Uniform-gravitational time dilation -- exact or approximate

[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?

 

[.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).

 

[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.)

 

[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

 

[PeterDonis]

The formula for uniform gravitational time dilation which I have learned

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

 

[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.

 

[Hiero]

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

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.

 

[Hiero]

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.

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

 

[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).

 

[PAllen]

@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).

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.

 

[PAllen]

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

yes.

 

[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.

yes.

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...?

 

[PAllen]

@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...?

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.

 

[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?

 

[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.

 

[PAllen]

@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?

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.

 

[Hiero]

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.

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

"That the difference in approximations e^gΔh-(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 << 1So to sum up,

"That the difference in approximations e^gΔh-(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.

 

[Hiero]

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.

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)

 

[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.

 

[PAllen]

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)

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.

 

[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).

 

[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?

 

[PAllen]

@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
So now say Δt is the time elapsed on the bottom-most clock. Then if we relate this time to the upper 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?

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.

 

[Hiero]

No, because g changes slightly along the way such as to make such an integration yield the simpler Rindler finite formula.

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.

 

[PAllen]

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.

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 ).

 

[Hiero]

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.

Can't we look at the other way around? That is, can't we choose how g varies and then say that g-profile uniquely defines a certain curvature (the curvature needed for the family of observers to be static)?

I would think we can look at this way, but then I guess the point is that there is no possible curvature to keep them static and with the same g?

 

[PAllen]

Can't we look at the other way around? That is, can't we choose how g varies and then say that g-profile uniquely defines a certain curvature (the curvature needed for the family of observers to be static)?

I would think we can look at this way, but then I guess the point is that there is no possible curvature to keep them static and with the same g?

Bingo. Your last statement is correct.

And the reason gets back to SR and the principle of equivalence. Whatever the curvature, we can take a sufficiently close group of observers that the geodesic deviation effectively vanishes. Then, the expansion tensor can be computed here as for SR, which would establish that close group of observers is not static.

 

[Hiero]

I still feel like I'm missing something. This may just be my lack of (precise) understanding of curvature in the spacetime context.

Anyway thanks for pointing out (and in various ways) that subtlety.

The most important thing I wanted to know when I made this thread, though, is that the clock rate ratios are precisely reciprocal.
I feel better knowing this is the case.

 

[PAllen]

I still feel like I'm missing something. This may just be my lack of (precise) understanding of curvature in the spacetime context.

Anyway thanks for pointing out (and in various ways) that subtlety.

The most important thing I wanted to know when I made this thread, though, is that the clock rate ratios are precisely reciprocal.
I feel better knowing this is the case.

I thought of another way to help you understand.

Looking at the wikipedia link you gave, which looks ok to me, note that H in the formula for the Rindler case is not just some arbitrary constant. It is distance from the Rindler horizon. As such, it is c²/g for an observer with constant acceleration g. Then, the Rindler formula given, ( H+ h)/H becomes 1 + gh/c², which you will note is identical to the weak field formula given.

Now look at g(h) function for the Rindler case. It becomes, to first order in h, g ( 1- hg/c²). Thus for flat spacetime, the static observer relationship forces change in g linear with h (with, of course, very small constant of proportionality). Now, the one thing you have to accept is that curvature can only introduce corrections of order h² and higher powers. There is no way to for such second and higher order corrections to exactly cancel the Rindler first order effect over a small finite region. Thus there cannot exist curvature such that a family of observers with exactly identical constant proper acceleration maintains static mutual distances (including no curvature, which is just the pure Rindler case).

 

[sweet springs]

The most important thing I wanted to know when I made this thread, though, is that the clock rate ratios are precisely reciprocal.
I feel better knowing this is the case.

Hi. Let say points A and B share simultaneity and at two common simultaneous moments standard clocks show
Clock A 0, Clock B 0 and
Clock A a, Clock B b
A finds the pace of clock B is b/a and B finds the pace of clock A is a/b. Reciprocal as you expect.

For an example point A is higher floor than floor B in a same building. A broadcast radio wave saying current time of clock A to B. It takes time for propagation but time delay is constant during experiment. So in comparing two time difference it does not affect, i.e.
Clock A 0+c, Clock B 0 and
Clock A a+c, Clock B b
c mentions correction to simultaneity due to transmission delay, but it does not affect comparison of time intervals. Best.

 

[pervect]

Let's try the super-simple version. People usually mean the Rindler metric when they talk about a uniform gravitational field. If you mean something else, it's up to you to describe what you mean exactly.

The mathematical techniques usually used to describe a gravitational field are the associated metric. Give the metric, one requires that the metric satisfy Einstein's field equations. So if one wants to use some other notion of what a "uniform gravitational field" might be, other than the Rindler metric, one should write down the metric to communicate what this idea is.

This presuposes that one knows what a metric is. The situation if one doesn't know what a metric is problematic. The problem becomes - how does one verify that the proposed idea of a "uniform gravitaitonal field" is actually consistent with Einstein's field equations? WIthout the proper tools, one can't really answer this question.

The problem of time dilation on the Earth is interesting as well, but of course the gravitational field of the Earth isn't uniform field - loosely speaking, one might say it gets weaker as you get further away from the center of the Earth. It's perfectly valid though to find the appropriate time dilations for this case, using the Schwarzschild metric, and to explore the limit of what happens when the distances are short enough that the deviations from the field are almost uniform.

The basic answer for what happens when one takes the limit has already been hinted at (1+gh)(1-gh) = 1 - (gh)^2, and to linear order (gh)^2 = 0. So to linear order the "time dilations" are reciprocal. So things work out to linear order when one uses approximations valid to the linear order.

 

[PAllen]

Let's try the super-simple version. People usually mean the Rindler metric when they talk about a uniform gravitational field. If you mean something else, it's up to you to describe what you mean exactly.

The mathematical techniques usually used to describe a gravitational field are the associated metric. Give the metric, one requires that the metric satisfy Einstein's field equations. So if one wants to use some other notion of what a "uniform gravitational field" might be, other than the Rindler metric, one should write down the metric to communicate what this idea is.

This presuposes that one knows what a metric is. The situation if one doesn't know what a metric is problematic. The problem becomes - how does one verify that the proposed idea of a "uniform gravitaitonal field" is actually consistent with Einstein's field equations? WIthout the proper tools, one can't really answer this question.

The problem of time dilation on the Earth is interesting as well, but of course the gravitational field of the Earth isn't uniform field - loosely speaking, one might say it gets weaker as you get further away from the center of the Earth. It's perfectly valid though to find the appropriate time dilations for this case, using the Schwarzschild metric, and to explore the limit of what happens when the distances are short enough that the deviations from the field are almost uniform.

The basic answer for what happens when one takes the limit has already been hinted at (1+gh)(1-gh) = 1 - (gh)^2, and to linear order (gh)^2 = 0. So to linear order the "time dilations" are reciprocal. So things work out to linear order when one uses approximations valid to the linear order.

In an early post in the thread, I already exactly worked out the Rindler case, showing the linear formula is exactly correct in reference to any given Rindler observer, producing exact reciprocals due to the precise way each observer's proper acceleration differs.

The wikipedia link posted a few posts ago also establishes the exactness of the linear formula for Rindler using a different method.

 

[PAllen]

I thought it might be nice to post order of magnitude numbers on the change of proper acceleration of stationary observers per altitude for Earth surface conditions. Assuming I have not made any arithmetic errors,

1) If the Earth surface situation were described exactly by the Rindler metric, the change in g per meter of altitude (using meters and seconds) would be about 10^-16 per meter (subtracted from surface g).

2) Assuming Newtonian gravity is exact, it would be about 3*10^-7 per meter (subtracted from surface g). Obviously, the Rindler effect is wholly insignificant against this.

3) The Schwarzschild correction to the Newtonian figure is of order 10^-6 (that is, given a precise Newtonian figure for per meter change, it would be modified by order of 1 part in 10⁶ for Schwarzschild metric). Thus the Schwarzschild correction to the Newtonian figure is still much larger than the Rindler figure.

Thus you can see, Rindler is about as close as you can possibly get to uniform gravity.

 

[pervect]

In an early post in the thread, I already exactly worked out the Rindler case, showing the linear formula is exactly correct in reference to any given Rindler observer, producing exact reciprocals due to the precise way each observer's proper acceleration differs.

No argument here. But sometimes people have funny ideas of what a "uniform gravitational field" really is. Working out the Rindler case exactly won't help if their mental idea of what a "uniform gravitational field" is some other metric than the Rindler metric. Unless both parties can agree on the interpretation of what a "uniform gravitational field" is, they talk past each other.

 

[sweet springs]

Hi. Uniform is not necessary but stationary is a more appropriate condition to deal with, I think. Re:>>30. Zero tx,ty and tz components of metric also.