# B Uniform-gravitational time dilation -- exact or approximate

1. Jul 24, 2017

### Hiero

If a clock runs n times faster than another clock due to flawed design, then it is logically necessary (and rather trivial) that the other clock runs 1/n times slower than the flawed clock.

I thought the same should be the case for gravitational time dilation; if a clock on a tower runs n times faster than a clock on the ground, then I expected the clock on the ground to run 1/n times slower than the one on the tower.

The formula for uniform gravitational time dilation which I have learned is the clock on top will run faster than the ground clock by a factor (1+gh), but also the clock on the ground will run slower by a factor (1-gh)

So we do not have the n=>1/n relationship .... although we do have that relationship for small h [if we neglect 2nd+ order terms of h, then 1/(1-gh)=1+gh].

I think the {(1+gh), (1-gh)} factors are exact not approximate, but this seems strange to me that they aren't reciprocal.

To make it more clear what bothers me: After 10 seconds on the ground clock, a ground observer might see 11 seconds passed on the tower clock. But then when 11 seconds pass on the tower clock, a tower observer will see only 9.9 seconds passed on the ground clock.

I suppose nothing is wrong with this apparent discrepancy?

2. Jul 24, 2017

### .Scott

In the situation you are describing, the relationship is 1:n or (1/n):n. Of course, n is so close to 1, that it is very nearly (1+d):1 and 1:(1-d).

3. Jul 24, 2017

### Hiero

So you're saying it is just an approximate formula... But then what is the exact form? Will the bottom clock see the top clock run at a rate (1+gh) or 1/(1-gh) (or something else of which these are both approximate forms)?

(Taking gravity to be exactly uniform still.)

4. Jul 24, 2017

### Hiero

It would be nice if it had a form like n = (1+gh)/(1-gh) so as to have the property that when h becomes -h, n becomes 1/n

5. Jul 24, 2017

### Staff: Mentor

From where? Please give a reference so we have context.

6. Jul 24, 2017

### PAllen

The formula 1 + gh is exactly correct for the local frame of a uniformly accelerating observer. (Precisely, it follows from the metric in Fermi-Normal coordinates for such an observer, which is readily obtained by shifting the origin from standard Rindler coordinates to a particular Rindler observer being the origin).

The mystery is solved by the understanding that if mutual measured distance between uniformly accelerating observers is to remain constant, they must have slightly different uniform acceleration.

Specifically, if acceleration is g at x=0, it will necessarily be g / (1 + g h ) for an observer at x=h (in units with c=1). If you then apply the 1 + a h formula using -h and this the modified acceleration for the observer at h, you will get 1/(1 + gh) for the dilation of the first observer per the second. Mystery solved.

7. Jul 24, 2017

### Hiero

The derivation involves using the equivalence principle and (perhaps unsurprisingly) neglecting 2nd+ order terms.

It's actually seems obvious now what the exact form should be, I should've thought about it more before making this thread.
(It seems exactly analogous to the approximations of thermal expansion {(1+kΔT),(1-kΔT)})

If we have the approximation (t is time) Δt ≈ t(1+gh)-t = tgh then the obvious thing to do is take it to be true on an infinitesimal scale [dt = tg(dh)] then integrate.

8. Jul 24, 2017

### Hiero

So you're saying my previous post (#7) is wrong? :\

9. Jul 24, 2017

### Hiero

@PAllen I understand the idea behind Rindler space (that the acceleration must differ at each point to appear static) but isn't a uniform gravitational field somehow different?

In other words I don't see why your comments on Rindler space is relevant to uniform gravitational dilation (sorry if I missed your point).

10. Jul 24, 2017

### PAllen

By the principle of equivalence, a uniformly accelerated frame in flat spacetime is the same as uniform gravitation. Further, in any situation in SR and GR, if two nearby observers were to have the same proper acceleration, then per each they would be moving apart. Thus, to have have two observers, each with constant proper acceleration staying the same distance apart so they can compare clocks over time, they need to have different constant proper accelerations. The Rindler case is the closest one can conceivably mean by uniform gravitational field.

11. Jul 24, 2017

### PAllen

yes.

12. Jul 24, 2017

### Hiero

@PAllen
It is my understanding that even a uniform gravitational field cannot be reproduced by an accelerating frame over finite distances (i.e. the "locally" part of the equivalence principle is about more than just tidal effects). A Rindler frame is as close as we could get, but over finite distances it is not the same as a uniform gravitational field.

My integral in post #7 implies that if a clock is separated from another clock by Δh, (positive being higher-up) then it will run at a rate e^(gΔh) compared to the first clock. This is the exact form I was looking for (where Δh→-Δh ⇒ n → 1/n). You are saying this result is wrong and the (1+gΔh) result is the exact one...?

13. Jul 24, 2017

### PAllen

The thing is that the only metric (SR or GR, noting that SR is just that flat special case of GR) that exactly produces time dilation of 1 + g h, if h is a conventionally measured distance, is the translated Rindler metric. This exact metric has no issue with the paradox you initially presented, because if you transform to coordinates with a different origin, the g changes, such that the time dilations are reciprocal. So there is no need to treat this as only approximate. Further, any small region of spacetime is approximated by this case.

If you want to talk about something different, you then need to introduce curvature (else the exact Rindler case is the only thing possible). But if you introduce curvature, it is not really possible talk about uniform gravity in any meaningful sense.

14. Jul 24, 2017

### Hiero

@PAllen you are confusing me. A tower on Earth is not a Rindler metric. My question is about the rates of two clocks at the bottom/top of a tower on Earth. I really do not care if "uniform gravity" cannot meaningfully exist in the context of spacetime-curvature... by "uniform gravity" I simply mean that 'the gravitational force does not vary appreciably over the height of the tower' (lest the integral get to messy).

I am just trying to ask a simple question about these two clocks between which is an essentially constant gravitational field.
That question is, is the rate of the top clock more accurately described by (1+gh) times the rate of the lower clock, or e^(gh) times the rate of the lower clock?

I am convinced the latter is the proper factor (so much so that I would not even have made this thread had I realized the exponential form a bit sooner) but you keep saying things that sound like arguments for the former factor being the correct one... Can you please clearly say which factor is more accurate for the situation I described?

15. Jul 24, 2017

### DrGreg

I suspect that if $h$ is small enough for the change in $g$ to be negligible, then the difference between $e^{gh}$ and $1+gh$ will also be negligible.

16. Jul 24, 2017

### PAllen

A tower on earth is the Rindler metric to reasonable precision. For higher precision it is the Schwarzschild metric. I am giving you the only simple answer that is reasonably correct, but you don't want to accept it. 1+ g h is what is observed until precision is extreme. When this isn't good enough, the Schwarzschild metric curvature corrections are what appear. Your proposed model does not appear.

17. Jul 24, 2017

### Hiero

I had the same thought, but it does not make sense to me.

"That the difference in approximations egΔh-(1+gh) is negligable" ↔ gΔh<<1 (=c^2)
"That Δh is small enough for the change in g to be negligble" ↔ km/R2-km/(R+Δh)2 << 1 (where R is the distance from the center of Earth to the tower bottom)

This second condition can be written (using km/R2=g and the taylor series expansion) as g - g(1-2Δh/R+O((Δh/R)^2) ≈ g - g(1-2Δh/R) = 2Δh/R << 1

So to sum up,

"That the difference in approximations egΔh-(1+gh) is negligable" ↔ gΔh<<1 (=c^2)
"That Δh is small enough for the change in g to be negligble" ↔ Δh/R << 1

But, we can increase g without changing Δh/R by increasing the mass of the planet. Thus we can violate the first condition while maintaining the second condition. Thus we can have a difference in the two approximation methods be significant, while still having insignificant changes in g.

This is the reason your suspicion seems false to me, but this train of thought it is open for criticism.

18. Jul 24, 2017

### Hiero

Fair enough. I only truly understand SR and don't intend to learn GR in the course of this thread. There is just one thing that bothers me, which is outlined at the (almost) end of the OP.

So I guess what I would like to ask you is, in this Schwarzschild metric, does the top clock see the bottom clock say 10 seconds when it says 11 seconds? (refer to OP)

19. Jul 24, 2017

### PAllen

Let me try again to explain why the Rindler metric is the correct explanation of reciprocal dilation up to curvature corrections. The principle of equivalence states ( for our purpose) that a local experiment on the surface of the earth is the same as an equivalent case in SR with no gravity up to the point curvature corrections are detectable. In SR, if two world lines have the same acceleration as measured by accelerometers, then their mutually measured distance increases steadily with time. To maintain constant mutually measured distance, they must have slightly different accelerations. Then, the principle of equivalence states this MUST be true on earth for a building that doesn't expand. The only observable corrections to the SR picture will be those due to curvature.

Last edited: Jul 24, 2017
20. Jul 24, 2017

### PAllen

I thought I explained this. The observed time dilations are exactly reciprocal. Let us call gl the acceleration at the bottom and gt the acceleration at the top. Then, until precision is such as to detect curvature corrections, the time dilation rate is given by 1 + gl h per the bottom and 1 - gt h per the top, wth gt related to gl so as to make these exact reciprocals: 1 - gt h = 1 / ( 1+ gl h). As h becomes small for a given g, the difference between gt and gl becomes negligible, and the reciprocal becomes indistinguishable from simple sign inversion.

Last edited: Jul 24, 2017
21. Jul 24, 2017

### PAllen

Oh, in the Schwarzschild metric, it would also be true that static observers at different altitude see exactly reciprocal clock rate ratios, but the function of distance is different from Rindler (by an amount impossible to detect for most experiments).

22. Jul 24, 2017

### Hiero

@PAllen Sorry if I seem to be stubborn, but it is because I am trying to understand what is wrong with using the exponential form?

You say in post #13 "Further, any small region of spacetime is approximated by this [translated Rindler metric]."

Suppose we have a string of clocks all the way up the tower, each dh apart. If we relate each clock to its neighbor by the Rindler dilation, then when Δt elapses on a certain clock, Δt(1+g dh) elapses on the clock above it. The change in the change in time (that is, the amount by which the upper clock is ahead) will be Δt(1+g dh) - Δt = Δtgdh
If we relate the time on the bottom-most clock to the time on the top-most clock by relating it through each neighboring clock by the Rindler dilation, then we have the differential equation [d(Δt)=g Δt d(h)] leading to the exponential form.

If small spacetime regions are approximated by the Rindler dilation, then isn't the above paragraph the most reasonable thing to do?

23. Jul 24, 2017

### PAllen

No, because g changes slightly along the way such as to make such an integration yield the simpler Rindler finite formula. As long as any observer uses their own exact acceleration, the simple formula applies, but the results are still exactly reciprocals for any two Rindler observers.

24. Jul 24, 2017

### Hiero

I don't understand this. Why can't g vary however the universe wants it to?

Also, I may be misunderstanding this: https://en.wikipedia.org/wiki/Gravitational_time_dilation but if you let g(h')=g (i.e. let g be constant over h') in the first formula on that page, then it seems to be the same as what I am saying.

25. Jul 24, 2017

### PAllen

Because if your region is small enough to ignore curvature, exactly constant g would tear the building apart over time ( see the bell spaceship paradox). If curvature is significant, it uniquely determines how g depends on position for static observers. Implicit in that formula is that the family of observers maintain static positions. You do not have arbitrary freedom to choose this. In flat spacetime Rindler or comoving inertial families are static. In Schwarzschild metric there is a unique such family. In many situations in GR, there is no such family, and gravitational time dilation cannot be defined at all (except for small spacetime regions that are then approximated by the Rindler metric ).

Last edited: Jul 24, 2017