Atomic clocks in gravitational field

In summary: SR? Does SR state that our head ages faster because it is blueshifted? I am concerned and confused that the effect of time-dilation is attributed to redshift? But I think the experiment by Chou reveals reality? Or is SR not applicaple here?
  • #106
tom.stoer said:
For weak fields / stationary spacetimes / appropriate symmetries (Killing vector fields) this is certainly OK. I don't think this is a coincidence, it's a restriction to these special cases.
For the relationship between time dilation and energies that I indicated I know no exception. Could you give an example where an equal increase in potential and kinetic energy results in time dilation?
 
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  • #107
jartsa said:
So, what's wrong with this reasoning:

[..]
Always when a gravitational frequency redshift factor x is observed, then also an energy redshift of gravitational kind is observed, where the factor of energies is x. (This is based on formula E=hf )

[..]
("energy redshift" means that phenomenom that is sometimes explained as light losing energy while climbing upwards)
The basis of formula E=hf is misleading in that context: although it's no doubt connected through conservation of energy, it's the wrong explanation. See https://www.physicsforums.com/showthread.php?p=3162897
 
  • #108
harrylin said:
For the relationship between time dilation and energies that I indicated I know no exception. Could you give an example where an equal increase in potential and kinetic energy results in time dilation?
In order to give you an example where this relationship fails you have to present it here explicitly. But I still miss the formula which demonstrates this relationship. I indicated above why I think that such a formula cannot be given in general.
 
  • #109
WannabeNewton said:
"Associated with" is meaningless in the way you have presented it. Unless there is a mathematical relation that you can provide, this is moot.


I'll declare some restrictions: The gravity field must be static. And static measuring instruments measure the redshift of energy and the time dilation. Those restriction should suffice.

Now we can say the time dilation factor and energy redshift factor are the same number.

Well, now there's a mathematical relation.
 
  • #111
Yes, we never had doubts that in special cases as described above (in several posts) there is this direct relation between proper times and energy or frequency.

The only thing we wanted to stress is that you do have redshift even if these restrictions do not apply, i.e. where a "gravitational potential" does no longer make sense. And b/c this is the case we should not say that differences in the gravitational potential energies cause redshift, simply b/c this does not apply in general.
 
  • #112
tom.stoer said:
The only thing we wanted to stress is that you do have redshift even if these restrictions do not apply, i.e. where a "gravitational potential" does no longer make sense.


Like scattering of light from a fast moving black hole?
 
  • #113
tom.stoer said:
Yes, we never had doubts that in special cases as described above (in several posts) there is this direct relation between proper times and energy or frequency.

The only thing we wanted to stress is that you do have redshift even if these restrictions do not apply, i.e. where a "gravitational potential" does no longer make sense. And b/c this is the case we should not say that differences in the gravitational potential energies cause redshift, simply b/c this does not apply in general.
Inversely, what I (we) wanted to stress is that as long as kinetic and potential energy are unambiguously defined, there is a clear relationship with time dilation (for undefined or ambiguously defined quantities we can't say of course!). Therefore we should not say that time dilation has nothing to do with energy, as, I think, for all the cases that those concepts make sense, that relationship does apply.
 
  • #114
jartsa said:
Like scattering of light from a fast moving black hole?
No, in this case the conditions do apply ;-) Instead of a fast moving black hole you study a black hole at rest. Outside the Schwarzschild radius this is exactly the same as near a normal star or a planet. As a last step you apply a transformation from the black hole rest frame to a different, fast-moving frame. That's all.
 
  • #115
harrylin said:
... we should not say that time dilation has nothing to do with energy, as, I think, for all the cases that those concepts make sense, that relationship does apply.
Yes, I think so, too.

The statement "that relationship does apply" is OK, of course. But there were statements that redshift is caused by differences in potential energies, and this is not OK b/c then we would have to answer the question "what caused redshift when no potential energy can be defined?"

There is another problem, namely the idea that the difference in kinetic energy (difference in frequency) is due to a difference in potential energy. This cannot be true in general, either. Kinetic energy and frequency of a photon is always defined locally w.r.t. to two independent, local reference frames. So the two kinetic energies are not related at all, a priori. The difference in potential energy relies on a unique definition of "gravitational potential", which is a global entity and which cannot be given in general.

I only wanted o make clear that redshift is a much more general phenomenon and applies to every spacetime geometry, whereas for the definition of potential energy rather special conditions must apply.

So yes, if these conditions do apply, the relation is there, and this is not a coincidence. But no, there are situations where these conditions to not apply, but there's still redshift (the best known example is cosmological redshift in an expanding, anisotropic universe we live in)
 
  • #116
tom.stoer said:
...anisotropic universe we live in)

You meant isotropic, right?
 
  • #117
It seems that the most recent question in this forum is if time dilation and energy are related in someway. It is said in the beginning of the forum that an atomic clock at someones feet would tick slower than an atomic clock at someones head. I wish I could quote this person but I am new to the forum and still have not figured out how to quote people so if anyone could assist me it would be appreciated. Now is is known that the closer something gets to a massive object like Earth, the gravitational pull that effects it increases. This is due to the space surrounding the object being more curved closer to the surface of a planet. General Relativity states that gravity is causes by the curvature of space due to matter displacing it.

What I am trying to relate this to is the spinning of a wheel. Say a bicycle wheel is spinning. The wheel is one object therefore all parts of the wheel have the same energy. However, the outer part of the wheel spins faster than the inner rim of the wheel. If the inner part of the wheel was taken out and made to be its own separate wheel, and the wheels took the same amount of time to make a full rotation, the larger wheel would have covered more distance. Now to relate it back to the clock, a person walking with an atomic clock at their head and feet would experience the clock at their head ticking faster. Their head is technically moving faster than their feet. If the world is thought of as a giant bicycle wheel, the head being the outermost part of the rubber and the feet being the innermost part of the rubber where it touches the metal rim, the persons whole body is moving as one but at different speeds.

Now the simple formula for time is s=d/t. All numbers are made up and are not necessarily accurate but say a person is walking along the Earth and the Earth's circumference is 15 ft. Now the persons head travels along a circle with the feet but the circumference of the path the head travels is 30 ft. The person moves at 5 feet per second. It should take the feet 3 seconds to makes the journey when plugged into the equation s=d/t. It should take the head 6 seconds to make the journey, but it moves with the feet so it really takes 3 seconds. I'm not really sure what all of this means but to me there must be some relation in this to time dilation and energy. Speed and energy are somewhat interchangable I believe as speed is the measurement of how much distance is covered in an amount of time. The more energy as object has the more distance it will cover in less time. The head traveling along the outer wheel must have more energy than the feet because it travels faster than the feet. This is only the case in curved space like that of a planet. In perfectly flat space, both the head and the feet would travel at the same speed covering the same distance in the same amount of time, but when traveling through curved space, the head somehow travels faster and possibly with more energy due to its increased speed.

Just some things to think about and possibly contribute to the discussion
Dan
 
  • #118
TrickyDicky said:
You meant isotropic, right?
no, I mean anisotropic; anisotropy is tiny, but it's there, as can be seen in the CMB (Planck data)
 
  • #119
tom.stoer said:
Yes, I think so, too.

The statement "that relationship does apply" is OK, of course. But there were statements that redshift is caused by differences in potential energies, and this is not OK b/c then we would have to answer the question "what caused redshift when no potential energy can be defined?" [..]
[rearranged] The difference in potential energy relies on a unique definition of "gravitational potential", which is a global entity and which cannot be given in general.
I can't imagine a situation where no definition is possible.

There is another problem, namely the idea that the difference in kinetic energy (difference in frequency) is due to a difference in potential energy. This cannot be true in general, either. Kinetic energy and frequency of a photon is always defined locally w.r.t. to two independent, local reference frames. [..]
That sounds like another definition issue, and I think even a wrong one. When comparing energies one must use the same reference system (compare my post #17).

[..]So yes, if these conditions do apply, the relation is there, and this is not a coincidence. But no, there are situations where these conditions to not apply, but there's still redshift (the best known example is cosmological redshift in an expanding, anisotropic universe we live in)
Of course Doppler redshift must be taken in account too; while that complicates things, I don't see it as a problem of principle.
 
  • #120
harrylin said:
I can't imagine a situation where no definition is possible.
FLRW. Or any spacetime that isn't asymptotically flat.
 
  • #121
DaleSpam said:
FLRW. Or any spacetime that isn't asymptotically flat.
Was that impossibility somewhere discussed, do you know? A quick search of PF did not give me something...

PS: using keywords from the here following post by WannabeNewton I found some discussion in the physics FAQ, and they comment on the question of conservation of energy with:
'it depends on what you mean by "energy", and what you mean by "conserved" '...
- http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
 
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  • #122
FLRW is not a stationary space-time. You cannot define a gravitational potential for non-stationary space-times because you need a time-like killing vector field to do so. Regardless there is still a redshift.
 
  • #123
harrylin said:
PS: using keywords from the here following post by WannabeNewton I found some discussion in the physics FAQ, and they summarized it at the start as follows:
'it depends on what you mean by "energy", and what you mean by "conserved" '...
- http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
Yep that sums it up quite brilliantly. Check out ch 11 of Wald if you want more details. The problem is that there is no asymptotic flatness hence no way to define a meaningful notion of the total global energy of space-time (this part doesn't care about being stationary necessarily) and there is no time-like killing vector field so you can't define a gravitational potential (the gravitational potential is defined as ##\varphi = \frac{1}{2}\ln(-\xi^{a}\xi_{a}) ## where ##\xi^{a}## is the time-like killing vector field).

Just as a side note, the FAQ keeps saying static but you only need stationary; static is sufficient but not necessary (static is a much stronger condition than stationary).
 
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  • #124
WannabeNewton said:
Just as a side note, the FAQ keeps saying static but you only need stationary; static is sufficient but not necessary (static is a much stronger condition than stationary).

Could you clarify the distinction?
 
  • #125
Nugatory said:
Could you clarify the distinction?
Yes certainly. A stationary space-time is one which has a time-like killing vector field ##\xi^{a}##. A static space-time is a stationary space-time with the additional condition that ##\xi^{a}## is hypersurface orthogonal i.e. there exist scalar fields ##\alpha, \beta## such that ##\xi^{a} = \alpha \nabla^{a}\beta##; the hypersurface orthogonality condition can also be stated as ##\xi_{[a}\nabla_{b}\xi_{c]} = 0## which is equivalent to what I said above because of Frobenius' theorem but is much easier to check because it is a straightforward computation. For example in Schwarzschild space-time, the time-like killing vector field can be written as ##\xi^{a} = \alpha \nabla^{a}t## where ##t## is the usual time coordinate, meaning that the time-like killing vector field is everywhere orthogonal to the surfaces of constant time. On the other hand, Kerr space-time is stationary but not static i.e. it has a time-like killing vector field ##\xi^{a}## but it can be shown that ##\xi_{[a}\nabla_{b}\xi_{c]}\neq 0## for Kerr space-time (the integral curves of ##\xi^{a}## start twisting around one another because of the source's rotation and as a result fails to be everywhere orthogonal to a family of hypersurfaces).

EDIT: To put it in physical terms, a stationary space-time is one which possesses a time translation symmetry that leaves the space-time invariant, which is basically like saying that the same thing happens at each instant of time, whereas in a static space-time nothing ever happens at all.
 
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  • #126
harrylin said:
I can't imagine a situation where no definition is possible.
As said, non-stationary, or non-asymptotically flat spacetime

harrylin said:
That sounds like another definition issue, and I think even a wrong one. When comparing energies one must use the same reference system
Yes, but in GR there is no global refence frame!

You have local frames which are not related a priori and which do not allow for a global definition of energy. In order to compare energies you have to define how to transport reference frames through spacetime (along null lines). But once yo have done that you get the redshift for free - w/o defining energy.
 
  • #127
Consider for example a FLRW space-time with ##k = 0##. If we let ##\xi^{a}## be the tangent field of the congruence defined by the comoving observers then the spatial metric on a given isotropic and homogenous time-slice will be given by ##h_{ab}(\tau) = g_{ab} + \xi_{a}\xi_{b}## hence ##\xi^{b}\nabla_{c}h_{ab} = -\nabla_{c}\xi_{a}##. From this we find that ## \nabla_{\mu}\xi_{\nu}= \Gamma ^{\gamma}_{\mu \tau}h_{\gamma\nu} = \frac{\dot{a}}{a}h_{\mu \nu}## in the comoving coordinates hence ##\nabla_{a}\xi_{b} = \frac{\dot{a}}{a}h_{ab}## in general.

Now the frequency of a light signal with wave 4-vector ##k^{a}## as measured by an observer with 4-velocity ##\xi^{a}## is ##\omega = -k_{a}\xi^{a}## so ##\frac{\mathrm{d} \omega}{\mathrm{d} \lambda} = k^{b}\nabla_{b}\omega = -k^{a}k^{b}\nabla_{b}\xi_{a} = -\frac{\dot{a}}{a}k^{a}k^{b}\xi_{b}\xi_{a} = -\frac{\dot{a}}{a}\omega^{2}##. Then ##\frac{\mathrm{d} \omega}{\mathrm{d} \tau}\frac{d\tau}{d\lambda}= \omega \dot{\omega} = -\frac{\dot{a}}{a}\omega^{2}\Rightarrow \frac{\dot{\omega}}{\omega} = -\frac{\dot{a}}{a}## thus ##\frac{\omega_2}{\omega_1} = \frac{a_1}{a_2}## which is the usual cosmological redshift for the aforementioned space-time. Here all we have used is the definition of the frequency of a light signal at a given event as measured by an observer passing through that event and the parallel transport of this frequency along the associated null geodesic using the derivative operator ##\nabla_{a}##. There was no use of any killing symmetries nor any global energies (there is no meaningful notion of global energy for this space-time anyways). This was to basically give an example of what Tom said above.
 
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  • #128
WannabeNewton said:
Yep that sums it up quite brilliantly. Check out ch 11 of Wald if you want more details. The problem is that there is no asymptotic flatness hence [..] you can't define a gravitational potential [..].

OK that remark made it immediately clear to me, thanks!

As far as I can see, the FAQ does not claim that it is impossible to define energies in GR; only that it's a can of worms. :tongue2:

For me, if no definition of gravitational potential is possible with a model, then that is an argument against that model. A quick search gave me a recent university paper about doing numerical calculations for asymptotic flatness, which suggests to me that this is still accepted as possible solution - http://eprints.ma.man.ac.uk/1535/01/covered/MIMS_ep2010_93.pdf
 
  • #129
You can define various kinds of total global energy easily in asymptotically flat space-times (calculating and physically interpreting it is a whole 'nother story); some standard examples are Komar energy (only for stationary asymptotically flat space-times), Bondi energy, and ADM energy. The problem is with non-asymptotically flat space-times wherein you simply cannot define any meaningful notion of total global energy.

Gravitational potential is a purely Newtonian concept. There is no reason to expect that it has any meaning for arbitrary space-times in GR. Stationary space-times have a time translation symmetry which allows us to define a GR analogue of the Newtonian potential but this is a very special case.
 
  • #130
harrylin said:
As far as I can see, the FAQ does not claim that it is impossible to define energies in GR; only that it's a can of worms.

For me, if no definition of gravitational potential is possible with a model, then that is an argument against that model.
An overview regarding non-local definitions of mass, energy etc. including open issues can be found here:
http://relativity.livingreviews.org/Articles/lrr-2009-4/
Quasi-Local Energy-Momentum and Angular Momentum in General Relativity
The present status of the quasi-local mass, energy-momentum and angular-momentum constructions in general relativity is reviewed. First, the general ideas, concepts, and strategies, as well as the necessary tools to construct and analyze the quasi-local quantities, are recalled. Then, the various specific constructions and their properties (both successes and deficiencies are discussed. Finally, some of the (actual and potential) applications of the quasi-local concepts and specific constructions are briefly mentioned.

Rejecting models w/o definition of gravitational potential means rejecting GR.
 
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  • #131
tom.stoer said:
An overview regarding non-local definitions of mass, energy etc. including open issues can be found here:
http://relativity.livingreviews.org/Articles/lrr-2009-4/
Quasi-Local Energy-Momentum and Angular Momentum in General Relativity
The present status of the quasi-local mass, energy-momentum and angular-momentum constructions in general relativity is reviewed. First, the general ideas, concepts, and strategies, as well as the necessary tools to construct and analyze the quasi-local quantities, are recalled. Then, the various specific constructions and their properties (both successes and deficiencies are discussed. Finally, some of the (actual and potential) applications of the quasi-local concepts and specific constructions are briefly mentioned.
Thanks for the link!

Rejecting models w/o definition of gravitational potential means rejecting GR.

Here you launch a similar strong opinion as fact like when this side discussion started; but instead of arguing about it, I simply take note that it does not match that of the FAQ nor of that article.
 
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  • #132
harrylin said:
Here you launch a similar strong opinion as fact like when this side discussion started; but instead of arguing about it, I simply take note that it does not match that of the FAQ nor of that article.
Which FAQ?

It is a mathematical fact that energy cannot be defined as usual in GR. Potentials cannot be defined, either (this means only in special cases). Note that this has nothing to do with physics, it's due to the mathematical structure underlying GR. There's no choice.
 
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  • #133
tom.stoer said:
Which FAQ? [..]
The one I cited just before.

Hmm... a logical and equally strong follow-up of your claims would be that accepting GR means rejecting the principle of conservation of energy - one of the pillars of modern physics. Michael Weiss and John Baez don't go that far.
 
  • #134
harrylin said:
The one I cited just before.
Sorry, can't find it.

harrylin said:
Hmm... a logical and equally strong follow-up of your claims would be that accepting GR means rejecting the principle of conservation of energy - one of the pillars of modern physics.
Yes and no.

There's still a locally conserved energy-momentum tensor of all non-gravitationally d.o.f.

But instead of a conserved vector current

##\partial_a j^a = 0##

which is only available when a space-time symmetry encoded in a Killing vector field does exist

##j^a = T^{ab}\,\xi_b##

in general situations w/o space-time symmetry only a conserved tensor density exists

##(DT)^a = 0##

Due the the covariant derivative = the Levi-Cevita Connection the usual trick

##\partial_0 j^0 \;\to\; \partial_0 \int_V dV\,j^0 = \partial_0 Q = 0##

with

##\oint_{\partial V} d\vec{S}\,\vec{j} = 0##

does not work.

So when no symmetry is present, one cannot constructed a conserved vector current j, therefore one cannot construct a conserved charge Q b/c the volume integral

##\int_V dV\,T^{00}##

is neither meaningfull (no well-defined transformation property) nor conserved

##\partial_0 \int_V dV\,T^{00} \neq 0##

That means that there is a local conservation law, but there's no way to derive a globally conserved quantity simply due to mathematical reasons.

A simple fact is that a redshifted photon looses energy w/o transferring this energy to another system, i.e. the gravitational field. Locally the Einstein-Maxwell equations guarantuees energy-momentum conservation of the el.-mag. field. Globally energy is not conserved.
 
  • #135
tom.stoer said:
Sorry, can't find it.
Post # 121 : 'it depends on what you mean by "energy", and what you mean by "conserved" '

Yes and no.

[...]

That means that there is a local conservation law, but there's no way to derive a globally conserved quantity simply due to mathematical reasons.

That sounds very much like a "yes" ...

A simple fact is that a redshifted photon looses energy w/o transferring this energy to another system, i.e. the gravitational field. Locally the Einstein-Maxwell equations guarantuees energy-momentum conservation of the el.-mag. field. Globally energy is not conserved.

If you talk about basic gravitational redshift, that's simply wrong - see my posts #14 and #107 .
 
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  • #136
harrylin said:
Post # 121 : 'it depends on what you mean by "energy", and what you mean by "conserved" '
OK. I checked John's articles. Very clear - as usual.

He says that there is no gravitational energy in general and he explains why globally defined energy doesn't exist. He writes "So the flux integral is not well-defined, and we have no analogue for Gauss's theorem". That's exactly my reasoning.

Of course you can play around with Theta, but there's one big issue (besides the fact that it's a pseudo-tensor): it has nothing in common with other definitions of energy-momentum density, there's no relation with Noether's Theorem, and there's no relation with a Hamiltonian as generator of time translations.

harrylin said:
That sounds very much like a "yes" ...
I would say, global energy conservation - in the sense we are used to - does not exist. Playing around with Theta is a trick, but it has nearly nothing in common with what we usually call energy.

harrylin said:
If you talk about basic gravitational redshift, that's simply wrong - see my posts #14 and #107 .
I don't see the relevance of your post. As I said a couple of times: in general (!) there is no potential energy, no globally defined energy, and no energy conservation. A photon in the CMB appears to have a different energy observed on Earth than observed at emission, so energy is not conserved, but b/c there is no potential energy in expanding spacetime, there is nothing where the missing energy went to. The whole concept becomes meaningless, and I think we (including John) explained why. It does not make sense to rely on concepts from Newtonian mechanics valid in very special cases of GR, but not being able to generalize these concepts to full GR.
 
  • #137
tom.stoer said:
OK. I checked John's articles. Very clear - as usual.

He says that there is no gravitational energy in general and he explains why globally defined energy doesn't exist. He writes "So the flux integral is not well-defined, and we have no analogue for Gauss's theorem". That's exactly my reasoning. [..]
Yes it appears to be very difficult. Perhaps it is even impossible to do. Once more, their "it depends" is quite not the same as your "no"; and "is not well-defined" and "we have no", is quite different from "there is no" and "doesn't exist". Thus I noticed the difference in opinions.

I don't see the relevance of your post. [..] A photon in the CMB appears to have a different energy observed on Earth than observed at emission, so energy is not conserved [..]
Of course features such as frequencies are measured differently with different reference standards; it is fundamentally erroneous to compute energies by mixing such reference systems - adding apples to oranges is a big No-No in physics. I also highlighted that point in post #77, and it turned out that the fact that energy is frame-dependent is just the point that PeterDonis tried to make too.

And although it is not the same problem as standard gravitational redshift, the threads that I linked bear out that same fact: a difference in measured frequency with different reference systems does not imply that "energy is not conserved".
 
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  • #138
ok; instead of saying that energy is not conserved in GR I should better say that in general energy cannot be defined in GR, so the question whether it is conserved becomes meaningless.
 
  • #139
tom.stoer said:
ok; instead of saying that energy is not conserved in GR I should better say that in general energy cannot be defined in GR, so the question whether it is conserved becomes meaningless.
That's more like it. :smile:
 

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