No redshift in a freely falling frame

[vin300]

Why is there no redshift in a freely falling frame? The photon in a freely falling frame also rises in the gravitational field, so isn't it supposed to be redshifted?

 

[PeterDonis]

From the viewpoint of an observer at rest in the freely falling frame, there is no "gravitational field". He is weightless, feeling no force, so to him, physics looks the same as in free space with no gravity. (We are assuming the freely falling frame is small enough that tidal effects are negligible.) Photons in free space with no gravity do not redshift.

From the viewpoint of an observer at rest in the gravitational field, the freely falling frame is accelerating downward. Suppose a photon is emitted upward towards a freely falling observer some distance above, who is at rest in the gravitational field at the instant the photon is emitted. By the time the photon reaches the observer, it will have redshifted, but the observer will have picked up just enough downward velocity so that when the observer receives the photon, there will be a Doppler blueshift that exactly cancels the gravitational redshift.

 

[TSny]

As you can see from PeterDonis' excellent answer, "redshift" (wavelength) is not a property belonging to a photon itself. It is a joint property of a photon and an observer (detector).

 

[Naty1]

Does the PeterDonis description also apply to the deBroglie wavelength of a free falling matter particle?

I'm thinking of the Tamara Davis article I posted where she notes such a deBroglie wavelength of a matter particle redshifts in the same proportion in cosmological expansion as the wavelength [redshift] of a photon.

I'm wondering if that analogy applies to gravitational redshift as well.

I'm having trouble understanding " but the observer will have picked up just enough downward velocity so that when the observer receives the photon..."

In what sense is the 'at rest' observer accelerating downward??

 

[PeterDonis]

Does the PeterDonis description also apply to the deBroglie wavelength of a free falling matter particle?

I'm thinking of the Tamara Davis article I posted where she notes such a deBroglie wavelength of a matter particle redshifts in the same proportion in cosmological expansion as the wavelength [redshift] of a photon.

I'm wondering if that analogy applies to gravitational redshift as well.

I would say yes, because the de Broglie formula is basically just the rest mass m > 0 version of the formula for light. I can't think of any literature offhand where I've seen it discussed, though.

I'm having trouble understanding " but the observer will have picked up just enough downward velocity so that when the observer receives the photon..."

In what sense is the 'at rest' observer accelerating downward??

In the sentence you quoted, "observer" refers to a freely falling observer. I was trying to describe how an observer static in the field would view the observations made by a freely falling observer. The "acceleration" of the free-falling observer is only coordinate acceleration, true, but from the viewpoint of the static observer it still changes the relative motion of the two during the time of flight of the photon.

 

[PeterDonis]

As you can see from PeterDonis' excellent answer, "redshift" (wavelength) is not a property belonging to a photon itself. It is a joint property of a photon and an observer (detector).

Hmm... this does bring up a good point relative to Naty1's question which I just posted an answer to. I need to qualify my answer somewhat.

In the case of a photon in free space, its velocity is constant; so the "dispersion relation" between frequency and wavelength is also constant (they are both the same in relativistic units where c = 1). Another way of saying this is that a Lorentz transformation on a photon does not change its speed; it just changes its frequency/wavelength, via the Doppler shift.

In the case of a massive particle, things are more complicated. The particle's velocity is not constant, since a Lorentz transformation changes it, so its dispersion relation is not constant either. That complicates the relationship between de Broglie frequency/wavelength and the particle's kinematics--energy and momentum. Falling or rising in a gravity field still has an effect, but I'm not sure it can be described as simply as the effect on a photon can be.

 

[TSny]

See attachment for experimental observation of effect of gravity on "matter waves" of neutrons. You can get the gist of the article from reading the first few paragraphs and the last few paragraphs.

[I have never tried attaching a file before, so I hope the attachment works. If not, you can view the article here

http://www.atomwave.org/rmparticle/ao%20refs/aifm%20refs%20sorted%20by%20topic/inertial%20sensing%20refs/gravity/COW75%20neutron%20gravity.pdf

]

 

[Dale]

By the way, vin300, in case you want to search on your own, the principle that PeterDonis described is called the "Equivalence Principle" and is one of the key concepts behind General Relativity.

 

[yuiop]

Just to be clear, the "no redshift" condition only occurs if the source and the observer are both free falling and reasonably close to each other. If the source is stationary relative to the gravitational field, then a stationary observer lower down in the gravitational well, will see the light as blue shifted and a free falling observer lower down will see the same light as red shifted.

 

[pervect]

Another way of saying this is that there is no redshift to first order. If you plot redshift vs distance, for example, using a reasonably intuitive notions of redshift and distance (this somewhat ambiguious description is defined exactly by using fermi-normal coordinates for the distances and the metric coeffficeint g_00 for the redshift), you'll see that at the origin the curve is a horizontal line, a curve having zero slope, but due to second and higher-order terms the redshift becomes non-zero as you get further away from the origin of the coordinate system.

 

[vin300]

Can it be shown mathematically?

 

[pervect]

Can it be shown mathematically?

What I said can be shown mathematically. If you have or can find MTW's Gravitation, see pg 332, eq 13.73

It gives the Fermi normal metric for a non-rotating observer with zero propagation to accurate to second order The expression for g_00 is

$$g_{00} \approx -1 -R_{\hat{0}\hat{l}\hat{0}\hat{m}}x^l x^m$$

where R is the Riemann curvature tensor in the frame-field basis of the falling observer. It's mentioned elsewhere in MTW that the components of the curvature tensor $R_{ijkl}$ for the Schwarzschild metric near a single massive body aren't affected by motion towards or away from the body. Furthermore, the radial component is given by $R_{0r0r}$ = -2GM/r^3[/itex], equal to the Newtonian tidal force. The transverse components are half that and of the opposite sign (and are also equal to the Newtonian tidal force). The off diagonal terms R_{0i0j} for i not equal to j are zero.

The biggest issues are 1) knowing that the metric g_00 represents time dilation and 2) thinking of Fermi-normal coordinates as representing "an intuitive notion of distance". The last is probably the most problematic - though they are a pretty straightforward generalization of coordinates from the flat space-time to curved. The biggest issue is that there is more than one way to make the transition.

Online, http://arxiv.org/pdf/gr-qc/0010096v1.pdf and http://arxiv.org/pdf/0901.4465.pdf may be helpful

The later reference also provides some justification for thinking of the Fermi Normal coordinates as having a "physical" meaning (though I prefer to say intuitive).

Nowadays the Fermi normal coordinates are usually - although improperly - called Fermi coordinates. In exper-
imental gravitation, Fermi normal coordinates are a powerful tool used to describe various experiments: since the
Fermi normal coordinates are Minkowskian to first order, the equations of physics in a Fermi normal frame are the
ones of special relativity, plus corrections of higher order in the Fermi normal coordinates, therefore accounting for
the gravitational field and its coupling to the inertial effects. Additionally, for small velocities v compared to light
velocity c, the Fermi normal coordinates can be assimilated to the zeroth order in (v/c) to classical Galilean coordi-
nates. They can be used to describe an apparatus in a “Newtonian” way (e.g. [1, 3, 8, 10]), or to interpret the outcome
of an experiment (e.g. [11] and comment [21], [5, 6, 15, 17]). In these approaches, the Fermi normal coordinates are
considered to have a physical meaning, coming from the principle of equivalence (see e.g. [18]), and an operational
meaning: the Fermi normal frame can be realized with an ideal clock and a non extensible thread [29].

 

[Naty1]

In the sentence you quoted, "observer" refers to a freely falling observer. I was trying to describe how an observer static in the field would view the observations made by a freely falling observer.

seemed clear to me after this morning coffee! Maybe I was drunk yesterday!??

In the case of a photon in free space, its velocity is constant. In the case of a massive particle, things are more complicated. The particle's velocity is not constant, since a Lorentz transformation changes

I should have posted that with my question: I have been puzzling about exactly that since I read the Tamara Davis article. She does not explain her conclusion just states it.

...I'm not sure it can be described as simply as the effect on a photon can be.

I think you are right: it can't be as simple. If a free falling photon exhibits one redshift, how could a matter particle exhibit the same redshift in the same spacetime? The changing speed of the matter particle does not affect the speed of the emitted deBroglie wavelength but it must affect it's redshift relative to a more rapidly receding photon.

 

[Naty1]

...I'm not sure it can be described as simply as the effect on a photon can be.

I think you are right:

oops now I am not so sure:

I came across this from Jonathan Scott, which I also thought correct:

Neither photons nor even massive objects change in frequency or energy
when moving in a static gravitational field as observed by anyone observer. In Newtonian terms this is because the combined effects of changes in kinetic energy (from motion) plus potential energy (from time dilation) result in constant total energy.

so I have more thinking to do.

edit: Upon reconsideration I think the above from Scott is incorrect regarding frequency: frequency corresponds to the KE of light so it should change as a photon moves in a non uniform static gravitational field. So that description does NOT seem to answer our question...

I'll leave this here in case others find that 'logic' of value...or in case I'm wrong again!

I return to agreeing with PeterDonis.

 

[GAsahi]

If the source is stationary relative to the gravitational field, then a stationary observer lower down in the gravitational well, will see the light as blue shifted and a free falling observer lower down will see the same light as red shifted.

It is considerably more complicated than the above. The first half of your statement is correct, the second one is not. Proof

Start with the (reduced) Schwarzschild metric:

$$d \tau^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)$$

1. If the two observer and the source are stationary, then $dr=0$ at radial coordinates $r_1<r_2$ , you can write:

$$d \tau_1^2=(1-r_s/r_1)dt^2$$
$$d \tau_2^2=(1-r_s/r_2)dt^2$$

It follows that :

$$\frac{d \tau_1}{d \tau_2}=\sqrt {\frac{1-r_s/r_1}{1-r_s/r_2}}$$

Since $r_1<r_2$ it follows
$$d \tau_1< d \tau_2$$
i.e $$f_1>f_2$$ (blueshift)

2. If the source is moving, things get more complicated.

$$d \tau_1^2=(1-r_s/r_1)dt^2$$
$$d \tau_2^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)=(1-r_s/r)dt^2 (1-\frac{(dr/dt)^2}{(1-r_s/r)^2})$$

Therefore:

$$\frac{d \tau_1^2}{d \tau_2^2}=\frac{1-r_s/r_1}{1-r_s/r}\frac{1}{1-\frac{v^2}{(1-r_s/r)^2}}$$

The second fraction is always larger than 1.
The first fraction is sometimes greater than 1 (for $r_1>r$) or smaller than 1, (for $r_1<r$), so you can get either blue or redshift (you don't always get redshift).
For $v^2=(1-r_s/r)(r_s/r_1-r_s/r)$ there is no shift whatsoever.

 

[PeterDonis]

Upon reconsideration I think the above from Scott is incorrect regarding frequency: frequency corresponds to the KE of light so it should change as a photon moves in a non uniform static gravitational field.

But "KE change" is not invariant, it's observer dependent; that was really Jonathan Scott's point. Whether or not the KE changes depends on who is measuring the change. The KE of freely falling light changes from the viewpoint of static observers; that's what is normally referred to as gravitational redshift/blueshift. But from the viewpoint of the freely falling photon itself, its energy does not change; more precisely, its 4-momentum does not change, it is parallel transported along the photon's worldline. The "change", from the photon's point of view, is in the 4-velocity vectors of static observers at different heights.

 

[vin300]

I've found an explanation that violates the equivalence principle. An observer in a uniform gravitational field sees objects accelerating downwards at a constant g. An observer with a constant upward acceleration g also sees the same thing, but there's something that's not equivalent in the two cases. An observer in a gravitational field says that a falling object has an unchaging total energy, all the gain in its kinetic energy is due to equivalent loss of potential energy, but the accelerating observer doesn't say that its potential energy is decreasing. The man in the field says the energy of the object doesn't change, but the accelerating one says that the energy changes at a constant rate.

 

[PeterDonis]

I've found an explanation that violates the equivalence principle. An observer in a uniform gravitational field sees objects accelerating downwards at a constant g. An observer with a constant upward acceleration g also sees the same thing, but there's something that's not equivalent in the two cases. An observer in a gravitational field says that a falling object has an unchaging total energy, all the gain in its kinetic energy is due to equivalent loss of potential energy, but the accelerating observer doesn't say that its potential energy is decreasing. The man in the field says the energy of the object doesn't change, but the accelerating one says that the energy changes at a constant rate.

First of all, the above is irrelevant to the EP because it's not based on local observations. The difference between the two observers (in so far as it is a difference--see below) only becomes evident when they make observations over time.

Second, even leaving out the above, you're using two different definitions of energy for the two accelerated observers; that's where the apparent difference comes from. The reason you can tell it's only apparent is that it doesn't affect the results of any actual experiments. Whatever experiment the observer accelerating in free space tries to run to show that the free-falling object's energy "changes at a constant rate", the observer at rest in the gravity field will be able to run the same experiment and get the *same* result. Similarly, whatever experiment the observer at rest in the field tries to run to show that the free-falling object's total energy is unchanged, the observer accelerating in free space will be able to run and get the same result. So the apparent "difference" between the two is only a matter of the definition of "total energy"; it doesn't reflect any actual physics.

Btw, what I said in the last paragraph is *not* true in general in GR; it's a special feature of Schwarzschild spacetime (strictly speaking, of Schwarzschild spacetime outside the horizon) that this extended correspondence with Rindler observers in Minkowski spacetime, which is what justifies the statements I made above, can be drawn. That's why those statements are not about the EP; the EP is true in any spacetime, but only locally.

 

[DrGreg]

Energy is always measured relative to an observer (or, more accurately, relative to some coordinate system). In those cases where it is possible to meaningfully define energy relative to a non-inertial observer so that energy is conserved, you need to define potential energy. So there is a "pseudo-gravitational" potential energy associated with Rindler coordinates, for example.

 

[stevendaryl]

It is considerably more complicated than the above. The first half of your statement is correct, the second one is not. Proof

Start with the (reduced) Schwarzschild metric:

$$d \tau^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)$$

1. If the two observer and the source are stationary, then $dr=0$ at radial coordinates $r_1<r_2$ , you can write:

$$d \tau_1^2=(1-r_s/r_1)dt^2$$
$$d \tau_2^2=(1-r_s/r_2)dt^2$$

It follows that :

$$\frac{d \tau_1}{d \tau_2}=\sqrt {\frac{1-r_s/r_1}{1-r_s/r_2}}$$

Since $r_1<r_2$ it follows
$$d \tau_1< d \tau_2$$
i.e $$f_1>f_2$$ (blueshift)

2. If the source is moving, things get more complicated.

$$d \tau_1^2=(1-r_s/r_1)dt^2$$
$$d \tau_2^2=(1-r_s/r)dt^2-dr^2/(1-r_s/r)=(1-r_s/r)dt^2 (1-\frac{(dr/dt)^2}{(1-r_s/r)^2})$$

Therefore:

$$\frac{d \tau_1^2}{d \tau_2^2}=\frac{1-r_s/r_1}{1-r_s/r}\frac{1}{1-\frac{v^2}{(1-r_s/r)^2}}$$

The second fraction is always larger than 1.
The first fraction is sometimes greater than 1 (for $r_1>r$) or smaller than 1, (for $r_1<r$), so you can get either blue or redshift (you don't always get redshift).
For $v^2=(1-r_s/r)(r_s/r_1-r_s/r)$ there is no shift whatsoever.

I think that your answers are correct, but I think that the computation is not quite correct. The reason why is because what you have calculated is the ratio of clock rates as measured using Schwarzschild coordinates. Relative clock rates for spatially separated clocks is a coordinate-dependent quantity; a different coordinate system might give a different answer.

The quantity that is directly observable, and is independent of coordinate systems, is this:
Let a photon be emitted from one observer. Let f₁ be its frequency, as measured by that observer. Let the photon travel to the other observer. Let f₂ be the frequency of that same photon, as measured by the second observer. Then the measured redshift or blueshift would be f₂/f₁.

I don't know right off the bat if that gives the same answer as the ratio you computed, or not, but conceptually, they are different.

 

[GAsahi]

I think that your answers are correct, but I think that the computation is not quite correct. The reason why is because what you have calculated is the ratio of clock rates as measured using Schwarzschild coordinates. Relative clock rates for spatially separated clocks is a coordinate-dependent quantity; a different coordinate system might give a different answer.

The quantity that is directly observable, and is independent of coordinate systems, is this:
Let a photon be emitted from one observer. Let f₁ be its frequency, as measured by that observer. Let the photon travel to the other observer. Let f₂ be the frequency of that same photon, as measured by the second observer. Then the measured redshift or blueshift would be f₂/f₁.

I don't know right off the bat if that gives the same answer as the ratio you computed, or not, but conceptually, they are different.

What I calculated is the ratio of proper periods. This is known to be coordinate independent. If you want the ratio of frequencies, then you know that $\frac{f_1}{f_2}=\frac{d \tau_2}{d \tau_1}$, as I have shown in the first half of my calculations.

 

[stevendaryl]

What I calculated is the ratio of proper periods. This is known to be coordinate independent.

No, in general, the ratio that you computed is not coordinate-independent. How are you defining a "proper period"?

What you computed was equivalent to the following:

1. Let e₁ be some event at the first observer (the one at radius r₁).
2. Let e₂ be some event at the second observer (the one at radius r₂) that is simultaneous with e₁.
3. Let e₃ be some event at the first observer at a time δt later than e₁.
4. Let e₄ be some event at the second observer at a time δt later than e₂.
5. Let d$\tau$₁ be the invariant interval between e₁ and e₃.
6. Let d$\tau$₂ be the invariant interval between e₂ and e₄.
7. Then define the relative rates to be the ratio d$\tau$₁/d$\tau$₂

The phrases in bold-face are coordinate-dependent. Specifically, in order to compare clock rates at different locations, you need to know that e₁ is simultaneous with e₂, and that e₃ is simultaneous with e₄. If you switch to a different coordinate system, with a different notion of simultaneity, then you get a different answer for the ratio.

The easiest way to see this is in the case of two identically accelerating rockets in flat spacetime, each equipped with an onboard clock. Using an inertial coordinate system, the ratio of the rates will be 1, because the time dilation effects will be the same for both clocks. Using the Rindler coordinate system in which the rear rocket is at rest, the front clock will be seen to be running faster than the rear clock.

 

[GAsahi]

No, in general, the ratio that you computed is not coordinate-independent. How are you defining a "proper period"?

What you computed was equivalent to the following:

[*]Let e₁ be some event at the first observer (the one at radius r₁).

Yes.

[*]Let e₂ be some event at the second observer (the one at radius r₂) that is simultaneous with e₁.

Wrong.

[*]Let e₃ be some event at the first observer at a time δt later than e₁.

Wrong, $d \tau_1$ later.

[*]Let e₄ be some event at the second observer at a time δt later than e₂.

Wrong.

[*]Let d$\tau$₁ be the invariant interval between e₁ and e₃.
[*]Let d$\tau$₂ be the invariant interval between e₂ and e₄.
[*] Then define the relative rates to be the ratio d$\tau$₁/d$\tau$₂

You got this part right. Now, given (by your own admission) that d$\tau$₁ and d$\tau$₂ are invariant, it follows that their ratio d$\tau$₁/d$\tau$₂ is also invariant.

 

[stevendaryl]

The easiest way to see this is in the case of two identically accelerating rockets in flat spacetime, each equipped with an onboard clock. Using an inertial coordinate system, the ratio of the rates will be 1, because the time dilation effects will be the same for both clocks. Using the Rindler coordinate system in which the rear rocket is at rest, the front clock will be seen to be running faster than the rear clock.

Let me do an explicit calculation to prove my point.

In Rindler coordinates (X,T), we have two clocks, one at X = X₁, and one at X₂. The Rindler interval is:

d$\tau$² = X² dT² - dX²

So for clocks at rest in the X,T coordinates, we have:
d$\tau$ = X dT

So the ratio of the rates is: d$\tau$₁/d$\tau$₂ = X₁/X₂

Conclusion: the "higher" clock (with greater X) runs faster.

Now, do the same calculation in the coordinate system (x,t) related to (X,T) through:

x = X cosh(gT)
t = X/c sinh(gT)

So d$\tau$² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d$\tau$ = $\sqrt{1-(v/c)^{2}}$ dt


The ratios of the rates in this coordinate system is given by:
d$\tau$₁/d$\tau$₂ = $\sqrt{1-(v_{1}/c)^{2}}$/$\sqrt{1-(v_{2}/c)^{2}}$

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

 

[stevendaryl]

You got this part right. Now, given (by your own admission) that d$\tau$₁ and d$\tau$₂ are invariant, it follows that their ratio d$\tau$₁/d$\tau$₂ is also invariant.

The invariant interval d$\tau$ is defined by a pair of events. If you fix a pair of (neighboring) events, then the corresponding d$\tau$ is an invariant. So we have a pair of events e₁ and e₃ taking place at the first observer, and there is a corresponding invariant interval d$\tau$₁, and there is a pair of events e₂ and e₄ taking place at the second observer, and a corresponding invariant interval d$\tau$₂. The issue is: how do you choose WHICH 4 events e₁, e₂, e₃ and e₄ to use for the comparison of clock rates? Give me a coordinate-independent way of choosing those 4 events.

What you wrote down was:

d$\tau$₁ = √(1-r/r₁) dt
d$\tau$₂ = √(1-r/r₂) dt

That corresponds to choosing the four events so that δt₁ = δt₂, where δt₁ = the coordinate time between e₁ and e₃, and δt₂ = the coordinate time between e₂ and e₄. That choice is coordinate-dependent. If you meant for a coordinate-independent choice, then what is your basis for choosing the events e₁, e₂, e₃ and e₄?

 

[GAsahi]

Let me do an explicit calculation to prove my point.

In Rindler coordinates (X,T), we have two clocks, one at X = X₁, and one at X₂. The Rindler interval is:

d$\tau$² = X² dT² - dX²

So for clocks at rest in the X,T coordinates, we have:
d$\tau$ = X dT

So the ratio of the rates is: d$\tau$₁/d$\tau$₂ = X₁/X₂

Conclusion: the "higher" clock (with greater X) runs faster.

Now, do the same calculation in the coordinate system (x,t) related to (X,T) through:

x = X cosh(gT)
t = X/c sinh(gT)

So d$\tau$² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d$\tau$ = $\sqrt{1-(v/c)^{2}}$ dt


The ratios of the rates in this coordinate system is given by:
d$\tau$₁/d$\tau$₂ = $\sqrt{1-(v_{1}/c)^{2}}$/$\sqrt{1-(v_{2}/c)^{2}}$

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

It is really simple, IF you did your calculations correctly, you SHOULD have obtained the same result. In fact, correct calculations are confirmed by experiment, in this case it is the Pound-Rebka experiment. Neither set of your calculations recovers the ones confirmed by experiment:

$$d \tau_1^2=(1-r_s/r_1)dt^2$$
$$d \tau_2^2=(1-r_s/r_2)dt^2$$
$$\frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-r_s/r_1}{1-r_s/r_2}}$$

 

[GAsahi]

What you wrote down was:

d$\tau$₁ = √(1-r/r₁) dt
d$\tau$₂ = √(1-r/r₂) dt

That corresponds to choosing the four events so that δt₁ = δt₂, where δt₁ = the coordinate time between e₁ and e₃, and δt₂ = the coordinate time between e₂ and e₄. That choice is coordinate-dependent. If you meant for a coordinate-independent choice, then what is your basis for choosing the events e₁, e₂, e₃ and e₄?

OK,

I understand what you are doing wrong. You totally misunderstand the problem statement (and, consequently, what I have been doing).
A distant observer measures a periodic phaenomenon of period $dt$.
Observer1, located at $r_1$ measures on HIS clock, the corresponding interval $d \tau_1$.
Observer2 , located at $r_2$ measures on HIS clock (identical to the one used by Observer1), obtaining the value $d \tau_2$. What is the relationship between $d \tau_2$ and $d \tau_1$? Since the observers aren't moving, you can write:

$$d \tau_i^2=(1-r_s/r_i)dt^2$$, i=1,2

where $dt$ is the interval as measured by a distant observer. Eliminate $dt$ and you get the correct answer.

 

[stevendaryl]

It is really simple, IF you did your calculations correctly, you SHOULD have obtained the same result.

No, your claim is just wrong. The relative rate of two clocks is coordinate-dependent. We know that from Special Relativity, and I gave you an explicit calculation proving it. You're just confused on this point.

In fact, correct calculations are confirmed by experiment, in this case it is the Pound-Rebka experiment.

My original post said that your answer might be correct. My complaint was your method of getting that answer. As I said, to make a comparison of "clock rates" at different spatial locations, you need 4 events: two at one clock, and two at the other clock. I've asked you what is the basis for choosing those 4 events. Your derivation used equal times between events, as measured in the Schwazschild coordinates. That is not a coordinate-independent choice.

Now, a coordinate-independent choice is to use light signals. At event e₁, a light signal is sent from the first observer. Event e₂ is the reception of that signal by the second observer. Event e₃ is a second light signal sent by the first observer, and e₄ is the reception of that signal by the second observer. Then, for those specific 4 events, we can do a comparison d$\tau$₁/d$\tau$₂, where d$\tau$₁ is the invariant interval between e₁ and e₃, and d$\tau$₂ is the time between e₂ and e₄.

That choice of the 4 events is coordinate-independent. But that's not what you said in your original calculation.

 

[stevendaryl]

OK, I understand what you are doing wrong. You totally misunderstand the problem statement (and, consequently, what I have been doing).
Observer1, located at $r_1$ measures on HIS clock, the interval $d \tau_1$.
Observer2 , located at $r_2$ measures remotely the SAME interval, on HIS clock (identical to the one used by Observer1), obtaining the value $d \tau_2$.

That's exactly what I meant when I said that your computation was a coordinate-dependent quantity. There is no coordinate-independent way to measure the length of a distant interval. A clock can only measure times between events AT THAT CLOCK. It cannot measure times between distant events. In order to measure times between distant events, you need a coordinate system.

 

[GAsahi]

My original post said that your answer might be correct.

The answer IS correct. You get the SAME answer from experimental measurement (see, once again, Pound-Rebka, for further details, read here). I do not plan to waste more time on this subject.

 

[stevendaryl]

The answer IS correct.

But your reasoning was incorrect.

 

[Austin0]

The easiest way to see this is in the case of two identically accelerating rockets in flat spacetime, each equipped with an onboard clock. Using an inertial coordinate system, the ratio of the rates will be 1, because the time dilation effects will be the same for both clocks. Using the Rindler coordinate system in which the rear rocket is at rest, the front clock will be seen to be running faster than the rear clock.

What do you mean by this?
There would not seem to be a Rindler system covering both rockets. Such a system is predicated on an acceleration differential which is absent here.
If you extend a virtual Rindler frame from the rear rocket to the front rocket then the front rocket would actually be running slow relative to the colocated Rindler observer/clock , yes?
Are you perhaps basing your assumption on observed Doppler shift?

 

[stevendaryl]

But your reasoning was incorrect.

Let me complete the argument here, if anyone is actually reading. In order to talk about relative clock rates for distant clocks in a coordinate-independent fashion, you have to have some basis for choosing corresponding events: You pick two events e₁ at one clock, and a corresponding event e₂ at the second clock, an event e₃ at the first clock, and an event e₄ at the second clock. Then for these 4 events, one can compute a ratio of invariant intervals:

d$\tau$₁/d$\tau$₂

where d$\tau$₁ is the invariant interval between e₁ and e₃, and d$\tau$₂ is the invariant interval between e₂ and e₄. You obviously get different answers for the ratio, depending on which 4 events you choose for the comparison.

Criterion 1: equal coordinate times
There is one way of choosing the 4 events, which is to choose them according to coordinate time: Choose e₂ and e₄ so that the coordinate time difference, dt = t₄ - t₂ is equal to the coordinate time difference between e₁ and e₃, dt = t₃ - t₁. That is a coordinate-dependent criterion, and that gives you a coordinate-dependent notion of relative clock rates. Which is actually fine, because relative clock rates IS a coordinate-dependent quantity.

Criterion 2: Null geodesics
Another way of choosing the 4 events is to use null geodesics (a null geodesic is the path taken by a light signal): Choose e₂ and e₄ such that there is a null geodesic connecting e₁ and e₂, and similarly a null geodesic connecting e₃ and e₄. This criterion is what you care about if you are going to experimentally test gravitational redshift.

Now, to complete the argument, you need to use the fact that for a coordinate system such that (1) the components of the metric are static (independent of the time coordinate), and (2) the two clocks are stationary (none of the spatial coordinates are changing), the two criteria are the same. That is, if two light signals are sent from one location at a time dt apart, then they will arrive at the destination at a time dt apart.

Since the Schwarzschild coordinates do have a static metric components, criterion 1 is good enough.

 

[GAsahi]

Let me do an explicit calculation to prove my point.

In Rindler coordinates (X,T), we have two clocks, one at X = X₁, and one at X₂. The Rindler interval is:

d$\tau$² = X² dT² - dX²

So for clocks at rest in the X,T coordinates, we have:
d$\tau$ = X dT

So the ratio of the rates is: d$\tau$₁/d$\tau$₂ = X₁/X₂

Conclusion: the "higher" clock (with greater X) runs faster.

This derivation is correct.

Now, do the same calculation in the coordinate system (x,t) related to (X,T) through:

x = X cosh(gT)
t = X/c sinh(gT)

So d$\tau$² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d$\tau$ = $\sqrt{1-(v/c)^{2}}$ dtThe ratios of the rates in this coordinate system is given by:
d$\tau$₁/d$\tau$₂ = $\sqrt{1-(v_{1}/c)^{2}}$/$\sqrt{1-(v_{2}/c)^{2}}$

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

This part of the derivation is in error. If you did it correctly, you would have gotten that the correct result is $\frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-v/c}{1+v/c}}$ where $v$ is the instantaneous speed of the rocket containing the two clocks wrt the launcher frame.
I can get into all the details of why the above is the correct result but I won't , the way to get the correct result is not simply using the equations of hyperbolic motion, you can simply use the equivalence principle and to observe the Doppler effect on the frequency emitted at one end of the rocket and received at the other end, the two ends being separated by a distance $h=X_1-X_2$. The bottom line is that there is always motion between the two ends of the rocket, so you cannot write

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

If the above were true, you would have found a way of disproving the principle of equivalence.

 

[Austin0]

Let me do an explicit calculation to prove my point.

In Rindler coordinates (X,T), we have two clocks, one at X = X₁, and one at X₂. The Rindler interval is:

d$\tau$² = X² dT² - dX²

So for clocks at rest in the X,T coordinates, we have:
d$\tau$ = X dT

So the ratio of the rates is: d$\tau$₁/d$\tau$₂ = X₁/X₂

Conclusion: the "higher" clock (with greater X) runs faster.

This derivation is correct.

Now, do the same calculation in the coordinate system (x,t) related to (X,T) through:

x = X cosh(gT)
t = X/c sinh(gT)

So d$\tau$² = dt² - 1/c² dx²
= dt² (1 - v²/c²)
where v = dx/dt = the speed of the clock. So

d$\tau$ = $\sqrt{1-(v/c)^{2}}$ dt


The ratios of the rates in this coordinate system is given by:
d$\tau$₁/d$\tau$₂ = $\sqrt{1-(v_{1}/c)^{2}}$/$\sqrt{1-(v_{2}/c)^{2}}$

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

This part of the derivation is in error. If you did it correctly, you would have gotten that the correct result is $\frac{d \tau_1}{d \tau_2}=\sqrt{\frac{1-v/c}{1+v/c}}$ where $v$ is the instantaneous speed of the rocket containing the two clocks wrt the launcher frame.
I can get into all the details of why the above is the correct result but I won't , the way to get the correct result is not simply using the equations of hyperbolic motion, you can simply use the equivalence principle and to observe the Doppler effect on the frequency emitted at one end of the rocket and received at the other end, the two ends being separated by a distance $h=X_1-X_2$. The bottom line is that there is always motion between the two ends of the rocket, so you cannot write

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

It would appear that if stevendaryl's calculation of relative rate as the relationship between instantaneous gammas is incorrect then a basic principle of SR falls. Specifically the Clock Hypothesis. Delta t' for either clock must be equal to an integration over that worldline interval based on instantaeous (infinitesimal) velocity gammas ,yes?
Or comparably a ratio of rear dt/$\gamma$₁ front dt/$\gamma$₂ as measured in the launch frame.

More explicitly:If there are individual coordinate charts for the two clocks ,say from v=0 to v=0.9c then the integrated proper times for these worldlines must agree with the calculated proper times for these clocks using the Rindler metric Yes?

At t=0, $v_{1}$ = $v_{2}$ = 0. So the ratio starts off equal to 1, not X₁/X₂.

If the above were true, you would have found a way of disproving the principle of equivalence.

How could it not start out at At t=0, $v_{1}$ = $v_{2}$ = 0 ? it does not instantaneously attain its final proper acceleration ,yes?? It would seem it would have to start out at 1 and over some finite time interval reach the relative ratio.