Atomic clocks in gravitational field

[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".

 

[jartsa]

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).

Seems to me that nothing happens to gravitational redshift.

Does the freely falling observer start to fall from infinity? Then the doppler blueshift and the gravitational redshift will cancel out, in which case there is gravitational redshift large enough to cancel the doppler shift, which gravitational redshift is the same gravitational redshift as the static observer's gravitational redshift.

 

[harrylin]

Seems to me that nothing happens to gravitational redshift.

Does the freely falling observer start to fall from infinity? Then the doppler blueshift and the gravitational redshift will cancel out, in which case there is gravitational redshift large enough to cancel the doppler shift, which gravitational redshift is the same gravitational redshift as the static observer's gravitational redshift.

Right. It is of course possible to temporarily mask gravitational redshift on a single object by means of an accelerating reference system. But it is not possible to do that for a plurality of bodies such as satellites orbiting the Earth.

 

[PAllen]

Right. It is of course possible to temporarily mask gravitational redshift on a single object by means of an accelerating reference system. But it is not possible to do that for a plurality of bodies such as satellites orbiting the Earth.

Well, it depends. Two (reasonably close) concentric 2-spheres of free falling bodies will will experience detect no 'gravitational' redshift or blue shift for radial light in either direction.

 

[Philosopha]

Right. It is of course possible to temporarily mask gravitational redshift on a single object by means of an accelerating reference system. But it is not possible to do that for a plurality of bodies such as satellites orbiting the Earth.

You picked something there I believe: "mask". - Very much the essence of what I am wondering about.

 

[harrylin]

Well, it depends. Two (reasonably close) concentric 2-spheres of free falling bodies will will experience detect no 'gravitational' redshift or blue shift for radial light in either direction.

I was thinking about communication satellites around the equator. Some of them are even at opposite sides. As they fall in opposite directions, it is not possible to mask the gravitational redshift of all of them with a single falling reference system.

You picked something there I believe: "mask". - Very much the essence of what I am wondering about.

This is how Einstein put it in 1916 (translated from German):

"It is, for instance, impossible to choose a body of reference such
that, as judged from it, the gravitational field of the Earth (in its entirety) vanishes."
- Relativity: The Special and General Theory
http://www.bartleby.com/173/20.html

 

[Philosopha]

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

What about the fact that atomic clocks after an experiment can be put next to each other on a table, be examined and leave no doubt, that they have recorded the absolute value of the relative difference between their prior frames? The clock at height of our foot experiences less oszillation periods (phenomenon known as time-dilation) than the clock at our head. But why is that so? Time-dialtion is just how that phenomenon is called. For what reason does one clock oszillate slower than the other? The oszillation velocity is a direct reflection of the clocks energy/mass state methinks. Which suggests the clock at the bottom to have a lower energy/mass content than the one on top.

 

[WannabeNewton]

If you are asking why as in "why does nature behave this why" then GR doesn't answer that question I'm afraid.

 

[tom.stoer]

For what reason does one clock oszillate slower than the other? The oszillation velocity is a direct reflection of the clocks energy/mass state ...

Here's what GR has to say - and it has nothing to do with energy but is a purely geometrical effect.

Assume we have two clocks located at (t,x) = (0,0) in one specific coordinate system. They will meet again at a later time T but at the same location x=0, i.e. at (T,0).

Assume one clock is traveling along a curve C from point A to point B in spacetime. The second clock is traveling along a different curve C' from point A to point B in spacetime. Of course we could introduce the coordinates for A and B, but that is not necessary for the next steps.

Now you have to believe me that the proper time tau measured by a clock along its curve between A and B is given by the "length" of the curve through spacetime.

\tau = \int_C d\tau

As the two curves C and C' through spacetime are different for the two clocks their proper times will differ.

\Delta\tau_{A\to B} = \int_{C_{A\to B}} d\tau - \int_{C^\prime_{A\to B}} d\tau

These generic formulas are rather formal. In order to calculate something one introduces a coordinate system (t,x), a spacetime-metric g which fully describes spacetime-geometry and a velocity v=dx/dt along a curve C. Then the above mentioned formula for the proper times can be expressed as

\tau = \int_C d\tau = \int_0^T dt\,\sqrt{g_{\mu\nu}\,v^\mu\,v^\nu}

Please note that time dilation due to geometry and due to velocity cannot be separated in general. Please note that we have not introduced any expression for energy.

Asking "why this formula explaines time dilation and differential aging" and "why nature behave this why" then GR doesn't answer that question b/c it's like asking "why GR?" which cannot be answered by GR (this is to stress what WannabeNewton said)

 

[Philosopha]

Thank you very much for everything - I will definitely come back to that topic once I studied my way through GR. It is good that you mentioned the "why". Lots of people are thinking about that. I believe there is a why and maybe that observations/ facts will give us an idea.

 

[Philosopha]

And if something calculates the right amount, maybe that would also give an idea.

 

[tom.stoer]

The problem is not the idea, but the " why this idea?" The idea is GR, but there's no deeper reason for GR. GR cannot answer "why GR?" If there would be a more fundamental theory X from which GR emerges in some way, then the same question "why X?" cannot be answered based on X.

So your questions are essentially meta-physically. They are interesting, but not as physical questions.

 

[WannabeNewton]

And if something calculates the right amount, maybe that would also give an idea.

Calculation isn't a problem. Tom gave you one way of calculating the effect. The problem is in answering the much deeper question of "why does this happen?" which as Tom stated is more along the lines of metaphysics than it is physics.

 

[harrylin]

[..] The oszillation velocity is a direct reflection of the clocks energy/mass state methinks. Which suggests the clock at the bottom to have a lower energy/mass content than the one on top.

Concerning mass it's a bit more complex, as also c varies from that non-local perspective.
There was an interesting and informative discussion on that topic in another forum (see in particular the first reply by Jonathan Scott): http://sci.physics.research.narkive.com/tSrjiEsH/mass-of-particles-in-gr-field.

 

[jartsa]

What about the fact that atomic clocks after an experiment can be put next to each other on a table, be examined and leave no doubt, that they have recorded the absolute value of the relative difference between their prior frames? The clock at height of our foot experiences less oszillation periods (phenomenon known as time-dilation) than the clock at our head. But why is that so? Time-dialtion is just how that phenomenon is called. For what reason does one clock oszillate slower than the other? The oszillation velocity is a direct reflection of the clocks energy/mass state methinks. Which suggests the clock at the bottom to have a lower energy/mass content than the one on top.

It seems pretty safe to say that an object hanging at low altitude has less energy than a similar object at higher location.

Because energy is released when two gravitating objects fuse together, and then a mass defect can be measured.

https://en.wikipedia.org/wiki/Binding_energy#Mass_change
https://en.wikipedia.org/wiki/Gravitational_binding_energy

 

[tom.stoer]

nice; but nevertheless time dilation has nothing to do with energy

 

[PeterDonis]

It seems pretty safe to say that an object hanging at low altitude has less energy than a similar object at higher location.

Except that, if the energies of the objects are each measured by observers at the same altitude, they will be the same. So you have to specify relative to what observer the energy of the lower object is less. (For example, the lower object's energy at infinity is less, assuming both are at rest at their respective altitudes.)

Because energy is released when two gravitating objects fuse together, and then a mass defect can be measured.

This is also a change in energy at infinity.

 

[harrylin]

Except that, if the energies of the objects are each measured by observers at the same altitude, they will be the same. [..]

Similarly it seems pretty safe to say that a high energy electron has more kinetic energy because of its high speed, upon which you might comment "Except that, if the energies of the objects are each measured by observers at the same velocity, they will be the same."

 

[harrylin]

nice; but nevertheless time dilation has nothing to do with energy

Time dilation can be written as the combination of the effects of kinetic and potential energy. For example, on the geoid potential energy balances kinetic energy, so that the time dilation effects cancel out. If time dilation has nothing to do with energy, then that must be pure coincidence - which seems highly unlikely to me.

 

[PeterDonis]

Similarly it seems pretty safe to say that a high energy electron has more kinetic energy because of its high speed, upon which you might comment "Except that, if the energies of the objects are each measured by observers at the same velocity, they will be the same."

Yes, exactly, because energy is frame-dependent, which was my point. When you use the term "energy", you have to specify what it's relative to; or else you are implicitly assuming a certain frame of reference. In jartsa's case, he implicitly assumed that "energy" meant "energy at infinity". In your case, you're implicitly assuming that "energy" means "energy relative to the laboratory in which the electron is moving at high speed".

There's nothing wrong with that as long as you realize that that's what you're doing, and don't try to claim that your particular choice of what to measure energy relative to is somehow absolute. (The same applies to other frame-dependent concepts like time dilation and redshift/blueshift.)

 

[PAllen]

Similarly it seems pretty safe to say that a high energy electron has more kinetic energy because of its high speed, upon which you might comment "Except that, if the energies of the objects are each measured by observers at the same velocity, they will be the same."

Yes, of course. There is nothing different about a 'fast' electron or a 'slow' electron except for the observer relative to whom it is fast versus slow. Unless one posits an objective, absolute, frame, there is no conceivable difference.

 

[tom.stoer]

Time dilation can be written as the combination of the effects of kinetic and potential energy. For example, on the geoid potential energy balances kinetic energy, so that the time dilation effects cancel out. If time dilation has nothing to do with energy, then that must be pure coincidence - which seems highly unlikely to me.

How do you translate the above mentioned formula

\tau = \int_C d\tau = \int_0^T dt\,\sqrt{g_{\mu\nu}\,v^\mu\,v^\nu}

into an expression containing energy?

 

[tom.stoer]

Let me explain where I see the problem: in my formula we have

$v^\mu = dx^\mu / dt = (1, dx^i/dt)$

expressed in some coordinates.

In order to introduce energy we hae to use the 4-velocity and the 4-momentum

$u^\mu = dx^\mu / d\tau$
$p^\mu = m u^\mu$

For a geodesic along which 4-momentum is conserved we have

$\tau = \int_C \sqrt{g_{\mu\nu}\,dx^\mu\,dx^\nu} = \int_C d\tau \, \sqrt{g_{\mu\nu}\,u^\mu\,u^\nu} = m^{-1} \int_C d\tau \sqrt{g_{\mu\nu} \, p^\mu \, p^\nu}$

But

$g_{\mu\nu} \, p^\mu \, p^\nu = m^2$

and therefore we arive at

$\tau = \int_C d\tau$

therefore 4-momentum trivially drops out.

That's why I doubt that

Time dilation can be written as the combination of the effects of kinetic and potential energy.

does not work

 

[Philosopha]

What seems not possible today might be possible tomorrow. No question should be surrendered to Metaphysics.

 

[tom.stoer]

What seems not possible today might be possible tomorrow. No question should be surrendered to Metaphysics.

This is not metaphysics but math.

In order to say that time dilation for proper times τ is caused by differences in energy E you have to provide a formula like τ = f(E), a function of E.

 

[Philosopha]

Calculation isn't a problem. Tom gave you one way of calculating the effect. The problem is in answering the much deeper question of "why does this happen?" which as Tom stated is more along the lines of metaphysics than it is physics.

Calculating experimental results however from the vantage point of changes in energy/mass content translated into frequency if achieved would allow for an interpretation as of the why. It must be rational and calculated, or it is not an answer.

 

[tom.stoer]

Calculating experimental results however from the vantage point of changes in energy/mass content translated into frequency if achieved would allow for an interpretation as of the why.

Before claiming to know the "why" you should present the calculation. Any (general covariant) result for proper time expressed in terms of energy and momentum is appreciated.

 

[WannabeNewton]

Gravitational time dilation can be related to the gravitational potential of the gravitational field but that's as close a relationship to "energy" as I can think of.

 

[tom.stoer]

Gravitational time dilation can be related to the gravitational potential of the gravitational field but that's as close a relationship to "energy" as I can think of.

I think this relation is limited to special cases where you can interpret the 00 coefficient of the metric tensor as gravitational potential. But in general this is not possible - and it's not in the spirit of GR and general covariance.

 

[WannabeNewton]

Well gravitational time dilation really only makes sense for stationary space-times (since we need a notion of clocks being "at rest" in the sense that they follow orbits of the time-like killing vector field) and for stationary space-times we can always define a gravitational potential using the time-like killing vector field.

 

[tom.stoer]

Well gravitational time dilation really only makes sense for stationary space-times and for stationary space-times we can always define a gravitational potential.

The general formula I presented makes sense for arbitrary spacetimes.

 

[WannabeNewton]

The general formula I presented makes sense for arbitrary spacetimes.

Oh I was thinking of gravitational time dilation in the sense of "two clocks at rest in the gravitational field ticking at different rates" with the emphasis on the "at rest" part. I don't disagree at all with what you're saying though; even if for stationary space-times we can relate things to gravitational potential I don't see any relation to kinetic energy (which is what I think was originally mentioned).

 

[tom.stoer]

Oh I was thinking of gravitational time dilation in the sense of "two clocks at rest in the gravitational field ticking at different rates" with the emphasis on the "at rest" part.

This case is covered as well. At rest means "at rest w.r.t. some coordinate system" and the gravitational field is the metric.

 

[WannabeNewton]

I was talking about the physical definition of "at rest" meaning following an orbit of a time-like killing vector field in a stationary space-time, which is a physical description of being "at rest". The existence of the time-like killing vector field picks out a preferred family of "static" observers for the space-time (a preferred vector field), including "static" clocks in a gravitational field.

 

[tom.stoer]

Of course you can do this as well. As said in restricted cases you can define a gravitational potential, but in general (no Killing vector field) it will not work

 

[WannabeNewton]

Right, I'm not disagreeing with that at all my friend :) I'm just saying if one wanted to really stretch things and find some special cases of relationships with "energy" then one could do something along the lines of the above.

 

[tom.stoer]

I fully agree ;-)

 

[Philosopha]

Before claiming to know the "why" you should present the calculation. Any (general covariant) result for proper time expressed in terms of energy and momentum is appreciated.

I do not claim to know the why nor that I got any calculations to work. So ?

By the guidelines of Physicsforum one is not ment to discuss any personal unpublished hypothesis I was told. So I was just presenting some definite facts with the atomic clocks. Facts are always good.

My sentence you quoted read "...if achieved would... ". Expressing the consideration of a possibility (by whoever would achieve it, if at all) is not claiming to know the answer. Hope you were not upset at anything :(.

 

[Philosopha]

Merely I'm saying facts are the best providers of answers. One doesn't even need to interpret anything into the clock experiment. The results are as they are. Equating frequency (here of clock) with energy content is just an old basic.

 

[tom.stoer]

Hi Philosopha, no, I am not upset at all. No problem. The only issue I wanted to clarify is that time dilation as a result of energy, gravitational potential etc. is limited to very special conditions (stationary spacetime) and not valid in general (even if sometimes presented that way). I was not asking for any new theory or formula, only for some existing expression unknown to me.

 

[harrylin]

Gravitational time dilation can be related to the gravitational potential of the gravitational field but that's as close a relationship to "energy" as I can think of.

Yes, that's what I meant. One can calculate time dilation for a certain mass as function of potential and kinetic energy. An equal increase of both such as with the equilibrium of the geoid results in zero time dilation, as the two exactly compensate each other.

How do you translate the above mentioned formula

\tau = \int_C d\tau = \int_0^T dt\,\sqrt{g_{\mu\nu}\,v^\mu\,v^\nu}

into an expression containing energy?

I doubt that potential and kinetic energy as commonly defined fit with such GR equations; instead I was referring to the weak field approximations, and surely you know those. Your claim that time dilation has nothing to do with energy implies that the aforementioned correlations are just a coincidence. I find that extremely unlikely.