Gravitational time dilation near Earth

[DanMP]

We know (measured) that a clock on a mountain "ticks" faster than a clock at sea level. At higher altitude, the clock runs even faster.

Now, if we go much higher, towards the Sun, on the line between the center of the Earth and the center of the Sun, the clock should begin, at some point, to run slower, influenced by Sun's gravitation. Where is that "point" (where the clock would stop increasing its tick rate and then start to decrease again), at what distance from the center of the Earth? It was calculated and/or tested experimentally?

PS I asked the same question on another forum and made some progress in finding the answer, but it seems that the value I got is not good enough ... Please help.

 

[Dale]

What assumptions are you making regarding the orbits? I.e. Do you want to ignore Earth's orbital motion? If not do you want to assume that the clock is in orbit around the sun? Or do you want to assume that the clock is using thrust to maintain some specified location?

 

[DanMP]

... do you want to assume that the clock is in orbit around the sun? Or do you want to assume that the clock is using thrust to maintain some specified location?

The clock(s) should be (using thrust) on the Sun-Earth line (rotating around the Sun with Earth's angular speed).

Comparing the difference in ticking/frequency between two clocks on that line (close enough to each other), we should detect the point of inversion.

 

[Dale]

I don't know the metric that would describe this system. Maybe someone else here does.

 

[PeterDonis]

I don't know the metric that would describe this system.

There isn't a known exact solution for this case. However, we can make a pretty good approximation if we assume weak fields and slow motion, ignore the rotation of the Sun and Earth, and just superpose the Schwarzschild solution for the Sun and the Schwarzschild solution for the Earth (in the weak field slow motion approximation). I'm pretty sure the nonlinear terms involved (i.e., the ones that would not be included in the superposition) are too small to matter.

For an even simpler first approximation, we could ignore the revolution of the Earth around the Sun and treat the system as static. I'll write down what I get in that case below.

I asked the same question on another forum and made some progress in finding the answer, but it seems that the value I got is not good enough

I don't think the suggestions and calculations in those other discussions are correct.

The approach I described just above in response to @Dale (where we ignore the Earth's revolution around the Sun and superpose the weak field slow motion approximations for the Earth and the Sun) would result in an equation that looks like this (I am using units in which $G = c = 1$):

$$
\phi(x) = 1 - \frac{M_E}{x} - \frac{M_S}{R - x}
$$

where $M_E$ is the mass of the Earth, $M_S$ is the mass of the Sun, $x$ is the distance from the center of the Earth along the Earth-Sun line, and $R$ is the distance between the centers of the Sun and Earth. It should be straightforward to show that $\phi(x)$ has a maximum in the interval $0 < x < R$ and determine what it is.

 

[Dale]

if we assume weak fields and slow motion, ignore the rotation of the Sun and Earth, and just superpose the Schwarzschild solution for the Sun and the Schwarzschild solution for the Earth

That would be my approach too, but I don't know the limitations or subtleties enough to be confident in it. I was not sure if there were other assumptions beyond weak field and slow motion required for a linearized approach.

 

[PeterDonis]

I don't know the limitations or subtleties enough to be confident in it

The weak field, slow motion limit is sufficient to make linear superposition of solutions a good approximation for this case. Weak field implies $M / r << 1$ for all $M$ and $r$ that are relevant; slow motion implies $v << 1$ for all $v$ that are relevant. Any nonlinear terms in the case under consideration will be products of two or more factors of one of those forms ($M / r$ or $v$), so they are all negligible in this limit.

 

[David Lewis]

We know (measured) that a clock on a mountain "ticks" faster than a clock at sea level.

It only appears to be fast when you observe it from sea level. If you were on top of the mountain, you wouldn't notice anything amiss. Both clocks tick at the same rate (in their own frame) but measure a different amount of time. I believe that's an equivalent way of saying the same thing.

 

[DanMP]

There isn't a known exact solution for this case. However, we can make a pretty good approximation if we assume weak fields and slow motion, ignore the rotation of the Sun and Earth, and just superpose the Schwarzschild solution for the Sun and the Schwarzschild solution for the Earth (in the weak field slow motion approximation).
...
The approach I described just above in response to @Dale (where we ignore the Earth's revolution around the Sun and superpose the weak field slow motion approximations for the Earth and the Sun) would result in an equation that looks like this (I am using units in which $G = c = 1$):

$$
\phi(x) = 1 - \frac{M_E}{x} - \frac{M_S}{R - x}
$$

where $M_E$ is the mass of the Earth, $M_S$ is the mass of the Sun, $x$ is the distance from the center of the Earth along the Earth-Sun line, and $R$ is the distance between the centers of the Sun and Earth. It should be straightforward to show that $\phi(x)$ has a maximum in the interval $0 < x < R$ and determine what it is.

Your solution/suggestion leads to what I obtained here, also ignoring the rotation (not acceptable, the rotation is present and important).

 

[DanMP]

It only appears to be fast when you observe it from sea level. If you were on top of the mountain, you wouldn't notice anything amiss. Both clocks tick at the same rate (in their own frame) but measure a different amount of time. I believe that's an equivalent way of saying the same thing.

If you have 2 accurate synchronized clocks at sea level, take one on a mountain and return after few days, you'll see that the sea level clock is behind the mountain clock, so the difference in ticking is real, although nothing is/was amiss.

See here:

Observers in relative motion or at different gravitational potentials measure disparate clock rates. These predictions of relativity have previously been observed with atomic clocks at high velocities and with large changes in elevation. We observed time dilation from relative speeds of less than 10 meters per second by comparing two optical atomic clocks connected by a 75-meter length of optical fiber. We can now also detect time dilation due to a change in height near Earth’s surface of less than 1 meter.

I think that with this technique it is possible to experimentally detect the point of inversion/transition. It should be done, because it may be important.

 

[phinds]

If you have 2 accurate synchronized clocks at sea level, take one on a mountain and return after few days, you'll see that the sea level clock is behind the mountain clock, so the difference in ticking is real, although nothing is/was amiss.

Which is exactly what David Lewis was pointing out. When you are on the mountain with the clock on the mountain, it is ticking at exactly one second per second. When you are at sea level with the sea level clock it is ticking at exactly one second per second. ALL clocks tick at one second per second in their local frame. Since the two clocks are on different space-time paths, they tick for a different AMOUNT of ticks but that's not because they tick faster or slower. They do not tick faster or slower, they just travel through different space-time paths.

 

[DanMP]

... Since the two clocks are on different space-time paths, they tick for a different AMOUNT of ticks but that's not because they tick faster or slower. They do not tick faster or slower, they just travel through different space-time paths.

According to Chou, C. W.; Hume, D. B.; Rosenband, T.; Wineland, D. J. - Science, Volume 329, Issue 5999, pp. 1630- (2010):

Observers in relative motion or at different gravitational potentials measure disparate clock rates.

Are they wrong?

 

[phinds]

According to Chou, C. W.; Hume, D. B.; Rosenband, T.; Wineland, D. J. - Science, Volume 329, Issue 5999, pp. 1630- (2010):

Are they wrong?

No, you are misunderstanding what is being said vs what I said. What they are saying is that FROM DIFFERENT FRAMES OF REFERENCE you see the clock ticking at a different rate. That is true. What I said is that in the FOR of the clock you do not see it ticking at anything but 1 second per second. You are confusing time dilation with differential aging due to different paths through space time. Time dilation is frame dependent whereas differential aging is not.

 

[PeterDonis]

Your solution/suggestion leads to what I obtained here,

Yes. The force is the gradient of the potential, so at an extremum of the potential (in this case a maximum, as is easy to show by taking the second derivative), the force is zero.

also ignoring the rotation (not acceptable, the rotation is present and important)

We are not really "ignoring" the rotation; we are just considering a hypothetical case in which, instead of having a clock that orbits the Sun with the same period as the Earth, we consider a clock that just moves along a radial line between the Sun and some point on the Earth's orbit, starting at the Earth's orbit and moving inwards, and asking at what point along that radial line its clock rate is maximized.

We could also consider a different, somewhat less hypothetical case, in which we take a clock that is moving around the Sun with the same period as the Earth, and then start it moving inward from the Earth towards the Sun while maintaining the same orbital period around the Sun (another way to think of this would be to imagine a 150 million km long tether between the Earth and the Sun, and imagine the clock moving inward along the tether). We could then ask at what point this clock's rate would be maximized. (See below for more on that.)

The point of all this is that there is not a single unique answer to the question you asked at the end of your scienceforums.net post: "where is the real point in which a clock going towards the Sun would reverse its ticking rate?" There is no "real point" in any absolute sense; there are different points for clocks with different motions. A more general formula that allows the clock to have an arbitrary velocity with respect to the Sun and the Earth, but still requires weak fields and slow motions (i.e., all velocities much less than $c$), is (in units where $G = c = 1$) [Edit--fixed $v^2$ terms]:

$$
\phi(r_E, r_S, v_E, v_S) = 1 - \frac{M_E}{r_E} - \frac{M_S}{r_S} - \frac{1}{2} v_E^2 - \frac{1}{2} v_S^2
$$

Note that this is a function of four variables, $r_E$ (the distance from the object to the Earth), $r_S$ (the distance from the object to the Sun), $v_E$ (the velocity of the object relative to the Earth), and $v_S$ (the velocity of the object relative to the Sun). In the very simple case I addressed before, I assumed $v_E = v_S = 0$ and used the obvious relationship between $r_E$ and $r_S$ (which I called $x$ and $R - x$, respectively) for that case.

For the case where the clock is orbiting the Sun with the same period as the Earth, the relationships are more complicated, but you can still express everything in terms of, say, $r_E$, since all of the other variables will have a known dependence on that one for this case. That allows you to express $\phi$ as a function of one variable and use the standard technique to find its maximum.

 

[DanMP]

No, you are misunderstanding what is being said vs what I said. ...

The important thing is that we can see/measure disparate clock rates for different heights. Comparing 2 optical clocks connected by optical fiber and "climbing" towards the Sun, one above the other, we should be able to find the point of inversion/transition.

 

[PeterDonis]

Comparing 2 optical clocks connected by optical fiber and "climbing" towards the Sun, one above the other, we should be able to find the point of inversion/transition

In principle, yes, this would work. (In practice it would require a very large investment in space propulsion technology in order to keep the "climbing" clock on the proper trajectory, since that trajectory will not be a free fall trajectory except at one point--the inversion point--and will require continuous thrust with very precise control, which we don't currently have the technology to do.) The predicted answer can be derived using the procedure I described in my previous post.

 

[phinds]

The important thing is that we can see/measure disparate clock rates for different heights. Comparing 2 optical clocks connected by optical fiber and "climbing" towards the Sun, one above the other, we should be able to find the point of inversion/transition.

Agreed, and the optical fiber would not be needed, you could just send a transmitter along with the clock. Appropriate calculations would compensate for the transmission delay just as they would in an optical fiber.

 

[DanMP]

$$
\phi(r_E, r_S, v_E, v_S) = 1 - \frac{M_E}{r_E} - \frac{M_S}{r_S} - v_E^2 - v_S^2
$$

Note that this is a function of four variables, $r_E$ (the distance from the object to the Earth), $r_S$ (the distance from the object to the Sun), $v_E$ (the velocity of the object relative to the Earth), and $v_S$ (the velocity of the object relative to the Sun). In the very simple case I addressed before, I assumed $v_E = v_S = 0$ and used the obvious relationship between $r_E$ and $r_S$ (which I called $x$ and $R - x$, respectively) for that case.

For the case where the clock is orbiting the Sun with the same period as the Earth, the relationships are more complicated, but you can still express everything in terms of, say, $r_E$, since all of the other variables will have a known dependence on that one for this case. That allows you to express $\phi$ as a function of one variable and use the standard technique to find its maximum.

Thank you, but it seems to lead to the solution I mentioned in OP.

$v_E=0$ and $v_S=\omega (R-x)$ ...

 

[PeterDonis]

it seems to lead to the solution I mentioned in OP.

$v_E=0$ and $v_S=\omega (R-x)$ ...

I'm not sure $v_E = 0$ is actually correct, because the velocities are supposed to be relative to a non-rotating frame, and we now have the Earth rotating around the Sun, so a clock that stays between the Earth and the Sun is moving in a non-rotating frame centered on the Earth. I think that results in $v_E = \omega x$.

 

[DanMP]

I'm not sure $v_E = 0$ is actually correct, because the velocities are supposed to be relative to a non-rotating frame, and we now have the Earth rotating around the Sun, so a clock that stays between the Earth and the Sun is moving in a non-rotating frame centered on the Earth. I think that results in $v_E = \omega x$.

You are right.

The solution is not the same. I'll do the math tomorrow.

Until then, you are sure that your equation/formula is correct? How you obtained it? What does it mean?

 

[PeterDonis]

you are sure that your equation is correct? How you obtained it?

It's a superposition of two copies of the weak field, slow motion approximation to the Schwarzschild solution, one centered on the Sun and one centered on the Earth. The Schwarzschild solution itself (no approximation) gives

$$
\phi = \sqrt{1 - \frac{2M}{r} - v^2}
$$

for an object moving at velocity $v$ at radius $r$ relative to a central mass $M$. For weak fields $M << r$ and slow motion $v << 1$ we can expand the square root and drop quadratic and higher terms to obtain

$$
\phi = 1 - \frac{M}{r} - \frac{1}{2} v^2
$$

Superposing the two copies means multiplying together the $\phi$ expressions for each one and then expanding out the product and again dropping quadratic and higher terms.

(Note that in my earlier post I left out the factor $1/2$ in front of the $v^2$ terms; I have edited that post to fix this.)

 

[DanMP]

It's a superposition of two copies ...
(Note that in my earlier post I left out the factor $1/2$ in front of the $v^2$ terms; I have edited that post to fix this.)

Ok :) with $1/2$ it is my OP solution (that's why I asked about the equation/formula ...), if we ignore the (weak $\omega ^2 x$) centrifugal force resulted from the "rotation" around the Earth. But if we don't ignore it, the point is closer to the Earth, as I expected. Perfect. Thank you!

 

[DanMP]

... A more general formula that allows the clock to have an arbitrary velocity with respect to the Sun and the Earth, but still requires weak fields and slow motions (i.e., all velocities much less than $c$), is (in units where $G = c = 1$) [Edit--fixed $v^2$ terms]:

$$
\phi(r_E, r_S, v_E, v_S) = 1 - \frac{M_E}{r_E} - \frac{M_S}{r_S} - \frac{1}{2} v_E^2 - \frac{1}{2} v_S^2
$$

Note that this is a function of four variables, $r_E$ (the distance from the object to the Earth), $r_S$ (the distance from the object to the Sun), $v_E$ (the velocity of the object relative to the Earth), and $v_S$ (the velocity of the object relative to the Sun). In the very simple case I addressed before, I assumed $v_E = v_S = 0$ and used the obvious relationship between $r_E$ and $r_S$ (which I called $x$ and $R - x$, respectively) for that case.

For the case where the clock is orbiting the Sun with the same period as the Earth, the relationships are more complicated, but you can still express everything in terms of, say, $r_E$, since all of the other variables will have a known dependence on that one for this case. That allows you to express $\phi$ as a function of one variable and use the standard technique to find its maximum.

In SI units, the above formula becomes:

$$
\phi(r_E, r_S, v_E, v_S) = 1 - \frac{GM_E}{c^2 r_E} - \frac{GM_S}{c^2 r_S} - \frac{v_E^2}{2 c^2} - \frac{v_S^2}{2 c^2} ?
$$

or

$$
\phi(r_E, r_S, v_E, v_S) = 1 - \frac{2GM_E}{c^2 r_E} - \frac{2GM_S}{c^2 r_S} - \frac{v_E^2}{c^2} - \frac{v_S^2}{c^2} ?
$$

or ?

 

[DanMP]

Agreed, and the optical fiber would not be needed, you could just send a transmitter along with the clock. Appropriate calculations would compensate for the transmission delay just as they would in an optical fiber.

You are sure that optical fiber is not needed? How they compared the optical clocks rates?

 

[phinds]

You are sure that optical fiber is not needed? How they compared the optical clocks rates?

They transmit via optical cable but I see no reason why that would be needed except perhaps that it's simpler than adding transmitters/receivers. Also, transmitters/receivers would have to have an additional compensation built into the calculations for the delay times through them. Yeah, it does look more complicated, but again, technically feasible.

 

[DanMP]

They transmit via optical cable but I see no reason why that would be needed ...

Maybe their technique involves optic methods to compare optical clock rates/frequencies ... How else it can be done?

 

[phinds]

Maybe their technique involves optic methods to compare optical clock rates/frequencies ... How else it can be done?

Whatever information they are sending via cable could just as well be sent via radio.

 

[DanMP]

Whatever information they are sending via cable could just as well be sent via radio.

How do you send by radio a beam of light of a certain frequency?
What other information do you think is transmitted?

 

[Ibix]

Optic fibres keep the FCC (or whatever your local regulatory body is) off your back and keep relatively rapidly varying atmospheric conditions from mucking around with signal travel times and keep external light pollution from interfering. Lots of reasons to use them - but they amount to engineering rather than the actual physics being tested. If you could do the experiment on the moon during lunar night (dark, airless, unregulated) then you probably wouldn't bother with the fibres.

 

[DanMP]

In principle, yes, this would work. (In practice it would require a very large investment in space propulsion technology in order to keep the "climbing" clock on the proper trajectory, since that trajectory will not be a free fall trajectory except at one point--the inversion point--and will require continuous thrust with very precise control, which we don't currently have the technology to do.)

I suggest 2 optical clocks linked with a cable and fiber optic. The upper clock would use thrust to climb in a spiral from the Earth, towing the second clock.

A transition should occur all around the Earth, not only on the Sun-Earth line. On Earth orbit line, after an increase in ticking (going up from the Earth), should follow a steady ticking zone.

 

[PeterDonis]

In SI units, the above formula becomes

The first one you gave is correct.

 

[DanMP]

Optic fibres keep the FCC (or whatever your local regulatory body is) off your back and keep relatively rapidly varying atmospheric conditions from ...

What means FCC?

Do you know their method? How they compared optical clock rates?

 

[Ibix]

What means FCC?

Federal Communications Commission. They object to unlicensed radio signals in the US - similar bodies exist elsewhere. There are plenty of ways of not annoying them - using light signals in fibres is one.

Do you know their method? How they compared optical clock rates?

No. The classic experiment is the Pound-Rebka experiment, which is a really clever way of comparing the frequency of light emitted at the top of a tower to the frequency received at the bottom. The details of that aren't paywalled and are easily Googleable.

 

[DanMP]

The first one you gave is correct.

Thank you!

So, we have:
$$
\phi(r_E, r_S, v_E, v_S) = 1 - \frac{GM_E}{c^2 r_E} - \frac{GM_S}{c^2 r_S} - \frac{v_E^2}{2 c^2} - \frac{v_S^2}{2 c^2}
$$

where: $r_E=x$, $r_S=R-x$ (R=Sun-Earth distance), $v_E=\omega x$, $v_S=\omega (R-x)$ ($\omega$ = Earth's angular speed around the Sun)

So, we can write:
$$
\phi(r_E, r_S, v_E, v_S) = \phi(x)= 1 - \frac{GM_E}{x c^2} - \frac{GM_S}{(R-x) c^2} - \frac{(\omega x)^2}{2 c^2} - \frac{(\omega (R-x))^2}{2 c^2}
$$
and
$$
\frac{d \phi(x)} {dx} = \frac{GM_E}{x^2 c^2} - \frac{GM_S}{(R-x)^2 c^2} - \frac{\omega^2 x}{c^2} + \frac{\omega^2 (R-x)}{c^2}
$$

We need $\frac{d \phi(x)} {dx} = 0$

so
$$
\frac{GM_E}{x^2 c^2} - \frac{GM_S}{(R-x)^2 c^2} - \frac{\omega^2 x}{c^2} + \frac{\omega^2 (R-x)}{c^2}=0
$$
or
$$
\frac{GM_E}{x^2} - \frac{GM_S}{(R-x)^2} - \omega^2 x+ \omega^2 (R-x)=0
$$
or
$$
\frac{GM_S}{(R-x)^2} + \omega^2 x = \frac{GM_E}{x^2} + \omega^2 (R-x)
$$

so x is the place where gravitational and centrifugal forces acting on the clock cancel each other ... Interesting.

The value is: 1,349,811,584 m

 

[PeterDonis]

x is the place where gravitational and centrifugal forces acting on the clock cancel each other

Yes, that is to be expected since, in this approximation, the force is the gradient of the potential, and the potential is what determines the clock rate. So the maximum of the potential, which is where the maximum clock rate will be, is also where the gradient is zero and therefore the net force is zero.