their own oscillators.

 

Then it will matter where the signal originator is relative to the clocks. They will not see 1 GHz if they are not at the same elevation as the source. If the source is remotely located, you may have to also account for a Sagnac term (which arises because the earth is not in an inertial frame; it rotates).

[J.C.MacSwell]

Near the earth's surface (where you can assume g is constant) the difference is gh/c^2, so clocks or signals will differ in frequency by 8.7 x 10^-13, which gives a timing difference of about 75 ns/day; the clock on top of the mountain will run faster.

 

 

For atomic clocks, 30 microseconds a day is huge. Atomic clocks can be remotely measured to better than 1 ns/day, and the limitation is in the signal, not the clocks.

 

 

 

Then it will matter where the signal originator is relative to the clocks. They will not see 1 GHz if they are not at the same elevation as the source. If the source is remotely located, you may have to also account for a Sagnac term (which arises because the earth is not in an inertial frame; it rotates).

 

These effects should cancel with that of the proper time elapse of each location.They would both be constantly "incorrected" to match each other, and never get far out of synch with each other.

Edited October 28, 2011 by J.C.MacSwell

[timo]

The question is: what difference of time these clocks will show per day (24h), according to the theory of relativity?

Assuming a sufficiently long separation time such that the effects caused by separating the clocks and bringing them back together for comparison can be ignored the theory of relativity predicts a difference of zero tics per 24 hours in your scenario. But I wonder: Why do you even care?

[swansont]

These effects should cancel with that of the proper time elapse of each location.They would both be constantly "incorrected" to match each other, and never get far out of synch with each other.

 

What is the source of the correction?

 

We know from the Pound-Rebka experiment that the frequency will change — it will not be in synch with an oscillator at the receiver location. That phase difference will continue to accumulate.

[J.C.MacSwell]

What is the source of the correction?

 

We know from the Pound-Rebka experiment that the frequency will change — it will not be in synch with an oscillator at the receiver location. That phase difference will continue to accumulate.

 

My interpretation of the proposed scenario:

 

The remote driver of the signal is driving the two clocks.Their time can vary but the difference cannot accumulate.

 

If the two clocks were proper time clocks for their respective locations the difference would accumulate, but they are constantly adjusted ("incorrected") to match within signal time and 1 cycle time differences by the remote driver.

[swansont]

My interpretation of the proposed scenario:

 

The remote driver of the signal is driving the two clocks.Their time can vary but the difference cannot accumulate.

 

If the two clocks were proper time clocks for their respective locations the difference would accumulate, but they are constantly adjusted ("incorrected") to match within signal time and 1 cycle time differences by the remote driver.

 

A clock measure the accumulated phase of the oscillator. If you have different frequencies, the time difference will continually grow. The description of the clocks (as I read it) is that the oscillator is the incoming signal; there's no local clock that's being continually re-synched.

[J.C.MacSwell]

A clock measure the accumulated phase of the oscillator. If you have different frequencies, the time difference will continually grow. The description of the clocks (as I read it) is that the oscillator is the incoming signal; there's no local clock that's being continually re-synched.

 

I don't see the difference. The incoming signal is received at different frequencies but the time elapsed at each location compensates. So the clocks are kept in synch.

 

After 10 (or 10,000,000) signals/cycles only the signal lag of the last signal, including any relativistic effects, will effect the difference in the clocks. Once received, each will have received 10 (or 10,000,000) signals/cycles

[swansont]

I don't see the difference. The incoming signal is received at different frequencies but the time elapsed at each location compensates. So the clocks are kept in synch.

 

After 10 (or 10,000,000) signals/cycles only the signal lag of the last signal, including any relativistic effects, will effect the difference in the clocks. Once received, each will have received 10 (or 10,000,000) signals/cycles

 

OK, I think I I see what you're saying. The upper signal will be the same as the lower one. But one (or both) will disagree with a local clock. (This is basically an analogue of what GPS does by using an oscillator that has been fractionally slowed down)

[J.C.MacSwell]

OK, I think I I see what you're saying. The upper signal will be the same as the lower one. But one (or both) will disagree with a local clock. (This is basically an analogue of what GPS does by using an oscillator that has been fractionally slowed down)

 

That's it. Not that the clocks are right. Essentially they are reading the signallers time, not there own. So the clocks tend to agree, even if they accumulate an increasing error for their location (+ or - depending on location)

 

OK, I think I I see what you're saying. The upper signal will be the same as the lower one. But one (or both) will disagree with a local clock. (This is basically an analogue of what GPS does by using an oscillator that has been fractionally slowed down)

 

That's it. Not that the clocks are right. Essentially they are reading the signallers time, not there own. So the clocks tend to agree, even if they accumulate an increasing error for their location (+ or - depending on location)

 

I guess the GPS gets tuned to "Earth time" (lapse) not it's own.

Edited October 31, 2011 by J.C.MacSwell

[Guest Rollin78]

Relative velocity time dilation is an effect ofspecial relativity, and can happen independently of gravitational fields. Bothobservers disagree about the difference between their clocks.

Gravitational time dilation is an effect ofgeneral relativity, and happens only in gravitational fields.

Both observers agree about the relative differencebetween their clocks.

Since special relativity is derivable form generalrelativity, they're really both effects of general relativity.

 

The simplified specific equations are differentfor each type of time dilation.