Atomic clocks in gravitational field

[jartsa]

So all effects regarding time-dilation, redshift etc. are due to different paths through spacetime probing different gravitational fields. The effect is non-local. Thefore it does not make sense to talk about "time-dilated feet", b/c the question is always "time-dilated with respect to what?"

In that sense I think A-D are quite confusing. I would pick none of them.

Here's an alternative formulation:

In this table "time dilated foot" means gravitationally time dilated foot, according to head that is located six feet above the foot.


A: Time dilated foot emits time dilated EM-waves
B: Time dilated foot emits redshifted EM-waves
C: Redshifted foot emits redshifted EM-waves
D: Time dilated foot emits EM-waves that are not redshifted, but may become redshifted, if climbing upwards



Two new ones:
E: low altidude foot is a low energy foot, and emits low energy photons
F: a foot whose energy is redshifted to 0.1 Joules may emit radiation that has 1 Joules energy

 

[PeterDonis]

The derivative of the parametrized curve representing the null geodesic with respect to the null affine parameter, which in tom's post shows up as $\dot{x}$.

Ok, but this $\dot{x}$ just seems to me to be another way of writing the photon's 4-momentum (modulo a normalization factor to correct the units), and as far as I can see, $\dot{x}$ is constant along the photon's worldline ($x$ itself isn't constant, yes, but $x$ isn't what appears in the redshift formula, $\dot{x}$ is); the change in inner product from emitter to detector comes entirely from the change in $u$.

I'll look through the living reviews article that tom.stoer linked to in more detail when I get a chance; I've only briefly glanced at it to look at equation (37) that he referenced. There may be subtleties that I'm missing.

 

[PeterDonis]

Here's an alternative formulation:

None of these look right to me, since none of them match the E that I posted earlier (including the two new ones you added).

 

[Philosopha]

None of these look right to me, since none of them match the E that I posted earlier (including the two new ones you added).

So you think 'low altitude foot is low energy foot' is wrong? And do you think none of the changes are absolute?

 

[WannabeNewton]

...But I’m not sure about this because I have always read about the gravitational frequencyshift of light as if it was absolute...

It is quite possible that you are using absolute in a different way from how I've usually seen it used but if the gravitational redshift of a light wave emitted between an emission point and detection point is absolute then all observers at the detection point should agree on the gravitational redshift of the light wave coming from the emission point. The way one usually derives gravitational redshift is by considering specifically the emission and detection of the light wave by two static observers, who constitute a privileged class of observers in stationary space-times. On the other hand, think about what happens to the gravitational redshift at the detection point if instead we replaced the static observer there with a freely falling observer (but kept the static observer at the emission point).

 

[Philosopha]

Thx Newton. Does the above consider that the two observers are also 'affected the same' by the same circumstance (i.e. potential) as the EM wave they are observing if they 'co-locate' with it? I thought, that because they are affected in the same way as the EM wave they observe, that when they see a difference this is absolute. Of course in case of the free falling observer there would be a relative difference on top of this absolute... - hope my questions don't appear nonsense

 

[WannabeNewton]

I'll look through the living reviews article that tom.stoer linked to in more detail when I get a chance; I've only briefly glanced at it to look at equation (37) that he referenced. There may be subtleties that I'm missing.

I've always learned that $\dot{x}^{\mu} = k^{\mu}$ as well, so that $\dot{x}^{\nu}\nabla_{\nu}\dot{x}^{\mu} = 0$. I too have to read the link, either now or after I finish watching random tv shows

 

[WannabeNewton]

...hope my questions don't appear nonsense

Not at all but I'm just having a bit of trouble interpreting your sentences. I think we may be using different definitions of the word "absolute". In the above scenario, the freely falling observer at the detection event would not detect any gravitational redshift (try to convince yourself of this) whereas the static observer at the exact same detection event would so the gravitational redshift is not absolute in the sense that there exist two different observers at the detection event who will disagree on the gravitational redshift; it is dependent upon the receiving observer (as well as the source).

 

[harrylin]

Well, if that's the particular thing he thought didn't make much sense (light waves changing as they go upwards), then his A, B, and C don't make much sense either, since they all talk about redshifted light waves. [..]

His A, B and C talk about light waves that are redshifted at emission (compared to light from similar emitters at the other location) - contrary to how I understood his D, the light waves do not change in flight according to A, B and C. Once more, the essential difference between those interpretations and interpretation D was very clearly explained in a paper and in the earlier thread.

 

[jartsa]

None of these look right to me, since none of them match the E that I posted earlier (including the two new ones you added).

Yes well, I might say it like this:

A freely flying photon does not change. I have questions:

1: Does the three-momentum of a free flying photon change?
2 How about a cannonball shot upwards in a gravity field, does it change? Particularly energy of it.Freely flying = free falling = moving along a geodesic.

 

[Philosopha]

You are right Newton, we use different definitions of absolute. In my head I have been introducing a god-like observer. I do not argue that two different observer will "perceive" things differently, and therefore the notion of absolute seems to become meaningless in our relative world. However I say, that when the detector freely falls - "of course" it does not see any redshift because it is influenced the same way, but that doesn't mean that there isn't any redshift as of a god-like observer - who is not the static observer, because even the static observer is somewhat influenced. so a god-like observer would see what is "really" happening, would see the absolute values, as I guess we have achieved with the atomic clock experiments.

 

[WannabeNewton]

But god-like observers don't exist in relativity. This is why we say something is absolute if and only if all intrinsic observers agree on it.

 

[harrylin]

You are right Newton, we use different definitions of absolute. In my head I have been introducing a god-like observer. I do not argue that two different observer will "perceive" things differently, and therefore the notion of absolute seems to become meaningless in our relative world. However I say, that when the detector freely falls - "of course" it does not see any redshift because it is influenced the same way, but that doesn't mean that there isn't any redshift as of a god-like observer - who is not the static observer, because even the static observer is somewhat influenced. so a god-like observer would see what is "really" happening, would see the absolute values, as I guess we have achieved with the atomic clock experiments.

That is commonly introduced as an inertial observer far away from heavy bodies. But you don't need such a "god-like" observer, if you reference to the centre of mass and consider its effect only. Just consider geostationary satellites around the Earth; for that case both the velocity and gravity effects (that is, the amount by which these modify f₁/f₀) are considered "absolute". The amount of redshift between emission and detection is a given that doesn't depend on our choice of reference system as they relate to "proper" local measurements. And the effect is "real" for a stationary Earth (choosing its proper frame) from a "gods-eye" view.

 

[Dale]

I have had discussions about the meaning of "absolute" before also. I think that it is probably just best to drop the term entirely. As far as I know it has no standard scientific meaning, so any question that requires the term is probably not answerable.

 

[WannabeNewton]

Observers at infinity are only available for asymptotically flat space-times whereas gravitational redshift shows up in any stationary space-time, asymptotically flat or not.

I have had discussions about the meaning of "absolute" before also. I think that it is probably just best to drop the term entirely. As far as I know it has no standard scientific meaning, so any question that requires the term is probably not answerable.

Yeah some words do get thrown around a lot without actually being well-defined first. Thankfully the mathematics speaks for itself. Yay math!

 

[TrickyDicky]

Observers at infinity are only available for asymptotically flat space-times whereas gravitational redshift shows up in any stationary space-time, asymptotically flat or not.

Hi WN! A little digression, just curious: are you aware of any relevant static space-time that is not asymptotically flat?

 

[WannabeNewton]

The Einstein static universe (although I don't know in what sense you meant the solution to be 'relevant').

 

[tom.stoer]

... are you aware of any relevant static space-time that is not asymptotically flat?

I am aware of relevant spacetimes which are neither static nor asymptoticsllx flat ;-)

 

[TrickyDicky]

The Einstein static universe (although I don't know in what sense you meant the solution to be 'relevant').

Thanks, It just came to mind, I was not taking into account spacetimes with lambda for some strange reason. But I cannot think of any other, right now except the NUT vacuum that is mentioned in wiki.

 

[Bill_K]

Hi WN! A little digression, just curious: are you aware of any relevant static space-time that is not asymptotically flat?

Obvious answer: The Weyl solutions, which represent the general static axisymmetric solution of the vacuum Einstein equations. They are given in terms of a single axisymmetric solution of the Laplace equation ψ in three-dimensional flat space, and will be asymptotically flat if and only ψ goes to zero at infinity.

 

[TrickyDicky]

Obvious answer: The Weyl solutions, which represent the general static axisymmetric solution of the vacuum Einstein equations. They are given in terms of a single axisymmetric solution of the Laplace equation ψ in three-dimensional flat space, and will be asymptotically flat if and only ψ goes to zero at infinity.

Ok, thanks.

 

[PeterDonis]

So you think 'low altitude foot is low energy foot' is wrong?

As it stands, yes, because energy is frame-dependent.

And do you think none of the changes are absolute?

I tend to agree with DaleSpam that the word "absolute" causes more problems than it solves; but the term "frame-invariant" is unambiguous so I'll use that term instead. The frequency of a given photon as measured by a given observer is frame-invariant; so, therefore, is the "gravitational redshift" of a photon traveling from an observer who is static at some altitude above a gravitating body, to an observer who is static at a higher altitude.

 

[PeterDonis]

A freely flying photon does not change.

Ok, good, at least we agree on that.

1: Does the three-momentum of a free flying photon change?

Three-momentum relative to what? Three-momentum is frame-dependent.

2 How about a cannonball shot upwards in a gravity field, does it change? Particularly energy of it.

Again, relative to what? Energy is frame-dependent. The energy at infinity of the cannonball does not change, but its energy relative to the static observers it passes does.

 

[tom.stoer]

I've always learned that $\dot{x}^{\mu} = k^{\mu}$ as well, so that $\dot{x}^{\nu}\nabla_{\nu}\dot{x}^{\mu} = 0$.

But $\dot{x}^{\mu} = k^{\mu}$ need not be constant along the null-line.

 

[WannabeNewton]

But $\dot{x}^{\mu} = k^{\mu}$ need not be constant along the null-line.

Indeed but I was just referring to Peter's statement that it is parallel transported along the null geodesic.

 

[tom.stoer]

OK.

All what I wanted to indicate is that there is no observation which allows you to distinguish between a change of the 4-momentum k and a change of the 4-velocity u, simply b/c all you observe is the quotient with <k,u> at P and Q. It does not make sense to say "it's due to a change in u", neither is it reasonable to say "it's due to a change in k". They both contribute, but in order to say "which one contributes what" you have to introduce an artificial split in terms of a global reference frame, which means the distinction is frame-dependent.

 

[jartsa]

Three-momentum relative to what?

I been pondering falling photon's possible momentum change relative to an observer standing straight below the photon, that is falling straight down.

Again, relative to what? Energy is frame-dependent. The energy at infinity of the cannonball does not change, but its energy relative to the static observers it passes does.

I disagree with this answer. Why would the observer say the energy of the cannonball relative to himself is changing?

 

[Dale]

I disagree with this answer. Why would the observer say the energy of the cannonball relative to himself is changing?

PeterDonis was talking about a family of static observers. As the cannonball falls, the velocity relative to each succeeding static observer is changing, and therefore the energy relative to each succeeding static observer is different than the previous.

 

[PeterDonis]

All what I wanted to indicate is that there is no observation which allows you to distinguish between a change of the 4-momentum k and a change of the 4-velocity u, simply b/c all you observe is the quotient with <k,u> at P and Q.

I agree that there is no way to directly measure the photon's 4-momentum vector itself; all you can measure is its contraction with some 4-velocity. However, that's not the only factor involved. See below.

It does not make sense to say "it's due to a change in u", neither is it reasonable to say "it's due to a change in k". They both contribute, but in order to say "which one contributes what" you have to introduce an artificial split in terms of a global reference frame, which means the distinction is frame-dependent.

I'm not sure I agree, because the fact that the photon's 4-momentum gets parallel transported along its worldline is not frame-dependent, so there is an invariant way of defining what it means for the photon's 4-momentum to not change. On this view, as long as the photon is moving freely (i.e., it's not in a waveguide or some other device that causes it to move on a non-geodesic worldline), it's 4-momentum *never* changes, so any change in its observed frequency *must* be due to a change in 4-velocity of the detector relative to the emitter. That will be true in any reference frame.

 

[PeterDonis]

Why would the observer say the energy of the cannonball relative to himself is changing?

Strictly speaking, it's the cannonball's kinetic energy that changes, as DaleSpam pointed out. I suppose if you want to count potential energy as well, you can say the cannonball's energy would be unchanged (since counting potential energy basically means you're looking at energy at infinity), but if that's your criterion, then the photon's energy doesn't change either, even though its frequency as measured by static observers changes as it changes altitude. So it really depends on how you want to define "energy".