Clock Ticking in Gravitational Well: Observation Rate

In summary, as the clock drops towards the mass, it would appear to tick more slowly for someone standing far away, while for someone standing on the planet, it would appear to tick more rapidly. This is due to the effects of gravitational time dilation and the relativistic Doppler shift. However, the actual observed rate would also be affected by the kinematic blueshift for the person standing below the clock.
  • #1
sqljunkey
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I was wondering what someone standing far away from a planet with mass would see if he drops a clock towards the mass. And then vice versa if I was standing on the planet, what would I see. would I see the clock tick fast and then slow as it approaches?

Thanks!
 
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  • #2
Although in reality one cannot totally ignore the time dilation due to speed, I think that (1) it's small enough to ignore and (2) it's not really intended to be part of your question. SO ... ignoring it you get:

As the clock drops towards the mass, you would see it tick more slowly and the person on the ground would see it tick more rapidly. This is the effect to gravitational time dilation.
 
  • #4
sqljunkey said:
Wouldn't the clock tick slower as it gets closer to the person standing on Earth though? https://en.wikipedia.org/wiki/Gravity_Probe_A
All clocks, unless they are broken, tick at one second per second regardless of where they are in gravity wells and how fast they are going relative to another object.
 
  • #5
The person standing far away would see just redshift, because the motion away causes redshift and the increasing time dilation causes redshift.

For the person below the clock, and sticking to weak field, the gravitational potential at radius ##r## is ##\phi(r)=GM/r## and the tick rate of a clock hovering at that radius as observed by someone at ##r=r_E## is ##1-\phi(r)/c^2+\phi(r_E)/c^2## ticks per second. At radius ##r## conservation of energy says that the clock has acquired velocity ##v(r)##, where ##v(r)^2=2GM/r##. Plugging that into the Lorentz ##\gamma## gets you the Lorentz factor as a function of ##r##, ##\gamma(r)##, and hence the kinematic time dilation. Therefore the total rate is ##(1-\phi(r)/c^2+\phi(r_E)/c^2)/\gamma(r)## ticks per second.

Plot graphs to check that the results for kinematic and gravitational time dilation are reasonable independently (it's early here, sign errors are a distinct possibility). Then plot the final graph and see what you get. Or differentiate wrt ##r## and look for zeroes.
 
  • #6
Oh yeah, and you may need to multiply by the naive Newtonian kinematic blueshift depending on what you want to know.
 
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  • #7
sqljunkey said:
I was wondering what someone standing far away from a planet with mass would see if he drops a clock towards the mass. And then vice versa if I was standing on the planet, what would I see. would I see the clock tick fast and then slow as it approaches?

Thanks!
Unless it was a really giant clock, I think it would be too small to see.
 
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  • #8
Ibix said:
The person standing far away would see just redshift, because the motion away causes redshift and the increasing time dilation causes redshift.
And the person standing below the clock would see just blueshift, because the motion towards causes blueshift and the decreasing gravitational time dilation (since the clock is higher up than the observer) causes blueshift.

Ibix said:
the kinematic time dilation
Is not what the observer actually sees. He sees the relativistic Doppler shift, which for motion towards the observer is ##\sqrt{(1 + v)/(1 - v)}##.

The correct formula for this case combines the gravitational blueshift with the Doppler blueshift, and since we have ##v(r) = \sqrt{2 \phi(r)}##, we have (rearranging things slightly to make it clear that the final answer is greater than ##1##):

$$
\left[ 1 + \left( \phi(r_E) - \phi(r) \right) \right] \sqrt{\frac{1 + \sqrt{2 \phi(r)}}{1 - \sqrt{2 \phi(r)}}}
$$
 
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  • #9
PeterDonis said:
Is not what the observer actually sees. He sees the relativistic Doppler shift, which for motion towards the observer is ##\sqrt{(1 + v)/(1 - v)}##.
Agreed - hence my additional comment #6.
 
  • #10
PeterDonis said:
The correct formula for this case combines the gravitational blueshift with the Doppler blueshift
And actually, dropping the weak field approximation makes the formula less messy for a change; the gravitational factor is ##\sqrt{(1 - 2M / r) / (1 - 2M / r_E)}##, and the final formula then becomes, using ##v(r)^2 = 2M / r##:

$$
\sqrt{\frac{1 - v(r)^2}{1 - 2M / r_E}} \sqrt{\frac{1 + v(r)}{1 - v(r)}} = \frac{1 + \sqrt{2M / r}}{\sqrt{1 - 2M / r_E}}
$$
 
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  • #11
so if I added the c^2 terms back into the final equations from peterdonis at post #10 I would get "ticks" per second? and then if I wanted to calculate time at r for the clock, standing on earth, I would take the doppler blueshift off, and just calculate the gravitational shift at r?
 
  • #12
If you are adding ##c^2## back in you need to put ##G## back in too.

The result is a dimensionless quantity that is the number of distant ticks per local tick seen (literally seen) by a person at ##r_E## looking at a clock at ##r## that has fallen from rest at infinity.

I don't know what you mean by "the time at ##r##". If you want to know the time the falling clock shows at ##r## then no, you need to do some integral and it's messy. Probably easier to work with the geodesic equation, messy as that is. If you mean that you want to know the instantaneous rate of the falling clock, assuming Schwarzschild coordinates, pwhen it's at ##r## then you need to decrease the rate by the Newtonian blueshift, yes. This isn't a particularly meaningful figure, as the coordinate dependence should hint. It isn't anything anyone measures, or anything you can use in the way a similar calculation for a satellite in a circular orbit can use.
 

FAQ: Clock Ticking in Gravitational Well: Observation Rate

What is the clock ticking rate in a gravitational well?

The clock ticking rate in a gravitational well is affected by the strength of the gravitational field. The closer an object is to a massive body, the slower its clock will tick compared to a clock in a weaker gravitational field.

How does the clock ticking rate change near a black hole?

Near a black hole, the clock ticking rate slows down significantly due to the extreme strength of the gravitational field. This phenomenon is known as gravitational time dilation.

Why is the clock ticking rate important in gravitational well observations?

The clock ticking rate is important in gravitational well observations because it allows us to measure the strength of the gravitational field. By comparing the ticking rate of a clock in a strong gravitational field to that of a clock in a weaker field, we can determine the strength of the gravitational well.

Can the clock ticking rate be affected by other factors besides gravity?

Yes, the clock ticking rate can also be affected by factors such as velocity and acceleration. However, in the context of gravitational well observations, the effects of these factors are negligible compared to the effects of gravity.

How does the clock ticking rate affect the concept of time in a gravitational well?

The clock ticking rate in a gravitational well can significantly alter our perception of time. Time appears to pass slower in stronger gravitational fields, meaning that an observer in a weaker field would perceive time passing faster for an object in a stronger field. This concept is known as time dilation.

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