High School Clock Ticking in Gravitational Well: Observation Rate

[sqljunkey]

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!

 

[phinds]

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.

 

[sqljunkey]

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

 

[phinds]

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.

 

[Ibix]

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.

 

[Ibix]

Oh yeah, and you may need to multiply by the naive Newtonian kinematic blueshift depending on what you want to know.

 

[PeroK]

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.

 

[PeterDonis]

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.

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 $1$):

$$
\left[ 1 + \left( \phi(r_E) - \phi(r) \right) \right] \sqrt{\frac{1 + \sqrt{2 \phi(r)}}{1 - \sqrt{2 \phi(r)}}}
$$

 

[Ibix]

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.

 

[PeterDonis]

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 $v(r)^2 = 2M / r$:

$$
\sqrt{\frac{1 - v(r)^2}{1 - 2M / r_E}} \sqrt{\frac{1 + v(r)}{1 - v(r)}} = \frac{1 + \sqrt{2M / r}}{\sqrt{1 - 2M / r_E}}
$$

 

[sqljunkey]

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?

 

[Ibix]

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.